Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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I somehow deduced that $\tan x=\iota$ for any real value of $x$ by equating the value of $\tan(\frac{\pi}{2}+x)$ obtained using two identities. Let's assume that we're familiar with the identity : $\tan \Bigg (\dfrac{\pi}{2} + x \Bigg ) = -\cot x$ which we have derived using the unit circle.
I was trying to equate the... | This is a cool "paradox," hadn't seen it before!
Even before the red line, the identity
$$\tan \Bigg ( \dfrac{\pi}{2} + x \Bigg ) = \dfrac{\tan\dfrac{\pi}{2} + \tan x}{1 - \tan \dfrac{\pi}{2} \tan x}$$
is objectionable.
This isn't true-- or, more accurately, it isn't even grammatically correct, since $\frac{\pi}{2}$ is... | {
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"url": "https://math.stackexchange.com/questions/3726639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What's going on with this Wolfram Alpha matrix multiplication? I was computing the matrix exponential of
$$X = \begin{pmatrix} 1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1 \end{pmatrix}$$
by diagonalization. Wolfram Alpha and Symbolab both solve this problem correctly. After the diagonalization and taking the exponential of... | As user1551 pointed out in a comment, the answer was as simple as a missing $1$: If
$$D = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{pmatrix}, \quad \textrm{then} \quad e^D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & e & 0 \\ 0 & 0 & e^3 \end{pmatrix}, \quad \textrm{not} \quad e^D = \begin{pmatrix} 0 & 0 & 0 \\ 0 ... | {
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"url": "https://math.stackexchange.com/questions/3727087",
"timestamp": "2023-03-29T00:00:00",
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Find points that are on two elliptic curves in $F_p$ I have some basic cryptography question but I don't know if my calculations are not overcomplicated and if there's some simpler solution. Following algorithm is explained here.
Problem
There are two elliptic curves in $F_5$ with equations:
a) $y^2 = x^3 + 2x + 1 \pmo... | Since you are working on the same field for both the curves. To get the common points $(x_0,y_0)$ that lie on both the curves, you just need to to equate the following (think in terms of intersection of two curves)
\begin{align*}
y_0^2 =x_0^3+2x_0+1 &\equiv x_0^3+x_0+1 \pmod{5}\\
x_0 & \equiv 0 \pmod{5}.
\end{align*}
T... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find area of rectangle inscribed in ellipse. In an ellipse $4x^2+9y^2=144$ inscribed is a rectangle whose vertices lies on the ellipse and whose sides are parallel with the ellipse axis.
Longer side which is parallel to major axis, relates to the shorter sides as $3:2$. Find area of rectangle.
I can find the val... | We're lucky that the ellipse is centered at the origin. :)
In this case, the inscribed rectangle is also centered at the origin. If $P = (x, y)$ is the vertex of the inscribed rectangle at the first quadrant, then the smaller rectangle spanned by the origin and point $(x, y)$ is similar to the inscribed rectangle.
Each... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Question about finding the value of an infinite sum What is the value of:
$$\sum_{k=0}^\infty \frac{1}{(4k+1)^2}?$$
I realised that
$$\sum_{n=2,4,6,8,...} \frac{1}{n^2} + \sum_{n=1,3,5,7,...} \frac{1}{n^2} = \sum_{n \geq 1 } \frac{1}{n^2}$$
$$\sum_{n \geq 1 } \frac{1}{4n^2}+\sum_{n \geq 0 } \frac{1}{(2n+1)^2} = \sum_{n... | Recall the definition of Catalan's Constant:
$$K=\sum_{k=0}^{\infty} \frac 1 {(4k+1)^2}- \sum_{k=0}^{\infty} \frac 1 {(4k+3)^2}$$
Combining that with the equation you have already derived we immediately get $$\sum_{k=0}^{\infty} \frac 1 {(4k+1)^2}=\frac K2+\frac {\pi^2}{16}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Continuity of this Piecewise function $f:\mathbb{R}^2\to \mathbb{R}$ I have shown that $f:\mathbb{R}^2\to\mathbb{R}$ given by $f(0,0) = 0$ and $\displaystyle f(x,y)=\frac{x|y|}{\sqrt{x^2+y^2}}$ if $(x,y)\ne (0,0)$ isn't differentiable at $(0,0)$, now I'm trying to show whether it is continuous or not.
My attempt: I mus... | Let us consider the directional derivative of $f$ in the direction $v = (a,b)$ at the point $(0,0)$. We have
\begin{align*}
D_{v}f(0,0) & = \lim_{t\to 0}\frac{f((0,0) + t(a,b)) - f(0,0)}{t}\\\\
& = \lim_{t\to 0}\frac{ta|tb|}{t\sqrt{t^{2}a^{2}+t^{2}b^{2}}} = \frac{a|b|}{\sqrt{a^{2}+b^{2}}}
\end{align*}
On the other hand... | {
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Integrate $\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx$ How can I evaluate this integral $$\int\dfrac{2\cos x-\sin x}{3\sin x+5\cos x } \mathop{dx}$$
I tried using the half angle formula
$$\sin x=\dfrac{2\tan\dfrac x2}{1+\tan^2\dfrac x2}, \cos x=\dfrac{1-\tan^2\dfrac x2}{1+\tan^2\dfrac x2}$$
substituted and simplif... | hint
Dividing by $\cos(x)$ and putting $\tan(x)=t$, it becomes
$$\int \frac{(2-t)dt}{(5+3t)(1+t^2)}$$
$$=\frac{1}{34}\int \Bigl(\frac{33}{5+3t}+\frac{-11t+7}{t^2+1}\Bigr)dt$$
$$=\frac{11}{34}\int \frac{3dt}{5+3t}-\frac{11}{68}\int \frac{2tdt}{t^2+1}+$$
$$\frac{7}{34}\int \frac{dt}{t^2+1}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Parametrization the curve of intersection of sphere $x^2+y^2+z^2=5$ and cylinder $x^2+\left(y-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2$. I've seen similar posts here but none of the answers helped me. I am trying to parametrize a curve of intersection of a (top half $z>0$) sphere $x^2+y^2+z^2=5$ and cylinder $x^... | Sphere
$$ x^2 +y^2 + z^2= 5 \tag1 $$
simplify given second Cylinder equation
$$ x^2+y^2 -y=0 $$
Subtract
$$z^2+y = 5 \; \tag2 $$
To satisfy this relation now introduce parameter $t$
$$z^2 = 5 \cos^2 t,\;\tag3 $$
so that
$$ y = 5 \sin ^2 t \tag4 $$
From (1) and (4)
$$ z^2= 5 -x^2 -25 \sin ^4 t \tag 5 $$
Equate RHS of $z... | {
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Geometry question: Find the area of blue-shared area inside this isosceles See below, looks a bit interesting, but I cannot find a solution.
I think a starting point might be the similarity of the lower white triangle and the larger triangle composed of lower blue, pink, and lower right triangle though..
| Expanding-upon a comment, as requested.
Let the triangle have vertex angle $2\theta$, let $a\leq b$ be the sides of the squares, and let $t$ denote the target area. We immediately find
$$\begin{align}
a+a\cot\theta = |PQ| = b+b\tan2\theta \tag{1} \\
t = \frac12\cdot a\cdot a\cot\theta + \frac12\cdot b \cdot b\tan2\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735895",
"timestamp": "2023-03-29T00:00:00",
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Evaluate $\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}}$ Evaluate $\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}}$
My attempt:
Let $\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}}=\sqrt{x}+\sqrt{y}$
Square both sides: $a+b+\sqrt{\left(2ab+b^2\right)}=x+2\sqrt{xy}+y$
Rearrange: $\sqrt{\left(2ab+b^2\right)}-2\sqrt{xy}=x+y-a-b$
That's where my ... | By conjugates $\sqrt{a+b\pm\sqrt{2ab+b^2}}=\sqrt{x}\pm\sqrt{y}$ i.e. $a+b=x+y,\,2ab+b^2=4xy$, so $x,\,y$ are roots of $t^2-(a+b)t+ab/2+b^2/4=(t-a-b/2)(t-b/2)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Can this inductive proof that $\sum_{i=0}^n2^{2i+1}=\frac23(4^n-1)$ be simplified? The general structure of equations I've used for the inductive step for proofs with a summation is something like:
We'll prove that $\sum_{i = 0}^{n + 1} (\text{something}) = (\text{closed form expression})$
\begin{align}
\sum_{i = 0}^{n... | Note that $2^{2n+1}=2\cdot 2^{2n}=2\cdot 4^n$. Now,
$$\frac{2}{3}(4^n-1)+2^{2n+1}=\frac{2}{3}(4^n-1)+2\cdot 4^n =\frac{8}{3}\cdot 4^n -\frac{2}{3} = \frac{2}{3}(4^{n+1}-1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit of a fraction involving square and third roots. $$\lim_{n\to\infty} \frac{\sqrt{n^2+1} - \sqrt{n^2+n}}{\sqrt[3]{n^3+1} - \sqrt[3]{n^3+n^2+1}}$$
Is it a good idea to substitute the numerator and denominator using that $a-b=\frac{a^2-b^2}{a+b}$ and $a-b=\frac{a^3-b^3}{a^2+ab+b^2}$, since I’ll end up having to multi... | Considering $$y= \frac{\sqrt{n^2+1} - \sqrt{n^2+n}}{\sqrt[3]{n^3+1} - \sqrt[3]{n^3+n^2+1}}$$ compose Taylor series and use binomial expansions
$$\sqrt{n^2+1}=n+\frac{1}{2 n}+\cdots$$
$$\sqrt{n^2+n}=n+\frac{1}{2}-\frac{1}{8 n}+\cdots$$
$$\sqrt[3]{n^3+1}=n+\frac{1}{3 n^2}+\cdots$$
$$\sqrt[3]{n^3+n^2+1}=n+\frac{1}{3}-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738751",
"timestamp": "2023-03-29T00:00:00",
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If $a=7!$, $b=_{13}P_k$, $\frac{ab}{\operatorname{lcm}(a,b)}=120$, find the $k$
Question:
$a=7!$, $b=_{13}P_k$, $\dfrac{ab}{\operatorname{lcm}(a,b)}=120$, then find the $k$.
My attempts:
$$\frac{ab}{\operatorname{lcm}(a,b)}=\gcd{(a,b)}=120$$
$\gcd{(7!, _{13}P_k)}=120 \Longrightarrow \begin{cases} \dfrac{13!}{(13-k)!×... | I think it'd be simpler to do this:
$\frac {ab}{\operatorname{lcm}(a,b)}=\gcd(a,b)$ so we have
$\gcd(7!, b) = 120$
So $\gcd(2^4\cdot 3^2\cdot 5\cdot 7, b) = 2^3\cdot 3\cdot 5$.
So $3|b$ but $9\not \mid b$ and $8|b$ but $16\not\mid b$ and $5|b$ and $7\not \mid b$.
$b =13*(13-1)*....*(13-(k-1))= \prod_{m=13-k+1}^{13} m$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that if $|z| < 1$, then $\displaystyle\frac{z}{1+z} + \frac{2z^{2}}{1+z^{2}} + \frac{4z^{4}}{1+z^{4}} + \frac{8z^{8}}{1+z^{8}} + \ldots$ Show that if $|z| < 1$, then
\begin{align*}
\frac{z}{1+z} + \frac{2z^{2}}{1+z^{2}} + \frac{4z^{4}}{1+z^{4}} + \frac{8z^{8}}{1+z^{8}} + \ldots
\end{align*}
converges
MY ATTEMPT
Le... | So far, you've found
$$
\sum_{k=0}^{n-1} \frac{2^{k}z^{2^k}}{1+z^{2^k}} = \frac{z}{1-z} - \frac{2^{n}z^{2^{n}}}{1-z^{2^{n}}}.
$$
As you said, you'll need to prove this by induction. To finish the proof, you need to show
$$
\lim_{n\rightarrow\infty}\frac{2^n z^{2^n}}{1-z^{2^n}} =0.
$$
You can then combine these to get
$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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distributing different balls into different boxes I have this question:
*
*Find in how many ways we can distribute $10$ similar balls to $4$ different boxes such that in no box there are exactly $3$ balls.
*Find in how many ways we can distribute $10$ different balls to $4$ different boxes such that no box has exact... |
In how many ways can we distribute $10$ indistinguishable balls to $4$ different boxes such that no box receives exactly three balls?
Your strategy is correct. However,
$$\binom{13}{10} - \binom{4}{1}\binom{9}{7} + \binom{4}{2}\binom{5}{4} - \binom{4}{3}\binom{1}{1} = 286 - 144 + 30 - 4 = 168$$
In how many ways can... | {
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"url": "https://math.stackexchange.com/questions/3741659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all $a$ such that $y=\log_\frac{1}{\sqrt3} (x-2a) = \log_3(x-2a^3-3a^2) $
Find all values of parameter $a\in \mathbb{Z}$ such that $$y= \log_\frac{1}{\sqrt3} (x-2a)$$ $$and$$ $$y = \log_3(x-2a^3-3a^2)$$ intersect at points with whole coordinates.
This is what I did:
$$\log_\frac{1}{\sqrt3}(x-2a) = \frac{log_3(x-... | After getting a common base, $3$, and exponentiating, we have
$$
\frac{1}{(x-2 a)^2} = -3 a^2 - 2 a^3 + x
$$Since we are requiring $a,x$ to be whole numbers, the RHS is an integer which means the LHS must be as well: then $x-2a=\pm 1$. However, since these came from logarithms, only the case $x-2a=1$ is allowed. Then t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\int _0^1\frac{\ln \left(x^2+x+1\right)}{x\left(x+1\right)}\:dx$ How can i evaluate this integral, maybe differentiation under the integral sign? i started expressing the integral as the following,
$$\int _0^1\frac{\ln \left(x^2+x+1\right)}{x\left(x+1\right)}\:dx=\int _0^1\frac{\ln \left(x^2+x+1\right)}{x}\... | I dont think Feynman's trick would work best here, following your path:
$$\int _0^1\frac{\ln \left(x^2+x+1\right)}{x\left(x+1\right)}\:dx=\int _0^1\frac{\ln \left(x^2+x+1\right)}{x}\:dx-\underbrace{\int _0^1\frac{\ln \left(x^2+x+1\right)}{x+1}\:dx}_{x=\frac{1-t}{1+t}}\:$$
$$=\int _0^1\frac{\ln \left(x^3-1\right)}{x}\:d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3744330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the relation of the Infinite power series with these fraction series? Take this infinite series:
$$S = 1 + \sum_{n=1}^\infty\prod_{i=1}^n\frac{2i+1}{4i} = 1 + \frac{3}{4} + \frac{3\times5}{4\times8} + \frac{3\times5\times7}{4\times8\times12} + ....$$
We want to find the sum of this series. I didn't know how to ... | Note that setting $nx=\frac{3}{4}$ and $\frac{n(n-1)}{2!}=\frac{3\cdot 5}{4\cdot 8}$ in order to get values for $n$ and $x$ is just a clever approach, but not a proof that $P=S$.
To make it a proof we have to additionally verify that the general terms evaluated at $n=\frac{3}{2}$ and $x=-\frac{1}{2}$ are equal.
Let's... | {
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"timestamp": "2023-03-29T00:00:00",
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Curve tangent line A curve with the equation $x^2-x+1$ has two tangent lines $a$ and $b$ that intersects at $x=1$, what is $y$? can I determine $y$ when $a$ and $b$ are perpendicular, or the gradient of $a$ and $b$ are $-1$
I tried using the second point, but knowing that they have the same gradient means they won't in... | Let $A$ be the $x$-coordinate of the point of tangency of the line $a$, then its equation is
$$y=(2A-1)(x-A)+A^2-A+1.$$
Similarly for $b$,
$$y=(2B-1)(x-B)+B^2-B+1.$$
Therefore for $x=1$ we have
$$(2A-1)(1-A)+A^2-A+1=y=(2B-1)(1-B)+B^2-B+1$$
and we finally find
$$(A-B)(A+B-2)=0.$$
It follows that if $a$ and $b$ are disti... | {
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"url": "https://math.stackexchange.com/questions/3750818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Another way to solve $\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx$ without the substitution $y=\tan\left(\frac{x}{2}\right)$? Is there another way to solve an integral $$\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx$$ without the substitution $y=\tan\left(\frac{x}{2}\right)$?
$\large \int \frac{\sin^3(x)}{1+\cos^2(x)}\ dx$ is easi... | $$\sin^4x=(\sin^2x+1)(\sin x+\sqrt{2})(\sin x-\sqrt{2})+(2+\sin^2x)$$
$$1+\cos^2x=(\sqrt{2}+\sin x)(\sqrt{2}-\sin x)$$
and so your integral becomes:
$$\int\frac{\sin^4x}{1+\cos^2x}dx=\int\frac{(\sin^2x+1)(\sin x+\sqrt{2})(\sin x-\sqrt{2})+(2+\sin^2x)}{(\sqrt{2}+\sin x)(\sqrt{2}-\sin x)}dx=-\int1+\sin^2xdx+\int\frac{2+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3751405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
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How to find $f\circ g$ when $f(y)$ and $g(x)$ are given? $f(y)=\frac4{y-2},\;g(x)=\frac5{3x-1}$
Find the graph of $f\circ g$?
I am stuck at this step. Since $f(x)$ is not given.
$f\circ g\;=\;f(g(x))\\f(\frac5{3x-1})=\frac4{y-2}$
I am thinking whether $f^{-1}(y)=f(x)$?
| Variables are just place holders.
If $f(y)=\frac4{y-2}$ then $f(x) = \frac 4{x-2}$ and $f(c) = \frac 4{c-2}$ and $f(\text{fred the tap-dancing goat}) = \frac 4{\text{fred the tap-dancing goat}-2}$
And $f(g(x)) = \frac 4{g(x)-2}$.
So $f\circ g$ is the function so that $f\circ g(x) = f(g(x)) = \frac 4{g(x)-2} = \frac 4{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3752936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How can I integrate $\int \frac{u^3}{(u^2+1)^3}du?$ How to integrate following
$$\int\frac{u^3}{(u^2+1)^3}du\,?$$
What I did is here:
Used partial fractions
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{Au+B}{(u^2+1)}+\dfrac{Cu+D}{(u^2+1)^2}+\dfrac{Au+B}{(u^2+1)^3}$$
After solving I got
$A=0, B=0, C=1, D=0, E=-1, F=0$
$$\dfrac{u^3}{... | Substitute $v=u^2+1$ so that $du=\frac{1}{2u}dv$ to turn the integral into:
$$\int{\frac{u(v-1)}{2uv^3}} dv$$
$$\frac{1}{2}\int{\frac{v-1}{v^3}} dv$$
$$=\frac{1}{2}(\int{\frac{1}{v^2} dv}-\int{\frac{1}{v^3}} dv)$$
$$=\frac{1}{2}(\frac{1}{2v^2}-\frac{1}{v})+C$$
$$=\frac{1}{4v^2}-\frac{1}{2v}+C$$
Substitute $v=u^2+1$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3753883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Unclear problem with $n$-th power matrix and limit
Find $$\lim\limits_{n \to \infty} \frac{A_n}{D_n}$$ where
$$\begin{pmatrix}
19 & -48 \\
8 & -21 \\
\end{pmatrix} ^{\! n} =
\begin{pmatrix}
A_n & B_n \\
C_n & D_n \\
\end{pmatrix}$$
$n$ - is the power of a matrix, but what is $A_n, B... | $$A:=\begin{pmatrix}19&-48\\8&-21\end{pmatrix}=\begin{pmatrix}2&3\\1&1\end{pmatrix}\begin{pmatrix}-5&0\\0&3\end{pmatrix}\begin{pmatrix}2&3\\1&1\end{pmatrix}^{-1}=PDP^{-1}$$
So $A^n=PDP^{-1}PDP^{-1}\cdots PDP^{-1}=PD^nP^{-1}$
That is, $$A^n=\begin{pmatrix}2&3\\1&1\end{pmatrix}\begin{pmatrix}(-5)^n&0\\0&3^n\end{pmatrix}\... | {
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"answer_id": 0
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Find the greatest integer less than $3^\sqrt{3}$ without using a calculator and prove the answer is correct. Find the greatest integer less than $3^\sqrt{3}$ without using a calculator and prove the answer is correct.
I'm puzzled on how to solve this problem, any help is appreciated. There was hints about turning the e... | Use the (easily verified) inequalities $\sqrt3\lt1.75=7/4$ and $\sqrt{7/4}\lt4/3$ to show that
$$3^\sqrt3\lt3^{1.75}=3\cdot3^{1/2}\cdot3^{1/4}\lt3\cdot{7\over4}\cdot{4\over3}=7$$
and $\sqrt3\gt1.7$ to show
$$3^\sqrt3\gt3^{1.7}=3\cdot3^{7/10}$$
Now show $3^7=3^4\cdot3^3=81\cdot27\gt1600\gt1024=2^{10}$ to conclude $3^{7/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3757754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Let $ABC$ be a triangle and $M$ be the midpoint of $BC$. Squares $ABQP$ and $ACYX$ are erected. Show that $PX = 2AM$.
$\textbf{Question:}$ Let $ABC$ be a triangle and $M$ be the midpoint of $BC$. Squares $ABQP$ and $ACYX$ are erected. Show that $PX = 2AM$.
I could solve this problem using computational techniques but... | Let's first give names to the sides of $\overset{\Delta}{PAX}$ triangle: $|PA| = k$, $|AX| = m$ and $|PX| = b$. Also for $\overset{\Delta}{ABC}$ triangle: $|AB| = k$, $|AC| = m$, $|BC| = 2l$ and the length of the median $|AM| = a$. Writing law of cosines for $\overset{\Delta}{PAX}$ triangle for angle $\widehat{PAX}$, w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to solve a non-linear system of 3 equations involving $x^2$, $y^2$ and $xy$ only? Let $b_1, b_2,b_3$ be real constants, how to solve the following non-linear system?
\begin{align}
(-3x^2 + y^2 -6xy) + b_1 &= 0 \qquad\text{(L1)}\\
(\phantom{-} x^2 -3 y^2 -6xy) + b_2 &= 0 \qquad\text{(L2)}\\
(\phantom{.}3x^2 + 3 y^2... | Since you have three equations and two unknowns, I should define $z=xy$ to make
$$\begin{align}
(-3x^2 + y^2 -6z) + a &= 0 \tag 1\\
(\phantom{-} x^2 -3 y^2 -6z) + b &= 0 \tag 2\\
(\phantom{.}3x^2 + 3 y^2 -2z) + c &= 0 \tag 3
\end{align}$$
So, four solutions
$$x=\pm\frac{\sqrt{3 a-2 b-3 c}}{2 \sqrt{5}}\qquad y=\pm \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Motivation confusion in floor and ceiling function algebra From HMMT https://hmmt-archive.s3.amazonaws.com/tournaments/2019/nov/team/solutions.pdf:
Compute the sum of all positive real numbers $x \le 5$ satisfying $$x=\frac{\left\lceil{x^2}\right\rceil+\left\lceil x\right\rceil\cdot\left\lfloor x\right\rfloor}{\left\l... | Since all floors and ceilings are integral, it is clear that the right hand side is a rational number. Because $n < x < n+1$, we have $\left\lfloor{x}\right\rfloor = n$ and $\left\lceil{x}\right\rceil = n+1$. Therefore the denominator of the right hand side is $(n+1) + n = 2n+1$. By hypothesis, this equals the left ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the parabola of the form $y=ax^2+b$ which best fits the points $(1,0), (4,4), (5,8)$ by minimizing the sum of squares, $S$, given by
$$S=(a+b)^2 + (16a+b−4)^2 + (25a+b−8)^2$$
My work so far
The variables are a and b, so we set
$$\frac{∂S}{∂a}=2(a+b)+32(16a+b−4)+50(25a+b−8)=0$$
and
$$\frac{∂S}{∂b}=2(a+b)+2(16a+b−4)... | Given set of $n$ points $(x_i,y_y)$, $i=1,\dots,n$,
the parameters of the best fit parabola $y(x)=a_2x^2+a_1 x+a0$
can be found as:
\begin{align}
\begin{bmatrix}
a_0 \\ a_1 \\ a_2
\end{bmatrix}
&=
\begin{bmatrix}
n & s_x & s_{x^2}\\
s_x & s_{x^2} & s_{x^3}\\
s_{x^2} & s_{x^3} & s_{x^4}
\end{bmatrix}
^{-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral: $\int \dfrac{dx}{(x^2-4x+13)^2}$?
How can I integrate $$\int \dfrac{dx}{(x^2-4x+13)^2}?$$
Here is my attempt:
$$\int \dfrac{dx}{(x^2-4x+13)^2}=\int \dfrac{dx}{((x-2)^2+9)^2}$$
Substitute $x-2=3\tan\theta$, $\ dx=3\sec^2\theta d\theta$
\begin{align*}
&=\int \dfrac{3\sec^2\theta d\theta}{(9\tan^2\theta+9)^2}\... | You can solve it by using induction formula: $\color{blue}{\int \frac{dt}{(t^2+a^2)^n}=\frac{t}{2(n-1)a^2(t^2+a^2)^{n-1}}+\frac{2n-3}{2(n-1)a^2}\int \frac{dt}{(t^2+a^2)^{n-1}}}$ as follows
$$\int \dfrac{dx}{(x^2-4x+13)^2}$$$$=\int \dfrac{d(x-2)}{((x-2)^2+3^2)^2}$$
$$=\frac{x-2}{2\cdot 3^2((x-2)^2+3^2)}+\frac{1}{2\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 1
} |
Check and comment my proof of $a+b \geq 2 \sqrt{ab}$ I want to prove
$$
a+b \geq 2 \sqrt{ab}, \quad \text{for} \quad a, b>0 \tag 1
$$
First question:
Isn't this wrong? Shouldn't it be
$$
a+b \geq 2 \sqrt{ab}, \quad \text{for} \quad a, b\geq0 \tag 2
$$
Or
$$
a+b > 2 \sqrt{ab}, \quad \text{for} \quad a, b>0 \tag 3
$$
... | You want to prove $a+b \geq 2 \sqrt{ab}, \quad \text{for} \quad a, b>0$.
You can use $a^2 > 0$ for all $a \in \mathbb{R}$ as an assumption and segue into (9), with the restriction that $a,b > 0$. Then, instead of working backwards, you're working forwards and the proof is nice.
Rewrite your proof so that it begins wit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3762541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to solve polynomial rational relations for $y$ (e.g $\sqrt{4-3y-y^2} = x(y+4)$)? From time to time, I struggle to solve polynomial relations for $y$.
A trivial example is :
$$ \frac{y}{x} = x \iff y = x^2$$
Easy.
But consider this relation:
$$ \sqrt{4-3y-y^2} = x(y+4)$$
No matter how much I mess around it, seems i... | \begin{align*}
\sqrt{4-3y-y^2} &= x(y+4)\\
\sqrt{(1-y)(y+4)} &= x(y+4)\Rightarrow y\in[-4,1],x\in[0,+\infty)\\
\sqrt{y+4}\left(\sqrt{1-y}-x\sqrt{y+4}\right) &=0\Rightarrow \boxed{y=-4}\\
\sqrt{1-y}&=x\sqrt{y+4}\\
\sqrt{\frac{5}{y+4}-1}&=x\\
\boxed{y=\frac5{x^2+1}-4}\in(-4,1]\,\forall\,x\in[0,+\infty)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Solving for positive reals: $abcd=1$, $a+b+c+d=28$, $ac+bc+cd+da+ac+bd=82/3$ $$a,b,c,d \in \mathbb{R}^{+}$$
$$ a+b+c+d=28$$
$$ ab+bc+cd+da+ac+bd=\frac{82}{3} $$
$$ abcd = 1 $$
One can also look for the roots of polynomial
$$\begin{align}
f(x) &= (x-a)(x-b)(x-c)(x-d) \\[4pt]
&= x^4 - 28x^3 + \frac{82}{3}x^2 - (abc+abd+a... | [Here's an unsatisfactory answer as I don't expect the discriminant to be calculated in a contest-math setting.]
Let $ C = abc + bcd + cda + dab > 0$.
The equation $ x^4 - 28x^3 + \frac{82}{3} x^2 - Cx + 1 =0 $ has 4 positive real roots.
The discriminant (Thanks Wolfram) is
$$\Delta = -5565901568/81 + (82093312 C)/9 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\int \frac{2-x^3}{(1+x^3)^{3/2}} dx$
Evaluate:
$$\int \frac{2-x^3}{(1+x^3)^{3/2}} dx$$
I could find the integral by setting it equal to $$\frac{ax+b}{(1+x^3)^{1/2}}$$
and differentiating both sides w.r.t.$x$ as
$$\frac{2-x^3}{(1+x^3)^{3/2}}=\frac{a(1+x^3)^{3/2}-(1/2)(ax+b)3x^2(1+x^3)^{-1/2}}{(1+x^3)}$$$$=\f... | Split the integral and integrate by parts:
$$\begin{align}I = \int\dfrac{2 - x^3}{\left(1 + x^3\right)^{3/2}}\,\mathrm dx &\equiv\int\dfrac2{\left(1 + x^3\right)^{3/2}}\,\mathrm dx - \int\dfrac{x^2}{\left(1 + x^3\right)^{3/2}}x\,\mathrm dx \\ &= \int\dfrac2{\left(1 + x^3\right)^{3/2}}\,\mathrm dx - \left(-\dfrac{2x}{3\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Finding $\lim_{n\to\infty}\frac1{n^3}\sum_{k=1}^{n-1}\frac{\sin\frac{(2k-1)\pi}{2n}}{\cos^2\frac{(k-1)\pi}{2n}\cos^2\frac{k\pi}{2n}}$
For all $n\ge 1$, let
$$
a_{n}=\sum_{k=1}^{n-1} \frac{\sin \left(\frac{(2 k-1) \pi}{2 n}\right)}{\cos ^{2}\left(\frac{(k-1) \pi}{2 n}\right) \cos ^{2}\left(\frac{k \pi}{2 n}\right)}
$$
... | This solution is based on one of the official solutions for the problem.
The dummy variable $k$ is a dummy variable in $a_n$. Since
\begin{align}
\sum_{k=1}^{n-1}f(k;n) &= f(1;n)+f(2;n)+\cdots+f(n-1;n)\\
&= f(n-1;n) + f(n-2;n) +\cdots f(n-(n-1);n)\\
&= \sum_{k=1}^{n-1}f(n-k;n)
\end{align}
we can substitute $n-k$ for $k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3772724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate $\sum_{n=1}^{\infty}\:\frac{2n+1}{2n(n+1)^2}$? Note: Similar questions have been asked here and here, but this is quite different.
I am trying to evaluate
$$\sum_{n=1}^{\infty}\:\frac{2n+1}{2n(n+1)^2} \quad (1)$$
I re-wrote the fraction as $$ \frac{2n+1}{2n(n+1)^2} = \frac1{2(n+1)} \cdot \frac{2n+1}{n(... | 1). $\frac{2n+1}{2n(n+1)^2}=\frac{n+n+1}{2n(n+1)^2}=\frac{1}{2(n+1)^2}+\frac{1}{2n(n+1)}$
So
\begin{align*}
\sum_{n\geq1}{}\frac{2n+1}{2n(n+1)^2}=\frac{1}{2}\sum_{n\geq1}\frac{1}{(n+1)^2}+\frac{1}{2}\sum_{n\geq1}{}\frac{1}{n(n+1)}\\
=\frac{1}{2}(\zeta(2)-1)+\frac{1}{2}\sum_{n\geq1}{}\frac{1}{n}-\frac{1}{n+1}\\
=\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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If $y = \frac{2}{5}+\frac{1\cdot3}{2!} \left(\frac{2}{5}\right)^2+\frac{1\cdot3\cdot5}{3!} \left(\frac{2}{5}\right)^3+\cdots$, find $y^2+2y$
If $$y = \frac{2}{5}+\frac{1\cdot3}{2!} \left(\frac{2}{5}\right)^2+\frac{1\cdot3\cdot5}{3!} \left(\frac{2}{5}\right)^3+\cdots$$ what is $y^2+2y$?
Attempt:
We know that for negat... | You can calculate $y$ using a "well known" binomial series expansion for
$$(1-4x)^{-\frac 12} = \sum_{n=0}^{\infty}\binom{2n}nx^n$$
To see this rewrite the coefficients
$$\prod_{k=1}^n\frac{2k-1}{k} = \prod_{k=1}^n\frac{(2k-1)\cdot 2k}{k\cdot 2k}= \frac 1{2^n}\binom{2n}n$$
So in your case the series becomes
$$y = \sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Proving a solution of a Bernoulli type equation Prove that
\begin{equation}
y(x) = \sqrt{\dfrac{3x}{2x + 3c}}
\end{equation}
is a solution of
\begin{equation}
\dfrac{dy}{dx} + \dfrac{y}{2x} = -\frac{y^3}{3x}
\end{equation}
All the math to resolve this differential equation is already done. The exercise simply asks to p... | @kira1985 you should not add the p(x).y on both side, here you have to find the answer .
But you are using the answer it self to solve the problem .As the answer is unknown to have to just find the value of the P(x) and Q(x).
which is
/ = /2 − (^3)/3
P(x) = - 1/(2x)
Q(x) = - 1/(3x)
Is the write answer for the given que... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3776217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How to prove this statement. "There are not odd integers $x,y,z$ according to $(x-z)^2+(y-z)^2=(x+y)^2$." I did try to prove this by contradiction(I guess it works). So I suppose this statement is false but I have confused about the meaning of it.
If this statement is false it means "There are odd integers $x,y,z$ ac... | Write $x, y,$ and $z$ as $2a+1, 2b+1,$ and $2c+1$ respectively.
Then, the equation becomes $(2(a-c))^2+(2(b-c))^2=(2(a+b)+2)^2$, or $4(a-c)^2+4(b-c)^2=(2a+2b+2)^2$.
Expanding both sides, one gets $4(a^2-2ac+c^2)+4(b^2-2bc+c^2)=4a^2+4b^2+4+8ab+8a+8b$.
Dividing both sides by $4$, one gets $a^2-2ac+c^2+b^2-2bc+c^2=a^2+b^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3776798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Show that the inequality $\left|\int_{0}^{1} f(x)\,dx\right| \leq \frac{1}{12}$ holds for certain initial conditions
Given that a function $f$ has a continuous second derivative on the interval $[0,1]$, $f(0)=f(1)=0$, and $|f''(x)|\leq 1$, show that $$\left|\int_{0}^{1}f(x)\,dx\right|\leq \frac{1}{12}\,.$$
My attemp... | Use $\text{Taylor}$ we can get:
$$
f\left( 0 \right) =f\left( x \right) -xf'\left( x \right) +\frac{f''\left( \xi _1 \right)}{2}x^2,\xi _1\in \left( 0,x \right)
$$
$$
f\left( 1 \right) =f\left( x \right) +\left( 1-x \right) f'\left( x \right) +\frac{f''\left( \xi _2 \right)}{2}\left( x-1 \right) ^2,\xi _2\in \left( x,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Under what conditions does $ \ (a+b)^{n}=a^{n}+b^{n}$ for a natural number $ n \geq 2$? Under what conditions does $ \ (a+b)^{n}=a^{n}+b^{n}$ holds for a natural number $ n \geq 2$?
My attempt at solving:
Using $(a+b)^2=a^2+2ab+b^2$; if $(a+b)^2=a^2+b^2$, $2ab=0$ therefore $a$ and/or $b$ must be $0$.
If $a$ and/or $b$ ... | For even $n$ you are right. For odd $n$ we have also $a+b=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Indefinite integral of $\frac{\sec^2x}{(\sec x+\tan x)^\frac{9}{2}}$ $$\frac{\sec^2x}{(\sec x+\tan x)^\frac{9}{2}}$$
My approach:
Since it is easy to evaluate $\int{\sec^2x}$ , integration by parts seems like a viable option.
Let $$I_n=\int{\frac{\sec^2x}{(\sec x+\tan x)^\frac{9}{2}}}$$
$$I_n=\frac{\tan x}{(\sec x+\tan... | $-77I_n=4 \times \frac{\tan x}{(\sec x+\tan x)^\frac{9}{2}} + 18 \times \frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}}$
$-77I_n= 11 \times \frac{\tan x}{(\sec x+\tan x)^\frac{9}{2}} - 7 \times \frac{\tan x}{(\sec x+\tan x)^\frac{9}{2}} + 11 \times \frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}} + 7 \times \frac{\sec x}{(\sec x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
The number $85^9 - 21^9 + 6^9$ is divisible by an integer between 2000 and 3000. Compute that integer.
The number $85^9 - 21^9 + 6^9$ is divisible by an integer between 2000 and 3000. Compute that integer.
I tried taking the expression mod 5, 7, etc. getting a few congruences, but nothing that led to an answer betwe... | To elaborate on the trial division approach, we can observe that $$85^9 - 21^9 = (85-21)(85^8 + 85^7 21^1 + \cdots + 21^8), \tag{1}$$ thus this part is divisible by $85 - 21 = 64 = 2^6$. Then $6^9 = 2^9 \cdot 3^9$, hence we know $2^6$ divides the entire number $N = 85^9 - 21^9 + 6^9$. We know $2$ does not divide the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3781456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Show that the transformation $w=\frac{2z+3}{z-4}$ maps the circle $x^2+y^2-4x=0$ onto the straight line $4u+3=0$ Question:
Show that the transformation $w=\frac{2z+3}{z-4}$ maps the circle $x^2+y^2-4x=0$ onto the straight line $4u+3=0$.
My try:
$$\begin{align}\\
&x^2+y^2-4x=0\\
&\implies (x-2)^2+y^2=4\\
&\implies |z-... | You can't replace modulus like that in the last line ( because complex numbers with equal magnitude may not be actually equal. For e.g., $2i$ and $-2i$ have magnitude $2$ but aren't equal.)
Use $w = u+iv$
So, $|2(u+iv)+7| = 2|(u+iv)-2|$
Evaluating and squaring both sides,
$\Rightarrow (2u+7)^2 + 4v^2 = 4(u-2)^2 + 4v^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3781558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Inverse Laplace Transform via Circuit Analysis [HELP] Inverse Laplace Transform $\frac{1}{s^2 + \sqrt{2}s + 1}$
so what I did it changed the denominator to complete the square format which is $\left(s+\frac{\sqrt{2}}{2}\right)^2 + \frac{1}{2}$, then I can solve for $s$, it will make it as
$$
\left(\left(s+ \frac{\sqrt{... | Let us denote by $a$ the value $1/\sqrt 2$ to have an easy typing. The the partial fraction decomposition over complex numbers of the given fraction is
indeed
$$
\begin{aligned}
\frac{1}{s^2 + \sqrt{2}s + 1}
&=
\frac{1}{(s^2 + 2as + a^2)+a^2}
=
\frac1{2ia}\left(\frac 1{s+a-ia}-\frac 1{s+a+ia}\right)
\\
&=\frac K{s+a-i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3781963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Range of Convergence of $\sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \ 3^n (x-5)^n}$ $$\sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \ 3^n (x-5)^n}$$
I am trying to use the alternating series test to find a range of $x$ for which $(1) b_n > b_{n+1}$ and $ (2) \lim_{n \to \infty} \frac{1}{n \ 3^n (x-5)^n} = 0$. If $... | Hints : by Cauchy Hadamard test , range will be $ |x-5|\gt\frac{1}{3} \implies x\gt \frac{16}{3},x\lt \frac{14}{3} $.
Only equality $x=\frac{16}{3} $ is possible because series $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} $ is converge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3782557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Finding the determinant of a $5\times 5$ matrix Let
$$A = \left[\begin{array}{rrrrr}6 & 2 & 2 & 2 & 2 \\ 2 & 6 & 2 & 2 & 2 \\ 2 & 2 & 6 & 2 & 2 \\2 & 2 & 2 & 6 & 2 \\ 2 & 2 & 2 & 2 & 6\end{array}\right] \in {M}_{5}(\mathbb{R})$$
Which of following options is $\det(A)$ ?
*
*$4^4 \times 14$
*$4^3 \times 14$
*$4^2 \... | Your generalization, with an obvious but unstated definition of $A_n$, can be proved as follows. Define $E_n$ as the $n\times n$ matrix whose entry are all $1$s. You can show its only nonzero eigenvalue is $2n$, and the associated eigenspace is $1$-dimensional. By the rank-nullity theorem, the eigenvalue $0$ has multip... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\lim_{x\to 0} \frac 1x -\frac{2}{e^{2x}-1}$ Factorizing the expression
$$\lim_{x\to 0} \frac{e^{2x}-1-2x}{x(e^{2x}-1)}$$
$$=\lim_{x\to 0} \frac{e^{2x}-(1+2x)}{(x)(e^x-1)(e^x+1)}$$
How do I proceed?
| Another approach using the Hopital's rule:
$$\lim_{x \to 0}\left(- \frac{2}{e^{2 x} - 1} + \frac{1}{x}\right)= \lim_{x \to 0} \frac{- 2 x + e^{2 x} - 1}{x \left(e^{2 x} - 1\right)}$$
$$\lim_{x \to 0} \frac{- 2 x + e^{2 x} - 1}{x \left(e^{2 x} - 1\right)} \stackrel{\,\,(0/0) \\\text{Hopital}}{=} \lim_{x \to 0} \frac{2 e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
$p\equiv 1\pmod 4\Rightarrow p=a^2+b^2$ and $p\equiv 1\pmod 8\Rightarrow p=a^2+2b^2$, what about for $p\equiv 1\pmod {2^n}$ in general Primes $p$ with $p\equiv 1\pmod 4$ can be written as $p=a^2+b^2$ for some integers $a,b$. For $p\equiv 1\pmod 8$ we have $p=a^2+2b^2$. Can primes that satisfy $p\equiv 1\pmod{2^n}$ for ... | Let prime $p=a^2+2^{n-2}\cdot b^2\equiv1\pmod{2^n}$, where $n>4$ and $a,b$ is integers.
Then sequence $(n, min(p))$=(5,97), (6,193), (7,257), (8,257), (9,18433), (10,18433), (11,18433), (12,65537), (13,1097729), (14,65537), (15,1179649), (16,65537), (17,1179649), (18,26214401), (19,117964801), (20,26214401), (21,169869... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
On proving $a^3+b^3+c^3-3abc \geq 2\left({b+c\over 2}-a\right)^3$. This problem was a "warm-up" problem by the author. Note: $a, b, c$ are non-negative numbers.
$$a^3+b^3+c^3-3abc \geq 2\left({b+c\over 2}-a\right)^3$$
I tried to remove the $2$ from ${b+c\over 2}$ and got this-
$$ 4(a^3+b^3+c^3-3abc) \geq (b+c-2a)^3 $$
... | Using algebra.
Consider that we look for the minimum value of
$$F=a^3+b^3+c^3-3abc -2\left({b+c\over 2}-a\right)^3$$
$$\frac{\partial F}{\partial a}=3 a^2+6 \left(\frac{b+c}{2}-a\right)^2-3 b c \tag 1$$
$$\frac{\partial F}{\partial b}=3b^2-3 \left(\frac{b+c}{2}-a\right)^2-3 a c\tag 2$$
$$\frac{\partial F}{\partial c}=3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Evaluating $\cos24^\circ-\cos84^\circ-\cos12^\circ+\sin42^\circ$ How to evaluate:
$$\cos24^\circ-\cos84^\circ-\cos12^\circ+\sin42^\circ$$
Can somebody help me handle it?
I have no idea what to do.
This is my attempt:
$$\cos24^\circ-\cos(60^\circ+24^\circ)-\cos12^\circ+\sin (12^\circ+30^\circ)$$
| $$-\cos84^\circ-\cos96^\circ$$
$$-\cos12^\circ=\cos192^\circ=\cos168^\circ$$
$$\sin42^\circ=\cos(?)$$
As $\cos5x=-\dfrac12$ for $x=120^\circ,240^\circ\equiv-120^\circ,480^\circ\equiv120^\circ,960^\circ\equiv-120^\circ\pmod{360^\circ}$
Now $\cos5x+\cos x=2\cos3x\cos2x$
$\cos5x=2(4\cos^3x-3\cos x)(2\cos^2x-1)=?$
So, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to prove $\frac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}$? Give $a,b,c>0$. Prove that: $$\dfrac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}.$$
My direction: (we have the equation if and only if $a=b=c$)
$a^{n+1}+a^nb+a^nc \ge 3a^n\sqrt[3]{abc}$
$b^{n+1}+b^na+b^nc \ge 3b^n\sqrt[3]{abc}$
$c... | By Chebyshev's sum inequality (https://en.wikipedia.org/wiki/Chebyshev%27s_sum_inequality),
we have
$$a^n\cdot a + b^n\cdot b + c^n \cdot c
\ge \frac{1}{3}(a^n + b^n + c^n)(a+b+c).$$
Thus,
$$\frac{a^{n+1}+b^{n+1} + c^{n+1}}{a^n + b^n + c^n} \ge \frac{a+b+c}{3} \ge \sqrt[3]{abc}.$$
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Why $8^{\frac{1}{3}}$ is $1$, $\frac{2\pi}{3}$, and $\frac{4\pi}{3}$ The question is:
Use DeMoivre’s theorem to find $8^{\frac{1}{3}}$. Express your answer in complex form.
Select one:
a. 2
b. 2, 2 cis (2$\pi$/3), 2 cis (4$\pi$/3)
c. 2, 2 cis ($\pi$/3)
d. 2 cis ($\pi$/3), 2 cis ($\pi$/3)
e. None of these
I think that ... | Here,
$$\begin{align*}
8^{1/3} &= (|8|e^{2\pi kj})^\frac{1}{3}, k = 0,1,2\\
&= |8|^\frac{1}{3} e^{\frac{2}{3}\pi kj}, k = 0,1,2\\
&= 2 e^{\frac{2}{3}\pi kj}, k = 0,1,2\\
\end{align*}$$
so,for $k=1$,$k=2$ we get $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$
Or take:
$$8^{1/3}=x$$
Then we get,
$$(x-2)(x^2+2x+4)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Using termwise differentiation and integration find the sum $\sum\limits_{n=0}^{\infty}(n+1)x^n = 1 + 2x + 3x^2 + ... +(n+1)x^n$ Using termwise differentiation and integration find the sum $$\sum\limits_{n=0}^{\infty}(n+1)x^n = 1 + 2x + 3x^2 + \cdots +(n+1)x^n$$
The answer can be achevied simply by setting $S = \sum\li... | We have
$$\sum_{n\geq 0}(n+1)x^n=\left(\sum_{n\geq 0}x^{n+1}\right)'=\left(\frac{x}{1-x}\right)'=\frac{1}{(1-x)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How were amplitudes of the $\cos$ and $\sin$ chosen? I don't understand why we use $\displaystyle\sqrt{1^2+\left(\frac{1}{2}\right)^2}$ in the below transformation. Can someone help to explain?
from
$$f(x)=\frac{3}{5}-\frac{3}{5}e^t\left(\cos(2t)+\frac{1}{2}\sin(2t)\right)$$
transform to
$$f(x)=\frac{3}{5}-\frac{3}{5}\... | This is a way to normalize the vector $v=(a,b)=\left(1,\frac12\right)$ that is
$$|v|=\sqrt{a^2+b^2} \implies \hat v=\frac{v}{|v|}$$
has length equal to $1$ and this allows to perform the subsequent transformation for $\cos \phi$ and $\sin \phi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Finding $\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ $\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ this limit according to wolframalpha is equal to $0$.
So this is my work thus far
$\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}... | $$A=x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3}=x + \frac{2x^{3}}{3} - \frac{2x^3}{3}\left(1+\frac{1}{x^2}\right)^{3/2}$$ For the very last term, let $\frac{1}{x^2}=\epsilon$ and use the binomial expansion
$$(1+\epsilon)^{3/2}=1+\frac{3 \epsilon }{2}+\frac{3 \epsilon ^2}{8}+O\left(\epsilon ^3\right)$$ Repla... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
} |
How to evaluate $\int_0^{\pi/2} x\ln^2(\sin x)\textrm{d}x$ in a different way? The following problem
$$\int_0^{\pi/2} x\ln^2(\sin x)\textrm{d}x=\frac{1}{2}\ln^2(2)\zeta(2)-\frac{19}{32}\zeta(4)+\frac{1}{24}\ln^4(2)+\operatorname{Li}_4\left(\frac{1}{2}\right)\tag1$$
was already solved in this solution.
The question here... | From here we have
$$\frac23\arcsin^4x=\sum_{n=1}^\infty\frac{H_{n-1}^{(2)}(2x)^{2n}}{n^2{2n\choose n}}=\sum_{n=1}^\infty\frac{H_{n}^{(2)}(2x)^{2n}}{n^2{2n\choose n}}-\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^4{2n\choose n}}$$
Set $x=1$ we get
$$\sum_{n=1}^\infty\frac{4^n}{n^4{2n\choose n}}=\sum_{n=1}^\infty\frac{4^nH_{n}^{(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3796229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Why is $|-a+\sqrt{a^2-1}|<1<|-a-\sqrt{a^2-1}|$ (where $a>1$) true? Why does $|-a+\sqrt{a^2-1}|<1<|-a-\sqrt{a^2-1}|$ (where $a>1$) hold? I understand that $a>1 \implies 1<|-a-\sqrt{a^2-1}|$ and that $|-a+\sqrt{a^2-1}|<|-a-\sqrt{a^2-1}|$
But I can't see why $a>1 \implies |-a+\sqrt{a^2-1}|<1$.
Does anyone see why? Thank y... | $$\begin{align}\left|-a-\sqrt{a^2-1}\right|=\left|-\left(a+\sqrt{a^2-1}\right)\right|=\left|a+\sqrt{a^2-1}\right|>a>1\end{align}$$
$$\begin{align}\left|-a+\sqrt{a^2-1}\right|=\left|-\left(a-\sqrt{a^2-1}\right)\right|=\left|a-\sqrt{a^2-1}\right|=\dfrac{1}{a+\sqrt{a^2-1}}<1\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3796517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Proving limit using the definition of a limit Heres the question (provided for context) about proving a limit by using the definition of a limit:
$\mathop {\lim }\limits_{x \to 4} \left({x^2} + x - 11\right) = 9$
So, let’s get started. Let $\varepsilon > 0$ be any number then we need to find a number $\delta > 0$ so... | You are working with the assumption $|x-4|<1$, that means $3<x<5$.
So $x+5=|x+5|$ because $x+5$ is always positive when $x \in (3,5)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find continuous $f$ such that $f(x+1)=f(x)+f(\frac{1}{x})$
Find all real continuous functions that verifies :
$$f(x+1)=f(x)+f\left(\frac{1}{x}\right) \ \ \ \ \ \ (x\neq 0) $$
I found this result $\forall x\neq 1 \ \ f(x)=f\left(\frac{x}{x-1} \right)$ and I tried to study the behaviour of the function $g$ defin... | Some results
Assume that $f$ is defined and continuous on all of $\mathbb{R}$ even if the equation $f(x+1) = f(x) + f(\frac{1}{x})$ is not defined for $x=0.$
From the given identity
$
f(x+1) = f(x) + f(\frac1x)
$
we get
$
f(\frac1x) = f(x+1) - f(x) \to f(1) - f(0)
$
as $x\to0$. Thus $\lim_{R\to\infty}f(R) = \lim_{R\to\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
If $f(x)=\sin^{-1} (\frac{2x}{1+x^2})+\tan^{-1} (\frac{2x}{1-x^2})$, then find $f(-10)$ Let $x=\tan y$, then
$$
\begin{align*}\sin^{-1} (\sin 2y )+\tan^{-1} \tan 2y
&=4y\\
&=4\tan^{-1} (-10)\\\end{align*}$$
Given answer is $0$
What’s wrong here?
| We can't bluntly take $\sin^{-1}(\sin 2y) = 2y$ and so with $\tan^{-1}(\tan 2y)$, because we don't know the value of $2y$ and the range in which it lies.
So, substitute directly.
$f(-10) = \sin^{-1}\left(\dfrac{-20}{101}\right)+\tan^{-1}\left(\dfrac{20}{99}\right) = -\sin^{-1}\left(\dfrac{20}{101}\right)+\tan^{-1}\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
How to solve $ (y+u)\dfrac{\partial u}{\partial x} + (x+u)\dfrac{\partial u}{\partial y} = x+y$ via method of characteristics? How to solve $ (y+u)\dfrac{\partial u}{\partial x} + (x+u)\dfrac{\partial u}{\partial y} = x+y$ via method of characteristics?
My attempt.
These are equations with which I begin:
$\dfrac{dx}{ds... | Your differential equations are linear, with constant coefficients, for which there are well-established methods of solution. First write them in matrix form as:
$$
\pmatrix{\frac{dx}{ds}\\
\frac{dy}{ds}\\
\frac{du}{ds}}=\pmatrix{0&1&1\\
1&0&1\\
1&1&0}\pmatri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3802816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\frac{1 - x^{n+1} }{n+1} \lt \frac{1-x^n}{n}$ given $n$ is a positive integer and $0 < x \lt 1$. Problem Statement: If $n$ is a positive integer and $0 < x \lt 1$, show that
$$ \frac{1 - x^{n+1} }{n+1} \lt \frac{1-x^n}{n}.$$
My Solution:
$$
\frac{ 1- x^{n+1} }{n+1} \lt \frac{1-x^n}{n} ~~~~\text{is true} \\
... | Another option:
$$
\frac{1-x^n}{n} = \int_x^1 t^{n-1} \, dt
$$
and the integrand $t^{n-1}$ decreases strictly on $(x, 1)$ if $n$ increases.
Yet another option: The inequality
$$
\frac{1 - x^{n+1} }{n+1} \lt \frac{1-x^n}{n}
$$
is equivalent to
$$
x^n < \frac{1}{n+1}x^0 + \frac{n}{n+1}x^{n+1}
$$
and that is true becaus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Compute the area of a surface, encountering a strange integral compute the area of this yellow surface, which is actually a paraboloid:
$x^2+y^2=2az$(yellow one), cutted by $(x^2+y^2)^2=2a^2xy$(blue one)
To compute the part be surrounded in the blue surface, use polor coordinates:
\begin{cases}
x=r\cos\theta\\
y=r\sin\... | I am coming from assumption, that to obtain desired answer you should have cylinder $(x^2+y^2)^2=2a^2xy$ in place of brought in title $x^2+y^2=2a^2xy$ (blue one).
Then we will have $\sqrt{1+z'_x{^2}+z'_y{^2}} = \frac{1}{a}\sqrt{a^2+x^2+y^2}$. After writing double integral it is enough to use usual polar coordinates, so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding remainder of $123^{456}$ divided by 88 using Chinese Remainder Theorem I tried using Chinese remainder theorem but I kept getting 19 instead of 9.
Here are my steps
$$
\begin{split}
M &= 88 = 8 \times 11 \\
x_1 &= 123^{456}\equiv 2^{456} \equiv 2^{6} \equiv 64 \equiv 9 \pmod{11} \\
y_1 &= 9^{-1} \equiv 9^9 \equ... | You used the inverse of $x_i$ instead of the inverse of $\frac{M}{m_i}$. So for example,
$$
9 \cdot \frac{88}{11} \cdot 5 + 1 \cdot \frac{88}{8} \cdot 1 \equiv 0 + 1 \cdot 11 \cdot 1 \equiv 3 \not \equiv 1\pmod{8}.
$$
If you use the inverse of $\frac{M}{m_i}$ instead, you would have
$$
x_1 \cdot \frac{88}{11} \cdot 8^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
} |
recurrence relation where $c_n = c_{n-1} + 2c_{n-2}$
A sequence $(c_n)$ is defined recursively as follows: $c_0 = 1, c_1 = 1, $ and $c_n = c_{n-1} + 2c_{n-2}$ for $n\geq 2$. We use $[x^n]g(x)$ to denote the coefficient of $x^n$ of the polynomial $g(x).$ Show that $c_{2n} = [x^{2n}] \dfrac{1}2\left(\dfrac{1}{1-x-2x^2}+... | Your attempt to equate coefficients in
$$(1-5x^2+4x^4)\sum_{n\ge 0}c_{2n}x^{2n}=1-2x^2$$
will work if you do it correctly. When you multiply out the lefthand side you should get
$$c_0+(c_2-5c_0)x^2+\sum_{n\ge 2}(c_{2n}-5c_{2n-2}+4c_{2n-4})x^{2n}=1-2x^2\,,$$
and you know that $c_0=1$ and $c_2=3$, so this reduces to
$$\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding an expression and proof for $\sqrt{1-4x}$ This is similar to $$\frac{1}{\sqrt{1-4x}}=\sum_{n\geq0}{2n\choose n}x^n$$
However I want to find an expression the same way for $$\sqrt{1-4x}$$ rather than $$\frac{1}{\sqrt{1-4x}}$$
Here's my thoughts so far:
$$(1-4x)^\frac{1}{2}=\sum_{n\geq0}{\frac{1}{2}\choose n}(-4)... | If the representation
\begin{align*}
\frac{1}{\sqrt{1-4x}}=\sum_{n\geq 0}\binom{2n}{n}x^n
\end{align*}
for $|x|<\frac{1}{4}$ is supposed to be known, we can also use
\begin{align*}
\sqrt{1-4x}&=\frac{1-4x}{\sqrt{1-4x}}\\
&=\sum_{n \geq 0}\binom{2n}{n}x^n-4\sum_{n\geq 0}\binom{2n}{n}x^{n+1}\\
&=1+\sum_{n\geq 1}\left(\bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find the inverse of $A$ Let
$$
A=
\begin{bmatrix}
n & n_1 & n_2 & \cdots & n_s & 0 \\
n_1 & n_1 & 0 & \cdots & 0 & 1 \\
n_2 & 0 & n_2 & \cdots & 0 & 1 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
n_s & 0 & 0 & \cdots & n_s & 1 \\
0 & 1 & 1 & \cdots & 1 & 0
\end{bmatrix},
$$
where $n=\sum_{i=1}^sn_i$.
My q... | Partial Answer: Here's a strategy you might find helpful.
Let $M$ denote the submatrix of $A$ given by
$$
M = \pmatrix{n & n_1 & n_2 & \cdots & n_s \\
n_1 & n_1 & 0 & \cdots & 0 \\
n_2 & 0 & n_2 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n_s & 0 & 0 & \cdots & n_s}.
$$
we see that $M$ can be exp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3810402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Expected number of days before all magical seeds become apple trees This is a question that I came across recently.
At the end of day $0$, $6$ magical seeds are planted. On each day following it, each seed has a chance to magically transform into an apple tree with a probability of $\frac{1}{2}$. The outcomes of seeds ... | Let $X_i$ the time of growth of the $i$th seed, $X$ the time until all grow.
$X_i\sim Geom (0.5)$ so $\Pr(X_i\leq t)=1-0.5^t$ for $t=1,2,\ldots$.
$X$ is the maximum of $X_1,\ldots,X_6$ so $\Pr(X\leq t)=\prod\limits_{i=1}^6\Pr(X_i\leq t)=(1-0.5^t)^6$ and $\Pr(X=t)=\Pr(X\leq t)-\Pr(X\leq t-1)=(1-0.5^t)^6-(1-0.5^{t-1})^6$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3810512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Limit of multivariable function $f(x,y) = {(x^2+y^2)}^{x^2y^2}$ $$f(x,y) = {(x^2+y^2)}^{x^2y^2}$$
I need to find the limit at (0,0) point
I applied the exponent rule and got $$e^{x^2y^2ln(x^2+y^2)}$$
and now with chain rule, I need to find the limit of $${x^2y^2ln(x^2+y^2)}$$
and how? :D
There isn't L'Hôpital's rule fo... | For $0 < x^2+y^2<1$ we have
$$1 \geqslant (x^2+y^2)^{x^2y^2} \geqslant (x^2+y^2)^{\frac{1}{4}(x^2+y^2)^2}$$
and
$$\lim\limits_{x \to 0 \\y \to 0}(x^2+y^2)^{\frac{1}{4}(x^2+y^2)^2}=\lim\limits_{t \to 0+}t^{\frac{1}{4}t^2} = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3812048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Diophantine equation: $x^4+4=py^4$
find all primes p for which $x^4+4=py^4$ is solvable for integers.
My try: I started out with $p=2$ . LHS must be even, thus $(x^4+4)\mod(16)=4$ similarly
$(2y^4)\mod (16)=${$0,2$}. Thus for $p=2$ there are no solutions.
Now I tried factoring i.e $x^4+4=(x^2+2x+2)(x^2-2x+2)=py^4.$ ... | @WhatsUp has the answer, but I wanted to show just for fun how to systematically derive the other case $(X,Y,P)=(0,\pm 1,4)$. It doesn't answer the question because $P=4$ is not prime, but to those who ask "How can we derive the other solution without some kind of trial and error?" well here it is.
Consider the equati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3813217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Number of non negative integer solutions of $x+y+2z=20$
The number of non negative integer solutions of $x+y+2z=20$ is
Finding coefficient of $x^{20}$ in
$$\begin{align}
&\left(x^0+x^1+\dots+x^{20}\right)^2\left(x^0+x^1+\dots+x^{10}\right)\\
=&\left(\frac{1-x^{21}}{1-x}\right)^2\left(\frac{1-x^{11}}{1-x}\right)\\
=&\... | If $z=0$, there are 21 ways $x+y=20$.
If $z=1$, there are 19 ways $x+y=18$.
$$\vdots\hspace{4in}$$
If $z=9$ there are 3 ways $x+y=2$.
If $z=10$ ther is 1 way $x+y =0.$
$$\begin{aligned}\text{number of ways } &= 1+3+\cdots+ 19+21 \\ &= \sum_{k=1}^{11} (2k-1)\\ &=2\sum_{k=1}^{11}k - 11 \\ &= 2\cdot 66 -11 \\ &= 121.\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3813339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Infinitely many solutions of the equation $\frac{x+1}{y}+\frac{y+1}{x} = 4$ Prove that there exists infinitely many positive integer solutions in $(x,y)$ to the equation :
$$\frac{x+1}{y} + \frac{y+1}{x} = 4$$
| $$\frac{x+1}{y} + \frac{y+1}{x} = 4$$
$$x^2+x+y^2+y=4xy$$
$$(x+y)^2+(x+y)=6xy$$
Let $X=x+y$ and $Y=x-y$. Then $X^2+X=\dfrac32\left(X^2-Y^2\right)$ or $(X-1)^2-3Y^2=1$.
That's a Pell equation, which has infinitely many solutions:
$X-1=$$1,2,7,26,97,...$ and $Y=$$0,1,4,15,56,...$.
Click on the blue numbers to see more. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3814407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $(a+b+c) \Big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Big) \leqslant 25$ For $a,b,c \in \Big[\dfrac{1}{3},3\Big].$ Prove$:$
$$(a+b+c) \Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Big) \leqslant 25.$$
Assume $a\equiv \text{mid}\{a,b,c\},$ we have$:$
$$25-(a+b+c) \Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Big) =\dfra... | I found a better estimation
$$ (a+b+c) \Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Big) \leqslant \frac{209}{9}.$$
Equality occur when $a=b=3,\,c=\frac 13$ or $a=b=\frac 13,\,c=3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3817104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Proving $6(x^3+y^3+z^3)^2 \leq (x^2+y^2+z^2)^3$, where $x+y+z=0$ Question :
Let $x,y,z$ be reals satisfying $x+y+z=0$. Prove the following inequality:$$6(x^3+y^3+z^3)^2 \leq (x^2+y^2+z^2)^3$$
My Attempts :
It’s obvious that $x,y,z$ are either one negative and two positive numbers or one positive and two negative num... | Thanks for all the answers you guys provided, and after spending a large amount of time wandering in this question and understanding these answers, I came up with my solution which I think is valid.
By AM-GM:
$x^2+xy+y^2\\=\frac{x(x+y)}{2}+\frac{y(x+y)}{2}+\frac{x^2+y^2}{2}\\ \geq3\cdot \sqrt[3]{\frac{xy(x+y)^2}{4}\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3817541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
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If $a^2+b^2-ab=c^2$ for positive $a$, $b$, $c$, then show that $(a-c)(b-c)\leq0$
Let $a$, $b$, $c$ be positive numbers. If $a^2+b^2-ab=c^2$. Show that
$$(a-c)(b-c)\leq0$$
I have managed to get the equation to $(a-b)^2=c^2-ab$, but I haven't been able to make any progress.
Can someone help me?
| Form condition we get $c = \sqrt{a^2-ab+b^2},$ therefore
$$(a-c)(b-c)=a^2+b^2-c(a+b)=a^2+b^2-(a+b)\sqrt{a^2-ab+b^2}$$
$$=-\frac{ab(a-b)^2}{a^2+b^2+(a+b)\sqrt{a^2-ab+b^2}} \leqslant 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3820511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Solve differential equation: $y' = \frac{y^2}{x^3} + 2\frac{y}{x} - x$ I need to solve this differential equation: $y' = \frac{y^2}{x^3} + 2\frac{y}{x} - x$.
My attempt
I found that $y = x^2$ is a solution. Then I tried to put $y = x^2f(x)$, and solved this way:
$$2xf(x) + x^2f'(x) = xf^2(x) + 2xf(x) - x \implies x^2f'... | With a slightly different approach,
$$\frac{y'}x=\left(\frac y{x^2}+1\right)^2-2$$
indeed hints the change of variable
$$y=x^2z,$$ which makes the equation separable,
$$\frac{2xz+x^2z'}x=(z+1)^2-2,$$
$$\frac{z'}{z^2-1}=\frac1x$$
and
$$\text{artanh } z=\log Cx,$$
$$z=\tanh \log Cx.$$
$$y=x^2\sqrt{\frac{1+Cx}{1-Cx}}.$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3821061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Finding all real $(a,b,c)$ satisfying $a+b+c=\frac1{a}+\frac1{b}+\frac1{c}$ and $a^2+b^2+c^2=\frac1{a^2}+\frac1{b^2}+\frac1{c^2}$ I have been trying to find my error in the following question for a while, but am yet to succeed:
Find all triples $(a,b,c)$ of real numbers that satisfy the system of equations:
$$\begin{a... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.
Thus, our conditions give: $$uw^3=v^2$$ and $$(9u^2-6v^2)w^6=9v^4-6uw^3.$$
Now, if $v^2=0$ so $u=0,$ which gives $$ab-(a+b)^2=0$$ or $$a^2-ab+b^2=0,$$ which gives $a=b=0,$ which is impossible.
Thus, $u\neq0$, $v^2\neq0$ and $w^3\neq0$, which ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3822452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Determine all zeros of the polynomial $X^4 - 2X^3 - X^2 + 2X + 1 \in \mathbb C[X]$. This is Exercise 14 on page 110 of Analysis I by Amann and Escher.
The hint given is as follows: multiply the polynomial by $1/X^2$ and substitute $Y = X - 1/X$.
If I attempt this, I get the following:
\begin{align*}
X^4 - 2X^3 - X^2 + ... | $y^2=x^2-2+1/x^2$, so the expression after multiplying by $1/x^2$ can be rewritten
$$y^2-2y+1=(y-1)^2=0$$
So $y=1$, or $x-1/x=1$ or $x^2-x-1=0$. Solving this quadratic gives $x=\frac{1\pm\sqrt5}2$, and it is easy to check that $x=0$ doesn't satisfy the original equation, so we have found all solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3822979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing $\lim_{x\to-5}\frac{x^2+2x-15}{|x+5|}$ The problem is: $$\lim_{x\to-5}\frac{x^2+2x-15}{|x+5|}$$
I factored the numerator to get: $\frac{(x-3)(x+5)}{|x+5|}$
How do i solve the rest?
| $$\frac{(x-3)(x+5)}{ |x+5|} = (x-3)\frac{(x+5)}{ |x+5|}=(x-3)\text{sign}(x+5)$$
As $x \to -\infty$, then $\text{sign}(x+5)=-1$, So $(x-3)\text{sign}(x+5)=-x+3 \to \infty$.
Addition.
If we take limit point $x=-5$, then of course we come from infinite limit to non existence of limit. First multiplier $x-3 \to -8, x \to -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3824902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$ \lim_{x \to 0}x \tan (xa+ \arctan \frac{b}{x})$ I have to evaluate the following limit
$$ \lim_{x \to 0}x \tan (xa+ \arctan \frac{b}{x})$$
I tried to divide tan in $\frac{sin}{cos}$ or with Hopital but I can't understand where I'm making mistakes.
The final result is:
$\frac{b}{1-ab}$ if $ab \ne 1$
$- \infty$ if $ab... | We have that by $\arctan x+\arctan \frac 1x= \pm \frac \pi 2$
$$x \tan \left(xa+ \arctan \frac{b}{x}\right)=x \tan \left(\pm\frac\pi 2+xa-\arctan \frac{x}{b}\right)=$$
$$=-\frac{x}{ \tan \left(xa-\arctan \frac{x}{b}\right)}=-\frac{x}{ xa-\arctan \frac{x}{b}} \frac{xa-\arctan \frac{x}{b}}{ \tan \left(xa-\arctan \frac{x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3828453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
$\lim_{(x,y,z) \to (0,0,0)} \frac{xyz}{x^2+y^2+z^2}=0$ How to show that
$$\lim_{(x,y,z) \to (0,0,0)} \frac{xyz}{x^2+y^2+z^2}=0,$$ where $x,y,z>0$.
My attempt:
$$||(x,y,z)|| < \delta \implies |x|, |y|, |z| < \delta$$
$$\left | \frac{xyz}{x^2+y^2+z^2} \right | < \left | \frac{xyz}{x^2}\right | < \frac{\delta^3}{x^2}.$$
N... | Note that $|x|\le \sqrt{x^2+y^2+z^2}$, $|y|\le \sqrt{x^2+y^2+z^2}$, and $|z|\le \sqrt{x^2+y^2+z^2}$.
Hence, given $\varepsilon>0$,
$$ \left| \frac{xyz}{x^2+y^2+z^2}\right|\le \sqrt{x^2+y^2+z^2}<\varepsilon$$
whenever $\sqrt{x^2+y^2+x^2}<\delta=\varepsilon$.
And we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3829296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Can I solve $\lim_{(x,y)\to\ (0,0)} \frac{x^2y^2}{x^2+x^2y^2+y^2}$ by converting to polar coordinates? Is it correct to solve this problem like this?
$$\lim_{(x,y)\to\ (0,0)} \frac{x^2y^2}{x^2+x^2y^2+y^2} $$
$$\lim_{(x,y)\to\ (0,0)} \frac{1}{1+\frac{x^2+y^2}{x^2y^2}}$$
$$\frac{x^2+y^2}{x^2y^2}=\lim_{r\to\ 0} \frac{1}{r... | You don't have to make it more complicated. Simply observe that $x^2 \le x^2+x^2y^2+y^2\implies \dfrac{x^2y^2}{x^2+x^2y^2+y^2} \le y^2\implies \text{limit} = 0$ .
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Minimize $(x+y)(y+z)(z+x)$ given $xyz(x+y+z) = 1$ $x,y,z$ are positive reals and I am given $xyz(x+y+z) = 1$. Need to minimize $(x+y)(y+z)(z+x)$. Here is my approach.
Using AM-GM inequality
$$ (x+y) \geqslant 2 \sqrt{xy} $$
$$ (y+z) \geqslant 2 \sqrt{yz} $$
$$ (z+x) \geqslant 2 \sqrt{zx} $$
So, we have
$$ (x+y)(y+z)(z... | Setting $t=xyz(x+y+z),$ then $xy+yz+zx \geqslant \sqrt{3t}=\sqrt 3.$ Using known inequality
$$(x+y)(y+z)(z+x) \geqslant \frac{8}{9}(x+y+z)(xy+yz+zx).$$
We have
$$(x+y)(y+z)(z+x) \geqslant \frac{8}{9}(x+y+z)(xy+yz+zx)$$
$$\geqslant \frac{8}{9} \cdot \sqrt{3(xy+yz+zx)} \cdot (xy+yz+zx) = \frac{8}{9}\sqrt{3(xy+yz+zx)^3}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3829530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\iint_{D} y^{3}\,dx\,dy$ $$\iint_{D} y^{3}\,dx\,dy$$ where $D$ is the domain between $x^2 + y^2 = 6$ circle and the parabola $y=x^2$
Edit: Also, I got the intersection of the curves $(\sqrt2,2)$ and $(\sqrt-3, -3)$
I draw the curves but I got confused at the bounds of the integrals.
| Double-check your work finding the points of intersection, as they should be $\left(-\sqrt{2},2\right)$ and $\left(\sqrt{2},2\right)$. These give $x=-\sqrt{2}$ and $x=\sqrt{2}$ as your lower and upper bounds for $x$, respectively.
Looking at the plot provided, it is clear that the lower and upper bounds for $y$ will b... | {
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"timestamp": "2023-03-29T00:00:00",
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Choose four integers from six integers such that $1\cdot \alpha_1 + 2\cdot \alpha_2 + 3\cdot \alpha_3 + 4\cdot \alpha_4 = 5\cdot \beta$. Could you help me with the following task?
Prove that for four numbers form a set $S = \{ \alpha_1, \alpha_2, \alpha_3, \alpha_4, \alpha_5, \alpha_6 \}$ of distinct integers it is po... | The key observation is that for every integer $k$ and every choice of $x_0,x_1,x_2,x_3,x_4\in S$ you have
$$k\sum x_i+\sum ix_i\equiv \sum ix_{i-k}\pmod{5},$$
if you take the indices mod $5$. So it suffices to find five elements $x_i\in S$ such that
$$\sum x_i\not\equiv0\pmod{5},$$
as then this sum is coprime to $5$, a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3833733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If complex number $a, b, c, d,$ and $|a|=|b|=|c|=|d|=1$, why $|a(c+d)|+|b(c-d)|\leq 2\sqrt{2}$? If we have 4 complex number $a, b, c, d,$ and $|a|=|b|=|c|=|d|=1$, So, how to prove that $|a(c+d)|+|b(c-d)|\leq 2\sqrt{2}$？
I try to separate $|a(c+d)|+|b(c-d)|$ to $|a||(c+d)|+|b||(c-d)|$ than I get $|(c+d)|+|(c-d)|$.... | $$|a(c+d)| +|b(c-d)|=|a||c+d| +|b||c-d|=|c+d| +|c-d|=\sqrt{(\cos \alpha +\cos \phi )^2 +(\sin\alpha +\sin\phi )^2 } +\sqrt{(\cos \alpha -\cos \phi )^2 +(\sin\alpha -\sin\phi )^2 }=\sqrt{2+ 2\cos \alpha \cos\phi +2\sin\alpha \sin \phi} +\sqrt{2- 2\cos \alpha \cos\phi -2\sin\alpha \sin \phi} =\sqrt{2+2\cos (\alpha -\phi ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3834332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Determine convergence of the sequence $x_0=1 , x_{n+1}=x_n (1+ 2^{-(n+1)})$ I want to check if the following sequence converges:
$$x_0=1 , x_{n+1}=x_n \left(1+ \frac{1}{2^{n+1}}\right)$$
I proved the sequence is increasing :
$\cfrac{x_{n+1}}{x_n}=1+ \cfrac{1}{2^{n+1}} \gt 1$
Now I should prove it is bounded above. le... | The Limit Exists
Cross-multiplication and comparison shows that
$$
\frac{1+\frac1{2^{k-1}}}{1+\frac1{2^k}}\le1+\frac1{2^k}\le\frac{1-\frac1{2^k}}{1-\frac1{2^{k-1}}}\tag1
$$
Then telescoping products lead to
$$
\underbrace{\prod_{k=1}^\infty\frac{1+\frac1{2^{k-1}}}{1+\frac1{2^k}}}_2
\le\prod_{k=1}^\infty\left(1+\frac1{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3836064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Definite Integral involving logarithm and tangent function
Show that
$$\mathcal{I}:=\int_0^{\frac{\pi}2} \log |1-a^2\tan^2\theta| d\theta= \pi\log\sqrt{a^2+1}.$$
I tried to use the substitution $\tan\theta=z$, to get that
$$\mathcal{I}:=\int_0^{\infty} \frac{\log|1-a^2z^2|}{z^2+1}dz$$
This integral is quite similar t... | Note that we'll need to take the Cauchy Principal Value. Observe that by DUTIS,
\begin{align*}
I(a) &= \int_0^{\frac{\pi}{2}} \ln\left|1-a^2\tan^2\theta\right|\,\mathrm{d}\theta\\
I'(a) &= \int_0^{\frac{\pi}{2}} \frac{\partial}{\partial a} \ln\left|1-a^2\tan^2\theta\right|\,\mathrm{d}\theta\\
&= \int_0^{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3839712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Show $\int_0^\infty \frac{\ln^2x}{(x+1)^2+1} \, dx=\frac{5\pi^3}{64}+\frac\pi{16}\ln^22$ Tried to evaluate the integral
$$I=\int_0^\infty \frac{\ln^2x}{(x+1)^2+1} \, dx$$
and managed to show that
\begin{align}
I &= \int_0^1 \frac{\ln^2x}{(x+1)^2+1} \, dx + \int_0^1 \frac{\ln^2x}{(x+1)^2+x^2} \, dx\\
&= \int_0^1 \frac{\... | Here is a solution based on real methods. Note that $\frac{1}{(x+1)^2+1}=\frac{x^2-2x+2}{x^4+4}$. Then
\begin{align}
&\int_0^\infty \frac{\ln^2x}{(x+1)^2+1}dx \\
= & \int_0^\infty {\frac{x^2\ln^2x }{x^4+4}}\overset{x^2=1/t^2}{dx}
-2\int_0^\infty {\frac{x\ln^2x }{x^4+4}} \overset{x^2=2t}{dx}
+ 2\int_0^\infty { \frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 0
} |
Limit of sequence involving reciprocal of the sum of the first squares Let $(x_n)$ a sequence defined by $x_n=\sum_{k=1}^n \frac{1}{1^2+2^2+\dots+k^2}.$ What is $\lim_{n\to \infty}x_n$?
My idea: $1^2+\dots+k^2=\frac{k(k+1)(2k+1)}{6}<\frac{k(k+1)(k+2)}{3}$. Thus
$$x_n>3\sum_{k=1}^n\frac{1}{k(k+1)(k+2)}=3\left(\frac 12 -... | As you properly wrote$$x_n=\sum_{k=1}^n \frac{1}{1^2+2^2+\dots+k^2}=\sum_{k=1}^n \frac{6}{k (k+1) (2 k+1)}$$ Using partial fraction decomposition, as @Ben Grossmann already did it,
$$x_n=\sum_{k=1}^n \frac{6}{k}+\sum_{k=1}^n \frac{6}{k+1}-\sum_{k=1}^n \frac{24}{2 k+1}$$
$$x_n=(6 H_n)+(6 H_{n+1}-6)-(12 H_{n+\frac{1}{2}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
expected value rule of Uniform random variable Let be a uniform random variable on the range $\{−1,0,1,2\}$. Let $=^4$. Use the expected value rule to calculate $\mathbb{E}[Y]$.
Why isn’t the answer $1/4 \cdot(-1)^4 +1/4\cdot (0)^4+ 1/4 \cdot(1)^4 + 1/4 \cdot(2)^4$
| The rv X is the following
$$X =
\begin{cases}
\frac{1}{4}, & \text{if $x=-1$} \\
\frac{1}{4}, & \text{if $x=0$} \\
\frac{1}{4}, & \text{if $x=1$} \\
\frac{1}{4}, & \text{if $x=2$}
\end{cases}$$
If you transform $Y=X^4$ when $X=\pm1$ you get that $Y=1$ in both cases, so you have
$$Y =
\begin{cases}
\frac{1}{4}, & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Show that the solutions to the equation $ax^2 + 2bx + c =0$ are given by $x = -\frac{b}{a} \pm \sqrt{\frac{b^2-ac}{a^2}}$
Show that the solutions to the equation $ax^2 + 2bx + c =0$ are given by $x = -\frac{b}{a} \pm \sqrt{\frac{b^2-ac}{a^2}}$
Hint: Start by dividing the whole equation by $a$
At first I have tried so... | As you did, complete the square by first taking out a factor of $a$, where $a\neq 0$ since the equation is a quadratic.
\begin{align}
ax^2+2bx+c&=0 \\
a(x^2+2\frac{b}{a}+\frac{c}{a})&=0 \\
a\left[(x+\frac{b}{a})^2-\frac{b^2}{a^2}+\frac{c}{a}\right]&=0 \\
(x+\frac{b}{a})^2&=\frac{b^2-ac}{a^2} \\
x+\frac{b}{a}&=\pm\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3841751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that $ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$ By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that
$$ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$$
Here's what I've done so far (starting from after expansion):
$\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x... | We have that
$$\cos^6x + \sin^6x =(\cos^2x +\sin^2x)(\cos^4x -\cos^2x\sin^2x +\sin^4x)=$$
$$=\cos^4x -\cos^2x\sin^2x +\sin^4x$$
and
$$\cos^4x -\cos^2x\sin^2x +\sin^4x=1-3\cos^2x\sin^2x=1-\frac34\sin^2(2x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 4
} |
$6(m + 1)(n − 1), 6 + (m − 1)(n + 1)$ and $(m − 2)(n + 2)$ are simultaneously perfect cubes.
Find all pairs of integers (m, n) such that the integers $6(m + 1)(n − 1), 6 +
(m − 1)(n + 1)$ and $(m − 2)(n + 2)$ are simultaneously perfect cubes.
I assumed $6(m + 1)(n − 1), 6 +
(m − 1)(n + 1)$ and $(m − 2)(n + 2)$ to be ... | Hint: If I am not mistaken, we have$$m=\frac{24-a^3+6b^3}{12} $$
$$n=\frac{12-a^3+18b^3-12c^3}{12} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $x+y+z=xyz$, prove $\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$
If $x+y+z=xyz$, prove
$\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$
given that $x^2~,~y^2~,~z^2\ne1$
I came across... | $$2x(1-y^2)(1-z^2)=2x-2xy^2-2xz^2+2xy^2z^2$$
$$2xy^2z^2=2yz(x+y+z)=2xyz + 2y^2z+2yz^2$$
$$\implies 2x(1-y^2)(1-z^2)=2x-2xy^2-2xz^2+2xyz+2y^2z+2yz^2$$
$$\implies 2y(1-z^2)(1-x^2)=2y-2yz^2-2yx^2+2xyz+2z^2x+2zx^2$$
$$\implies 2z(1-x^2)(1-y^2)=2z-2zx^2-2zy^2+2xyz+2x^2y+2xy^2$$
Summing up gives $2(x+y+z+3xyz)=8xyz$, which r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 6,
"answer_id": 5
} |
Compute $\sum_{k\geq0}\frac{2^k}{5^{2^k}+1}$ Problem Statement:
Compute:
$$\sum_{k\geq0}\frac{2^k}{5^{2^k}+1}$$
First we note that $\sum_{k\geq0}\frac{2^k}{5^{2^k}+1}< \sum_{k\geq0}\frac{2^k}{5^{2^k}}< \sum_{k\geq0}\frac{2^k}{5^{k}}< \infty$, so that this sum converges. We can now do the following manipulations:
\begin... | This is a beautiful sum. Note:
$$\frac 1 {n-1} -\frac1{n+1} =\frac {2}{n^2-1}$$
Thus:
$$\begin{align}\frac 1 {n-1} -\frac1{n+1}-\frac {2}{n^2+1} &=\frac {2}{n^2-1}-\frac {2}{n^2+1}\\& = \frac {2^2}{n^{2^2}-1}\\\frac 1 {n-1} -\frac1{n+1}-\frac {2}{n^2+1}-\frac {2^2}{n^{2^2}+1} &= \frac {2^2}{n^{2^2}-1}-\frac {2^2}{n^{2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3847562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Question about complex integrals I have two questions about integrals.
1.) $$\frac{1}{2\pi i} \int_{\gamma} \frac{dw}{\sin{\frac{1}{w}}}$$ where $\gamma$ is the circle $|w| = \frac{1}{5}$.
2.) $$\int_{\gamma_{a}} \frac{z^{2} + e^{z}}{z^{2}(z-2)}dz$$
where $\gamma_{a}$ is the positively oriented circle $|z| = a, a >0, a... | For the first problem, we enforce the transformation $z\mapsto \frac1w$ to obtain
$$\oint_{|z|=1/5}\frac1{\sin(1/w)}\,dw=\oint_{|z|=5}\frac1{\sin(z)}\frac1{z^2}\,dz\tag1$$
There are 3 poles of $\frac{1}{z^2\sin(z)}$ that are inside the circle $|z|=5$. These include a third order pole at $z=0$ and simple poles at $z=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3851233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Understand how to evaluate $\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ We are given this limit to evaluate:
$$\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
In this example, if we try substitution it will lead to an indeterminate form $\frac{0}{0}$. So, in order to evaluate this limit, we can multiply this exp... | In this case, the rationalisation of the denominator not only simplifies the expression a bit, but reveals a rate of change, which makes the determination of the limit quite fast:
$$\frac{(\sqrt{6-x}-2)\dfrac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}}{2-x}=\underbrace{\frac{\sqrt{6-x}-2}{2-x}}_{\textstyle\downarrow\atop \text... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3858398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 4
} |
The equation $x^4-x^3-1=0$ has roots $\alpha,\beta,\gamma,\delta$. Find $\alpha^6+\beta^6+\gamma^6+\delta^6$ The equation $x^4-x^3-1=0$ has roots $\alpha,\beta,\gamma,\delta$.
By using the substitution $y=x^3$, or by any other method, find the exact value of
$\alpha^6+\beta^6+\gamma^6+\delta^6$
This is a problem from F... | Making the given substitution yields $y^{4/3}=y+1$, or $y^4=(y+1)^3$ or $y^4-y^3-3y^2-3y-1=0$. The roots of this polynomial are $\alpha^3,\beta^3,\gamma^3,\delta^3$, so we can work out
$$\alpha^6+\beta^6+\gamma^6+\delta^6=(\alpha^3+\beta^3+\gamma^3+\delta^3)^2-2(\alpha^3\beta^3+\cdots)=1^2-2(-3)=7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
if $\frac{\sin(2a-b)}{\sin b}=1+x$ prove that...
if $\frac{\sin(2a-b)}{\sin b}=1+x$ prove that $$\frac{\cos b}{ \cos a}=1+\frac{1}{2}x\tan^2 a$$ approximately .(Take $x$ to be very small)
Attempt by given eq. $$\sin 2a \cos b-\sin b \cos 2a=\sin b +x\sin b$$
or $$2\sin a \cos a \cos b -\sin b (1-2\sin^2 a)=\sin b +x... | $\begin{align}\frac{\cos b}{\cos a} &= \frac{\sec^2a}{2}\left[1-2\sin^2a+(1+x)\right] = \frac{\sec^2a}{2} - \tan^2a +\frac{1+x}{2}\sec^2a
\\&= \left(1+\frac x2\right)\sec^2a -\tan^2a = \left(1+\frac x2\right)(1+\tan^2a)-\tan^2a = \left(1+\frac x2\right)+\frac x2\tan^2a \\&\approx1+\frac x2\tan^2a\end{align} $
As $x$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3863627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"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.