Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Integration Trig Substitution After making the correct trig substitution what does the integral of
$\dfrac{1}{\sqrt{9-x^2}} dx$ reduce to without solving the equation?
I reduced it down to the integral of $3\cos(\theta)d\theta$ is this correct?
| Performing trig. substitution of $x=3 \sin \theta$ (so $dx=3\cos \theta \, d\theta$) gives $$\int\frac{1}{\sqrt{9-x^2}} \, dx=\int\frac{1}{\sqrt{9-9 \sin^2 \theta}} \, (3 \cos \theta \, d\theta)=\int\frac{1}{|3 \cos \theta|} \, (3 \cos \theta \, d\theta) = \int 1 \, d\theta$$
which is equal to $\theta + C$, or $\displa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/875029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For which $a$ does $\lvert x+1\rvert+\lvert 2-x\rvert=a^2 -1$ have exactly two solutions? If it is not a problem, I would really appreciate if someone could explain to me how to solve and graph the following equation:
For which real numbers $a$ does the equation
$$\lvert x+1\rvert +\lvert 2-x\rvert=a^2 -1$$
Have exactl... | This solution does not use graph or distance between points. We have $4$ cases to look at:
Case 1:
If $x < -1$, then $x+1 < 0$, and $2-x > 0$, so $LHS = -x-1 + 2-x = 1 - 2x = a^2 - 1 \to x = \dfrac{2-a^2}{2}$. This means the equation either has only one solution, or no solution depending on $a$. In particular it has on... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/875185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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If $a+b+c+d=4$, then $\sum\frac{1}{a+3}\le \frac{1}{abcd}$ This is somehow related to this problem but I don't have any idea about it.
Let $a$, $b$, $c$ and $d$ be positive reals such that $a+b+c+d=4$. Prove that:
$$\frac{1}{a+3}+\frac{1}{b+3}+\frac{1}{c+3}+\frac{1}{d+3}\le \frac{1}{abcd}$$
Now I also tried to prov... | $\sum$ will be used to denote the cyclic sum.
For the second inequality, let $t=ab+bc+ca$. Note that $t\le 3$, and also:
$$3t=(a+b+c)(ab+bc+ca)\ge 9\sqrt[3]{(abc)(a^2b^2c^2)}=9abc$$
Now, by Cauchy-Schwarz:
$$\sum\frac{2bc}{a+2}=2\sum\frac{(bc)^2}{abc+2bc}\ge \frac{2t^2}{3abc+2t}\ge \frac{2t}{3}$$
Note further that $\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/876031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Evaluation of $\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx$
Compute the indefinite integral
$$
\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx
$$
My Attempt:
$$
\begin{align}
\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx &= \int\frac{\cos 2x}{\sin^2 x\sqrt{\cos 2x}}\sin xdx\\
&= \int\frac{2\cos^2 x-1}{(1-\cos^2 x)\sqrt{2\cos^2 x-1} }\sin ... | If in the last integral you substitute $t=\frac{1}{\sqrt{2}}\cosh z$, you end with:
$$ I = \frac{1}{\sqrt{2}}\int\frac{1}{\frac{\cosh^2 z}{2}-1}dz=-\operatorname{arctanh}(\sqrt{2}\tanh z)=-\operatorname{arctanh}\left(\sqrt{2-\frac{1}{t^2}}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/876175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Calculate $\frac{1}{5^1}+\frac{3}{5^3}+\frac{5}{5^5}+\frac{7}{5^7}+\frac{9}{5^9}+\cdots$ I'm an eight-grader and I need help to answer this math problem.
Problem:
Calculate $$\frac{1}{5^1}+\frac{3}{5^3}+\frac{5}{5^5}+\frac{7}{5^7}+\frac{9}{5^9}+\cdots$$
This one is very hard for me. It seems unsolvable. How to calcul... | Informally:
You're taking the sum of the row sums of
$
\ \ \ \displaystyle{1\over 5^{\phantom 1}}
$
$
\ \ \ \displaystyle{1\over 5^{ 3}} \ \ \ \ \displaystyle{1\over 5^{ 3}}\ \ \ \ \displaystyle{1\over 5^{ 3}}
$
$
\ \ \ \displaystyle{1\over 5^{ 5}} \ \ \ \ \displaystyle{1\over 5^{ 5}}\ \ \ \ \displaystyle{1\over 5^{ 5}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/876893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
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"answer_id": 3
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Solve for $x$, $\tan x +\sec x = 2\cos x$ ; $−∞ < x < ∞$ Solve for $x$, $\tan x +\sec x = 2\cos x$ ; $−∞ < x < ∞$
$$\tan x + \sec x = 2\cos x$$
I tried changing it all to sin and cos
$$\frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2\cos x$$
then I made it to one fraction
$$\frac{\sin x + 1}{\cos x} = 2 \cos x$$
Then I don... | $$\sec x+\tan x=2\cos x\iff \sec x-\tan x=\frac1{2\cos x}$$
Adding we get $$2\sec x=2\cos x+\frac1{2\cos x}$$
Multiplying by $2\cos x,$
$$4=4\cos^2x+1\iff2\cos^2x=\frac32\iff\cos2x=2\cos^2x-1=\frac12=\cos60^\circ$$
$$\iff 2x=360^\circ m\pm60^\circ$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/877539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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prove $\sum\limits_{cyc} \frac {a^3} {b+c+d} \geq \frac {1} {3}$
Show that if $a$, $b$, $c$ and $d$ are non-negatives and $ab+bc+cd+da=1$ then:
$$\sum\limits_{cyc} \frac {a^3} {b+c+d} \geq \frac {1} {3}$$
yet again it should be solved with Cauchy inequality.
thing i have done so far:
$(\sum\limits_{cyc} \frac {a... | By Holder an AM-GM we obtain:
$$\sum_{cyc}\frac{a^3}{b+c+d}\geq\frac{(a+b+c+d)^3}{4\sum\limits_{cyc}(b+c+d)}=\frac{1}{12}(a+b+c+d)^2\geq$$
$$\geq\frac{1}{12}\left(2\sqrt{(a+c)(b+d)}\right)^2=\frac{1}{3}(ab+bc+cd+da)=\frac{1}{3}.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/877621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Antiderivative of $\frac{1}{1+\sin {x} +\cos {x}}$ How do we arrive at the following integral
$$\displaystyle\int\dfrac{dx}{1+\sin {x}+\cos {x}}=\log {\left(\sin {\frac{x}{2}}+\cos {\frac{x}{2}}\right)}-\log {\left(\cos {\frac{x}{2}}\right)}+C\ ?$$
| $$1+\cos x=2\cos^2\frac x2$$
$$1+\cos x+\sin x=2\cos^2\frac x2+2\sin\frac x2\cos\frac x2=2\cos\frac x2\left(\sin\frac x2+\cos\frac x2\right)$$
$$=\sqrt2\cos\frac x2\cos\left(\frac\pi4-\frac x2\right)$$
For numerator, $$\cos\frac\pi4=\cos\left[\frac x2-\left(\frac\pi4-\frac x2\right)\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/877681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Trigonometric Arithmetic Progression If $a$, $b$, $c$ are in arithmetic progression, prove that
$$\cos A \cot\frac{A}{2} \qquad \cos B \cot \frac{B}{2} \qquad \cos C \cot\frac{C}{2}$$
are in arithmetic progression, too.
Here, $a$, $b$, $c$ represent the sides of a triangle and $A$, $B$, $C$ are the opposite angles of t... | $$\cos B\cot\frac B2=\left(1-2\sin^2\frac B2\right)\cot\frac B2=\cot\frac B2-\sin A$$
Clearly using Law of Sines, $\sin A,\sin B,\sin C$ are in AP
So, it is sufficient to show that $\displaystyle\cot\frac A2,\cot\frac B2,\cot\frac C2$ are also in AP
We have $$a+c=2b\iff \sin A+\sin C=2\sin B$$
Now using Prosthaphaere... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/877784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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In a triangle ABC, prove that cot(A/2)+cot(B/2)+cot(C/2) =cot(A/2)cot(B/2)cot(C/2) In a triangle ABC, prove that $\cot \left ( \frac{A}{2} \right )+\cot \left ( \frac{B}{2} \right )+\cot \left ( \frac{C}{2} \right )=\cot \left ( \frac{A}{2} \right )\times \cot \left ( \frac{B}{2} \right )\times \cot \left ( \frac{C}{2}... | Applying $\cot(x+y)=\dfrac{\cot x\cot y-1}{\cot x+\cot y}$ twice,
$$\cot(x+y+z)=\frac{\sum\cot x\cot y-1}{\prod\cot x-\sum\cot x}$$
Here $x=\dfrac A2$ etc. so that $x+y+z=\dfrac\pi2\implies\cot(x+y+z)=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/878577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove ${\large\int}_{-1}^1\frac{dx}{\sqrt[3]{9+4\sqrt5\,x}\ \left(1-x^2\right)^{2/3}}=\frac{3^{3/2}}{2^{4/3}5^{5/6}\pi }\Gamma^3\left(\frac13\right)$ Here is one more conjecture I discovered numerically:
$${\large\int}_{-1}^1\frac{dx}{\sqrt[3]{9+4\sqrt5\,x}\ \left(1-x^2\right)^{\small2/3}}\stackrel{\color{#808080}?}=\f... | I've found a generalization of your conjecture:
If $k>0$ real number, then
$${\large\int}_{-1}^1\frac{dx}{\sqrt[3]{k\pm\tfrac{4\sqrt5k}{9}\,x}\
\left(1-x^2\right)^{\small2/3}}\stackrel{?}{=}
\frac{1}{\sqrt[3]{k}}\cdot\frac{3^{\small13/6}}{2^{\small4/3}5^{\small5/6}\pi}\Gamma^{3}\left(\tfrac{1}{3}\right).$$
For $k=9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/879854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 2,
"answer_id": 0
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Understanding the logic behind particular derivative I have
$\frac{\partial (f(x) g(x))}{\partial x}$=$g(x) f'(x)+f(x) g'(x)$,
I need to differentate this function with respect to x.
$f(x)=(x+1) (x+2)^2 (x+3)^3 (x+4)^4$
However I do not see the logic using the product rule.
| HINT
You must do that step by step.
$f(x)=(x+1) (x+2)^2 (x+3)^3 (x+4)^4$
$$f'(x)=((x+1))' ((x+2)^2 (x+3)^3 (x+4)^4)+((x+1)) ((x+2)^2 (x+3)^3 (x+4)^4)'$$
$$((x+2)^2 (x+3)^3 (x+4)^4)'=((x+2)^2)' ((x+3)^3(x+4)^4)+((x+2)^2) ((x+3)^3(x+4)^4)'$$
$$((x+3)^3(x+4)^4)'=((x+3)^3)'((x+4)^4)+((x+3)^3)((x+4)^4)'$$
now you go from th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/880175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Middle binomial coefficient mod 4 It is known that the middle binomial coefficient is always even. Show that $\binom{2n}{n}= 2 \mod 4$ if and only if $n$ is a power of 2.
| The power of $2$ in $\binom{2n}{n}$ is
$$\lfloor \frac{2n}{2} \rfloor +\lfloor \frac{2n}{2^2} \rfloor +\lfloor \frac{2n}{2^3} \rfloor + ... -2 \left(\lfloor \frac{n}{2} \rfloor +\lfloor \frac{n}{2^2} \rfloor +\lfloor \frac{n}{2^3} \rfloor +...\right) \\
=\lfloor \frac{2n}{2} \rfloor - \left(\lfloor \frac{n}{2} \rfloor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/880614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find depth of a half-filled parabolic cross-section Given a cross-section of an object that is parabolic in shape, how do you find the depth of the object when it is "half full".
A full example given in an exam:
A long trough whose cross-section is parabolic is $1\frac{1}{2}$ metres wide at the top and $2$ metres deep... | The way to do this is to find the area of the parabola as a function of $x$. Let's assume for simplicity that function is: $x = c*y^2$. This is equivalent to $y = \sqrt{\frac{x}{c}}$ Since the parabola is symmetric, we can double this to get the total width at any point $x$.
$$ w(x) = 2 \sqrt{\frac{x}{c}}$$
Now we can ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Integrate a division of polynomials Hi I have the following integral:
$$\int \frac{2x}{x^2+6x+3}\, dx$$
I made some changes like:
$$\int \dfrac{2x+6-6}{x^2+6x+3}\, dx$$
then I have:
$$\int \dfrac{2x+6}{x^2+6x+3}\, dx -\int\dfrac{6}{x^2+6x+3}\, dx$$
and thus: $$\ln(x^2+6x+3)-\int\dfrac{6}{x^2+6x+3}\, dx$$
Ok, I have de... | HINT:
As
$\displaystyle x^2+6x+3=(x+3)^2-(\sqrt6)^2,$
using Trigonometric substitution, set $x+3=\sqrt6\sec\theta$
or use $\#1$ of this
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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"answer_id": 3
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Series for logarithms This is more of a challenge than a question, but I thought I'd share anyway. Prove the following identities, and prove that the pattern continues.
\begin{equation*}
\sum_{n=0}^\infty\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)=\ln2
\end{equation*}\begin{equation*}
\sum_{n=0}^\infty\left(\frac{1}{3n+... | Let $m\geq 2$. Put:
$$S_m=\sum_{n\geq 0}(\frac{1}{mn+1}+\cdots+\frac{1}{mn+m-1}-\frac{m-1}{mn+m}) $$
As $\displaystyle \frac{1}{mn+r}=\int_0^1x^{mn+r-1}dx$, We have
$$S_m=\int_0^{1}\frac{1+x+\cdots+x^{m-2}-(m-1)x^{m-1}}{1-x^m} dx$$
But
$$1+x+\cdots+x^{m-2}-(m-1)x^{m-1}=(1-x^{m-1})+\cdots+(x^{m-2}-x^{m-1})$$
Hence
$$1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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prove $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq 3(a^2+b^2+c^2)$
If $a,b,c$ are positive real numbers and $a+b+c=1$,Prove: $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq 3(a^2+b^2+c^2)$$
Additional info:We can use AM-GM and Cauchy inequalities mostly.We are not allowed to use induction.
Things I have done so... | Homogenization gives $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge \frac{3(a^2+b^2+c^2)}{a+b+c}$$
We'll prove a stronger one $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge 6\cdot \frac{a^2+b^2+c^2}{a+b+c}-(a+b+c)$$
$$\Leftrightarrow (a+b+c)\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)\ge 6(a^2+b^2+c^2)-(a+b+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/883436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Why is $f(n) =\frac{n(n+1)(n+2)}{(n+3)}$ in $O(n^2)$? Let:
$$f(n) = n(n+1)(n+2)/(n+3)$$
Therefore :
$$f∈O(n^2)$$
However, I don't understand how it could be $n^2$, shouldn't it be $n^3$? If I expand the top we get $$n^3 + 3n^2 + 2n$$ and the biggest is $n^3$ not $n^2$.
| $$f(n)=\frac{n(n+1)(n+2)}{n+3}=\frac{(n^2+n)(n+2)}{n+3}=\frac{n^3+2n^2+n^2+2n}{n+3}=\frac{n^3+3n^2+2n}{n+3} \\ =n^2-\frac{6}{n+3}+2$$
Let $f(n)=O(n^2)$.Then, $\exists c>0 \text{ and } n_0 \geq 1 \text{ such that } \forall n \geq n_0: \\ f(n) \leq cn^2 \Rightarrow n^2-\frac{6}{n+3}+2 \leq cn^2 \Rightarrow c \geq 1+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/883972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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Prove $\lim_{n\to\infty} \frac{(2^{2^n}+1)(2^{2^n}+3)(2^{2^n}+5)\cdots (2^{2^n+1}+1)}{(2^{2^n})(2^{2^n}+2)(2^{2^n}+4)\cdots (2^{2^n+1})}=\sqrt{2}$ Question:
Prove or disprove
$$I=\lim_{n\to\infty} \frac{(2^{2^n}+1)(2^{2^n}+3)(2^{2^n}+5)\cdots (2^{2^n+1}+1)}{(2^{2^n})(2^{2^n}+2)(2^{2^n}+4)\cdots (2^{2^n+1})}=\sqrt{2}$$... | $$\log \left(1+\tfrac{1}{2^{2^n}}\right)\left(1+\tfrac{1}{2^{2^n}+2}\right)\cdots\left(1+\tfrac{1}{2^{
2^n+1}}\right)
= \log \left(1+\tfrac{1}{2^{2^n}}\right)+
\log \left(1+\tfrac{1}{2^{2^n}+2}\right)+\cdots + \log\left(1+\tfrac{1}{2^{
2^n+1}}\right)$$
Expanding on Alex' idea let $t = 2^{2^n}$ which is getting very lar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/885448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Trigonometric identity, simplifying an expression to $(1-\sin^2 a\cos^2a)/(2+\sin^2a\cos^2a)$ Question:
$$\left(\frac{1}{\sec^2A-\cos^2A}+\frac{1}{\csc^2A -\sin^2A}\right)\sin^2A\cos^2A=\frac{1-\sin^2A \cos^2A}{2+\sin^2A\ \cos^2A}$$
Prove L.H.S. = R.H.S.
My Efforts:
$$=\left(\frac{1}{\frac{1}{\cos^2 A}-\cos^2A}+\frac{1... | $$\left(\frac{\cos^2A}{1-\cos^4A}+\frac{\sin^2A}{1-\sin^4A }\right)\sin^2A\cos^2A$$
$$=\left(\frac{\cos^2A}{(1+\cos^2A)(1-\cos^2A)}+\frac{\sin^2A}{(1+\sin^2A)(1-\sin^2A) }\right)\sin^2A\cos^2A$$
$$=\left(\frac{\cos^2A}{(1+\cos^2A)\sin^2A}+\frac{\sin^2A}{(1+\sin^2A)\cos^2A}\right)\sin^2A\cos^2A$$
$$=\frac{\cos^4A}{1+\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/886342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Does every integer $n > 2$ have a "reciprocating" pseudoprime $a$? Does every integer $n > 2$ have a "reciprocating" pseudoprime $a$ such that $\gcd(a, n) = 1$ and satisfies both $a^{n - 1} \equiv 1 \mod n$ and $n^{a - 1} \equiv 1 \mod a$? (And of course allowing for the idea of even pseudoprimes).
From Fermat's little... | For any composite $n$, take $a = \dfrac{n^n - 1}{n - 1}$. It is easy to prove that $a$ is also composite and $(a, n) = 1$. Note that$$
a = \frac{n^n - 1}{n - 1} = n^{n - 1} + \cdots + n + 1 \equiv 1 \pmod{n} \Longrightarrow a^{n - 1} \equiv 1 \pmod{n}.
$$
Also,$$
n^n = (n - 1)a + 1 \equiv 1 \pmod{a} \Longrightarrow n^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/887070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Trigonometric simplification for limits: $\lim_{x\to 0} \frac{1-\cos^3 x}{x\sin2x}$ Have to evaluate this limit, but trigonometry part is :(
$$\lim_{x\to 0} \dfrac{1-\cos^3 x}{x\sin2x}.$$
Had written the denominator as $2x\sin x\cos x$, no idea what to do next. Please help...
| Otherwise using:
$$a^3-b^3=(a-b)(a^2+ab+b^2)$$
$$\eqalign{
\lim_{x\to 0} \dfrac{1-\cos^3 x}{x\sin2x}
&=\lim_{x\to 0} \dfrac{(1-\cos x)(1+\cos x+\cos^2x)}{2x\sin x\cos x}\\
&=\frac32\lim_{x\to 0} \dfrac{(1-\cos x)}{x\sin x\cos x}\\
&=\frac32\lim_{x\to 0} \dfrac{\frac{(1-\cos x)}x}{x\frac{\sin x}x\cos x}\\
&=\frac32\lim_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/888081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Why does $\lfloor\frac{n}{x}\rfloor$ have at most $2\sqrt{n}$ values when $x=1,2,\dots n$? The question is very short and clear:
Why does $\lfloor\frac{n}{x}\rfloor$ (floor of $\frac nx$) have at most $2\sqrt{n}$ values when $x = 1, 2,\dots, n $?
I saw this statement at tutorial of 449A at codeforces, I really want to ... | Over the range $1 \le x \le \sqrt{n}$, $x$ can take on only $\sqrt{n}$ distinct integer values. Thus, $\left\lfloor\dfrac{n}{x}\right\rfloor$ can only take on $\sqrt{n}$ distinct values, one for each distinct value of $x$.
Over the range $\sqrt{n} \le x \le n$, we have that $1 \le \dfrac{n}{x} \le \sqrt{n}$. Thus, $\l... | {
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How many different positive integer factors does have? How many different positive integer factors does $(2^7)(3^4)(7^3)(23^5)$ have?
Do we have to do any combinations between the powers here?
| Note the following:
*
*$\gcd(2,3)=1$
*$\gcd(2,7)=1$
*$\gcd(2,23)=1$
*$\gcd(3,7)=1$
*$\gcd(3,23)=1$
*$\gcd(7,23)=1$
Therefore, the number of different positive integer divisors of $(2^7)(3^4)(7^3)(23^5)$ is:
$$(7+1)\cdot(4+1)\cdot(3+1)\cdot(5+1)=960$$
| {
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Evaluating a limit. What makes the equality right? I'm reading a proof of a limit calculation. The limit is:
$$\lim\limits_{x\to 0}\left(\frac{a^x+b^x}{2}\right)^\frac{1}{x}$$
where $a,b>0$.
The aother claims that:
$$\lim\limits_{x\to 0}\left(\frac{a^x+b^x}{2}\right)^\frac{1}{x} =
\exp\left( \lim\limits_{x\to 0}\frac{\... | If we try with
$$
\lim_{x\to 0} \frac{\log(a^x+b^x)-\log 2}{x}
$$
and apply l'Hôpital's theorem, we get
$$
\lim_{x\to 0}\frac{a^x\log a+b^x\log b}{a^x+b^x}=\frac{\log a+\log b}{2}=
\log\sqrt{ab}.
$$
It's just the derivative of $x\mapsto (a^x+b^x)/2$ at $0$, of course.
However,
$$
\lim_{x\to0}\frac{\log(1+x)}{x}=1
$$
so... | {
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"source": "stackexchange",
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Complex analysis integral (maybe using residue theory?) Trying to compute the following integral:
$$
\int_{-\pi}^{\pi} \frac{d\theta}{1+\sin^2(\theta)}.
$$
This is for a complex analysis course, so I'm trying to find a way to use residue theory of something of that nature to solve the problem. I can't think of a substi... | You can solve the integral also without the help of complex analysis. For instance
\begin{align*}
\int\dfrac{1}{1+\sin^2(x)}\, dx &= \int\dfrac{1}{1+\sin^2(x)}\dfrac{\frac{1}{\sin^2(x)}}{\frac{1}{\sin^2(x)}} \ dx \\
&=\int\dfrac{\frac{1}{\sin^2(x)}}{\frac{1}{\sin^2(x)}+1}\, dx \\
&=\int\dfrac{\frac{1}{\sin^2(x)}}{2+\co... | {
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Find a real numbers $a,b$ such $a^n+b^n$ is rational Question:
prove or disprove :there exsit real numbers $a,b$ such follow two condition:
(1):$a+b$ is irrational
(2): for any postive integer $n\ge 2$, then $a^n+b^n$ is rational.
I have know if
$n=2k$ case is true,because I let $a=\sqrt{2}+1,b=\sqrt{2}-1$,so
$$a^{... | Theorem. There are no real numbers $a,b$ such that
$$\hbox{$a+b$ is irrational}$$
and
$$\hbox{$a^2+b^2$, $a^3+b^3$, $a^4+b^4$ and $a^6+b^6$ are rational.}$$
Proof. Suppose that there are such $a,b$; clearly they are non-zero. Then
$$2a^2b^2=(a^2+b^2)^2-(a^4+b^4)$$
and
$$2a^3b^3=(a^3+b^3)^2-(a^6+b^6)$$
are rational, ... | {
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"source": "stackexchange",
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area formed by a box and line suppose we have a box defined by coordinates $(1,1)$, $(-1,1)$, $(-1,-1)$, $(1,-1)$.
Suppose, a line $y=m(x+b)$ crosses the box with $m>0$ and $b>0$. What is the area of left upper triangle. Assume that the line crosses the box.
Thank you very much.
Partial Answer:
the line intersects a... | The box that is defined by coordinates $(1,1), (-1,1), (-1,-1), (1,-1)$ is the following:
To draw also the line $y=m(x+b) $ :
For $x=-1: y=m(-1+b)$
For $y=1: x=\frac{1}{m}-b$
Let's suppose that the line intersects the box as followed:
So we want to find the area of the yellow region.
$A(-1,1), B(\frac{1}{m}-b,1), C(-... | {
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How prove that $ x+y+z>4$ for $ a+b+c=4$ and $ ax+by+cz=xyz$? Given positive reals $ a,b,c,x,y,z$ such that $ a+b+c=4$ and $ ax+by+cz=xyz$.
How prove that $ x+y+z>4$?
| Suppose there exist $a, b, c, x, y, z > 0$ such that
$$a+b+c=4,$$
$$ax+by+cz = xyz,$$
$$x+y+z \leq 4.$$
Observe that we have
$$axyz + bxyz + cxyz = 4xyz,$$
so that
$$axyz + bxyz + cxyz = 4ax + 4by + 4cz,$$
and hence
$$ax(yz-4) + by(xz-4) + cz(xy-4) = 0.$$
Since
$x+y+z \leq 4$
and
$a, b, c, x, y, z > 0,$
what follow are... | {
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Do the medians (or other cevians) form all the triangles? I want to know whether set of medians of all triangles, or some other class of cevians, can form the set of all the triangles?
For example, in the case of altitudes, $(4,7,10)$ is an counterexample. While $(4,7,10)$ can form a triangle, there is no triangle wit... | From the Wikipedia article on Median, the lengths of the medians in terms of the sidelengths are:
$m_a = \dfrac{1}{2}\sqrt{2b^2+2c^2-a^2}$
$m_b = \dfrac{1}{2}\sqrt{2c^2+2a^2-b^2}$
$m_c = \dfrac{1}{2}\sqrt{2a^2+2b^2-c^2}$
Solving for $(m_a,m_b,m_c)$ in terms of $(a,b,c)$ yields:
$a = \dfrac{2}{3}\sqrt{2m_b^2+2m_c^2-m_... | {
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Is this real number an integer? Is this real number :
$$\Big(2+\frac{10}{9}\sqrt{3}\Big)^{1/3}+\Big(2-\frac{10}{9}\sqrt{3}\Big)^{1/3}$$
an integer ?
I've tried different factorization, but nothing seems to work.
| Let $a = \sqrt[3]{2+\frac{10}{9}\sqrt{3}}$ and $b = \sqrt[3]{2-\frac{10}{9}\sqrt{3}}$. Then:
$a^3+b^3 = (2+\frac{10}{9}\sqrt{3}) + (2-\frac{10}{9}\sqrt{3}) = 4$
$ab = \sqrt[3]{(2+\frac{10}{9}\sqrt{3})(2-\frac{10}{9}\sqrt{3})} = \sqrt[3]{4-\frac{10^2}{9^2}\cdot 3} = \sqrt[3]{\frac{8}{27}} = \frac{2}{3}$.
Now, let $x ... | {
"language": "en",
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"source": "stackexchange",
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Finding the sum to infinity Question:
Find the sum to infinity for the following series
$$1, -\frac{1}{2}, \frac{1}{2^2}, -\frac{1}{2^3},\cdots$$
What would be the technique used to find such a sum?
| $(1+\frac{1}{2^2}+\frac{1}{2^4}+...)-(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+...)$
Both the brackets above contain GP of $n$ terms with $r=\frac{1}{2^2}=\frac{1}{4}$
$\Large(\frac{(\frac{1}{4})^n-1}{-3/4})-(\frac{1}{2}\frac{(\frac{1}{4})^n-1}{-3/4})$
$\Large\frac{\frac{1}{2}(\frac{1}{4})^n-\frac{1}{2}}{\frac{-3}{4}}=\... | {
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"url": "https://math.stackexchange.com/questions/901672",
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"question_score": "4",
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Counting the roots of a polynomial over a finite field Let $\mathbb{F}_{11}$ be the field of 11 elements and let $\mathcal{K}$ be the splitting field of $x^{3} - 1$ over $\mathbb{F}_{11}$. How many roots does $(x^{2} - 3)(x^{3} - 3)$ have in $\mathcal{K}$?
First, note that $x^{3} - 1 = (x- 1)(x^{2} + x + 1)$ and the qu... | First, note that $x^{3} - 1 = (x- 1)(x^{2} + x + 1)$ and the quadratic factor is irreducible over
$\mathbb{F}_{11}$ so that $\mathcal{K} = \mathbb{F}_{11}/(x^{2} + x + 1) \cong \mathbb{F}_{11^{2}}$.
Note that $3^{6} = 3$ in $\mathbb{F}_{11^{2}}$ so that $(x^{2} - 3)(x^{3} - 3) = (x - 27)(x+ 27)(x-9)(x^{2} + 9x + 81)$... | {
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What is the first step to solving $\cos3x - \sin x = \sqrt{3}(\cos x - \sin 3x)$? My calculus BC teacher has given us some trig "review".
$$\cos3x - \sin x = \sqrt{3}(\cos x - \sin 3x).$$
How do I get rewrite the cos3x and sin3x? Do I just use sum and difference, because it makes everything really complicated.
can so... | $\cos3x - \sin x = \sqrt{3}(\cos x - \sin 3x)$
$\implies \cos3x+\sqrt3\sin3x=\sqrt3\cos x+\sin x$
Now divide both sides by 2 to get :
$\frac{1}{2}\cos3x+\frac{\sqrt3}{2}\sin3x=\frac{\sqrt3}{2}\cos x+\frac{1}{2}\sin x$
$\implies \cos \frac{\pi}{3} \cos 3x + \sin \frac{\pi}{3} \sin 3x=\cos \frac{\pi}{6} \cos x + ... | {
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Calculate $\int_{S^2}\frac{1}{\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}}dS$ where $a^2+b^2+c^2<1$.
Let $a^2+b^2+c^2 < 1$ and $S^2$ be unit sphere in $R^3$. Calculate $$\int_{S^2}\frac{1}{\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}}dS$$
Let $(x,y,z)=(\cos\theta \cos\phi,\cos\theta \sin\phi, \sin\theta)$. By definition,
$$\int_{S^2}\frac{1}{\... | Let $(x,y,z)=(a+\cos\theta \cos\phi,b+\cos\theta \sin\phi,c+ \sin\theta)$
$$\int_{S^2}\frac{1}{\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}}dS\\
=\int_0^{2\pi}\int_0^{\pi}\frac{1}{\sqrt{(\cos\theta\cos\phi)^2+(\cos\theta \sin\phi)^2+(\sin\theta)^2}}\sin\theta d\theta d\phi=\int_0^{2\pi}\int_0^{\pi}\sin\theta d\theta d\phi\\=4\pi$$
| {
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"source": "stackexchange",
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Prove $\sum\limits_{cyc} \frac{\sqrt{xy}}{\sqrt{xy+z}}\le\frac{3}{2}$ if $x+y+z=1$
if $x,y,z$ are positive real numbers and $x+y+z=1$ Prove:$$\sum_{cyc} \frac{\sqrt{xy}}{\sqrt{xy+z}}\le\frac{3}{2}$$
where $\sum_{cyc}$ denotes sums over cyclic permutations of the symbols $x,y,z$.
Additional info:I'm looking for solu... | $$\sum_{cyc} \frac{\sqrt{xy}}{\sqrt{xy+z}}\le\frac{3}{2} $$
$$ \sum_{cyc} \frac{\sqrt{xy}}{\sqrt{xy+z}} = \sum_{cyc} \frac{\sqrt{xy}}{\sqrt{xy+1-x-y}} = \sum_{cyc} \frac{\sqrt{xy}}{\sqrt{(1-x)(1-z)}} = \sum_{cyc} \frac{\sqrt{xy}}{\sqrt{(y+z)(x+z)}} $$
And now, by AM-GM
$$ \sum_{cyc} \frac{\sqrt{xy}}{\sqrt{(y+z)(x+z)... | {
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If $\frac{5x}{2x^2+5x+1}=\frac13$ then the value of $x+\frac{1}{2x}$ is If $\frac{5x}{2x^2+5x+1}=\frac13$ then the value of $x+\frac{1}{2x}$ is
note $5x=1$
$2x^2+5x+1=3$
or
$15x=2x^2+5x+1$
so $2x^2-10x+1$
so $\frac{5+\sqrt{23}}2$ or $\frac{5-\sqrt{23}}2$
| HINT:
As $x\ne0,$
$$3=\frac{5x}{2x^2+5x+1}=\frac{5/2}{x+5/2+1/2x}$$
$$\iff\frac{x+5/2+1/2x}{5/2}=\frac13$$
Hope you can take it from here
| {
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Factorization of $z^4 +1 = (z^2 - \sqrt 2z+1)(z^2 + \sqrt 2 z+1)$ for complex $z$ How can I get this equation from LHS to RHS by using the four roots of $z^4 +1 = 0$ are $z=\pm\sqrt{\pm i}$
$$z^4 +1 = (z^2 - \sqrt2 z+1)(z^2 + \sqrt2 z+1)$$
| You don't need to know the roots of $P(z)=z^4+1$ for the factorization: $$z^4+1=z^4+2z^2+1-2z^2=(z^2+1)^2-(\sqrt{2}z)^2=(z^2-\sqrt{2}z+1)(z^2+\sqrt{2}z+1)$$
If you want to use that the roots of $P$ are $\omega=e^{i\frac{\pi}{4}}$, $i\omega$, $-\omega$ and $-i\omega$ (I tend to avoid to use the notation $\sqrt{z}$ for ... | {
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Proving determinants using properties of determinants $$\begin{vmatrix}
1 & a^2+bc & a^3\\
1 & b^2+ca & b^3\\
1 & c^2+ab & c^3
\end{vmatrix}
= (a-b)(b-c)(c-a)(a^2+b^2+c^2)$$
we have to solve this by using the properties of determinants without actually expanding the determinant. I am Unable to think which calcula... | The first trick is to get as much zeroes as you can in the first row. That makes multiplication easier.
$\begin{vmatrix} 1 & a^2+bc & a^3 \\ 1 & b^2+ca & b^3\\ 1 & c^2+ab &c^3 \end{vmatrix}$
subtracting second row from first row and third row from second row:
$(a-b)(b-c)\begin{vmatrix} 0 & a+b-c & a^2+ab+b^2 \\ 0 & b+... | {
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Finding the value of $\frac{\cos^4\beta}{\cos^2\alpha} + \frac{\sin^4\beta}{\sin^2\alpha}$. Trigonometry
$\dfrac{\cos^4 \alpha}{\cos^2 \beta}+ \dfrac{\sin^4\alpha}{\sin^2\beta} = 1$
then the value of
$\dfrac{\cos^4\beta}{\cos^2\alpha}+ \dfrac{\sin^4\beta}{\sin^2\alpha}$ is?
NOTE: can somebody help me
$\cos^2\alpha \... | $\cos^2 \alpha \left(\frac{\cos^2\alpha}{\cos^2 \beta}\right) + \sin^2 \alpha \left(\frac{\sin^2 \alpha}{\sin^2 \beta}\right)=1$.
It can be noticed that $\frac{\cos^2\alpha}{\cos^2 \beta}$ and $\frac{\sin^2 \alpha}{\sin^2 \beta}$ must be $1$ (Why?).
This tells us that $\cos^2 \alpha = \cos^2 \beta$ and $\sin^2 \alpha ... | {
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How to describe 4-8+12-16+20-24+28 using summation ($\Sigma$) notation? Can anyone find the ∑ summation for this please?
4-8+12-16+20-24+28
It seems to be going up by steps of 4, but I can't seem to get how I should write it down, since it uses both + and -
| $$4-8+12-16+20-24+28=4(1-2+3-4+5-6+7)=4((-1)^{1+1}\cdot 1+(-1)^{2+1}\cdot 2+(-1)^{3+1}\cdot 3+(-1)^{4+1} \cdot 4+(-1)^{5+1} \cdot 5+(-1)^{6+1} \cdot 6+(-1)^{7+1} \cdot 7)=4 \sum_{i=1}^{7}(-1)^{i+1}i=4 \left ( \sum_{i=0}^{3} (2i+1)-\sum_{i=1}^{3} 2i\right )=4 \left( 2 \sum_{i=0}^{3} i +\sum_{i=0}^{3} 1-2\sum_{i=1}^{3} i... | {
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Sum of N terms of a series Perhaps this is a silly question but how can I simplify this to find the sum of $N$ terms?
$$\sum_{n=1}^N \frac{2n - 1}{n(n+1)(n+2)}$$
I know how to find the sum of cubes and squares etc but how with the denominator like that?
Any help would be appreciated thanks.
The questions in the book ha... | Checking if there is a pair of constants $(A,B)$ such that
$$\frac{2n-1}{n(n+1)(n+2)}=A\cdot\color{red}{\frac{1}{(n+1)(n+2)}}+B\cdot\color{blue}{\frac{1}{n(n+1)}}$$
gives us $(A,B)=(5/2,-1/2)$.
Hence, since we can have
$$\frac{2n-1}{n(n+1)(n+2)}=\frac{5}{2}\color{red}{\left(\frac{1}{n+1}-\frac{1}{n+2}\right)}-\frac 12... | {
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Integrate $\int \frac{6x-15}{x^2 +8x+101}dx$ I've been trying this one for days and I can't seem to get it. Any ideas?
$\int \frac{6x-15}{x^2 +8x+101}dx$
Thanks,
Nick
| $\displaystyle\int\dfrac{6x-15}{x^2+8x+101} dx=\int\dfrac{6x-15}{(x+4)^2+85} dx$.
Now let $u=x+4$, so $x=u-4$ and $dx=du$ to get
$\displaystyle\int\frac{6u-39}{u^2+85} du=3\int\frac{2u}{u^2+85} du-39\int\frac{1}{u^2+85} du$
$=3\ln(u^2+85)-39\left(\frac{1}{\sqrt{85}}\arctan\frac{u}{\sqrt{85}}\right)+C$.
(Now substitute... | {
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Taylor polynomial of $\frac{1}{2-x}$ Can someone show how to find the Taylor polynomial of $\frac{1}{2-x}$?
I tried this: $\frac{1}{2-x}=\frac{1}{1-(x-1)}$ and then use that $ \ T_n(\frac{1}{1-x})=1+x+\dots +x^n.$ But this gives $1+(x-1)+\dots +(x-1)^n$ which is the wrong answer. Why?
| We want to find $a_0, a_1, a_2,\ldots$ so that $$\frac1{2-x} = a_0 + a_1x + a_2x^2 + \ldots.$$
Multiply both sides by $2-x$: $$
\begin{align}
1 & = (2-x)a_0 + (2-x)a_1x + (2-x)a_2x^2 + \cdots\\
& = (2a_0-a_0x)+(2a_1x-a_1x^2) + (2a_2x^2 - a_2x^3) + \cdots \\
\end{align}$$
Combine like terms:
$$ 1 = 2a_0 + (2a_1-a_0)x +... | {
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Sum of the following series I'm stuck with this series:$$\frac12 + \frac{1\cdot 4}{2 \cdot 5} + \frac{1\cdot 4 \cdot 7}{2 \cdot 5 \cdot8} + \frac{1 \cdot 4 \cdot 7 \cdot 10}{2 \cdot 5 \cdot 8 \cdot 11}+\ldots$$
I cant even find the $n$-th th term here...
I have to prove that $$\sum_{k=1}^n A_k = \frac12 \left[ \... | Notice that the $n$-th term of the sum is given by:
$$A_n=\prod_{j=1}^{n}\frac{3j-2}{3j-1}.$$
Let now $B_n$ be:
$$ B_n = \prod_{j=1}^{n}\frac{3j+1}{3j-1}=(3n+1)\prod_{j=1}^{n}\frac{3j-2}{3j-1}=(3n+1)A_n=(3n-1)A_n+2A_n.$$
Since:
$$ (3n-1)A_n = (3n-2)A_{n-1} = B_{n-1}$$
it happens that:
$$ B_n = B_{n-1} + 2 A_n = B_{n-2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/919519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality: $x^2+y^2+xy\ge 0$ I want to prove that $x^2+y^2+xy\ge 0$ for all $x,y\in \mathbb{R}$.
My "proof": Suppose wlog that $x\ge y$, so $x^2\cdot x\ge x^2\cdot y\ge y^2\cdot y=y^3$ (because $x^2\ge 0$ so we can multiply both sides by it without changing the inequality sign) giving $x^3\ge y^3$. Substracting we ha... | One (simple) way:
let t = x2 + y2 + xy
then t - xy ≥ 0 since it is the sum of two real squares x2 + y2
and t + xy ≥ 0 since it is the square of the real (x + y) [since (x + y)2 = x2 + y2 + 2xy]
adding these, we get 2t ≥ 0, therefore t ≥ 0
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/920605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 2
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The expression $1 + x^2 +(-T_px+y)^2 +z^2$ is bounded below by a constant multiple of $(1+x^2+y^2+z^2)$ Suppose $T_p > 0$. Is there a simply way to show that
$1 + x^2 +(-T_px+y)^2 +z^2 \geq C (1+x^2+y^2+z^2)$, for all $(x,y,z) \in \mathbb R^3$, where $C>0$.
| We can show that for a certain positive constant $C$
$$ x^2+(-T_p x+y)^2+z^2+w^2 \geq C(x^2+y^2+z^2+w^2) \tag{1} $$
holds for any $(x,y,z,w)\in\mathbb{R}^4$. We just need that the eigenvalues of the symmetric matrix
$$ M=\left(\begin{array}{cccc}T_p^2+1-C & -T_p & 0 & 0 \\ -T_p & 1-C & 0 & 0 \\ 0 & 0 & 1-C & 0 \\ 0 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/921654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How to Integrate $\int\frac{(x^2)}{\sqrt{7-x^2}}dx$. I am trying to Integrate $\int\frac{(x^2)}{\sqrt{7-x^2}}dx$ and I have worked this problem a couple times and keep getting the same answer. So I will show my process and please point my errors out.
$$a=\sqrt{7}\hspace{10pt} x=\sqrt{7}\sin\theta\hspace{10pt}dx=\sqrt{... | $$\frac{x^2}{\sqrt{7-x^2}}=\frac{7-(7-x^2)}{\sqrt{7-x^2}}=7\cdot\frac1{\sqrt{(\sqrt7)^2-x^2}}-\sqrt{(\sqrt7)^2-x^2}$$
Now, $$\int\frac{dx}{\sqrt{a^2-x^2}}=\arcsin\frac xa$$
and $$\int\sqrt{a^2-x^2}dx=x\frac{\sqrt{a^2-x^2}}2+\frac{a^2}2\arcsin\frac xa$$
Reference : http://www.sosmath.com/tables/integral/integ13/integ13.... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Does the sequence converge, and to what? We have a sequence $\{a_n\}$
$$a_0 = 0$$
$$a_{n+1} = \frac{a_{n}}{2} + 1$$
Does it converge? And to what?
| $$a_{0}=0\\a_{n+1}=\frac{a_{n}}{2}+1\\a_{1}=\frac{a_{0}}{2}+1=1\\a_{2}=\frac{a_{2}}{2}+1=1+\frac{1}{2}\\a_{3}=\frac{\frac{3}{2}}{2}+1=1+\frac{3}{4}\\a_{4}=\frac{\frac{7}{4}}{2}+1=1+\frac{7}{8}\\a_{5}=\frac{\frac{15}{8}}{2}+1=1+\frac{15}{16}\\a_{n}=\frac{2^{n-1}-1}{2^{n}}+1\\...\\a_{n}<2\\so\\it-converge
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/922738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 5
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On a recursive sequence (exercise 8.9 Apostol) The exercise states: show convergence of the sequence ${a_n}$ knowing that:
$$|a_n| \le 2, \ \ \ |a_{n+2}-a_{n+1}| \le \frac{1}{8}|a_{n+1}^2 - a_{n}^2|.$$
The solution states:
$$|a_{n+2}-a_{n+1}| \le \frac{1}{8}|a_{n+1}^2 - a_{n}^2| = \frac{1}{8}|a_{n+1} - a_{n}||a_{n+1} +... | Since $-2\leq a_n\leq 2$
$$|a_{n+2}-a_{n+1}| \le \frac{1}{8}|a_{n+1}^2 - a_{n}^2| = \frac{1}{8}|a_{n+1} - a_{n}||a_{n+1} + a_{n}| \leq \frac{1}{8}|a_{n+1} - a_{n}||4|= \frac{1}{2}|a_{n+1} - a_{n}| $$
Next, call $b_n=a_{n+1}-a_n$ and rewrite the above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/922923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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For the following system to be consistent, what must k not be equal to? $6x - 4y + 4z = 5$
$9x - 6y + kz = -4$
$12x - 8y = -10$
Originally I just multiplied the first row by $\frac{3}{2}$ and subtracted it from the second, which gives you a value of $6$ for the answer. However, this is not the correct answer. Any ... |
A system of linear equations is called inconsistent if it has no solutions. A
system which has a solution is called consistent.
Now the augmented matrix
$
\left[\begin{array}{rrr|r}
6 & -4 & 4 & 5 \\
9 & -6 & k & -4 \\
12 & -8 & 0 & -10
\end{array}\right]$
Replace $R_3 $ by $R_3-2R_1$ and $R_2$ by $ R_2 - \frac{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/927651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Lucas Numbers Proof $L_n = \alpha^n + \beta^n$ Proof by Induction:
Lucas numbers are recursively defined as:
$L_n = L_{n-1} + L_{n-2}$ where $L_1 = 1$ and $ L_2 = 3 $for $n \ge 3$
Show that: $L_n = \alpha^n + \beta^n$ for $\alpha = {1+\sqrt{5}\over2}$ and $\beta = {1-\sqrt{5}\over2}$
Is this the correct approach?
bas... | Note that for any $\alpha$, $\beta$, $k$, $$(\alpha^k + \beta^k)(\alpha + \beta) = \alpha^{k+1} + \alpha \beta^k + \alpha^k \beta + \beta^{k+1} = \alpha^{k+1} + \beta^{k+1} + \alpha \beta(\alpha^{k-1} + \beta^{k-1}).$$
Now, if $\alpha = (1+\sqrt{5})/2$, $\beta = (1-\sqrt{5})/2$, what are the sum $\alpha + \beta$ and pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/933157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove combination identity $ \sum_{k=0}^n {2k \choose k} {2n-2k \choose n-k} = 4^n $
I tried with mathematical induction only to fail.
Is this formula related to some special function like Beta, Gamma, etc?
| Consider the generating function method. It is seen that for
\begin{align}
S_{n} = \sum_{k=0}^{n} \binom{2k}{k} \binom{2n-2k}{n-k}
\end{align}
yields
\begin{align}
\sum_{n=0}^{\infty} S_{n} t^{n} &= \sum_{n=0}^{\infty} \sum_{k=0}^{n} \binom{2k}{k} \binom{2n-2k}{n-k} t^{n} \\
&= \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} ... | {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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$[\mathbb{Q}(\sqrt [3] {2}+\sqrt {5}):\mathbb{Q}]$ What is $[\mathbb{Q}(\sqrt [3] {2}+\sqrt {5}):\mathbb{Q}]$? A straight forward way will be to just set $x=\sqrt [3] {2}+\sqrt {5}$, take powers and reach at a polynomial in $x$ and show the polynomial is irreducible. But the method is very tedious and seems to be hopel... | Denote $\alpha = \sqrt[3]{2}+ \sqrt{5}$.
Let $\zeta = - \frac{1}{2} + i \frac{\sqrt{3}}{2}$.$\ $
The product
\begin{eqnarray}
P(X) =\prod_{0\le k \le 2, 0 \le l \le 1} ( X - (\sqrt[3]{2}\cdot \zeta ^k + \sqrt{5}\cdot(-1)^l)\ )
\end{eqnarray}
is a polynomial with integer coefficients. We calculate:
\begin{eqnarray}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/937453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Do we simplify the Proof by Contradiction? Prove the following by contradiction:
Suppose $a,b\in\mathbb{Z}$. If $4|\left(a^2+b^2\right)$, then $a$ and $b$ are not both odd (in other words, $a$ and $b$ are even)
So, I did this:
Assume $a$ and $b$ are odd
Let $a={2k+1}$
Let $b={2l+1}$
$4|\left((2k+1\rig... | Because
$$
a^2+b^2=(a-b)^2+2ab,
$$
if $a^2+b^2$ is even, then $a$ and $b$ must share parity. But if $a$ and $b$ are both odd, then $(a-b)^2$ is divisible by $4$ whereas $2ab$ is not, so it must be that $a$ and $b$ are both even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/939707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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If $\int \frac{f(x)}{x^2(x+1)^3}\hspace{1mm}dx$ is rational and $f$ is quadratic with $f(0)=1$, then find $f'(0)$
If $\int \dfrac{f(x)}{x^2(x+1)^3}\hspace{1mm}dx$ is a rational function, and $f$ is quadratic function, such that $f(0)=1$. Then Find $f'(0)$
This looks like an interesting problem with an elegant solutio... | Let $f(x):=ax^2+bx+1$
$$\int \frac{ax^2+bx+1}{x^2(x+1)^3}dx=\int\left(\frac{a-b+1}{(x+1)^3}+\frac{2-b}{(x+1)^2}+\frac{b-3}{x}+\frac{3-b}{x+1}+\frac1{x^2}\right)dx$$
So $b=3$
Now $f'(x)=2ax+b,f'(0)=b=3$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is $5^2x^3-x^5 = x^3(x-5)(x+5)$ or $-x^3(5-x)(5+x)$ Geogebra's Factor function says that
$5^2x^3-x^5$
is
$-x^3(x-5)(x+5)$
but from what I do, it is positive, $x^3(5+x)(5-x)$
Note the x isnt in the same position
Am I wrong?
| Look:
$$5^2x^3-x^5=x^3(5^2-x^2)=x^3(5-x)(5+x)x^3(-1)(x-5)(x+5)=-x^3(x-5)(x+5).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Calculate $\int_0^1\frac{\log^2x\log(1+x^2)}{1-x^2}dx$ Evaluating $$\int_0^1\frac{\log^2(x)\log(1+x^2)}{1-x^2}dx$$
I found $- \dfrac{\pi^4}{32}+2G^2+\dfrac74 ζ(3)\log2 $ where $G$ is the Catalan's constant.
| \begin{align}J&=\int_0^1\frac{\log^2(x)\log(1+x^2)}{1-x^2}dx\\
&\overset{\text{IBP}}=\left[\left(\int_0^x\frac{\ln^2 t}{1-t^2}dt\right)\ln(1+x^2)\right]_0^1 -\int_0^1 \frac{\color{red}{2x}}{1+x^2}\left(\int_0^x\frac{\ln^2 t}{1-t^2}dt\right)dx\\
&=\frac{7}{4}\zeta(3)\ln 2-\int_0^1\int_0^1 \frac{2x^2\ln^2(tx)}{(1+x^2)(1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/941994",
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"source": "stackexchange",
"question_score": "3",
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Prove or disprove $ p^{r+s}\mid q^{ke} - 1 \iff p^s \mid k$. Let $p$ be an odd prime and $q$ be a power of prime. Suppose $e := \min\{\, e \in \mathbb{N} : p \mid q^e - 1 \,\}$ exists. Put $r := \nu_p(q^e - 1)$ (that is, $p^r \mid q^e - 1$ and $p^{r+1} \nmid q^e - 1$). What I want to prove is the following:
$ \forall ... | Suppose that $p$ divides $q^n-1$. Let $t=q^n$. We have that $p \mid (t-1)$, and we want to show that $\nu_p (t^k-1) = \nu_p (t-1) + \nu_p (k)$. We may assume that $k$ is prime, as the statement follows from this case by induction on the number of prime factors.
What we need to check is that $\nu_p (\frac{t^k-1}{t-1}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof that $|\sqrt{x}-\sqrt{y}| \leq \sqrt{|x-y|},\quad x,y \geq 0$ Any hints on how I can prove the inequality:
$$|\sqrt{x}-\sqrt{y}| \leq \sqrt{|x-y|},\quad x,y \geq 0$$
Thank you.
| Case 1: If $x = y$, both sides equal zero.
Case 2: If $x \neq y$, both sides of the inequality are positive, so we can square both sides of
$$|\sqrt{x} - \sqrt{y}| \leq \sqrt{|x - y|}$$
to obtain
$$|\sqrt{x} - \sqrt{y}|^2 \leq |x - y| = |(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})| = |\sqrt{x} + \sqrt{y}||\sqrt{x} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/946804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 6
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Is the sequence $x_n = \frac{\sin 1}{2} + \frac{\sin 2}{2^2} + \frac{\sin 3}{2^3} +\dots + \frac{\sin n}{2^n}$ Cauchy? $$x_{n} = \frac{\sin 1}{2} + \frac{\sin 2}{2^2} + \frac{\sin 3}{2^3} + ... + \frac{\sin n}{2^n}$$
I came across this sequence while studying, and while it is convergent, I'm curious as to whether or no... | You could be interested to notice that quite simple manipulations of the sines of multiple angles lead to $$x_n=\sum_{i=1}^n\frac{\sin (i)}{2^i}= \frac{2^{-n} \left(\sin (n)-2 \sin (n+1)+2^{n+1} \sin (1)\right)}{5-4 \cos (1)}$$ Similarly $$x_n=\sum_{i=1}^n\frac{\sin (i)}{3^i}=\frac{3^{-n} \left(\sin (n)-3 \sin (n+1)+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/947829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How can I find $\lim_{n\to\infty}\int_0^\infty\frac{n\cos^2(x/n)}{n+x^4}dx$? I am trying to find the value of this integral:
$\displaystyle{\lim_{n\to\infty}\int_0^\infty\frac{n\cos^2(x/n)}{n+x^4}dx}$.
The integrand tends to 1 as $n$ goes to infinity. So if some convergence theorem holds, the integral would tend to inf... | By a direct integration, we may obtain more than the value of the desired limit.
We have, as $n$ tends to $+\infty$:
$$
\int_0^{+\infty}n\:\frac{\cos^2(x/n)}{n+x^4}{\rm d}x =\color{#008B8B}{\frac{\pi\sqrt{2}}{4}} \color{#006699}{\:n^{1/4}}+\color{#008B8B}{\frac{\pi\sqrt{2}}{4}}\color{#006699}{\frac{1}{n^{5/4}}}+\color... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/949179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Find the order of the elements in the given groups I have to find the order of the following elements in the given groups:
*
*$(1 \ \ 2 \ \ 3) \ (1 \ \ 2\ \ 4) \text{ in } S_5$
$$\begin{pmatrix}
1 & 2 & 3 & 4 & 5\\
3 & 4 & 1 & 2 & 5
\end{pmatrix}=\begin{pmatrix}
1 & 3
\end{pmatrix}\begin{pmatrix}
2 & 4
\end{pmatri... | Your first two questions are indeed correct.
Note that your calculation of $A^3$ is "off":
$$A^3= A^2 A=\begin{bmatrix}
-1 & 0\\
0 & -1
\end{bmatrix} \begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix} =\begin{bmatrix}
0 & 1\\
-1 & 0
\end{bmatrix}$$
And $A^4 = A^2A^2 = -I\cdot -I = I$. Hence the order of $A = 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/949349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Proving terms are rational using Mathematical Induction
I was able to do the first part of the question, in the second part (Proof by Induction),
I showed it holds for $n=1$:
Then I Assumed its true for $n=k$.
$$\begin{align}I_{k+1}=\frac{e^2}{2}-\frac{(k+1)}{2}I_k\end{align}$$
Then for $n=k+1$:
$$\begin{align}I_{... | You wrote you showed it holds for $n=1$.
Assume that $I_k=A_ke^2+B_k$ where $A_k,B_k$ are rational numbers.
Then, $$\begin{align}I_{k+1}&=\frac 12e^2-\frac 12(k+1)I_k\\&=\frac 12e^2-\frac 12(k+1)(A_ke^2+B_k)\\&=\left(\frac{1}{2}-\frac{(k+1)A_k}{2}\right)e^2+\left(-\frac{(k+1)B_k}{2}\right).\end{align}$$
Since $$\frac{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that the equation Ax=x can be rewritten as (A-I)x = 0 and use this result to solve Ax=x for x. Given matrix A = \begin{bmatrix}2 & 1 & 2 \\ 2 & 2 & -2 \\ 3 & 1 & 1\end{bmatrix}
and x = \begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix}
Answer: Given $Ax = x$. Subtract from x from both sides to get $Ax - x = 0$. We kn... | One has
\begin{equation}
A-I
=
\left(\begin{matrix}2 & 1 & 2 \\ 2 & 2 & -2 \\ 3 & 1 & 1\end{matrix}\right)
-
\left(\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right)
=
\left(\begin{matrix}1 & 1 & 2 \\ 2 & 1 & -2 \\ 3 & 1 & 0\end{matrix}\right)
\end{equation}
With the rule of Sarrus
(see http://en.wik... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $z^{10}+\frac{1}{z^{10}}$ given $z^2+z+1=0$ $z$ is a complex number and $z^2+z+1=0$.
$$z^{10}+\frac{1}{z^{10}}=?$$
For the solution:
*
*the roots of $z^2+z+1$ are: $z_1=-\frac12+\frac{\sqrt3}{2}i$ and $z_2=-\frac12-\frac{\sqrt3}{2}i$
*converting these to their trigonometrical forms, we get: $z_1=\cos\frac{2\p... | Hint: $z^2+z+1=0$ implies that $0=(z-1)(z^2+z+1)=z^3-1$, so $z^3=1$. Also $z^{10}=z^9\cdot z$ and $z^{-10}=z^2\cdot z^{-12}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/952968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
$\lim_{x\to-\infty}\frac{\sqrt{6x^2 - 2}}{x+2}$ May I know how can I calculate the following expression?
$$
\lim\limits_{x\to-\infty}\frac{\sqrt{6x^2 - 2}}{x+2}
$$
From calculator, the answer is $-\sqrt{6}$ , my approach is by using dividing numerator and denominator by using $x$. Which is,
$$
\lim\limits_{x\to-\infty... | $$\lim_{x\to -\infty} \frac{\sqrt{6x^2-2}}{x+2}$$
First note that as $x\to -\infty$, then $x+2\lt 0$ and
$$ x+2=-|x+2|=-\sqrt{(x+2)^2} $$
Therefore
$$ \lim_{x\to -\infty} \frac{\sqrt{6x^2-2}}{x+2} = \lim_{x\to -\infty} \frac{\sqrt{6x^2-2}}{-\sqrt{(x+2)^2}} $$
$$ = -\lim_{x\to -\infty} \sqrt{\frac{6x^2-2}{x^2+4x+4}} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/956767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Chinese Remainder Theorem RSA I want to solve the following modular quadratic equation:
$x^2 \equiv 188 \pmod {437}$ using the fact that $437$ can be factorized by the primes as: $19⋅23$.
So far I have done:
$$x^2 \equiv 188 \pmod{19} \Rightarrow x^2 \equiv 17 \pmod {19}\\
x^2 \equiv 188 \pmod {23} \Rightarrow x^2 \equ... | You can find the solution using this calculator just fill in $$ x=6 \mod 19$$ and $$
x=2 \mod 23
$$
Your solution is $25$ and indeed $188 \equiv 25^2 \pmod {437} $
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 1
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How is $\lim_{x \to 0}\frac{\sqrt {2-x}-\sqrt {2}}{x} = \frac{-1}{2\sqrt{2}}$? How is
$$\displaystyle\lim_{x \to 0}\frac{\sqrt {2-x}-\sqrt {2}}{x} = \frac{-1}{2\sqrt{2}}$$
computable?
| Multiplying$$\frac{\sqrt{2-x}-\sqrt 2}{x}$$
by$$\frac{\sqrt{2-x}+\sqrt 2}{\sqrt{2-x}+\sqrt 2}$$
gives us
$$\begin{align}\lim_{x\to 0}\frac{\sqrt{2-x}-\sqrt 2}{x}&=\lim_{x\to 0}\frac{(\sqrt{2-x}-\sqrt 2)(\sqrt{2-x}+\sqrt 2)}{x(\sqrt{2-x}+\sqrt 2)}\\&=\lim_{x\to 0}\frac{(2-x)-2}{x(\sqrt{2-x}+\sqrt 2)}\\&=\lim_{x\to 0}\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/959064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Confused by the solution of $x^3+bx^2+cx+d=0$ From $x^3 + bx^2 + cx + d = 0$, we have $(x-x_1)(x-x_2)(x-x_3)=0$ for some roots $x_1$, $x_2$ and $x_3$.
Expanding this second expression gives us
$$x^3 + \left(x_1+x_2+x_3\right)x^2 + \left(x_1x_2 + x_1 x_3 + x_2 x_3\right) x + x_1 x_2 x_3 = 0$$
and comparing coefficients ... | If $$x^3+bx^2+cx+d=(x-x_1)(x-x_2)(x-x_3),$$
then
$$x_1+x_2+x_3=-b$$
$$x_1x_2x_3=-d$$
instead of
$$x_1+x_2+x_3=b$$
$$x_1x_2x_3=d$$
because
$$(x-x_1)(x-x_2)(x-x_3)$$
$$=x^3-(x_1+x_2+x_3)x^2+(x_1x_2+x_2x_3+x_3x_1)x-x_1x_2x_3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/959158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding the inverse of $f(x) = x\sqrt{2+x^2}$ For a function $f(x) = x\sqrt{2+x^2}$ find out if it's bijective and if so, find its inverse.
The function is surjective because $x^2 > 0:\forall x\in\mathbb{R}$. I'm having difficulties proving that the function is injective.
I tried the following: $$f(x) = f(y) \iff x\sqr... | Hint:
To find the inverse let $y= f^{-1}(x)$ then since $f(f^{-1}(x)) = x$ we have
$$f(f^{-1}(x)) = x = y\sqrt{2+y^2}$$
which gives
$$x^2 = y^2(2+y^2)\to y^4 + 2y^2 - x^2 = 0$$
This is a quadratic equation in $y^2$: $(y^2)^2 + 2y^2 - x^2 = 0$ for which you can use the standard formula on. To determine which root to pic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/959742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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general way to get formula as multiplication. We assume:
$$ 1^n + 2^n + 3^n + .. + k^n$$
where k and n are natural numbers.
Are there a general way to get it as multiplication?
For example:
$$ 1^3 + 2^3 + 3^3 + .. + k^3 = \binom {k+1} 2 ^2 $$
| Suppose that for $0\le j\lt m$, there is a polynomial $P_j$, with degree $j+1$, so that
$$
\sum_{k=0}^nk^j=P_j(n)\tag{1}
$$
The binomial theorem says that
$$
(k+1)^{m+1}-k^{m+1}=\sum_{j=0}^m\binom{m+1}{j}k^j\tag{2}
$$
Sum $(2)$ in $k$ using $(1)$:
$$
\begin{align}
(n+1)^{m+1}
&=\sum_{k=0}^n\sum_{j=0}^m\binom{m+1}{j}k^j... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/960824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$ \lim\limits_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$ Find the limit: $$ \lim_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$$
I did the following:
\begin{align}
(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x}) = \frac{(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})}{1} \cdot \frac{(\sqrt{x^2 + 2x} + \sqrt... | Expanding the rightmost term, we get:
$$
\frac{(x^2+2x) - (x^2-7x)}{\sqrt{x^2+2x} + \sqrt{x^2-7x}} = \frac{9x}{\sqrt{x^2+2x} + \sqrt{x^2-7x}}
$$
now just divide the numerator and denominator by $x = \sqrt{x^2}$:
$$
\frac{9}{\sqrt{1+\frac{2}{x}} + \sqrt{1-\frac{7}{x}}}.
$$
We have that $\sqrt{1+\frac{2}{x}} + \sqrt{1-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/962458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
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$\Sigma_{k=m+1}^{\infty}\frac{1}{10^k} \leq \frac{1}{10^m}$ Is it true that $\Sigma_{k=m+1}^{\infty}\frac{1}{10^k} \leq \frac{1}{10^m}$ for $m \in \mathbb{N}$?
| $\Sigma_{k=m+1}^{\infty}\frac{1}{10^k}=\frac{1}{10^{m+1}}+\frac{1}{10^{m+2}}+\frac{1}{10^{m+3}}+.... =\frac{1}{10^m}(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+...)\leq \frac{1}{10^m}\cdot1=\frac{1}{10^m}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/966046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Proving $R +r\le h_{max} $ If $R$ is the circumradius , $r$ the inradius and $ h_{max}$ is the largest altitude of acute angled triangle $ABC$, then prove that $$R +r\le h_{max}. $$
I tried this using Euler's inequality but I did not succeed.
| By assuming $a\leq b\leq c$ we have that the greatest altitude is $h_a$ and:
$$ R = \frac{abc}{4\Delta},\quad r=\frac{2\Delta}{a+b+c},\quad h_a=\frac{2\Delta}{a} $$
so, by Heron's formula, we have to prove that:
$$ 2abc + (-a+b+c)(a-b+c)(a+b-c) \leq \frac{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}{a} $$
or:
$$2a^2 bc\leq(b+c)(-a+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/967866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Closed-form of $\sum_{k=1}^{\infty}\arctan(1/k^3)$ Wolfram said that
$$\sum_{k=1}^{\infty}\arctan\left(\frac{1}{k^2}\right)=\arctan\left(\frac{1-\cot\left(\frac{\pi}{\sqrt 2}\right)\tanh\left(\frac{\pi}{\sqrt 2}\right)}{1+\cot\left(\frac{\pi}{\sqrt 2}\right)\tanh\left(\frac{\pi}{\sqrt 2}\right)}\right).$$
I was wonderi... | Notice for any positive number $x$, we have
$$\tan^{-1}(x) = \frac{1}{2i}\log\left(\frac{1+ix}{1-ix}\right) = \Im\log(1 + ix)$$
We can rewrite the sum at hand as
$$\sum_{k=1}^\infty \tan^{-1}\frac{1}{k^3} = \Im\left[ \sum_{k=1}^\infty \log\left(1 + \frac{i}{k^3}\right) \right]
$$
For each $k$, we have the factorization... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Help with using Master Theorem on $T(n)=9T(n/3) + \Theta(n^2/\operatorname{lg}(n))$ I want to use the Master theorem to solve the following recurrence.
$$T(n)=9T(n/3) + \Theta(n^2/\operatorname{lg}(n))$$
We can easily see that $a=9$ and $b=3$ and $f(n) = n^2/\operatorname{lg}(n)$.
I am very new to using this theorem, a... | By way of enrichment we solve another closely related recurrence that
admits an exact solution. Suppose we have $T(0)=0$ and $T(1)=T(2)=1$
and for $n\ge 3$
$$T(n) = 9 T(\lfloor n/3 \rfloor) +
\frac{n^2}{\lfloor \log_3 n \rfloor}.$$
Furthermore let the base three representation of $n$ be
$$n = \sum_{k=0}^{\lfloor \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/972275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Difference of inverse squares Given that the positive number $a$ is the difference of inverse squares:
$$a = \frac{1}{n^2} - \frac{1}{m^2}, m, n \in \mathbb{N},$$
could it well be that the $pa$ is also a difference of inverse squares , when p - some natural number ?
| If
$$a = \frac{1}{n^2} - \frac{1}{m^2},$$
where $m, n \in \mathbb{N}$ and $\mathbb{N}$ is the set of natural numbers, then
$$2a = \frac{2}{n^2} - \frac{2}{m^2}.$$
The OP is asking for solutions to the Diophantine equation
$$2a = \frac{2}{n^2} - \frac{2}{m^2} = \frac{1}{r^2} - \frac{1}{s^2},$$
where $r, s \in \mathbb{N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/973120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Evaluating Telescopic Sum $ \sum\frac{n}{1+n^2+n^4} $ How to evaluate following
$$ \sum_{n=1}^{\infty}\frac{n}{1+n^2+n^4}$$
I posted my way as an answer, Is there another Interesting approach to evaluate this sum of series?
| I did it this way
$x$th Term of this sum can be written as
$$T_x=\frac{x}{1+x^2+x^4}=\left(\frac {1}{2\cdot(x^2-x+1)}\right)-\left(\frac{1}{2\cdot(x^2+x+1)}\right)$$
Similarly $(x+1)$th Term can be written as
$$T_{x+1}=\frac{x+1}{1+(x+1)^2+(x+1)^4}=\left(
\frac{1}{2\cdot(x^2+x+1)}\right)-\left(\frac{1}{2\cdot(x^2+3x+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/973217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Quadratic equation $4x^2+4x=7$ using quadratic formula Solve using quadratic formula. $4x^2+4x=7$
So $4x^2+4x-7=0$
$A=4$ $b=4$ $c=-7$
$$x=\frac{-4\pm\sqrt{(4)^2-4(4)(-7)}}{2(4)}=\frac{-4\pm\sqrt{16+112}}{8}=\frac{-4\pm\sqrt{128}}{8}$$
What's next?
| I can't see what you have written but it should look like:
$$\begin{align}
4x^2+4x-7&=0
\\ \Rightarrow x_{\pm}&=\frac{-4\pm\sqrt{(4)^2-4(4)(-7)}}{2(4)}
\\ &=\frac{-4\pm{\sqrt{16+112}}}{8}
\\&=\frac{-4\pm\sqrt{128}}{8}
\\ &=\frac{-4+\sqrt{128}}{8}\,\,\,\text{ or }\frac{-4-\sqrt{128}}{8}
\\ &=\frac{-4+\sqrt{8^2\cdot 2}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/975016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof for $|1 - z| \geq 1 - |z|$ for $|z| < 1$, $z \in \mathbb{C}$ I can prove it "by picture" by drawing a picture of a circle of radius $|z|$ centered at $(0, 1)$. Then $1 - |z|$ is the length from the origin to the intersection of the circle with the x-axis (to the left). $|1 - z|$ is length from the origin to any p... | let $z = a + bi$ with $a^2 + b^2 < 1$, then $|1-z| = |1 - (a+bi)| = |1-a + (-b)i| = \sqrt{(1-a)^2 + b^2}$. So we need to prove: $\sqrt{(1-a)^2 + b^2} \geq 1 - \sqrt{a^2 + b^2}$.
Squaring both sides of the above inequality we have: $(1-a)^2 + b^2 \geq 1 - 2\sqrt{a^2 + b^2} + a^2 + b^2 \iff -2a \geq -2\sqrt{a^2 + b^2} \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/976168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Triplets satisfying $(a^3+b)(b^3+a)=2^c$
Find the number of triplets $(a,b,c)$ satisfying $(a^3+b)(b^3+a)=2^c$, where $a,b,c\in \mathbb{N}$
A trivial solution is $(1,1,2)$. I think there aren't any other such triplets, so I've been trying to prove that but haven't been able to do so.
How can other solutions be foun... | shaurya gupta already noted the solution $(a,b,c) = (1,1,2)$;
rogerl found another pair, $(3,5,12)$ and $(5,3,12)$, and conjectured
that there are no others. I prove that this conjecture is correct.
[The question required that $a,b,c \in \mathbb N$; this notation
"$\mathbb N$" is sometimes used for nonnegative integer... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/976257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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An inequality, which is supposed to be simple Let $x,y,z\in\mathbb{R}$.Let $xy+yz+xz=1$.
Prove:$\displaystyle \frac{x}{\sqrt{x^2+1}}+\frac{y}{\sqrt{y^2+1}}+\frac{z}{\sqrt{z^2+1}}\leq \frac{3}{2}$
| Let $a,b,c\in(0,\pi)$ be such that $x=\cot a$, $y=\cot b$, $z=\cot c$.
Using the addition formula for cotangent, one can show that
$$ \cot(a+b+c)
= \frac{\cot a\cot b\cot c - \cot a - \cot b - \cot c}
{\cot a\cot b + \cot b\cot c + \cot c\cot a - 1}
\tag{$\ast$} $$
By hypothesis, the denominator on the RHS is 0,... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof of: if $x^2+y^2=2xy$ then $x=y$ I am trying to prove $x^2+y^2=2xy$ then $x=y$
What I have done is suppose $x^2+y^2=2xy$
then $x^2+y^2+(-2xy)=0\iff
x^2+(-xy)+(-xy)+y^2=0
\iff (x+(-y))\cdot x+(x+(-y))\cdot-y=0
\iff (x+(-y))^2=0$
i then square root both sides but i'm not sure if that's mathematically correct?
which ... | This isn't true: Consider $$x = \cos \frac{\pi}{8}, \qquad y = \sin \frac{\pi}{8}. \qquad(\ast)$$ Then, substituting gives
$$x^2 - y^2 = \left(\cos \frac{\pi}{8}\right)^2 - \left(\sin \frac{\pi}{8}\right)^2 = \cos\left[2 \left( \cos \frac{\pi}{8}\right)\right] = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}.$$ On the other h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/977653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Boolean Simplification of $(\overline{a+b+c})+a\cdot(b+ \overline{c})$ I'm lost, when checking my answer via truth tables, my simplified form does not match the original equation. My work, with reasoning step by step is below. Can you help me figure out where I'm wrong, and what it should be?
$(\overline{a+b+c})+a\cdot... | Your application of what you call "commutativity" is incorrect. Variables commute only over the same operations.
Instead: $$\overline{a} \cdot \overline{b} \cdot \overline{c} +a\cdot b+ a \cdot \overline{c}= \overline{a} \cdot \overline{b} \cdot \overline{c} + a\cdot (b+\overline c) $$
$$= \overline{a} \cdot \overline{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/979636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Define continuity for $f(x)=\arctan(2x^3)/x^2$ at $x=0$. $$f(x)=\dfrac{\arctan(2x^3)}{x^2}.$$
*
*How are we allowed to define $f(x)$ at $x=0$ for it to be continuous there?
*Find the derivative for all $x$ real numbers.
I can't see this work out since $x=0$ is not defined in the denominator.
Thanks beforehand if a... | Compute the limit $$\lim_{x\to0} f(x)=\lim_{x\to0}\frac{\arctan ( 2x^3 )}{x^2}=\lim_{x\to0}\frac{2x^3}{x^2}=0.$$ Hence we define $f(0)=0$, and $f(x)$ is continuous.
Now compute the derivative. For $x\ne 0$, we have \begin{align}
& f'(x)=\frac{{( \arctan ( 2x^3 ))^{\prime }}\cdot x^2-\arctan( 2x^3)\cdot{ (x^2)^{\prime... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/980893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Assume that 495 divides the integer $\overline{273x49y5}$ where $x,y \in \{0,1,2...9\}$. Find $x$ and $y$. So, I know that $495 = 5\times 9\times 11$. So then, if that's the case, then the number $\overline{273x49y5}$ must be divisible by $495$ if and only if it is divisible by $5$ and $9$ and $11$.
Then, I know that $... | Notice $495 = 5\times 99$. It will be easier to look at everything modulo $99$.
$$0 \equiv \overline{273x49y5}
\equiv 27 + \overline{3x} + 49 + \overline{y5}
\equiv \overline{yx} + 111
\equiv \overline{yx} + 12 \pmod{99}
$$
This leads to $\overline{yx} = 87 \implies (x,y) = (7,8)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/981544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Cool little system of equations. Solving the system of equations for integers:
$2^a+3^b=5^b$
$3^a+6^b=9^b$
How is it done? I tried substituting the $2^a$ from the first equation into the second, and dividing the two equations by $2^a$.
| The second equation gives you $b\le a\lt2b$, because $3^b+6^b\le9^b\lt3^{2b}+6^b$ when $b\gt0$ (and the equations obviously have no solutions when $b\le0$).
By diving it by $3^b$, you get $3^{a-b}+2^b=3^b$.
It means $2^b=3^b-3^{a-b}=3^{a-b}(3^{2b-a}-1)$.
By unicity of decomposition in prime factors, we must have $a-b=0... | {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\frac{x^2+2xy+y^2}{x^2-y^2} >x+y$
Find the set of integer solutions $(x,y)$ to
$$\frac{x^2+2xy+y^2}{x^2-y^2} >x+y$$
I can't seem to multiply both sides by the expression in the denominator.
Nor can I simplify and cancel any terms. How should I it?
| Hint
$$\frac{x^2+2xy+y^2}{x^2-y^2}=\frac{(x+y)^2}{(x+y)(x-y)}=\frac{x+y}{x-y}$$
Edit
So, the initial inequality is equivalent to
$$\frac{x+y}{x-y}>x+y.$$
If $x+y>0$ it is
$$\frac{1}{x-y}>1,$$ which has no solution. (If $x-y<0$ then $\frac{1}{x-y}<0$ can't be greater than $1.$ If $x-y>0$ then $\frac{1}{x-y}\le 1$ can't... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Remainder of $3^7/8$ I read here that the remainder of $\frac{ab}{c}$ is equal to the remainder of $\frac{a}{c}\frac{b}{c}$ implying that the remainder of $\frac{a^b}{c}$ is equal to the remainder of $[\frac{a}{c}]^b$. However, when I apply this here, I would get remainder of $[\frac{3}{8}]^7=3^7$ (while the correct an... | You may use congruences:
$$3^2=9\equiv1\pmod{8}$$
So
$$3^7=(3^2)^3\times 3\equiv 1\times 3\equiv3\pmod{8}$$
Remember that $a\equiv b\pmod{m}$ means that $a$ and $b$ have same remainder when divided by $m$. It's also equivalent to $m|a-b$ (you read it $m$ divides $a-b$).
And with congruences you can write that if $a\equ... | {
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Factoring the quartic polynomial $x^4-2x^2+1$ Okay, I am practicing factoring for an upcoming assignment and I know that this is basic algebra, but I forgot how to attack this polynomial. Every method that I have used so far from simply guessing to using the quadratic formula to long division has failed me in replicati... | $x^4-2x^2+1=0$ clearly has a root of $1$. Since $\dfrac{x^4-2x^2+1}{x-1}= x^3+x^2-x-1$ you now have $$(x-1)(x^3+x^2-x-1)=0.$$
$x^3+x^2-x-1=0$ clearly has a root of $1$. Since $\dfrac{x^3+x^2-x-1}{x-1}= x^2+2x+1$ you now have $$(x-1)(x-1)(x^2+2x+1)=0.$$
And so on.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $\sin{A}:\sin{B}:\sin{C}$ how to find $\sin{(2A)}:\sin{(2B)}:\sin{(2C)}$? Question:
Given $x,y,z>0$, and the positive number $k$ such
$$\dfrac{x^2}{x^2+k}+\dfrac{y^2}{y^2+k}+\dfrac{z^2}{z^2+k}=1$$
in $\Delta ABC$,
$$(1):\sin{A}:\sin{B}:\sin{C}=\dfrac{x}{x^2+k}:\dfrac{y}{y^2+k}:\dfrac{z}{z^2+k}$$
$$(2): \... | Let
$$p = \frac{x^2}{x^2 + k}, \quad q = \frac{y^2}{y^2 + k}, \quad r = \frac{z^2}{z^2 + k}.$$
Then by assumption $p + q + r = 1$.
Letting $a$, $b$ and $c$ be the sides of the triangle, we have by the sine law
$$\begin{align*}
a^2 : b^2 : c^2 &= \sin^2 A : \sin^2 B : \sin^2 C \\
&= \frac{x^2}{(x^2 + k)^2} : \frac{y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/991328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Ellipse $3x^2-x+6xy-3y+5y^2=0$: what are the semi-major and semi-minor axes, displacement of centre, and angle of incline? Given the ellipse
$$3x^2-x+6xy-3y+5y^2=0$$
find the following:
*
*semi-major axis, $a$
*semi-minor axis, $b$
*displacement of centre from origin (or coordinates of centre of ellipse $(h,k)$)
... | Here's another method suggested by a friend of mine.
Keep the solution of the centre of the ellipse identified earlier as $(-\frac13,\frac12)$ .
Change the centre of the ellipse such to the origin. Equation of translated ellipse is:
$$E':\qquad 3x^2+6xy+5y^2-\frac7{12}=0\qquad \cdots (1)$$
Consider a circle at the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/993625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
How to prove $\int_0^{2\pi} \ln(1+a^2+2a\cos x)\, dx=0$? How can I prove $\int_0^{2\pi} \ln(1+a^2+2a\cos x)\, dx=0$, where $a<1$?
Thanks.
| Let $a$ be a real number.
*
*$\color{blue}{\text{Case 1.}}$ $\quad|a|<1$
$$
\int_0^{2\pi}\log \left(1+2a\cos x+ a^2\right){\rm d}x=0
$$
Observe that
$$
\left(1+ae^{ix}\right)\left(1+ae^{-ix}\right)=1+2a\cos x+ a^2, \quad x \in [0,2\pi],
$$
and that
$$
\begin{align}
\log \left(1+2a\cos x+ a^2\right)
&=\log \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/994310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
The minimum value of $\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+x^3+\frac{1}{x^3}}$ Problem :
The minimum value of $$\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+x^3+\frac{1}{x^3}}$$
Can I use this in numerator and denominator :
The minimum value of $a +\frac{1}{a}$
Using A... | Here is one way to go about it. We want to express$\left(x^6+\frac1{x^6}\right)$ in terms of $\left(x+\frac1x\right)$.
$$\left(x+\frac1x\right)^6={x}^{6}+6{x}^{4}+15{x}^{2}+20+\frac{15}{x^2}+\frac{6}{x^4}+\frac1{x^6}$$
We can remove the $x^4$ and $\frac1{x^4}$ terms by subtracting $6\left( x+\frac1x\right)^4$, which gi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/994458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
How do I know if two vectors with $n$ components are parallel? How do I know if two vectors with $n$ components are parallel?
For example
$$\begin{pmatrix}5\\2\\1\\3\\4\end{pmatrix} \text{, and } \begin{pmatrix}4\\1\\2\\3\\6\end{pmatrix}.$$
| Check to see if they are constant multiples of one another. In your example, if there were a $c$ such that
$$c\begin{pmatrix}
5\\2\\1\\3\\4
\end{pmatrix}=
\begin{pmatrix}
4\\1\\2\\3\\6
\end{pmatrix}$$
then by equating the first components, we see that $c$ would have to be $\frac{4}{5}$. However, when we check the secon... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/998001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
If $n = 2^k - 1$ for $k \in \mathbb{N}$, then every entry in row $n$ of pascal's triangle is odd. Prove: If $n = 2^k - 1$ for $k \in \mathbb{N}$, then every entry in row $n$ of pascal's triangle is odd.
I know that the $n$th row in pascal's triangle correspond to the coefficients of $(x+y)^n$:
$$\begin{align}(x+y)^n = ... | Counting the Number of Factors of $\boldsymbol{2}$
In Corollary $(7)$ of this answer, it is shown that the number of factors of $p$ in $\binom{n}{k}$ is
$$
\frac{\sigma_p(k)+\sigma_p(n-k)-\sigma_p(n)}{p-1}\tag{1}
$$
where $\sigma_p(n)$ us the sum of the digits in the base-$p$ representation of $n$.
In base-$2$, each di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/999328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Tough inequality problem, splitting into cases Find the set of all $x$ for which $\dfrac{2x}{2x^2 + 5x + 2} > \dfrac{1}{x+1}$.
I am new to solving such problems and am not able to appropriately break up the problem into cases, so any help would be thoroughly appreciated.
| $$\dfrac{2x}{2x^2+5x+2}>\dfrac{1}{x+1}\iff\frac{2x}{2x^2+5x+2}-\frac{1}{x+1}>0 \\ \iff\frac{2x(x+1)-(2x^2+5x+2)}{(2x^2+5x+2)(x+1)}>0 \\ \iff-\frac{3x+2}{(2x+1)(x+2)(x+1)}>0 \\ \iff \frac{3x+2}{(2x+1)(x+2)(x+1)}<0$$
We can see that the numerator vanishes when $x=-\frac{2}{3}$, also the fraction is discontinuous for $x\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Minimum of $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}$ What is the minimum of $$f(a,b,c):=\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}$$ where $a,b,c$ are positive real numbers?
When $a=b=c$, we have $f(a,b,c)=\dfrac{3}{\sqrt{2}}\approx 2.12$
When $a=1,b=c\rightarrow\infty$, we... | If $a=b=1$ and $c\rightarrow0^+$ we get $f(a,b,c)\rightarrow2$.
We'll prove that
$$\sum_{cyc}\sqrt{\frac{a}{b+c}}\geq2.$$
Indeed, by AM-GM
$$\sum_{cyc}\sqrt{\frac{a}{b+c}}=\sum_{cyc}\frac{2a}{2\sqrt{a(b+c)}}\geq\sum_{cyc}\frac{2a}{a+b+c}=2.$$
Done!
Also we can use Holder:
$$\left( \sum_{cyc}\sqrt{\frac{a}{b+c}}\right)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Poisson Integral is equal to 1
Show
$$
\int_{-\pi}^{\pi}P(r, \theta)d\theta = 1
$$
Let $\alpha(r) = \frac{r^2 - 1}{2r}$ and $\gamma(r) = -\big(\frac{r^2 + 1}{2r}\big)$.
Then
$$
\frac{1}{2\pi}
\int_{-\pi}^{\pi}\frac{1 - r^2}{1 - 2r\cos(\theta) + r^2}d\theta
= \frac{\alpha}{2\pi}
\int_{-\pi}^... | For $0 \le r < 1$,
$$
\begin{align}
\sum_{n=-\infty}^{\infty}r^{|n|}e^{in\theta}
& =\sum_{n=0}^{\infty}r^{n}e^{in\theta}+\sum_{n=1}^{\infty}r^{n}e^{-in\theta} \\
& =\frac{1}{1-re^{i\theta}}+\left(\frac{1}{1-re^{-i\theta}}-1\right) \\
& = \frac{1}{1-re^{i\theta}}+\frac{re^{-i\theta}}{1-re^{-i\theta}} \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
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.