Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Every positive integer can be written as form of $ab-cd$?
Show that: every positive integer can be written as form of $ab-cd$,where $a-b=2(c-d),a,b,c,d\in Z$
for example
(1):$a=1,b=1,c=d=0$ then we have
$\color{red}1=1\cdot 1-0\cdot 0,$ such $a-b=2(c-d)$
(2):$a=4,b=2,c=3,d=2$ then we have
$\color{red}2=4\cdot 2-3\c... | For every $n \in \Bbb N$ you can simply choose $a = 2n+1$, $c=n+1$
and $b = d = 1$.
Then $ab - cd = (2n+1) - (n+1) = n$ and $a-b = 2n = 2(c-d)$.
| {
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"url": "https://math.stackexchange.com/questions/1275761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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solution for complex number equation with power i have the following equation $$z^3 = 2 + 2i$$
and I need to find the solutions of this equation in trigonometric form.
how do I find multiple solutions of such an equation?
| $$z^3=2+2i$$
$$z^3=|2+2i|e^{\arg(2+2i)i}$$
$$z^3=2\sqrt{2}e^{\tan^{-1}\left(\frac{2}{2}\right)i}$$
$$z^3=2\sqrt{2}e^{\tan^{-1}(1)i}$$
$$z^3=2\sqrt{2}e^{\frac{\pi}{4}i}$$
$$z=\left(2\sqrt{2}e^{\left(\frac{\pi}{4}+2\pi k\right)i}\right)^{\frac{1}{3}}$$
$$z=\sqrt{2}e^{\left(\frac{1}{12}\pi +\frac{2}{3}\pi k\right) i}$$
A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1280977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify the expression (combination and factorial) Simplify the following expression:
$\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!}$
My attempt:
$\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n+1-3)!} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n-2)!} * \frac{(n-1)! + (n-2)!}{(n+1)!}$
a... | You must have made a typo with your second entrance into WA. To confirm your first answer:
\begin{align}\require{cancel}
\binom{n+1}{3}\cdot\frac{(n-1)!+(n-2)!}{(n+1)!}&= \frac{\cancel{(n+1)!}}{3!(n-2)!}\cdot\frac{(n-1)!+(n-2)!}{\cancel{(n+1)!}}\\[1em]
&= \frac{(n-1)!+(n-2)!}{3!(n-2)!}\\[1em]
&= \frac{\cancel{(n-2)!}\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1283618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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A limit involving nested trigonometric functions and logarithms
Evaulate
$$ L = \lim_{x \to 0} \frac{1-\cos(\sin x)+\ln(\cos x)}{x^4}. $$
I can solve it using Maclaurin series, but I'm trying to figure out a way to the solution without using it. L'Hopital would probably work but needs to be applied 4 times, and the... | Instead of using the usual limits
$$
\lim_{y\rightarrow 0}\frac{1-\cos y}{y^{2}}=\frac{1}{2},\ \ \ \ \ \ \ \
\lim_{y\rightarrow 0}\frac{\log (1+y)}{y}=1.
$$
which lead to
$$
L=\lim_{x\rightarrow 0}\frac{\frac{1-\cos (\sin x)}{\sin ^{2}x}\frac{\sin
^{2}x}{x^{2}}+\frac{\ln (1+(\cos x-1))}{\cos x-1}\frac{\cos x-1}{x^{2}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1284141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Show that $\sum_{k = 0}^{4} (1+x)^k = \sum_{k=1}^5 \binom{5}{k}x^{k-1}$ Question:
Show that:
$$\sum_{k = 0}^{4} (1+x)^k = \sum_{k=1}^5 \binom{5}{k}x^{k-1}$$
then go on to prove the general case that:
$$\sum_{k = 0}^{n-1} (1+x)^k = \sum_{k=1}^n \binom{n}{k}x^{k-1}$$
Attempted solution:
It might be doable to first prove... | For $\sum_{k = 0}^{n-1} (1+x)^k$ simplify with the geometric summation formula
For $\sum_{k=1}^n \binom{n}{k}x^{k-1}$ use the binomial theorem
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1284853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
solving $a = \sqrt{b + x} + \sqrt{c + x}$ for $x$ I'm trying to solve a very simple looking square root equation but nothing seems to work. The equation has this form (solve for $x$):
$$
a = \sqrt{b + x} + \sqrt{c + x}
$$
Squaring both sides obviously doesn't help since it will still give me a square root. Rearranging ... | HINT
If we agree that $\alpha = \sqrt{b + x}$ and $\beta = \sqrt{c + x}$ we obtain that $\alpha + \beta = a$ and $\alpha^{2} -\beta^{2} = b - c$. Hence it can be claimed $(\alpha + \beta)(\alpha - \beta) = a(\alpha - \beta) = b-c$ which implies the following system of equations
\begin{align*}
\begin{cases}
\alpha - \be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Formal power series coefficient problem Find the coefficient of:
$[x^{33}](x+x^3)(1+5x^6)^{-13}(1-8x^9)^{-37}$
I have figured out that I need to use this identity:
$(1-x)^{-k} = \sum\limits_{i>=0} \binom {n+k-1} {k-1} x^n $
But I have no clue how to proceed with this, I have been stuck with this for hours please help... | You know that
$$(1+5x^6)^{-13}=\sum_{n\ge 0}\binom{n+12}{12}\left(-5x^6\right)^n=\sum_{n\ge 0}\binom{n+12}{12}(-1)^n5^nx^{6n}\tag{1}$$
and
$$(1-8x^9)^{-37}=\sum_{n\ge 0}\binom{n+36}{36}\left(8x^9\right)^n=\sum_{n\ge 0}\binom{n+36}{36}8^nx^{9n}\;.\tag{2}$$
The coefficient of $x^{33}$ in $(x+x^3)(1+5x^6)^{-13}(1-8x^9)^{-... | {
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"url": "https://math.stackexchange.com/questions/1289014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Convergence of $\sum_{n=0}^\infty \frac{\ln(1+2^n)}{n^2+x^{2n}}$ Let $x \in \mathbb{R}$. Define the series:
$$\sum_{n=0}^\infty \frac{\ln(1+2^n)}{n^2+x^{2n}}.$$
For what $x$ does it converge?
It clearly has positive terms. The ratio and root tests seem ineffective, and I am personally not able to study the confrontati... | For $|x|\le 1$ we have
$$
\frac{\ln(1+2^n)}{n^2+x^{2n}}>\frac{\ln2^n}{n^2+n^2}=\frac{\ln2}{2}\cdot\frac{1}{n}\quad \forall n\ge 1.
$$
Since the Harmonic series $\sum_{n=1}^\infty\frac{1}{n}$ is divergent, so is the series $\sum_{n=0}^\infty\frac{\ln(1+2^n)}{n^2+x^{2n}}$.
For $|x|>1$ we have:
$$
\frac{\ln(1+2^{n+1})}{(n... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\alpha$ lies between $0$ and $4$.
Let $a,b,c$ be the length of the sides of the triangle $ABC$ .
Given $(a+b+c)(b+c-a)=\alpha bc$.Then Prove that the value of $\alpha$ lies in between $0$ and $4$.
$\begin{align}(a+b+c)(b+c-a)&=\alpha bc\\
\implies \alpha&=\dfrac{b^2+c^2-a^2}{bc}+2\\
\alpha&=2\cos A+2\... | We have
$$(b+c)^2-a^2 = \alpha bc \implies b^2+c^2-a^2 = bc(\alpha-2) \implies \alpha = 2 + \dfrac{b^2+c^2-a^2}{bc} = 2+2\cos(A)$$
Note that $\cos(A) \in [-1,1]$. Hence, we have
$$\alpha \in \left[2-2,2+2\right] = \left[0,4\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1291780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Need help with taylor series.
Evaluate the limit
$$\lim\limits_{x \to 1} \frac{1-x + \ln x}{1+ \cos πx}$$
The limit im trying to get is $-\frac{1}{π^2}$ as I've solved from l'Hopitals rule.
Now I need to solve the limit by using Taylor Series and this is what i did so far
$$\begin{align*}
f(x) &= 1-x + \ln x = 1 -x... | First, let $h=x-1$ then $\cos \pi x=\cos (\pi +\pi h)=-\cos (\pi h).$
Next re-write the original fraction as the following product
\begin{equation*}
\frac{1-x+\ln x}{1+\cos \pi x}=\frac{-h+\ln (1+h)}{1-\cos (\pi h)}=\frac{%
-h+\ln (1+h)}{h^{2}}\times \frac{(\pi h)^{2}}{(1-\cos (\pi h))}\times \frac{1%
}{\pi ^{2}}.
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1292372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Creative way to find this area Let's say We have a circle with center at $(0,0)$ with radius $r$
and we have the line $y=a$ where $0 \leq a \leq r$.
the question is what is the area that between the circle and the line $y=a$(the area that above the line).
illustration for $r=5$ and $a=4$: http://www.wolframalpha.com/in... | Using, geometry we have the equation of the circle centered at the origin as:
$$x^2+y^2=r^2$$
Now, substitute $y=a$ in the above equation
$$x^2+a^2=r^2 \implies x=\pm \sqrt{r^2-a^2}$$
Now, join the points of intersection $(\sqrt{r^2-a^2}, a),\,(-\sqrt{r^2-a^2}, a)$ to the origin, to get a sector with radius $r$ and an ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1292813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Finding $\int\frac{1}{x^{11}+4x^6}dx$ I wanted to find out if there is an easy way to evaluate $$\displaystyle\int\frac{1}{x^{11}+4x^6}dx.$$
I substituted $u=x^5$ and then used partial fractions, but maybe there is a simpler way to find this.
| $$\int\frac{1}{x^{11}+4x^6}dx=$$
$$\int \left(\frac{1}{4x^6}+\frac{x^4}{16(x^5+4)}-\frac{1}{16x}\right)dx=$$
$$\frac{1}{16}\int \frac{x^4}{x^5+4}dx-\frac{1}{16}\int \frac{1}{x}dx+\frac{1}{4}\int \frac{1}{x^6}dx=$$
$$\frac{1}{80}\int \frac{1}{u}du-\frac{1}{16}\int\frac{1}{x}dx+\frac{1}{4}\int\frac{1}{x^6}dx=$$
$$\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Method to solve equation to find the value of each variable Ok lets say I am given a question:
There are 17 males students and 19 female students.
out of the 17 male students 11 of them do hip hop while the other 6 do drag racing
while in the female side 11 of them do hip hop while the other 8 drag racing.
out of the... | Assuming e.g. mhe represents a single variable (i.e., not $m \times h \times e$), we can perform Gauss-Jordan Elimination to obtain the reduced row echelon form:
$$\left(\begin{array}{cccccccc|c} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 36 \\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 17 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 19 \\ 1 & 1 & 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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A not so hard basic calculus problem? But it appears to be very lengthy Find the coordinates of the two points on the curve $y=4-x^2$ whose tangents pass through the point $(-1,7)$.
My work:
Let the two points be $(a,b)$ and $(c,d)$. And $\frac{dy}{dx}=-2x$, so the gradients of the tangents are $-2a$ and $-2c$. We get ... | You made it complicated. Let the one point be $(a,\ 4 - a^2)$ and $\frac{\text{d}y}{\text{d}x} = −2x$, so the gradient of the tangent is $−2a$; we get $\frac{7 - (4 - a^2)}{(-1-a)} = -2a$, so you can get the answer easily: $(1,3)$, $(-3,-5)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove this is an isosceles triangle In a $\triangle ABC$, $$\qquad \sin B\cdot\sin C=\cos^2\left(\frac{A}{2}\right)$$
Prove that this is an isosceles triangle.
Can anyone guide me to prove this? Thanks!
| Note: in a triangle ABC, we have $A+B+C=\pi=180^o$
Given $$\sin B\sin C=\cos^2\frac{A}{2} \implies 2\sin B\sin C=2\cos^2\frac{A}{2} $$$$\implies \cos(B-C)-\cos(B+C)=2\cos^2\frac{A}{2} $$$$\implies \cos(B-C)-\cos(180^o-A)=2\cos^2\frac{A}{2} $$ $$\implies \cos(B-C)+\cos A=2\cos^2\frac{A}{2} $$$$\implies \cos(B-C)+2\cos^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1298165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 0
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Question in regard to solving for inverse laplace transform I am having some confusion when it comes to solving for the inverse laplace transform. ( We are allowed the tables with the common values by the way).
Il give an example.
Take,
$$y''+4y'+8y=-6e^{-2t}cos(3t)$$ with $y(0)=-2$ and $y'(0)=-5$
Now, by invoking the ... | Your work is correct. Gather everything under common denominator:
$$\frac{-2 s^3-21 s^2-84 s-181}{\left(s^2+4 s+8\right) \left(s^2+4 s+13\right)}=\frac{A s+B}{s^2+4 s+8}+\frac{C s+D}{s^2+4 s+13}$$
This will be super messy, I will only calculate $A$ and $B$. Multiply everything by ${s^2+4 s+8}$
$$\frac{-2 s^3-21 s^2-84 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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taking the limit $\lim\limits_{n\rightarrow \infty} {\frac{(3^{n+1} + 4)(7^n-47)}{(7^{n+1}-47)(3^n +4)} }$ I need help with a guide on how i will deal with this kind of problem.. This a part of my solution in series convergence. I find it hard taking the limit of this: $$\lim_{n\rightarrow \infty} {\frac{(3^{n+1} + 4)(... | You can factor out the exponentials to get:
$$
\frac{3^{n + 1}7^n\left(1 + \frac{4}{3^{n + 1}}\right)\left(1 - \frac{47}{7^n}\right)}{7^{n+1}3^n\left(1 - \frac{47}{7^{n + 1}}\right)\left(1 + \frac{4}{3^n}\right)}
$$
Which leaves us with:
$$
\frac{3}{7}\cdot\frac{\left(1 + \frac{4}{3^{n + 1}}\right)\left(1 - \frac{47}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simple trigonometry equation The previous class we were doing trigonometry exercises. Before the class finished, our teacher wrote exercises on the table. I am stuck with the following one:
$$
\cos(2x) + 1 + 3\sin x = 0
$$
I have come up with this:
$$
1= \sin^2 x + \cos^2 x
$$
$$
\cos(2x) = \cos^2 x - \sin^2 x
$$
Whe... | $$\cos(2x)+1+3\sin x=0$$
now let's substitute $\cos (2x)=1-2\sin^2x$
$2-2\sin^2x+3\sin x=0$
And then, if we denote $y=\sin x$ we get a quadratic equation
$$2-2y^2+3y=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1301550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Inequality and Trigonometric Substitution Prove that for all positive real $a,b,c$, we have
$$(a^2+2)(b^2+2)(c^2+2) \geq 9(ab+bc+ca).$$
Because of the term $a^2+2$, this motiveates me to substitute $a=\sqrt{2}\tan A, b=\sqrt{2}\tan B, c=\sqrt{2}\tan C$ and the fact that $1+\tan^2x=\sec^2x$, then the desired inequality... | Ok,if you use trigonometry to solve it
as you
$$\Longleftrightarrow \cos{A}\cos{B}\cos{C}(\cos{A}\sin{B}\sin{C}+\sin{A}\cos{B}\sin{C}+\sin{A}\sin{B}\cos{C})\le\dfrac{4}{9}$$
$$\cos{A}\cos{B}\cos{C}(\cos{A}\cos{B}\cos{C}-\cos{(A+B+C)})\le\dfrac{4}{9}$$
By Jenson inequality we have
$$\cos{A}\cos{B}\cos{C}\le\cos^3{\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1302554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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For which values of $a$ does the matrix can be diagnolized? Given $$A=\begin{pmatrix}
2 & 0 & 0\\
a & 2& 0\\
a+3 & a &-1
\end{pmatrix}$$
For which values of $a$ can $A$ be diagonal?
I found that $p_A(x)=(x-2)^2(x+1)$ and tried to find the eigen subspace of 2, to see if the geomtric multiplicity of the eigenvalue $2... | You'll get for $a \neq 0$, that the nullspace of $A-2I$ is less than $2$. To see why, you have that
\begin{align*} A - 2I = \left(\begin{matrix} 0 & 0 & 0 \\ a & 0 & 0 \\ a+3 & a & - 3 \end{matrix}\right) \end{align*}
is row equivalent to
\begin{align*}\left(\begin{matrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ a+3 & a & - 3 \e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1302932",
"timestamp": "2023-03-29T00:00:00",
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How can I show this inequality: $-2 \le \cos \theta (\sin \theta +\sqrt{\sin ^2 \theta +3})\le 2$
Show that $$-2 \le \cos \theta ~ (\sin \theta +\sqrt{\sin ^2 \theta +3})\le 2$$ for all value of $\theta$.
Trial: I know that $0\le \sin^2 \theta \le1 $. So, I have $\sqrt3 \le \sqrt{\sin ^2 \theta +3} \le 2 $. After tha... | Let $\cos \theta ~ (\sin \theta +\sqrt{\sin ^2 \theta +3})=y$
$\iff \sin \theta +\sqrt{\sin ^2 \theta +3}=y\sec\theta$
$\iff \sqrt{\sin ^2 \theta +3}=y\sec\theta-\sin \theta$
Squaring we get $\sin ^2 \theta +3=y^2(1+\tan^2\theta)+\sin^2\theta-2y\tan\theta$
$\iff y^2(\tan^2\theta)-2y(\tan\theta)+y^2-3=0$
As $\tan\theta$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1303772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Is the argument I used to evaluate the convergence of the series $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{n+a}{(n+b)(n+c)}$ right?
If $a,b,c$ be real constants, analyze the convergence of $$\sum_{n=1}^{\infty} (-1)^{n-1}\frac{n+a}{(n+b)(n+c)}$$
What I tried to:
I compared the general term of my series to $\frac{1}{n}$: $... | As Julian Aguirre pointed, we are not allowed to use a comparison with an alternating series. However, Dirichlet's test is enough to ensure convergence, since the partial sums of $(-1)^n$ are bounded and $\frac{(n+a)}{(n+b)(n+c)}$ is eventually decreasing to zero as $n\to +\infty$.
We may notice that:
$$\frac{(n+a)}{(n... | {
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"url": "https://math.stackexchange.com/questions/1303968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Find the Matrix of T with respect to basis B. A linear transformation $T : P_2 \rightarrow P_2$ is given by
$T(a+bx+cx^2) = (a−b+c)+(b+c)x+(2b−a)x^2$.
It is given that the set $B=\{1+x+x^2, x+x^2, x^2\}$
is a basis for $P_2$.
(a) Find the matrix of $T$ with respect to the basis $B$ in both the domain and the
codomain.
... | The matrix $T$, with respect to the basis $v_1= 1 + x + x^2$, $v_2= x + x^2$, and $v_3=x^2$ can be computed directly as follows
$T(v_1)=T(1+x+x^2)=(1-1+1) + (1+1)x + (2-1)x^2=1 + 2x + x^2= v_1+ v_2 - v_3$,
$T(v_2)=T(x+x^2)= 2x + 2x^2 = 2v_2$,
$T(v_3)=T(x^2)= 1 + x = v_1-v_3$
Therefore, the matrix $T=(a_{ij})$ is the m... | {
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"timestamp": "2023-03-29T00:00:00",
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Simple argument for $\frac{(x+y)^2}{x^2+xy+y^2}\le 4/3$ I would like to show that $\forall x,y\in\mathbb R^+:\frac{(x+y)^2}{x^2+xy+y^2}\le 4/3$.
The inequality is indeed true as the maximum of $\frac{(x+y)^2}{x^2+xy+y^2}$ is reached for $x=y$ and its value is $4/3$.
Except for the standard way of computing partial deri... | Since $x^2+xy+y^2=\left(x+\frac y2\right)^2+\frac{3y^2}{4}\gt 0$, one has$$\begin{align}\frac{(x+y)^2}{x^2+xy+y^2}\le\frac 43&\iff (x+y)^2\le\frac{4}{3}(x^2+xy+y^2)\\&\iff \frac 13x^2-\frac 23xy+\frac 13y^2\ge0\\&\iff \frac 13(x-y)^2\ge 0\end{align}$$
This is true.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to distribute $x \times \sqrt{x^2 + 1}$ How do I distribute the x in this problem?
How do I "gain access" so to say. Does it become $x^\frac{1}{2}(x^2 + 1)^\frac{1}{2}$ or $(x^3+x)^\frac{1}{2}$ ?
Or do I need to even do that in order to integrate $\int x(x^2+1)^\frac{1}{2}$ ?
| You don't need to expand in order to integrate. Just use a $u$ substitution with $u = x^2 + 1$ and so $du = 2x \space dx$ and so $dx = \frac{du}{2x}$ and so
your integral becomes $$\large{\int \color{red}{x}u^\frac{1}{2} \times \frac{du}{2\color{red}{x}}}$$ and so you cancel out the $x$ to end up with $$ \large{\int u^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim\limits_{x\to+\infty}\frac{\left(x+\sqrt{x^2-1}\right)^n+\left(x-\sqrt{x^2-1}\right)^n}{x^n},n\in \mathbb{N}$ If we use the following
$$a^n-b^n=\left(a-b\right)\left(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}\right)=u\times t$$
$$u=x+\sqrt{x^2-1}-x+\sqrt{x^2-1}=2\sqrt{x^2-1}=2x\sqrt{1-\frac{1}{x}}$$
Now, th... | Let $a= x+\sqrt{x^2-1}$, then we have $x-\sqrt{x^2-1}=\frac{1}{a}$ and $a+ \frac{1}{a}= 2x$. Next, we can write $$\lim\limits_{x\to+\infty}\frac{(x+\sqrt{x^2-1})^n+(x-\sqrt{x^2-1})^n}{x^n}
=\lim\limits_{a\to+\infty}\frac{a^n+\frac{1}{a^n}}{\frac{1}{2^n}(a+\frac{1}{a})^n}=\lim\limits_{a\to+\infty}\frac{a^n}{\frac{1}{2^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:
Theorem: $\sqrt{2}$ is irrational.
Proof:
$3^2-2\cdot 2^2 = 1$.
(That's it)
That is a corollary of
this result:
Theorem:
If $n$ is a positive integer... | The irrationality of $\sqrt{2}$ can be deduced from the following
Theorem (Fermat, 1640): The number $1$ is not congruent.
Reasoning: If $\sqrt{2}$ were rational then $\sqrt{2},\sqrt{2}$,and $2$ would be the sides of a rational right triangle with area $1$. This is a contradiction of $1$ not being a cong... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "114",
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Using an inverse operator to find a particular solution to a differential equation. I am just learning about inverse operators in solving a differential equation, but I don't understand exactly how they work. For example, find a particular solution to
$$4y''-3y'+9y=5x^2$$
using inverse operators.
The above equation is ... | Notice that
$$4D^2 - 3D + 9 = \left [D - \left ( \frac{3}{8} (1 - i \sqrt{15} )\right)\right ] \left [D - \left ( \frac{3}{8} (1 + i \sqrt{15} )\right)\right ] $$
Also note that
$$ \frac{1}{-a( 1 - D /a )} = -\frac{1}{a} \left ( 1 + \frac{D}{a} + \frac{D^2}{a^2} + \ldots \right) $$
Thus to find $y_p$, we simply have ... | {
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"timestamp": "2023-03-29T00:00:00",
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An asymptotic formula for the bounded sum of primes. How to prove the following asymptotic formula?:
$$
\sum\limits_{p\leq x} p \sim \frac{x^2}{2 \log x}
$$
I'm stuck and I don't know where to start. I've been suggested the use of $\pi (x) = \frac{x}{\log x} + O\left( \frac{x}{\log ^2 x} \right) $. But I'm unsure about... | If $\displaystyle S(n) = \sum\limits_{p\leq n} p$, (sum of primes not exceeding $n$),
then by Abel's summation formula, we may write:
$$S(n) = n\pi(n) - \sum\limits_{j=2}^{n-1} \pi(j)$$
So, using the estimate $\displaystyle \pi (x) = \frac{x}{\log x} + O\left( \frac{x}{\log ^2 x} \right)$,
$\begin{align}S(n) &= \frac{n... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Homework question: Represent function as power series Represent the following function as a power series of powers of x-2.
$$f(x)=\frac{x-5}{3x^2+5x-2}$$
Now, the Taylor series got me thinking. Knowing my professor, she wouldn't have us solving huge derivatives like this one. So I'm wondering, is there any other way to... | Yes, there is another way of approaching this.
First, it will make things a bit easier to let $y=x-2$.
Then, write
$$\begin{align}
3x^2+5x-2&=(3x-1)(x+2)\\\\
&=(3(x-2)+5)((x-2)+4)\\\\
&=(3y+5)(y+4)
\end{align}$$
Next, using partial fraction expansion reveals that
$$\begin{align}
\frac{x-5}{3x^2+5x-2}&=\frac{y-3}{(... | {
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For what $k$ does $\lim_{x \to-\infty} \frac{5^{kx}-1}{5^{-2x} + 1}$ exist? For what values of $k$ does this limit exist?
$$\lim_{x \to-\infty} \frac{5^{kx}-1}{5^{-2x} + 1}$$
Progress: I worked it by dividing everything by $5^{-2x}$ and now have $5^{x(k+2)}$ since the bottom half ended up being $1$ and the top ends up ... | One more hint:
$$\frac{5^{kx}-1}{5^{-2x} + 1} = \frac{5^{kx}}{5^{-2x} + 1} - \frac{1}{5^{-2x} + 1} = \frac{1}{5^{-(2+k)x} + 5^{-kx}} - \frac{1}{5^{-2x} + 1} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Taylor expansion of $f(x) = \frac{x}{x+3}\frac{1}{x-2}$ near $x=2$. I am trying to Taylor expand the function $$f(x) = \frac{x}{x+3}\frac{1}{x-2}$$ around the point $x_0 = 2$. Clearly, the last factor explodes around this point, so I will try and expand that term. However, terms on the form $\frac{1}{x-2}$ are a geomet... | I think here we may rather talk about a Laurent series expansion around $x=2$.
We have, as $x \to 2$,
$$
\begin{align}
f(x) &= \frac{1}{x-2}\frac{x}{x+3}\\\\
&= \frac{1}{5(x-2)}\frac{2+(x-2)}{\left(1+\dfrac{(x-2)}5\right)}\\\\
&= \frac{2+(x-2)}{5(x-2)}\sum_{n\geq0}\dfrac{(-1)^n}{5^n}\left(x-2\right)^n,\quad 0<|x-2|<5,\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Asymptotic expansion of integral Find the first two terms in asymptotic expansion (as $x \rightarrow \infty$) of the following integral
$$\int_{0}^{\frac{\pi}{2}}e^{-x \sin^{2} t}dt.$$
| As Lucian pointed out in a comment $$f(x)=\int_{0}^{\frac{\pi}{2}}e^{-x \sin^{2} t}dt=\frac{ \pi}{2} e^{-x/2}\, I_0\left(\frac{x}{2}\right)$$ At this point, one could use the asymptotic expansion given by Hankel for the modified Bessel function of the first kind $$I_{\alpha}(z)=\frac{e^z}{\sqrt{2\pi z}}\Big(1-\frac {4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1318546",
"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int_0^{\pi/2} \ln(\sin(x)) dx$ using the Beta, Gamma and Digamma Functions. Evaluate
$$I(\theta)=\int_0^{\pi/2} \ln(\sin(\theta))d\theta$$
$$$$
My incorrect attempt:$$$$
Consider $$J(a,0)=\int_0^{\pi/2} \sin^a(\theta)cos^0(\theta) d\theta$$
$\dfrac{d}{da}J(0,0)=I(\theta)$$$$$
Now, $J(a,0)=\beta\bigg ... | It does give the correct answer, indeed. I think you only need the powerful identity:
$$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}\tag{1}$$
that follows from the series definition of the digamma function. That gives, for instance:
$$\log 2 = \sum_{n\geq 1}\frac{(-1)^{n+1}}{n} = \sum_{n\geq 0}\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1318608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove Alternating Series Approximation Prove if $S=\sum_{n=1}^{\infty}a_{n}$ is an alternating series with $\left | a_{n+1}\right | < \left | a_{n} \right |$, and $\lim_{n\to\infty}a_{n}=0$, then $\left |S-(a_{1}+a_{2}+\cdots+a_{n}) \right | \leq \left | a_{n+1} \right | $ I'm supposed to group the terms in the error a... | It is I think best to look at a particular example, or two, or three. Consider the alternating series
$$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23}+\cdots.\tag{1}$$
This is the famous Leibniz series. It converges to $\frac{... | {
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"timestamp": "2023-03-29T00:00:00",
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Show an $\arctan$ and $\arcsin$ function is constant Show that for every $x\geq1$ the following is true:
$2\arctan x + \arcsin \frac{2x}{1+x^2} = \pi$
One way (mentioned in the link at the bottom) would be to calculate the derivative of the left side, show that it is always $0$ then show that for $x=1$ the equation is ... |
\begin{align} 2\arctan(x)+\arcsin\left(\frac{2x}{1+x^2}\right)
&=\pi
\end{align}
Let's $\chi=\arctan(x)\ge\pi/4$ since $x\ge1$.
Consider a unit circle
Then in a $\triangle OAB$ we have $|OA|=|OB|=1$,
$\angle OAB=\angle OBA=\chi$,
$\angle AOB=\alpha=\pi-2\chi$,
$|AH|=\sin(\alpha)=a$.
In the $\triangle ABH$
\begin{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1321351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is $y^2=x^3+7$ unsolvable modulo some $n$? The equation $y^2=4x^3+7$ has no integral solution since $y^2\equiv4x^3+7\pmod4$ has no solution (i.e. has no solution in $\Bbb{Z}/4\Bbb{Z}$).
It is well known that $y^2=x^3+7$ has no integral solution, but is there an integer $n (n>1)$ such that $y^2\equiv x^3+7\pmod n$ ha... | The discriminant of $x^3+7$ is $-27\cdot49$. Given a prime power $q$, this discriminant is non-zero in the field $F_q$ unless $q$ is $3,9,27,7,$ or $49$. For other values of $q$, $y^2=x^3+7$ is an elliptic curve over $F_q$, and we can use Hasse's bound
$$|N-(q+1)| \le 2\sqrt q$$
where $N$ is the number of points on the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1321423",
"timestamp": "2023-03-29T00:00:00",
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The algebra generated by the set $\{1,x^2\}$ is dense in $C\left[0,1\right]$ with the supremum norm but fails to be dense in $C\left[-1,1\right]$. Show that the algebra generated by the set $\{1,x^2\}$ is dense in $C\left[0,1\right]$ with the supremum norm but fails to be dense in $C\left[-1,1\right]$.
I have know that... | To show that the algebra generated by $\{1,x^2\}$ is not dense in $C[-1,1]$ consider $x\in C[-1,1]$. Let $f$ be in the algebra, since $1$ and $x^2$ are both even functions, $f$ is also an even function.
Consider the cases where $x=\frac{1}{2}$ and $x=-\frac{1}{2}$. At these points, $f(-\frac{1}{2})=f(\frac{1}{2})$. ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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On representation of even integers as $a^x+b^y$ Does there exist some $k \in \mathbb N$ such that for every even integer $n \ge k $ we can find positive integers $a,b,x,y$ such that $n=a^x+b^y$ , where $\gcd (a,b)=1 ; a,b>1$ and at least one of $x,y$ is more than $1$ ? I know that for every even integer $n>6$ there are... | Let $n \ge 6$. Choose a prime $p$ such $p^2 \le n-2$ and $p$ does not divide $n$. Then $n=p^x+(n-p^2)^y$ with $x=2$ and $y=1$, and $p$ and $n-p^2$ are greater than $1$ and relatively prime. This fails only if there is no such prime, which happens only if $n$ is divisible by every prime between $2$ and $\sqrt{n-2}$. ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$x^2-y^2=196$, can we find the value of $x^2+y^2$? $x$ and $y$ are positive integers.
If $x^2-y^2=196$, can we know what the value of $x^2+y^2$ is?
Can anyone explain this to me? Thanks in advance.
| Rewrite the equation to get $x^2=y^2+196$. Now write $x=y+k$, then $$x^2=(y+k)^2=y^2+2yk+k^2=y^2+196$$
Therefore $2yk+k^2=196$
Now note that $k \mid 196$ and $k$ must be even.
This leaves the possibilities $2,4,14,28,98,196$.
$k=2$ gives $y=48$.
$k=4$ gives $y=22.5$, which is not an integer.
$k=14$ gives $y=0$, but t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to integrate $\int x\sqrt { \frac {a^2-x^2}{a^2+x^2} } \, dx$? How to integrate $\displaystyle\int x\sqrt {\frac{a^2 - x^2}{a^2+x^2}} \, dx$ ?
I tried this for quite some time, but not being able to solve. Help!
| Hint:
$$x\sqrt{\frac{a^2-x^2}{a^2+x^2}}=x\sqrt{\frac{1-(\frac{x}{a})^2}{1+(\frac{x}{a})^2}}$$
and use the trigonometric identity
$$\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=|\tan\left(\frac{\theta}{2}\right)|$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Theorem about prime numbers which are one more than a multiple of $8$ Assume the following:
A prime $p>2$ can be written as $p=m^2+n^2$ for integral $m,n$ iff $p$ is $1$ more than a multiple of $4$.
Now, prove that every prime which is one more than a multiple of $8$ be written as $x^2+16y^2$ where $x$ and $y$ are in... | Let $p\equiv 1\pmod{\! 8}$. Since $p\equiv 1\pmod{\! 4}$, exist $x,y\in\Bbb Z$ giving you $p=x^2+y^2$.
Exactly one of $x,y$ is even, wlog $y$ is even.
$x^2\equiv 1\pmod{\! 8}$, $y^2\equiv \{0,4\}\pmod{\! 8}$ and $p\equiv 1\pmod{\! 8}$.
So we must have $y^2\equiv 0\pmod{\! 8}$.
$8\mid y^2\iff 4\mid y$. Let $y... | {
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"timestamp": "2023-03-29T00:00:00",
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Determine when value will be greater than x I want every time to add '$(b-x)$' to the value of $a$ and i want to know how many times i will repeat this operation to be the value of $a$ greater or equal to $x$.
$a = b+a - x$
I know that $b > x$ and $ a < x $
Any ideas?
| So you have an arithmetic sequence
$t_n = a + n(b - x)$, (Starting at $n = 0$).
where $a < x < b$, and you want to know the first time that $t_n > x$
\begin{align}
t_n & > x\\
a + n(b - x) & > x\\
n(b - x) & > x - a\\
n & > \dfrac{x - a}{b - x}\\
n & = \left \lceil \dfrac{x - a}{b - x} \right \rceil
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1325063",
"timestamp": "2023-03-29T00:00:00",
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Prove that the sum of the Lagrange (interpolation) coefficients is equal to 1 Prove that the sum of the Lagrange (interpolation) coefficients is equal to 1.
Please suggest me a book-reference or give a solution for me. Thanks a lot in advance.
If $f = \sum_{i=0}^nf(x_i)L_i(x)$ then one has to prove $\sum_{i=0}^nL_i(x... | Let me propose one more method of proof.
"Beginner's guide to mapping simplexes affinely", section "Lagrange interpolation", describes a determinant form of Lagrange polynomial that interpolates $(a_0;b_0)$, $\dots$, $(a_n;b_n)$
$$
P(x) = (-1)
\frac{
\det
\begin{pmatrix}
0 & b_0 & b_1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1325292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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converting a differential equation to polar coordinates I have the following family of autonomous systems, I'm having trouble with part b):
$$x'=x(1-\sqrt {x^2+y^2})-y-\epsilon y$$
$$y'=y(1-\sqrt {x^2+y^2})+x+\epsilon(x+x^2+y^2)$$
a) Convert the system to polar coordinates $(r(t;r_0,\theta_0,\epsilon),\theta(t;r_0,\the... | $$\left ( \begin{array}\\ \cos{\theta} & -r \sin{\theta} \\ \sin{\theta} & r \cos{\theta} \end{array} \right )^{-1}= \left ( \begin{array}\\ \cos{\theta} & \sin{\theta} \\ \frac{-1}{r}\sin{\theta} & \frac{1}{r} \cos{\theta} \end{array} \right )$$
Multiplied on the left side, simplified and obtained:
$$\left ( \begin{a... | {
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"url": "https://math.stackexchange.com/questions/1325594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integers as a sum of $\frac{1}{n}$ Say $\sum_{i \in I} \frac{1}{n_i} = 2$, where $(n_i)_{i \in I}$ is a finite sequence of positive integers (not necessarily distinct). Is there a subsequence $(n_i)_{i \in J}$ of $(n_i)_{i \in I}$ such that $\sum_{i \in J} \frac{1}{n_i} = 1$?
| Here is a counterexample even when all fractions are required to be different!
$\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{11}+\frac{1}{13}+\frac{1}{16}+\frac{1}{17}+\frac{1}{21}+\frac{1}{1105}+\frac{1}{55692}+\frac{1}{1361360}$
This came from:
$\frac{1}{2}+\frac{1}{3}+... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find area bounded by $(\frac{x}{3}+\frac{y}{6})^3-9xy=0$ $C = \{(x,y) \mid (\frac{x}{3}+\frac{y}{6})^3-9xy=0\}$.
I want to find the area bounded by $C$ in first quadrant.
Can you tell me how to solve exercises like this?
I have no idea what integral I need to solve, because my curve is implicitly defined.
| Let's use (slightly modified) polar coordinates. Set $x = 3r\cos\theta$ and $y = 6r\sin\theta$.
The Jacobian of this transformation is $\left|\begin{matrix}\tfrac{\partial x}{\partial r} & \tfrac{\partial y}{\partial r} \\ \tfrac{\partial x}{\partial \theta} & \tfrac{\partial y}{\partial \theta}\end{matrix}\right| = \... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the half range cosine fourier series expansion for $f(x)=(x-1)^2,\quad 0Find the half range cosine fourier series expansion for
$$f(x)=(x-1)^2,\quad 0<x<1$$
and hence deduce that
$$\pi^2=8\left(\frac 1 {1^2}+\frac 1 {3^2}+\frac 1 {5^2}+\ldots\right)\tag{1}$$
My work
I have derived that the expansion is
$$f(x)=\fra... | Define $f$ as a periodic function of $1$ on $\mathbb{R}$ as
$$
f(x)=(x−1)^2, \hspace{5 mm} 0\le x<1
$$
Set $x=0$, then
$$
f(0)=1=\frac 1 3+\sum\limits_{n=1}^\infty \frac 4 {n^2\pi^2}
$$
So we have
$$
\dfrac{2}{3}=\sum\limits_{n=1}^\infty \frac 4 {n^2\pi^2} \hspace{4 mm} \text{and} \hspace{4 mm} \dfrac{\pi^2}{6}=\sum\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1329752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Variation on Pythagoras: If $a^2 + b^2 = c^2$, then $a + b \leq c\sqrt{2}$ This is a very interesting word problem that I came across in an old textbook of mine. So I know its got something to do with derivative of the Pythagorean Theorem using calculus, trigonometry, geometry, or plain old algebra, which yields the sh... | We solve this derivative of the Pythagorean Theorem using calculus, trigonometry,
geometry, or plain old algebra, which yields the shortest, simplest proofs!
Calculus:
The heavy hand of calculus does not provide a short
or elegant proof, but I always like seeing more than one approach.
$a^2 + b^2 = c^2$
$a^2 = c^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 9,
"answer_id": 4
} |
How to evaluate $\int \frac { \sin x+\cos x }{ \sin^4 x+\cos^4x}\, dx$? How can one find $$\int \frac { \sin x+\cos x }{ \sin^4 x+\cos^4x}\, dx?$$
| Split the integral as
$$
\int\frac {\cos x}{\sin^4 x + (1 - \sin^2 x)^2} dx
+ \int \frac{\sin x}{(1 - \cos^2 x)^2 + \cos^4 x} dx
$$
Compute the integrals with the substitutions $u = \sin x$ and $u = \cos x$ respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Nesbitt's Inequality for 4 Variables I'm reading Pham Kim Hung's 'Secrets in Inequalities - Volume 1', and I have to say from the first few examples, that it is not a very good book. Definitely not beginner friendly.
Anyway, it is proven by the author, that for four variables $a, b, c$, and $d$, each being a non-negati... | The simplest is to use the AM-HM inequality; which is a consequence of the AM-GM inequality, but can be proved independently.
Remember the harmonic mean of $x$ and $y$ is the number whose inverse is the arithmetic mean of the inverses of $x$ and $y$. Explicitly:
$$\frac1H=\frac12\Bigl(\frac1x+\frac1y\Bigr)\ge \frac1A=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Complex integration with trigonometric and logarithm
Show that $\int_0^{2\pi}\log\sin^22\theta dx=4\int_0^\pi\log\sin \theta d\theta=-4\pi
\log2$
I did $$\int_0^{2\pi}\log\sin^22\theta d\theta=4\int_0^{\frac{\pi}{4}}\log\sin^22\theta d\theta$$
taking $u=2\theta$
$$2\int_0^\frac{\pi}{2}\log\sin^2\theta d\theta$$
but ... | Consider what has already been demonstrated
\begin{align}
\int_{0}^{2\pi}\log(\sin^{2}(2\theta)) \, d\theta &= \frac{1}{2} \, \int_{0}^{\pi} \log(\sin^{2}(\theta)) \, d\theta = \frac{I}{2}
\end{align}
Now let $2 \sin^{2}(\theta) = 1 - \cos(2\theta)$ to obtain
\begin{align}
I &= - \ln(2) \, \int_{0}^{\pi} d\theta + \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1335164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Why does $\cos(x) + \cos(y) - \cos(x + y) = 0$ look like an ellipse? The solution set of $\cos(x) + \cos(y) - \cos(x + y) = 0$ looks like an ellipse. Is it actually an ellipse, and if so, is there a way of writing down its equation (without any trig functions)?
What motivates this is the following example. The solution... | we have $$\begin{align} 0 &= \cos x +\cos y -\cos(x + y)\\
&=\cos x +\cos y -\cos x \cos y+ \sin x \sin y\\
&=(1-\cos x)\cos y+\sin x \sin y+\cos x\\
&=2\sin^2 (x/2)\cos y+2\sin(x/2)\cos(x/2)\sin y + \cos x\\
&=2\sin(x/2)\left(\sin (x/2)\cos y+\cos(x/2)\sin y)\right)+\cos x\\
&=2\sin(x/2)\sin (x/2 + y)+\cos x\\
\end{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1336698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
"answer_count": 9,
"answer_id": 3
} |
Two different trigonometric identities giving two different solutions Using two different sum-difference trigonometric identities gives two different results in a task where the choice of identity seemed unimportant. The task goes as following:
Given $\cos 2x =-\frac {63}{65} $ , $\cos y=\frac {7} {\sqrt{130}}$, unde... | In your calculation of $x+y$ via $\sin(x+y)$, you wrote that $x+y=\frac{\pi}{4}$ is a possibility. It is not, since already $\sin y=\frac{9}{\sqrt{130}}\gt\frac{1}{\sqrt{2}}$, so $y\gt \frac{\pi}{4}$.
Remark: In this case, computing $\cos(x+y)$ is the better strategy, since the cosine here unambiguously identifies $x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
How to solve this system of equations for $x^2+y^2+z^2$? For the complex numbers $x,y,z$, the system of equations
$x^2-yz=i~~~~~
y^2-zx=i~~~~~
z^2-xy=i$
It is not easy for me to get $x^2+y^2+z^2$ from the above. I don't need the values of $x,y,z$
I'm stuck in what to do at first. Any advice would be helpful.
My attem... | Subtracting the equations in pairs we get
$$(x-y)(x+y+z) = 0$$
$$(x-z)(x+y+z) = 0$$
$$(y-z)(x+y+z) = 0$$
If $x+y+z \not = 0$ then $x=y=z$, but then $x^2-yz = 0 \not = i$ so we must have $x+y+z = 0$.
Next we can use the identity
$$x^2 + y^2 + z^2 = (x+y+z)^2 - 2(xy+yz+xz)$$
togeather with the equation system to arrive a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Expand $\binom{xy}{n}$ in terms of $\binom{x}{k}$'s and $\binom{y}{k}$'s Motivated by this question, I want to find a complete set of relations for the ring of integer-valued polynomials, where the generators are the polynomials $\binom{x}{n}$ for $n\in \mathbb{N}$. The best way to do this is would be to describe how t... | A beautiful combinatorial answer was found by Gjergji Zajmi. For an expanded version, see the solution to Exercise 3.9 in my Notes on the combinatorial fundamentals of algebra. (At least, it is Exercise 3.9 in the version of 10 January 2019. In future versions, the numbering can shift.)
Unlike Martin's answer, this one... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Finding $\int_{0}^{\frac{\pi}{2}}\frac{1}{\cos (x-\frac{\pi}{3}).\cos (x-\frac{\pi}{6})}\mathrm{d}x$ How can I find $$\int_{0}^{\frac{\pi}{2}}\frac{1}{\cos (x-\frac{\pi}{3}).\cos (x-\frac{\pi}{6})}\mathrm{d}x$$ ? I suspect this has something simple to do with the basic definite integral properties; but can't find a way... | Let $$\displaystyle y = \frac{x-\frac{\pi}{3}+x-\frac{\pi}{6}}{2} = x-\frac{\pi}{4}\Rightarrow x= \left(y+\frac{\pi}{4}\right)$$ and $dx = dy$ and changing limit
Put into $$\displaystyle I = \int_{0}^{\frac{\pi}{2}}\frac{1}{\cos \left(x-\frac{\pi}{3}\right)\cdot \cos \left(x-\frac{\pi}{6}\right)}dx = \int_{-\frac{\pi}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to solve the integral $\int\tan^{3}x \sec^{3/2}x\; dx$? How to solve the following indefinite integral
$$\int \tan^{3}x \sec^{3/2}x \; dx$$
to get the solution in the form of
$$\large\frac{2}{7}\sec^{7/2}x - \frac{2}{3}\sec^{3/2}x +c$$
I tried taking
$$u = \sec^{2}x \implies du = \tan x \; dx$$
$$\large \int \ta... | Your $du$ is wrong. $(\tan x)'=\sec^2 x$ (not the other way around).
Instead, write the integrand as $\sec x \tan x \tan^2 x \, (\color{maroon}{\sec x})^{1/2}$ and let $u=\color{maroon}{\sec x}$. Write $\tan^2 x$ in terms of $\color{maroon}{\sec x}$ and note $du$ then is $\sec x\tan x\, dx$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1347207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The Laurent series around $z=0$ of the function $f(z) = \frac{z}{(z-i)(z-2)}$ in the annulus $A(0,1,2)$ What I got so far:
$$
\frac{z}{(z-i)(z-2)} = \frac{z}{(2-i)(z-i)} + \frac{z}{(2-i)(z-2)}
$$
which is equal to
$$
\frac{z}{(2-i)(z-i)} + \frac{z}{(2-i)(z-2)} = \frac{z}{(2-i)z + 1-2i} + \frac{z}{(2-i)z + 2i - 4}
$$
an... | (z/((z-i)(z-2)))=(z/((2-i)(z-i)))+(z/((2-i)(z-2))),1<∣z∣<2. The expansion in powers of z is
(z/((2-i)z(1-i/z)))-(z/((2-i)2(1-z/2)))
=(1/(2-i))∑₀((i/z))ⁿ-(1/(4-2i))∑₀((z/2))ⁿ.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral of $\frac{x^2+1}{(1-x^2)\sqrt{1+x^4}}$ So we have to evaluate $\int\frac{x^2+1}{(1-x^2)\sqrt{1+x^4}}dx$.
My work-
We can write the integrand as $\frac{(x+1)^2-2x}{(1-x)(1+x)\sqrt{1+x^4}}dx$.
So we wish to deduce $\int\frac{(x+1)}{(1-x)\sqrt{1+x^4}}dx-\int\frac{2x}{(1-x^2)\sqrt{1+x^4}}dx$
So lets write it a... | If we replace $x$ by $\sqrt{u}$ we have:
$$ \int\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}\,dx=\int\frac{(1+u)}{2(1-u)\sqrt{u(1+u^2)}}\,du$$
Then, if we replace $\frac{1+u}{1-u}$ by $v$ we have:
$$ \int\frac{(1+u)}{2(1-u)\sqrt{u(1+u^2)}}\,du =\frac{1}{2\sqrt{2}}\int\frac{v}{\sqrt{v^4-1}}\,dv=\frac{1}{2\sqrt{2}}\,\log\left(v^2+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
A limit problem: $\lim\limits_{n\to\infty}\frac{1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n} }{1+\frac{1}{3}+\frac{1}{9}+\cdots+\frac{1}{3^n} }$ I need help in solving the limit below:
$$\lim_{n\to\infty}\frac{1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n} }{1+\frac{1}{3}+\frac{1}{9}+\cdots+\frac{1}{3^n} }$$
What I... | If you multiply the top by $2$, you get $2x = 2+x$, whence $x = 2$
Likewise, the bottom multiplied by $3$ gives $3y = 3+y$, gives $y = \frac{3}{2}$.
So you then find $\left(\frac{2}{1.5}\right) = \frac{4}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
If det $A = 0$ and $\det B \neq 0$ then show that $abc = -1$ This has been hurting my head for a while now....
If
$$
\det\begin{bmatrix}a&a^2&1+a^3\\b&b^2&1+b^3\\c&c^2&1+c^3\end{bmatrix}=0
$$
And
$$
\det\begin{bmatrix}a&a^2&1\\b&b^2&1\\c&c^2&1\end{bmatrix} ≠0
$$
Then show that $abc=-1$.
| The hard part is to compute the determinants and factor them nicely:
$
\det\begin{bmatrix}a&a^2&1\\b&b^2&1\\c&c^2&1\end{bmatrix} = (ab^2+bc^2+ca^2)- (b^2c+c^2a+a^2b) = (a-b)(b-c)(c-a)
$
and
$\det\begin{bmatrix}a&a^2&1+a^3\\b&b^2&1+b^3\\c&c^2&1+c^3\end{bmatrix}$
$= (ab^2(1+c^3)+bc^2(1+a^3)+ca^2(1+b^3)) - ((1+a^3)b^2c+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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Find the cubic equation of roots $α, β, γ$. Taken from Fitzpatrick $4$ unit course textbook. The question says:
If the cubic equation $\ ax^3+bx^2+cx+d$ has roots $α, β, γ$. Find the cubic equation who's roots are $α^2, β^2, γ^2$
I keep getting a $±$ sign that I can't get rid of. The answer in the back is $x(ax+c)^2=(b... | $$ax^3+bx^2+cx+d=a^3(x-\alpha)(x-\beta)(x-\gamma)$$
Set $x=y$ to get $$ay^3+by^2+cy+d=a^3(y-\alpha)(y-\beta)(y-\gamma)\ \ \ \ (1)$$
Set $x=-y$ to get $$a(-y)^3+b(-y)^2+c(-y)+d=a^3(-y-\alpha)(-y-\beta)(-y-\gamma)$$
$$\iff ay^3-by^2+cy-d=a^3(y+\alpha)(y+\beta)(y+\gamma)\ \ \ \ (2)$$
Multiply $(1),(2)$ to get $$a^6(y^2-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Straight line is tangent to the curve. The straight line $y=mx+1$ is tangent to the curve $x^2+y^2-2x+4y=0$. Find the possible values of $m$.
My attempt
Substitute the $y=mx+1$ into the equation $x^2+y^2-2x+4y=0$. $$x^2+(mx+1)^2-2x+4(mx+1)=0$$ $$x^2+m^2x^2+2mx+1-2x+4mx+4=0$$ $$(1+m^2)x^2+6mx-2x+5=0$$ $$(1+m^2)x^2+(6m-... | Hint: Differentiating implicitly yields $$2x + 2y \frac{\mathrm{d}y}{\mathrm{d}x} - 2 + 4\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \iff \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1-x}{y+2}$$
What can you say about gradient of tangent lines and derivatives of functions?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Uniform Convergence of $\sum_{n=0}^\infty (1-x^2)^2x^n$ on $[0,1]$; subsequent integral Let $a_n = (1-x^2)^2 x^n$. Show that $\sum_{n=0}^\infty a_n$ converges uniformly on $[0,1]$ and deduce that $\int_0^1 \frac{(1-x^2)^2}{1-x} dx = \sum_{n=1}^\infty \frac{8}{n(n+2)(n+4)}$.
Attempt: Denoting partial sums by $S_n$, we h... | To show uniform convergence of this series we can use the Weierstrass M-test. How big can $f_n(x)=(1-x^2)x^n$ be on $[0,1]?$ Set $f_n'(x) = 0.$ We get $x=\sqrt {n/(4+n)}.$ At this $x$ we have
$$ f_n(\sqrt {n/(4+n)}) = [1-n/(4+n)]^2\cdot[\sqrt {n/(4+n)}\,]^n.$$
The first factor on the right is $[4/(4+n)]^2.$ The second ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1355046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Generate all multisets of length k for n symbols I am trying to generate a list of all multisets of length $k$ in a set with $n$ symbols. For example, if I had the set
$S = {A, B, C}$
I would expect the following output for $k = 2$ and $n = 3$:
$O = {(A,A),(A,B),(A,C),(B,B),(B,C),(C,C)}$
What is the proper way to go a... | I presume you want to do this algorithmically. In Mathematica:
Union[Sort /@ Tuples[{A, B, C}, 2]]
{{A, A}, {A, B}, {A, C}, {B, B}, {B, C}, {C, C}}
More generally:
myMultiSet[n_Integer, k_Integer] := Union[Sort /@ Tuples[Range[n], k]]
So for instance myMultiSet[5,3] yields
{{1, 1, 1}, {1, 1, 2}, {1, 1, 3}, {1, 1, 4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluate $ \lim_{(x,y)\to(0,0)} \frac {e^{x+y^2}-1-\sin \left ( x + \frac{y^2}{2} \right )}{x^2+y^2} $
$$ \lim_{(x,y)\to(0,0)} \frac {e^{x+y^2}-1-\sin \left ( x +
\frac{y^2}{2} \right )}{x^2+y^2} $$
I've a few doubts about this limit. I mean, if I take polar coordinates, I get that the limit doesn't exist. And Wolfr... | For $(x,y)$ near $(0,0),$ we have
$$e^{x+y^2} = 1 + (x+y^2) +(x+y^2)^2/2 + O((x+y^2)^3), \sin (x+y^2/2) = (x+y^2/2) +O((x+y^2/2)^3).$$
So the numerator of our expression is
$$(y^2/2 + x^2/2) + xy^2 + y^4/2 + O((x+y^2)^3) + O((x+y^2/2)^3)= (y^2/2 + x^2/2) + r(x,y).$$
Dividing by $x^2 + y^2$ we get $1/2 + r(x,y)/(x^2+y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1357224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Proving $\sin 20^\circ \, \sin 40^\circ \, \sin 60^\circ \, \sin 80^\circ =\frac{3}{16}$ The problem is to prove that
$$\sin 20^\circ \, \sin 40^\circ \, \sin 60^\circ \, \sin 80^\circ =\frac{3}{16}$$
All my attempts were to get them in $\sin (2A)$ form after eliminating $\sin 60^\circ$ in both sides. Unfortunately, al... | Denote $Q = \sin 20 \sin 40 \sin 60 \sin 80$. Observe first that
\begin{align}
\sin 60 & = \sin (20+40) \\
& = \sin 20 \cos 40 + \cos 20 \sin 40 \\
& = \sin 20 (2\cos^2 20 - 1) + \cos 20 (2 \sin 20 \cos 20) \\
& = (4\cos^2 20 - 1) \sin 20
\end{align}
We can now write
\begin{align}
Q & = \sin 20... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1357880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Evaluate this limit at infinity $\lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2}$ Problem: Find the limit of \begin{align*} \lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2}
\end{align*}
Attempt at solution. The back of my textbook gives the answer as $-\frac{1}{4} \sqrt{2}$. Here's wh... | The $x\sqrt{2/x+\cdots}$ term does not go to zero, but to infinity. Even when you multiply out the numerator and get a $\bigl(\sqrt{1/x+\cdots}\bigr)x\sqrt{2/x+\cdots}$ term, this goes to $\sqrt2$ rather than to $0$.
The term that arises from the $1$ when you multiply out does go to zero.
What you should probably do in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Find the Sum using bijection Find the sum of $S=\displaystyle\sum_{i,j,k \ge 0, i+j+k=17} ijk$.
I am looking for a solution that uses some bijection.
I couldn't find any bijection.
I am able to do the problem by other method by observing that,
$S$ is the coefficient of $x^{17}$ in,
$(x+2x^2+3x^3+4x^4........)^3$ and th... | For this you'll require the sums up to fourth powers:
$\begin{align}
\sum_{0 \le k \le n} k
&= \frac{n (n + 1)}{2} \\
\sum_{0 \le k \le n} k^2
&= \frac{n (n + 1) (2 n + 1)}{6} \\
\sum_{0 \le k \le n} k^3
&= \frac{n^2 (n + 1)^2}{4} \\
\sum_{0 \le k \le n} n^4
&= \frac{n^2 (6 n^3 + 15 n^2 + 10 n - 1)}{30}
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1361746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Fast way to get a position of combination (without repetitions) This question has an inverse: (Fast way to) Get a combination given its position in (reverse-)lexicographic order
What would be the most efficient way to translate a combination of $k$-tuple into its positions within the $\left(\!\!\binom{n}{k}\!\!\right)... | Let us denote your tuple [a b c] as [1 1 1 0 0 0] and so on.
Define $\binom{n}{r}=0$ for $n<r$
For your tuple: $[a d f] = [1 0 0 1 0 1]$
$$P = 1\cdot \binom{0}{1}+0\cdot \binom{1}{1}+0\cdot \binom{2}{1}+1\cdot \binom{3}{2}+0\cdot\binom{4}{2}+1\cdot\binom{5}{3} + 1$$
$$P=0 + 0 +0 +3+0+10+0+1 = 14$$
General Algorithm:
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 0
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Solve this exponential equation: $3^{2x}+\left(\frac{1}{2}\right)^{-x} \cdot 3^{x+1}-2^{2x+2}=0$ I tried solving this equation $$3^{2x}+\left(\frac{1}{2}\right)^{-x} \cdot 3^{x+1}-2^{2x+2}=0$$ by taking the log of both sides, but with no results, what do I do? Sorry if this equation is very easy, I couldn't solve it...... | we have $$(3^x)^2+2^x\cdot3^x\cdot 3-(2^x)^2\cdot2^2=0$$ divided by $(2^x)^2$ we have $$\left(\frac{3^x}{2^x}\right)^2+3\cdot \frac{3^x}{2^x}-4=0$$
setting $$u=\frac{3^x}{2^x}$$ we have a quadratic equation in $$u$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How find all complex numbers such that: $|\,1 - z\,| < k\ (1 - |\,z\,|\, )$? Let $k > 1$ be a real number. How may one find all complex numbers such that:
$|\,1 - z\,| < k\ (1 - |\,z\,|\, )$?
................................................................................................................................... | This never can be a circle. However :
$$ z= x +iy $$
$$ |1-z| =\sqrt{ (1-x)^2 +y^2 } $$
$$ 1- |z|=1-\sqrt{x^2 +y^2} $$
$$ (1-x)^2 +y^2 <(K-K\sqrt{x^2 +y^2})^2$$
$$ (1-x)^2 +y^2 -K^2 -K^2(x^2+y^2)<-2K^2\sqrt{x^2 +y^2}$$
$$ 1+x^2-2x+y^2-K^2-K^2x^2-K^2y^2<-2K^2\sqrt{x^2 +y^2}$$
$$1-K^2+x^2(1-K^2) + y^2(1-K^2)-2x<-2K^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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The converges of $ \sqrt 2 +\sqrt { 2-\sqrt 2} +\sqrt { 2-\sqrt { 2+\sqrt 2} } + \cdots =$ I would like to know wheather this series converge or diverge?
$\sqrt 2 +\sqrt { 2-\sqrt 2} +\sqrt { 2-\sqrt { 2+\sqrt 2} } +\sqrt { 2-\sqrt { 2+\sqrt { 2+\sqrt 2} } } +\cdots$
My work:
multiplyIing both demominator and nume... | Let:
$$ c_1=\sqrt{2},\quad c_2=\sqrt{2+\sqrt{2}},\quad c_{n+1}=\sqrt{2+c_n}$$
and $d_n=c_n/2$. Then $d_1=\cos\frac{\pi}{4}$ and $d_{n+1}=\sqrt{\frac{1+d_n}{2}}$, by recognizing the cosine duplication formula, give:
$$ c_n = 2 \cos\frac{\pi}{2^{n+1}} $$
so:
$$ \sqrt{2} = 2\sin\frac{\pi}{4},\quad \sqrt{2-\sqrt{2}}=2\sin\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $(\sqrt2 − 1)^n, \forall n \in \mathbb{Z^+}$ can be represented as $\sqrt{m} − \sqrt{m−1}$ for some $m \in \mathbb{Z^+}$ (no induction). From the 1994 Canada National Olympiad:
Prove that $(\sqrt2 − 1)^n, \forall n \in \mathbb{Z^+}$ can be represented as $\sqrt{m} − \sqrt{m−1}$ for some $m \in \mathbb{Z^+}$... | Step 1
Let us define
$$
\psi(p,q) = p \sqrt{2} + q.
$$
Let us define
$$
\Psi = \big\{ \psi(p,q) | p, q \in \mathbb{Z} \big\}.
$$
We have
$$
\big(p \sqrt{2} + q \big) \big(r \sqrt{2} + s \big)
= \big( p s + q r \big) \sqrt{2} + \big( 2 p r + q s \big).
$$
Therefore
$$
\forall \psi_1, \psi_2 \in \Psi : \psi_1 \psi_2 \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
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simplify and evaluate $\frac{\tan80^\circ-\tan20^\circ}{1+\tan80^\circ\tan20^\circ}$ How do you simplify and evaluate $\dfrac{\tan80^\circ-\tan20^\circ}{1+\tan80^\circ\tan20^\circ}$?
What is the problem asking?
| Hint:
Try to reduce this problem to one that makes use of the more familiar $\sin$ and $\cos$ sum of angle identities.
$$\begin{array}{lll}
\displaystyle\frac{\tan 80^\circ - \tan 20^\circ}{1+\tan80^\circ\tan20^\circ} &=&\displaystyle\frac{\tan 80^\circ + \tan (-20^\circ)}{1-\tan80^\circ\tan(-20^\circ)}\\
&=&\displayst... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Help needed to solve combinatorics problem. I have been revisiting my old probability courses and I found a problem, which I can't figure out how to solve or at least what I get differs from the answer in the book.
The problem reads as follows:
Given 8 balls i.e. 5 white 1 black, 1 blue and 1 red, find how many sequen... | Use exponential generating functions. The white balls give:
$$
1 + \frac{z}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \frac{z^4}{4!} + \frac{z^5}{5!}
$$
Each of the other colors gives:
$$
1 + \frac{z}{1!}
$$
In all, we want:
$$
6! [z^6] \left( 1 + \frac{z}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \frac{z^4}{4!} + \frac{z^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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How to solve $z^3 + \overline z = 0$ I need to solve this:
$$z^3 + \overline z = 0$$
how should I manage the 0?
I know that a complex number is in this form: z = a + ib so:
$$z^3 = \rho^3\lbrace \cos(3\theta) + i \sin (3\theta)\rbrace$$
$$\overline z = \rho\lbrace \cos(-\theta) + i \sin (-\theta)\rbrace$$
but how about... | One solution is $z=0$. When $z\ne0$, multiplying by $z$ will not alter the solutions of the equations; so we can look at $z^4+|z|^2=0$. If we write $z=re^{it}$, the equation becomes
$$
r^4e^{4it}+r^2=0.
$$
As we are assuming $r\ne0$, this reduces to $-r^2e^{4it}=1$. So $r=1$ and $e^{4it}=-1$, that is (since $-1=e^{i\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 7
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Third point of intersection is also a point of inflection? Let $C \subset \mathbb{P}_2$ be a nonsingular cubic. If $L$ is a line through two distinct points of inflection on $C$, how do I show that the third point of intersection is also a point of inflection?
| This is Problem 2 in section 5.7, "Elliptic Curves," of I. Niven, H. S. Zuckerman, H. L. Montgomery, An Introduction to the Theory of Numbers, 5th ed., Wiley (New York), 1991. I use the nomenclature from that book.
After I worked out this answer, I realized that it is the same as Lubin's.
$C$ is elliptic. Let $\mathb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Finding Linear Combination of Polynomials I am stuck on a question involving finding the greatest common divisor of polynomials and then solving to find the linear combination of them yielding the greatest common divisor. My work thus far is as follows:
Define: $f(x) = x^3-3x+3$, $g(x) = x^2 - 4$.
Execute the Euclidea... | We chose polynomials $f(x)$ and $g(x)$ such that $$f(x)(x^2-4) + g(x)(x^3-3x+3)=1$$ and for this it is necessary the equality of degrees.
So $d(f)+2=d(g)+3\iff d(f)=d(g)+1$.
Chose the simplest possibilities
$f(x)=ax^2+bx+c$ and $g(x)=dx+e$ and calculate the coefficients.It follows
$(a+d)x^4+(b+e)x^3+(-4a+c-3d)x^2+(-4b... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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First principle derivative of a square root and conjugates I'm trying to find the derivative of the equation: $$g(x)=\sqrt {x+2}-3x^2$$.
I can find the solution just fine using the power rule but am finding trouble with First Principles.
Essentially, I understand getting as far as $$\displaystyle\lim_{h\to 0}\frac{\sqr... | $$\lim_{h\to0}{\frac{g(x+h)-g(x)}{h}}$$
$$\lim_{h\to0}{\frac{\sqrt{x+h+2}+3(x+h)^2-\sqrt{x+2}-3x^2}{h}=\frac{\sqrt{x+h+2}-\sqrt{x+2}-3x^2-6xh-3h^2+3x^2}{h}}$$
$3x^2$ is eliminated, and $3h^2$ is negligible:
$$=\frac{\sqrt{x+h+2}-\sqrt{x+2}-6xh}{h}=\frac{\sqrt{x+h+2}-\sqrt{x+2}}{h}-\frac{6xh}{h}$$
now using the conjug... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given the differential equation, how to solve the y function with x as the independent variable? $y\frac{dy}{dx} = x(y^4 + 2y^2 + 1)$
$y = 1$ when $x = 4$
I tired to integrate by substitution, but it doesn't seem to work out.
| Notice, we have $$y\frac{dy}{dx}=x(y^4+2y^2+1)$$ $$\frac{y}{y^4+2y^2+1}dy=xdx$$ $$\frac{y}{(y^2+1)^2}dy=xdx$$ Now, integrating both the sides we have $$\int \frac{y}{(y^2+1)^2}dy=\int xdx$$ $$\frac{1}{2}\frac{(y^2+1)^{-1}}{-1}=\frac{x^2}{2}+c$$ $$-\frac{1}{2(y^2+1)}=\frac{x^2}{2}+c$$ Substituting $y=1$ & $x=4$, we get
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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How to integrate a function of form f (x)/g (x) $\frac {(ax^n+b) dx}{cx^m+d}$ How to find $\int \frac {(ax^n+b) dx}{cx^m+d}$ (where m>n and m,n are natural numbers,and a,b,c,d are integers) ?
| The general approach will involve:
*
*If $n\geq m$, use long division to get a polynomial plus a fraction of two polynomials where the degree of the numerator is smaller than $m$.
*Factoring $cx^m+d$ in factors with linear and quadratic terms.
*Applying partial fraction decomposition.
*Seperating all partial frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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New Idea to prove $1+2x+3x^2+\cdots=(1-x)^{-2}$
Given $|x|<1 $ prove that $\\1+2x+3x^2+4x^3+5x^4+...=\frac{1}{(1-x)^2}$.
1st Proof: Let $s$ be defined as
$$
s=1+2x+3x^2+4x^3+5x^4+\cdots
$$
Then we have
$$
\begin{align}
xs&=x+2x^2+3x^3+4x^4+5x^5+\cdots\\
s-xs&=1+(2x-x)+(3x^2-2x^2)+\cdots\\
s-xs&=1+x+x^2+x^3+\cdots\\
s... | Let $S=1+2x+3x^2+4x^3+\dotsb$.
\begin{align}
\phantom{-x^2}S&=1+2x+3x^2+4x^3+\dotsb\\
\phantom{^2}-xS&=\phantom1-\phantom2x-2x^2-3x^3-\dotsb\\
\phantom{^2}-xS&=\phantom1-\phantom2x-2x^2-3x^3-\dotsb\\
\phantom{-}x^2S&=\phantom{1+2x+2}x^2+2x^3+\dotsb
\end{align}
Adding them together:
\begin{align}
(1-2x&+x^2)S\\
&=1+0x+0... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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"answer_id": 14
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Trigonometric equation cos sin and power The problem is $2\cos t - 3\sin^2t +2 = 0$.
I get to $2\cos t -3\sin^2t =-2$
I think that I need to use a trigonometric identity like $\cos(x+y)$ and to divide $2\cos t -3\sin^2t$ with the $\sqrt{2^2+3^2}$
Do you know how to solve this? It should be $\sqrt{2^2 + 3^2}$
| $$2 \cos t - 3 \sin^2t +2 = 0\\.$$ Write $$\sin^2 t=1-\cos^2 t, $$ then you have a quadratic equation. Solve for $\cos t$ $$ 2\cos t -3(1-\cos^2 t)+2=0\\3\cos ^2 t+2\cos t-3+2=0\\\cos t=-1,\;\frac{1}{3}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Trigonometric equation with sine and cosine So the equation is $3\cos ^2t + 5\sin t = 1$
Now I have simplified this to $$3(1-\sin ^2t) + 5\sin t -1 = 0$$
which leads to $$-3\sin ^2t + 5\sin t + 2 = 0$$
Then I get $$-3t^2 + 5 t +2 = 0$$
Is this the correct way to go with this equation then use $t = t/2 \pm \sqrt {(t/2)^... | $$3\cos ^2t + 5\sin t = 1$$
$$3(1-\sin ^2t) + 5\sin t -1 = 0$$
$$-3\sin ^2t + 5\sin t + 2 = 0$$
$$3\sin ^2t - 5\sin t - 2 = 0$$
Now , Let $\sin t =x$, then equation reduces to,
$$3x^2-5x-2=0$$
$$(3x+1)(x-2)=0$$, Now can you finish from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve $2\sin^3x + \sin3x +3\sin^2x \cos x + \cos^3x=0$ $2\sin^3x + \sin3x +3\sin^2x\cos x + \cos^3x=0$
My try:
$$2\sin^3x +3\sin x - 4\sin^3x +\cos x(3\sin^2x+\cos^2x)=0 $$
$$ \cos x(2\sin^2x+1) - 2\sin^3x+3\sin x=0.$$
And then i have no idea.
| Notice, $$2\sin^3x + \sin3x +3\sin^2x\cos x + \cos^3x=0$$
$$2\sin^3x +3\sin x - 4\sin^3x +3\cos x(1-\cos^2x)+cos^3x=0 $$
$$3\sin x - 2\sin^3x +3\cos x-3\cos^3x+cos^3x=0 $$
$$ 3(\cos x+\sin x)-2(\cos^3x+\sin^3x)=0$$ $$ 3(\cos x+\sin x)-2(\cos x+\sin x)(\cos^2x+\sin^2x-\cos x\sin x)=0$$ $$(\cos x+\sin x)(3-2(1-\cos x\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding the horizontal and vertical tangents of a parametric equation.
Find the points at which the polar curve $r=2+2\sin{(\theta)}$ has a
horizontal or vertical tangent line.
Translate the parametric equation to Cartesian coordinates:
$$
r^2=2r+2r\sin{(\theta)} \\
x^2+y^2=2\sqrt{x^2+y^2}+2y \\
x^2+y^2-2y+1=2\sqrt... | Mistake in taking derivatives:
$$
\frac{\mathrm{d}}{\mathrm{d}x}\left[x^2+(y-1)^2\right]
=\frac{\mathrm{d}}{\mathrm{d}x}\left[2\sqrt{x^2+y^2}+1\right] \\
2x+2(y-1)y'=\frac{2x+2yy'}{\sqrt{x^2+y^2}}
$$
and you dropped the $y'$ on the RHS...
UPDATE
Another approach is to let $x=x(\theta)$ and $y = y(\theta)$ and comp... | {
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"url": "https://math.stackexchange.com/questions/1377436",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Why does $\sqrt{6} + \sqrt{10} + \sqrt{15}$ have four conjugates? I am having trouble understanding how algebraic number $\sqrt{6} + \sqrt{10} + \sqrt{15}$ has four conjugates.
Minimal polynomial is $x^4-62 x^2-240 x-239$ according to Wolfram Alpha.
Factorized:
$$\left(x-2\sqrt{15 (4-\sqrt{15})}-8\sqrt{4-\sqrt{15}}-\sq... | $$\sqrt{6}+\sqrt{10}+\sqrt{15}=\sqrt{2\cdot 3}+\sqrt{2\cdot 5}+\sqrt{3\cdot 5} = \frac{1}{2}\left(\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2-(2+3+5)\right)$$
where $\sqrt{2}+\sqrt{3}+\sqrt{5}$ is an algebraic number of degree $8$ over $\mathbb{Q}$, having conjugates $\pm\sqrt{2}\pm\sqrt{3}\pm\sqrt{5}$, whose minimal pol... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why is $\frac{1}{4/3} - \frac{1}{3/2}$ not the same as $\bigl(\frac{4}{3} - \frac{3}{2}\bigr)^{-1}$ If you have the problem:$$\frac{1}{\frac{4}{3}} - \frac{1}{\frac{3}{2}} =?$$
Why can't you change the problem to $(\frac{4}{3} - \frac{3}{2})^{-1}$ and get the same answer?
In the first scenario, the answer is $1/12$
I... | It looks like based on your post that you want to know why $\frac{1}{\frac{4}{3}} - \frac{1}{\frac{3}{2}}$ is not equal to $\frac{1}{\frac{4}{3} - \frac{3}{2}}$.
In general, you can only say a property holds like, in your case, $(a + b)^{-1} = a^{-1} + b^{-1}$ if you can show it holds for every $a$ and $b$. But as you... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Help in finding the integral function. Can somebody provide a hint in finding the following integral? $$\displaystyle \int \dfrac{1}{(x^3+1)^3} \text{ d}x$$ I thought of using partial fractions but that isn't making any sense.
| This following trick, which is essentially the same as @Daniel Fischer's suggestion, reduces your burden of calculating partial fraction decomposition. First, write
$$ \int \frac{dx}{(1+x^3)^3} = \int \frac{1 + x^3}{(1+x^3)^3} \, dx - \int x \cdot \frac{x^2}{(1+x^3)^3} \, dx. $$
Then by integration by parts, we have
$$... | {
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"timestamp": "2023-03-29T00:00:00",
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Solve $2+ \cos{\frac{3x}{2}} + \sqrt{3} \sin{\frac{3x}{2}} = 4\sin^2{\frac{x}{4}}$ $$2+ \cos{\frac{3x}{2}} + \sqrt{3} \sin{\frac{3x}{2}} = 4\sin^2{\frac{x}{4}}$$
My try:
$$ \cos{\frac{3x}{2}} + \sqrt{3} \sin{\frac{3x}{2}} = \sqrt{4}\left(\frac{\sqrt{3}}{2} \sin{\frac{3x}{2}} + \frac{1}{2}\cos{\frac{3x}{2}}\right) = 2\... | $2+2\sin(\frac{3x}{2}+\frac{\pi}{6})=4 \sin^2(\frac{x}{4})$ simplifies to $2\sin(\frac{3x}{2}+\frac{\pi}{6})+2 cos(\frac{x}{2})=0$
$\sin(\frac{3x}{2}+\frac{\pi}{6})=-cos(\frac{x}{2})$
$\sin(\frac{3x}{2}+\frac{\pi}{6})=sin(\frac{3\pi}{2}-\frac{x}{2})$
I hope you can solve further.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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New idea to solve this equation I was teaching $\left \lfloor x \right \rfloor$ function properties and equation . I solved this equation in my class . My works are show below. Some students ask me for new Idea...,and now I am looking for various method to solve this (like this) equation .$$\left \lfloor x\right \rf... | $x-1<\lfloor x\rfloor \leq x$, so $3x-1<\lfloor x\rfloor +2x=1\leq 3x$, this implies $x\in [1/3, 2/3)$, so $\lfloor x\rfloor=0$, then $2x=1$, $x=1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
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ODE $x'' + 2x' +5x = \sin3t$, $x(0)=1,\ x'(0)=-1$, Solve using Laplace Transform While solving the differential equation
$$x'' + 2 x' + 5 x = \sin3t, \quad x(0) = 1, \quad x'(0) = -1$$
by use of Laplace transform I got to
$$X(s^2 +2s+5)=\frac{(3)}{s^2 +9} +s +1$$
$$X=\frac{\left(\:s^3+s^2+9s+12\right)}{\left(s^2+9\rig... | Notes:
*
*The first part of this solution is for an equation that contained an error. It remains here as a demonstration. The corrcted equation is the second half of this solution.
*The process of the proposer is correct in process but does need some guidance in finding a nice result.
Consider the differentia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Substitution to solve an initial value problem By using the substitution $y(x) = v(x)x$, how can I solve the initial value problem
$$
\frac{dy}{dx} = \frac{x^2+y^2}{xy - x^2},\quad y(1)=1
$$
And also keep my answer in the form $g(x,y)= 4e^{-1} xe^\frac{y}{x}$
| $$\frac { dy }{ dx } =\frac { { x }^{ 2 }+{ y }^{ 2 } }{ xy-{ x }^{ 2 } } =\frac { 1+\frac { { y }^{ 2 } }{ { x }^{ 2 } } }{ \frac { y }{ x } -1 } \\ \frac { y }{ x } =t\Rightarrow y=xt\Rightarrow dy=t+xdt\\ t+xdt=\frac { 1+{ t }^{ 2 } }{ t-1 } \\ xdt=\frac { 1+{ t }^{ 2 } }{ t-1 } -t=\frac { 1+{ t }^{ 2 }-{ t }^{ 2 }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$A+B+C=2149$, Find $A$ In the following form of odd numbers
If the numbers
taken from the form where $A+B+C=2149$
Find $A$
any help will be appreciate it, thanks.
| If $A$ were on the $n^{th}$ row, then $B=A+2n$ and $C=A+2(n+1)$.
Out of laziness, we will try to figure out what row this would need to be to be in the right ballpark through estimation. $\frac{2149}{3}\approx 716$, so $A<716<B<C$. Which row would $715$ fit into?
By changing your figure some by first adding one to ea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
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$\frac{CE}{DE}=\frac{(a+b)^2}{kc^2}$ If bisector of angle C of an acute triangle ABC cuts the side AB in D and the circumcircle of triangle ABC in E,then $\frac{CE}{DE}=\frac{(a+b)^2}{kc^2}$.Then what is value of k?
Since angle bisector of C,bisects the arc BC.And circumcenter O will fall on line CE.And $\angle AOB=2\a... |
$\frac{CE}{DE}=\frac{(a+b)^2}{kc^2}$. What is value of $k$?
Note that
\begin{align}
\angle EAB&=\angle EBA=\tfrac\gamma2
\end{align}
Assuming that
$|AB|=c$, $|BC|=a$, $|CA|=b$,
$\angle BAC=\alpha$,
$\angle ABA=\beta$,
$\angle ACB=\gamma$,
we can find $|CD|$ and $|CE|$ from the area relations:
\begin{align}
S_{\trian... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $ (a+b) \cos(x) = a-b$, if $ \sin(x) + \tan(x) = \frac{4a\sqrt{ab}}{a^2-b^2},$ $ \tan(x) - \sin(x) = \frac{4a\sqrt{ab}}{a^2-b^2}.$ Prove that $ (a+b) \cos(x) = a-b$, if $$ \sin(x) + \tan(x) = \frac{4a\sqrt{ab}}{a^2-b^2},$$
$$ \tan(x) - \sin(x) = \frac{4a\sqrt{ab}}{a^2-b^2}.$$
I tried solving it with system, ... | grand_chat has already pointed out that you may want
$$\tan x-\sin x=\frac{4\color{red}{b}\sqrt{ab}}{a^2-b^2}\tag 1$$
With
$$\tan x+\sin x=\frac{4a\sqrt{ab}}{a^2-b^2}\tag 2$$
now $(1)+(2)$ gives
$$2\tan x=\frac{4(a+b)\sqrt{ab}}{(a+b)(a-b)}\Rightarrow \tan x=\frac{2\sqrt{ab}}{a-b}$$
Also, $(2)-(1)$ gives
$$2\sin x=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Closed form for an infinite product The following fascinating formula appears in the paper "On gamma quotients and infinite products" by M.Chamberland and A.Straub (see page 9):
$$\prod_{n=2}^\infty\left(1-\frac{1}{n(n-1)}\right)=-\frac1\pi\cos\left(\frac{\sqrt5}{2}\pi\right).$$
Unfortunately, it's given without any re... | We will be only dealing with the following identities:
$$
\Gamma (z) = \frac{ e^{ -\gamma z }}{ z } \, \prod_{ k=1 }^{ \infty } \left\{ \left( 1+\frac{ z }{ k } \right)^{ -1 } \, e^{ \frac { z }{ k } } \right\}
\hspace{10mm}
\Gamma (-z)\Gamma (z)\quad =\quad \frac { -\pi }{ z \sin { (\pi z) } }
$$
Let's start ... | {
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"url": "https://math.stackexchange.com/questions/1385953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 3
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prove that $\int_0^\frac{\pi}{4} \frac{1-\cos x}{x} \,dx < \frac{\pi^2}{64}$ prove that $\int_0^\frac{\pi}{4} \frac{1-\cos x}{x} \,dx < \frac{\pi^2}{64}$
I showed that in
$$ \forall x \in [0,\frac{\pi}{4}] \quad \int_0^\frac{\pi}{4} \frac{1-\cos x}{x} \,dx \le \int_0^\frac{\pi}{4} \frac{1-\cos(\frac{\pi}{4})}{\frac{... | Since:
$$\frac{\sin x}{x}=\prod_{n\geq 1}\left(1-\frac{x^2}{n^2\pi^2}\right) $$
over the interval $\left[0,\frac{\pi}{4}\right]$ we have:
$$ \frac{1-\cos x}{x}=\left(\frac{\sin(x/2)}{x/2}\right)^2\cdot\frac{x}{2}\leq\left(1-\frac{x^2}{4\pi^2}\right)^2\frac{x}{2} $$
so:
$$ \int_{0}^{\frac{\pi}{4}}\frac{1-\cos x}{x}\,dx ... | {
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"url": "https://math.stackexchange.com/questions/1387076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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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.