Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
$\lim_{n \rightarrow \infty} n ((n^5 +5n^4)^{1/5} - (n^2 +2n)^{1/2})$ $$\lim_{n \rightarrow \infty} n ((n^5 +5n^4)^{1/5} - (n^2 +2n)^{1/2})$$
Please, help me to find the limit.
| Use the general fact
$$n(n+a-\sqrt[k]{n^k +akn^{k-1}})\rightarrow \frac{k+1}{2}a^2$$ as $n\to \infty$
to get a limit of
$$\frac{2+1}{2}-\frac{5+1}{2}=-\frac{3}{2}$$
For proof of the above fact.
If we let $A=\sqrt[k]{n^k +akn^{k-1}}$
then $$n((n+a)-A)=n\frac{(n+a)^k-A^k}{(n+a)^{k-1}+\cdots +A^{k-1}}
=\frac{\binom{k}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/867115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some integers $a,b$. Prove that $a,b$ are both divisible by $p$. Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some integers $a,b$. Prove that $a,b$ are both divisible by $p$.
My attempt:
$p\mid a^2+ab+b^2 \implies p\mid (a-b)(a... | Suppose that $p=3k+2$ is prime and
$$
\left.p\ \middle|\ a^2+ab+b^2\right.\tag1
$$
then, because $a^3-b^2=(a-b)\left(a^2+ab+b^2\right)$, we have
$$
\left.p\ \middle|\ a^3-b^3\right.\tag2
$$
Case $\boldsymbol{p\nmid a}$
Suppose that $p\nmid a$, then $(2)$ says $p\nmid b$. Furthermore,
$$
\begin{align}
a^3&\equiv b^3&\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/867413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Number theory proofs regarding perfect squares How do you prove that $3n^2-1$ is never a perfect square
| Let $3n^2-1=b^2, \text{ for a } b \in \mathbb{Z}$
$$3n^2-1 \equiv -1 \pmod 3 \equiv 2 \pmod 3$$
$$b=3k \text{ or } b=3k+1 \text{ or } b=3k+2$$
Then:
$$b^2=9k \equiv 0 \pmod 3 \text{ or } b^2=3n+1 \equiv 1 \pmod 3 \text{ or } b=3n+1 \equiv 1 \pmod 3$$
We see that it cannot be $b^2 \equiv 2 \pmod 3$,so the equality $3n^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/867476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Double integral of a rational function Consider the region $D$ given by $1\leq x^2+y^2\leq2\land0\leq y\leq x$. Compute $$\iint_D\frac{xy(x-y)}{x^3+y^3}dxdy$$
Attempt: The region $D$ is part of a ring in the first quadrant below the line $y=x$
Any hints are wellcome.
| Changing to polar coordinates, $x=\rho \cos\theta$, $y=\rho \sin\theta$, and the Jacobian of the transformation is $J=\rho$. Then:
$$\int_1^\sqrt2 \rho d\rho\int_0^\frac{\pi}{4}\frac{\sin\theta\cos\theta(\cos\theta-\sin\theta)}{\cos^3\theta+\sin^3\theta}d\theta$$
The first integral is immediate and yields $\frac{1}{2}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/867856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Surface area of sphere $x^2 + y^2 + z^2 = a^2$ cut by cylinder $x^2 + y^2 = ay$, $a>0$ The cylinder is given by the equation $x^2 + (y-\frac{a}{2})^2 = (\frac{a}{2})^2$.
The region of the cylinder is given by the limits $0 \le \theta \le \pi$, $0 \le r \le a\sin \theta$ in polar coordinates.
We need to only calculate t... | Given the equations
$$
x^2+y^2+z^2=a^2,
$$
and
$$
x^2+y^2 = ay,
$$
we obtain
$$
ay + z^2 = a^2.
$$
Using
$$
\begin{eqnarray}
x &=& a \sin(\theta) \cos(\phi),\\
y &=& a \sin(\theta) \sin(\phi),\\
z &=& a \cos(\theta),\\
\end{eqnarray}
$$
we obtain
$$
a^2 \sin(\theta) \sin(\phi) + a^2 \cos^2(\theta) = a^2
\Rightarrow \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/867961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Evaluate $\int \frac{1}{(2x+1)\sqrt {x^2+7}}\,\text{d}x$. How to do this indefinite integral (anti-derivative)?
$$I=\displaystyle\int \dfrac{1}{(2x+1)\sqrt {x^2+7}}\,\text{d}x.$$
I tried doing some substitutions ($x^2+7=t^2$, $2x+1=t$, etc.) but it didn't work out.
| Using Euler substitution by setting $t-x=\sqrt{x^2+7}$, we will obtain
$x=\dfrac{t^2-7}{2t}$ and $dx=\dfrac{t^2+7}{2t^2}\ dt$, then the integral turns out to be
\begin{align}
\int \dfrac{1}{(2x+1)\sqrt {x^2+7}}\ dx&=\int\frac{1}{t^2+t-7}\ dt\\
&=\int\frac{1}{\left(t+\dfrac{\sqrt{29}+1}{2}\right)\left(t-\dfrac{\sqrt{29... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/871595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Minimum Value of $x_1+x_2+x_3$ For an Acute Triangle $\Delta ABC$
$$\begin{align}x_n=2^{n-3}\left(\cos^nA+\cos^nB+\cos^nC\right)+\cos A\,\cos B\,\cos C\end{align}$$ Then find the least value of $$x_1+x_2+x_3$$
My Approach: I have found $x_1$, $x_2$ and $x_3$
$$\begin{align}x_1=\frac{1}{4}\left(\cos A+\cos B+\cos C\rig... | Use AM-GM inequality,we have
$$\cos^3{x}+\dfrac{\cos{x}}{4}\ge 2\sqrt{\cos^3{x}\cdot\dfrac{\cos{x}}{4}}=\cos^2{x}$$
then we have
$$x_{1}+x_{3}\ge\cos^2{A}+\cos^2{B}+\cos^2{C}+2\cos{A}\cos{B}\cos{C}=2x_{2}$$
so
$$x_{1}+x_{2}+x_{3}\ge 3x_{2}=\dfrac{3}{2}$$
because we have use this follow well know
$$\cos^2{A}+\cos^2{B}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/872592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Finding Cauchy principal value for: $ \int_1^\infty \frac{ a x^2 + c }{x^4 - b x^2 - c} \mathrm{d}x $ I need to solve the integral
$ \displaystyle \mathcal{P} \int_1^\infty \frac{ a x^2 + c }{x^4 - b x^2 - c} \mathrm{d}x $,
where $\mathcal{P}$ is the Cauchy principal value, $ - 1 \leq c \leq 1$ and $a, b$ are both real... | I actually ended up using a different solution, I found more direct and intuitively.
It is completely equivalent with @Yves method, but I just state it for completeness.
$$ \mathcal{P} \int_{1}^{\infty} \mathrm{d} x
\frac{a x^{2} + c }
{
x^{4} - b x^{2} - c
}
=
\mathcal{P} \int_{1}^{\infty} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/873682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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A transcendental number from the diophantine equation $x+2y+3z=n$ Let $\displaystyle n=1,2,3,\cdots.$ We denote by $D_n$ the number of non-negative integer solutions of the diophantine equation
$$x+2y+3z=n$$
Prove that
$$
\sum_{n=0}^{\infty} \frac{1}{D_{2n+1}}
$$
is a transcendental number.
| If I have not mistaken something,
$$ [x^{2n+1}]\frac{1}{(1-x)(1-x^2)(1-x^3)} = \frac{(n+1)(n+3)}{3}, $$
when $n\equiv 0,2\pmod{3}$, and
$$ [x^{2n+1}]\frac{1}{(1-x)(1-x^2)(1-x^3)} = \frac{(n+1)(n+3)+1}{3} $$
when $n\equiv 1\pmod{3}$, hence:
$$\begin{eqnarray*}\sum_{n=0}^{+\infty}\frac{1}{D_{2n+1}}&=&\sum_{n=0}^{+\infty}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/874158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
if $\frac{1}{(1-x^4)(1-x^3)(1-x^2)}=\sum_{n=0}^{\infty}a_{n}x^n$,find $a_{n}$ Let
$$\dfrac{1}{(1-x^4)(1-x^3)(1-x^2)}=\sum_{n=0}^{\infty}a_{n}x^n$$
Find the closed form $$a_{n}$$
since
$$(1-x^4)(1-x^3)(1-x^2)=(1-x)^3(1+x+x^2+x^3)(1+x+x^2)(1+x)$$
so
$$\dfrac{1}{(1-x^4)(1-x^3)(1-x^2)}=\dfrac{1}{(1-x)^3(1+x+x^2+x^3)(1+x+... | Hints :
*
*First, prove that
$$\frac{1}{(1-x^2)(1-x^3)(1-x^4)} = \frac{7}{32(x+1)}-\frac{59}{288(x-1)}+\frac{1}{8(x-1)^2}+\frac{1}{16(x+1)^2}-\frac{1}{24(x-1)^3}+\frac{x+2}{9(x^2+x+1)}+\frac{1-x}{8(x^2+1)}.$$
*Then, use that
$$ \frac{1-x}{x^2+1} = -\frac{1+i}{2(x-i)}+\frac{-1+i}{2(x+i)}.$$
and
$$ \frac{x+2}{x^2+x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/875792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Need help with this Geometric sequence problem First, sorry if Im not using the right syntax, im translating the problem and im not sure if im supposed to say "Sequence" or "series", and also thanks to who ever tries to help.
The sum of a geometric sequence is 20, and the sum of its squared terms is 205.
find how many... | You have found that $q^n+1=20.5(q+1)$.
But since the sum of the terms is $20$, we have $q^n-1=40(q-1)$. (In essence you had written down this equation also.)
Subtract, and find $q$, and then $n$. The numbers are disappointingly small, $3$ and $4$ respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/876061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\frac{1}{2}(x^2-xy+y^2)}dx\, dy$ I need to evaluate the following integral:
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\frac{1}{2}(x^2-xy+y^2)}dx\, dy$$
I thought of evaluating the iterated integral $\displaystyle\int_{-\infty}^{\infty}dx\int_{-\infty... | Recall:
$$x^2-xy+y^2=(x-\frac{1}{2}y)^2+\frac{3}{4}y^2$$
With that you get:
$$\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1}{2}(x^2-xy+y^2)}dxdy=\\\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1}{2}\left((x-\frac{1}{2}y)^2+\frac{3}{4}y^2\right)}dxdy=\\
\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/877711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Is there a way to calculate the area of this intersection of four disks without using an integral? Is there anyway to calculate this area without using integral ?
| Let $R$ be its radius and $D$ its diameter: $R = 5$, $D = 10$.
$$\begin{align}
\text{Area of big square} &= D^2 = 100 \\
\text{Area of circle} &= \frac{\pi D^2}{4} \approx 78.54 \\
\text{Area outside circle} &= 100 - 78.54 = 21.46 \\
\text{Area of 4 petals} &= 78.54 - 21.46 = 57.08 \\
\text{Area of single petal} &= \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/878457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 5,
"answer_id": 3
} |
If one number is thrice the other and their sum is $16$, find the numbers If one number is thrice the other and their sum is $16$, find the numbers.
I tried,
Let the first number be $x$ and the second number be $y$
Acc. to question
$$
\begin{align}
x&=3y &\iff x-3y=0 &&(1)\\
x&=16-3y&&&(2)
\end{align}
$$
| Let the first number be $x$.
Let the second number be $y$.
According to question
$$ \tag{1}
x+y=16
$$
$$
\tag{2}
x=3y
$$
So, $x-3y=0 \tag{2}$
Multiply equation $(1)$ by $3$.
Solve both equations:
$$\tag{1} 3x+3y=48$$
$$\tag{2} x-3y=0$$
$$\tag{1) + (2}4x=48$$
$$\tag{3}x=12$$
Putting in equation $(1)$:
$$\tag{1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/879886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Summation of Infinite Geometric Series Determine the sum of the following series:
$$\sum_{n=1}^{\infty } \frac{(-3)^{n-1}}{7^{n}} $$
My work:
$$\sum_{n=1}^{\infty } \frac{(-3)^{n-1}}{7^{n}} = \sum_{n=1}^{\infty } \frac{-1}{7} (\frac{3}{7})^{n-1}$$
$$\sum_{n=1}^{\infty } ar^{n-1} = \frac{a}{1-r} = \frac{\frac{-1}{7}}{1-... | $$\begin{align}
\sum_{n=1}^{\infty } \frac{-3^{n-1}}{7^{n}}
& = - \frac{1}{7} \sum_{n=1}^{\infty } (\frac{3}{7})^{n-1}
\\
& = - \frac{1}{7} \frac{1}{1-\frac 3 7}
\\
& = - \frac 1 4
\\[2ex]
\sum_{n=1}^{\infty } \frac{(-3)^{n-1}}{7^{n}}
& = \frac{1}{7} \sum_{n=1}^{\infty } (-\frac{3}{7})^{n-1}
\\
& = \frac{1}{7} \frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/880019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How find this sum $\sum\limits_{i=0}^{2n}\binom{2n}{2i}\binom{2i}{i}y^{2i}$
Find the sum close form
$$f(x)=\sum_{i=0}^{2n}\dfrac{\binom{2n}{2i}\binom{2i}{i}x^{2i}}{2^{2i}}$$
if we let
$$\dfrac{x}{2}=y$$
then
$$f(y)=\sum_{i=0}^{2n}\binom{2n}{2i}\binom{2i}{i}y^{2i}$$
this PDF have this page 5
$$\sum_{k=j}^{n}\binom... | Given any formal Laurent series $\;(???) = \sum \alpha_{k_1 k_2 \ldots k_n} t_1^{k_1} t_2^{k_2} \cdots t_n^{k_n}$, we will use the notation $[ t_1^{k_1} t_2^{k_2} \cdots t_n^{k_n} ](???)$ to denote the coefficient $\alpha_{k_1 k_2 \cdots k_n}$ in front of corresponding monomial.
Instead of $f(y)$, let us denote the po... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/880295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Algebraic proof of $\tan x>x$ I'm looking for a non-calculus proof of the statement that $\tan x>x$ on $(0,\pi/2)$, meaning "not using derivatives or integrals." (The calculus proof: if $f(x)=\tan x-x$ then $f'(x)=\sec^2 x-1>0$ so $f$ is increasing, and $f(0)=0$.) $\tan x$ is defined to be $\frac{\sin x}{\cos x}$ where... | Here is a sketch of what you might be looking for:
Showing $\tan x > x$ is equivalent to showing $\sin x - x \cos x > 0$, since $\cos x > 0$ on $(0,\pi/2$).
The series for $\sin x - x \cos x$ is $\displaystyle\sum_{j=1}^{\infty} \dfrac{(2j)x^{2j+1}}{(2j+1)!} = x^3/3 - x^5/30 + x^7/840 - x^9/45360 \ldots$
Group the term... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/881668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
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Finding the perimeter of the room If the length and breadth of a room are increased by $1$ $m$, the area is increased by $21$ $m^2$. If the length is increased by $1$ $m$ and breadth is decreased by $1$ $m$ the area is decreased by $5$ $m^2$. Find the perimeter of the room.
Let the length be $x$ and the breadth be $y$
... | $A=xy$
$(x+1)(y+1)=xy+21$
$(x+1)(y-1)=xy-5$
Foil out both equations to get:
$xy+x+y+1=xy+21 \quad \to \quad x+y=20 \quad \to \quad y=20-x$
$xy-x+y-1=xy-5 \quad \to \quad -x+y=-4 \quad \to \quad y=-4+x$
Set them equal to each other:
$20-x=-4+x$
$2x=24 \to x=12$
Since we know $x+y=20, y=8$.
You can verify this solution b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/881954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Prove $\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2} \geq \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}$
If $a$, $b$ and $c$ are positive real numbers, prove that:
$$\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2} \geq \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}$$
Additional info:We can use AM-GM and Cauchy inequaliti... | Another way to do this would be the following (I'm doing Liu Gang's suggested generalization):
We have to show
$$\frac{a^{n+1}}{b^n} + \frac{b^{n+1}}{c^n} + \frac{c^{n+1}}{a^n} - \frac{a^n}{b^{n-1}} - \frac{b^n}{c^{n-1}} - \frac{c^n}{a^{n-1}} \ge 0.$$
The left hand side equals
$$\frac{a^n(a - b)}{b^n} + \frac{b^n(b-c)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/883384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Recurrence of the form $2f(n) = f(n+1)+f(n-1)+3$ Can anyone suggest a shortcut to solving recurrences of the form, for example:
$2f(n) = f(n+1)+f(n-1)+3$, with $f(1)=f(-1)=0$
Sure, the homogenous solution can be solved by looking at the characteristic polynomial $r^2-2x+1$, so that in general a solution for the homogen... | Let $f_p = A + Bn + Cn^2$.
$$
\begin{cases}
2f(n) = 2A +2Bn + 2Cn^2\\
-f(n+1) = -A - B(n+1) - C(n+1)^2\\
-f(n-1) = -A - B(n-1) - C(n-1)^2
\end{cases}
\quad \Rightarrow
$$
$$
3 = Cn^2 - C(2n^2 + 2) = -2C \quad \Rightarrow \quad C = -\frac{3}{2}
$$
Thus,
$$
f(n) = C_1 + C_2n -\dfrac{3n^2}{2}
$$
Now use the given initial... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/886753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Evaluating $ \int x \sqrt{\frac{1-x^2}{1+x^2}} \, dx$ I am trying to evaluate the indefinite integral of
$$\int x \sqrt{\frac{1-x^2}{1+x^2}} \, dx.$$
The first thing I did was the substitution rule:
$u=1+x^2$, so that $\displaystyle x \, dx=\frac{du}2$ and $1-x^2=2-u$. The integral then transforms to
$$\int \sqrt{\fr... | $\text {Let } x^{2}=2 \sin ^{2} \theta-1 \textrm{ for } \frac{\pi}{4} \leqslant \theta \leqslant \frac{\pi}{2}, \text {then } x d x=\sin \theta \cos \theta d \theta $.
\begin{aligned}\int x \sqrt{\frac{1-x^{2}}{1+x^{2}}} d x&=\int \sqrt{\frac{2-2 \sin ^{2} \theta}{2 \sin ^{2} \theta} }\sin \theta \cos \theta d \theta \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/887784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 4
} |
Solving the logarithimic inequality $\log_2\frac{x}{2} + \frac{\log_2x^2}{\log_2\frac{2}{x} } \leq 1$ I tried solving the logarithmic inequality:
$$\log_2\frac{x}{2} + \frac{\log_2x^2}{\log_2\frac{2}{x} } \leq 1$$
several times but keeping getting wrong answers.
| Here are the steps
\[
\log_2 \frac{x}{2}+\frac{\log_2 x^{2}}{\log_2 \frac{2}{x}} \le 1
\]
\[
\log_2 x -\log_2 2+\frac{2\log_2 x}{\log_2 2-\log_2 x} \le 1
\]
\[
\log_2 x -1+\frac{2\log_2 x}{1-\log_2 x} \le 1
\]
Let $\alpha= \log_2 x$, then
\[
\alpha -1+\frac{2\alpha}{1-\alpha} \le 1
\]
\[
\alpha -2+\frac{2\alpha}{1-\alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/888174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
How many $3$ digit numbers with digits $a$,$b$ and $c$ have $a=b+c$ My question is simple to state but (seemingly) hard to answer. How many $3$ digit numbers exist such that $1$ digit is the sum of the other $2$. I have no idea how to calculate this number, but I hope there is a simple way to calculate it. Thank you in... | Assuming a digit is an element of $\{0,1,2,3,4,5,6,7,8,9,10\}$ we have three cases for $a,b,c$ to see:
*
*$a=b=c=0$. All easy here, yields $1$ combination.
*$b=c\ne 0$. $a=2b$, so $b<5$ giving us $4$ choices (digits $1$ to $4$). The position of $a$ uniquely determines the code, so multiply b $3$ to get $4\cdot 3 = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/889687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
A nice trignometric identity How to prove that:
$$\cos\dfrac{2\pi}{13}+\cos\dfrac{6\pi}{13}+\cos\dfrac{8\pi}{13}=\dfrac{\sqrt{13}-1}{4} $$
I have a solution but its quite lengthy, I would like to see some elegant solutions. Thanks!
| I add another answer although it is not different in principle from some of the others.
We have
$$t=\cos \frac{2\pi}{13}+\cos \frac{6\pi}{13}+\cos \frac{8\pi}{13}$$
it is natural to then consider the other even divisions,
$$s=\cos \frac{4\pi}{13}+\cos \frac{10\pi}{13}+\cos \frac{12\pi}{13}$$
Now
$$t+s=\cos \frac{2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/890052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Error in proving of the formula the sum of squares Given formula
$$
\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}
$$
And I tried to prove it in that way:
$$
\sum_{k=1}^n (k^2)'=2\sum_{k=1}^n k=2(\frac{n(n+1)}{2})=n^2+n
$$
$$
\int (n^2+n)\ \text d n=\frac{n^3}{3}+\frac{n^2}{2}+C
$$
But
$$
\frac{n^3}{3}+\frac{n^2}{2}+C $$ i... | Here's an interesting approach using the summation of binomial coefficients.
First, note that
$$\sum_{i=1}^n {i+a\choose b} = {{n+a+1}\choose {b+1}}$$
and also that
$$i^2=2\cdot \frac{(i+1)i}{1\cdot 2}-i=2{{i+1}\choose 2}-{i\choose 1}$$
Hence
$$\begin{align}
\sum_{i=1}^ni^2 &=\sum_{i=1}^n \left[2{{i+1}\choose 2}-{i\ch... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/890254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
} |
If $ \frac {z^2 + z+ 1} {z^2 -z +1}$ is purely real then $|z|=$? If z is a complex number and $ \frac {z^2 + z+ 1} {z^2 -z +1}$ is purely real then find the value of $|z|$ .
I tried to put $ \frac {z^2 + z+ 1} {z^2 -z +1} =k $ then solve for $z$ and tried to find |z|, but it gets messy and I am stuck.
The answer giv... | Let's come up with a more interesting question (which might be what you intended to ask).
Find an $\alpha \in \mathbb{R}$ such that $\forall z \in \mathbb{C}: |z| = \alpha \implies \dfrac{z^2+z+1}{z^2-z+1} \in \mathbb{R} \cup \{\infty\}$.
I'll show that $\alpha = 1$ works. If $|z| = 1$, then $z = e^{i\theta}$ for s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/890723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Proving or disproving inequality $ \frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y} \ge x + y + z $
Given that $ x, y, z \in \mathbb{R}^{+}$, prove or disprove the
inequality
$$ \dfrac{xy}{z} + \dfrac{yz}{x} + \dfrac{zx}{y} \ge x + y + z $$
I have rearranged the above to:
$$ x^2y(y - z) + y^2z(z - x) + z^2x(x - y) \ge 0 ... | Holder's inequality:
$$u\cdot v \leq |u||v|$$
for any vectors $u,v$.
Let $\mathbf u=(1/x,1/y,1/z)$ and $\mathbf v=(xz,xy,yz)$. Then show $|v| = xyz |u|$ and thus $$|\mathbf u||\mathbf v| = xyz\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)= \dfrac{xy}{z} + \dfrac{yz}{x} + \dfrac{zx}{y} $$
and:
$$\mathbf u\cdot\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/891997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solving exponential equation $e^{x^2+4x-7}(6x^2+12x+3)=0$ How would you find $x$ in:
$e^{x^2+4x-7}(6x^2+12x+3)=0$
I don't know where to begin. Can you do the following?
$e^{x^2+4x-7}=1/(6x^2+12x+3)$
and then find $ln$ for both sides?
| $e^{x^2+4x-7}(6x^2+12x+3)=0 \Rightarrow e^{x^2+4x-7}=0 \text{ or } \ 6x^2+12x+3=0$
$$\text{It is known that } e^{x^2+4x-7} \text{ is non-zero }$$
therefore,you have to solve :
$$6x^2+12x+3=0$$
The solutions are:
$$x=-1-\frac{1}{\sqrt{2}} \\ x=-1+\frac{1}{\sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/892484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to solve the recurrence relation $T(n) = T(\lceil n/2\rceil) + T(\lfloor n/2\rfloor) + 2$ I'm trying to solve a recurrence relation for the exact function (I need the exact number of comparisons for some algorithm). This is what i need to solve:
$$\begin{aligned}
T(1) &= 0 \\
T(2) &= 1 \\
T(n) & = T(\lceil n/2\rcei... | As mentioned in comment, the first two conditions $T(1) = 0, T(2) = 1$ is incompatible with
the last condition
$$\require{cancel}
T(n) = T(\lfloor\frac{n}{2}\rfloor) + T(\lceil\frac{n}{2}\rceil) = 2\quad\text{ for }
\color{red}{\cancelto{\;\color{black}{n > 2}\;}{\color{grey}{n \ge 2}}}
\tag{*1}$$
at $n = 2$. We will a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/893251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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Elementary algebra problem Consider the following problem (drawn from Stanford Math Competition 2014): "Find the minimum value of $\frac{1}{x-y}+\frac{1}{y-z}+ \frac{1}{x-z}$ for for reals $x > y > z$ given $(x − y)(y − z)(x − z) = 17.$"
Method 1 (official solution): Combining the first two terms, we have
$\frac{x−z}{(... | In the second solution, when $=$ in $\ge$ is satisfied, $a=b=c$. But then, $abc=17$ and $a+b=c$ makes the contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/893793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How prove $(\ln{\frac{1-\sin{xy}}{1+\sin{xy}}})^2 \geq \ln{\frac{1-\sin{x^2}}{1+\sin{x^2}}}\ln{\frac{1-\sin{y^2}}{1+\sin{y^2}}}$ How prove that if $x, y \in (0,\sqrt{\frac{\pi}{2}})$ and $x \neq y$, then $(\ln{\frac{1-\sin{xy}}{1+\sin{xy}}})^2 \geq \ln{\frac{1-\sin{x^2}}{1+\sin{x^2}}}\ln{\frac{1-\sin{y^2}}{1+\sin{y^2}}... | the right question is :
$(\ln{\dfrac{1-\sin{xy}}{1+\sin{xy}}})^2 \le \ln{\dfrac{1-\sin{x^2}}{1+\sin{x^2}}}\ln{\dfrac{1-\sin{y^2}}{1+\sin{y^2}}}$
$f(x)=\ln{\dfrac{1-\sin{x}}{1+\sin{x}}}\le 0 , f(0)=0$
WLOG $y=ax,a\ge1 \implies $the inequality $\iff (f(ax^2))^2 \le f(x^2)f(a^2x^2) \iff (f(ax))^2 \le f(x)f(a^2x) \iff \d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/894759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determine variables that fit this criterion... There is a unique triplet of positive integers $(a, b, c)$ such that $a ≤ b ≤ c$.
$$
\frac{25}{84} = \frac{1}{a} + \frac{1}{ab} + \frac{1}{abc}
$$
Just having trouble with this Canadian Math Olympiad question. My thought process going into this, is:
Could we solve for $\fr... | Factoring, we see
$\displaystyle \frac{25}{84} = \frac{1}{a}(1+\frac{1}{b}(1+\frac{1}{c}))$
And we know the prime factoring of 84 gives $2\times2\times3\times7$ So we know $a,b,$ and $c$ are each going to be multiples of these primes. So we start with finding $a$:
$\displaystyle \frac{25}{84}a = 1+\frac{1}{b}(1+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/895556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Inequality involving a finite sum this is my first post here so pardon me if I make any mistakes.
I am required to prove the following, through mathematical induction or otherwise:
$$\frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} < 2{\sqrt{n}}$$
I tried using mathematical induction th... | If you want to take a look on the Riemann's sum method :
$\forall n > 1$, we have $$\int_{n-1}^n \frac{dt}{\sqrt{t}} \ge (n-(n-1))\cdot \underset{x \in [n-1,n]}{\min} \frac{1}{\sqrt{x}} = \frac{1}{\sqrt{n}} $$
Hence, $$\frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} \le 1+ \int_{1}^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/896259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Proof by Induction: $(1+x)^n \le 1+(2^n-1)x$ I have to prove the following by induction:
$$(1+x)^n \le 1+(2^n-1)x$$
for $n \ge 1$ and $0 \le x \le 1$.
I start by showing that it's true for $n=1$ and assume it is true for one $n$.
$$(1+x)^{n+1} = (1+x)^n(1+x)$$
by assumption:
$$\le (1+(2^n-1)x)(1+x)$$
$$= 1+(2^n-1)x+x+(... | $$(1+x)^{n+1}=(1+x)(1+x)^{n}\le(1+x)(1+(2^{n}-1)x)=1+(2^{n}-1)x+x(1+(2^{n}-1)x)$$
$$\underbrace{\le}_{x\le1}1+(2^{n}-1)x+x(1+2^{n}-1)=1+(2^{n}-1)x+2^{n}x$$
$$=1+(2\cdot2^{n}-1)x=1+(2^{n+1}-1)x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/896720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Consecutive Prime Gap Sum (Amateur) List of the first fifty prime gaps:
1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2, 6, 4, 6, 8, 4, 2, 4, 2, 4, 14, 4, 6, 2, 10, 2, 6, 6, 4, 6, 6, 2, 10, 2, 4, 2, 12, 12, 4, 2, 4.
My conjecture is that the sum of consecutive prime gaps is always prime whenever a prime gap ... | Set $g_n=p_{n+1}-p_n$, where $p_n$ is the series of prime numbers, with $p_1=2$.
Then
$$
p_1+\sum_{i=1}^n g_i=\sum_{i=1}^n g_i+2=p_{n+1}.
$$
So the conjecture is obviously true, but not useful.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/896802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
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Evaluation of a sum of $(-1)^{k} {n \choose k} {2n-2k \choose n+1}$ I have some question about the paper of which name is Spanning trees: Let me count the ways. The question concerns about $\sum_{k=0}^{\lfloor\frac{n-1}{2} \rfloor} (-1)^{k}
{n \choose k} {2n-2k \choose n+1}$. Could you recommend me how to prove $\dis... | Since $\binom{m-2k}{n+1}$ is a degree $n+1$ polynomial in $k$ with lead term $\frac{(-2k)^{n+1}}{(n+1)!}$, we get that the multiple forward difference $\Delta_k^{n+1}\binom{m-2k}{n+1}=(-2)^{n+1}$. Multiply both sides by $(-1)^{n+1}$ to get
$$
\begin{align}
2^{n+1}
&=\sum_{k=0}^{n+1}(-1)^k\binom{n+1}{k}\binom{m-2k}{n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/897948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Finding the asymptotes of a general hyperbola I'm looking to find the asymptotes of a general hyperbola in $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ form, assuming I know the center of the hyperbola $(h, k)$. I came up with a solution, but it's too long for me to be confident that I didn't make a mistake somewhere, so I wa... | For a conic, $$ax^2+2hxy+by^2+2gx+2fy+\color{blue}{c}=0$$
which is a hyperbola when $ab-h^2<0$.
Its asymptotes can be found by replacing $\color{blue}{c}$ by $\color{red}{c'}$ where
$$\det
\begin{pmatrix}
a & h & g \\
h & b & f \\
g & f & \color{red}{c'}
\end{pmatrix}
=0$$
That is
$$\color{red}{c'}=\frac{af^2+bg^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/898005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Integrate $\int \left(A x^2+B x+c\right) \, dx$ I am asked to find the solution to the initial value problem:
$$y'=\text{Ax}^2+\text{Bx}+c,$$
where $y(1)=1$,
I get:
$$\frac{A x^3}{3}+\frac{B x^2}{2}+c x+d$$
But the answer to this is:
$$y=\frac{1}{3} A \left(x^3-1\right)+\frac{1}{2} B \left(x^2-1\right)+c (x-1)+1.$$
Co... | $$\frac{A x^3}{3}+\frac{B x^2}{2}+c x+d$$ where $d$ is a cosntant to fix is equivalent to $$\frac{1}{3} A \left(x^3-1\right)+\frac{1}{2} B \left(x^2-1\right)+c (x-1)+d$$ where $d$ is a constant to fix.
And the second one is surely more handy to apply the initial condition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/898363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
find formula to count of string I have to find pattern of count of series.
Lenght of series is $2n$. It is neccessary to use exaclty double times every number in range $[1...n]$ and all of neighboring numbers are different.
Look at example.
$a_0 = 0$
$a_1 = 0$
$a_2 = 2$ (because of $1212$ and $2121$)
And my idea is:... | For $a_3$, you'll add two $3$'s somewhere to each one of those so they're not next to each other. There are $5$ possible places: the two ends, and three in between two numbers, so there are ${5 \choose 2}$ ways of putting the two $3$'s in so they're not next to each other.
For case $a_n$: There are $2n+1$ possible pl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/898527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Simple use of log I am struggling to see how we can go from the first expression to the second:
$$\begin{align}
2\log_3 12 - 4\log_3 6 &= \log_3 \left ( \frac{4^2 \cdot 3^2}{2^4 \cdot 3^4} \right )\\
&= \log_3 (3^{-2}) = -2
\end{align}$$
| First note that
\[ \log_{b}(x^{n})=n\log_{b}(x) \]
\[ \log_{b}(x)-\log_{b}(y)=\log_{b}\left(\frac{x}{y}\right) \]
Here are the steps
\[
2\log_{3} 12 - 4\log_{3} 6 = 2\log_{3} (4\cdot 3) - 4\log_{3} (2\cdot 3)
\]
\[
=\log_{3} (4\cdot 3)^{2} - \log_{3} (2\cdot 3)^{4} = \log_{3} (4^{2}\cdot 3^{2}) - \log_{3} (2^{4}\cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/903168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How to find $\int \frac{x\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\mathrm dx$ $$I=\int x.\frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\mathrm dx$$
Try 1:
Put $z= \ln(x+\sqrt{1+x^2})$, $\mathrm dz=1/\sqrt{1+x^2}\mathrm dx$
$$I=\int \underbrace{x}_{\mathbb u}\underbrace{z}_{\mathbb v}\mathrm dz=x\int zdz-\int (z^2/2)\mathrm dz\tag{... | Hint:
$$
\int x\frac{\ln({x+\sqrt{1+x^2})}}{\sqrt{1+x^2}}dx=\int\ln({x+\sqrt{1+x^2})}d\sqrt{1+x^2}
$$
and
$$
(\ln({x+\sqrt{1+x^2})})'=\frac{1}{\sqrt{1+x^2}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/904296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Indefinite integral of $\frac{2x^3 + 5x^2 +2x +2}{(x^2 +2x + 2)(x^2 + 2x - 2)}$ How do I find $$\int\frac{2x^3 + 5x^2 +2x +2}{(x^2 +2x + 2)(x^2 + 2x - 2)}\mathrm dx$$
I used partial fractions by breaking up $x^2 + 2x - 2$ into $(x+1)^2 - 3$ and split it into $(a+b)(a-b)$ but as u can see it's extreme tedious. I was won... | HINT:
Write $$\frac{2x^3 + 5x^2 +2x +2)}{(x^2 +2x + 2)(x^2 + 2x - 2)}=\frac{Ax+B}{x^2+2x+2}+\frac{Cx+D}{x^2+2x-2}$$
For the ease of calculation, we can write
$$\frac{Ax+B}{x^2+2x+2}=\frac A2\cdot\frac{\left(\dfrac{d(x^2+2x+2)}{dx}\right)}{x^2+2x+2}+\frac C{(x+1)^2+1^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/905084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How can I show why this equation has no complex roots? I've been asked to show why an equation has no complex roots but i'm at a complete loss.
The equation is
$F_{n+2}=F_n$
Where $F_n=(x-1)(x-2)...(x-n)$ and n is a positive integer.
I'd really appreciate if someone could explain how I could go about showing this becau... | $F_{n}=(x-1)(x-2) \cdots (x-n)$
$F_{n+2}=(x-1)(x-2) \cdots (x-n)(x-(n+1))(x-(n+2))$
$F_{n} = F_{n+2}$
$(x-1)(x-2) \cdots (x-n)=(x-1)(x-2) \cdots (x-n)(x-(n+1))(x-(n+2))$
$(x-1)(x-2) \cdots (x-n)[(x-(n+1))(x-(n+2)) - 1] = 0$
$(x-1)(x-2) \cdots (x-n)[(x^2-[(n+2)+(n+1)]x+(n+1)(n+2) - 1] = 0$
$(x-1)(x-2) \cdots (x-n)[x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/907395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Sum the series $\sum_{n = 1}^{\infty}\{\coth (n\pi x) + x^{2}\coth(n\pi/x)\}/n^{3}$ This sum is from Ramanujan's letters to G. H. Hardy and Ramanujan gives the summation formula as
\begin{align} &\frac{1}{1^{3}}\left(\coth \pi x + x^{2}\coth\frac{\pi}{x}\right) + \frac{1}{2^{3}}\left(\coth 2\pi x + x^{2}\coth\frac{2\pi... | Recall the well known Mittag-Leffler expansion of hyperbolic cotangent function (denote $\mathbb{W}=\mathbb{Z}/\{0\}$) :
$$\sum_{m\in\mathbb{W}}\frac{1}{m^2+z^2}=\frac{\pi\coth\pi z}{z}-\frac{1}{z^2}\tag{ML}$$
Hence, your sum is by its symmetry :
$$\begin{align}
S&=\frac{1}{2}\sum_{n \in \mathbb{W}}\{\coth (n\pi x) +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/907480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
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"answer_id": 0
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How to express $\log_2 (\sqrt{9} - \sqrt{5})$ in terms of $k=\log_2 (\sqrt{9} + \sqrt{5})$? If $$k=\log_2 (\sqrt{9} + \sqrt{5})$$
express $\log_2 (\sqrt{9} - \sqrt{5})$ in terms of $k$.
| Adding the logarithms $\log_2{(\sqrt{9}-\sqrt{5})}$ and $\log_2{(\sqrt{9}+\sqrt{5})}$ we get the following:
$$\log_2{(\sqrt{9}-\sqrt{5})}+\log_2{(\sqrt{9}+\sqrt{5})}=\log_2{(\sqrt{9}-\sqrt{5})(\sqrt{9}+\sqrt{5})}=\log_2{(9-5)}=\log_2{4}=\log_2{2^2}=2 \cdot \log_2{2}=2$$
Knowing that $k=\log_2{(\sqrt{9}+\sqrt{5})}$ we h... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to deduce a closed formula given an equivalent recursive one? I know how to prove that a closed formula is equivalent to a recursive one with induction, but what about ways of deducing the closed form initially?
For example:
$$ f(n) = 2 f(n-1) + 1 $$
I know how to use induction to prove that $\forall n \ge1$:
$$ f(... | $$f(n) = 2f(n-1) + k \\ f(n-1) = 2f(n-2) + k \\ f(n-2) = 2f(n-3) + k \\ \dots \\ f(1) = 2f(0) + k$$
$$$$
$$\Rightarrow f(n) = 2f(n-1) + k= \\ 2(2f(n-2) + k)+k= \\ 2^2f(n-2)+(2+1)k=2^2(2f(n-3) + k)+(2+1)k= \\ 2^3f(n-3)+(2^2+2+1)k= \\ \dots \overset{*}{=} \\ 2^nf(0)+(2^{n-1}+2^{n-2}+\dots +2+1)k$$
Therefore, $$f(n)=2^nf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/908568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 1
} |
Check if two vector equations of parametric surfaces are equivalent Give the vector equation of the plane through these lines:
$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\1\\1\end{pmatrix}+\lambda\cdot\begin{pmatrix}0\\2\\1\end{pmatrix}\,\,\,$ and $\,\,\,\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\... | But the really simple way is to pick three arbitrary points (not on the same line) from the first plane and check if they all lie in the second one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/912048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Equation $3x^4 + 2x^3 + 9x^2 + 4x + 6 = 0$ Solve the equation
$$3x^4 + 2x^3 + 9x^2 + 4x + 6 = 0$$
Having a complex root of modulus $1$.
To get the solution, I tried to take a complex root $\sqrt{\frac{1}{2}} + i \sqrt{\frac{1}{2}}$ but couldn't get the solution right. Please help me.
| Let the root be $\cos y+i\sin y,$
Using Complex conjugate root theorem, $\cos y-i\sin y$ must be another root
So, if the four roots are $\cos y\pm i\sin y,u, v$
using Vieta's formula, $(\cos y+i\sin y)(\cos y-i\sin y)u\cdot v=\dfrac63\implies v=\dfrac2u$
So we have $$3[x-(\cos y+i\sin y)][x-(\cos y-i\sin y)](x-u)\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/913039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
general of partial sum of sequence I am trying to find the limit of an infinite series given as
$$\sum\frac{1}{n^2-1}.$$
I came across the following general term of the sequence of partial sums
$$3/4-\left(\frac{1}{2n}-\frac{1}{2(n+1)}\right).$$ I would appreciate assistance to understand how this expression is arrived... | You have
$$
\frac{1}{n^2-1} = \frac 1 2 \left(\ \underbrace{\frac{1}{n-1}-\frac{1}{n+1}\ }_{\text{Call this $\{$A$\}$}} \right)
$$
Consequently
\begin{align}
& \frac12\left( \left(\frac{1}{2-1} - \frac{1}{2+1}\right) + \left(\frac{1}{3-1} - \frac{1}{3+1}\right) + \left( \frac{1}{4-1} - \frac{1}{4+1} \right) + \cdots ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/914853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Simplifying the sum of powers of the golden ratio I seem to have forgotten some fundamental algebra. I know that:
$(\frac{1+\sqrt{5}}{2})^{k-2} + (\frac{1+\sqrt{5}}{2})^{k-1} = (\frac{1+\sqrt{5}}{2})^{k}$
But I don't remember how to show it algebraicly
factoring out the biggest term on the LHS gives
$(\frac{1+\sqrt{5}... | $$ \left (\frac{1+\sqrt{5}}{2} \right )^{k-2} + \left (\frac{1+\sqrt{5}}{2} \right )^{k-1} = \left ( \frac{1+ \sqrt{5}}{2}\right )^{k-2} \left ( 1+ \frac{1+ \sqrt{5}}{2}\right)$$
It is known that the Greek letter phi (φ) represents the golden ratio,which value is:
$$\phi=\frac{1+ \sqrt{5}}{2}$$
One of its identities is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/915456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluating Summation of $5^{-n}$ from $n=4$ to infinity The answer is $\frac1{500}$ but I don't understand why that is so.
I am given the fact that the summation of $x^{n}$ from $n=0$ to infinity is $\frac1{1-x}$. So if that's the case then I have that $x=\frac15$ and plugging in the values I have $\frac1{1-(\frac15)}... | The problem you have is that you do not know why the formula works to begin with. If you did the situation would be clear. Here's the thing:
$$\sum_{n=0}^{\infty}r^n=\frac{1}{1-r} \; \;\;\;\;\; |r|<1.$$
Let $r=\frac{1}{5}$, then you really have the following situation:
$$\left(\frac{1}{5}\right)^0+\left(\frac{1}{5}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/920050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Evaluating $\lim_{x \to 0}\frac{(1+x)^{1/x} - e}{x}$ How to evaluate the following limit? $$\lim_{x \to 0}\frac{(1+x)^{1/x} - e}{x}.$$
| You may write
$$
\begin{align}
\frac{(1+x)^{1/x} - e}{x} &= \frac{\large e^{\large\frac{\log (1+x)}{x}} - e}{x}\\\\
&= \frac{e^{\large \frac{x-\frac{x^2}{2}+{\mathcal{O}}(x^3)}{x}} - e}{x}\\\\
&= \frac{ e^{1-\frac{x}{2}+{\mathcal{O}}(x^2)} - e}{x}\\\\
&= \frac{ e \:e^{-\frac{x}{2}+{\mathcal{O}}(x^2)} - e}{x}\\\\
&= \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/920132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Are these proofs logically equivalent? Here are two proofs, firstly:
x = 0.999...
10x = 9.999...
= 9 + 0.999...
= 9 + x
9x = 9
x = 1
And secondly:
x = 1 - 1 + 1 - 1 + 1 - 1 ...
= 1 - (1 - 1 + 1 - 1 + 1 ...
= 1 - x
2x = 1
x = 1/2
The fourth line in the first and the third line in the second use the same tr... | No. The first series is absolutely convergent, so its terms can be rearranged without changing the sum, while the second series is divergent.
The number $0.\overline{9}$ can be expressed in the form
\begin{align*}
0.\overline{9} & = \sum_{k = 1}^{\infty} \frac{9}{10^k}\\
& = \frac{9}{10} \sum_{k = 0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/921172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Show that $\frac{(b+c)^2} {3bc}+\frac{(c+a)^2}{3ac}+\frac{(a+b)^2}{3ab}=1$ If $a^3+b^3+c^3=3abc$ and $a+b+c=0$ show that $\frac{(b+c)^2} {3bc}+\frac{(c+a)^2}{3ac}+\frac{(a+b)^2}{3ab}=1$
| $a+b+c=0\iff b+c=-a\implies (b+c)^2=a^2$
$\implies\dfrac{(b+c)^2}{3bc}=\dfrac{a^3}{3abc}$
Actually, $b+c=-a\implies(b+c)^3=(-a)^3$
$\implies -a^3=b^3+c^3+3bc(b+c)=b^3+c^3+3bc(-a)\iff\sum a^3=3abc$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/927310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to show that $a+b> \sqrt{a^2+b^2-ab}, \qquad a, b >0$ How do you show that
$$a+b> \sqrt{a^2+b^2-ab}, \qquad a, b >0$$
I could write $\sqrt{a^2+b^2-ab}=\sqrt{(a+b)^2-3ab}$, but this seems to lead nowhere.
| $$a+b> \sqrt{a^2+b^2-ab} \iff (a+b)^2>\left (\sqrt{a^2+b^2-ab}\right )^2$$ $$ \iff\quad a^2+2ab+b^2>a^2+b^2-ab$$ $$\iff\quad 3ab>0 \quad \iff\quad ab>0$$ which is correct because of the hypothesis $a,b >0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/927765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How to solve the integral $\int \frac {(x^2 +1)}{x^4- x^2 +1} dx$ I have started this problem but I'm not completely sure I'm going down the right path with it.
So far I have completed the square in the denominator.
$x^4-x^2+1= (x^2-1/2)^2+\frac{3}{4}$
Then, let $u=x^2-\frac{1}{2}$ so $x=\sqrt(u+\frac{1}{2})$
$\int\f... | $$\frac{x^2+1}{x^4-x^2+1}=\frac{1+\dfrac1{x^2}}{x^2-1+\dfrac1{x^2}}$$
Now $\displaystyle\int\left(1+\dfrac1{x^2}\right)dx=x-\dfrac1x$
and $\displaystyle x^2-1+\dfrac1{x^2}=\left(x-\dfrac1x\right)^2+2-1$
Hope you can take it from here
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/928040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
$\nabla \cdot f + w \cdot f = 0$ Let $w(x,y,z)$ be a fixed vector field on $\mathbb{R}^3$. What are the solutions of the equation
$$
\nabla \cdot f + w \cdot f = 0 \, ?
$$
Note that if $w = \nabla \phi $, then the above equation is equivalent to
$$
\nabla \cdot (e^\phi f) = 0,
$$
for which the solutions are of the for... | You can try to proceed along the following lines:
\begin{equation*}
(\partial _{\mathbf{x}}+\mathbf{w})\cdot \mathbf{f}=0
\end{equation*}
Special case $\mathbf{w}$ is constant. Then
\begin{equation*}
\exp [-\mathbf{w\cdot x}]\partial _{\mathbf{x}}\exp [+\mathbf{w\cdot x}
]=\partial _{\mathbf{x}}+\mathbf{w}
\end{equatio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/929032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
If $a+b+c=1$ and $abc>0$, then $ab+bc+ac<\frac{\sqrt{abc}}{2}+\frac{1}{4}.$ Question:
For any $a,b,c\in \mathbb{R}$ such that $a+b+c=1$ and $abc>0$, show that
$$ab+bc+ac<\dfrac{\sqrt{abc}}{2}+\dfrac{1}{4}.$$
My idea: let
$$a+b+c=p=1, \quad ab+bc+ac=q,\quad abc=r$$
so that
$$\Longleftrightarrow q<\dfrac{\sqrt{r}}{2... | Edit: Incomplete approach. Only works if $a,b,c\geq 0$.
By replacing $\frac{1}{4}$ on the RHS with $\frac{(a+b+c)^2}{4}$, the inequality you seek is equivalent to
$$
a^2+b^2+c^2+2\sqrt{abc}>2(ab+bc+ca).\tag{I}
$$
To prove (I), we use the following result
$$
a^2+b^2+c^2+3(abc)^{2/3}\geq 2(ab+bc+ca)\tag{II}
$$
the proof ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/930427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
The area not covered by six pointed star In a circle with radius $r$, two equi triangles overlapping each other in the form of a six pointed star touching the circumference is inscribed! What is the area that is not covered by the star?
Progress
By subtracting area of the star from area of circle , the area of the surf... | The length of a side of an equilateral triangle is
$$\sqrt{r^2+r^2-2\cdot r\cdot r\cdot \cos (120^\circ)}=\sqrt 3r.$$
The distance between the center of the circle and each side of an equilateral triangle is $$\sqrt 3r\cdot \frac{\sqrt 3}{2}\cdot \frac{1}{3}=\frac 12r.$$
Hence, the length of a side of the smaller equil... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/932172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\frac{1}{1^4}+\frac{1}{2^4}+\cdots+\frac{1}{n^4} \le 2-\frac{1}{\sqrt{n}}$ I have to show: $\displaystyle\frac{1}{1^4}+\frac{1}{2^4}+...+\frac{1}{n^4} \le 2-\frac{1}{\sqrt{n}}$ for natural $n$
I tried to show it by induction (but I think it could be possible to show it using some ineqaulity of means) so for... | You should show its:
$\displaystyle2-\frac{1}{\sqrt{n}}+\frac{1}{(n+1)^4}\le2-\frac{1}{\sqrt{n+1}}$
The other ways does not help your proof (think about it for a while)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/932427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Can someone help me to find a counter example that shows that $a \equiv b \mod m$ does not imply $(a+b)^m \equiv a^m +b^m \mod m$ Can someone help me to find a counter example that shows that $a \equiv b \mod m$ does not imply $(a+b)^m \equiv a^m +b^m \mod m$. I have tried many different values but I can't seem to find... | Take $a=1$, $b=2$ and $m=6$.
$(a+b)^6 = a^6 + 6 a^5 b + 15 a^4 b^2 + 20 a^3 b^3 + 15 a^2 b^4 + 6 a b^5 + b^6$
$(a+b)^6 \equiv a^6 - a^3 b^3 + b^6 \bmod 3$
If $(a+b)^6 \equiv a^6 + b^6 \bmod 6$, then $3$ must divide $a$ or $b$.
It is enough to take $a=1$ and $b=2$ to have $(a+b)^6 \not\equiv a^6 + b^6 \bmod 6$.
In ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/933915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating an indefinite integral $\int\sqrt {x^2 + a^2} dx$ indefinite integral $$\int\sqrt {x^2 + a^2} dx$$
After some transformations and different substitution, I got stuck at this
$$a^2\ln|x+(x^2+a^2)| + \int\sec\theta\tan^2\theta d\theta$$
I am not sure I am getting the first step correct. Tried substituting $ x=... | Here we have another way to see this:
$$
\int \sqrt{x^2+a^2} dx
$$
using the substitution
$$
t=x+\sqrt{x^2+a^2}\\
\sqrt{x^2+a^2}=t-x
$$
and squaring we have
$$
a^2 =t^2-2tx\\
x=\frac{t^2-a^2}{2t}.
$$
Finally we can use:
$$
dx=\frac{2t(t)-(t^2-a^2)(1)}{2t^2}dt = \frac{t^2+a^2}{2t^2}dt\\
\sqrt{x^2+a^2}=t-\frac{t^2-a^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/935519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Question of trigonometry If $\cos^2 A=\dfrac{a^2-1}{3}$ and $\tan^2\left(\dfrac{A}{2}\right)=\tan^{2/3} B$. Then find $\cos^{2/3}B+\sin^{2/3}B $.
I tried componendo and dividendo to write the second statement as cos A but i couldnt simplify it
| $$\cos^{2/3}B+\sin^{2/3}B=\cos^{2/3}B\left(1+\tan^{2/3}B\right)=\cos^{2/3}B\left(1+\tan^2\frac{A}{2}\right)=\left(\cos^2B\right)^{1/3}\left(1+\tan^2\frac{A}{2}\right)=\left(\frac{1}{1+\tan^6\left(\frac{A}{2}\right)}\right)^{1/3}\left(1+\tan^2\frac{A}{2}\right)=\left(1+\frac{3\tan^2 \left(\frac{A}{2}\right)\left(1+\tan^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/936253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How do I solve $\lim_{x\to -\infty}(\sqrt{x^2 + x + 1} + x)$? I'm having trouble finding this limit:
$$\lim_{x\to -\infty}(\sqrt{x^2 + x + 1} + x)$$
I tried multiplying by the conjugate:
$$\lim_{x\to -\infty}(\frac{\sqrt{x^2 + x + 1} + x}{1} \times \frac{\sqrt{x^2 + x + 1} - x}{\sqrt{x^2 + x + 1} - x}) = \lim_{x\to -\i... | Little mistake:
$$
\lim_{x\to -\infty}\frac{x + 1}{\sqrt{x^2 + x + 1} - x} \times \frac{\frac{1}{x}}{\frac{1}{x}} = \lim_{x\to -\infty}\frac{1 + \frac{1}{x}}{{\color{red}-}\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} - 1}=-\frac{1}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/937182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How find this integral $\iint_{D}(x^2y+xy^2+2x+2y^2)dxdy$ let $$D=\{(x,y)|y\ge x^3,y\le 1,x\ge -1\}$$
Find the integral
$$I=\dfrac{1}{2}\iint_{D}(x^2y+xy^2+2x+2y^2)dxdy$$
My idea:
$$I=\int_{0}^{1}dx\int_{x^3}^{1}(x^2y+2y^2)dy+\int_{-1}^{0}dx\int_{0}^{-x^3}(xy^2+2x+2y^2)dy$$
so
$$I=\int_{0}^{1}[\dfrac{1}{2}x^2y^2+\dfrac... | $\frac{67}{90}$ doesn't look correct. Here is what wolfram computes
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/937955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Closed Form of Recursion $a_n = \frac{6}{a_{n-1}-1}$ Given that $a_0=2$ and $a_n = \frac{6}{a_{n-1}-1}$, find a closed form for $a_n$.
I tried listing out the first few values of $a_n: 2, 6, 6/5, 30, 6/29$, but no pattern came out.
| Start with $a_n=\frac6{a_{n-1}-1}$, and replace $a_{n-1}$ with $\frac6{a_{n-2}-1}$. We obtain
$$a_n=\frac6{\frac6{a_{n-2}-1}-1}=\frac{6(1-a_{n-2})}{a_{n-2}-7}$$
Doing this again with $a_{n-2}=\frac6{a_{n-3}-1}$ and so forth, we get
$$a_n=\frac{6(7-a_{n-3})}{7a_{n-3}-13}=\frac{6(13-7a_{n-4})}{13a_{n-4}-55}=\frac{6(55-13... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/939725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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How to simplify $(\sin\theta-\cos\theta)^2+(\sin\theta+\cos\theta)^2$? Simplify: $(\sin \theta − \cos \theta)^2 + (\sin \theta + \cos \theta)^2$
Answer choices:
*
*1
*2
*$ \sin^2 \theta$
*$ \cos^2 \theta$
I am lost on how to do this. Help would be much appreciated.
| $$\begin{align}
&\phantom{=}\left(\sin x-\cos x\right)^2+\left(\sin x+\cos x\right)^2\\
&=\sin^2x-2\sin x\cos x+\cos^2x+\sin^2x+2\sin x\cos x+\cos^2x\\
&=2\sin^2x+2\cos^2x\\
&=2\left(\sin^2x+\cos^2x\right)\\
&=2
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/940738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Infinitely Many Circles in an Equilateral Triangle
In the figure there are infinitely many circles approaching the vertices of an equilateral triangle, each circle touching other circles and sides of the triangle. If the triangle has sides of length 1, find the total area occupied by the circles.
I need to find the ... | Look at the following figure carefully,
As the triangle is equilateral ($AC$ is the angle bisector). So, $\angle ACD = 30^{\circ}$
$$\tan 30^{\circ} = \frac{AD}{DC} = 2AD\ (\because DC = 1/2) $$
$$\therefore AD = \frac{1}{2\sqrt{3}}$$
This is the radius of the bigger circle, let its area be $A_1$
$$\therefore A_1 = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/945123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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If the product $(x+2)(x+3)(x+4)\cdots(x+9)(x+10)$ expands to $a_9x^9+a_8x^8+\dots+a_1x+a_0$, then what is the value of $a_1+a_3+a_5+a_7+a_9$?
When expanded, the product $(x+2)(x+3)(x+4)\cdots(x+9)(x+10)$ can be written as $a_9x^9+a_8x^8+\dots+a_1x+a_0$. What is the value of $a_1+a_3+a_5+a_7+a_9$?
| Let $P(x) = (x+2)(x+3)..(x+10)$. Then what is $\dfrac{P(1)-P(-1)}2=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/945590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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finding a function from given function here is a function for:
$f(x-\frac{\pi}{2})=\sin(x)-2f(\frac{\pi}{3})$
what is the $f(x)$?
I calculate $f(x)$ as follows:
$$\begin{align}
x-\frac{\pi}{2} &= \frac{\pi}{3} \Rightarrow x= \frac{5\pi}{6} \\
f(\frac{\pi}{3}) &=\sin\frac{5\pi}{6}-2f(\frac{\pi}{3}) \\
3f(\frac{\pi}{3}) ... | Assuming $f$ is defined for all $x\in\mathbb{R}$. First, note that for any $x$,
$$
f(x) = \sin\!\left(x+\frac{\pi}{2}\right)-2f\!\left(\frac{\pi}{3}\right) = \cos x -2f\!\left(\frac{\pi}{3}\right)
$$
so it only remains to compute $f\!\left(\frac{\pi}{3}\right)$. From the expression above
$$
f\!\left(\frac{\pi}{3}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/947010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Closed form of $\sum_{n=1}^{\infty}(-1)^{n+1} n (\log(n^2+1)-\log(n^2))$ How would you start computing this series?
$$\sum_{n=1}^{\infty}(-1)^{n+1} n (\log(n^2+1)-\log(n^2))$$
One of the ways to think of would be Frullani integral with the exponential function , but it's troublesome due to the power of $n$ under logar... | I'm getting the same answer as you.
$$ \sum_{n=1}^{\infty} (-1)^{n+1} n \log \left( \frac{n^{2}+1}{n^{2}}\right) = \sum_{n=1}^{\infty} (-1)^{n+1} n \int_{0}^{1} \frac{1}{n^{2}+x} \ dx$$
Then since $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}n}{n^{2}+x}$ converges uniformly on $[0,1]$,
$$ \begin{align} &\sum_{n=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/947187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Prove $\frac{n^2+2}{(2 \cdot n^2)-1} \to \frac{1}{2}$
Prove $\frac{n^2+2}{(2 \cdot n^2)-1} \to \frac{1}{2}$ for $n \to \infty$.
I've been looking at this for hours! Also, sorry I don't have the proper notation.
This is where I'm at:
$$
\left| \frac{n^2 + 2}{2 \cdot n^2 - 1} - \frac{1}{2}\right| = \left| \frac{5}{4 ... | $\left|\frac{n^2+2}{2n^2-1} - \frac {1}{2}\right| = \frac{5}{4n^2-2}$
Let $\epsilon>0$ be given. Let $n_0$ be the smallest integer such that $n\geq n_0> \sqrt{\frac{5}{4\epsilon} + \frac 12}$. Equivalently, $\epsilon>\frac{5}{4n^2-2}$.
Thus, $\left|\frac{n^2+2}{2n^2-1} - \frac {1}{2}\right|< \epsilon$, for all $n\geq ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/947452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
3 Variable Diophantine Equation Find all integer solutions to $$x^4 + y^4 + z^3 = 5$$ I don't know how to proceed, since it has a p-adic and real solution for all $p$.
I think that the only one is (2, 2, -3) and the trivial ones that come from this, but I can't confirm it.
| After a careful investigation I present some results which might be helpful in the final resolution of this problem. Lets start with the original equation i.e.
$$x^4+y^4+z^3=5$$
It is easy to see that there is no solution of this equation where $x,y$ and $z$ are all positive. Second if $(x,y,z)$ is a solution then so i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/947680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
2nd order homogeneous ODE I am trying to solve a system of differential equations (for the full system see below) and I am stuck the following 2nd order ODE (with $a$ and $b$ being constants and $\dot x = \frac{dx}{dt}$):
$$\ddot x - \frac34 a\dot x^2 -b \dot x + 2 a b = 0$$
I tried to substitute $v := \dot x$, which l... | As I said in a comment, start defining $z=x'$; so the differential equation becomes $$\frac{dz}{dt} - \frac34 a z^2 -b z + 2 a b = 0$$ that is to say $$\frac{dz}{dt} = \frac34 a z^2 +b z - 2 a b $$ then, as Semsem suggested, it is separable; so $$\frac{dt}{dz} = \frac{1}{\frac34 a z^2 +b z - 2 a b}$$ so $$t+C=-\frac{2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/948030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How do I solve this geometric series I have this geometric series $2+1+ \frac{1}{2}+ \frac{1}{4}+...+ \frac{1}{128}$to solve. So I extract the number two and get $2(\frac{1}{2}^0+ \frac{1}{2}^1+...+ \frac{1}{2}^7)$
I use the following formula $S_n= \frac{x^{n+1}-1}{x-1}$ so I plug in the values in this formula and get ... | Hint: Its $$2 + (1+\frac12+ \frac14+ \cdots + \frac{1}{128})$$ not multiplied with $2$.
You can also think of it as follows: The first term is $a_1=2$ and the common ratio is $r=1/2$ and then you sum it using the formula where you last term is $a_9=1/128$.
Edit: If you do want to factor out a $2$, then you get $$2(1+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/948146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Finding cartesian equation for trigonometric parametric forms I'm trying to find the cartesian equation for these parameteric forms:
$$
x = sin\theta + 2 cos \theta \\
y = 2 sin\theta + cos\theta
$$
I tried:
$$\begin{align}
x^2 & = sin^2\theta + 4cos^2\theta \\
& = 1 - cos^2\theta + 4cos^2\theta \\
& = 1 + 3cos^2\the... | In general, $(a+b)^2\ne a^2+b^2$ unless $ab=0$
Solve for $\sin\theta,\cos\theta$ in terms of $x,y$
Then use $\sin^2\theta+\cos^2\theta=1$ to eliminate $\theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/949290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Finding Solutions to Trigonometric Equation Find all $x$ in the interval (0, $\frac{\pi}{2}$) such that $$\frac{\sqrt{3}-1}{\sin x} + \frac{\sqrt{3}+1}{\cos x} = 4\sqrt{2}.$$
| Rewrite it in the form
$$2\sqrt2\left(\frac{\sqrt3-1}{2\sqrt2}\cos x+\frac{\sqrt3+1}{2\sqrt2}\sin x\right)=2\sqrt2\sin 2x.$$
For $\phi=\arcsin\frac{\sqrt3-1}{2\sqrt2}$ it implies
$$\sin(x+\phi)=\sin 2x,$$
i.e. $x+\phi=2x+2\pi n$ or $x+\phi=\pi-2x+2\pi n$, $n\in\Bbb Z$. Therefore, the only solutions in $(0,\pi/2)$ are $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/949390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\int\frac{2x+1}{x^2+2x+5}dx$ by partial fractions $$\int\frac{2x+1}{x^2+2x+5}dx$$
I know I'm supposed to make the bottom a perfect square by making it $(x+1)^2 +4$ but I don't know what to do after that. I've tried to make $x+1= \tan x$ because that's what we did in a class example but I keep getting stuck.
| $$
\int\frac{2x+1}{x^2+2x+5}dx
$$
I would first write $w=x^2+2x+5$, $dw=(2x+2)\,dx$, and then break the integral into
$$
\int\frac{2x+2}{x^2+2x+5}dx + \int\frac{-1}{x^2+2x+5}dx.
$$
For the first integral I would use that substitution. Then
$$
\overbrace{\int\frac{-dx}{x^2+2x+5} = \int\frac{-dx}{(x+1)^2 + 2^2}}^{\text{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/950354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Convergence of $\sum \frac{(2n)!}{n!n!}\frac{1}{4^n}$ Does the series
$$\sum \frac{(2n)!}{n!n!}\frac{1}{4^n}$$
converges?
My attempt: Since the ratio test is inconclusive, my idea is to use the Stirling Approximation for n!
$$\frac{(2n)!}{n!n!4^n} \sim (\frac{1}{4^n} \frac{\sqrt{4\pi n}(\frac{2n}{e})^{2n}}{\sqrt{2 n \p... | A much simpler way:
$$
\frac{a_{n+1}}{a_n}=\frac{(2n+2)(2n+1)}{4(n+1)(n+1)}=\frac{2n+1}{2n+2}\ge
\sqrt{\frac{n}{n+1}},
$$
since
$$
\left(\frac{2n+1}{2n+2}\right)^2=\left(1-\frac{1}{2n+2}\right)^2\ge 1-\frac{2}{2n+2}
=\frac{n}{n+1},
$$
and hence
$$
a_n=a_1\prod_{k=2}^n\frac{a_{k}}{a_{k-1}}\ge a_1\prod_{k=2}^n\sqrt{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/951171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Given $a+b+c=4$ find $\max(ab+ac+bc)$ $a+b+c = 4$.
What is the maximum value of $ab+ac+bc$?
Could this be solved by a simple application of Jensen's inequality? If so, I am unsure what to choose for $f(x)$. If $ab+ac+bc$ is treated as a function of $a$ there seems no easy way to express $bc$ in terms of $a$.
EDIT: The ... | Jensen's Inequality gives
$$
\left[\frac13(a+b+c)\right]^2\le\frac13(a^2+b^2+c^2)
$$
and we know that
$$
\begin{align}
ab+bc+ca
&=\frac12\left[(a+b+c)^2-(a^2+b^2+c^2)\right]\\
&\le\frac12\left[(a+b+c)^2-\frac13(a+b+c)^2\right]\\
&=\frac13(a+b+c)^2
\end{align}
$$
and equality can be achieved when $a=b=c$. Therefore, if ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/951997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Integration without substitution How to i integrate this with out substitutions or Partial fraction decomposition ?
($3x^2$+$2$)/[$x^6$($x^2$+1)]
I've got to : 2/x^6(x^2+1),but after this i haven't been able to eliminate the 2.
| You can use $\displaystyle\frac{3x^2+2}{x^6(x^2+1)}=\frac{2(x^2+1)}{x^6(x^2+1)}+\frac{x^2}{x^6(x^2+1)}=\frac{2}{x^6}+\frac{1}{x^4(x^2+1)}$
$\displaystyle=\frac{2}{x^6}+\left(\frac{x^2+1}{x^4(x^2+1)}-\frac{x^2}{x^4(x^2+1)}\right)=\frac{2}{x^6}+\frac{1}{x^4}-\frac{1}{x^2(x^2+1)}$
$\displaystyle=\frac{2}{x^6}+\frac{1}{x^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/953936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Find the value of $\sqrt{10\sqrt{10\sqrt{10...}}}$ I found a question that asked to find the limiting value of $$10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{...}}}}}$$If you make the substitution $x=10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{...}}}}}$ it simplifies to $x=10\sqrt{x}$ which has solutions $x=0,100$. I don't unde... | Denote the given problem as $x$, then
\begin{align}
x&=10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{\cdots}}}}}\\
&=10\cdot10^{\large\frac{1}{2}}\cdot10^{\large\frac{1}{4}}\cdot10^{\large\frac{1}{8}}\cdot10^{\large\frac{1}{16}}\cdots\\
&=10^{\large1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots}\\
&=10^{\large y}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/955190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
"answer_count": 7,
"answer_id": 4
} |
Prove that $ \int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x =\int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x $ In a related question the following integral was evaluated
$$
\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x
=\int_0^{\pi} \frac{\mathrm{d}x/2}{1 +... | $$
\frac{\cos(\pi/2-x)^2}{1+\cos(\pi/2-x)\sin(\pi/2-x)} =
\frac{\sin(x)^2}{1+\sin(x)\cos(x)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/955294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 4
} |
Any shorter way to solve trigonometric problem? If $10 \sin^4\theta + 15 \cos^4 \theta=6$, then find value of $27 \csc^2 \theta + 8\sec^2 \theta$
I know the normal method o solve this problem in which we need to multiply L.H.S. of $10 \sin^4\theta + 15 \cos^4 \theta=6$ by $(\sin^2\theta + \cos \theta^2)^2$ and then sim... | How can $20\sin^2\theta + 15\cos^2\theta = 6$? $20\sin^2\theta + 15 \cos^2\theta = 15 + 5\sin^2\theta$, implying that $5 \sin^2\theta = -9$, but $\sin^2\theta$ should be nonnegative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/955698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Sum the series (real analysis) $$\sum_{n=1}^\infty {1 \over n(n+1)(n+2)(n+3)(n+4)}$$
I tried to sum the above term as they way I can solve the term $\sum_{n=1}^\infty {1 \over (n+3)}$ by transforming into ${3\over n(n+3)} ={1\over n}-{1\over(n+3)}$ but I got stuck while trying to transform $12\over n(n+1)(n+2)(n+3)(n+4... | By working it out, by various methods, it is seen that
\begin{align}
\frac{1}{n(n+1)(n+2)(n+3)(n+4)} = \frac{1}{4!} \, \sum_{r=0}^{4} (-1)^{r} \binom{4}{r} \, \frac{1}{n+r}
\end{align}
Now,
\begin{align}
S &= \sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)(n+3)(n+4)} \\
&= \frac{1}{4!} \, \sum_{r=0}^{4} (-r)^{r} \binom{4}{r} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/956221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Sums with squares of binomial coefficients multiplied by a polynomial It has long been known that
\begin{align}
\sum_{n=0}^{m} \binom{m}{n}^{2} = \binom{2m}{m}.
\end{align}
What is being asked here are the closed forms for the binomial series
\begin{align}
S_{1} &= \sum_{n=0}^{m} \left( n^{2} - \frac{m \, n}{2} - \frac... | HINT:
As $\displaystyle n\binom mn=m\binom{m-1}{n-1},$
$\displaystyle\sum_{n=0}^mn\binom mn^2=m\sum_{n=0}^m\binom mn\binom{m-1}{n-1}=m\binom{2m-1}m$
comparing the coefficient of $x^{2m-1}$ in $(1+x)^m(1+x)^{m-1}=(1+x)^{2m-1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/957605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove by contradiction that $(x-y)^3+(y-z)^3+(z-x)^3 = 30$ has no integer solutions By factorizing it I found that
$(x-y)(y-z)(z-x) = 10$
| Just as Karvens comments:
Let $a=x-y$ and $b=y-z$. Then $-(a+b)=z-x$. Clearly, $a,b,c \in \Bbb Z$. So
$$30=a^3+b^3-(a+b)^3=(a+b)(a^2-ab+b^2)-(a+b)^3=-3ab(a+b)$$
And hence
$$10=-ab(a+b).$$
Therefore $a=\pm 1$ or $a=\pm 2$ or $a=\pm 5$ or $a=\pm 10$. Claerly they cannot be the solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/957875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Distance between powers of 2 and 3 As we know $3^1-2^1 = 1$ and of course $3^2-2^3 = 1$. The question is that whether set $$ \{\ (m,n)\in \mathbb{N}\quad |\quad |3^m-2^n| = 1 \} $$
is finite or infinite.
| Note first that if $3^{2n}-1=2^r$ then $(3^n+1)(3^n-1)=2^r$. The two factors in brackets differ by $2$ so one must be an odd multiple of $2$, and this is only possible if $n=1$ (the only odd number we can allow in the factorisation is $1$)
Now suppose that $3^n-1=2^r$ and $n$ is odd. Now $3^n\equiv -1$ mod $4$ so $3^n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/958304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculation of $\int\frac1{\tan \frac{x}{2}+1}dx$ Calculation of $\displaystyle \int\frac{1}{\tan \frac{x}{2}+1}dx$
$\bf{My\; Try}::$ Let $\displaystyle I = \displaystyle \int\frac{1}{\tan \frac{x}{2}+1}dx$, Now let $\displaystyle \tan \frac{x}{2}=t\;,$ Then $\displaystyle dx=\frac{2}{1+t^2}dt$
So $\displaystyle I = 2\... | Hint:
Multiply the denominator and nominator of the fraction by $\cos x/2(\sin x/2-\cos x/2)$. At the bottom you get $-\cos x$, on the top you get $-\cos^2 x/2+1/2\sin x$.
What can you do with $\cos^2 x/2$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/958679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
How to derive the closed form of this recurrence? For the recurrence, $T(n) = 3T(n-1)-2$, where $T(0)= 5$, I found the closed form to be $4\cdot 3^n +1$(with help of Wolfram Alpha). Now I am trying to figure it out for myself. So far, I have worked out: $T(n-1) = 3T(n-2)-2, T(n-2) = 3T(n-3)-2, T(n-3) = 3T(n-4)-2$ leadi... | By making use of the generating function method the difference equation $T_{n} = 3 T_{n-1} - 2$, $T_{0} = 5$, can be seen as follows
\begin{align}
T(x) &= \sum_{n=0}^{\infty} T_{n} x^{n} = 3 \sum_{n=0}^{\infty} T_{n-1} x^{n} - 2 \sum_{n=0}^{\infty} x^{n} \\
T(x) &= 3 \left( T_{-1} + \sum_{n=1}^{\infty} T_{n-1} x^{n} \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/958781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Integral of $\int \frac{\cos \left(x\right)}{\sin ^2\left(x\right)+\sin \left(x\right)}dx$ What is the integral of $\int \frac{\cos \left(x\right)}{\sin ^2\left(x\right)+\sin \left(x\right)}dx$ ?
I understand one can substitute $u=\tan \left(\frac{x}{2}\right)$
and one can get (1) $\int \frac{\frac{1-u^2}{1+u^2}}{\left... | Consider the integral
\begin{align}
I = \int \frac{\cos(x) \, dx}{\sin^{2}(x) + \sin(x) } .
\end{align}
Method 1
Make the substitution $u = \tan\left(\frac{x}{2}\right)$ for which
\begin{align}
\cos(x) &= \frac{1-u^{2}}{1+u^{2}} \\
\sin(x) &= \frac{2u}{1+u^{2}} \\
dx &= \frac{2 \, du}{1+u^{2}}
\end{align}
and the int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/959992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Can't get this implicit differentiation I've been working at this implicit differentiation problem for a little over an hour now, and I, nor my friends can figure it out. The question reads "Find the equation of the tangent line to the curve (a lemniscate) $2(x^2+y^2)^2=25(x^2−y^2)$ at the point (3,1). Write the equati... | We have
$2(x^2+y^2)^2=25(x^2−y^2)$.
Since
$(x^2)' = 2 x \ dx$,
I would differentiate this as
$2(2(x^2+y^2)(x^2+y^2)')
=25(2x\ dx - 2y\ dy)
$
or
$4(x^2+y^2)(2x\ dx-2 y\ dy)
=25(2x\ dx - 2y\ dy)
$
or
$8(x(x^2+y^2)dx-y(x^2+y^2)dy)
=50x\ dx-50y\ dy
$
or
$(8x(x^2+y^2)-50x)dx
=(8y(x^2+y^2)-50y)dy
$
or
$\dfrac{dy}{dx}
=\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/964720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Proving $\frac{\sin x + \sin 2x + \sin3x}{\cos x + \cos 2x + \cos 3x} = \tan2x$ I need to prove:
$$
\frac{\sin x + \sin 2x + \sin3x}{\cos x + \cos 2x + \cos 3x} = \tan2x
$$
The sum and product formulae are relevant:
$$
\sin(A + B) + \sin (A-B) = 2 \sin A \cos B \\
\sin(A + B) - \sin (A-B) = 2 \cos A \sin B \\
\cos(A + ... | \begin{align}
\frac{\sin x + \sin 2x + \sin3x}{\cos x + \cos 2x + \cos 3x}&=\frac{\sin x + \sin 3x + \sin2x}{\cos x + \cos 3x + \cos 2x}\\
&=\frac{2\sin 2x\cos x+ \sin 2x}{2\cos 2x\cos x + \cos 2x}\\
&=\frac{\sin2x(2\cos x+1)}{\cos 2x(2\cos x + 1)}\\
& = \tan2x
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/965329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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implicit equation for elliptical torus I just wondering what the implicit equation would be if an ellipse with major axis a and minor axis b, rotating about the Z axis with a distance of $R_0$. The $R_0$>a and $R_0$>b which means the rotation will result in a non-degenerate torus.
My aim is to determine if some points... | You can obtain this as follows.
If you start with a slice where $y = 0$, you begin with the equation
$$
\frac{z^2}{a^2} + \frac{(x - R_0)^2}{b^2} - 1 = 0
$$
However, this doesn't give you the rotated version; to rotate it about the $z$-axis, simply replace the $x$ by $\sqrt{x^2 + y^2}$, yielding
$$
\frac{z^2}{a^2} + \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/966503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
How to find all solutions of $\tan(x) = 2 + \tan(3x)$ without a calculator? Find all solutions of the equation $\tan(x) = 2 + \tan(3x)$ where $0<x<2\pi$.
By replacing $\tan(3x)$ with $\dfrac{\tan(2x) + \tan(x)}{1-\tan(2x)\tan(x)}$
I've gotten to $\tan^3 (x) - 3 \tan^2 (x) + \tan(x) + 1 =0$.
I am not sure how to procee... | Let $\tan x=t$. Then, we have
$$t=2+\frac{3t-t^3}{1-3t^2}\Rightarrow (t-2)(1-3t^2)=3t-t^3$$$$\Rightarrow t^3-3t^2+t+1\Rightarrow (t-1)(t^2-2t-1)=0.$$
Hence, we have
$$\tan x=t=1,1\pm\sqrt 2\Rightarrow x=\frac{\pi}{4}+n\pi,\frac{3}{8}\pi+n\pi,-\frac{\pi}{8}+n\pi$$
where $n\in\mathbb Z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/970284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Given the following derivatives, find the integrals
Find the derivatives of $\ln(x+\sqrt{x^2+1})$ and $\arcsin(x)$, and use the result to find the integrals of the following functions:
*
*$$ \dfrac{1}{ \sqrt{ \pm x^2 \pm a^2 }} $$
*$$ \sqrt{\pm x^2 \pm a^2} $$
Except for the cases where both are minuses. $a$ is a ... | Hint. You may use an integration by parts for the second family:
$$
\begin{align}
\int \sqrt{\pm x^2 \pm a^2} \:{\rm{d}}x&=x\sqrt{\pm x^2 \pm a^2}-\int \frac{x \times\pm x }{\sqrt{\pm x^2 \pm a^2}}\:{\rm{d}}x\\
&=x\sqrt{\pm x^2 \pm a^2}-\int \frac{\pm x^2}{\sqrt{\pm x^2 \pm a^2}}\:{\rm{d}}x\\
&=x\sqrt{\pm x^2 \pm a^2}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/972735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Sum of roots of an equation $\sqrt{x-1}+\sqrt{2x-1}=x$ Find the sum of the roots of the equation $\sqrt{x-1}+\sqrt{2x-1}=x$
My attempt: Squaring the equation: $(x-1)+(2x-1) +2\sqrt{(x-1)(2x-1)}=x^2$
$\implies x^2-3x+2=2\sqrt{(x-1)(2x-1)} $
$\implies (x-1)(x-2)=2\sqrt{(x-1)(2x-1)} $
$\implies (x-2)=2\sqrt{\displaystyle ... | The problem is that you divided by $x-1$, so, you lost a root.Strating from $$ (x-1)(x-2)=2\sqrt{(x-1)(2x-1)}$$ as you properly wrote and squaring $$(x-1)^2(x-2)^2=4{(x-1)(2x-1)}$$ Expanding and grouping leads to $$x^4-6 x^3+5 x^2=0$$ so the sum of the roots is $6$ (you can check that the roots are $0,0,1,5$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\int\frac{dx}{(x^2+1)\sqrt{x^2+2}}$ I would like some guidance regarding the following integral:
$$\int\frac{dx}{(x^2+1)\sqrt{x^2+2}}$$
EDIT: The upper problem was derived from the following integral $$\int\frac{\sqrt{x^2+2}}{x^2+1}dx$$
Where I rationalized the numerator which followed into: $$\int\frac{dx}{... | There is a general formula for it. $$\int \frac{dx}{(x^2+1)\sqrt{x^2+a}}=\frac{1}{\sqrt{a-1}}\tan^{-1}\left(\frac{\sqrt{a-1}x}{\sqrt{x^2+a}}\right)+C\tag{1}$$
$a=2$ gives $$\int \frac{dx}{(x^2+1)\sqrt{x^2+2}}=\tan^{-1}\left(\frac{x}{\sqrt{x^2+2}}\right)+C$$
Formula $(1)$ can be proven by substitution : $t=1/x$, $s=\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/978177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Estimating the behavior for large $n$ I want to find how these coefficients increase/decrease as $n$ increases:
$$ C_n = \frac{1}{n!} \left[(n+\alpha)^{n-\alpha-\frac{1}{2}}\right]$$
with $\alpha=\frac{1}{br-1}$ and $0\leq b,r \leq 1$.
I used the Stirling's Approximation factorial $n!\sim \sqrt{2\pi n} n^n e^{-n}$ a... | Taking the reciprocal of Stirling's Asymptotic expansion as derived in this answer:
$$
n!=\frac{n^n}{e^n}\sqrt{2\pi n}\left(1+\frac{1}{12n}+\frac{1}{288n^2}-\frac{139}{51840n^3}+O\left(\frac{1}{n^4}\right)\right)
$$
we get
$$
\frac1{n!}=\frac{e^n}{n^n}\frac1{\sqrt{2\pi n}}\left(1-\frac{1}{12n}+\frac{1}{288n^2}+\frac{13... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/978422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$
If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$
Hint: solve for $b^2$ in terms of $a^2$ and then solve for $a$
I've attempted the question but I don't think I've done it correctly:
$$
\begin{align*}
b^2 &= 4 - a^2\\
b &= \sqrt{4-a^2}
\end{align*}
$$
Therefore,
$$
\... | Let $z=a+ib,$ then it is given that $z^3=8.$ Therefore taking the modulus of both sides $|z|^3=8.$ Hence $|z|=2$ and $|z|^2=a^2+b^2=4.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/979252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.