Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Finding solutions to $2^x+17=y^2$
Find all positive integer solutions $(x,y)$ of the following equation:
$$2^x+17=y^2.$$
If $x = 2k$, then we can rewrite the equation as $(y - 2^k)(y + 2^k) = 17$, so the factors must be $1$ and $17$, and we must have $x = 6, y = 9$.
However, this approach doesn't work when $x$ is ... | A largely complete answer:
Starting from $2^x+17=y^2$, we obtain $2^x+2^4=(y-1)(y+1)$. By observation, $x=4$ fails as $33\ne y^2$. Note that $(y-1),(y+1)$ are two consecutive even numbers, so their difference is $2$, and one of them contains a single factor of $2$ while the other contains more than one factor of $2$.
C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2298402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 2
} |
Proving Trigonometric Equality I have this trigonometric equality to prove:
$$\frac{\cos x}{1-\tan x}-\frac{\sin x}{1+\tan x}=\frac{\cos x}{2\cos^2x-1}$$
I started with the left hand side, reducing the fractions to common denominator and got this:
$$\frac{\cos x+\cos x\tan x-\sin x+\sin x\tan x}{1-\tan^2x}\\=\frac{\cos... | $$\frac { cosx }{ 1-tanx } -\frac { sinx }{ 1+tanx } =\frac { cosx }{ 2cos^{ 2 }x-1 } \\ \frac { \cos { x } \left( 1+tanx \right) }{ \left( 1-tanx \right) \left( 1+tanx \right) } -\frac { \sin { x } \left( 1-tanx \right) }{ \left( 1-tanx \right) \left( 1-tanx \right) } =\\ =\frac { \cos { x } +\sin { x } -\sin { x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2299015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 1
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Find limit $a_{n + 1} = \int_{0}^{a_n}(1 + \frac{1}{4} \cos^{2n + 1} t)dt,$ Find the limit of the sequence:
$$a_{n + 1} = \int_{0}^{a_n}(1 + \frac{1}{4} \cos^{2n + 1} t)dt,$$
such that $a_0 \in (0, 2 \pi)$
That was one of the tasks in the Olympiad.
Here is my approach.
First, I wanted to simplify the integral:
$\int... | You can easily verify that $$I_(2n+1) \to 0$$,$$ as n\to \infty$$
And $$\lim_{n\to\infty} a_(n+1)-a_n =0$$
And a_1=\int_{0}{a_0} (1+cost)dt=2π
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2300001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Using De Moivre's Theorem for summation
I am able to get the required denominator but I keep ending with 4 terms in numerator. Can somebody help me out in this question?( I am using geometric sum identity)
| $$\Im\sum_{n=1}^{10} \left(\frac{1}{2}e^{i\pi/10}\right)^n=\Im\left(\frac{1}{2}e^{i\pi/10}\frac{1-(\frac{1}{2}e^{i\pi/10})^{10}}{1-\frac{1}{2}e^{i\pi/10}}\right)$$
$$=\Im\left(\frac{1}{2}e^{i\pi/10}\frac{1-\frac{1}{1024}e^{i\pi}}{1-\frac{1}{2}e^{i\pi/10}}\right)$$
$$=\Im\left(\frac{1}{2}e^{i\pi/10}\frac{1+\frac{1}{1024... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2300201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find power series of $f(z)=\frac{e^z}{1-z}$ at $z_0=0$? I tried to calculate few derivatives, but I cant get $f^{(n)}(z)$ from them. Any other way?
$$f(z)=\frac{e^z}{1-z}\text{ at }z_0=0$$
| $$
g(z) = a_0+a_1z+a_2z^2+...,\\
\frac{g(z)}{1-z}=a_0\frac{1}{1-z}+a_1\frac{z}{1-z}+a_2\frac{z^2}{1-z}+...
$$
Using the power series for $\frac{1}{1-z}$ gives
$$
g(z)=a_0(1+z+z^2+...)+a_1z\cdot(1+z+z^2+...)+a_2z^2\cdot(1+z+z^2+...),\\
g(z)=a_0(1+z+z^2+...)+a_1\cdot(z+z^2+z^3+...)+a_2\cdot(z^2+z^3+z^4...),\\
g(z)=a_0+(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2300613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Calculating $\sum_{k=1}^\infty \frac{k^2}{2^k}=\frac{1}{2}+\frac{4}{4}+\frac{9}{8}+\frac{16}{16}+\frac{25}{32}+\cdots+\frac{k^2}{2^k}+\cdots$ I want to know the value of $$\sum_{k=1}^\infty \frac{k^2}{2^k}=\frac{1}{2}+\frac{4}{4}+\frac{9}{8}+\frac{16}{16}+\frac{25}{32}+\cdots+\frac{k^2}{2^k}+\cdots$$
I added up to $k=5... | Alternatively, note that if $T = \sum_{k \geq 0}k^2 2^{-k}$ then $$T = 2T-T = \sum_{k\geq 0}\frac{2k+1}{2^k} = 2\sum_{k \geq 0}\frac{k}{2^k} + \sum_{k \geq 0}2^{-k} $$
But we can let $S= \sum_{k \geq 0}k2^{-k}$ and note that $$S = 2S-S = \sum_{k\geq 0}2^{-k} = 2.$$
Hence $T = 2\cdot 2 + 2 = 6.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2300889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Solve the Inequality $x+\frac{x}{\sqrt{x^2-1}} \gt \frac{35}{12}$ Solve the Inequality $$x+\frac{x}{\sqrt{x^2-1}} \gt \frac{35}{12}$$
First of all the Domain of LHS is $(-\infty \:\: -1) \cup (1 \:\: \infty)$
So i assumed $x=\sec y$ since Range of $\sec y$ is $(-\infty \:\: -1) \cup (1 \:\: \infty)$
So
$$\sec y+ |\cs... | We need only concerntrate on the interval $(1,\infty)$ ( the expression is clearly negative in the other interval) ... now change the inequality into an equation, in order to find the points where the expression will change sign
\begin{eqnarray*}
x+\frac{x}{\sqrt{x^2-1}}=\frac{35}{12} \\
\frac{x^2}{(x^2-1)}= \left( \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2301181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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How to find the area of the surface enclosed between a sphere $x^2+y^2+z^2=4a^2$ and a cylinder $x^2+(y-a)^2=a^2$?
Find the area of the surface enclosed between a sphere $x^2+y^2+z^2=4a^2$ and a cylinder $x^2+(y-a)^2=a^2$. The correct answer should be $(8\pi-16)a^2$.
This is an illustration:
First, there's the follo... | After some chat and subsequent corrections, all is clear now. Continuing from the end of the question,
$$S/2=\int_0^{\pi}\int_0^{2a\sin\theta}r\sqrt{\dfrac{r^2}{4a^2-r^2}+1}drd\theta=$$
$$=\int_0^{\pi}\int_0^{2a\sin\theta}\frac {2ar}{\sqrt{4a^2-r^2}}drd\theta=2a\int_0^{\pi}\left(-\sqrt{4a^2-4a^2\sin^2\theta}+2a\right)d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2305680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to find the number of solution pairs for this equation? Question: Consider the equation $(1+a+b)^2=3(1+a^2+b^2)$ where $a$ and $b$ are real numbers. How many solution pair(s), ($a$,$b$), are possible for this equation?
My attempt:
$(1+a+b)^2=3(1+a^2+b^2)$
=>$1+a^2+b^2+2(a+ab+b)=3+3a^2+3b^2$
=>$1+a^2+b^2+2a+2ab+2b=... | Hint: after expanding and collecting, the equation can be written as:
$$a^2 - a b - a + b^2 - b + 1 = 0 \;\;\iff\;\; a^2 - (b+1)\,a + b^2-b+1=0$$
Considering it as a quadratic in $\,a\,$, its discriminant is:
$$
\Delta=(b+1)^2-4(b^2-b+1)=-3b^2+6b-3=-3(b-1)^2
$$
For the quadratic to have real roots the discriminant mu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2306016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Sketch the graph of $F_4^n(x)$ on the unit interval, where $F_4(x) = 4x(1-x)$. Conclude that $F_4$ has at least $2^n$ periodic points of period $n$. Sketching the graph I found that $F_4$ has exactly $2^{n-1}$ point wich prime period $n$. My $F_4^n$ graph look like a $|sin(x)|$ graph with $2^n$ intersections on $Id$, b... | Incomplete and with typos ... but contains a few hints and references.
Given the following functions:
$$\varphi(x)=-4x+2,\varphi^{-1}(x)=\frac{x-2}{-4}, g(x)=x^2-2$$
we have:
$$F_4(x)=(\varphi^{-1} \circ g\circ \varphi)(x)$$
or
$$F_4^{\circ n}(x)=(\varphi^{-1} \circ g^{\circ n} \circ \varphi)(x)$$
Further to this, fun... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2306233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Trouble understanding this Equality of Complex number question. I have this question here
$$(3-2j)(x+yj)=4+j^9$$
I know that $a + bj$ and $x + yj$ are equal when $a = x$ and $b = y$.
This question is confusing me because I have had three different answers but they are all wrong.
What feels like my closest attempt was... | \begin{align*}(3-2i)(x+yi)=4+i^9&\Longleftrightarrow x+yi=\frac{4+i}{3-2i}\\&\Longleftrightarrow x+yi=\frac{(4+i)(3+2i)}{(3-2i)(3+2i)}\\&\Longleftrightarrow x+yi=\frac{10+11i}{13}=\frac{10}{13}+\frac{11}{13}i\\&\Longleftrightarrow x=\frac{10}{13}\text{ and }y=\frac{11}{13}\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
How to calculate $\int{\frac {x^2}{ ( x\cos x -\sin x ) ^2}}\,{\rm d}x$? When I want to calculate
$$\int{\frac {x^2}{ ( x\cos x -\sin x ) ^2}}\,{\rm d}x$$
I have tested with software and get
$${\frac {x\sin \left( x \right) +\cos \left( x \right) }{x\cos \left( x
\right) -\sin \left( x \right) }}$$
But I can not come ... | We can rewrite $$\begin{align}\int \frac{x^2}{(\sin x - x\cos x)^2} \, \mathrm{d}x &= \int \frac{x\sin x(x\sin x + \cos x)}{(\sin x - x\cos x)^2} \, \mathrm{d}x - \int \frac{x\cos x}{\sin x - x\cos x} \, \mathrm{d}x \\ & = -\frac{x\sin x + \cos x}{\sin x - x\cos x} \color{green}{+} \int \frac{x\cos x}{\sin x- x\cos x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Ellipse in a Rectangle What is the equation for an ellipse (or rather, family of ellipses) which has as its tangents the lines forming the rectangle $$x=\pm a, y=\pm b\;\; (a,b>0)$$?
This question is a modification/extension of this other question here posted recently.
| Here's a purely algebraic approach without any use of trigonometric functions.
By symmetry, the center of the ellipse is at the origin, so its
equation is just
$$
Ax^2 + Bxy + Cy^2 = 1
$$
with $A>0,$ $C>0,$ and $B^2 < 4AC.$
The tangents at $y = \pm b$ will occur at values of $y$ for which the
quadratic equation in $x,$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2308198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Calculate the value of the series $\,\sum_{n=1}^\infty\frac{1}{2n(2n+1)(2n+2)}$ Calculate the infinite sum
$$\dfrac{1}{2\cdot 3\cdot 4}+ \dfrac{1}{4\cdot 5\cdot 6}+\dfrac{1}{6\cdot 7\cdot 8}+\cdots$$
I know this series is convergent by Comparison Test, but I can't understand how can I get the value of the sum.
Is th... | In the style suggested by Michael Rozenberg's comment (solution 2), let us take one and two terms out of Mercator series
$$\begin{align}
S_0=\log(2)&=\sum_{k=0}^\infty\left(\frac{1}{2k+1}-\frac{1}{2k+2}\right)\\
&=1-\sum_{k=0}^\infty \left(\frac{1}{2k+2}-\frac{1}{2k+3}\right)\\
&=1-\frac{1}{2}+\sum_{k=0}^\infty \left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Proof of limit using epsilon-L proof We have $\lim \frac{n+6}{n^2-6}=0$. I need to prove this limit using the epsilon-L proof method.
I wrote down the following proof:
$\dfrac{n+6}{n^2-6}<\dfrac{2n}{\frac{1}{2}n^2}=\dfrac{4}{n}<\epsilon$ (why are we allowed to say this?) so $n>\dfrac{4}{\epsilon}$.
Proof. Let $\epsilon... | For $n>6$, we have
$$n^2-6 >n^2-36>0$$
$$n^2-36=(n+6)(n-6) $$
thus
$$\frac {n+6}{n^2-6}<\frac {1}{n-6} $$
To satisfy $\frac {n+6}{n^2-6}<\epsilon $, it sufficient to have
$n>6$ and $\frac {1}{n-6}<\epsilon $
or $n>6$ and $n>6+\frac {1}{\epsilon} $
so we will take $N=6+\frac {1}{\epsilon} $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Linear combination of others Write one of the vectors as a linear combination of the others.
$$ \left[
\begin{array}{cc|c}
0\\
1\\
1
\end{array}
\right],\left[
\begin{array}{cc|c}
1\\
0\\
-1
\end{array}
\right],\left[
\begin{array}{cc|c}
4\\
5\\
... | If one can be expressed as a linear combination of the others, they are linearly dependent. This means that we can find constants $x,y,z $ s.t.
$$ \left[
\begin{array}{cc|c}
0\\
1\\
1
\end{array}
\right]x+
\left[
\begin{array}{cc|c}
1\\
0\\
-1
\end{array}
\right]y + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Show that if $\frac1a+\frac1b+\frac1c = a+b+c$, then $\frac1{3+a}+\frac1{3+c}+\frac1{3+c} \leq\frac34$
Show that if $a,b,c$ are positive reals, and
$\frac1a+\frac1b+\frac1c = a+b+c$, then
$$\frac1{3+a}+\frac1{3+b}+\frac1{3+c} \leq\frac34$$
The corresponding problem replacing the $3$s with $2$ is shown here:
How... | This is a follow-on to Michael Rozenberg's answer, generalizing to all real $k>2$, given only the theorem that under the given constraint,
$$
\sum_{\mbox{cyc}}\frac1{2+a}\leq 1
$$
Use (C-S)
$$ \sum \left(\frac{\alpha_i^2}{x_i}\right) \geq \frac{\left(\sum \alpha_i\right)^2}{\sum x_i}
$$
for the $n=2$ case with
$$
\mat... | {
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"url": "https://math.stackexchange.com/questions/2312672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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For a polynomial with integer coefficients, is it true that if constant term is prime then it cannot be the root of the polynomial.
For a polynomial with integer coefficients, is it true that if constant term is prime then it cannot be the root of the polynomial.
Let $p$ be a polynomial with constant term $a_0$ and... | Consider the polynomial $$f(x)=x^2 - 6 x + 5$$
The constant term here is $5$, which is prime.
However, \begin{align}f(5)&=5^2-6\times 5+5\\
&=25-30+5\\
&=0\end{align}
And thus $5$ is a root of the equation.
Therefore, I have disproved your hypothesis through contradiction
For higher degree polynomials, anything of the... | {
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"url": "https://math.stackexchange.com/questions/2313539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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How can I find maximum and minimum modulus of a complex number? I have this problem. Let be given complex number $z$ such that
$$|z+1|+ 4 |z-1|=25.$$
Find the greastest and the least of the modulus of $z$.
I tried with minimum.
Put $A(-1,0)$, $B(1,0)$ and $M(x,y)$ present of $z$.
We have $O(0,0)$ is the midpoint of th... | Put $A(-1,0)$, $B(1,0)$, $M(x,y)$ is the point present for $z$. Note that $O(0,0)$ is midpoint of the segment $AB$ and $AB=2$. We have $AM+4BM=25$. We need to find the least and the greastest of the segment $OM$.
Put $a=AM$, $b=BM$ $(a,\, b >0)$, we have $a+4b=25$ or $a=25-4b$.
Because of $|MA-MB|\leqslant AB$ or
$$|a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Integration of $1/(1+a \csc^2(x))$
Integration of $$\int_{0}^{\frac{(M-1)\pi}{M} }\frac{1}{1+\alpha \csc^2(x)} dx,$$ where $ \alpha $ is a constant.
I tried taking $\cot(x) = t$, then differentiating it w.r.t $dx$ we get, $-\csc^2(x)dx = dt$. And as we know that, $\csc^2(x)= \cot^2(x) +1$, so tried substituting the... | $$\int^{ }_{} \frac{1}{1+\alpha \csc^2(x)}dx=\int \frac{\csc^2(x)}{\csc^2(x)+\alpha \csc^4(x)}dx$$
and $$\csc^2(x)=\cot^2(x)+1$$
$\int \frac{\cot^2 (x) +1}{\cot^2 (x)+1+\alpha (\cot^2 (x) +1)^2}dx$
$\int \frac{\csc^2x}{\cot^2 (x)+1+\alpha (\cot^2 (x) +1)^2}dx$
Then substitute $t=\cot(x)=> dt = -\csc^2 x \,dx.$
$\int \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Solve for $x$ in $\cos(2 \sin ^{-1}(- x)) = 0$
Solve $$\cos(2 \sin ^{-1}(- x)) = 0$$
I get the answer $\frac{-1}{ \sqrt2}$
By solving like this
\begin{align}2\sin ^{-1} (-x )&= \cos ^{-1} 0\\
2\sin^{-1}(- x) &= \frac\pi2\\
\sin^{-1}(- x) &= \frac\pi4\\
-x &= \sin\left(\frac\pi4\right)\end{align}
Thus $x =\frac{-1}{\... | Let $\sin^{-1}(-x)=u\implies\sin u=-x$
$$\cos(2\sin^{-1}(-x))=\cos2u=1-2\sin^2u=1-2(-x)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Very hard hyperbolic inequality I would like to solve this
Your inequality is equivalent to :
$$\sum_{cyc}\frac{a}{13}\sin(\arctan(\sqrt{\frac{13}{5}}\frac{a}{b}))^2\geq \frac{a+b+c}{18}$$
Each side is divided by $b$
We get :
$$\frac{a}{13b} \sin(\arctan(\sqrt{\frac{13}{5}}\frac{a}{b}))^2+\frac{1}{13}\sin(\arctan(\sqrt... | So, my attempt is based on putting: $\tfrac{b}{a}=x$ and $\tfrac{c}{b}=y$, so $x,y>0$. Then our OP inequality:
$$\frac{a^3}{13a^2+5b^2}+\frac{b^3}{13b^2+5c^2}+\frac{c^3}{13c^2+5a^2}\geq\frac{a+b+c}{18}$$
$$\frac{a}{13+5x^2}+\frac{b}{13+5y^2}+\frac{c}{13+5(\tfrac{1}{xy})^2}\geq\frac{a+b+c}{18}$$
$$\frac{\tfrac{b}{x}}{13... | {
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"url": "https://math.stackexchange.com/questions/2317313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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How to evaluate $\int \frac {dx}{\cos x+C}$? I want to evaluate -
$$\int \frac {dx}{\cos x+C}$$
Where $C$ is an arbitrary constant.I tried substitution and parts but could not do it.
Note that for $C=1$ one can simply do this by using compound angle formulas.But what about other values of $C $?
Thanks for any help!!... | Hint:
Make the substitution $ \theta = \frac{x}{2}$.
\begin{align}
\frac{1}{C + \cos 2 \theta} &= \frac{1}{C -1 + 2\cos ^2 \theta} \\
&= \frac{\sec^2 \theta}{(C-1)\sec^2 \theta + 2} \\
&= \frac{\sec^2 \theta}{C + 1 + (C-1)\tan^2 \theta}
\end{align}
Hence
\begin{align}
\int \frac{1}{C + \cos x} \text{d}x &= \int \frac... | {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 1
} |
Integrating rational functions. $\int\frac{x^2-3x+1}{x^4-x^2+1}\mathrm{d}x$ Is there really another way of solving this problem $$\int\frac{x^2-3x+1}{x^4-x^2+1}\mathrm{d}x$$
avoiding partial fractions decomposition? I have tried the partial fractions decomposition, but it is not serving me right.
| The integral can be split into two pieces:
$$\begin{aligned}
\int {\frac{x}{{{x^4} - {x^2} + 1}}dx} &= \frac{1}{2}\int {\frac{1}{{{u^2} - u + 1}}du} \\
&= \frac{1}{{\sqrt 3 }}\arctan \left( {\frac{{2u - 1}}{{\sqrt 3 }}} \right) + C \\&= \frac{1}{{\sqrt 3 }}\arctan \left( {\frac{{2{x^2} - 1}}{{\sqrt 3 }}} \right) + C
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2322455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\sum_{n=1}^{\infty}\frac{n}{3^n-1}.$ Evaluate $$\sum_{n=1}^{\infty}\frac{n}{3^n-1}.$$
Solution(Partial):
$|x|<1$
$$\sum_{n=1}^{\infty}\frac{nx^{n-1}}{3^n-1}=\frac{d}{dt}\sum_{n=1}^{\infty}\frac{t^n}{3^n-1}\big{|}_{t=x}$$
$$\sum_{n=1}^{\infty}\frac{t^n}{3^n-1}=\sum_{n=1}^{\infty} t^n\sum_{k=1}^{\infty}\frac{1}... | Even though $\frac{n-1}{3^n-1}>\frac{1}{(3^n-1)^2}$ for all $n\geq 2$, your sum
$$\sum_{n=1}^{\infty}\frac{n-1}{3^n-1}=\sum_{k=1}^{\infty}\frac{1}{(3^k-1)^2}$$
starts with $n=1,$ so that first term may compensate the others. In fact, a quick numerical check shows that both sides essentially coincide (well, 0.267279737... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Equation of the line through the point $(\frac{1}{2},2)$ and tangent to the parabola $y=\frac{-x^2}{2}+2$ and secant to the curve $y=\sqrt{4-x^2}$ Find the equation of the line through the point $(\frac{1}{2},2)$ and tangent to the parabola $y=\frac{-x^2}{2}+2$ and secant to the curve $y=\sqrt{4-x^2}$
Let the required... | Problem
$$
\color{red}{y = 2 - \frac{x^{2}}{2}}
\tag{1}
$$
$$
\color{blue}{y = \sqrt{4-x^2}}
\tag{2}
$$
Equation of the line
Find the equation of the tangent line
$$
y = m x + b
\tag{3}
$$
tangent to $\color{red}{y(x)}$
We are given the a point
$$
p = \left( \frac{1}{2}, 2 \right)
$$
Find the slope, $m$, and th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Simplification of Trigo expression
Simplify $$\frac{\tan^2 x+\cos^2 x}{\sin x+ \sec x}$$
My attempt,
$$=\frac{(\cos^2x+\frac{\sin^2x}{\cos^2x})}{\frac{\sin x \cos x+1}{\cos x}} $$
$$=\frac{\cos^4 x+\sin^2 x}{\cos^2 x}\cdot \frac{\cos x}{\sin x \cos x+1}$$
$$=\frac{\cos^4 x+\sin^2 x}{\cos x(\sin x \cos x+1)}$$
I'm st... | $$\scriptsize\frac {\tan^2x+\color{blue}{\cos^2x}}{\sin x+\sec x}=\frac {\overbrace{\tan^2x+\color{blue}1}^{\sec^2 x}\color{blue}{-\sin^2x}}{\sin x+\sec x}=\frac {\sec^2x-\sin^2x}{\sec x+\sin x}=\frac{(\sec x -\sin x)(\sec x+\sin x)}{\sec x+\sin x}=\sec x-\sin x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find the minimal value of $\sum\limits_{n=1}^8|x-n|$
Find the minimum value of $( |x-1|+|x-2|...+|x-8|) $
My attempt
Using triangle inequality i.e.
$|x-y|≤|x|+|y|$ $|x-1|=|x-1-0|≤|x-1|+0$ $|x-2|=|x-1-1|≤|x-1|+1$ $|x-3| =|x-1-2| ≤|x-1|+2$
...
...
...
$|x-8| =|x-1-7| ≤|x-1|+7$
Adding all these inequalities, $( |x-1|+... | Use the triangle inequality: $|x+y|\le |x|+|y|$.
$|x-1|+|8-x|\ge |x-1+8-x|=7$,
$|x-2|+|7-x|\ge |x-2+7-x|=5$,
$|x-3|+|6-x|\ge |x-3+6-x|=3$,
$|x-4|+|5-x|\ge |x-4+5-x|=1$,
Summing we get min $16$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
If $x^3 - 5x^2+ x=0$ then find the value of $\sqrt {x} + \dfrac {1}{\sqrt {x}}$ If $x^3 - 5x^2+ x=0$ then find the value of $\sqrt {x} + \dfrac {1}{\sqrt {x}}$
My Attempt:
$$x^3 - 5x^2 + x=0$$
$$x(x^2 - 5x + 1)=0$$
Either,
$x=0$
And,
$$x^2-5x+1=0$$
??
| If $y=\sqrt x + 1/\sqrt x$ then your value of $y$ is given as a solution of $y$ for the system
\begin{align}
x^2&-5x+1=0\\
t^2&=x\\
t^2&=ty-1\\
\end{align}
Putting the second and third equation together $x=ty-1$ so
$$(ty-1)^2-5(ty-1)+1=t^2y^2-2ty+1-5ty+5+1=(ty-1)y^2-7(ty-1)=0$$
If $ty-1\neq 0$ then $y^2=7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
how to compute $E(X^4)$ when $X$ follow the normal law $N(0,1)$ Suppose $X$ follow the normal law $N(0,1)$
We have the density $f= \frac{1}{\sqrt{\sigma^2 2\pi}} e^{- \frac{(x-m)^2}{2\sigma^2}}$
We want to compute $E(X^4)$
We have by definition that $\displaystyle E(X^4) = \frac{1}{\sqrt{2\pi}} \int_\infty^\infty{e^{-... | \begin{align}
& \frac 1 {\sqrt{2\pi}} \int_{-\infty}^\infty x^4 e^{-x^2/2}\,dx \\[10pt]
= {} & 2\cdot\frac 1 {\sqrt{2\pi}} \int_0^\infty x^4 e^{-x^2/2} \, dx & & \text{since the integral is of an even} \\ & & & \text{function over an interval that} \\[8pt]
& & & \text{is symmetric about 0} \\[10pt]
= {} & \sqrt{\frac 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
How to integrate $\int \frac {\cos^4 x}{\sqrt {1- \sin x}} $ I frankly don't know where to start. Looked for some identities to get rid of the square root below.
$$\int \frac {\cos^4 x}{\sqrt {1- \sin x}} \, dx $$
| Write $\cos(x)^4 = \cos(x)(1-\sin^2(x))^{3/2} = \cos(x)(1-\sin(x))^{3/2}(1+\sin(x))^{3/2}.$ Thus $$\int \frac{\cos(x)^4}{\sqrt{1-\sin(x)}}dx = \int (1-\sin(x))(1+\sin(x))^{3/2}\cos(x) dx$$ Put $u = \sin(x)$. Then $$\int\frac{\cos(x)^4}{\sqrt{1-\sin(x)}}dx = \int(1-u)(1+u)^{3/2}du.$$ Now put $z = 1+u$ so $$\int\frac{\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Given $u_n=au_{n-1}+b$ show that $u_n=A\cdot a^n + d$ where $d$ is a fixed point of $u_n$ Given $u_n=au_{n-1}+b$ show that $u_n=A\cdot a^n + d$ where:
*
*$d$ is a fixed point of the recurrence relation $u_n=au_{n-1}+b$;
*$A$ is a constant.
My attempt:
I have tried evaluating the recurrence relation and I've come ... | Just for illustration, a brute force approach yields
$$
\begin{split}
u_n &= au_{n-1} + b \\
&= a(au_{n-2} + b) + b
= a^2 u_{n-2} + b(a+1)\\
&= a^2 (au_{n-3} + b) + b(a+1)
= a^3 u_{n-3} + b(a^2+a+1)\\
&= a^nu_0 + b \sum_{k=0}^{n-1}a^k\\
&= a^nu_0 + b \frac{1-a^n}{1-a} \\
&= a^n \left( u_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\frac{1}{2\pi i}\int_\mathcal{C} |1+z+z^2+\cdots+z^{2n}|^2~dz =2n$ where $\mathcal{C}$ is the unit circle On the generalization of a recent question, I have shown, by analytic and numerical means, that
$$\frac{1}{2\pi i}\int_\mathcal{C} |1+z+z^2+\cdots+z^{2n}|^2~dz =2n$$
where $\mathcal{C}$ is the unit circ... | For $z\in\mathcal C$, we have $$|1+...+z^{2n}|^2 = (1+z+...+z^{2n})(1+z^{-1}+...+z^{-2n}) = \sum_{j=0}^{2n}\sum_{k=0}^{2n}z^jz^{-k} = \sum_{j=0}^{2n}\sum_{k=0}^{2n}z^{j-k}.$$
Therefore $$\frac{1}{2\pi i}\int_{\mathcal C}|1+z+...+z^{2n}|^2 dz = \sum_{j=0}^{2n}\sum_{k=0}^{2n}\frac{1}{2\pi i}\int_{\mathcal C}z^{j-k}dz.$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Prove $5^n +5 <5^{n+1}$ $∀n∈N$ using induction Prove $5^n +5 <5^{n+1}$ $∀n∈N$
Base Case: $n=1$
$\implies 5^1 +5 <25$
$\implies 10<25$ ; holds true
Induction hypothesis: Suppose $5^k +5 < 5^{k+1}$ is true for k∈N
Then;
$\implies 5^{k+1} +5 < 5^{k+2}$
$\implies 5\cdot 5^k +5 < 25*5^k$
I don't know how to proceed after t... | $$5^{k+1}+5=5\times 5^k+5<5(5^{k+1}-5)+5=5^{k+2}-25+5<5^{k+2}$$
Notice that, by the induction step, we have $$5^k+5<5^{k+1}\implies 5^k<5^{k+1}-5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Find $\tan(\frac{1}{2}\arcsin(\frac{5}{13}))$ I wanted to solve it by using this formula: $\tan(\frac{x}{2})=\pm\sqrt{\frac{1-x^2}{1+x^2}}.$ I thought it wouldn't work (because there are $\pm$). Then used the right triangle method: $$\frac{1}{2}\arcsin(\frac{5}{13})=\alpha\Rightarrow\frac{5}{26}=\sin\alpha$$ $$a^2+b^2=... | Let $\arcsin(\frac{5}{13})=x$, thus $\sin(x)=\frac{5}{13}$
So, this problem reduced to find value of $\tan(\frac{x}{2})=u$,
According to $\sin(x)=\frac{2u}{u^2+1}$ identity for $u=\tan(\frac{x}{2})$;
$\frac{2u}{u^2+1}=\frac{5}{13}$
$5*(u^2+1)=13*2u$
$5u^2+5=26u$
$5u^2-26u+5=0$
$(u-5)*(5u-1)=0$
Due to $\tan(x)>0$ constr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Calculating integral $\oint\limits_{|z|=4} \frac{\sin^2(z)}{(z-\frac{\pi}{6})^2(z+\frac{\pi}{6})}dz$
Calculating $$\oint\limits_{|z|=4} \frac{\sin^2(z)}{(z-\frac{\pi}{6})^2(z+\frac{\pi}{6})}dz$$
Due to the fact of the denominator I've looked for the partial fractions:
$$1=A\left(z-\frac{\pi}{6}\right)^3+B\left(z+\fra... | It happens that$$\frac1{\left(z-\frac\pi6\right)^2\left(z+\frac\pi6\right)}=\frac9{\pi ^2\left(z+\frac\pi6\right)}-\frac9{\pi ^2\left(z-\frac\pi6\right)}+\frac3{\pi\left(z-\frac\pi6\right)^2}$$Therefore\begin{multline*}\oint_{|z|=4}\frac{\sin^2(z)}{\left(z-\frac\pi6\right)^2\left(z+\frac\pi6\right)}dz=\\=\frac9{\pi^2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\tan^2{\frac{\pi}{5}}+\tan^2{\frac{2\pi}{5}}$ without a calculator The question is to find the exact value of:
$$\tan^2{\left(\frac{\pi}{5}\right)}+\tan^2{\left(\frac{2\pi}{5}\right)}$$
without using a calculator.
I know that it is possible to find the exact values of $\tan{\left(\frac{\pi}{5}\right)}$ and $... | $$\tan^2\frac{\pi}{5}+\tan^2\frac{2\pi}{5}=\frac{1-\cos\frac{2\pi}{5}}{1+\cos\frac{2\pi}{5}}+\frac{1-\cos\frac{4\pi}{5}}{1+\cos\frac{4\pi}{5}}=$$
$$=\frac{2-2\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}}{1+\cos\frac{2\pi}{5}+\cos\frac{4\pi}{5}+\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}}=$$
$$=\frac{2-\frac{4\sin\frac{2\pi}{5}\cos\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2336186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
} |
Finding number of asymptotes in $[ 0 , 2\pi]$ interval Consider $f(x) = \frac{2\tan 2x + 1}{2\cos x + 1}$ . Now find number of asymptotes in $[ 0 , 2\pi]$ interval . I know we can simplify it to $\frac{2\sin 2x + 1}{\cos 2x (2\cos x + 1)}$ and then solve $\cos 2x (2\cos x + 1) = 0$ but is there any way for doing it str... | $\cos2x\neq0$ gives asymptotes $x=\frac{\pi}{4}$, $x=\frac{3\pi}{4}$, $x=\frac{5\pi}{4}$ and $x=\frac{7\pi}{4}$.
$\cos{x}\neq-\frac{1}{2}$ gives asymptotes $x=\frac{2\pi}{3}$ and $x=\frac{4\pi}{3}$.
Id est, we have six asymptotes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2336459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving trigonometric identity $\cos^6A+\sin^6A=1-3 \sin^2 A\cos^2A$ Show that
$$\cos^6A+\sin^6A=1-3 \sin^2 A\cos^2A$$
Starting from the left hand side (LHS)
\begin{align}
\text{LHS} &=(\cos^2A)^3+(\sin^2A)^3 \\
&=(\cos^2A+\sin^2A)(\cos^4A-\cos^2A\sin^2A+\sin^4A)\\
&=\cos^4A-\cos^2A\sin^2A+\sin^4A
\end{align}
Can a... | Let $a=\cos^2 (A),b=\sin^2 (A) $ . Now use $a^3+b^3=(a+b)^3-3ab (a+b) $ also note that $a+b=1$. Hence the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Proving trigonometric identity $\frac{\sin(A)}{1+ \cos(A )}+\frac{1+ \cos(A )}{\sin(A)}=2 \csc(A)$
$$
\frac{\sin(A)}{1+\cos(A)}+\frac{1+\cos(A)}{\sin(A)}=2\csc(A)
$$
\begin{align}
\mathrm{L.H.S}&= \frac{\sin^2A+(1+\cos^2(A))}{\sin(A)(1+\cos(A))} \\[6px]
&= \frac{\sin^2A+2\sin(A)\cos(A)+\cos^2(A)+1}{\sin(A)(1+\cos(A))... | $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\i}{\mathrm{i}} \newcommand{\text}[1]{\mathrm{#1}} \newcommand{\root}[2][]{^{#2}\sqrt[#1]} \newcommand{\derivative}[3]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\abs}[1]{\left\vert\,{#1}\,\right\vert}$
$$\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
} |
Solve three equations with three unknowns Solve the system:
$$\begin{cases}a+b+c=6\\ab+ac+bc=11\\abc=6\end{cases}$$
The solution is:
$a=1,b=2,c=3$
How can I solve it?
| By method of substitution:
$$\begin{cases} b+c=6-a \\ a(6-a)+\frac{6}{a}=11 \\ bc=\frac{6}{a}\end{cases} \Rightarrow$$
$$a^3-6a^2+11a-6=0 \Rightarrow (a-1)(a-2)(a-3)=0 \Rightarrow a=1; 2; 3.$$
Substituting these and solving $$\begin{cases} b+c=6-a \\ bc=\frac{6}{a}\end{cases}$$
six solutions will be found:
$$(a,b,c)=(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Proving inequality $\ 1+\frac14+\frac19+\cdots+\frac1{n^2}\le 2-\frac1n$ using induction Question:
Prove $$\ 1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{n^2}\le 2-\frac{1}{n},
\text{ for all natural } n$$
My attempt:
Base Case: $n=1$ is true:
I.H: Suppose $1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{k^2}\le 2-\frac{1}{k}... | You are right!
We need to prove that:
$$\frac{1}{(k+1)^2}<\frac{1}{k}-\frac{1}{k+1}$$ or
$$\frac{1}{(k+1)^2}<\frac{1}{k(k+1)},$$
which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2340164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Binomial vs Probability
A shipment of 12 microwave ovens contains 3 defective units. A vending company has ordered 4 of these units, and because all are packaged identically, the selection will be at random. What is the probability that at least 2 units are good?
There are two ways to solve this problem:
*
*Using ... | You cannot use the binomial distribution, as this is an experiment without replacement: once you pick a defective oven, there are only two defective ovens left. Using combinations, the probability of selecting at least two good units indeed equals:
$$\frac{{9 \choose 4}{3 \choose 0} + {9 \choose 3}{3 \choose 1} + {9 \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What is the value of this logarithmic/trigonometric expression? I need the value of this expression, when simplified.
$$ \log_{10}(\cot(1)°) + \log_{10}(\cot(2°)) + \cdots + \log_{10}(\cot(89°))$$
All the $\log$ have base $10$.
| The sum is essentially
\begin{align}
S &= \sum_{k=1}^{89} \log_{10}\left(\cot\left(\frac{k \pi}{180}\right)\right) \\
&= \sum_{k=1}^{44} \log_{10}\left(\cot\left(\frac{k \pi}{180}\right)\right) + \log_{10}\left(\cot\left(\frac{\pi}{4}\right)\right) + \sum_{k=46}^{89} \log_{10}\left(\cot\left(\frac{k \pi}{180}\right)\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Hyperbolas: Deriving $\frac{x^2}{a^2} + \frac{y^2}{a^2 - c^2} = 1$ from $\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = \pm2a$ My textbook's section on Hyperbolas states the following:
If the foci are $F_1(-c, 0)$ and $F_2(c, 0)$ and the constant difference is $2a$, then a point $(x, y)$ lies on the hyperbola if an... | Let's do the case
$$
\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = 2a
$$
which holds provided $x>0$.
Move the second radical to the right-hand side and square:
$$
(x + c)^2 + y^2=4a^2+(x - c)^2 + y^2 + 4a\sqrt{(x - c)^2 + y^2}
$$
Lots of terms cancel:
$$
2cx=4a^2-2cx+4a\sqrt{(x - c)^2 + y^2}
$$
which becomes
$$
cx-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2342742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find $\frac{BF}{FC}$ in terms of $b,c$ If $AD$ is the bisector of angle $BAC$ and $E$ is the reflection of $D$ to point $M$(midpoint of $BC$).We construct point $F$ such that angles $BAF=EAC$.Then find $\frac{BF}{FC}$ in terms of $b,c$.
My attempt:We have:
$\frac{BF}{FC}=\frac{AB}{AC}*\frac{\sin BAF}{\sin CAF}$
But t... | By the bisector theorem $BD=\frac{c}{b+c}a$ and $CD=\frac{b}{b+c}a$.
Since $BM=CM=\frac{1}{2}a$, $BE=\frac{b}{b+c}a$ and $CE=\frac{c}{b+c}a.$
If $\widehat{BAF}=\widehat{EAC}$, the lines $AE$ and $AF$ have to be symmetric with respect to $AD$.
The trilinear coordinates of $E$ are $[0,c^2,b^2]$, hence the trilinear coo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2347471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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An identity form: $x_1^3+x_2^3+x_3^3+x_4^3=y_1^3+y_2^3+y_3^3+y_4^3$ I found an identity form: $x_1^3+x_2^3+x_3^3+x_4^3=y_1^3+y_2^3+y_3^3+y_4^3$ as follows:
$$(a+b+c)^3+a^3+b^3+c^3=(a+b)^3+(b+c)^3+(c+a)^3+(6y)^3$$
Where $abc=36y^3$
Poof of this identity is very simple. But the identity nice.
Which reference of the iden... | Equation (shown below) has parametrization:
$(a+b+c)^3+a^3+b^3+c^3=(a+b)^3+(b+c)^3+(c+a)^3+(6y)^3$
$(y,a,b,c)=((8)(3k-4),(256),(24)(3k-4),(3)(3k-4)^2)$
For $k=2$, we get: $(3,12,64,79)^3=(15,24,67,76)^3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2347537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Riccati D.E., vertical asymptotes
For the D.E.
$$y'=x^2+y^2$$
show that the solution with $y(0) = 0$ has a vertical asymptote at some point $x_0$. Try to find upper and lower bounds for $x_0$:
$$y'=x^2+y^2$$
$$x\in \left [ a,b \right ]$$
$$b> a> 0$$
$$a^2+y^2\leq x^2+y^2\leq b^2+y^2$$
$$a^2+y^2\leq y'\leq b^2+y^... | From numerical solution comes out that $x=2$ and $x=-2$ are vertical asymptotes. Trying to solve as a Bernoulli equation gives a mess and the substitution $w=\frac{1}{y}$ gives problems as
$$w=\frac{1}{y};\;w'=-\frac{y'}{y^2}$$
Divide the original equation by $y^2$
$$\frac{y'}{y^2}=\frac{x^2}{y^2}+1\rightarrow -w'=x^2w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Continuity of $f(x,y) = \dfrac{\sin(x^2+y^2)}{x^2+y^2}$ at $(x,y) = (0,0)$ Continuity of $f(x,y) = \dfrac{\sin(x^2+y^2)}{x^2+y^2}$ at $(x,y) = (0,0)$
I need to specify the value of $f(x,y)$ which makes the given function continuous.
I had tried to give a arbitral relationship between $y$ and $x$, but as $(x, y)$ gets c... | Recalling that $|x|\,|\cos(x)|\le |\sin(x)|\le |x|$ for $0\le |x|\le \pi/2$, we assert that for $x^2+y^2\le \pi/2$
$$|\cos(x^2+y^2)|\le \left|\frac{\sin(x^2+y^2)}{x^2+y^2}\right|\le 1$$
whence application of the squeeze theorem reveals
$$\lim_{(x,y)\to(0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Common complex roots
If the equations $ax^2+bx+c=0$ and $x^3+3x^2+3x+2=0$ have two common roots then show that $a=b=c$.
My attempts:
Observing $-2$ is a root of $x^3+3x^2+3x+2=0\implies x^3+3x^2+3x+2=(x+2)(x^2+x+1)=0$
Hence $ax^2+bx+c=0$ can have complex roots in common, comming from $(x^2+x+1)=0$
Both the roots of $... | We have that
$$
\eqalign{
& x^{\,3} + 3x^{\,2} + 3x + 2 = 0\quad \Rightarrow \quad x^{\,3} + 3x^{\,2} + 3x + 1 = - 1\quad \Rightarrow \cr
& \Rightarrow \quad \left( {x + 1} \right)^{\,3} = - 1\quad \Rightarrow \quad \left( {x + 1} \right) = e^{\,i\;\,\left( {1 + 2k} \right)\pi /3} = e^{\, \pm \,i\;\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
If $a + \frac{1}{a} = -1$, then the value of $(1-a+a^2)(1+a-a^2)$ is?
If $a + \frac{1}{a} = -1$ then the value of $(1-a+a^2)(1+a-a^2)$ is?
Ans. 4
What I have tried:
\begin{align}
a + \frac{1}{a} &= -1 \\
\implies a^2 + 1 &= -a \tag 1 \\
\end{align}
which means
\begin{align}
(1-a+a^2)(1+a-a^2) &=(-2a)(-2a^2) \\
&=4a^... | $a^3=1$.
Thus, $$(1-a+a^2)(1+a-a^2)=-2a\cdot(-2a^2)=4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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If a is a non real root of $x^7 = 1$, find the equation whose roots are $ a + a^6 , a^2 + a^5, a^3 + a^4$ If a is a non real root of $ x^7 = 1$, find the equation whose roots are $a + a^6 , a^2 + a^5, a^3 + a^4$. This is one of the questions I have encountered while preparing for pre rmo. I feel the question requires t... | Think of the roots as $a+a^{-1}$, $a^2+a^{-2}$ and $a^3+a^{-3}$. Then $a$ satisfies $a^6+a^5+a^4+a^3+a^2+a+1=0$ or equivalently,
$a^3+a^2+a+1+a^{-1}+a^{-2}+a^{-3}=0$. Can you write the expression
$x^3+x^2+x+1+x^{-1}+x^{-2}+x^{-3}$ as a linear combination of $(x+x^{-1})^3$, $(x+x^{-1})^2$, $x+x^{-1}$ and $1$? Putting in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Another limit involving an integral How to find the following limit (if exists ) : $\lim _{x \to \infty} \dfrac 1 x \int_0^x \dfrac {t}{1+x^2 \cos^2 t}dt$ ?
This looks quite similar to the previous To find a limit involving integral , but the huge difference is , there is apparently no closed form formula for the in... | I am using the same method as in this answer you linked. Let
$$
F(x)=\frac{1}{x}\int_0^x \frac{t}{1+x^2\cos^2{t}}\mathrm{d}t.
$$
If $n$ is a positive integer, we have
$$
F(n\pi)=\frac{1}{n\pi}\sum_{k=0}^{n-1}\int_{k\pi}^{(k+1)\pi} \frac{t}{1+(n\pi)^2\cos^2{t}}\mathrm{d}t,
$$
so we can control the $t$ in the integrand :... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Quadratic polynomial with integer roots. The quadratic polynomial $ax^2+bx+c$ has positive coefficients $a,b,c$ in A.P. in the given order. If it has integer roots $\alpha,\beta,$ find $\alpha+\beta+\alpha \beta.$
I tried with Vieta's theorem and putting $b=\frac{a+c}{2}$ to get $\alpha+\beta+\alpha \beta=\frac{b}{a}-1... | Unpacking J.G.'s answer for the masses:
$ax^2+bx+c$
$$\text{if} \quad (a,b,c) \quad \text{are in "AP", then} \quad $$
$$\begin{align}\begin{cases}
a&=a \\
b&=a+d \\
c&=a+2d
\end{cases}
\ \ &\underbrace{\implies}_{\text{b is the average of a and c}} \ \
\left[\frac{a+c}{2} =\frac{2a+2d}{2}=a+d=b\right]
\\
&\qquad \qu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2354951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$ then what is the value of $a^2-ax$ is equal to
If $$ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$$ then
the value of $a^2-ax$ is equal to:
a)2 b)1 c)0 d)-1
Ans. (d)
My attempt:
Rationalizing $a$ we get,
$ x+ \sqrt {x^2-4}$
$a^2=(x+\sqrt{... | You should check your rationalization of $a$ again. I believe you are missing a factor of $\frac{1}{2}$.
Additionally, you could find the answer by choosing some easily computable value of $x$, say $x = 2$, so $a = 1$ and $a^2 - ax = -1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How does the trigonometric identity $1 + \cot^2\theta = \csc^2\theta$ derive from the identity $\sin^2\theta + \cos^2\theta = 1$? I would like to understand how would the original identity of $$ \sin^2 \theta + \cos^2 \theta = 1$$ derives into
$$ 1 + \cot^2 \theta = \csc^2 \theta $$
This is my working:
a) $$ \frac{\... | Just as you divide $\sin^2(\theta)+\cos^2(\theta)=1$ by $\sin^2(\theta)$ to get $ 1 + \cot^2 (\theta) = \csc^2 (\theta)$,
you can divide $\sin^2(\theta)+\cos^2(\theta)=1$ by $\cos^2(\theta)$ to get $\tan^2 (\theta) + 1 = \sec^2(\theta)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Why does $\frac{\cos\Delta x - 1 }{\Delta x} \to 0$ I'm watching Lecture 3 in MIT single variable calculus.
https://www.youtube.com/watch?v=kCPVBl953eY&list=PL590CCC2BC5AF3BC1&index=3
And at one point the instructor does the following:
I was under the impression that when evaluating limits we need to avoid having $0/... | Since you know that
$$\lim_{x \rightarrow 0}\frac{\cos x -1}{x}=\lim_{x \rightarrow 0}\frac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}-1}{x} $$
$$\lim_{x \rightarrow 0}\frac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}-1}{x} =\lim_{x \rightarrow 0}\frac{-\sin^2\frac{x}{2}-\sin^2\frac{x}{2}}{x}$$
$$\lim_{x \rightarrow 0}\frac{-\sin^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2361672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Sequence $a_n = a_{n - 1} + a_{\lfloor n/2 \rfloor}$ Given a sequence with $a_1 = 1$. And $a_n = a_{n - 1} + a_{\lfloor n/2 \rfloor}$ for $n \geq 2$, prove that $a_n$ is not divisible by 4 for any $n \in \mathbb{N}$.
Don't have any conjectures.
| A few hints
Notice that $a_{2n+1}=a_{2n-1}+2a_n$. Since $a_1$ is odd, all $a_n$ are odd for odd $n$.
Then $a_{4n+2}=a_{4n+1}+a_{2n+1}$ even.
And $$a_{4n+2}=a_{4n+1}+a_{2n+1}=a_{4n-1}+2a_{2n}+a_{2n-1}+2a_n=a_{4n-2}+a_{2n-1}+3a_{2n-1}+4a_n=a_{4n-2}+4a_{2n-1}+4a_n$$ which shows by recurrence that $a_{4n+2}$ is not divisib... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2361824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
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Find number of real solutions of $3x^5+2x^4+x^3+2x^2-x-2=0$
Find a number of real roots of $$f(x)=3x^5+2x^4+x^3+2x^2-x-2$$
I tried using differentiation:
$$f'(x)=15x^4+8x^3+3x^2+4x-1=0$$ and I found number of real roots of $f'(x)=0$ by drawing graphs of $g(x)=-15x^4$ and $h(x)=8x^3+3x^2+4x-1$ and obviously from graph... | $LHD=x^3(3x^2+2x+1)=x^3(3(x+\frac13)^2+\frac23)$
$RHD=-2x^2+x+2=-2(x-\frac14)^2+\frac{17}8$
if $x>\frac54$, obviously this equality has no solution.
if $\frac14<x<\frac54$, since LHD is monotoniously increasing, RHD is its opposite,
so there is one solution.
if $0<x<\frac14$ ,$LHD<1<RHD$ there is no solution.
if $-1<... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Need help calculating a simple inverse function Quick question.
$$y=\sqrt{2x-x^2}$$
I need the inverse function for some other problem, but I just can't find it.
Could you please point me the steps to solve this?
Thanks
| The domain of $f(x)=\sqrt{2x-x^2}$ is $$\mbox{dom}(f)=\{x\in\mathbb{R} \ | \ 0\le x\le 2\}$$ moreover its image is $$\mbox{Im}(f)=\{y\in\mathbb{R} \ | \ 0\le y\le 1 \}$$ Its graph is a semicircumpherence with centre $C(1,0)$ and radius $r=1$ infact
$\\ y=\sqrt{2x-x^2}\implies y^2=2x-x^2\implies \\ \\ \\ x^2+y^2-2x=0\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2364290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Evaluate $\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$
Evaluate
$$\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$$
I assumed $x=\frac{1}{y}$ we get
$$L=\lim_{y \to 0}\frac{\left( \tan^{-1}\left(\frac{1+y}{1+4y}\right)-\frac{\pi}{4... | It would be much simpler to proceed directly without any change of variables. Note that the expression under limit can be written as $$x\arctan\dfrac{\dfrac{1+x}{4+x}-1}{\dfrac{1+x}{4+x}+1}$$ which is same as $$-\frac{3x}{5+2x}\cdot\dfrac{\arctan\dfrac{3}{5+2x}}{\dfrac{3}{5+2x}}$$ and first factor tends to $-3/2$ and s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2365914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 5
} |
What is $-\arctan(\frac{s}{c}) + \pi\Theta(c)$? I know that
$$-\arctan\left(\frac{1}{c}\right)+\pi\Theta(c)=\arctan(c)+\frac{\pi}{2}$$
where $\Theta(x)$ is the Heaviside step function.
I was wondering if it is possible to find a similar expression for
$$-\arctan\left(\frac{s}{c}\right) + \pi\Theta(c)$$
| Consider a right angled triangle:
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$
The sum of the angles $\measuredangle CAB$ and $\measuredangle ABC$ is $\frac{\pi}{2}$. We know that $\tan(\measuredangle CAB) = \frac{a}{b}$ and $\tan(\measuredangle ABC) = \frac{b}{a}$ so for $a,b>0$ we have
$$\frac{\pi}{2} = \arct... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2366167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
integral $\int_{}^{}\frac{dx}{1+x^4+x^8} $ looking for help for the following integral -
$$
\int_{}^{}\frac{dx}{1+x^4+x^8}
$$
what I tried to do:
$$\int_{}^{}\frac{dx}{1+x^4+x^8} = \int_{}^{}\frac{dx}{\frac14+x^4+x^8 + \frac{3}{4}}= \int_{}^{}\frac{dx}{\left(x^4+\frac{1}{2}\right)^2 + \frac{3}{4}} $$
and now I am stu... | Factorise the denominator $x^8+x^4+1=(x^4+x^2+1)(x^4-x^2+1)=(x^2+x+1)(x^2-x+1)(x^2+\sqrt{3}x+1)(x^2-\sqrt{3}x+1)$. Now do partial fractions
\begin{eqnarray*}
\frac{1}{1+x^4+x^8}=\frac{1}{4(1+x+x^2)}+\frac{1}{4(1-x+x^2)}+\frac{2x+\sqrt{3}}{4\sqrt{3}(1+\sqrt{3}x+x^2)}-\frac{2x-\sqrt{3}}{4\sqrt{3}(1-\sqrt{3}x+x^2)}
\end{e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2368798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Inclusion-Exclusion Principle; why is this wrong? I'm unsure where I'm going wrong with this.
In a class of 40 people studying music: 2 play violin, piano and recorder, 7 play at least violina nd piano, 6 play at least piano and recorder, 5 play at least recorder and violin, 17 play at least violin, 19 play at least ... | As previously stated, take the information you know. Your events for more than 1 instrument should be intersections, not unions. That is $\cap$ not $\cup$
$|A\cap B\cap C| = 2 \\ |A\cap B| = 7 \\ |A\cap C| =5 \\ |B \cap C| =
6 \\ |A| = 17 \\ |B|
= 19 \\ |C| = 14.$
You want $|A^c \cap B^c \cap C^c|$ which is simply ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2370822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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solution set of the inequality $\frac{x^2-1}{(x+2)(x+3)}>2$
Question: Find the solution set of the inequality $$\frac{x^2-1}{(x+2)(x+3)}>2$$
From the answer given in the previous problem I got this:
First $x\neq -2,-3$. solving the equation I get $-(2\sqrt{3}+5)<x<(2\sqrt{3}-5)$.
Is this ok?
| $$\frac { x^{ 2 }-1 }{ \left( x+2 \right) \left( x+3 \right) } >2\\ \frac { x^{ 2 }-1 }{ \left( x+2 \right) \left( x+3 \right) } -2>0\\ \frac { -{ x }^{ 2 }-10x-13 }{ \left( x+2 \right) \left( x+3 \right) } >0\\ \frac { { x }^{ 2 }+10x+13 }{ \left( x+2 \right) \left( x+3 \right) } <0\\ \frac { \left( x+2\sqrt { 3 }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2372513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What is an intuitive approach to solving $\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)$?
$$\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)$$
I managed to get the answer as $1$ by standard methods of so... | Notice that:
$$\lim_{n\rightarrow\infty}\dfrac{1}{n^2} + \dfrac{2}{n^2} + \dfrac{3}{n^2}+\cdots+\dfrac{n}{n^2}=\lim_{n\rightarrow\infty} \frac{1}{n^2}\left(1+2+\dots n\right)$$
and this last sum can be replaced by the Gauss formula, so it becomes:
$$\lim_{n\rightarrow\infty}\frac{1}{n^2}\left(\frac{n(n+1)}{2}\right)$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2373357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 11,
"answer_id": 7
} |
Minimum of the given expression
For all real numbers $a$ and $b$ find the minimum of the following expression.
$$(a-b)^2 + (2-a-b)^2 + (2a-3b)^2$$
I tried expressing the entire expression in terms of a single function of $a$ and $b$. For example, if the entire expression reduces to $(a-2b)^2+(a-2b)+5$ then its mini... | \begin{eqnarray*}
(a-b)^2+(2-a-b)^2+(2a-3b)^2=6a^2-12ab+11b^2-4(a+b)+4 \\
=6\left(a-b-\frac{1}{3}\right)^2+5\left(b-\frac{4}{5}\right)^2+\color{red}{\frac{2}{15}}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2374768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
How can I achieve this transformation? I calculate a integration which results in the below.
$$
\log{\left|\frac{x+\sqrt{x^2+1}-1}{x+\sqrt{x^2+1}+1}\right|}
$$
I've checked the integration with Wolfram Alpha and it give me this result.
$$
\log{\left|\frac{\sqrt{x^2+1}-1}{x}\right|}
$$
These expressions seems to have th... | $$
\begin{align}
\frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}\cdot\frac{x+1-\sqrt{x^2+1}}{x+1-\sqrt{x^2+1}}
&=\frac{(x^2-1)+2\sqrt{x^2+1}-(x^2+1)}{(x+1)^2-(x^2+1)} \\
&= \frac{2\sqrt{x^2+1}-2}{2x} \\
&= \frac{\sqrt{x^2+1}-1}{x} \\
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Recurrence relation $a_n = 11a_{n-1} - 40a_{n-2} + 48a_{n-3} + n2^n$ I didn't do a lot of maths in my career, and we asked me to solve the following recurrence relation:
$$a_{n} = 11a_{n-1} - 40a_{n-2} + 48a_{n-3} + n2^n$$
with
$a_0 = 2$, $a_1 = 3$ and $a_2 = 1$
What is the procedure to solve such relation? So far, I ... | The general way of solving this kind of problem is to define relation $G(n)$ as
$$G(n) = (a_n - 11a_{n - 1} + 40a_{n - 2} - 48a_{n - 3} = n2^n)$$
Then write out:
$$\begin{array} {rrl}
G(n) = &(a_n + \dots &= n2^n) \\
G(n + 1) = &(a_{n + 1} + \dots &= 2n2^n + 2 \cdot 2^n) \\
G(n + 2) = &(a_{n + 2} + \dots &= 4 n 2^n + 8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Computing the composition of a piecewise function Question:
$$ \text{Define } f:\mathbb{Z}\rightarrow\mathbb{Z}\text{ by }f(x)=\begin{cases}x+3\text{ if }x\text{ is ODD}\\ x-5\text{ if }x\text{ is EVEN}\end{cases} $$
Compute $ \ f \circ f$
My attempt:
$ \ (f \circ f)(x) = f(f(x))$
From here do I have to create anothe... | Yes, you could regard the composition as a piecewise function.
$$(f\circ f)(n) = f( f(n) ) = \begin{cases}
f(n) + 3 & \text{if $f(n)$ is odd, and} \\
f(n)-5 & \text{if $f(n)$ is even}. \\
\end{cases}$$
But notice that $f(n)$ is $n$, plus some odd integer (either $+3$ or $-5$). Thus $f(n)$ is even when $n$ is odd, and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove this inequality $2(a+b+c)\ge\sqrt{a^2+3}+\sqrt{b^2+3}+\sqrt{c^2+3}$ For $a,b,c$ are positive real numbers satisfy $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Prove that $$2\left(a+b+c\right)\ge\sqrt{a^2+3}+\sqrt{b^2+3}+\sqrt{c^2+3}$$
We have:$a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{9}{a+b+c}\Leftrigh... | We need to prove that
$$\sum_{cyc}\left(2a-\sqrt{a^2+3}\right)\geq0$$ or
$$\sum_{cyc}\frac{a^2-1}{2a+\sqrt{a^2+3}}\geq0$$ or
$$\sum_{cyc}\left(\frac{a^2-1}{2a+\sqrt{a^2+3}}+\frac{1}{4}\left(\frac{1}{a}-a\right)\right)\geq0$$ or
$$\sum_{cyc}(a^2-1)\left(\frac{1}{2a+\sqrt{a^2+3}}-\frac{1}{4a}\right)\geq0$$ 0r
$$\sum_{cy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2378941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Rationalizing the denominator having square roots and cube roots In middle-school mathematics, the teachers always tell you that if you have radicals on the denominator of a fraction, then it isn't fit to be a final answer - you have to rationalize the denominator, or get rid of all of the radicals in the denominator b... | It's always possible.
You want to multiply top and bottom by $M$ to get that $denominator*M$ has not radical.
As you have figured out: If the denominator is $a + b\sqrt{c}$ you mulitply by the conjugate to get $(a + b\sqrt{c})(a - b\sqrt{c}) = a^2 - b^2*c$.
This will also work with $(\sqrt a + \sqrt b)(\sqrt a - \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2379383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 1
} |
Pythagorean Triple: $\text{Area} = 2 \cdot \text{perimeter}$ Find the unique primitive Pythagorean triple whose area is equal to twice the perimeter.
So far I set the sides of the triangle to be $a, b,~\text{and}~c$ where $a$ and $b$ are the legs of the triangle and c is the hypotenuse.
I came up with 2 equations whic... | Hint: formula for primitive triples is
$a = mn; b = \frac {m^2 -n^2}2; c = \frac {m^2 + n^2}2$ for $\gcd(m,n) = 1$. And as $m^2 - n^2 $ is even $m$ and $n$ must both be odd.
So we want $\frac {mn(m^2 -n^2)}4 = 2(mn + \frac {m^2 - n^2}2 + \frac {m^2 - n^2} 2) = 2(mn + m^2)$
So $mn(m-n)(m+n) = 8m(m+n)$
So $n(m-n)=8$
For... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2381494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Formula for consecutive residue of primitive modulo n. \begin{align*}
3^0 \equiv 1\mod 7\\
3^1 \equiv 3\mod 7\\
3^2 \equiv 2\mod 7\\
3^3 \equiv 6\mod 7\\
3^4 \equiv 4\mod 7\\
3^5 \equiv 5\mod 7\\
3^6 \equiv 1\mod 7\\
3^7 \equiv 3\mod 7\\
\end{align*}
Now just focusing on 1, 3, 2, 6, 4, 5, 1....
How to devise a formula ... | If you don't like modulus calculations, you could use:
For odd $a_k:\quad a_{k+1} = (7-a_k)/2$
For even $a_k:\quad a_{k+1} = 7-a_k/2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2381921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Is there a simple proof of this inequality? I want to prove that
$$ (a-1)^2+(b-1)^2 \ge 2(\sqrt{ ab}-1)^2 $$ for any positive real numbers $a,b$, and that equality holds iff $a=b$.
Edit: My "proof" for this inequality was wrong! It turns out the inequality holds iff $a=b$ or $\sqrt{a}+\sqrt{b} \ge \sqrt{2}$. (See the a... | $$ (a-1)^2+(b-1)^2 - 2(\sqrt{ab}-1)^2 = a^2+b^2-2ab-2(a+b)+4\sqrt{ab} = (\sqrt{a}-\sqrt{b})^2((\sqrt{a}+\sqrt{b})^2-2), $$
which is nonnegative if and only if $\sqrt{a}+\sqrt{b} \geq \sqrt{2}$ or $a=b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Why should we have $\sin^2(x) = \frac{1-\cos(2x)}{2}$ knowing that $\sin^2(x) = 1 - \cos^2(x)$?
Why should we have $\sin^2(x) = \frac{1-\cos(2x)}{2}$ knowing that $\sin^2(x) = 1 - \cos^2(x)$?
Logically, can you not subtract $\cos^2(x)$ to the other side from this Pythagorean identity $\sin^2(x)+\cos^2(x)=1?$
When I ... | Let's address two things, the question in the post and the confusion others have pointed out.
Part 1: Proof
Here's a proof that $\cos^2(\theta)+\sin^2(\theta)=1$, from which you can show that $\sin^2(\theta)=1-\cos^2(\theta)$. It requires a bit of set up using the Euler identity and that $i=\sqrt{-1}$. If you are not ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Is there a mistake in my evaluation of $\int\frac{3x+1}{\sqrt{5x^2+1}}dx$? I want to evaluate the following integral:
$$\int\frac{3x+1}{\sqrt{5x^2+1}}dx$$
The answer given in my textbook is $\frac{3}{5}\sqrt{5x^2+1}+\frac{1}{\sqrt{5}}\ln(x\sqrt{5}+\sqrt{5x^2+1})$.
I think the author has excess factor $\sqrt{5}$ in l... | Thanks to all comments, the difference between two forms was exactly in additive constant. My problem is solved and answered now.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Partial fractions and linear vs quadratic factors I was watching some videos on partial fraction decompistion and I got confused on one of the examples:
Say for example you have $$\frac{x+4}{x^2(x^2 +3)^2}.$$
The partial fraction equation of this is apparently:
$$\frac{A}{x} + \frac{B}{x^2} + \frac{Cx+E}{x^2 +3} ... | hint
if
$$\frac {x+4}{x^2(x^2+3)^2}=\frac {ax+b}{x}+\frac {cx+d}{x^2}+$$
$$\frac {Ax+B}{x^2+3}+\frac {Cx+D}{(x^2+3)^2} $$
$$=a+\frac {b+c}{x}+\frac {d}{x^2}+... $$
then $x\to +\infty $ gives $a=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Definition of convergence of a series My text book gives the following definition of convergence of a series:
A sequence of real numbers is said to be convergent if it has a finite real number $L$ as its limit. We then say that the sequence $x$ converges to $L$.
But I find it to be confusing, as from the above defini... | There are two different kinds of objects that you are studying: sequences, and series.
A sequence $(a_n)$ of real numbers converges if there is some finite real number $L$ such that $\lim_{n\to\infty} a_n = L$. What this means is that we can make the difference between $a_n$ and $L$ as small as we like by choosing a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find the real and imaginary part of z let $z=$ $$ \left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \right)^n$$
Rationalizing the denominator:
$$\frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta}\cdot\left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta + i\cos\theta}\right) =... | It is convenient to use $e^{i\theta}=\cos\theta + i\sin\theta$.
We obtain
\begin{align*}
\color{blue}{z}&\color{blue}{=\left(\frac{1+\sin \theta +i\cos \theta}{1+\sin\theta-i\cos \theta}\right)^n}\\
&=\left(\frac{1+ie^{-i\theta}}{1-ie^{i\theta}}\right)^n\\
&=\left(\frac{1+ie^{-i\theta}}{-ie^{i\theta}(1+e^{-i\theta})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 2
} |
rotation about $x$ and $y$ axis on the Bloch sphere $$R_x(\theta)= \begin{bmatrix} \cos \left( \frac{\theta}{2} \right) & -i\sin \left( \frac{\theta}{2} \right) \\ -i\sin \left( \frac{\theta}{2} \right) & \cos \left( \frac{\theta}{2} \right)\end{bmatrix}$$
$$R_y(\theta)= \begin{bmatrix} \cos \left( \frac{\theta}{2} \r... | I will leave $R_y(\theta)$ as an exercise for you.
Below is a verification of $R_x(\theta)$ is a rotation about the $x$-axis.
Denote the Pauli matrices as
$$X=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, Y=\begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}, Z=\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} $$
We have the fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solving $\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}$. Why is my solution wrong? I'm following all hitherto me known rules for solving equations, but the result is wrong. Please explain why my approach is not correct.
We want to solve:
$$\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \f... | The error is in the third group of parentheses in the "multiplying factors again" step. What you have as $-12x$ should be positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2388701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 0
} |
Find all positive integers such that: $n^2-3|3n^3-5n^2+7$
Find all positive integers such that: $$n^2-3|3n^3-5n^2+7.$$
I did the following:
$$n^2-3|3n^3-5n^2+7,n^2-3\Rightarrow n^2-3|3n^3-5n^2+7,3n^3-9n$$
$$\Rightarrow n^2-3|5n^2-9n-7,5n^2-15\Rightarrow n^2-3|9n-8$$
Now we must have: $9n-8\geq n^2-3$.We should deter... | Continuing from $\,n^2\!-\!3\mid 9n\!-\!8\ $ we can avoid quadratic inequalities and checking $n = 1,\ldots,8$
$\!\bmod n^2\!-3\!:\ \color{#c00}{n^2\equiv 3}\,\Rightarrow\, 0 \equiv \color{#c00}{n}(9\color{#c00}{n}\!-\!8)\equiv {27}\!-\!8n\,\Rightarrow\,0\equiv 9n\!-\!8+27\!-\!8n \equiv n\!+\!19\,$ hence $\,n\equiv -19... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2389546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solve a system of non-linear equations: $2(a-b)=29+4ab$, $2(c-b)= 11+4bc$, $2(c+a) = 9-4ac$
So I've solved this system of equations: $$\begin{array}{lcl}2(a-b) & =& 29+4ab \\2(c-b)& = & 11+4bc \\2(c+a) & = & 9-4ac\end{array}$$
by simply solving for each variable in terms of the others, separately and plugging them in... | It's $$(2a+1)(2c+1)=10,$$ $$(2b-1)(2c+1)=-12$$ and $$(2a+1)(2b-1)=-30.$$
The rest is smooth:
$$(2a+1)^2(2c+1)^2(2b-1)^2=3600,$$
which gives
$$(2a+1)(2c+1)(2b-1)=60$$ or $$(2a+1)(2c+1)(2b-1)=-60.$$
If $(2a+1)(2c+1)(2b-1)=60$ then we obtain:
$2a+1=\frac{60}{-12}$, $2c+1=\frac{60}{-30}$ and $2b-1=\frac{60}{10}$, which i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2389859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
how to show limit of $\frac{x^2-y^2}{x^3-y^3}$ as (x,y) goes to $(0,0)$ does not exist I know that $\lim\limits_{(x,y)\to (0,0)} \frac{x^2-y^2}{x^3-y^3}$ does not exist, but i'm not sure how to show it.
What I have done is let $y=kx$
and then,
$\lim\limits_{x\to 0} \frac{x^2-(k x)^2}{x^3-(kx)^3} = \lim\limits_{x\to 0} ... | Using polar coordinates, $x=r\cos\theta, y=r\sin\theta,$ we have $$\frac{r^2(\cos^2\theta - \sin^2\theta)}{r^3(\cos^3\theta-\sin^3\theta)}=\frac{\cos\theta + \sin\theta}{r(\cos^2\theta+\cos\theta\sin\theta+\sin^2\theta)}=\frac{\cos\theta+\sin\theta}{r(1+\cos\theta\sin\theta)}$$
Choose any $\theta$ such that the numerat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $x$, $y$ and $z$ are distinct positive integers and $x+y+z=11$ then what is the maximum value of $(xyz+xy+yz+zx)$?
If $x$, $y$ and $z$ are distinct positive integers and $x+y+z=11$ then what is the maximum value of $(xyz+xy+yz+zx)$?
We know that product is maximum when difference between $x$, $y$ and $z$ is minimu... | Note that $(x+1)(y+1)(z+1)=1+(x+y+z)+(xy+yx+zx)+xyz$ so the sum you want is $$(x+1)(y+1)(z+1)-12$$ which justifies your comment about the product.
If $a\gt a-1\ge b+1\gt b$ we have $$(a-1)(b+1)=ab+(a-b)-1\gt ab$$ since $a-b\ge 2$.
The best selection of distinct integers is (as others have noted) $5+4+2=11$. And the p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Evaluate the integral $\int\frac{dx}{x^2\sqrt{4x^2-1}}$ I solved the question, but Wolfram Alpha and Symbolab both give me two completely different answers.
Here's my work:
Let $u = 2x$ and $a = 1$
Then $du = 2dx$ and $dx = \frac{du}{2}$
Then $\int\frac{dx}{x^2\sqrt{4x^2-1}}$ = $\int\frac{1}{u\sqrt{u^2-a^2}}$
$\therefo... | $\int\frac{dx}{x^2\sqrt{4x^2-1}}$
take $x=\dfrac{\sec \theta}{2}$, then $dx=\dfrac{\sec \theta\tan \theta}{2}d\theta$
Now $\sin\theta=\pm\dfrac{\sqrt{4x^2-1}}{2x}$
When $x>\dfrac{1}{2}$, then $\sqrt{4x^2-1}=\tan \theta$
$$\int\dfrac{dx}{x^2\sqrt{4x^2-1}}=2\int \dfrac{\sec \theta\tan \theta}{\sec^2\theta\sqrt{\sec^2\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int_0^{\pi/2}\int_0^{\pi/2} \sin x \sin^{-1}(\sin x \sin y) \ dx \ dy$
Evaluate $$\int_0^{\pi/2}\int_0^{\pi/2} \sin x \sin^{-1}(\sin x \sin y) \ dx \ dy$$
My attempt. Taking $\sin x \sin y=t$, $\iint\sin x \ dx \ dy$.
I am not sure but I think the limits would change to $x=0$ to $\pi/2$ and $t=0$ to $1$.
... | Here is an approach using power series expansions.
It is known that for $t\in (-1,1)$
$$\frac{1}{\sqrt{1-t}} =(1-t)^{-1/2}=\sum_{n=0}^\infty
\binom{-1/2}{n} (-t)^{n}=
\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} t^{n},$$
and for $t\in [-1,1]$,
$$\arcsin(t) =\int_0^t\frac{ds}{\sqrt{1-s^2}}= \sum_{n=0}^\infty \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2394075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
Tried (without sucess) two different approaches: (a) finding $x^3$ by raising the right expression to power 3, but was not able to find something useful in the result t... | Let $a=\sqrt[3]{2+\sqrt{5}}, b=\sqrt[3]{2-\sqrt{5}}\,$, then:
$$\require{cancel}
a^3+b^3 = 2+\cancel{\sqrt{5}} + 2-\cancel{\sqrt{5}} = 4 \\
ab = \sqrt[3]{(2+\sqrt{5})(2-\sqrt{5})} = \sqrt[3]{2^2 - 5} = -1
$$
From $\,(a+b)^3 = a^3+b^3+3ab(a+b)\,$ and given that $\,x=a+b\,$ it follows that $\,x^3=4-3x\,$ $\iff 0 = x^3+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 2
} |
Calculate $\lim_{x \rightarrow a} \frac{x^2 + ax - 2a^2}{\sqrt{2x^2 - ax} -a}$ I need to calculate:
$$\lim_{x \rightarrow a} \frac{x^2 + ax - 2a^2}{\sqrt{2x^2 - ax} -a}$$
I get $0/0$ and can then use l'hopital's rule to find the limit, I can do this but someone asked me how I can do this without using l'hopital's rule.... | You want
$\lim_{x \rightarrow a} \frac{x^2 + ax - 2a^2}{\sqrt{2x^2 - ax} -a}
$.
Let
$x = y+a$.
Then
$\begin{array}\\
\dfrac{x^2 + ax - 2a^2}{\sqrt{2x^2 - ax} -a}
&=\dfrac{(y+a)^2 + a(y+a) - 2a^2}{\sqrt{2(y+a)^2 - a(y+a)} -a}\\
&=\dfrac{y^2+2ay+a^2 + ay+a^2 - 2a^2}{\sqrt{2y^2+4ya+2a^2 - ay-a^2} -a}\\
&=\dfrac{y^2+3ay}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Partial fractions - integration $$\int \frac{4}{(x)(x^2+4)} $$
By comparing coefficients,
$ 4A = 4 $,
$A = 1$
$1 + B = 0 $,
$B= -1 $
$xC= 0 $,
$C= 0 $
where $\int \frac{4}{x(x^2+4)}dx =\int \left(\frac{A}{x} + \frac{Bx+C}{x^2 + 4}\right)dx$.
So we obtain $\int \frac{1}{x} - \frac{x}{x^2+4} dx$.
And my final an... | Yes, $4 = A(x^2+4)~+~(Bx+C)x \implies A=1, B=-1, C=0$
So
$$\require{enclose}\int \dfrac{4}{x(x^2+4)}~\mathrm d x ~{= \int \dfrac{1}{x}+\dfrac{\enclose{circle}{~-x~}}{(x^2+4)}~\mathrm d x \\ = \ln\lvert x\rvert -\int\dfrac{\tfrac 12\mathrm d (x^2+4)}{(x^2+4)} \\ = \ln\lvert x\rvert -\tfrac 12\ln\lvert x^2+4\rvert+D}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Difference of two complex polynomials Prove that for any integer $m>1$, $(z+a)^{2m}-(z-a)^{2m}=4maz\prod_{k=1}^{m-1}z^2+a^2\cot^2{\frac{k\pi}{2m}}$
I started off by expanding the left hand side using the binomial therem and noticed that some terms cancelled out and I ended up with the expression below for the LHS
$2\su... | We follow the comments and consider the polynomial
\begin{align*}
u^{2m}=1\qquad\text{ with roots }\qquad \exp\left(\frac{k\pi i}{m}\right), 1\leq k\leq 2m
\end{align*}
The function
\begin{align*}
u(z)=\frac{z-a}{z+a}
\end{align*}
has the inverse function
\begin{align*}
z(u)=-a\frac{u+1}{u-1}
\end{align*}
We conclude ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove or disprove : If $a\equiv b$ mod $m$, when $a,b,m\in \mathbb{Z}$ , then $a^3\equiv b^3$ mod $m^2$. Prove or disprove : If $a\equiv b$ mod $m$, when $a,b,m\in \mathbb{Z}$ , then
$a^3\equiv b^3$ mod $m^2$
My attempt:
since $a\equiv b$ mod $m$ then $a=b+mk$ for some $k$
Now $a^3=m^3k^3+b^3+3m^2k^2b+3mkb^2$
$a^3-b^... | $$1\equiv 5\pmod 4$$
but $$5^3\equiv 13\not\equiv 1\pmod{16}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Orthogonal projection of point onto line not through origin
What is the orthogonal projection of the point $(8,3)$ onto the line $y = -2x - 3$?
Using basic analytical methods, I get the point $(-4/5,-7/5)$, but I want to obtain this point by seeing what happens to the standard basis vectors. But I always end up with... | $y = -2x-3$ is denoted by $<(1,-2)>$ named A
The orthogonal projection of point $<(8,3)>$ named $\bar x$ subtract (-3) from the y vector to obtain a new $\bar x$ to give you $<(8,6)>$ is given by
$ A(A^{T}A)^{-1}A^{T}\bar x$
$(1,-2)\left((1,-2).(1,-2)\right)^{-1}\times\left((1,-2).(8,6)\right) = (1,-2)(\frac{1}{5}\ti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
3-D geometry: line intersecting 2 lines and parallel to plane Find a surface generated by line intersecting lines $$y=a=z$$ and $$x+3z=a=y+z$$ and parallel to plane $$x+y=0$$
I tried to form a line equation which intersects the given two lines i.e.
$(y-a)+k1(z-a)=0$ and $(x+3z-a)+k2(y+z-a)=0$. But don't know how to us... | Reordering the equation of the line
$$\alpha: y+k_1 z-a k_1-a=0;\;\beta:x+k_2 y+(k_2+3) z -ak_2-a=0$$
The normal vector of plane $\alpha$ is $n_1=(0,1,k_1)$ and the normal vector of $\beta$ is $n_2=(1,k_2,k_2+3)$
the direction vector of the line is the cross product $m=n_1\times n_2=(3+k_2-k_1 k_2,k_1,-1)$
The line $(y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\frac{\tan x}{x}>\frac{x}{\sin x}, x\in(0,\pi/2)$
Prove that $$\frac{\tan x}{x}>\frac{x}{\sin x},\;\;\; x\in(0,\pi/2).$$
My work
I formulated $$f(x)=\tan x \sin x - x^2$$ in hope that if $f'(x)>0$ i.e. monotonic then I can conclude for $x>0, f(x)>f(0)$ and hence, prove the statement.
However, I got $$f'(x... | We can prove a stronger inequality
$$\frac{\tan x}{x} > \left(\frac{x}{\sin x}\right)^2$$ for $x \in (0, \pi/2)$. Indeed the function
$\tan x \sin^2 x - x^3$ has derivative $\tan^2 x + 2 \sin^2 x - 3 x^2$. Now, from this answer we see that $\frac{\tan x + 2 \sin x }{3} > x$ for $x \in (0, \frac{\pi}{2})$ so
$\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 3
} |
How to find $\lim_{x \to \infty} \frac{ex^{x+1}-x(x+1)^x}{(x+1)^x}$ I came across this problem a few days ago and I have not been able to solve it. Wolfram Alpha says the answer is 1/2 but the answer I came up with is 0. Can anyone see what is wrong with my work and/or provide the correct way of solving this problem?
$... | This solution is not as elegant as @farruhotas, but it's more direct. This solution relies on Taylor approximations and handwaving. I realize it's not rigorous; the purpose of this is providing intuition as to why the limit is $\frac{1}{2}$. (I'll be assuming that $x$ is an integer. Assuming the limit exists, since $x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Show that $a^4+b^4\geq \frac18$ given that $a+b=1$
Show that $a^4+b^4\geq \frac18$ given that $a+b=1$
$$b=1-a\Rightarrow a^4+b^4=2a^4-4a^3+6a^2-4a+1$$
If we try to find the minimum of a one-variable function,we must solve a 3rd degree equation,on the other hand making perfect squares seems somewhat difficult!
Please ... | Hint: by the generalized mean inequality: $\;\displaystyle \sqrt[4]{\frac{a^4+b^4}{2}} \ge \frac{a+b}{2}\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Equation of locus of points satisfied by $\frac{\left|z+3i\right|}{\left|z-6i\right|}=1$ Equation of locus of points satisfied by $\frac{\left|z+3i\right|}{\left|z-6i\right|}=1$
The answer I got is $y=\frac{3}{2}$, but the answer given in my book is $y=-0.5x+2.25$
Can anyone please confirm which is the right answer
Edi... | Given two points,
the set of points
equidistant from them
is the perpendicular bisector
of the line joining them.
If the points are
$(a, b)$ and
$(c,d)$,
the midpoint is
$((a+c)/2, (b+d)/2)$.
The slope of the line is
$\dfrac{d-b}{c-a}$,
so the slope of the normal
is the negative reciprocal
of this or
$-\dfrac{c-a}{d-b}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Approximating a series through successive telescopic series I am trying to approximate $\sum \frac{1}{n^2}$ with telescopic series. So far, I've understood up to this:
$$\begin{eqnarray*} \sum_{n\geq m}\frac{1}{n^2}&=&\sum_{n\geq
m}\left(\frac{1}{n}-\frac{1}{(n+1)}\right)+\frac{1}{2}\sum_{n\geq
m}\left(\frac{1}{n^2}-\f... | $$\frac{1}{n^3(n+1)^3}=\frac{1}{n^6}-\frac{3}{n^7}+\frac{6}{n^8}+\ldots \tag{1}$$
$$\frac{1}{n^5}-\frac{1}{(n+1)^5} = \frac{5}{n^6}-\frac{15}{n^7}+\frac{35}{n^8}\ldots\tag{2} $$
hence
$$ \frac{1}{n^3(n+1)^3}-\color{blue}{\frac{1}{5}}\left(\frac{1}{n^5}-\frac{1}{(n+1)^5}\right) = -\frac{1}{n^8}+\ldots \tag{3} $$
where t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2403098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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.