Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove that there are no integer solutions to: $x^4+6x^2+1=8y^4$
Prove that there are no integer solutions to:
$$x^4+6x^2+1=8y^4$$
where $x>1$.
My attempts:
Let $x^2=u, y^2=v$
$$u^2+6u+(1-8v^2)=0$$
$$\Delta=36-4(1-8v^2)=w_0^2$$
$$32v^2+32=(4w_1)^2$$
$$2v^2+2=w_1^2=(2w_2)^2$$
$$v^2+1=2w_2^2$$
$$v=w_2=1\implies x=y=1$$
... | $x^4 + 6x^2 + 1 = (x^2 + 3)^2 + 1 - 9 = 8y^4 \implies (x^2 + 3)^2 = 8 + 8y^4 = (2y^2 + 2)^2 + (2y^2 - 2)^2$
So the three numbers $a = 2y^2 + 2, b = 2y^2 - 2, c = x^2 + 3$ must be a pythagorean triple and for this to be true, $a,b,c$ should be in the form $m^2 - n^2, 4mn, m^2 + n^2$ which is not the case.
| {
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"url": "https://math.stackexchange.com/questions/4563040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Calculate $\sum_{n=2}^{\infty}\left (n^2 \ln (1-\frac{1}{n^2})+1\right)$ I am interested in evaluating
$$\sum_{n=2}^{\infty}\left (n^2 \ln\left(1-\frac{1}{n^2}\right)+1\right)$$
I am given the solution for the question is $\,\ln (\pi)-\frac{3}{2}\,.$
$$\sum_{n=2}^{\infty}\left(n^2\ln\left(\!1\!-\!\frac{1}{n^2}\!\right)... | Here's an elementary proof, assuming that we're allowed to use Stirling's formula:
$$\ln (n!) = \sum_{k=2}^n \ln(k) = n\ln n -n + \frac 1 2 \ln n + \frac 1 2 \ln 2\pi + \mathcal O(\frac 1 n)$$
Let us study the partial sums, denoting $u_k=k^2 \ln \left(1-\frac 1 {k^2} \right) + 1$:
$$\begin{align}
\sum_{k=2}^n u_k &=\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4565763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 0
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How to Evaluate the Integral? $\int_{0}^{1}\frac{\ln\left( \frac{x+1}{2x^2} \right)}{\sqrt{x^2+2x}}dx=\frac{\pi^2}{2}$ I am trying to find a closed form for
$$
\int_{0}^{1}\ln\left(\frac{x + 1}{2x^{2}}\right)
{{\rm d}x \over \,\sqrt{\,{x^{2} + 2x}\,}\,}.
$$
I have done trig substitution and it results in
$$
\int_{0}^{1... | You can "simplify" the problem using a first integration by parts to get rid of the logarithm
$$u=\log \left(\frac{x+1}{2 x^2}\right)\quad \implies \quad du=-\frac{x+2}{x^2+x}$$
$$dv=\frac{1}{\sqrt{x^2+2 x}}\quad \implies \quad v=2 \tanh ^{-1}\left(\sqrt{\frac{x}{x+2}}\right)$$
Using the bounds $u\,v=0$ and we are left... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
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Find the asymptotes of $xy^2 - y^2 - x^3 = 0$ I am lost with this question because there is no 3rd degree $y$ term and no 2nd degree $x$ term.
Can $y - x - 1/2 = 0$ and $y + x + 1/2 = 0$ be its asymptotes?
or $y - x - 1/3 = 0$ and $y + x + 1/2 = 0$
| Now, $y=\sqrt{\frac{x^3}{x-1}}$. Apart from the vertical asymptote $x=1$, we have two oblique asymptotes (OAs) in the form $y=ax+b$. To find these let us make the following limit zero:
$\begin{align}
\lim_{x\rightarrow \pm\infty}\sqrt{\frac{x^3}{x-1}}-(ax+b)&=\lim_{x\rightarrow \pm\infty}\frac{\frac{x^3}{x-1}-(ax+b)^2}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Volume of the region determined by $x^2 + y^2 + z^2 \leq 10, z \geq 2$ This is question 13 on page 294 of Vector Calculus by Marsden and Tromba.
Find the volume of the region determined by $x^2 + y^2 + z^2 \leq 10, z \geq 2.$
I have attempted it as follows.
The region can be described by using spherical polar coordin... | From $z=r\cos\phi=2$ you get $r=\frac{2}{\cos\phi}$. The bounds of the first integral which is with respect to $r$ must have been $\frac{2}{\cos\phi}\leq r\leq\sqrt{10}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4574446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $xyz = x+y+z$ for $x,y,z > 0$, then $\sqrt{1+x^2} + \sqrt{1+y^2} + \sqrt{1+z^2} \ge 6$
If $x,y,z \in \Bbb R_{>0}$ satisfy $xyz = x+y+z$, prove that $$\sqrt{1+x^2} + \sqrt{1+y^2} + \sqrt{1+z^2} \ge 6$$
We can express $1+x^2$ as
$$1+x^2 = (1-xy)(1-xz) = \frac{(y-xyz)(z-xyz)}{yz} = \frac{(x+z)(x+y)}{yz}$$
since $x + ... | We have: $$xyz = x+y+z \ge 3\sqrt[3]{xyz} \Longrightarrow xyz \ge 3^{3/2}$$
and for $a\in \{x,y,z \}$:
$$\sqrt{1+a^2} = \sqrt{1+\frac{a^2}{3}+\frac{a^2}{3}+\frac{a^2}{3}}\ge\sqrt{4\sqrt[4]{\frac{a^6}{3^3}}}=\frac{2}{3^{3/8}}\cdot a^{3/4}$$
Then
$$\begin{align}
LHS &\ge \frac{2}{3^{3/8}}\cdot \left(x^{3/4}+y^{3/4}+z^{3/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4575712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integral Involving Harmonic Numbers: $\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx$ (Motivation) In an attempt to answer this question, I got stuck on evaluating a certain integral. I have made up the following conjecture:
$$\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx = \frac{... | This is not a complete solution but an extended comment.
I was particularly interested to find out what makes the lower integration border $\sqrt{3}$ so special as a prerequisite for an appreciable simplification of the final expression of the integral.
Hence I started with the same decomposition of our integral into t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4576184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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Show $a^2+b^2+c^2 \equiv 0 \pmod {3}$ if $a, b, c$ are *not* multiples of $3$. I was given the following problem:
Show $a^2+b^2+c^2 \equiv 0 \pmod {3}$ if $a, b, c$ are not multiples of $3$.
I would like a. verification of my proof (I self-study); b. alternative proofs, to enrich my appreciation of the problem. Here... | You're not wrong but simply observe that for any $a$ such that $a \not \equiv 0$ mod $3$, then we have that $a^2 \equiv 1$ mod $3$. This is easy enough to prove and follows from your lemma for instance.
Then $x^2, y^2, z^2 \equiv 1$ mod $3$ each, and so $x^2 + y^2 + z^2 \equiv 0 $ mod $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4576791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How is $\ln (x-2) - \frac{1}{2} \ln (x-1) = \frac{1}{2} \ln \frac{(x-2)^2}{x-1}$
How is $\ln (x-2) - \frac{1}{2} \ln (x-1) = \frac{1}{2} \ln \frac{(x-2)^2}{x-1}$
Can someone enlighten me on how is these 2 actually equals and the steps taken? the left hand side is actually the answer for $\int \frac{x}{2 (x-2)(x-1)} d... | $$\ln (x-2) - \frac{1}{2} \ln (x-1)$$
$$=\frac{2\ln (x-2) -\ln (x-1)}{2}$$
$$=\frac{\ln (x-2)^2 -\ln (x-1)}{2}$$
$$=\frac{\ln\frac{ (x-2)^2}{ x-1}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4578029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Find the volume between the regions $x^2 + y^2+ z^2 = 4$ and $x = 4-y^2$ I want the volume of the sphere $x^2 + y^2 + z^2 = 4$ from $x = 0$ to $x = 4-y^2$. The integral that gives this volume is
$$\int\limits_{-2}^2 \int\limits_0^{4-y^2} \int\limits_{-\sqrt{4-x^2-y^2}}^{\sqrt{4-x^2-y^2}}\ 1\ dz\ dx\ dy$$
I don't find ... | Nobody commented on my scalar in the comments section so far.
The projection of the solid on the $xy$-plane is determined by curves, $x^2+y^2=4$, $x=4-y^2$ and $x=0$. The common solution of the circle and the parabola gives $(1,\pm \sqrt{3})$. Hence the quarter volume integral due to symmetries w.r.t $xy$-plane and $x$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4578958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there any better way of finding the required value If $$z=\cos\theta+i\sin\theta$$ find the value of $$\frac{1+z}{1-z}$$
The solution that I have is this
$$z=\cos\theta+i\sin\theta \implies$$
$$\frac{1+z}{1-z}=\frac{1+(\cos\theta+i\sin\theta)}{1-(\cos\theta+i\sin\theta)}=\frac{(1+\cos\theta)+i\sin\theta}{(1-\cos\th... | I don't think the question is really accurate in the first place, the value of $\frac{1+z}{1-z}$ can just be determined by plugging in the expression for $z$. To bring it in the final form is a matter of taste, I believe it mostly comes from the experience and knowing which result you want to get. Multiplying by $1$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4580285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show that all solutions of $z^5 - z + 16 = 0$ satisfy $1 \lt |z| \lt 2$ I have to show that all solutions of $z^5 - z + 16 = 0$ satisfy $1 \lt |z| \lt 2$.
My attempt: By using Euler's formula I can rewrite the equation into $r^5e^{5i\phi} - re^{i\phi} = -16$ and then rewrite this into $$r^5 \cos(5\phi) - r \cos(\phi) +... | We have $|z^5-z|=|-16|=16.$
If $|z|=2$ then $|z^5-z|\ge |z^5|-|z|=|z|^5-2=30>16.$
If $|z|>2$ then $|z-1|\ge |z|-|1|>1$ and $|z|^4>0,$ so $|z^5-z|=|z|^4\cdot |z-1|> |z|^4\cdot 1 > 2^4=16.$
If $|z|\le 1$ then $|z^5-z|\le |z^5|+|z|=|z|^5+|z|\le 1^5+1=2<16.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4582084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why can't I use the form $\frac1{x^2+1}$ from the derivative of $\arctan(x)$ to convert the integral form in this situation? In the question that solves the integral $\displaystyle\int\frac1{6x^2 + 36x + 78} \,\mathrm{d}x$, I first tried to solve it by changing the denominator in a form of $\dfrac1{x^2 + 1}$ to apply $... | A-ha! I think I found it.
You reduce your integral to
$$ \int \frac{1}{6x^2+36x+78} \ dx = \frac{1}{24} \int \frac{1}{\left( \frac{x+3}{2} \right)^2 + 1} \ dx $$
and wish to use $u = \frac{x+3}{2}$ so that $du = \frac{1}{2} \ dx.$ This makes $dx = 2 \ du$. There's your factor of two!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4585030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Find the value of $\int_0^1f(x)dx$
If $$f(x)=\binom{n}{1}(x-1)^2-\binom{n}{2}(x-2)^2+\cdots+(-1)^{n-1}\binom{n}{n}(x-n)^2$$
Find the value of $$\int_0^1f(x)dx$$
I rewrote this into a compact form.
$$\sum_{k=1}^n\binom{n}{k}(x-k)^2(-1)^{k-1}$$
Now,
$$\int_0^1\sum_{k=1}^n\binom{n}{k}(x-k)^2(-1)^{k-1}dx$$
$$=\sum_{k=1}^... | Solution 1. Here is another approach. Consider the shift operator $\Delta$ defined for functions on $\mathbb{R}$ by
$$ \Delta f(x) = f(x-1). $$
Then
\begin{align*}
\sum_{k=1}^{n} (-1)^{k-1} \binom{n}{k} (x-k)^2
&= \sum_{k=1}^{n} (-1)^{k-1} \binom{n}{k} \Delta^k x^2 \\
&= [\operatorname{id} - (\operatorname{id} - \Delta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4600131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
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Evaluation of $~\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta$ $$
I:=\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta
$$
My tries
$$\begin{align}
s&:=\sin\theta\\
c&:=\cos\theta\\
I&=\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\the... | $$I=\int_0^{2\pi}\frac{\cos^2\theta-\sin^2\theta}{\cos^4\theta+\sin^4\theta}d\theta$$
Substitute $\theta→\pi-\theta$.
$$I=\int_{-\pi}^\pi\frac{\cos^2\theta-\sin^2\theta}{\cos^4\theta+\sin^4\theta}d\theta$$
Since the integrand is even, we can use even-odd properties.
$$I=2\int_{0}^\pi\frac{\cos^2\theta-\sin^2\theta}{\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4603090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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On generalizing $\frac{17-\sqrt{17}}8 =\sin^2(t)+\sin^2(2t)+\sin^2(4t)+\sin^2(8t)$? Given $\color{blue}{t = 2\pi/p}$ for the appropriate prime $p=4m+1$.
I. Sine
$$\begin{align}
\frac{5+\sqrt{5}}8 &=\sin^2(t)\\
\frac{13+\sqrt{13}}8 &=\sin^2(t)+\sin^2(3t)+\sin^2(4t)\\
\frac{17+\sqrt{17}}8 &=\sin^2(3t)+\sin^2(5t)+\sin^2(... | The ones that have all even or all odd quadratic residues relative to p, will do the trick.
The proof for this lies is polygon isomorphisms. The numbers are invariant if taken to an e power (ie an even quadratic), but not an odd power. So you end up with two numbers a, b whose sum would give p, and the difference is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4603836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Where did I go wrong with this integration? $$\int_0^{\frac {1}{\sqrt 3}} \frac {1}{\sqrt {2-3x^2}}$$
$$\int_0^{\frac {1}{\sqrt 3}} \frac {1}{\sqrt {2(1-\frac32x^2)}}$$
$$\frac 1{\sqrt 2}\int_0^{\frac {1}{\sqrt 3}} \frac {1}{\sqrt {1-\frac32x^2}}$$
$$\bigg(\frac 1{\sqrt 2}\sin^{-1}{\sqrt {\frac 32 x}}\bigg)\bigg|_0^\fr... | In case you're not required to rely on trigonometric substitutions, we can make an alternative substitution of
$$t = \frac{\sqrt{2-3x^2}-\sqrt2}x \implies x=-\frac{2\sqrt2\,t}{3+t^2} \implies dx = -\frac{2\sqrt2(3-t^2)}{(3+t^2)^2} \, dt$$
Then
$$\begin{align*}
I &= \int_0^{\frac1{\sqrt3}} \frac{dx}{\sqrt{2-3x^2}} \\[1e... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Contest Math Question: simplifying logarithm expression further I am working on AoPS Vol. 2 exercises in Chapter 1 and attempting to solve the below problem:
Given that $\log_{4n} 40\sqrt{3} = \log_{3n} 45$, find $n^3$ (MA$\Theta$ 1991).
My approach is to isolate $n$ and then cube it. Observe:
\begin{align*}
\frac{\l... | $\displaystyle\frac{\ln(40\sqrt{3})}{\ln(4n)} = \frac{\ln(45)}{\ln(3n)}$
$\displaystyle \frac{\ln(3n)}{\ln(4n)}
= \frac{\ln(45)}{\ln(40\sqrt{3})} × \frac{k}{k}
= \frac{\ln(45^k)}{\ln[(40\sqrt{3})^k]}$
Find k, such that ratio inside ln/ln matched
$\displaystyle \frac{3n}{4n} = \left(\frac{45}{40\sqrt{3}}\right)^k = \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4610313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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For $x,y,z∈ℝ^{+}$,without using Hölder's inequality prove that $\frac{x}{\sqrt{x^2+8yz}}+\frac{y}{\sqrt{y^2+8xz}}+\frac{z}{\sqrt{z^2+8xy}}\geq1$.
For $x,y,z∈ℝ^{+}$, prove that $\frac{x}{\sqrt{x^2+8yz}}+\frac{y}{\sqrt{y^2+8xz}}+\frac{z}{\sqrt{z^2+8xy}}\geq1$.
In this question solution used Hölder's inequality, but I a... | Another way.
By AM-GM, C-S and AM-GM again we obtain:$$\sum_{cyc}\frac{x}{\sqrt{x^2+8yz}}=\sum_{cyc}\frac{2x(x+y+z)}{2\sqrt{(x+y+z)^2(x^2+8yz)}}\geq\sum_{cyc}\frac{2x(x+y+z)}{(x+y+z)^2+x^2+8yz}=$$
$$=\sum_{cyc}\frac{2x^2(x+y+z)}{x(x+y+z)^2+x^3+8xyz}\geq\frac{2(x+y+z)^3}{\sum\limits_{cyc}(x(x+y+z)^2+x^3+8xyz}=$$
$$=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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A $n \times n$ linear system Solve the linear system
$
\begin{array}{ c c c c c c c c c c c c c c }
& & x_{2} & + & x_{3} & + & \dotsc & + & x_{n-1} & + & x_{n} & = & 2 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( 1)\\
x_{1} & & & + & x_{3} & + & \dotsc & + & x_{n-1} & + & x_{n} & = & 4 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (... | Let $A_n\in M_n(K)$ be the matrix with $0$ on the diagonal and entry $1$ everywhere else. Then $\det(A)\neq 0$ for $n\ge 2$. In fact, there is an easy formula for the determinant on this site, namely $\det(A)=(-1)^{n-1}\cdot (n-1)$, see here:
Determinant of a matrix with diagonal entries $a$ and off-diagonal entries $b... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $ \frac{x^2}{1+x} + \frac{y^2}{1+y} + \frac{z^2}{1+z} \geq \frac{1}{2} $
Assume that positive numbers a, b, c, x, y, z satisfy $cy+bz =a;
az + cx = b$$bx + ay = c$.
Prove that $ \frac{x^2}{1+x} + \frac{y^2}{1+y} + \frac{z^2}{1+z} \geq \frac{1}{2} $
I've tried appling A.M.-G.M. inequality but it didn't ... | Observations towards a solution
*
*As suggested by Siddharth, show that $ x = \frac{ -a^2 + b^2 + c^2 } { 2bc }$.
*As remarked by Abastro, $x, y, z$ are cosines of angles of the a-b-c triangle. In particular $x, y, z \leq 1$ (though we won't need this).
*In fact, the inequality holds for any triangle (with positi... | {
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"url": "https://math.stackexchange.com/questions/4616087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Sketching an arbitrary ellipse from its parametric equation I was solving a question, in which I was asked to solve the following system of differential equations:
$$
\dot x = 3x + 2y, \quad \dot y = -5x - 3y.
$$
I got the general solution in $\mathbb R^2$ to be
$$
\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix}... | First write $\cos t$ and $\sin t$ in terms of $x$ and $y$:
$$
\begin{pmatrix} x \\ y\end{pmatrix}=\begin{pmatrix} C & D \\ E & F \end{pmatrix}\begin{pmatrix} \cos t \\ \sin t \end{pmatrix}\\
\begin{pmatrix} \cos t \\ \sin t \end{pmatrix}=\begin{pmatrix} C & D \\ E & F \end{pmatrix}^{-1}\begin{pmatrix} x \\ y\end{pmatri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4617870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Find $x$ such that $\sqrt{x+1} - \sqrt{1-x} = 1$ To solve this equation, I started by putting the condition $x\in [-1, 1]$, then squared a few times: $\sqrt{x+1} - \sqrt{1-x} = 1 \iff x + 1 +1-x-2\sqrt{1-x^2} =1 \iff 2\sqrt{1-x^2}=1 \iff 4(1-x^2)=1 \iff 4x^2=3 \iff x=\pm \frac{\sqrt{3}}{2}$
This, however, is not the ri... | Using a trigonometric substituition:
As $-1 \le x \le 1$, then $x = \cos \theta$:
$$\sqrt{1+x} = \sqrt{1-\cos \theta} = \sqrt{2\sin^2 \frac{\theta}{2}} = \sqrt{2}\left|\sin \frac{\theta}{2}\right|$$
$$\sqrt{1+x} = \sqrt{1+\cos \theta} = \sqrt{2\cos^2 \frac{\theta}{2}} = \sqrt{2}\left|\cos \frac{\theta}{2}\right|$$
Then... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integral in two variables I have to prove that
$$\int_D xy\; dxdy=\frac{3}{8}$$
with $D=\{(x,y)\in \mathbb{R}^2: x^2+y^2 \geq 1,\ \frac{x^2}{4}+y^2 \leq 1,\ x \geq 0,\ y \geq 0 \}$.
So I have the intersection of ellipse $\frac{x^2}{4}+y^2 \leq 1$ and the ball $x^2+y^2 \leq 1$.
If I use the sets
$$\begin{eqnarray}
P&=&... | It's better do the integral in Cartesian anyway
$$\int_0^1 \int_{\sqrt{1-y^2}}^{2\sqrt{1-y^2}}xy\:dxdy = \int_0^1\frac{3}{2}\left(y-y^3\right)\:dy = \frac{3}{4} - \frac{3}{8} = \frac{3}{8}$$
Don't always immediately go to polar coordinates whenever you see conic sections. The integrand made this nice since we had odd p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4619032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do we prove $x^6+3x^3+2x^2+x+1 \geq 0$ Question
How do we prove
$$x^6 + 3x^3+2x^2+x+1\geq0$$
My progress
$$x^6+3x^3+2x^2+x+1=(x+1)^2(x^4-2x^3+3x^2-x+1)$$
I appreciate your interest
| Case 1: $x$ is nonnegative.
The last term is greater than or equal to $(x^2-x+1)^2,$ since it is $(x^2-x+1)^2 + x.$ By the Trivial Inequality, both factors are positive or zero, so the product is also positive or zero.
We can also see this directly from the question. It is clear that the $x^6+3x^3+2x^2+x+1 \geq 0,$ sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4619243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the largest/least value of the function $f$ in $\mathbb{R}^2$? What is the largest/least value of the function $f=x^2ye^{-x^2-2y^2}$ in $\mathbb{R}^2$?
We have that $f'_x=y\left(2e^{-x^2-2y^2}x-2e^{-x^2-2y^2}x^3\right)$ and $f'_y=x^2\left(e^{-x^2-2y^2}-4e^{-x^2-2y^2}y^2\right)$. If $x=0$ then $f'_x=f'_y=0$ so $... | The idea is to make a circle(ball) that contains the critical points. By the extreme value theorem there will be a greatest and least value within that circle. If we can show that the boundary points' f-value approaches 0 as the radius get larger then we have shown that for an arbitrary large circle that contains the c... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $\lim_{x\to\infty}(x-x^2 \ln{\frac{1+x}{x}})$ without L'Hopital I've seen the solution to $\lim_{x\to\infty}(x-x^2 \ln{\frac{1+x}{x}})$ using L'Hopital and I was wondering if there's a way to find out the result without it. My initial attempt was outright stupid of me because I tried to substitute the limit of $\... | Use substitution $e^t = \frac{x}{1 + x}$;
$$x - x^2\ln\left(\frac{1+x}{x}\right) \quad \overset{e^t = \frac{x}{1 + x}}{\longrightarrow} \quad \frac{1}{e^{-t} - 1} + \frac{t}{\left( e^{-t} - 1 \right)^2} = \frac{t + e^{-t} - 1}{\left( e^{-t} - 1 \right)^2} \\ \quad \\ \quad \\ = \frac{te^{t} + 1 - e^{t}}{e^{t}\left( e^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4622401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Can $\,9\!\cdot\!10^n+4\,$ be a perfect square? I think $\,9\!\cdot\!10^n+4\,$ can be a perfect square, since it is $0 \pmod 4$ (a quadratic residue modulo $4$), and $1 \pmod 3$ (also a quadratic residue modulo $3$).
But when I tried to find if $\;9\!\cdot\!10^n+4\,$ is a perfect square, I didn’t succeed. Can someone ... | Comment:
Lets try construction such a number. Suppose we have:
$k^2-9\times 10^n=4$
We use following known Pell's equation:
$x^2-Dy^1=1$
For $D=10$ we have $x=19$ and $y=6$ such that:
$19^2-10\times 6^2=1$
multiplying both sides by $2^2$ we get"
$38^2-10\times 12^2=4$
we rewrite this as:
$38^2-9\times(4^2\times 10)=4$
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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In $pqr$ method, if $q+r=4$, show that $p^2-2q+3r\ge2p$
Positive numbers $x$, $y$ and $z$ satisfy $xy+yz+zx+xyz=4$, show that
\[x^2+y^2+z^2+3xyz\ge2(x+y+z).\]
I thought of $pqr$ method, let $\begin{cases}p=x+y+z,\\q=xy+yz+zx,\\r=xyz.\end{cases}$ so we have $q+r=4$, and needed to prove
\[p^2-2q+3r\ge2p.\]
Transform in... | pqr method:
It suffices to prove that
$$p^2 - 2q + 3r - 2p + (4 - q - r) \ge 0$$
or
$$p^2 - 2p + 4 + 2r - 3q\ge 0.$$
Degree three Schur's inequality
yields $p^3 - 4pq + 9r \ge 0$
which results in
$q \le \frac{p^3 + 9r}{4p}$.
It suffices to prove that
$$p^2 - 2p + 4 + 2r - 3\cdot \frac{p^3 + 9r}{4p}\ge 0$$
or
$$p(p-4)^2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How far from home, can my robot roam? It has constant step size, and turns by increasing amounts. A robot's step size is always $1$. Between steps it turns right, by increasing amounts: $\frac{1\pi}{2},\frac{2\pi}{3},\frac{3\pi}{4},\frac{4\pi}{5},...$
What is the robot's maximum distance from the origin?
Here are the... | Identify Euclidean plane $\mathbb{R}^2$ with complex plane $\mathbb{C}$.
For each $k \in \mathbb{Z}_{+}$, let
$u_k = i(-1)^k e^{i \pi H_k}$ where $H_k = \sum\limits_{\ell=1}^k \frac{1}{\ell}$ are the Harmonic numbers.
In terms of $u_k$, the position after $n^{th}$ move, $(x_n,y_n)$, is given by the formula:
$$a_n \stac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4624148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to solve a system of equations over a finite field? I need to solve a system of equations over $\mathbb{Z}_{11}$.
My system is:
$$ \left\{
\begin{array}{l}
2x + 5y + z = 8 \\
7x + 6y + 8z = 10 \\
10x + 3y + 4z = 6
\end{array}
\right.
$$
In matrix form:
$$
\begin{bmatrix}
2 & 5 & 1\\
7 & 6 & 8\\
10 & 3 & 4\\
\end{b... | You made a calculation error when calculating $R_3-10* R_1$. The last row in the matrix should be $[0,0,10|10]$ rather than $[0,0,1|10]$. With this fix, you get that the solution to the system is $x=y=z=1$, which you can easily check actually solves the system.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $7^a+b!=13^c$ over $\mathbb{Z}^{+}$
Solve the following Diophantine equation $7^a+b!=13^c$ over positive
integers.
Clearly $7^1+3!=13^1$ and $7^2+5!=13^2$. Are there any more solutions? Here are some thoughts.
$b!$ must be even, so $b>1$.
Case 1: $c$ is even: let $c=2k$. Then, $b!\equiv13^c\equiv13^{2k}\equiv1... | The Dipohatine equation
$7^a+6=13^c$
has no solutions when $a\ge2$. We prove this with some fairly creative modular arithmetic.
Modulo 49
If $a\ge 2$, then $7^a\overset{\text{forced}}{\equiv}0\bmod49$ and so $13^c\equiv6$. From calculating powers of $13\bmod 49$ we then infer that $c\equiv11\bmod14$. The modulus on th... | {
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Proof that $(x+y)^n = x^n + y^n$ iff. $x = 0 \lor y = 0 \lor x = -y$ for $n$ odd and $\geq 3$ I was going through Calculus by Spivak when in the first chapter I encountered problem 16, which in the end recites
you should know make a good guess as to when $\left(x+y\right)^n = x^n + y^n$
quoting the results of previou... | Like you mentioned, solving $(1+z)^n = 1+z^n$ for $0<z<1$ is easy by binomial or Bernoulli. For when $z<0,$ let $z\to -z$ and for $0<z<1$ and odd $n$, look at:
$$z^n+(1-z)^n = 1.$$ But this cannot happen since:
$$1 = z^n + (1-z)^n < z + (1-z) = 1.$$
| {
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Solve $\log_{4}3x+1,+\log_{3}16x+1=\log_{2}12x+4$ Here is a basic logarithmic problem that I am struggling to solve:
$\log_{4}(3x+1)+\log_{3}(16x+1)=\log_{2}(12x+4)$
Here is my step:
$\log_{4}(3x+1)+\log_{4}(\frac{16x+1}{3})=\log_{2}(12x+4)$
Since the bases are the same now, we can multiply the arguments.
$\log_{4}(3x+... | The issue with your solution, as already noted in the comments, is when you converted $$\log_3(16x+1)$$ to $$\log_4\left(\frac{16x+1}3\right)$$
The log base rule states that:
$$\log_ba=\frac{\log_da}{\log_db}$$
In this case, $$\log_3(16x+1)$$
would convert to $$\frac{\log_4(16x+1)}{\log_43}$$
which is not equal to $$\l... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $(g \circ f)^{-1}(x) = {\left( {\frac{{x - 7}}{2}} \right)^{\frac{1}{3}}}$
If $g\left( x \right) = ax + c$, $f\left( x \right) = {x^b} + 3$, and $(g \circ f)^{-1}(x) = {\left( {\frac{{x - 7}}{2}} \right)^{\frac{1}{3}}}$ what is the value of $a+b+c$?
My approach is as follows
Given $g\left( x \right) = ax + c... | Notice that $$(g\circ f)(x) = 2x^3+7$$
But by definition of $g$ and $f$ we have also $$(g\circ f)(x) =ax^b+3a+c$$
Now if $x=0$ then we get $7 =3a+c$ and then $ax^b = 2x^3$. So if $x=1$ we get $a=2$ and $c=1$ and $x^b=x^3$. So if $x=2$ we get $2^b = 8$ and thus $b=3$.
Notice: Since we don't know apriori that $b$ is posi... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The most elementary proof of divisibility of sum of powers I am wondering how one can prove that for an arbitrary odd natural number $n$ and an arbitrary natural number $a$ the power-sum $S_{(a,n)}=1^n+2^n+\ldots +a^n $ is divisible by $S_{(a,1)}=1+2+\ldots + a$ in the most elementary way (possible) which comes to your... | Note that $r^n+(P-r)^n$ is divisible by $P$ by Binomial expansion and cancelling of the only terms without factors of $P$ (as $n$ is odd). If $a$ is odd, then all the terms in $S_{(a,n)}$ pair up in this way, with $P=a$, except for $a^n$, which is certainly a multiple of $a$, so $S_{(a,n)}$ is a multiple of $a$. If $a$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $a$, $b$, $c$ be positive integers such that $\frac{bc}{b+c}$, $\frac{ca}{c+a}$, $\frac{ab}{a+b}$ are integers. Show that $\gcd(a,b,c) >1$. Let $a$, $b$, $c$ be positive integers such that $\frac{bc}{b+c}$, $\frac{ca}{c+a}$, $\frac{ab}{a+b}$ are integers. Show that $\gcd(a,b,c) >1$.
I can see that $\gcd(b,c)$, $\gc... | Let's assume $(a,b,c)=1$. Since $\frac{ab}{a+b}$ is a positive integer, we must have $(a,b)>1$. The reason is clear. If $(a,b)=1$, then:
$$(a+b,a)=(a+b,b)=(a,b)=1 \implies (a+b, ab)=1.$$
Now, suppose $a=dm$ and $b=dn$, where $d>1$ and $(m,n)=1$. We will have:
$$\frac{ab}{a+b}=\frac{dmn}{m+n};$$
but $(m+n, mn)=1$ becaus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4632054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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For $c>b>a$ with $c,b,a\in\Bbb N,c-b=b-a=1$, prove $\frac{a^c}{b^b}+\frac{b^c}{c^b}+\frac{c^c}{a^b}\ge\frac{a^b}{b^a}+\frac{b^b}{c^a}+\frac{c^b}{a^a}$
Given $c>b>a$, also $c-b=b-a=1$ where $c,b,a$ are Natural numbers, prove that
$$
\frac{a^c}{b^b}+\frac{b^c}{c^b}+\frac{c^c}{a^b} \geqslant \frac{a^b}{b^a}+\frac{b^b}{c^... | Hint for $a,b,c>3$ we have :
$$\frac{a^c}{b^b}+\frac{b^c}{c^b}+\frac{c^c}{a^b}\geq \frac{(a+b+c)^{c-b}}{3^{c-b-1}}$$
And :
$$\frac{a^b}{b^a}+\frac{b^b}{c^a}+\frac{c^b}{a^a}<3\frac{c^b}{a^a}$$
We have for $c\geq 2(a+b)$:
$$\frac{(a+b+c)^{c-b}}{3^{c-b-1}}-3\frac{c^{b}}{a^{a}}>0$$
For the case $a+b\leq c\leq 2(a+b)$ :
We ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $\frac{x-8}{x-10}+\frac{x-4}{x-6}=\frac{x-5}{x-7}+\frac{x-7}{x-9}$ Solve $\frac{x-8}{x-10}+\frac{x-4}{x-6}=\frac{x-5}{x-7}+\frac{x-7}{x-9}$
$\Rightarrow \frac{(x-10)+2}{x-10}+\frac{(x-6)+2}{x-6}=\frac{(x-7)+2}{x-7}+\frac{(x-9)+2}{x-9} \ \ \ ...(1)$
$\Rightarrow 1+\frac{2}{x-10}+1+\frac{2}{x-6}=1+\frac{2}{x-7}+1+\... | Look closesly at step $(7)$
$$\frac{2x-16}{(x-10)(x-6)}=\frac{2x-16}{(x-9)(x-7)}$$
The numerators are equal for any value of $x$, but is the same true for the denominator too i.e. for any $x\in\mathbb{R}/\{6,7,9,10\}$? No!
What can you conclude from this? One way to look at it is equating this to $0$, as mentioned in t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4633508",
"timestamp": "2023-03-29T00:00:00",
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Show that $a^2+b^2+c^2=x^2+y^2+z^2$
Let $a,b,c,x,y,z$ be real numbers such that
$$ a^2+x^2=b^2+y^2=c^2+z^2=(a+b)^2+(x+y)^2=(b+c)^2+(y+z)^2=(c+a)^2+(z+x)^2 $$
Show that $a^2+b^2+c^2=x^2+y^2+z^2$.
Progress: Note that $$0=\sum_{\text{cyc}} (a + b)^2 + (x + y)^2 - \sum_{\text{cyc}} a^2 + x^2 = (a + b + c)^2 + (x + y + z)... | Writing this using complex numbers makes it a little easier. The equations mean that these following complex numbers have the same distance from the origin of the complex plane.
$$a+ix=re^{i\theta_1}\mbox{ with } a=r\cos(\theta_1),x=r\sin(\theta_1).$$
$$b+iy=re^{i \theta_2}\mbox{ with } b=r\cos(\theta_2),y=r\sin(\theta... | {
"language": "en",
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How do I solve the equation $(x - 1)(x - 2)(x - 3) = (x - 2)(x - 3)(x - 4)$? Problem: $(x - 1)(x - 2)(x - 3) = (x - 2)(x - 3)(x - 4)$
Heres my question with this problem: why do I end up with a wrong answer when I divide both sides by $(x-2)(x-3)$ to cancel out the $(x-2)(x-3)$ on both sides. Is this not allowed and wh... | You cannot divide both sides by $(x-2)(x-3)$ if it equals $0$. When you divide both sides by it, you are ignoring the solutions $x=2,3$
$$(x-1)(x-2)(x-3)=(x-2)(x-3)(x-4)$$
A proper way to solve this is as follows
$$(x-1)(x-2)(x-3)-(x-4)(x-2)(x-3)=0$$$$[(x-1)-(x-4)](x-2)(x-3)=0$$
$$3(x-2)(x-3)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4634658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Hard geometry problem: given a triangle with sides 16, 30, 34, find the area of the triangle introduced by the inscribed circle You are given a triangle $\triangle PQR$ with sides $16, 30, 34$. Let the incircle touch the sides of $\triangle PQR$ at $X,Y,$ and $Z$. Given that the ratio $[XYZ]/[PQR]$ can be written as $\... | John Omeilan has provided an excellent solution. However the answer can be simpler if we observe that $\Delta PQR$ is a right angled triangle.
Since $16^2+30^2=34^2$, $\Delta PQR$ is a right angled triangle and hence $ [PQR] = \frac{16 \times 30}{2}=240$
$s=\frac{p+q+r}{2}=\frac{30+16+34}{2}=40$
Let the radius of the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4635374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can I parametrize the intersection curve of cone $z=\sqrt{(x^2+y^2)}$ and plane $z=1+x+y$ Usually, I equalize both equation, then complete the square in order to find something that looks like $\cos^2 t+\sin^2 t=1$ to get a value of $x(t)$ and $y(t)$. I then substitute to get $z(t)$. I am unable to apply that techn... | We have $z^2=x^2+y^2$ and $z^2=(1+x+y)^2$, which gives $$x^2+y^2=1+2x+2y+2xy+x^2+y^2$$ which is equivalent to $1+2x+y(2+2x)=0$.
Now we can set $x=t$ and $y=-\frac{1+2x}{2+2x}=-\frac{1+2t}{2+2t}$ and $z=1+x+y=1+t-\frac12\frac{1+2t}{1+t}$.
Furthermore, $z=\sqrt{x^2+y^2}$ gives $z>0$, removing the problem that the curve i... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $ a, b, c $ are in arithmetic sequence for the given condition. A question on my mathematics textbook, Mathematics—Textbook for Class XI, goes thus:
If $a\left(\frac{1}{b} + \frac{1}{c}\right), b \left(\frac{1}{c} + \frac{1}{a}\right)$ and $c \left(\frac{1}{a} + \frac{1}{b}\right)$ are in arithmetic sequenc... | Hint: Let $x_1=a\bigg(\frac{1}{b}+\frac{1}{c}\bigg)$, $x_2=b\bigg(\frac{1}{a}+\frac{1}{c}\bigg)$ and
$x_3=c\bigg(\frac{1}{a}+\frac{1}{b}\bigg)$.
Then
$$
x_3-2x_2+x_1=\frac{(ab+ac+bc)(a-2b+c)}{abc}
$$
Can you finish from here ?
| {
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Closed form expression for $\psi_{e^{\pi}}^{(3)}(1-i)$ Let $\psi_q(z)$ be the q-DiGamma function defined for a real variable $\Re(z)>0$ as $$\psi_q(z)=\frac{1}{\Gamma_q(z)}\frac{\partial}{\partial z} (\Gamma_q(z))$$
where $\Gamma_q(z)$ is the q-Gamma function defined as $$\Gamma_q(z)=(1-q)^{1-z}\prod_{n=0}^{\infty}\fra... | I am not an expert in such sums, this derivation can probably be done in a simpler way. From the link provided by Tyma Gaidash, we have
\begin{align*}
\psi _{{\rm e}^\pi }^{(3)} (1 - {\rm i}) = & - \pi ^4 \sum\limits_{n = 1}^\infty {\frac{1}{{{\rm e}^{\pi n} + 1}}} + 7\pi ^4 \sum\limits_{n = 1}^\infty {\frac{1}{{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4642004",
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$\int \frac{x^2}{\sqrt{5-x^2}}dx$ using substitution $u = \sqrt{5-x^2}$? I was trying to solve $$\int\frac{x^2}{\sqrt{5-4x^2 }} dx$$ by using the substitution
$$
u=\sqrt{5-4x^2}\\
dx = \frac{\sqrt{5-4x^2}}{-4x}du$$ So the integral can be written as $$\int \left(-\frac{x}{4}\right) du$$ and
$$
x= \begin{cases}
\sqrt{(5-... | Note that
$$\arcsin\sqrt{\frac{5-4x^2}5}= \arccos\sqrt{1-\frac{5-4x^2}5}
=\arccos\frac{2x}{\sqrt5}=\frac\pi2-\arcsin\frac{2x}{\sqrt5}
$$
Therefore, the two results differs by the constant $-\frac\pi2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4644624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Derivative of $y = \sin^3(\frac\pi 3(\cos(\frac\pi{3\sqrt2}(-4x^3 + 5x^2 + 1)^{3/2})))$ at $x=1$ Today I came across a problem:
If $y = \sin^3\left(\frac\pi 3\left(\cos\left(\frac\pi{3\sqrt2}\left(-4x^3 + 5x^2 + 1\right)^{3/2}\right)\right)\right)$, then at $x=1$ which of the following option is correct?
*
*$2y'+\... | It is good to know the answer and solve. You never see your mistakes. I tried to be quick:
$y=\sin^3A$ and $A=\frac\pi 3\cos B$ and $B=\frac{\pi}{3\sqrt 2}C^{3/2}$ and $C=-4x^3+5x^2+1$.
$y'=3\sin^2A\cos A\frac\pi 3(-\sin B)\frac{\pi}{3\sqrt2}\frac32C^{1/2}(-12x^2+10x).$
At $x=1$, $C=2$, $B=\frac{2\pi}3$, $A=-\frac\pi 6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4646083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Computing the limit of $ \frac{1^3+2^3+\cdots+n^3}{\sqrt{4n^8 +1}} $ I had this exercise:
Compute the limit
$$ \lim_{n\to\infty} \frac{1^3+2^3+\cdots+n^3}{\sqrt{4n^8 +1}} $$
I tried two different approaches and got different answers.
Approach 1:
$$\begin{split}
\lim_{n\to\infty} \frac{1^3+2^3+\cdots+n^3}{n^4\sqrt{4 +... | One more method is to see the Laurent series of the expression. It is equal to $$\frac{1}{8}+\frac{1}{4n}+\frac{1}{8n^2}-\frac{1}{64n^8}+O\left(\left(\frac{1}{n}\right)^9\right)$$
You can see that it confirms your second approach.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Prove $\dfrac{n^2}{\ln n^2-1} - \dfrac{\dfrac{n^2}{2}}{\ln \dfrac{n^2}{2}-1.1} > n$ for $n \ge 347$ Prove $\dfrac{n^2}{\ln n^2-1} - \dfrac{\dfrac{n^2}{2}}{\ln \dfrac{n^2}{2}-1.1} > n$ for $n \ge 347$. It seems to be true for all $n \ge 11$, but I only need it to be true for $n \ge 347$. I tried to manipulate the equat... | I tried to manipulate the equation in all sorts of ways but I couldn't get it to work
In fact, there is an exact solution to the equation
$$\dfrac{n^2}{\log( n^2)-1} - \dfrac{\dfrac{n^2}{2}}{\log \left(\frac{n^2}{2}\right)-a} = n$$ which can rewrite
$$n=4\frac{\left(\log (n)-\frac{1}{2}\right) (\log
(n)-\alpha )}{\l... | {
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"url": "https://math.stackexchange.com/questions/4647326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do I integrate $\int_0^{\infty} \frac{\log(t+1)}{t^2+a^2} \mathrm{d}t$? How do I integrate the following integral:
$$\int_0^{\infty} \frac{\log(t+1)}{t^2+a^2} \mathrm{d}t$$
Where $a$ is some parameter?
I know that the solution includes Lerch Transcendents and logs (which is what I'm trying to arrive at); however, I... | let's generalize your integral
$$ \mathcal{I}(b) = \int_0^\infty \frac{\ln(1+b \, x)}{a^2+x^2} \, \mathrm{d}x \hspace{5mm} \cdots (1) $$
Using Feynmann's Trick we get (differentiating w.r.t b)
\begin{align*}
\mathcal{I}'(b) &= \int_0^\infty \frac{x}{(1+b\,x) \, (x^2 + a^2)} \, \mathrm{d}x \\
&= \frac{1}{1+a^2 \, b^2} ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
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Distance between the point of touching in three touching circles $\textbf{Question : }$ Say you have three touching circles $\Gamma_1,\Gamma_2,\Gamma_3$ with radii $x,y,z$ and centers $A,B,C$ as per the diagram, then prove the following $$|DE|=\frac{2}{\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x}+\dfr... | Partial solution
Set $\gamma=\frac{z+x}{y+z}$ and $\mu=\sin^2\alpha$. You end up with $$1=\sqrt{1-\gamma^2\mu}+\sqrt{\gamma^2-\gamma^2\mu}$$
Squaring gives $$1=(1+\gamma^2)-2\gamma^2\mu+2\sqrt{(1-\gamma^2\mu)(\gamma^2-\gamma^2\mu)}$$
Which can be re-written to $$-\gamma^2+2\gamma^2\mu=2\sqrt{\gamma^2-(1+\gamma^2)\mu+\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4648790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find the coefficients of product Given the following product,
$$(1+ax)(1+a^2x)(1+a^3x)\cdots (1+a^mx) $$
where $a$ is some real number which will be taken to be unity in the end. I want to know the coefficient of general term of expansion $x^p$. I know that the coefficient of $x^p$ comes from choosing $p$ terms from $m... | Products of this type are known to be generalisations of the binomial
\begin{align*}
(1+x)^m=\sum_{k=0}^m\binom{m}{k}x^k
\end{align*}
Often such generalisations are written with $q$ instead of $a$ and here I'll follow this convention. We denote with $[x^p]$ the coefficient of $x^p$ in a series and want to find
\begin{a... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is there an integral that proves $\pi > 333/106$? The following integral,
$$ \int_0^1 \frac{x^4(1-x)^4}{x^2 + 1} \mathrm{d}x = \frac{22}{7} - \pi $$
is clearly positive, which proves that $\pi < 22/7$.
Is there a similar integral which proves $\pi > 333/106$?
| From the relationship
$$\frac{333}{106}=\frac{377-2·22}{120-2·7}$$
and integrals
$$\pi = \frac{22}{7} - \int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx$$
and
$$\pi = \frac{377}{120} - \frac{1}{2}\int_0^1 \frac{x^5(1-x)^6}{1+x^2}dx$$
we obtain
$$\int_0^1 \frac{14x^4(1-x)^4-60x^5(1-x)^6}{1+x^2}dx = \int_0^1 \frac{2x^4(1-x)^4(7-30x(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "129",
"answer_count": 6,
"answer_id": 3
} |
Lateral surface area by hand I need to find the exact solution for the lateral surface area of the solid generated by revolving the region bounded by $y=x^2$, $y=0$, $x=0$, and $x=\sqrt 2$ about the X axis.
I have come up with my own solution, but I'm not sure if it's right.
Before substituting values back in, I got
$$... | I assume you have in your earlier steps:
\[ \sigma = 2\pi \int_0^\sqrt{2} x^2 \sqrt{1+4x^2} dx \]
\[ \tan \theta = 2x \; \mathrm{and} \; \sec \theta = \sqrt{1+4x^2} \]
\[ \sigma = \frac{\pi}2 \int_{x=0}^{x=\sqrt{2}} \sec^3\theta \tan^2\theta d\theta. \]
So you have a messy antiderivative you worked out, and want to che... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Funny identities Here is a funny exercise
$$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$
(If you prove it don't publish it here please).
Do you have similar examples?
| $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3) = \pi$
(using the principal value),
but if you blindly use the addition formula
$\tan^{-1}(x) + \tan^{-1}(y)
= \tan^{-1}\dfrac{x+y}{1-x y}$
twice, you get zero:
$\tan^{-1}(1) + \tan^{-1}(2) =
\tan^{-1}\dfrac{1+2}{1-1*2}
=\tan^{-1}(-3)$;
$\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/8814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "281",
"answer_count": 63,
"answer_id": 21
} |
Inequality of three variable polynomial I read that one can prove by AM-GM-inequality that for all $a,b,c\in\mathbb{R}_+$ we have that
$$11(a^6 + b^6 + c^6) + 40abc(ab^2 + bc^2 + ca^2) \ge 51abc(a^2b + b^2c + c^2a)$$
How this can be done? Is it possible to prove stronger inequalities like
$$10(a^6 + b^6 + c^6) + 41abc... | Each of the inequalities below is easily verified by AM-GM
\begin{eqnarray*}
&29(& a^6 & & & + a^3bc^2 & +3a^2b^3c & & & & & \geq & 5a^3b^2c & & &) \\
&29(& & b^6 & & & + a^2b^3c & +3ab^2c^3 & & & & \geq & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/11189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proof of the formula $1+x+x^2+x^3+ \cdots +x^n =\frac{x^{n+1}-1}{x-1}$
Possible Duplicate:
Value of $\sum x^n$
Proof to the formula
$$1+x+x^2+x^3+\cdots+x^n = \frac{x^{n+1}-1}{x-1}.$$
| Since $$\frac1{1-x}=1+x+x^2+x^3+\cdots,$$ we have
$$
\frac{1-x^{n+1}}{1-x}=(1+x+x^2+x^3+\cdots)-(x^{n+1}+x^{n+2}+x^{n+3}+x^{n+4}+\cdots)
$$
And on the right hand side every thing cancels except $1+x+x^2+\cdots+x^n$.
(This argument is probably circular! :) )
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/11703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 6
} |
Nice expression for minimum of three variables? As we saw here, the minimum of two quantities can be written using elementary functions and the absolute value function.
$\min(a,b)=\frac{a+b}{2} - \frac{|a-b|}{2}$
There's even a nice intuitive explanation to go along with this: If we go to the point half way between two... | Based on Christian Blatter's answer and the trigonometric solution of the cubic equation, we can derive the following unusual solution.
Let $a$, $b$ and $c$ be real numbers. Then they are the roots of the equation
$$(x-a)(x-b)(x-c)=0$$
which can also be written as
$$x^3-(a+b+c)x^2+(ab+bc+ca)x-abc=0.$$
The trigonometric... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/13253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "66",
"answer_count": 5,
"answer_id": 0
} |
A general formula for $\sum (k-1)(k-2)(k-3)$? What is a "simpler" formula for
$$\sum_{3}^{n} \frac{(k-1)(k-2)(k-3)}{6}$$
| Perhaps a messy and boring way, we can use the generating function.
$$\sum_{k=3}^{n}\frac{(k-1)(k-2)(k-3)}{6}=\frac{1}{6}\sum_{k=2}^{n-1}k(k-1)(k-2)$$
In addition, the generating function for $k(k-1)(k-2)$ is $x^3\left(\frac{1}{1-x}\right)^{(3)}.$
Hence, the sum is the coefficient of $x^{n-1}$ in $\frac{1}{6}\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/15145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 4
} |
Algebra Problem The expression $x^2-4x+5$ is a factor of $ax^3+bx^2+25$. Express the sum a+b as an integer.
Please give an explanation of how the answer
| Since $x^2-4x+5=0$ has the roots $2 \pm i$ they must also be roots of $ax^3+bx^2+25\,$ therefore:
$$
0 = a(2+i)^3+b(2+i)^2+25=8a -6a + 4b - b + 25 + i\,(12 a -a + 4b) \\
\iff
\begin{cases}
\begin{align}
2a + 3b + 25 = 0 \\
11 a + 4b = 0
\end{align}
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/16284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Finding Value of the Infinite Product $\prod \Bigl(1-\frac{1}{n^{2}}\Bigr)$ While trying some problems along with my friends we had difficulty in this question.
*
*True or False: The value of the infinite product $$\prod\limits_{n=2}^{\infty} \biggl(1-\frac{1}{n^{2}}\biggr)$$ is $1$.
I couldn't do it and my friend... | We have
\begin{align*}
p_{k} &= \prod_{n = 2}^{k} \left( 1 - \frac{1}{n^{2}} \right) = \prod_{n=2}^{k} \frac{(n-1)(n+1)}{n^{2}} \\ & = \frac{1 \cdot 3}{2 \cdot 2} \cdot \frac{2\cdot 4}{3 \cdot 3} \cdot \frac{3\cdot 5}{4 \cdot 4} \cdot \cdots \cdot \frac{(k-2)\cdot k}{(k-1)\cdot(k-1)} \cdot \frac{(k-1)(k+1)}{k\cdot k}\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/18179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
"answer_count": 6,
"answer_id": 5
} |
Recurrence equation: $u_n = 4u_{n−1} + 4u_{n−2}$ ; is $4x+4 = 4$ the characteristic equation? Given this recurrence equation:
$u_1 = 0, u_2 = 1$
$u_n = 4u_{n−1} + 4u_{n−2}$
Is the correct characteristic equation:
$4x+4 = 4$
EDIT:
Complete solve:
The characteristic equation is
$x^2-4x-4=0$
We solve the quadratic equatio... | Give Wilf's techniques a try... define $U(z) = \sum_{n \ge 0} u_{n + 1} u^z$,
and write the recurrence as:
$$
u_{n + 2} = 4 u_{n + 1} + 4 u_n \qquad u_1 = 0, u_2 = 1
$$
By properties of generating functions:
$$
\frac{U(z) - u_1 - u_2 z}{z^2} = 4 \cdot \frac{U(z) - u_1}{z} + 4 \cdot U(z)
$$
Maxima solves this as:
$$
U(z... | {
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"url": "https://math.stackexchange.com/questions/19626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Evaluate $\int \cos^3 x\;\sin^2 xdx$ Is this correct? I thought it would be but when I entered it into wolfram alpha, I got a different answer.
$$\int (\cos^3x)(\sin^2x)dx = \int(\cos x)(\cos^2x)(\sin^2x)dx
= \int (\cos x)(1-\sin^2x)(\sin^2x)dx.$$
let $u = \sin x$, $du = \cos xdx$
$$\int(1-u^2)u^2du = \int(u^2-u^4... | Take $\sin x=t, \cos x dx=dt$
\begin{align*}
\therefore \int \cos^{3}x\sin^{2}x dx &=\int \cos^{2}x\sin^{2} x\cos x dx\
&=\int (1-\sin^{2}x) \sin^{2} x \cos x dx\
&= \int (1-t^{2})t^{2} dt\
&=\int (t^{2}-t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/21589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Counting orbits under $\operatorname{Sym}(3)$ I'd like a closed form expression $x(n)$ for the number of orbits of the symmetric group on $3$ points acting on the triples in $\{ (a,b,c) \mid a,b,c \in \Bbb{Z}, 1 \leq a,b,c \leq n, c = 2n−a−b \}$.
I feel like this should be a really basic problem, but my standard method... | This one can also be done using the Polya Enumeration Theorem. Recall the cycle index of the symmetric group on three elements which can be computed using pen and paper by factorizing the six constituent permutations and has the value
$$Z(S_3) = \frac{1}{6}\left(a_1^3 + 3 a_1 a_2 + 2 a_3 \right).$$
Then the desired val... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/27506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Prove inequality: When $n > 2$, $n! < {\left(\frac{n+2}{\sqrt{6}}\right)}^n$ Prove: When $n > 2$,
$$n! < {\left(\frac{n+2}{\sqrt{6}}\right)}^n$$
PS: please do not use mathematical induction method.
EDIT: sorry, I forget another constraint, this problem should be solved by
algebraic mean inequality.
Thanks.
| For $a_1, a_2, \ldots, a_n \geqslant 0$, we have:$$\sqrt [ n ]{ { a }_{ 1 }\cdot { a }_{ 2 }\cdots { a }_{ n } } \leqslant \frac { { a }_{ 1 }+{ a }_{ 2 }+\cdots +{ a }_{ n } }{ n } $$
Consider ${ \left( n! \right) }^{ 2 }=\left( n\cdot 1 \right) \left( \left( n-1 \right) \cdot 2 \right) \cdots \left( 1\cdot n \right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/28084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Nice proofs of $\zeta(4) = \frac{\pi^4}{90}$? I know some nice ways to prove that $\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \pi^2/6$. For example, see Robin Chapman's list or the answers to the question "Different methods to compute $\sum_{n=1}^{\infty} \frac{1}{n^2}$?"
Are there any nice ways to prove that $$\z... | Let $ f(x)=x^2-\frac{1}{12} $ defined on $ [-\frac{1}{2},-\frac{1}{2}) $, and extend $ f $ to be periodic on $ \mathbb{R} $.
We compute that
\begin{align*}
\hat f(0)=\int_{-1/2}^{1/2}f(x)dx=\int_{-1/2}^{1/2}\left(x^2-\frac{1}{12}\right)dx=0.
\end{align*}
And for $ k\ne 0 $:
\begin{align*}
\hat f(k)&=\in... | {
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"url": "https://math.stackexchange.com/questions/28329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "119",
"answer_count": 15,
"answer_id": 12
} |
Finding the first 5 terms of the Maclaurin Series of $\frac{\sin x}{e^x}$ Question.
Find the first 5 terms of the Maclaurin Series of $\frac{\sin x}{e^x}$.
I'm trying to find it using division. Is there possibly a shorter way because every time I try to do it, I get the wrong answer:
$$
x-2x^3/3+\cdots$$
I don't reall... | Here are the first two terms:
$x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \cdots$
$=x\left(1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!}+ \dfrac{x^5}{5!}+ \dfrac{x^6}{6!}\cdots\right) -x^2 -\dfrac{2x^3}{3}-\dfrac{x^4}{6}-\dfrac{x^5}{30}-\dfrac{x^6}{120}-\dfrac{x^7}{5040}-\cdots$
$= (x-x^2)\... | {
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"url": "https://math.stackexchange.com/questions/29416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Can this integral $\int_0^{2\pi} \frac{d\theta}{(a^2 \cos^2 \theta +b^2\sin^2\theta)^{3/2}}$ be written in the form of a elliptic integral I am trying to find the magnetic field due to an elliptic loop of wire. How to do integrals of the type $$\int_0^{2\pi} \frac{d\theta}{(a^2 \cos^2 \theta +b^2\sin^2\theta)^{3/2}}$$... | Wow, this really worked out incredibly nicely!
The elliptic integral which gives $L$ is
$$
\frac{L}{4} = J(a,b) := \int_0^{\pi/2} \sqrt{a^2 \cos^2 \theta + b^2 \sin^2 \theta} \, d\theta
$$
and it is clearly symmetric, $J(a,b)=J(b,a)$.
The change of variables $t=b \tan\theta$ gives
$$
\sqrt{a^2 \cos^2 \theta + b^2 \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/34516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 2
} |
How to find roots of $X^5 - 1$? How to find roots of $X^5 - 1$? (Or any polynomial of that form where $X$ has an odd power.)
| Edit 2: Since the equation $x^n-1=0$ is equivalent to $x=\sqrt[n]{1}$, it has the following $n$ solutions:
$$e^{i\dfrac{2\pi (1+k)}{n}}=\cos\left( \dfrac{2\pi (1+k)}{n}\right)+i\sin\left( \dfrac{2\pi (1+k)}{n}\right),\qquad k=0,1,\ldots ,n-1,$$
where $n$ is a positive integer, odd or even.
See comment by Gerry Myerso... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/39202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Solving $\sqrt{x+5} = x - 1$ I'm currently learning about radicals and simplifying them, and I came across this problem on the internet and tried to solve it:
$$\sqrt{x+5} = x - 1$$
So I used this logic:
$$
\begin{align}
\sqrt{x+5} &= x - 1 \\
x + 5 &= (x-1)^2 \\
x + 5 &= (x-1)(x-1) \\
x + 5 &= x^2 - 2x + 1 \\
0 ... | *
*Function on the left hand side is strictly increasing (i.e always increasing).
*Function on the right hand side is strictly increasing (i.e always increasing) too.
*We draw a graph to find out where too look for the roots. You can use this or any other tool to draw two graphs for left and right hand side function... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/41152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 7,
"answer_id": 6
} |
Asymptotic approximation for confluent hypergeometric function I have the following nasty expression that I would like to expand in powers of $\frac{1}{N}$:
\begin{align}
\frac{2^{\frac{3}{2}} 3^{\frac{1}{2}} \Biggl[ \sqrt{u} \cdot \Gamma\left(\frac{2+N}{4}\right)
\cdot {}_1F_1 \left( \frac{2+N}{4},\f... | Perhaps not a big help: For ${}_1F_1((2+N)/4,3/2,3r^2/2u)$ we want asymptotics of ${}_1F_1(1/2+x,3/2,a)$ as $x \to 0$. Using the series, I get
$$
{}_1F_1\left(\frac{1}{2}+x,\frac{3}{2},a\right)=
\frac{\sqrt{\pi} \mathrm{erf} (i \sqrt{a})}{2 \sqrt{-a}} + \sum_{k = 0}^{\infty} \frac{\Bigl(\Psi \Bigl(\frac{1}{2} + k\B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/47121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Getting the inverse of a lower/upper triangular matrix For a lower triangular matrix, the inverse of itself should be easy to find because that's the idea of the LU decomposition, am I right? For many of the lower or upper triangular matrices, often I could just flip the signs to get its inverse. For eg: $$\begin{bmatr... | Given that the diagonal elements of a triangular matrix $A$ are all $1$. It is immediate to see that the characteristic polynomial of $A$ is
\begin{align*}
f(\lambda) = \det(\lambda I_{(n)} - A) = (\lambda - 1)^n =
\lambda^n - \binom{n}{1}\lambda^{n - 1} + \binom{n}{2}\lambda^{n - 2} + \cdots + (-1)^n.
\end{align*}
By... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/47543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "51",
"answer_count": 6,
"answer_id": 1
} |
Inequality: $(x + y + z)^3 \geq 27 xyz$ Edit: $a,b,c$ and $x,y,z$ are positive, real numbers.
Since $(a-b)^2 \geq 0~$, $a^2 + b^2 - 2ab\geq0~$ and $a^2 + b^2 \geq 2ab~$. Similarly, $a^2 + c^2 \geq 2ac~$ and $b^2 + c^2 \geq 2bc~$.
Adding these inequalities together, $2(a^2 +b^2 + c^2) \geq 2(ab + ac +bc)~$ and accord... | Here are two one-line-formula proofs:
*
*By AM-GM:
$$
\frac{x+y+z}{3} \geq \sqrt[3]{x y z}
$$
Now take the cubic value on both sides of this inequality.
*Use the following identity which also gives you the exact deviation in positive terms from $27 x y z$: (from which you can derive tighter bounds of the LHS)
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/48621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 8,
"answer_id": 5
} |
Prove that $3(a^3+b^3+c^3) > (a+b+c)(a^2+b^2+c^2)$ Let $a,b,c$ be positive, not equal. Prove that $3(a^3+b^3+c^3) > (a+b+c)(a^2+b^2+c^2)$.
I know the proof by subtracting LHS by RHS and then doing some arrangement.
But isn't there any inequality which can be used in $(1+1+1)(a^3+b^3+c^3) >(a+b+c)(a^2+b^2+c^2)$, or an e... | There is a fairly general way to prove inequalities like this using rearrangement, but just for fun, let me show you how you can prove a large class of inequalities like this by repeated use of AM-GM. First subtract $a^3, b^3, c^3$ from the RHS, then symmetrically sum the inequalities
$$a^3 + a^3 + b^3 \ge 3 a^2 b.$$
E... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/49211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
Find all pairs of positive integers $(x,y)$ such that $x^2-y^2=a^3$ and $x^3-y^3=b^2$ for integral $a, b$? How to find ALL pairs of positive integers $(x,y)$ such that the difference in their squares is a perfect cube and the difference in their cubes is a perfect square.
i.e.,
Positive integers $(x,y)$ such that
$x^2-... | Not an answer, just thinking out loud about the question of whether we can ever take $x,y$ relatively prime.
We solve $x^2-y^2=a^3$ by writing $a^3=a_1a_2$ with $a_1,a_2$ of the same parity, then $x-y=a_1$, $x+y=a_2$, $2x=a_2+a_1$, $2y=a_2-a_1$, and we see any common divisor of $a_1,a_2$ other than perhaps $2$ is a co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/53292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to divide polynomial matrices If I am given two $2\times2$ polynomial matrices and I need to divide them, what are the steps I need to follow?
I know I need to do right division and left division, and that the answer will have the right quotient and remainder and the left quotient and remainder.
Please show me how ... | Dividing matrices is multiplying by the inverse, i.e if you want to divide $A$ by $B$ then you calculate $A\cdot B^{-1}$ (right) or $B^{-1}\cdot A$ (left). In order to find the inverse you use the Gauss reduction or the adjoint formula. For $2\times2$ matrices you have:
$$\left( \begin{array}{cc} a&b \\ c&d \end{array}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/54728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Determining the Jordan form of a matrix given the invariant factors I am trying to recover the Jordan normal form of a matrix given a list of invariant factors and was wondering if I am proceeding correctly in constructing the Jordan blocks.
Let $F = \mathbb{C}$ and let $V$ be a finite dimensional vector space over $F$... | Edit: I'm sorry, but my first answer was definitely incorrect and I really hope I didn't cause any confusion. Won't speedread problems in the future :)
Since we have $V = F[x] / (x) \oplus F[x]/ (x(x-1)(x-2)^2) \oplus F[x]/(x^2(x-1)^2(x-3)^3)$, we can look at our three summands separately.
Our first summand, $F[x]/(x)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/56820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Rational parameterization of surface The surface $$ (x^{2} + y^{2} - 2 y + 1) \cdot (x^{2} + y^{2} + 2 y + 1) \cdot (x^{2} + y^{2} - 2 x + 1) \cdot (x^{2} + y^{2} + 2 x + 1) - z^2 = 0$$
has the points $\left(\dfrac{t}{s},\dfrac{t}{s},\dfrac{s^{4} + 4 t^{4}}{s^{4}}\right)$, $ \left(\dfrac{- s^{2} + t^{2}}{s^{2} + t^{2}}... | I may have some results on this in due course. So bookmark this page and check back in a week or two.
But meanwhile, I'm curious to know how this relates to Euler bricks, unless you'd rather keep that under your hat.
Your equation is clearly equivalent over $\mathbb{Q}$ to the pair:
$ p^2 + q^2 + r^2 = 1 $
$ (1 - q^2) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/57094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Solving an Exponential Equation $$3^{x-1}+3^{x-2}+3^{x-3}=3159$$
Another exponential equation I'm having a hard time with, the answer is given and equals to : $8$. I'm absolutely sure I'm making a wrong step somewhere along the way. Any help is appreciated.
EDIT:
$$3^{x-3}(\frac{1}{8}+\frac{1}{243}+1)=3159$$
This is wh... | I would try factoring $3^x$ first, and isolating:
$$\begin{align*}
3^{x-1} + 3^{x-2} + 3^{x-3} &= 3159\\
3^x\left(3^{-1} + 3^{-2} + 3^{-3}\right) &= 3159\\
3^x &= \frac{3159}{3^{-1}+3^{-2}+3^{-3}}
\end{align*}$$
For an extra flourish,
$$\begin{align*}
\frac{3159}{3^{-1}+3^{-2}+3^{-3}} &= \frac{3159}{\frac{1}{3} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/60373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Is this asymptotic equation correct? Is this equation correct?
$$ \frac {1 + \Theta(\frac 1 {2n})} {(1 + \Theta(1/n))^2} = 1 + O(1 / n) $$
I need this equation to prove that
$$ \binom {2n} n = \frac {2 ^ {2n}} {\sqrt {\pi n}} (1 + O(1 / n)) $$
which could be calculated from Stirling's Approximation:
$$ n! = \sqrt {2\p... | Suppose $Bx\le\Theta(x)\le Cx$ for each $\Theta$.
$$
\begin{align}
\frac{1+\Theta(\frac{1}{2n})}{(1+\Theta(\frac{1}{n}))^2}
&\le\frac{1+\frac{C}{2n}}{(1+\frac{B}{n})^2}\\
&=(1+\frac{C}{2n})(1-2\frac{B}{n}+O(\frac{1}{n^2}))\\
&=(1+(\frac{C}{2}-2B)\frac{1}{n}+O(\frac{1}{n^2}))\\
&=1+O(\frac{1}{n})
\end{align}
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/62996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Difficult Integral: $\int\frac{x^n}{\sqrt{1+x^2}}dx$ How to calculate this difficult integral: $\int\frac{x^2}{\sqrt{1+x^2}}dx$?
The answer is $\frac{x}{2}\sqrt{x^2\pm{a^2}}\mp\frac{a^2}{2}\log(x+\sqrt{x^2\pm{a^2}})$.
And how about $\int\frac{x^3}{\sqrt{1+x^2}}dx$?
| I would first try the substitution $x=\tan(\theta)$, so that $\sqrt{1+x^2}=\sec(\theta)$. That gives
$$
\begin{align}
\int\frac{x^n}{\sqrt{1+x^2}}\;\mathrm{d}x
&=\int \tan^n(\theta)\sec(\theta)\;\mathrm{d}\theta\\
&=\tan^{n-1}(\theta)\sec(\theta)-(n-1)\int\tan^{n-2}(\theta)\;\sec^3(\theta)\;\mathrm{d}\theta\\
&=\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/64450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
How to prove $a^2 + b^2 + c^2 \ge ab + bc + ca$? How can the following inequation be proven?
$$a^2 + b^2 + c^2 \ge ab + bc + ca$$
| From Cauchy-Schwarz
$ab+bc+ac=\sqrt{a^2}\sqrt{b^2}+\sqrt{b^2}\sqrt{c^2}+\sqrt{a^2}\sqrt{c^2} \leq \sqrt{a^2+b^2+c^2}\sqrt{a^2+b^2+c^2}$
Moving on;
$ab+bc+ac \leq a^2+b^2+c^2$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/64868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 12,
"answer_id": 9
} |
Solving Radical Equations $x-7= \sqrt{x-5}$ This the Pre-Calculus Problem:
$x-7= \sqrt{x-5}$
So far I did it like this and I'm not understanding If I did it wrong.
$(x-7)^2=\sqrt{x-5}^2$ - The Square root would cancel, leaving:
$(x-7)^2=x-5$ Then I F.O.I.L'ed the problem.
$(x-7)(x-7)=x-5$
$x^2-7x-7x+14=x-5$
$x^2-14x+1... | We can avoid squaring both sides. Let $x-5=u^2$, where $u \ge 0$. Then $\sqrt{x-5}=u$. Also, $x=u^2+5$, so $x-7=u^2-2$. Thus our equation can be rewritten as
$$u^2-2=u, \quad\text{or equivalently}\quad u^2-u-2=0.$$
But
$$u^2-u-2=(u-2)(u+1).$$
Thus the solutions of $u^2-u-2=0$ are $u=2$ and $u=-1$. Since $u \ge 0$, we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/66255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Convergence of Series, problems with intermediate steps I have problems with two exercises. I know the answer of those two but both of them have a step which I don't understand.
1) I have to prove that $\prod_{k=2}^n \frac{k^3-1}{k^3+1}$ is convergent.
My first steps were:
$\prod_{k=2}^n \frac{k^3-1}{k^3+1}= \prod_{k=2... | 1) The author uses a trick to compute the exact value of the infinite product but one does not need it to show the convergence only.
Namely, use that $\frac{k^3-1}{k^3+1}=1-\frac2{k^3+1}$, that $\frac2{k^3+1}<1$ for every $k\ge2$ and that the series $\sum\limits_{k}\frac2{k^3+1}$ converges absolutely.
2) Use the tric... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/67399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solution to a system of linear equations in GF(2) Denote the system in $GF(2)$ as $Ax=b$, where:
$$
\begin{align}
A=&(A_{ij})_{m\times m}\\
A_{ij}=&
\begin{cases}
(1)_{n\times n}&\text{if }i=j\quad\text{(a matrix where entries are all 1's)}\\
I_n&\text{if }i\ne j\quad\text{(the identity matrix)}
\end{cases}
\en... | Some observations, too long for a comment:
If $n$ is odd, your matrix is not invertible, and so there is no solution for arbitrary $b$ (and a solution will not be unique if it exists). First, do some row operations to rewrite the constituent blocks to $$\pmatrix{1&1&1&1\\0&0&0&0\\0&0&0&0\\0&0&0&0}, \pmatrix{1&0&0&0\\1&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/67460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Linear diophantine equation $100x - 23y = -19$ I need help with this equation: $$100x - 23y = -19.$$ When I plug this into Wolfram|Alpha, one of the integer solutions is $x = 23n + 12$ where $n$ is a subset of all the integers, but I can't seem to figure out how they got to that answer.
| The continued fraction solution goes as follows:
Expand 100/23 into a continued fraction (I'm essentially using the GCD algorithm):
\begin{align*}
\frac{100}{23} & = 4 + \frac{8}{23}\\
& = 4 + \frac{1}{\frac{23}{8}}\\
& = 4 + \cfrac{1}{2 + \cfrac{7}{8}}\\
& = 4 + \cfrac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/67969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 8,
"answer_id": 3
} |
The $3 = 2$ trick on Google+ I found out this on Google+ yesterday and I was thinking about what's the trick. Can you tell?
How can you prove $3=2$?
This seems to be an anomaly or whatever you call in mathematics. Or maybe I'm just plain dense.
See this illustration:
$$ -6 = -6 $$
$$ 9-15 = 4-10 $$
Adding $\frac{25}{4... | $2-(5/2)$ is not a positive square root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/68913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
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Probability of collecting all 4 different items while picking 1 random item from the set a.) There are $4$ distinct items in the set. What is the probability of picking all $4$ items after picking $n\ge4$.
b.) How many items do you need to pick to collect all four with a probability of at least $.9$.
My answer for part... | The original problem is incompletely specified. And one problem with a "stars and bars" approach is that not all solutions are equally likely, making the calculation of probabilities very difficult.
We consider the following equivalent situation. An experiment consists of picking one of the digits $1$, $2$, $3$, $4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/69570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to prove those "curious identities"? How to prove
$$ \prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = \frac{n}{2^{n-1}}$$
and
$$ \prod_{k=1}^{n-1} \cos\left(\frac{k\pi}{n}\right) = \frac{\sin(\pi n/2)}{2^{n-1}}$$
| Denote $w = e^{i \pi/n}$. We have
$$\prod_{k = 1}^{n-1} \sin \left(\frac{k\pi}{n}\right)= \prod_{k = 1}^{n-1} \frac{w^k - w^{-k}}{2i} = \frac{1}{2^{n-1}} \prod_{k = 1}^{n-1} \frac{w^k}{i} (1-w^{-2k})$$
Since we have
$$\sum_{k = 0}^{n-1} x^k = \prod_{k = 1}^{n-1} (x-w^{2k})$$
Setting $x=1$ yields
$$\prod_{k = 1}^{n-1} (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/70231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 4,
"answer_id": 2
} |
Wrong sign on the final answer, where did I go wrong? Algebra $12x^2-12x = 0$
This was suppose to be 18, but it is -18.
I think it is because of how I modified one of the factors.
$\sqrt{x^2+x^2}$
$\sqrt{2x^2}$
$\sqrt{2}\cdot \sqrt{x^2}$
$x\sqrt{2}$
If I let wolfram do this, it gives the same but it says "assuming X... | As you point out, by the Pythagorean Theorem, the area of the rectangle can be written as
$$\sqrt{2x^2}\sqrt{2(6-x)^2}.$$
Note that from the geometry, we have $0 < x < 6$.
It follows that $\sqrt{2x^2}=\sqrt{2}\;x$ and $\sqrt{2(6-x)^2}=\sqrt{2}\;(6-x)$.
Thus, if $A(x)$ is the area, then
$$A(x)=2x(6-x).$$
(Incidentally,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Did I tackle this implicit differentiation problem correctly? $$2x^3 + x^2y-xy^3 = 2$$
$$\frac{\mathrm{d}}{\mathrm{d}x} [2x^3+x^2y -xy^3 ] = \frac{\mathrm{d}}{\mathrm{d}x}(2)$$
$$6x^2 + \left(2xy + x^2\frac{\mathrm{d}y}{\mathrm{d}x}\right) - \left( 1 y^3 + 3y6^2 \frac{\mathrm{d}y}{\mathrm{d}x}\right ) = 0$$
$$\frac{\m... | $2x^3 + x^2y-xy^3 = 2$
$$\frac{\mathrm{d}}{\mathrm{d}x} [2x^3+x^2y -xy^3 ] = \frac{\mathrm{d}}{\mathrm{d}x}(2)$$
$$6x^2 + \left(2xy + x^2\frac{\mathrm{d}y}{\mathrm{d}x}\right) - \left( 1 y^3 + 3y^2x \frac{\mathrm{d}y}{\mathrm{d}x}\right ) = 0$$
$$\frac{\mathrm{d}y}{\mathrm{d}x} (x^2 - 3y^2x) + (6x^2+2xy-y^3) = 0$$
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/75596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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A "fast" way for computing $ \prod \limits_{i=1}^{45}(1+\tan i^\circ) $?
Which is the fastest paper-pencil approach to compute the product $$ \prod
\limits_{i=1}^{45}(1+\tan i^\circ) $$
| Using
$$
1+\tan x = \frac{\sin x + \cos x}{\cos x} = \frac{\sqrt{2} \cos (45^{\circ} - x)}{\cos x},
$$
the product can be written as:
$$
\prod_{x=1}^{45}(1+\tan x^\circ) = 2^{45/2} \prod_{x=1}^{45} \frac{\cos (45 - x)^{\circ}}{\cos x^{\circ}} \stackrel{(1)}{=} 2^{45/2} \cdot \frac{\prod\limits_{x=0}^{44} \cos x^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/75825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 3,
"answer_id": 0
} |
Problem with generating functions and binary recurrences I am considering the following recurrence:
$a_0 = 1$;
$a_1 = 2$
$a_{n} = 2 (a_{n - 1} + a_{n - 2})$
Then I proceeded with the generating function:
$F(x) = \displaystyle\sum_{n = 0}^\infty a_n x^n = 1 + 2x + \displaystyle\sum_{n = 2}^\infty a_{n} x^{n} = 1 + 2x +... | Let be $a(n)$ a geometric progression, or $a(n)=q^n$
So,
$a(n)=2a(n-1)+2a(n-2)$ becomes
$q^n=2q^{n-1}+2q^{n-2}$
so $q=0$ or
$q^2=2q+2$
So, $q_1=\frac{2+\sqrt{12}}{2}=1+\sqrt{3}$ or $q_2=\frac{2-\sqrt{12}}{2}=1-\sqrt{3}$.
and
$a(n)=bq_1^n+cq_2^n$
$a(0)=1$, $b+c=1$
$a(1)=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/76363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding smallest root of equation By trial and modification, I want to find the smallest root of this equation:
$-2\tan\left(\frac{a}{2}\right)=\tan(a)$
and the following as well:
$-\tan\left(\frac{a}{\sqrt2}\right)=\sqrt2\tan({a/2})$
| Hint:
If $\frac{a}{2}=b$, then you equation became:
$-2\tan(b)-\tan(2b)=0$
But,
$\tan(2b)=\frac{2\tan(b)}{1-\tan^2(b)}$
so we have:
$-2\tan(b)-\frac{2\tan(b)}{1-\tan^2(b)}=0 \iff$
$\frac{-2tan(b)(1-tan^2(b))}{1-tan^2(b)}-\frac{2tan(b)}{1-tan^2(b)}=0$
The denominator couldn't be 0, so:
$1-\tan^2(b) \neq 0 \iff \tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/77158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Tricky congruence question Let $p>5$ be a prime,
$$n\triangleq\frac{4^p+1}{5}\;\quad\text{and}\;\quad b\triangleq\frac{n-1}{4}\quad.$$
It can be shown that both are integers, and also that $n$ is composite, $b$ is odd and $p$ divides $b$.
I'd like to prove that
$$4^b\equiv-1\pmod n\;\quad.$$
I tried rephrasing the cong... | We will show the slightly stronger $$ 4^b \equiv -1 \pmod{4^p + 1}. $$ Fermat's theorem implies that $4^{p-1} \equiv 1 \pmod{p}$. Since $b = (4^{p-1}-1)/5$, it follows that $p|b$ (since $p > 5$). Suppose that $b = Xp$. Clearly $b$ is odd, therefore so is $X$. We have
$$\begin{align*} 4^{Xp} + 1 &= (4^{Xp} + 4^{(X-1)p})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/77331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How to prove a fact about the sum of three squares? How would I go about proving the following?
If $a$, $b$, $c$, $n$ are positive integers, then
$a^2+b^2+c^2 \neq 2^nabc$
I tried doing something similar to the proof for Adrien-Marie Legendre's Three Square theorem: $a^2+b^2+c^2=n$ iff there are not integers $k$, and ... | Since the right-hand side is even, either exactly one or all three of $a,b,c$ must be even.
The former case is impossible, as you can easily see by taking both sides mod 4.
In the latter case, let $2^k$ be the greatest power of 2 in the GCD of $a,b,c$. Then
$$\left(\frac{a}{2^k}\right)^2 + \left(\frac{b}{2^k}\right)^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/78122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Finding the slope of the tangent line to $\frac{8}{\sqrt{4+3x}}$ at $(4,2)$ In order to find the slope of the tangent line at the point $(4,2)$ belong to the function $\frac{8}{\sqrt{4+3x}}$, I choose the derivative at a given point formula.
$\begin{align*}
\lim_{x \to 4} \frac{f(x)-f(4)}{x-4} &=
\lim_{x \mapsto 4}... | HINT:
$$
\begin{eqnarray}
\frac{1}{x-4} \left ( \frac{8}{\sqrt{4+3x}}-2\right) &=&\frac{1}{x-4} \left ( \frac{8}{\sqrt{4+3x}}-2\right) \left ( \frac{8}{\sqrt{4+3x}}+2\right) \left ( \frac{8}{\sqrt{4+3x}}+2\right)^{-1} \\
&=& \frac{1}{x-4} \left( \frac{64}{4+3x} -4 \right) \left ( \frac{8}{\sqrt{4+3x}}+2\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/80010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Why this limit can't be computed like this? This is the limit:
$$\lim_{x \to 2} \frac{3x^2-x-10}{x^2-4}$$
The solution manual says it is: $\frac{11}{4}$
I've tried to solve it like a polynomial like this:
$$ \frac{(\frac{3x^2}{3x^2}-\frac{x}{3x^2}-\frac{10}{3x^2})*3x^2}{(\frac{x^2}{x^2}-\frac{4}{x^2})*x^2}=$$
$$= \frac... | For this problem, you'll want to factor the numerator and the denominator, (which both tend to $0$ as $ x \to 2$), cancel a factor of $x-2$, and then try direct substitution again.
$\frac{3x^2 -x - 10}{x^2-4} = \frac{(3x +5)(x - 2)}{(x + 2)(x - 2)} = \frac{3x + 5}{x + 2}$ (when $x \neq 2$)
Now letting $x \to 2$, we ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/81145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding Complex Number $z$ in $\frac{z+2i}{z-2i}=\frac{7-6i}{5}$
What I did: Cross Multiply, try to expand out the mod and args, but they all seem to lead to dead end (probably I am not seeing something)
| All your equations are correct but apparently your main difficulty was how
to find $z$ in the algebraic form $a+bi$ from the given equation, which you
have shown is equivalent to the linear equation in $z$
$$
2z-12-(24+6z)i=0.
$$
To solve this equation add the terms in $z$ and separately the independent
terms.
$$
\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/84947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Two Taylor expansions of $\frac{1}{1+\sqrt{2-z}}$ about $z=0$
How do you start expanding this function $$f(z)= \frac{1}{1+\sqrt{2-z}}$$ into two Taylor expansions about $z=0$?
The best I came up is to let $u=\sqrt{2-z}$ and then expand $f(z)$ as a geometric series.
| Alternatively, with some algebraic manipulations you get at:
$$
f(z) = \frac{1}{1+\sqrt{2-z}} = \frac{1-\sqrt{2-z}}{\left(1+\sqrt{2-z}\right)\left(1-\sqrt{2-z}\right)} = \frac{1-\sqrt{2-z}}{z-1} = \frac{\sqrt{2-z}-1}{1-z}
$$
Suppose, you worked out $\sqrt{2-z}-1 = \sum_{n=0}^\infty c_n z^n$, then
$$
\frac{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/85524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Fractions in Questions and Answers
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