Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to evaluate the following integral involving hyperbolic functions? I've been thinking about using contour integration, but I can't seem to find the right combination of function and contour. Thanks for your attention. :)
$$
\int_{0}^{\infty} \frac {\sinh ax \sinh bx}{\cosh cx} \ dx
$$
| Consider the contour integral
$$\oint_C dz \frac{\sinh{a z} \sinh{b z}}{\sinh{c z}} $$
where $C$ is the rectangle $-R-i \pi/(2 c), R-i \pi/(2 c), R+i \pi/(2 c), -R+i \pi/(2 c)$. Note that
$$\sinh{c (x \pm i \pi/2 c)} = \pm i \cosh{c x} $$
Therefore the contour integral is
$$\int_{-R}^R dx \, \frac{\sinh{a(x-i \pi/(2 c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/781522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Three Congruent Incircles of a divided Equilateral triangle Take an equilateral triangle with sides of unit length and choose a vertex from which to draw two cevians to the opposite side.These cevians divide the equilateral triangle into three subtriangles.
If these three subtriangles all have congruent incircles, can ... | Label elements of equilateral $\triangle ABC$ as shown:
Using the fact that
$$\text{area of } \triangle = \frac{1}{2} \cdot \text{inradius} \cdot \text{perimeter}$$
we have
$$\frac{2|\triangle APB|}{1+x+y} = r = \frac{2|\triangle APQ|}{1-2x+2y}$$
$$\frac{x \cdot \frac{\sqrt{3}}{2}}{1+x+y} = r = \frac{(1-2x) \cdot \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/781810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\int_{0}^{\infty}\frac{x}{x^3+1}dx$ =? So guys I have this improper integral $\int_{0}^{\infty}\frac{x}{x^3+1}dx$. I checked that it converges by $ \int_{0}^{1}\frac{x}{x^3+1}dx + \int_{1}^{\infty}\frac{x}{x^3+1}dx $ and using the $\frac{c}{(x-a)^\lambda} $ and $\frac{1}{x^\lambda} $ criteria. But for finding the val... | Continuing Daniel Fischer's answer, we have:
$$\int_{0}^{1}\frac{dx}{x^2-x+1}=\int_{-1/2}^{1/2}\frac{dx}{x^2+3/4}=\frac{\sqrt{3}}{2}\cdot\frac{4}{3}\int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{dx}{x^2+1}=\frac{4\sqrt{3}}{3}\arctan\frac{1}{\sqrt{3}}=\frac{2\pi}{3\sqrt{3}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/782531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
If $y =\sqrt{5+\sqrt{5-\sqrt{5+ \cdots}}}$, what is the value of $y^2-y$? If $$y =\sqrt{5+\sqrt{5-\sqrt{5 + \cdots}}},$$
what is the value of $y^2-y$ ?
I am unable to get the clue due to these alternative signs of plus and minus please help on this thanks...
| Indeed, there is another way:
Let $x =\sqrt{5-\sqrt{5+\sqrt{5 - \cdots}}}$,
then it's easy to see
$y^2 = 5 + x$, $x^2 = 5 - y$
then
$y^2 - x^2 = x+y$, therefore $y-x = 1$
Thus,
$$y^2 - y = y^2 - 1 - x = 5 + x - (1+x) =4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/783624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Inverse Identity + Constant Matrix I need to invert a square symmetric matrix $$ C = c\, I+cs\, B $$
Where: (1) $B$ is a constant matrix of 1 for each entry. (2) $c$ and cs are just positive real numbers. (3) $I$ is the identity.
However, the $\det(B) = 0$ and $B$ by itself does not have an inverse, but I am sure th... | You could use the Neumann series which is the inverse, if it converges:
$$A^{-1} = \sum_{k=0}^\infty(I-A)^k\qquad\text{(if convergent)}$$
Now we apply this to $\frac1c C = I + \frac{c_s}c B$ and get using $B^k = n^{k-1} B$ for $k>0$ (since $B^2=nB$) the following
\begin{align*}
\sum_{k=0}^\infty\left(I - \frac1c C\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/785264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Points on $(x^2 + y^2)^2 = 2x^2 - 2y^2$ with slope of $1$ Let the curve in the plane defined by the equation: $(x^2 + y^2)^2 = 2x^2 - 2y^2$
How can i graph the curve in the plane and determine the points of the curve where $\frac{dy}{dx} = 1$.
My work:
First i found the roots of this equation with a change of variable ... | I won't do the graph, just the algebra.
Implicit differentiation gives
$$2(x^2+y^2)(2x+2yy')=4x-4yy'$$
so setting $y'=1$ and simplifying leads to
$$(x^2+y^2)(x+y)=x-y\qquad(*)$$
If we now rewrite the equation for the curve as
$$(x^2+y^2)^2=2(x-y)(x+y)$$
we see that multiplying both sides of equation $(*)$ by $2(x+y)$ g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/785900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Verifying that the determinant is equal to $1!2!3!...(n-1)!$ Verifying that the determinant is equal to $1!2!3!...(n-1)!$
$$|A|=
\begin{vmatrix}
1 &1 & \dots &1\\
1 &2 & \dots &2^{n-1}\\
1 &3 & \dots &3^{n-1}\\
& & \dots\\
1 &n & \dots &n^{n-1}\\
\end{vmatrix}=1!2!3!...(n-1)!
$$
I used the definition of a determinan wi... | We can evaluate this determinant as follows.
Let $A_n$ be the $n \times n$ matrix in question.
First, for each $j = n-1, n-2, \ldots, 1$, subtract $n$ times column $j$ from column $j+1$.
This results in the following:
\begin{align*}
|A_n| &=
\begin{vmatrix}
1 & 1-n & 1-n & \ldots & 1-n & \ldots & 1-n \\
1 & 2 - n & 2^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/786065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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In how many ways can ww choose $10$ cards so there are $3$ exact matches? Suppose there are $20$ cards- $10$ reds numbered $1,2,\cdots,10$, and $10$ blues numbered $1,2, \cdots, 10$. In how many ways can 10 cards be picked to so that there are EXACTLY 3 MATCHES- where a match means a red card and blue card have the sam... | There are $\binom{10}{3}$ ways to pick the cards that will make up the three matching pairs. That is a component of your answer, so I will assume that part is clear to you. Now we count the number of ways to pick the remaining $4$ cards.
There remain $7$ "couples." We pick $4$ of these, and for each couple we choose t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/788804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving a metric with absolute value I need to prove that function $\mathbb R × \mathbb R → \mathbb R $ : $f(x,y) = \frac{|x-y|}{1 + |x-y|}$ is a metric on $\mathbb R$. First two axioms are trivial; it's the triangle inequality which is pain. $\frac{|x-y|}{1 + |x-y|}$ + $\frac{|y-z|}{1 + |y-z|} ≥ \frac{|x-z|}{1 + |x-z... | Since $1 + |x-y| + |y -z| \ge 1 + |x-z|$,
$$1 - \frac{1}{1 + |x-z|} \le 1 - \frac{1}{1 + |x-y| + |y -z|} \\
\implies \frac{|x-z|}{1 +|x-z|} \le \frac{|x-y| + |y-z|}{1 + |x-y| + |y-z|} \le \frac{|x-y|}{1 + |x-y|} + \frac{|y-z|}{1 + |y-z|}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/788995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Last 3 digits of $3^{999}$ I know that it's $3^{999} \mod 1000$ and since $\varphi(1000) = 400$ and $3^{400}\equiv1 \mod1000$ it will be equivalent to $3^{199} \mod 1000$ but what should I do from then? Or am I wrong about this from the start?
| Simple computation, and the fact that $\phi(8)=4$, gives
$$
\begin{align}
3^{999}
&\equiv3^3\\
&\equiv3\pmod8\tag1
\end{align}
$$
The square and multiply algorithm, and the fact that $\phi(125)=100$, gives
$$
\begin{align}
3^{999}
&\equiv3^{99}\\
&\equiv42\pmod{125}\tag2
\end{align}
$$
The Extended Euclidean Algorithm,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/789050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
} |
Solve $\sin(2\theta) -\tan(\theta) = 0 \ $ for $ 0\leq \theta \leq 2\pi $ I want to use the fact that $$\sin(2\theta) = \frac{2\tan(\theta)}{1 + \tan^2(\theta)}$$
to solve $\sin(2\theta) -tan(\theta) = 0 \ $ for $ 0\leq \theta \leq 2\pi$
My solution:
$\frac{2tan(\theta)}{1 + tan^2(\theta)} - tan(\theta) = 0 $
so $\fra... | Try this
\begin{align}
\sin2\theta&=\tan\theta\\
\frac{2\tan\theta}{1 + \tan^2\theta}&=\tan\theta\\
\frac{2}{1 + \tan^2\theta}&=1\\
\tan^2\theta-1&=0\\
(\tan\theta-1)(\tan\theta+1)&=0.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/789707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
How prove this $\frac{r}{R}\ge\frac{AM}{AX}$ Let acute $\triangle ABC$ have largest angle $\angle A$, and let $R,r$ be the circumradius and inradius respectively. Let $M$ be the midpoint of $BC$, and $X$ be the intersection of the tangents at $B,C$ to the circumcircle of $\triangle ABC$
Show that:
$$\frac{r}{R}\ge\frac... | Let $(a,b,c) = (\angle A,\angle B,\angle C)$ in $\triangle ABC$
Let $Γ$ be the circumcircle of $\triangle ABC$ with centre $O$
Then $A,O$ are on the same side of $BC$ because $\triangle ABC$ is acute
Let $D,E$ be the intersections of the perpendicular bisector of $BC$ with $Γ$ such that $A,D$ are on the same side of $B... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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fractional part of Riemann zeta $\sum_{s=2}^\infty \{\zeta (s)\}=1$
$$\sum_{s=2}^\infty \{\zeta (s)\}=1$$
where $\zeta (s)$ is Riemann zeta, $\{x\}$ denotes the fractional part of the real number $x$
The problem was proposed by Michael Th. Rassias
$\{\zeta(2)\}=\frac{\pi^2}6-1,$ How to go on?
Thanks a lot!
| Since $ 1 < \zeta(s) <2$ for $s \ge 2$, it's equivalent to showing that $ \displaystyle\sum_{s=2}^{\infty} \left( \zeta(s)-1 \right)= 1$.
In which case,
$$ \sum_{s=2}^{\infty} \left( \zeta(s)-1 \right)= \sum_{s=2}^{\infty}\sum_{n=2}^{\infty} \frac{1}{n^{s}} = \sum_{n=2}^{\infty} \sum_{s=2}^{\infty} \frac{1}{n^{s}}$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/792624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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proving $\tan^{-1}(x)+\tan^{-1}(y) = \pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)\;,$ when $x>0,y>0,xy>1$ (1) How ca we prove $\displaystyle \tan^{-1}(x)+\tan^{-1}(y) = \pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)\;,$ when $x>0,y>0,xy>1$
(2) How ca we prove $\displaystyle \tan^{-1}(x)+\tan^{-1}(y) = -\pi+\tan^{-1}\left(\fr... | It is quite simple ;
lets consider tan(θ) = x and let it be so that tan(ϕ) = y , then we know , θ = tan⁻¹(x) and ϕ = tan⁻¹(y)
$\displaystyle\Rightarrow$ tan(θ+ϕ)= $\displaystyle(\frac{tanθ+tanϕ}{1-tanθ*tanϕ})\;$
$\displaystyle\Rightarrow$ tan(θ+ϕ) = $\displaystyle(\frac{x+y}{1-x*y})\;$ as [Note : tan(θ) = x and tan(ϕ) ... | {
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"url": "https://math.stackexchange.com/questions/795394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral $\int_0^1 \log \frac{1+ax}{1-ax}\frac{dx}{x\sqrt{1-x^2}}=\pi\arcsin a$ Hi I am trying to solve this integral $$
I:=\int_0^1 \log\left(\frac{1+ax}{1-ax}\right)\,\frac{{\rm d}x}{x\sqrt{1-x^2}}=\pi\arcsin\left(a\right),\qquad
\left\vert a\right\vert \leq 1.
$$
It gives beautiful result for $a = 1$
$$
\int_0^1 \lo... | The integral in question,
\begin{align}
I = \int_{0}^{1} \ln \left( \frac{1+ax}{1-ax} \right) \ \frac{dx}{x \sqrt{1-x^{2}}}
\end{align}
can be separated into the two integrals
\begin{align}
I = \int_{0}^{1} \frac{ \ln(1+ax)}{x \sqrt{1-x^{2}}} \ dx - \int_{0}^{1} \frac{ \ln(1-ax)}{x \sqrt{1-x^{2}}} \ dx
\end{align}
whic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/795493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 3
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Gradient formula of volume of tetrahedron involving the scalar triple product Let $a,b,c,d \in \mathbb{R}^3$ be the vertices of a tetrahedron (I’m unsure whether or not the order of the vertices is important for what follows). The volume of the tetrahedron is
$$
\begin{align}
\operatorname{vol}(a,b,c,d)
... | This comes down to a straightforward calculation. We have for $a,b,x \in \mathbb{R}^3$
$$\begin{align}
\nabla_x(a \times b)\cdot x
&= \nabla_x \big( (a_2b_3 - a_3b_2)x_1 + (a_3b_1 - a_1b_3)x_2 + (a_1b_2 - a_2b_1)x_3 \big) \\
&= a \times b
\end{align}$$
and
$$\begin{align}
\nabla_x(a \times x) \cdot b
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/797845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Finding a real root in a cubic equation? I am trying to find a real root of the following cubic.
$x^3-6x^2+14x-15=0$
I did $x=y-\frac{b}{3a}$
which is $x=y+2$
I plugged my substitution to get the depressing cubic.
$(y+2)^3-6(y+2)^2+14(y+2)-15=0$
which is
$y^3+2y=3$
then I set up the method for solving.
$y=s-t$ , first... | Note that $x=3$ is a solution. We used the Rational Roots Theorem, since cubics in exercises often have a rational root. Exercises and real life tend to differ.
When the depressed cubic was reached (correctly) the root $y=1$ was available by inspection.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/798230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Gram-Schmidt process on complex space
Let $\mathbb{C}^3$ be equipped with the standard complex inner product. Apply the
Gram-Schmidt process to the basis: $v_1=(1,0,i)^t$, $v_2=(-1,i,1)^t$, $v_3=(0,-1,i+1)^t$ to find an orthonormal basis $\{u_1,u_2,u_3\}$.
I have found $u_1 = \dfrac{1}{\sqrt{2}} (1,0,i)^t$ and $u_2... | Your $u_1$ is correct, but your $u_2$ is incorrect; as noted in the comments, $u_1\cdot u_2 \neq 0$.
Recall that the standard inner product on $\mathbb{C}^n$ is given by $\langle u, v\rangle = u\cdot\bar{v}$. With this in mind, let's calculate $u_2$. First we have
$$\langle v_2, u_1\rangle = (-1, i, 1)\cdot\overline{\... | {
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"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Using Taylor expansion to find $\lim_{x \rightarrow 0} \frac{\exp(2x)-\ln(1-x)-\sin(x)}{\cos(x)-1}$
Use Taylor expansion to find $\displaystyle \lim_{x \rightarrow 0} \frac{\exp(2x)-\ln(1-x)-\sin(x)}{\cos(x)-1}$
I know that:
$\exp(2x) = 1 + (2x) + \frac{(2x)^2}{2!}+ O(x^3)$
$\ln(1-x) = 1-(-x) + (-x)^2 + O(x^3)$
$\sin... | Your Taylor expansion of $\ln(1-x)$ is wrong
It should be $\displaystyle \ln(1-x)=-x-\frac{x^2}{2}+O(x^3)$
Therefore, $$\frac{\exp(2x)-\ln(1-x)-\sin(x)}{\cos(x)-1}=\frac{1+2x+\frac{5}{2}x^2+O(x^3)}{\frac{-x^2}{2}+O(x^3)}$$
The last expression goes to $-\infty$
Further explanation of this last fact:
$$\frac{\exp(2x)-\l... | {
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"source": "stackexchange",
"question_score": "2",
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Find all real solutions of $6^x+1=8^x-27^{x-1}$
Find all real solutions of $6^x+1=8^x-27^{x-1}$.
Things I tried:
We want solutions of
$$2^x3^x+1 = (2^x)^3-\frac{(3^x)^3}{27}.$$ Write $a=2^x$ and $b=3^x$. This gives
$$ab+1 = a^3-\frac{b^3}{27}$$
or
$$27ab+27=27a^3-b^3$$
How to continue?
| The equation $27ab+27=27a^3-b^3$ you have found may be written as
$$27a^3-27ab-27-b^3=0,$$
which may be viewed as a cubic in $b$. Its discriminant is
$$\Delta_b(a)=-4\cdot(-1)\cdot(-27a)^3-27\cdot(-1)^2=-(a^3+1)^2,$$
which is negative for all $a$ except for $a=-1$, when it is zero. This does not correspond to a real va... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/800319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Help me calculating chromatic polynomial of this subgraph
This is the subgraph. I just wanted to do a couple of exercises with chromatic polynomials, but around the web are pretty standard ones, so i came up with this one myself. I thought it's pretty straightforward.
It is: i color $A$ with $x$ colors, automatically... | The problem here is that you haven't accounted for the fact that $E$ and $B$ might have the same color, and $E$ and $C$ might have the same color.
There are cases here that need considering:
Case 1: $E$ has the same color as $B$.
In this case, there are $x$ choices for $A$; $x-1$ for $B$ and $E$; $x-2$ for $C$; $x-1$ f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/801596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
The $n$'th derivative of $x^x$ I want to know the $n$'th derivative of $f(x)=x^x$.
Then, I'll calculate $f(0)$ with Taylor expansion of $f(x)$ on $a=1$.
Here is my answer, but it is unfinished.
The derivative of $f(x)=x^x$
$$\begin{align}
f'(x)&=x^x(\log x+1)\\
f''(x)&=x^x(\log x+1)^2+x^{x-1}\\
f'''(x)&=x^x(\log x+1)^3... | $$\frac{d^n x^x}{dx^n}=\frac{d^ne^{x\ln(x)}}{dx^n}$$
Expanding $e^y$ as a series and then $\ln(x)$ as a limit:
$$\frac{d^n}{dx^n}\sum_{k=0}^\infty \frac{x^k \ln^k(x)}{k!}=\lim_{c\to0}\sum_{k=0}^\infty \frac1{k!}\frac{d^n}{dx^n}\frac{x^k(x^c-1)^k}{c^k}$$
Now use the general Leibniz rule:
$$\frac{d^n}{dx^n}x^k(x^c-1)^k=\... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
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Evaluate $\int_0^1\frac{x\ln x}{(1+x^2)^2}\ dx$ $$\int_0^1\frac{x\ln x}{(1+x^2)^2}\ dx
$$
Help me please. I don't know any ways of solution. Thank you.
| Integrating by Parts,
$$I=\int \ln x\cdot\frac x{(1+x^2)^2}dx$$
$$=\ln x\int\frac x{(1+x^2)^2}dx-\int\left(\frac{d(\ln x)}{dx}\cdot \frac x{(1+x^2)^2}dx\right)dx $$
For $\displaystyle J=\int\frac x{(1+x^2)^2}dx$ set $1+x^2=u\implies2x\ dx=du$
$\displaystyle J=\int\frac{du}{2u^2}=-\frac1{2u}=-\frac1{2(1+x^2)}$
$\display... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/805617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 1
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How prove $\left(\frac{b+c}{a}+2\right)^2+\left(\frac{c}{b}+2\right)^2+\left(\frac{c}{a+b}-1\right)^2\ge 5$ Let $a,b,c\in R$ and $ab\neq 0,a+b\neq 0$. Find the minimum of:
$$\left(\dfrac{b+c}{a}+2\right)^2+\left(\dfrac{c}{b}+2\right)^2+\left(\dfrac{c}{a+b}-1\right)^2\ge 5$$
if and only if $$a=b=1,c=-2$$
My idea: Since ... | Note that "if and only if" is clearly not true, since any multiples will work as your expression is homogenous. In fact, the equality case is more general than what you listed.
Differentiating with respect to $c$ (which only appears in the numerator, hence is not too ugly), you can show that (Thanks Wolfram) the minimu... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\int_{0}^{\infty}\int_{1}^{\infty}\frac{x^2-y^2}{(x^2+y^2)^2}dxdy$ diverges? I want someone to review my proof that $$\int_{1}^{\infty}\int_{0}^{\infty}\frac{x^2-y^2}{(x^2+y^2)^2}dxdy$$ does not converge.
To make things easier, I said let's look at the entire first quadrant and then subtract the integral over the smal... | Since we are considering the region in the first quadrant above $y=1$, in polar coordinates we may write it as:
\begin{align*}
\int_{1}^{\infty}\int_{0}^{\infty}\frac{x^2-y^2}{(x^2+y^2)^2}\, dx\, dy &= \lim_{N\to\infty} \int_{\arcsin(1/N)}^{\pi/2} \, \int_{1/\sin\theta}^{N} \, \frac{\cos\left(2\, \theta\right)}{r}\, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/808085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Probability of picking three marbles in order A bag contains 9 red marbles, 5 blue marbles and 7 green marbles. Need to select one marble at a time without replacement. What is the probability that the first marble is blue, the second is blue or green an the third is red.
My Solution:
$= P(\text{blue}) * P(\text{blue} ... | That looks like the right answer. These are conditional probabilities, so
\begin{align*}
& \mathrm{Pr}(\text{marble 1 is blue} \cap \text{marble 2 is blue or green} \cap \text{marble 3 is red}) \\
&= \mathrm{Pr}(\text{marble 1 is blue}) \times \mathrm{Pr}(\text{marble 2 is blue or green} \mid \text{marble 1 is blue}) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/808345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the following equation in positive integers $x$ and $y$ What are the solutions in positive integers of the equation:
$${1+2^x+2^{2x+1}=y^2}$$
I tried to factorize the equation but it didn't help much.
Clearly $y $ is an odd integer. Substituting $y =2n+1$, we get
$2^x+2^{2x+1}=(2n)\cdot{(2n+2)}$
$\Rightarrow (2... | Let $2^x = z$, then: $2z^2 + z + y^2 - 1 = 0$.
Calculate $\triangle = 1^2 - 4\cdot 2\cdot (y^2 - 1) = 9 - 8y^2$.
$\triangle > 0 \iff 9 - 8y^2 > 0 \iff y^2 < \dfrac{9}{8} = 1.11$, so $y = 1$, and $\triangle = 1$. So $z = \dfrac{-1 \pm 1}{2\cdot 2} = 0$ or $-0.5$, but $z =2^x > 2$. So there is no solution.
In fact $\tria... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/808509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluation of $\int_{0}^{1}4x^3\cdot \left\{\frac{d^2}{dx^2}\left(1-x^2\right)^5\right\}dx$ The value of $\displaystyle \int_{0}^{1}4x^3\cdot \left\{\frac{d^2}{dx^2}\left(1-x^2\right)^5\right\}dx = $
$\bf{My\; Try::}$ Let $\displaystyle I = \int 4x^3\cdot \left\{\frac{d^2}{dx^2}\left(1-x^2\right)^5\right\}dx$
Using Int... | I wonder if it would not be simpler to establish that $$\frac{d^2}{dx^2}\left(1-x^2\right)^5=80 x^2 \left(1-x^2\right)^3-10 \left(1-x^2\right)^4=-10 \left(x^2-1\right)^3 \left(9 x^2-1\right)$$ and to use the fact that $4x^3=2x^2\frac{d(x^2)}{dx}$ which makes the change of variable $x^2=t$ quite clear. Then expansion of... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of non-decreasing sequences How do I find the number of non-decreasing sequences of length $N$, such that all number in the sequences lie in the range $[a, b]$.
Also, the frequency of the most frequently occurring element should be unique.
For example, 1 1 2 3 is a valid sequence, but 1 1 2 2 3 is not a valid se... | The maximum frequency, say $k$, for a sequence can range from $1$ to
$N$. We will count the number of valid sequences for a fixed $k$
then sum over $k = 1, ... ,N$.
Let $d = b - a$.
The number, say $c$, having (uniquely) the maximum frequency $k$ can
be any number from $a$ to $b$. So we will fix $c$, count the number
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/808690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A pair of dice rolls and outcome A pair of fair dice is rolled three times and each time the two digits are added. What is the probability that a sum that is greater than or equal to 9 occurs exactly once?
My solution:
The ways to get the various totals is:
9 = 5 4, 4 5
9 = 3 6, 6 3
10 = 5 5
10 = 6 4, 4 6
11 = 5 6, 6 5... | In words:
*
*Choose $1$ out of $3$ events for which the sum is greater than or equal to $9$
*Choose $2$ out of the remaining $2$ events for which the sum is smaller than $9$
Mathematically, it can be expressed as $\binom{3}{1} \cdot P(sum\geq9) \cdot \binom{2}{2} \cdot P(sum<9)$
*
*$\binom{3}{1} = \frac{3!}{1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve this weird inequality? $\frac{x-1}{x+1} < x$
Thanks!
I did the following.
$\frac{x-1}{x+1} - x< 0 /-x$
$\frac{x-1 - x(x+1)}{x+1} < 0$
$\frac{-x^2-1}{x+1} < 0$
What to do next?
| Note that the inequality can be written as
$$-(x^2 + 1) \cdot \frac{1}{x + 1} < 0$$
Now $x^2 + 1$ is always positive, so this is equivalent to studying
$$\frac{1}{x + 1} > 0$$
Can you solve this?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Finding an exact solution to a difference equation How would I solve an equation of the form:
$u(n+1)=1/2u(n)+(1/3)^n$ when $u(0)=1$?
I got an answer of the form $u(n)= c + \sum(1/3)^j*2^{j-1}$ but believe this is incorrect?
| Consider $u_{n+1} = \dfrac{1}{2}\cdot u_n$. This gives: $u_n = 2^{-n}$, $n\geq 0$. So the general solution would be: $u_n = a\cdot 2^{-n} + b\cdot 3^{-n}$.
$u_0 = 1 \to 1 = a + b$, and $u_1 = 0.5u_0 + 1 = \dfrac{3}{2} \to \dfrac{a}{2} + \dfrac{b}{3} = \dfrac{3}{2} \to 3a + 2b = 9 \to 3a + 2(1 - a) = 9 \to a = 7$, and ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Show that $f\in \mathcal{R}[0,1]$ Let $f$ be defined by
$$f(x) =\left\{\frac{1}{n}, \frac{1}{n+1} \lt x \le \frac{1}{n} \right\}$$ and $f(x) = 0$ when $x=0$ and $n \in N$. Show that $f$ is integrable and $$\int_{0}^{1}f(x)dx=\frac{\pi^2}{6}-1.$$
| Hint :
Suppose it is integrable can you relate that integral (Draw a picture) to following sum
$$\frac{1}{2}+(\frac{1}{2}-\frac{1}{3})\cdot \frac{1}{2}+(\frac{1}{3}-\frac{1}{4})\cdot \frac{1}{3}+(\frac{1}{4}-\frac{1}{5})\cdot \frac{1}{4}+\cdots$$
This should immediately give you required result.
| {
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Evaluating the integral with trigonometric integrand While solving another problem I have come across this integral which I am unable to evaluate. Can someone please evaluate the following integral? Thank you.
$$\int_0^{2\pi}\frac{1}{(2+\cos\theta)^2}\,d\theta.$$
| Here is an ugly way of doing it. Pranav's solution is much more elegant.
First substitute $u = \tan(\tfrac{\theta}{2})$ and $\mathrm{d}u = \tfrac{1}{2} \sec^2(\tfrac{\theta}{2})\mathrm{d}\theta$. Then, transform the integrand using the substitution $\sin(\theta) = \frac{2u}{u^2+1}$, $\cos(\theta) = \frac{1-u^2}{u^2+1}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/810087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 1
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Simplifying $(1/x-1/5 )/( 1/x^2-1/25)$ How do I get $\frac{5x}{x+5}$ from simplifying the following?
$$\frac{(\frac{1}{x}-\frac{1}{5} )}{( \frac{1}{x^2}-\frac{1}{25})}$$
My work:
I multiplied the top and bottom by the LCD: $25x^2$ (to get the same denominators).
Then I got: $\frac{25x-5x}{25-x^2}$
Then I got this for... | $$\frac{\frac1x-\frac15}{\frac1{x^2}-\frac1{25}}=\frac{\frac{5-x}{5x}}{\frac{25-x^2}{25x^2}}$$
$$=\frac{25x^2(5-x)}{5x(25-x^2)}=\frac{5x}{5+x}$$ assuming $x(5-x)\ne0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Eigenvalues and Eigenvectors for matrix. Complex Eigenvalues How can I find out the eigenvectors for this matrix:
$$A=
\begin{pmatrix}
-3 &0&0\\
0&3&-2\\
0&1&1
\end{pmatrix}
$$
I found the eigenvalues: $\lambda_{1}=-3$, $\lambda_{2}=2-i$, $\lambda_{3}=2+i$.
The eigenvector for $\lambda_{1}=-3$ is $$\begin{pmatrix}
1\... | I do not recommend separating the eigenvector into real and imaginary parts. In my opinion it only makes it more troublesome, complex numbers are perfectly good scalars to work with.
Note the following
$$\begin{align}
Av=\lambda _2v&\iff (A-\lambda _2I)v=0_{3\times 1}\\
&\iff \begin{pmatrix} -5+i & 0 & 0\\ 0 & 1+i & -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/810366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove: prove $\frac{1+\tan^2\theta}{1+\cot^2\theta} = \tan^2\theta$ I need to prove that $\frac{1+\tan^2\theta}{1+\cot^2\theta}= \tan^2\theta.$
I know that $1+\tan^2\theta=\sec^2\theta$ and that $1+\cot^2\theta=\csc^2\theta$, making it now $$\frac{\sec^2\theta}{\csc^2\theta,}$$ but I don't know how to get it do... | Know that: $\tan\theta=\dfrac{1}{\cot\theta}$. Therefore:
$$\dfrac{(1+\tan^2\theta)}{(1+\cot^2\theta)}=\dfrac{(1+\tan^2\theta)}{\bigg(1+\dfrac{1}{\tan^2\theta}\bigg)}=\dfrac{(1+\tan^2\theta)}{\bigg(\dfrac{1+\tan^2\theta}{\tan^2\theta}\bigg)}=\tan^2\theta.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 0
} |
Series Question: $\sum_{n=1}^{\infty}\frac{n+1}{2^nn^2}$ How to compute the following series:
$$\sum_{n=1}^{\infty}\frac{n+1}{2^nn^2}$$
I tried
$$\frac{n+1}{2^nn^2}=\frac{1}{2^nn}+\frac{1}{2^nn^2}$$
The idea is using Riemann zeta function
$$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$$
but the term $2^n$ makes complicate... | The series
\begin{align}
S = \sum_{n=1}^{\infty} \frac{n+1}{2^{n} \ n^{2}}
\end{align}
can be expressed as
\begin{align}
S = \sum_{n=1}^{\infty} \frac{1}{2^{n} \ n} + \sum_{n=1}^{\infty} \frac{1}{2^{n} \ n^{2}}
\end{align}
and is seen to be
\begin{align}
S = - \ln\left( 1 - \frac{1}{2} \right) + Li_{2}\left(\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/810514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
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How to solve this equation with three variables with an unknown parameter using Gaussian elimination? If I've got three equations:
$$\begin{array}{ccccccc} x & + & y & + & z & = 3 \\ 2x & + & ay & - & 2z & = 4 \\ x & + & 2y & - & az & = 1 \end{array}$$
How do I solve them using Gaussian elimination. I get to this poin... | This will get you to the step you asked about:
$\left[\begin{array}{ccc|c}
1 & 1 & 1 & 3\\
2 & a & -2 & 4\\
1 & 2 & -a & 1\end{array}\right]\Longrightarrow
\left[\begin{array}{ccc|c}
1 & 1 & 1 & 3\\
0 & a-2 & -4 & -2\\
0 & 1 & -a-1 & -2\end{array}\right]\Longrightarrow
\left[\begin{array}{ccc|c}
1 & 1 & 1 & 3\\
0 & 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/810628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Diophantine equation in four variables I would like to find a parametric solution for the following diophantine equation:
$-4 (1-a_1^2)(1-a_2^2) + (1+a_3^2 -a_1^2 -a_2^2)^2 = a_4^2$
Does such a solution exist?
How does one go about solving such questions systematically?
| Equation:
$q^2=(z^2+d^2-x^2-y^2)^2-4(d^2-x^2)(d^2-y^2)$
Has the solution:
$x=(p-s)sa^2+(p^2-ps+s^2)t^2-psk^2$
$y=-(p-s)sa^2+(p^2-ps+s^2)t^2-psk^2+2psak-2(p^2-ps+s^2)at$
$d=(p-s)sa^2-(p^2-ps+s^2)t^2-psk^2-2(p-s)sat+2pskt$
$z=(p-s)sa^2+(p^2-ps+s^2)t^2+psk^2-2(p^2-ps+s^2)kt-2(p-s)sak$
$q=4((p^2-ps+s^2)((p^2-s^2)k-(p^2-2ps... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$4 \sin 72^\circ \sin 36^\circ = \sqrt 5$ How do I establish this and similar values of trigonometric functions?
$$
4 \sin 72^\circ \sin 36^\circ = \sqrt 5
$$
| Hint: Notice that $72 = 2\cdot 36$ and $180 = 36\cdot 5$; use formulaes for halfed or doubled argument.
$$
\sin^2\frac{x}{2} = \frac{1-\cos x}{2}
$$
or
$$
\sin2x = 2\sin x\cos x
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/816195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Link between a cubic polynomial and a trig identity Alright, so I am told to prove that:
$$\tan (3A) = \frac{3\tan(A)-\tan^3(A)}{1-3\tan^2(A)}$$
This can be pretty easily done by applying the $\tan$ addition formula, taking the angles $2A$ and $A$, upon which one then applies the $\tan$ double angle formula. Simplifyin... | It's a lot of work to proof the identity so ;) :
Verify the following identity:
$$\tan(3a)=\frac{3\tan(a)-\tan^3(a)}{1-3\tan^2(a)}=>$$
$$\tan(3a)\left(1-3\tan^2(a)\right)=3\tan(a)-\tan^3(a)=>$$
$$\left(1-3\left(\frac{\sin(a)}{\cos(a)}\right)^2\right)\frac{\sin(3a)}{\cos(3a)}=3\frac{\sin(3a)}{\cos(3a)}-\left(\frac{\sin(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/816351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Applications of the identity $ab + \left(\frac{a+b}{2} - b\right)^2 = \left(\frac{a+b}{2}\right)^2$ I am reading Euclid's elements
I found the algebraic identity
$ab + \left(\frac{a+b}{2} - b\right)^2 = \left(\frac{a+b}{2}\right)^2$
I ponder on usage of this identity for $2$ hours.
but I can't click anything.
$a^2 + b^... | You can use it to find all ways of writing $c=a^2-b^2$ with $a$ and $b$ positive natural numbers (or any two real numbers) for fixed $c$. To see this, your identity says:
$$ab=\left ( \frac{a+b}{2} \right )^{2} - \left ( \frac{a-b}{2} \right )^{2}$$
Now pick a number, $c=16$. Then we can factorize $c=2*8$
, so that $a=... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
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$ 0< a,b,c <1\implies a+b+c-abc<2$ If $a,b,c$ are positive real numbers , all being less than $1$ , then how to prove that
$a+b+c-abc<2$ ?
| $0 \leq a,b,c \leq 1 \Rightarrow 0 \leq abc \leq 1$ and $0 \leq a+b+c \leq 3$.
It follows that $\sqrt[3]{abc} \geq abc$ and by the AM-GM inequality we know
$\dfrac{a+b+c}{3} \geq \sqrt[3]{abc} \geq abc$.
Finally
$a+b+c -abc \leq \dfrac{2}{3}(a+b+c) \leq 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/816965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
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Another inequality to prove involving arbitrary reals Let $a,b$ be non-zero reals such that $ab\ge \frac{1}{a}+\frac{1}{b}+3$ then prove the following inequality :
$$ \sqrt[3]{ab}\ge \frac{1}{\sqrt[3]{a}}+\frac{1}{\sqrt[3]{b}}$$
This one stumped me completely as usual. A solution would be welcome.
| I found another proof that establishes the equivalence between the two inequalities:
[As in my other solution, I will substitute all the variables with their cubes.]
$a^3 b^3 - \frac{1}{a^3}-\frac{1}{b^3} - 3 = ( a b - \frac{1}{a} - \frac{1}{b} ) ( a^2 b^2 + a + b + \frac{1}{a^2} + \frac{1}{b^2} - \frac{1}{ab} )$
$ = (... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate a limit (probably involving L'Hôpital rule) Evaluate the limit:
$$\mathop {\lim }\limits_{x \to \infty } x\left( {{{\left( {1 + {1 \over x}} \right)}^x} - e} \right)$$
My attempts didn't yield a result. I'd be glad for a guidance.
Thanks!
| You just have to persist in the L'Hopital rule.
Start with
$$\frac{(1+\frac{1}{x})^x-e}{\frac{1}{x}}$$ top and bottom go to zero,
by L'Hopital we get
$$\frac{(\ln(1+\frac{1}{x})-\frac{1}{x+1})(1+\frac{1}{x})^x}{-\frac{1}{x^2}}
=(\frac{x^2}{x+1}-x^2\ln(1+\frac{1}{x}))(1+\frac{1}{x})^x$$
Now $(1+\frac{1}{x})^x\to e$ so ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/819282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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Trying to evaluate integral$\int_0^\infty x \sqrt{1-e^{-x}}\,e^{-x}dx$ I am trying to integrating
$$
\int_0^\infty x \sqrt{1-e^{-x}}\,e^{-x}dx\equiv I
$$
but cannot get the answer, I would like a proof not a numerical answer. My attempt at proof:
$$
y=\sqrt{1-e^{-x}}\\
y(0)=0, \ y(\infty)=1\\
y^2=1-e^{-x}\\
2ydy=e^{... | You can continue with your method too. Use partial fraction decomposition this way:
$$\frac{4}{3}\sum_{k=1}^{\infty} \left(\frac{1}{2k}-\frac{1}{2k+3}\right)=\frac{4}{3}\int_0^1 \sum_{k=1}^{\infty} \left(x^{2k-1}-x^{2k+2}\right)\,dx=\frac{4}{3}\int_0^1 \,dx\left(\frac{1}{x}-x^2\right)\sum_{k=1}^{\infty} x^{2k}$$
where ... | {
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"url": "https://math.stackexchange.com/questions/820635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 4
} |
Find the limit of $ \lim_{x \to 7} \frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2} $ I need to evaluate the limit without using L'Hopital's rule.
$$\Large
\lim_{x \to 7} \frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}
$$
| Letting $x = 7 + h$, then we find: $$\displaystyle \lim_{h \to 0} \dfrac{\sqrt{h+9} - \sqrt[3]{h+27}}{\sqrt[4]{h+16} - 2} = \displaystyle \lim_{h \to 0} \dfrac{3\left(1+\dfrac{h}{18}\right) - 3\left(1+\dfrac{h}{81}\right)}{2\left(1+\dfrac{h}{64}\right) - 2} = \dfrac{112}{27},$$
where $(1+h)^{\alpha} \sim_0 1 + \alpha h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/821855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Series Question: $\sum_{n=0}^\infty\frac{2n!}{(2n+1)!!}=\pi$ How to prove:
$$\sum_{n=0}^\infty\frac{2n!}{(2n+1)!!}=\pi$$
I tried to use property of double factorial
$$(2n+1)!!=\frac{(2n+1)!}{2^nn!}$$
then
$$\frac{2n!}{(2n+1)!!}=\frac{2^{n+1}(n!)^2}{(2n+1)!}$$
The next step I am stuck. Any help would be appreciated. Tha... | Consider the series
\begin{align}
S = \sum_{n=0}^{\infty} \frac{ 2 \ n!}{(2n+1)!!} = 2 \sum_{n=0}^{\infty} \frac{ 2^{n} (n!)^{2} }{(2n+1)!}
\end{align}
which can quickly be cast into the form
\begin{align}
S = 2 \sum_{n=0}^{\infty} \frac{ (1)_{n} (1)_{n} }{ n! (3/2)_{n} \ 2^{n} }
\end{align}
which yields the hypergeom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/821931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Limit $\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}$ $$\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}=\lim_{x\rightarrow\infty}1-x+x\sqrt{\frac{2}{x^2}+\frac 2x+1}=\lim_{x\rightarrow\infty}1=1\neq2$$ as Wolfram Alpha state. Where I miss something?
| You have
\begin{align}
\lim_{x\rightarrow\infty}\left(1-x+\sqrt{2+2x+x^2}\right)
&=\lim_{x\rightarrow\infty}\left(1-x+x\sqrt{\frac{2}{x^2}+\frac 2x+1}\right) \\
&= \lim_{x\to\infty} \left(1+x\left(\sqrt{\frac{2}{x^2}+\frac{2}{x}+1}-1\right)\right).
\end{align}
The second term is of the form $\infty\cdot 0$; you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/822017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Evaluating e using limits What algebraic operations can I use on the $RHS$ to show $RHS = LHS$
$$e=\lim_{k\to\infty}\left(\frac{2+\sqrt{3+9k^2}}{3k-1}\right)^k$$
| Hint
For large values of $k$, the following Taylor expansion can be built $$\frac{2+\sqrt{3+9k^2}}{3k-1}=1+\frac{1}{k}+\frac{1}{2 k^2}+\frac{1}{6 k^3}+\frac{1}{24 k^4}+\frac{1}{72
k^5}+O\left(\left(\frac{1}{k}\right)^6\right)$$ So $$\log\Big(\frac{2+\sqrt{3+9k^2}}{3k-1}\Big)=\frac{1}{k}+\frac{1}{180 k^5}+O\left(\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/822469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Proving by induction: $ \frac{1\cdot3\cdot5\cdot \ldots \cdot (2n-1)}{1\cdot2\cdot3\cdot\ldots\cdot n} \leq 2^n $ WTS $ \frac{1\cdot3\cdot5\cdot \ldots \cdot (2n-1)}{1\cdot2\cdot3\cdot\ldots\cdot n} \leq 2^n $ for all natural $n$.
Have checked $P_1$, and assumed $P_k$.
Trying the following argument:
$P_{k+1} = \frac{... | You've basically got the right idea, except your last inequality is too weak (and not at all needed - you've basically just added an extra factor of $2$ without need). You've shown that
$$P_{k + 1} = P_k \cdot \frac{2k + 1}{k + 1}$$
Now we have that $P_k \le 2^k$ by hypothesis, and it's easy to see that
$$\frac{2k + 1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/822526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integration by parts - $\int \ln (2x+1) \text{dx}$ Use integration by parts to find
$\int \ln (2x+1) \text{dx}$.
So far I have:
$$x\ln(2x+1)-\int\dfrac{2x}{2x+1}dx+c$$
Using integration by substitution to find the integral
$$u=2x+1\Rightarrow\text{du}=2\text{dx}$$
$$\int\dfrac{2x}{2x+1}\cdot\dfrac{1}{2}\text{du}=\int x... | Using these steps
$u=2x+1$, $(1/2)du=dx$ and $2x=u-1$ so:
$$
\eqalign{
-\int\dfrac{u-1}u \frac12 du &= -\int\dfrac12du+\int\dfrac12u\,du \\
&= -\dfrac12(2x+1)+(1/2)\ln(2x+1).
}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/823162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
If $\alpha = \frac{2\pi}{7}$ then the find the value of $\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha.$ If $\alpha = \frac{2\pi}{7}$ then the find the value of $\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha$
My 1st approach :
$\tan(\alpha +2\alpha +4\alpha) = \f... | Not using Euler's formula which I don't think the best way for this
Let $\displaystyle a=\tan A,b=\tan2A,c=\tan4A$ where $A+2A+4A=n\pi$ where $7\nmid n$
As $\displaystyle\tan(n\pi-rA)=-\tan rA,\tan6A=-\tan A=-a$ etc.
Using Prove that $\tan A + \tan B + \tan C = \tan A\tan B\tan C,$ $A+B+C = 180^\circ$,
$\displaystyle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/823819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Expansion of $\sin^5 \theta$ using the Complex Exponential How do I expand $\sin^5\theta$ using the complex exponential, in order to obtain:
$$\frac{1}{16}\sin 5\theta - \frac{5}{16}\sin 3\theta + \frac{5}{8}\sin\theta$$
Thank you.
| Using this formula.
$$\exp({i\theta})^n = \exp(ni
\theta)$$
$$(\cos(\theta) + i\sin(\theta))^n = \cos(n\theta) + i\sin(n\theta)$$
$$(\cos(\theta) + i\sin(\theta))^5 = \cos(5\theta) + i\sin(5\theta)$$
$$\cos^5(\theta) + 5i\cos^4(\theta)\sin(\theta) -10\cos^3(\theta)\sin^2(\theta) -10i \cos^2(\theta)\sin^3(\theta) + \cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/825411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculate the product of these $98$ rational numbers $$\left( 1 + \frac{1}{2} \right) \left( 1 + \frac{1}{3} \right) \left( 1 + \frac{1}{4} \right) ... \left( 1 + \frac{1}{99} \right) $$
Of course I could do it factor for factor but there has got to be a more efficient way. Does anyone know it?
| We have
$$\left( 1 + \frac{1}{2} \right) \left( 1 + \frac{1}{3} \right) \left( 1 + \frac{1}{4} \right) ... \left( 1 + \frac{1}{99} \right) =\frac32\times\frac43\cdots\frac{100}{99}=\frac{100}2=50$$
by telescoping.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/827243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
} |
Prove the sum $\sum_{n=1}^\infty \frac{\arctan{n}}{n}$ diverges. I must prove, that sum diverges, but...
$$\sum_{n=1}^\infty \frac{\arctan{n}}{n}$$
$$\lim_{n \to \infty} \frac{\arctan{n}}{n} = \frac{\pi/2}{\infty} = 0$$
$$\lim_{n \to \infty} \frac{ \sqrt[n]{\arctan{n}} }{ \sqrt[n]{n} } = \frac{1}{1} = 1$$
Cauchy's co... | We have
$$\frac{\arctan n }n\sim_\infty \frac{\pi}{2n}$$
and the harmonic series $\sum\limits\frac1n$ is divergent so the given series is also divergent by the asymptotic comparison.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/827854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Show that $\sum_{r-k=0}^m\left(\sum_{k=0}^n\binom nk\binom{m}{r-k}x^r\right)=\sum_{r=0}^{m+n}\left(\sum_{k=0}^r\binom nk\binom{m}{r-k}\right)x^r$
How does this hold?
$$\sum_{r-k=0}^m\left(\sum_{k=0}^n\binom nk\binom{m}{r-k}x^r\right)=\sum_{r=0}^{m+n}\left(\sum_{k=0}^r\binom nk\binom{m}{r-k}\right)x^r$$
| Consider the relation to be proved, namely,
\begin{align}
\sum_{r=0}^{m+n} \left( \sum_{k=0}^{r} \binom{n}{k} \binom{m}{r-k} \right) \ x^{r} = \sum_{k=0}^{n} \left( \sum_{r=k}^{m} \binom{n}{k} \binom{m}{r-k} \right) \ x^{r}.
\end{align}
Now consider just the left hand side, labeled $S_{L}$,
\begin{align}
S_{L} &= \sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/829152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$a,b,c>0$ , $a^a b^b c^c =1$ then $a+b+c \le 3$? Let $a,b,c$ be positive real numbers such that $a^a b^b c^c =1$ , then is it true that $a+b+c \le 3$ ?
| Let $f(x)=x \log x$ , for $x>0$. Clearly $f''(x)=\frac{1}{x}>0~$, so by Jensen's inequality we have
$$\frac{a+b+c}{3}\log\left(\frac{a+b+c}{3}\right)\leq\frac{1}{3}(a\log a+b\log b+c\log c)=
\frac{\log(a^ab^bc^c)}{3}\leq 0$$
it follows that $\log\left(\frac{a+b+c}{3}\right)\leq0$ that is
$a+b+c\leq 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/831357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Partial fraction (doubt) I have this partial fraction
$$\displaystyle\frac{1}{(2+x)^2(4+x)^2}$$
I tried to resolve using this method:
$$\displaystyle\frac{A}{2+x}+\displaystyle\frac{B}{(2+x)^2}+\displaystyle\frac{C}{4+x}+\displaystyle\frac{D}{(4+x)^2}$$
$$1=A(2+x)(4+x)^2+B(4+x)^2+C(4+x)(2+x)^2+D(2+x)^2$$
When x=-2
$$1=... | Heaviside cover-up is an alternative technique worth exploring.
It directly gives $\begin{align}B = \frac{1}{(4+x)^2}\Bigg|_{x=-2} = \frac{1}{4} \text{ and } D = \frac{1}{(2+x)^2}\Bigg|_{x=-4} = \frac{1}{4}\end{align}\tag{1}$
Next, it requires that the $B$ and $D$ terms be moved to the other side.
The partial fractio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/833170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Showing $(a^{2}+2)(b^{2}+2)(c^{2}+2)\geq 9(ab+bc+ca)$
Let $a$, $b$, $c$ be nonnegative real numbers.
Prove $(a^{2}+2)(b^{2}+2)(c^{2}+2)\geq 9(ab+bc+ca)$.
| I will prove the stronger inequality:
$$(a^2+2)(b^2+2)(c^2+2)\ge 3(a+b+c)^2$$
because
$$(a^2+2)(b^2+2)=(a^2+1)(b^2+1)+a^2+b^2+3\ge (a+b)^2+\dfrac{1}{2}(a+b)^2+3
=\dfrac{3}{2}[(a+b)^2+2]$$
so
$$(a^2+2)(b^2+2)(c^2+2)\ge \dfrac{3}{2}[(a+b)^2+2](c^2+2)\ge\dfrac{3}{2}
[\sqrt{2}(a+b)+\sqrt{2}c]^2=3(a+b+c)^2$$
so
$$(a^2+2)(b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/834025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that $3^n > 2n^2 + 3n$ for $n \in [4,\infty) \cap\mathbb{N}$ If $n$ is a natural number $n\ge 4$, prove that $3^n > 2n^2 + 3n$
I assume I am supposed to use induction. The base $n=4$ step is clear, but how do I prove the inductive step. I tried several things including comparing $f(n+1)/f(n)$ and $f(n+1)-f(n)$ bu... | Write
$$ f(n) = 3^n - 2 n^2 - 3 n$$
Then
$$ f(n+1) = 3^{n+1} - 2 \big(n+1\big)^2 - 3 \big(n+1) $$
which can be written as
$$ f(n+1) = 3 \big( 3^n - 2 n^2 - 3 n \big) + 6 n^2 + 9 n - 2 \big(n+1\big)^2 - 3\big(n+1)$$
So
$$ f(n+1) = 3 f(n) + 4 n^2 + 2 n - 5$$
Note that
$$ f(3) = 0 $$
And note that
$$ 4 n^2 + 2 n - 5 > 0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/834107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Calculate $\int_{0}^{1}\frac{\arctan(x)}{x\sqrt{1-x^2}}dx$ I am preparing for a calculus exam and I was asked to calculate $$\int_{0}^{1}\frac{\arctan(x)}{x\sqrt{1-x^2}}dx$$
Using the hint that $$\frac{\arctan(x)}{x}=\int_{0}^{1}\frac{dy}{1+x^2y^2}$$
I ran into some trouble and would appreciate help.
What I did:
I used... | The following is a different approach that doesn't use that hint.
$$ \begin{align} \int_{0}^{1} \frac{\arctan (x)}{x\sqrt{1-x^{2}}} \ dx &= \int_{0}^{1} \frac{1}{x \sqrt{1-x^{2}}} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} x^{2n+1} \ dx \tag{1}\\ &= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \int_{0}^{1} \frac{x^{2n}}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/834205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
} |
Let $a_1=1$, $a_2=3$ , and for $n \ge 2$ let $a_n=a_{n-1}+a_{n-2}$. Show that $a_n < \left(\frac{7}{4}\right)^n$ for all natural numbers. Let $a_1=1$, $a_2=3$ , and for $n \ge 2$ let $a_n=a_{n-1}+a_{n-2}$. Show that $a_n <
\left(\frac{7}{4}\right)^n$ for all natural numbers.
I assume I'm supposed to use induction. base... | Here is the inductive step for anyone reading this...
Assuming $P(k-2)$: $a_{k-2} < \left(\frac{7}{4}\right)^{k-2}$
Assuming $P(k-1)$: $a_{k-1} < \left(\frac{7}{4}\right)^{k-1}$
Definition of $a_k$: $a_k=a_{k-1}+a_{k-2}$
Combining with inductive assumptions: $a_k < \left(\frac{7}{4}\right)^{k-2} + \left(\frac{7}{4}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/835321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Easier Proof of $\sin{3\theta} + \sin\theta = 2\sin{2\theta}\cos\theta$ I am curious to see whether anybody can give me a proof that takes less steps.
Here is how I did it:
$$\sin{3\theta} + \sin\theta = 2\sin{2\theta}\cos\theta$$
LHS $$\eqalign{\sin(2\theta + \theta) + \sin\theta &= \sin2\theta\cos\theta + \cos2\theta... | $$\sin(3\theta)+\sin(\theta)=\Im[e^{3i\theta}]+\Im[e^{i\theta}]=\Im[e^{3i\theta}+e^{i\theta}]=$$
$$=\Im[e^{2i\theta}(e^{i\theta}+e^{-i\theta})]=2\cos(\theta)\Im[e^{2i\theta}]=2\cos(\theta)\sin(2\theta)\ .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/837430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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$\iint_{\mathbb R^2} \frac{dx \, dy}{1+x^{10}y^{10}}$ diverges or converges? Question I'm trying to solve to prepare for an exam.
I need to find out if $\displaystyle\iint_{\mathbb R^2} \frac{dx\,dy}{1+x^{10}y^{10}}$ diverges or converges.
What I did:
I switched to polar coordinates, $x=r\cos \theta$, $y=r\sin\theta$, ... | Using a tip a friend gave me: It is enough to work on the first quadrant due to symmetry.
We will instead use the transform $v=xy, u=x$, the jacobian of said transform is $\frac{1}{u}$ which can be demonstrated.
$$\iint_{\mathbb R^2} \frac{1}{1+x^{10}y^{10}}dx\,dy = \int_0^\infty \int_{0}^{\infty} \frac{1}{1+v^{10}} \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/838680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
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$a^{12} \equiv 1 \pmod{35}$,knowing that $(a,35)=1$ Prove that $\forall a \text{ with } (a,35)=1:$
$$a^{12} \equiv 1 \pmod{35}$$
$$35 \mid a^{12}-1 \Leftrightarrow 5 \cdot 7 \mid a^{12}-1 \overset{(5,7=1)}{ \Leftrightarrow} 5 \mid a^{12}-1 \text{ and } 7 \mid a^{12}-1$$
Therefore, $\displaystyle{ a^{12} \equiv 1 \pmod{... | Yes we can tell. This is because we have $a^{12} \equiv 1 \pmod 5 \iff 5|(a^{12}-1).$ Similarly $a^{12} \equiv 1 \pmod 7 \iff 7|(a^{12}-1)$. Since ${\rm gcd}(5,7)=1$, we have $5\cdot 7 | (a^{12}-1)$, i.e., $a^{12} \equiv 1 \pmod {35}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/842741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
When is $(a^2+b)(b^2+a)$ a power of $2$? When is $(a^2+b)(b^2+a)$ a power of $2$?
($a$, $b$ positive integers)
I tried some values on the computer and it seems the only solutions are $a=1$, $b=1$. But I am not sure how to prove it?
Thanks for any help.
| If $a$ has lower power of $2$ than $b$:
$a+b^2$ will be an odd multiple of a power of $2$ and hence won't be a power of $2$
Similarly if $b$ has lower power of $2$ than $a$
Therefore $a,b$ must have the same power $p$ of $2$
Let $a = 2^p x$ where $x$ is odd
Let $b = 2^p y$ where $y$ is odd
Then $(a^2+b)(b^2+a) = 2^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/844764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Inequality for harmonic means Prove that for real numbers $a_1 ,a_2 ,...,a_n >0$ the following inequality holds
$$\frac{1}{a_1 } +\frac{2}{a_1 +a_2 } +...+\frac{n}{a_1 +a_2 +...+a_n }\leq 4\cdot \left(\frac{1}{a_1} +\frac{1}{a_2 } +...+\frac{1}{a_n} \right).$$
| The linked proof of Carleman's Inequality (in the comment) indicates the method of balancing coefficients in weighted mean inequalities. In the same spirit we can show, $$\frac{1}{a_1 } +\frac{2}{a_1 +a_2 } +...+\frac{n}{a_1 +a_2 +...+a_n } < 2 \left(\frac{1}{a_1} +\frac{1}{a_2 } +...+\frac{1}{a_n} \right)$$ for positi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/844995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
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Sketch the set $\{ z \in \mathbb{C} | \left|\frac{z-i}{z+i}\right|<1 \}$ My question is to sketch the set $\{ z \in \mathbb{C} | \left|\frac{z-i}{z+i}\right|<1\}$ in the complex plane.
I substituted $z$ for $a+bi$, but did not get anywhere:
$\left|\frac{a+(b-1)i}{a+(b+1)i}\right|<1\\
\left|\frac{(a+(b-1)i)(a-(b+1)i)}{... | Rewrite the inequality as $|z-i|<|z+i|$. Then think geometrically in terms of distances from the point $z$ to the points $\pm i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/846341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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To find the right most non zero digit When expanded 30! ends in 7 zeroes. Find the first non zero digit from right?
| First, consider the product of elements after dividing everything by all powers of 2 and 5.
$1\cdot 2 \cdot 3 \cdot \ldots \cdot 30$
becomes:
$1 \cdot 1 \cdot 3 \cdot 1 \cdot 1 \cdot 3 \cdot 7 \cdot 1 \cdot 9 \cdot 1 \cdot
11 \cdot 3 \cdot 13 \cdot 7 \cdot 3 \cdot 1 \cdot 17 \cdot 9 \cdot 19 \cdot 1 \cdot
21 \cdot 11 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/848501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that all real roots of the polynomial $P (x) = x^5 − 10x + 35$ are negative. I got this problem out of Andreescu's Putnam and Beyond. I solved it differently from the given solution and could not understand the solution. Can you explain what is happening in the last step of the solution?
Because P (x) has odd degr... | That should say $P(r) = r^5+1+1+1+2^5-5\cdot 2 \cdot r$.
The AM-GM inequality applied to $\{r^5,1,1,1,2^5\}$ gives us:
$\dfrac{r^5+1+1+1+2^5}{5} \ge \sqrt[5]{r^5 \cdot 1 \cdot 1 \cdot 1 \cdot 2^5}$
$\dfrac{r^5+35}{5} \ge 2r$
$r^5+35 \ge 10r$
$r^5-10r+35 \ge 0$
Equality only holds if $r^5 = 1 = 1 = 1 = 2^5$, which giv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/849977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
How to calculate $\int\frac{1}{x + 1 + \sqrt{x^2 + 4x + 5}}\ dx$? How to calculate $$\int\frac{1}{x + 1 + \sqrt{x^2 + 4x + 5}}dx?$$ I really don't know how to attack this integral. I tried $u=x^2 + 4x + 5$ but failed miserably. Help please.
| $$\frac1{x+1+\sqrt{x^2+4x+5}}=-\frac{x+1-\sqrt{x^2+4x+5}}{2(x+2)}$$
$$=-\frac{x+2-1-\sqrt{x^2+4x+5}}{2(x+2)}$$
$$=-\frac12+\frac1{2(x+2)}+\frac{\sqrt{(x+2)^2+1}}{2(x+2)}$$
Setting $x+2=\tan y,$
$$\int\frac{\sqrt{(x+2)^2+1}}{(x+2)}\ dx=\int\frac{\sec y}{\tan y}\sec^2y\ dy$$
$$=\int\frac{dy}{\cos^2y\sin y}=\int\frac{\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/850500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Interesting dilemma, answer not matching with stewart, My work is Included Question :
Compute flux through the upper hemisphere of $x^2+y^2+z^2 = 1$ .
Where $$\textbf{F} = \left( z^2x\right)\textbf{ i }+\left[\dfrac{1}{3}y^3+ \tan z\right]\textbf{ j } + \left(x^2z+y^2 \right)\textbf{ k }$$
ANSWER GIVEN AT BACK OF STEW... | Your volume element for spherical coordinates is not correct. It should be $$dV=\rho^2 \sin(\varphi) \, d \rho \, d \varphi \, d \theta.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/851007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the minimum of $\displaystyle \frac{1}{\sin^2(\angle A)} + \frac{1}{\sin^2(\angle B)} + \frac{1}{\sin^2(\angle C)}$ Is it possible to find the minimum value of $E$ where
$$E = \frac{1}{\sin^2(\angle A)} + \frac{1}{\sin^2(\angle B)} + \frac{1}{\sin^2(\angle C)}$$for any $\triangle ABC$.
I've got the feeling that $\... | $$E=\csc^2A+\csc^2B+\csc^2C$$
Since $f(x)=\csc^2x$ is a convex function in $(0,\pi)$, we have from Jensen's inequality:
$$\csc^2\left(\frac{A+B+C}{3}\right) \leq \frac{1}{3}\csc^2A+\frac{1}{3}\csc^2B+\frac{1}{3}\csc^2C$$
Since $A+B+C=\pi$
$$\Rightarrow \frac{\csc^2A+\csc^2B+\csc^2C}{3}\geq \frac{4}{3}$$
$$\Rightarrow \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/851285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Verification of Stokes Theorem I want to verify Stokes Theorem for the surface
$$
\Phi = \{ (x,y,z) \in \mathbb R^3 : z = x^2 - y^2, x^2 + y^2 \le 1 \}
$$
and the vector field $F(x,y,z) := (y,z,x)$. For this I use the parametrisation
$$
\varphi(r, \theta) = \begin{pmatrix} r\cos \theta \\ r \sin \theta \\ r^2(\cos^2 ... | The main thing is that for your line integral you should have (remember that $r=1$)
$$\int_0^{2\pi} \big((\sin\theta)(-\sin\theta) + (2\cos^2\theta-1)(\cos\theta) + (\cos\theta)(-4\cos\theta\sin\theta)\big)\,d\theta = -\pi,$$
and you lost a minus sign in the surface integral at the end. I think you just made a copying ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/851649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the exact value of $\sin (\theta)$ and $\cos (\theta)$ when $\tan (\theta)=\frac{12}{5}$ So I've been asked to find $\sin(\theta)$ and $\cos(\theta)$ when $\tan(\theta)=\cfrac{12}{5}$; my question is if $\tan (\theta)=\cfrac{\sin (\theta) }{\cos (\theta)}$ does this mean that because $\tan (\theta)=\cfrac{12}{5}$... | Most of the above solutions assume, wrongly, that $\theta$ is acute, i.e. $0^{\circ} < \theta < 90^{\circ}.$ Here I give all the cases for $0^{\circ}<\theta <360^{\circ}.$ I like to use simple identities to answer these questions. We know that $1+\tan^2\theta \equiv \sec^2\theta$. If $\tan\theta = \frac{12}{5}$ then we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/852337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Solution verification: Find the orthogonal trajectories of the family of curves for $x^2 + 2y^2 = k^2$ I need help with the following question:
Find the orthogonal trajectories of the family of curves for $x^2 + 2y^2 = k^2$
I have taken the following steps, are they correct? From what I understand, I have to take th... | From the family of ellipses
$$x^2+2y^2=k^2 ......(1)$$
you obtained the family of parabolas
$$y=Ax^2......(2)$$
Setting $x=Y\sqrt{2}$ and $y=X/\sqrt{2}$ in (1), we obtain:
$$2Y^2+X^2=k^2 ......(3)$$
So we obtain another family of parabolas
$$Y=B_1X^2......(4)$$
Which is equivalent to
$$x=By^2......(5)$$
This is beca... | {
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"source": "stackexchange",
"question_score": "6",
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Max. and Min. value of $|z|$ in $\left|z+\frac{2}{z}\right| = 2\,$ If $z$ is a complex no. such that $\displaystyle \left|z+\frac{2}{z}\right| = 2\,$ Then find max. and min. value of $\left|z\right|$.
$\bf{My\; Try:}$ Given $\displaystyle \left|z+\frac{2}{z}\right| = 2\Rightarrow \left|z+\frac{2}{z}\right|^2 = 2^2=4$.... | Let $z=re^{i\theta}$, then your last step can be written as
$$r^2+2\cos (2\theta)+\frac{1}{r^2}=4.$$
From this we can get
\begin{align*}
\left(r+\frac{1}{r}\right)^2 & = 6-2\cos (2\theta)
\end{align*}
Thus $2 \leq \left(r+\frac{1}{r}\right) \leq 2\sqrt{2}.$ Now try to find the range of $r$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/853992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Finding $\int_0^{2\pi}\frac{\sin t + 4}{\cos t + \frac{5}3} dt$ I'd like to ask something about the following integral:
$$
\int_0^{2\pi}\frac{\sin t + 4}{\cos t + \frac{5}3} dt
$$
I rewrote and took another variable.
$$
-i\int_0^{2\pi}\frac{e^{is}-e^{-is}+8i}{e^{is}+e^{-is} + \frac{10}3} dt
\quad = \quad
-i\int_{C(0,... | We first separate the integration into 4 intervals:
$$I=\int_0^{2\pi}\frac{\sin t + 4}{\cos t + \frac{5}3} dt=\int_0^{\pi/2}\frac{\sin t + 4}{\cos t + \frac{5}3} dt+\int_0^{\pi/2}\frac{\sin (t+(\pi/2)) + 4}{\cos (t+(\pi/2)) + \frac{5}3} dt+\int_0^{\pi/2}\frac{\sin (t+\pi) + 4}{\cos (t+\pi) + \frac{5}3} dt+\int_0^{\pi/2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/860003",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the Value of Integral Find the Value of $$\begin{align}I=\int_{0}^{1}\frac{\ln(x)\,dx}{1-x^2}\end{align}$$
I have tried like this: We have $$\begin{align}2I=\int_{0}^{1}\frac{\ln(x^2)\,dx}{1-x^2}=\int_{0}^{1}\frac{\ln(1-(1-x^2))\,dx}{1-x^2}\end{align}$$ So
$$2I=\begin{align}\int_{0}^{1}\frac{-(1-x^2)-\frac{(1-x^2)... | Use the series expansion for $\dfrac{1}{1-x^2}$ i.e
$\displaystyle \frac{1}{1-x^2}=\sum_{k=0}^{\infty} x^{2k}\,dx$
Hence,
$$I=\int_0^1 \ln(x)\left(\sum_{k=0}^{\infty} x^{2k}\right)\,dx=\sum_{k=0}^{\infty} \int_0^1 x^{2k}\ln x\,dx=-\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}$$
where the final integral can be evaluated using t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Solve $\dfrac{x}{x-2}>2$ by first rewriting it in the form $\dfrac{P(x)}{Q(x)}>0$ Edit: So then is this the correct final solution? $x<4,(\infty,4), x\ne2$
I am asked to do this:
Solve $\dfrac{x}{x-2}>2$ by first rewriting it in the form $\dfrac{P(x)}{Q(x)}>0$
$$\dfrac{x}{x-2}>2$$
$$\dfrac{x}{x-2}-\dfrac{2(x-2)}{1(x-2)... | Your method not is correct because: if
$$\frac{4-x}{x-2}>0\equiv-1\cdot\frac{(4-x)}{x-2}<(-1)\cdot0\equiv\frac{x-4}{x-2}<0$$
then $(x-4)>0$ and $(x-2)<0$, or $(x-4)<0$ and $(x-2)>0$
solutions $2<x<4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/861486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find expression for : $ S_n =\sum_{i=1}^{n} \frac{i}{i^4+i^2+1} $ I want to find a formula for the sum of this series using its general term.
How to do it?
Series
$$
S_n = \underbrace{1/3 + 2/21 + 3/91 + 4/273 + \cdots}_{n \text{ terms}}
$$
General Term
$$
S_n = \sum_{i=1}^{n} \frac{i}{i^4+i^2+1}
$$
| Set $$\color{blue}{a_k= \frac{1}{k^2 -k +1} \implies a_{k+1}= \frac{1}{(k+1)^2 -k } =\frac{1}{k^2+ k +1}} $$
But since $ (k^2 -k +1)(k^2 +k +1) =k^4 +k^2 +1$ we have,
$$ a_{k+1} -a_k = \frac{1}{k^2 +k +1}- \frac{1}{k^2 -k +1} = \frac{2k}{(k^2 -k +1)(k^2 +k +1)} =\frac{2k}{k^4 +k^2 +1}$$
Then
$$a_n-a_0 = \sum_{k=0}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/865442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Is this function $y=-\ln\left(1+\frac{\sin x -\cos x}{2}\right)$ convex? Is this function convex for $x\in[0,\frac{\pi}{4}]$?
$$y=-\ln\left(1+\frac{\sin x -\cos x}{2}\right)$$
without use the derivative.
| Disclaimer: it is quite a long exercise in trigonometry to avoid the use of derivatives to prove the convexity.
Since the function is continuous, you only need the midpoint-convexity in order to prove the convexity. If we take $[x,y]\subset[0,\pi/4]$, we need to prove:
$$-2\log\left(1+\frac{1}{\sqrt{2}}\sin\left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/865933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Simplify expression $(x\sqrt{y}- y\sqrt{x})/(x\sqrt{y} + y\sqrt{x})$ I'm stuck at the expression: $\displaystyle \frac{x\sqrt{y} -y\sqrt{x}}{x\sqrt{y} + y\sqrt{x}}$.
I need to simplify the expression (by making the denominators rational) and this is what I did:
$$(x\sqrt{y} - y\sqrt{x}) \times (x\sqrt{y} - y\sqrt{x}) ... | Given the question, we can write
\begin{eqnarray}
\frac{ x\sqrt{y} - y\sqrt{x} }{ x\sqrt{y} + y\sqrt{x} }
&=& \frac{ 1 - \sqrt{y/x} }{ 1 + \sqrt{y/x} }\\
&=& \frac{ \Big( 1 - \sqrt{y/x} \Big)^2 }
{ \Big( 1 + \sqrt{y/x} \Big) \Big( 1 - \sqrt{y/x} \Big)}\\
&=& \frac{ 1 + y/x - 2\sqrt{y/x} }{ 1 - y/x }\\
&=& \frac{ x + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/869037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
find a polynomial whose roots are inverse of squares of roots of $x^3+px+q$ Question is :
Given a polynomial $f(x)=x^3+px+q\in \mathbb{Q}[x]$ find a polynomial whose roots are inverse of sqares of roots of $f(x)$
Supposing $a,b,c$ as roots of $f(x)$ we have :
*
*$a+b+c=0$
*$ab+bc+ca=p$
*$abc=-q$
Now i need to kn... | Hint: start with: $\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} = \dfrac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2} = \dfrac{(ab+bc+ca)^2 - 2abc(a+b+c)}{(abc)^2} = \dfrac{p^2 - 0}{(-q)^2} = \dfrac{p^2}{q^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/872264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find the $n$th term of $1, 2, 5, 10, 13, 26, 29, ...$ How would you find the $n$th term of a sequence like this?
$1, 2, 5, 10, 13, 26, 29, ...$
I see the sequence has a group of three terms it repeats: Double first term to get second term, add three to get third term, repeat.
What about: $2, 3, 6, 7, 14, 15, 30,... $?... | For the first sequence:
$$S_n = {2}^{(2+\lfloor\frac{n}{2}\rfloor)}+6(\frac{n}{2}-\lfloor\frac{n}{2}\rfloor-1)$$
For example:
*
*$S_1 = 2^2-3 = 1$
*$S_2 = 2^3-6 = 2$
*$S_3 = 2^3-3 = 5$
*$S_4 = 2^4-6 = 10$
*$S_5 = 2^4-3 = 13$
*$S_6 = 2^5-6 = 26$
*$S_7 = 2^5-3 = 29$
For the second sequence:
$$S_n = {2}^{\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/872485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 4
} |
Symmetric matrix multiplication Let $A$ and $B$ be symmetric matrices.
Prove:
*
*$AB=BA$
*$AB$ is a symmetric matrix
As for 1. due to the axiom $(AB)^T=B^T A^T$ so $AB=BA$
As for 2. I did not find any axiom that can support the claim, but from test I found that it is true for symmetric matrices when the entries o... | Both claims are false and almost any $A$ and $B$ are counterexamples. For a specific example, you can see $$\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 3 & 5 \\ 3 & 5 \end{pmatrix}$$ while $$\begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix} \cdot \begin{p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/874469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 2,
"answer_id": 0
} |
Evaluation of $\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx$
Compute the indefinite integral
$$
\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx
$$
My Attempt:
$$
\begin{align}
\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx &= \int\frac{\cos 2x}{\sin^2 x\sqrt{\cos 2x}}\sin xdx\\
&= \int\frac{2\cos^2 x-1}{(1-\cos^2 x)\sqrt{2\cos^2 x-1} }\sin ... | $\displaystyle I=\int\frac{\sqrt{2t^2-1}}{t^2-1}dt$, so $\displaystyle t=\frac{1}{\sqrt{2}}\sec\theta, dt=\frac{1}{\sqrt{2}}\sec\theta\tan\theta d\theta$ gives
$\displaystyle I=\frac{1}{\sqrt{2}}\int\frac{\sec\theta\tan^2\theta}{\frac{1}{2}\sec^2\theta-1}d\theta=\sqrt{2}\int\frac{\sec\theta\tan^2\theta}{\sec^2\theta-2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/876175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 1
} |
How to factorize $x^4+2x^2+4$ to a product of polynomials with real coefficients? How do you factor
$$x^4+2x^2+4 $$
so it can be written as
$$ (x^2+2x+2)(x^2-2x+2) $$
| The key idea is to complete $\,x^4\!+4\,$ to a square $\,(x^2\!+2)^2-4x^2,\,$ yielding a difference of squares
$$ x^4\!+4\, =\, (x^2\!+2)^2 - (\color{#c00}{2x})^2 =\, (x^2\!+2-\color{#c00}{2x})(x^2\!+2 + \color{#c00}{2 x})$$
The same idea works for $\,x^4+4 + 2x^2\,$ if you meant that instead of $\,x^4+4.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/876244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve for $x$, $\tan x +\sec x = 2\cos x$ ; $−∞ < x < ∞$ Solve for $x$, $\tan x +\sec x = 2\cos x$ ; $−∞ < x < ∞$
$$\tan x + \sec x = 2\cos x$$
I tried changing it all to sin and cos
$$\frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2\cos x$$
then I made it to one fraction
$$\frac{\sin x + 1}{\cos x} = 2 \cos x$$
Then I don... | Multiply across by $\cos x$ to get $\sin x + 1 = 2 \cos^2 x= 2 - 2 \sin^2 x$,
or $2 \sin^2 x +\sin x -1 = 0$.
Since the solutions of $2y^2+y-1 = 0$ are $\{-1, {1 \over 2} \}$, the possible solutions to the original equations are $\sin x = -1$ and $\sin x = {1 \over 2} $. However, if $\sin x = -1$, then $\cos x = 0$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/877539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Prove $\frac{\sin A}{\sec A+\tan A-1}+ \frac{\cos A}{\csc A+\cot A-1}=1$ $$\frac{\sin A}{\sec A+\tan A-1}+ \frac{\cos A}{\csc A+\cot A-1}=1$$
Prove that L.H.S.$=$R.H.S.
This type of questions always creates problem when in right hand side some trigonometry function is given then it is bit easy to think how to proceed ... | $$\frac{\sin A}{\sec A+\tan A-1}+ \frac{\cos A}{\csc A+\cot A-1}= \frac{\sin A}{\frac{1}{\cos A}+\frac{\sin A}{\cos A}-\frac{\cos A}{\cos A}}+ \frac{\cos A}{\frac{1}{\sin A}+\frac{\cos A}{\sin A}-\frac{\sin A}{\sin A}} =$$ $$= \frac{\sin A \cos A}{1 + \sin A - \cos A} + \frac{\sin A \cos A}{1 + \cos A - \sin A} =$$ $$=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/877936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
} |
In a triangle ABC, prove that cot(A/2)+cot(B/2)+cot(C/2) =cot(A/2)cot(B/2)cot(C/2) In a triangle ABC, prove that $\cot \left ( \frac{A}{2} \right )+\cot \left ( \frac{B}{2} \right )+\cot \left ( \frac{C}{2} \right )=\cot \left ( \frac{A}{2} \right )\times \cot \left ( \frac{B}{2} \right )\times \cot \left ( \frac{C}{2}... | Geometric Proof
Let $a,b,c$ be the length of the sides $BC,CA,AB$ respectively, $p$ be the semi-perimeter, $r$ be the radius of the incircle, and $S$ be the area.
Notice that $$\cot\frac{A}{2}=\frac{p-a}{r},~~~\cot\frac{B}{2}=\frac{p-b}{r},~~~\cot\frac{C}{2}=\frac{p-c}{r}.$$
Hence, $$\cot\frac{A}{2}+\cot\frac{B}{2}+\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/878577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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Prove ${\large\int}_{-1}^1\frac{dx}{\sqrt[3]{9+4\sqrt5\,x}\ \left(1-x^2\right)^{2/3}}=\frac{3^{3/2}}{2^{4/3}5^{5/6}\pi }\Gamma^3\left(\frac13\right)$ Here is one more conjecture I discovered numerically:
$${\large\int}_{-1}^1\frac{dx}{\sqrt[3]{9+4\sqrt5\,x}\ \left(1-x^2\right)^{\small2/3}}\stackrel{\color{#808080}?}=\f... | As suggested by Chen Wang, this integral is related to an integral of the form
$$
\int_0^1 \frac{dx}{ \sqrt{1-x} \; x^{2/3} (1-zx)^{1/3} }
$$
which appears in this forum's
Question 879089
by the same author.
Here's a more direct route than given by Kirill, without
conversions between definite integrals and hypergeometr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/879854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 2,
"answer_id": 1
} |
Find modulus of $z$ given modulus of $(z-3w)/(3-z\overline{w})$ Question:
(23) If $z_1$, $z_2$ are complex numbers such that $\left|\dfrac{z_1-3z_2}{3-z_1\overline{z}_2}\right|=1$ and $|z_2|\neq 1$, then find $|z_1|$.
How would I attempt this question? I tried using values for $z_1$ and $\overline{z}_2$ but it is com... | $$\frac{z_1 - 3z_2}{3 - z_1 \overline{z_2}} = \frac{\frac{z1}{3} - z_2}{1 - \frac{z1}{3} \overline{z_2}}$$
$b = \overline{z_1} / 3, a = z_2 \Rightarrow |a - \overline{b}| = |1 - ab|$ and we should find $b$.
Let's look this problem geometrically and consider complex numbers $a, b$ as two-dimensional vectors (common appr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/880453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A strange trigonometric equation Today,in our class, we received a trigonometric equation
$$\sin^{10}{x}+\cos^{10}{x}=\frac{29}{16}\cos^4{2x}$$
and the question was to find the general solution of this equation. My approach was, at First, trying to show that there were no solutions using inequalities, but I failed. So,... | We know that $X^5+Y^5=(X+Y)(X^4-X^3Y+X^2Y^2-XY^3+Y^4)$; if we set $X=\sin^2x$ and $Y=\cos^2x$, the left hand side becomes
$$
(\sin^2x+\cos^2x)(\sin^8x-\sin^6x\cos^2x+\sin^4x\cos^4x-\sin^2x\cos^6x+\cos^8x)
$$
or
$$
\sin^8x+\cos^8x+\sin^4x\cos^4x-\sin^2x\cos^2x(\sin^4x+\cos^4x)
$$
The first three terms can be written
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/880723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How do you prove this integral involving the Glaisher–Kinkelin constant According to wikipedia on the page Glaisher–Kinkelin constant
$$\int_0^{1/2} \ln\Gamma(x) dx=\frac32\ln \text{A}+\frac5{24}\ln 2+\frac14\ln\pi$$
I got interested in how you possibly could prove something like that, but couldn't find any citations a... | Recall Kummer´s fourier expansion for LogGamma $0<x<1$
$$\ln\left(\Gamma(x)\right)=\frac{\ln\left(2 \pi\right)}{2}+ \sum_{k=1}^{\infty}\frac{1}{2k}\cos\left(2 \pi k x\right) + \sum_{k=1}^{\infty}\frac{\gamma+\ln\left(2 \pi k\right)}{ \pi k}\sin\left(2 \pi k x \right) \tag{1}$$
Integrating $(1)$ we obtain
$$\int_0^x\ln\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/882623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Series for logarithms This is more of a challenge than a question, but I thought I'd share anyway. Prove the following identities, and prove that the pattern continues.
\begin{equation*}
\sum_{n=0}^\infty\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)=\ln2
\end{equation*}\begin{equation*}
\sum_{n=0}^\infty\left(\frac{1}{3n+... | For any $m \in \mathbb{N},$
$$\sum_{k=0}^{\infty} \Big(\frac{1}{mk+1} + \frac{1}{mk+2} + ... + \frac{1}{mk+m-1}- \frac{m-1}{mk+m}\Big)$$ $$= \lim_{n \rightarrow \infty} \Big(\sum_{k=0}^{mn} \frac{1}{mk+1} + ... + \frac{1}{mk+m-1} - \frac{m-1}{mk+m}\Big)$$$$= \lim_{n \rightarrow \infty} \Big( \sum_{k=1}^{mn} \frac{1}{k}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/883348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Prove that the expression is Independent of theta Prove that $2\sin^2 \theta +4 \cos(\theta + \alpha ) \sin \alpha \sin \theta +\cos(2(\theta + \alpha))$ is independent of $\theta$.
How do we solve such problems ?
| $$2\sin^2\theta+4\sin\alpha\sin\theta\cos(\theta+\alpha)+\cos(2\theta+2\alpha)$$
$$=2\sin^2\theta+4\sin\alpha\sin\theta(\cos\theta\cos\alpha-\sin\theta\sin\alpha)+\cos2\theta\cos2\alpha-\sin2\theta\sin2\alpha$$
$$=2\sin^2\theta+4\sin\alpha\cos\alpha\sin\theta\cos\theta-4\sin^2\alpha\sin^2\theta+(1-2\sin^2\theta)(1-2\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/883538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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A double sum with combinatorial factors Let $n$, $p$ and $j$ be integers. As a byproduct of some other calculations I have discovered the following identity:
\begin{equation}
\sum\limits_{p=0}^{j} \sum\limits_{p_1=0}^j \binom{p+p_1}{p_1} \binom{j}{p_1} \binom{2n-j}{j-p} \binom{n-j+p}{1+p+p_1} (-1)^p = \frac{1}{2} \bin... | This is not going to be a complete answer but since this approach seems to me intuitive i will show it. This is a sort of ``by induction proof''. Let us look at the lhs for the biggest possible $j$ and then go down in $j$. We hope to see a pattern as a function of $j$ and thus guess the result.
So we take $j=n-1$.We h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/883802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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