Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
$x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$ Determine all the possibilities for rational roots of the polynomial $x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$. Then determine how many of the real roots of the polynomial may be positive and how many may be negative. Factor the polynomial to confirm your results.
The answer is possible rational... | You can factor it as
$$x^{ 4 }-4x^{ 3 }+6x^{ 2 }-4x+1={ x }^{ 4 }-2{ x }^{ 3 }+{ x }^{ 2 }-2{ x }^{ 3 }+4{ x }^{ 2 }-2x+{ x }^{ 2 }-2x+1={ x }^{ 2 }\left( { x }^{ 2 }-2x+1 \right) -2x\left( { x }^{ 2 }-2x+1 \right) +{ x }^{ 2 }-2x+1=\\ =\left( { x }^{ 2 }-2x+1 \right) \left( { x }^{ 2 }-2x+1 \right) ={ \left( x-1 \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
Determinant of a $4 \times 4$ matrix $A$ and $(\det(A))^5$ Calculate $\det(A)$ and $\det(A)^5$:
$$A= \begin{bmatrix}a&a&a&a\\a&b&b&b\\a&b&c&c\\a&b&c&d\end{bmatrix}$$
I found $\det A$ with Laplace expansion: $$a(-abc+b^2c+a c^2-b c^2+a b d-b^2 d-a c d+b c d) .$$
But how can I determine $\det A^5$ easily/fast? I know tha... | You can much easier get your expression by subtracting the third line from the fourth, then the second from the third and finally, the first from the second:
$$
\det A=
\det \begin{pmatrix}
a&a&a&a\\
a&b&b&b\\
a&b&c&c\\
0&0&0&d-c
\end{pmatrix}=
\det \begin{pmatrix}
a&a&a&a\\
a&b&b&b\\
0&0&c-b&c-b\\
0&0&0&d-c
\end{pmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Prove the Inequality proves true, three variables Prove that for all positive real numbers $a,b,c$ we have:
$$\frac{1}{4a}+\frac{1}{4b}+\frac{1}{4c}+\frac{1}{2a+b+c}+\frac{1}{2b+c+a}+\frac{1}{2c+a+b}\geq\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}$$
So far I have solved for $a=b=c=1$ and $a=1$, $b=2$, $c=3$ to show that t... | For all real numbers it's wrong, of course.
For positives $a$, $b$ and $c$, we can use an Integral method:
We see that for all positives $x$, $y$ and $z$ the following inequality holds.
$$\sum\limits_{cyc}(x^4-2x^2y^2+x^2yz)\geq0$$
Indeed, by Schur
$$\sum\limits_{cyc}(x^4-2x^2y^2+x^2yz)=\sum\limits_{cyc}(x^4-x^3y-x^3z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing $\int \frac{1}{(1+x^3)^3}dx$ I tried various methods like trying to break it into partial fractions after factorizing, applying substitutions but couldn't think of any. How will we integrate this?
| Hint. By writing
$$
1+x^3=(1+x)(1-x+x^2), \qquad (1+x^3)^3=(1+x)^3(1-x+x^2)^3,
$$ one may observe that there is a partial fraction decomposition of the form
$$
\begin{align}
\frac{1}{(1+x^3)^3}&=\frac{a_1}{(1+x)}+\frac{a_2}{(1+x)^2}+\frac{a_3}{(1+x)^3}
\\\\&+\frac{\alpha_1x+\beta_1}{1-x+x^2}+\frac{\alpha_2x+\beta_2}{(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
A nice problem on divisor and sum of divisor function. Find all $n$ such that $\sigma(n)+d(n)=n+100$.
My Try : Let $n=\prod\limits_{k=1}^{m}p_k^{\alpha_k};\quad\sigma(n)=\prod\limits_{k=1}^{m}\tfrac{p_k^{\alpha_k+1}-1}{p_k-1}=n\cdot\prod\limits_{k=1}^{m}\tfrac{p_k-p_k^{-\alpha_k}}{p_k-1}>n;\quad d(n)=\prod\limits_{k=1}... | If $\tau(n)+\sigma(n)=n+100$ then the sum of the proper divisors of $n$ must be less than $100$. Notice that if $n$ has $s$ prime divisors then we can consider all the products of $s-1$ primes, and these are proper divisors.
If $n\geq 4$ then the sum of these divisors is at least $(2\times3\times 5)+ (2\times 5\times 7... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$3\times3$ Magic Square and hidden formula [Solved] I was doing the $3 \times3$ Magic Square of Squares problem -- found here: http://www.multimagie.com/English/SquaresOfSquaresSearch.htm -- and I figured out that if such Magic square exists with $9$ distinct numbers, then it must satisfy the following equation:
$$\sqr... | From what I understand in your post and the comments, you would like to find -- as JMoravitz said -- $a,b,c,d,e$ such that
$$a^2 + b^2 = c^2 + d^2=e^2$$
We necessarily have
$$a^2+b^2-d^2=c^2$$
And if I may explain how this relates to magic squares problems, as most people won't know. The existence of a 3x3 magic square... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If y=y(x) and $\frac{2+sin(x)}{y+1}\frac{dy}{dx}=-cosx,y(0)=1$, then find $y(\frac{\pi}{2})$. Question:
If y=y(x) and $\frac{2+sin(x)}{y+1}\frac{dy}{dx}=-cosx,y(0)=1$, then find $y(\frac{\pi}{2})$.
My attempt:-
$$\frac{2+sin(x)}{y+1}\frac{dy}{dx}=-cosx$$
$$\frac{dy}{1+y}=-\frac{cos(x)}{2+sin(x)}dx$$
Integrating both s... | $$
\frac{2+\sin(x)}{y+1}\frac{\mathrm{d}y}{\mathrm{d}x}=-\cos(x)
$$
Separating variables gives
$$
\int\frac1{y+1}\mathrm{d}y=-\int\frac{\cos(x)}{2+\sin(x)}\,\mathrm{d}x
$$
Integrating yields
$$
\log(y+1)=C_1-\log(2+\sin(x))
$$
or
$$
y=\frac{C}{2+\sin(x)}-1
$$
Since $y(0)=1$, we get $C=4$. Thus, $y\,\left(\frac\pi2\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Area enclosed by the curve $5x^2+6xy+2y^2+7x+6y+6=0$ We have to find the area enclosed by the curve
$$5x^2+6xy+2y^2+7x+6y+6=0.$$
I tried and I got that it is an ellipse, and I know its area is $\pi ab$ where $a$ and $b$ are the semiaxis lengths of the ellipse.
But I am unable to find the value of $a$ and $b$.
| Hint
Let
$$ax^2+bxy+cy^2+dx+ey+f=0$$
Note
$$\tan 2\theta=\frac{b}{a-c}$$
or
$$\cot \theta=\frac{a-c}{b}+\sqrt{1+\left(\frac{a-c}{b}\right)^2}$$
we have
$$\cot \theta=\frac{1+\sqrt{5}}{2}$$
find $\sin\theta$ and $\cos\theta$ and apply
\begin{cases}
x=X\cos\theta-Y\sin\theta\\
y=X\sin\theta+Y\cos\theta
\end{cases}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to evaluate $\int \frac{1+x}{1+\sqrt x} dx$ I picked this exercice from B. Demidovitch and i started solving, but when it's complete, i get only the half of the solution (the first two fractions and the ln, doesn't show up).
$$\int \frac{1+x}{1+\sqrt x} dx$$
The Solution is:
$$2 \left[\frac{\sqrt{x^3}}{3}-\frac{x}{... | Let $u=\sqrt x+1$, so that $\mathrm{d}x=2(u-1)\,\mathrm{d}u$.
$$\int\frac{1+x}{1+\sqrt x}\,\mathrm{d}x=\int\frac{1+(u-1)^2}{u}(2(u-1))\,\mathrm{d}u$$
Expanding makes the integral trivial. (added some more details below)
$$\begin{align*}2\int\frac{u-1+(u-1)^3}{u}\,\mathrm{d}u&=2\int\frac{u^3-3u^2+4u-2}{u}\,\mathrm{d}u\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
Let a, b, c be positive real numbers. Prove that Let a,b,c be positive real numbers. Prove that
$$\frac{a^3+b^3+c^3}{3}\geq\frac{a^2+bc}{b+c}\cdot\frac{b^2+ca}{c+a}\cdot\frac{c^2+ab}{a+b}\geq abc$$
I will post what I had solved originally, however it is unfortunately incorrect. Please help solve and/or aid in findi... | The right inequality.
By AM-GM we obtain:
$$\prod\limits_{cyc}(a^2+bc)=\sqrt[3]{\frac{\prod\limits_{cyc}((a^2+bc)(b^2+ac))^2}{\prod\limits_{cyc}(a^2+bc)}}=\sqrt[3]{\frac{\prod\limits_{cyc}(ab(c^2+ab)+c(a^3+b^3))^2}{\prod\limits_{cyc}(a^2+bc)}}\geq$$
$$\geq\sqrt[3]{\frac{\prod\limits_{cyc}(4(a^3+b^3)(c^2+ab)abc)}{\prod\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893626",
"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 $,= 5 and $ab = 6$ . Find $\frac{1}{a}$ + $\frac{1}{b}$ $a +b $ = 5 and $ab = 6$ . Find $\cfrac{1}{a}$ + $\cfrac{1}{b}$
Any Ideas on how to begin?
Many Thanks
| Since $a+b = 5$ and $ab = 6$, then $a,b$ can be the roots of polynomial $x^2-5x+6=0$
By fraction, $(x-2)(x-3) = 0$.
Out of symmetry, let $a = 2,b=3$, so $\frac{1}{a}+\frac{1}{b}= \frac{5}{6}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1895021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable? Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable? I think it diverges although:
$$ \sqrt{n+1} - \sqrt{n} \approx \frac{1}{2\sqrt{n}}$$
for example by the Mean Value Theorem $f(x+1)-f(x) \approx f'(x)$ and then I might... | Hopefully I'm not copying Gottfried Helms' Euler sum, but the Euler summation formula allows you to sum this divergent series.
$$\sum_{k=1}^\infty(-1)^{k+1}\sqrt k=\sum_{n=0}^\infty\frac1{2^n}\sum_{k=0}^n\binom nk(-1)^k\sqrt{k+1}$$
This converges extremely quickly to the Riemann zeta function,
$$\zeta(-1/2)=\frac1{1-\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Infimum of $\frac{(1+2a)(1+2b)(1+3a-b-ab)}{(1-b)(1+2b)(1+2a-a^2)+3a(1+2a)(1+2b-b^2)}$ Let $0<a,b<1$. What is the infimum of $$f(a,b)=\frac{(1+2a)(1+2b)(1+3a-b-ab)}{(1-b)(1+2b)(1+2a-a^2)+3a(1+2a)(1+2b-b^2)}?$$
It is possible to make the term $1+3a-b-ab$ in the numerator arbitrarily close to $0$ by letting $a\rightarrow ... | To find the extrema inside the region $]0,1[ \times ]0,1[$, we should solve
$$\frac{\partial f(a,b)}{\partial a}=0$$
$$\frac{\partial f(a,b)}{\partial b}=0$$
After many algebraic operators, we obtain
$$(1+2b)(b-1)(2a^2b^2-b^2+b+2ab-3a^2b-2a-5a^2)=0$$
$$a(1+2a)(5a-2b+1-10ab+b^2-ab^2-4a^2b+2a^2b^2+8a^2)=0$$
We are only i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A Trig Identity There are many trig identities that can be proven with known identities, but often the identities are tricky in finding the right methods to use. This is one of those identities that seems easy, but elusive. Consider the cubic equation
$$ x^3 - 14 x^2 + 56 x - 56 = 0$$
for which one solution is $8 \sin... | Writing $7x=\pi$
We need to prove $$7-6\cos x+6\cos2x-2\cos3x=8\sin^22x=4(1-\cos4x)\ \ \ \ (1)$$
Now as $\cos2x=\cos(\pi-5x)=-\cos5x,\cos4x=\cdots=-\cos3x$
$(1)$ reduces to $$\cos x+\cos3x+\cos5x=\dfrac12$$
Can you use Prove that $\cos\frac {2\pi}{7}+ \cos\frac {4\pi}{7}+ \cos\frac {8\pi}{7}=-\frac{1}{2}$ to prove thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find $\frac{x^2}{y^2}$ + $\frac{y^2}{x^2}$ If $\frac{x}{y}$ + $\frac{y}{x}$ = 3 Find $\frac{x^2}{y^2}$ + $\frac{y^2}{x^2}$
Any Ideas on how to begin ?
| \begin{align*}
\frac{x}{y} + \frac{y}{x} = 3 &\Longrightarrow \left(\frac{x}{y} + \frac{y}{x}\right)^2 = 9 \\
&\Longrightarrow \frac{x^2}{y^2} + \frac{y^2}{x^2} = 9-2=7
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there a neater proof for $\lim_{x \rightarrow 1} 1/x = 1$? What I need to prove:
For all $\epsilon>0$, there exist a $\delta>0$, such that for all $x$ satisfying $0<|x-1|<\delta$, we have $|1/x-1|<\epsilon$.
My proof:
Let $\delta=\min \{ 1/2 ,\epsilon/2 \}$. Then when $\epsilon\geq 1$,
$$
|x-1| < \frac{1}{2} \\
(*... | If $|x - 1| \leqslant 1/2$, then $1 - x \leqslant |1 - x| = |x - 1| \leqslant 1/2$, therefore $x \geqslant 1/2$.
Since $x > 0$, the expression $1/x$ is well-defined.
Also because $x > 0$, we have $|x| = x \geqslant 1/2$.
If you must use the Triangle Inequality, you can argue (using it in the form $|u - v| \geqslant ||u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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If $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1$, what can we say about $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}$?
Suppose that $a,b,c$ are three real numbers such that $\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1$. What are the possible values for $\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}$?
Aft... | Hint. You have
$$ \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} - 1 = \frac{a^3+b^3+c^3+abc}{(a+b)(b+c)(c+a)} $$
as well as
$$ \frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} = \frac{(a^3+b^3+c^3+abc)(a+b+c)}{(a+b)(b+c)(c+a)}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Prove two vectors are parallel Let $ABCD$ be a parallelogram and $P$, $Q$ two points so that $\overrightarrow{PC} = \frac{1}{3}\overrightarrow{AC}$ and $\overrightarrow{BQ} = 2\overrightarrow{QD}$.
Find $\alpha$, $\beta$ $\in \mathbb{R}$ with the following property: $\overrightarrow{AP} + \overrightarrow{BQ} = \alpha\... | $\overrightarrow {AC} = \overrightarrow {AB} + \overrightarrow {AD}\\
\overrightarrow {AP} = \frac 23 \overrightarrow {AC}\\
\overrightarrow {BD} = -\overrightarrow {AB} + \overrightarrow {AD}\\
\overrightarrow {BQ} = \frac 23 \overrightarrow {BD}$
$\overrightarrow {AP}+\overrightarrow {BQ} =$$ \frac 23 \overrightarro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1903464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of digits in row of Pascal's triangle is $O\left(n^2\right)$
Let $a_n$ be the number of decimal digits in the $n$-th row of Pascal's Triangle
(so $a_0=1, a_1=2, a_2=3, a_3=4, a_4=5, a_5=8,\dots$).
Prove that $\frac{a_n}{n^2}$ converges and find the limit.
It's very easy to see that it converges. Indeed, it's... | The quantity you're investigating is $\displaystyle \sum_{k=0}^n \left( \lfloor \frac{\ln \binom{n}{k} }{\ln 10}\rfloor +1\right) = n + O(n) + \frac{1}{\ln 10}\sum_{k=0}^n \ln \binom{n}{k} $
A simple computation yields $\displaystyle \frac{1}{n^2}\sum_{k=0}^n \ln \binom{n}{k} = \frac 2n \sum_{k=1}^n \frac kn \ln\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Diophantine equation: $x^2 + 4y^2- 2xy -2x -4y -8 = 0$ How many integer pairs $ (x, y) $ satisfy $$x^2 + 4y^2- 2xy -2x -4y -8 = 0 ?$$
As it seems at first glance, and hence, my obvious attempt was to make perfect squares but the $2xy$ term is causing much problem.
| Too long for a comment.
$$x^2 + 4y^2- 2xy -2x -4y -8 = (x-y)^2 + 3(y-1)^2 -2x + 2y - 11\\ = (x-y-1)^2 + 3(y-1)^2 - 12 = 0,$$
$$(x-y-1)^2+3(y-1)^2=12,$$
$$(x-y-1,y-1)\in\{(0,2), (0,-2), (3,1), (3,-1), (-3,1), (-3,-1)\},$$
$$(x,y)\in\{(4,3), (0,-1), (6,2), (4,0), (0,2), (-2,0)\}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1908682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Show that $\frac{1}{2}\le \sum\limits_{k=0}^n\frac{1}{n+k}\le1$ for $n\in \mathbb N^+$
$$\frac{1}{2n}\le \frac{\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{n+n}}{n}\le\frac{1}{n}$$
I tried math induction and I tried take integral but I want to solve this with most elementary methods please give me hint or just show that. Th... |
We show the inequality chain
\begin{align*}
\frac{1}{2}\leq\sum_{k=0}^n\frac{1}{k+n}\leq 1\qquad\qquad\qquad n\geq 1\tag{1}
\end{align*}
is not valid for $n=1,2$ and valid for $n\geq 3$.
We denote the sum with $A(n):=\sum_{k=0}^n\frac{1}{k+n}$.
Case $n=1,2,3$ :
\begin{align*}
A(1)&=\sum_{k=0}^1\frac{1}{k+1}=1+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1912458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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Closed form of the following Recurrence Relation Let $L\colon\mathbb{N}^3 \to \mathbb{N}$ satisfy the following recurrence relationship,
$$
L(a,b,c) = 1 + \sum_{i=0}^{a-1} \sum_{j=0}^{b-1} \sum_{k=0}^{c-1} L(i,j,k),
$$
With "initial conditions" $L(0,a,b) = L(c,0,d) = L(0,e,f) = 0$. I am interested in knowing a closed ... | Continuing according to Fred's very clever deduction of the multiple z-tranform
( the credit should go to him, so I invite him to post it, and votes to be casted on it as well)
$$
F(x,y,z) = \frac{{\frac{{x\,y\,z}}
{{\left( {1 - x} \right)\left( {1 - y} \right)\left( {1 - z} \right)}}}}
{{1 - \frac{{x\,y\,z}}
{{\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1913600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find m, n, p of given expression with some conditions Find $m, n, p$ such that in the expansion of the expression $ \left( x^m + \frac{1}{x^p} \right)^n$ the 12th and the 24th terms contain $x$, respectively $x^5$, and furthermore, this expansion also contains one free term (without $x$).
By using what we are given, I ... |
Expanding the binomial we obtain
\begin{align*}
\left( x^m + \frac{1}{x^p} \right)^n&=\sum_{k=0}^n\binom{n}{k}x^{mk}x^{-p(n-k)}\\
&=\sum_{k=0}^n\binom{n}{k}x^{(m+p)k-pn}
\end{align*}
According to the requirements we derive three equations:
\begin{align*}
11(m+p)-pn&=1\tag{1}\\
23(m+p)-pn&=5\tag{2}\\
k(m+p)-pn&=0\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1916312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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For what value $k$ is $f(x) = \begin{cases} \frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2} & x \neq 2 \\ k & x = 2 \end{cases}$ continuous at $x=2$? For what value $k$ is the following function continuous at $x=2$?
$$f(x) = \begin{cases}
\frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2} & x \neq 2 \\
k & x = 2
\end{cases}$$
All those square roo... | Note that $$\frac { \sqrt { 2x+5 } -\sqrt { x+7 } }{ x-2 } =\frac { \left( \sqrt { 2x+5 } -\sqrt { x+7 } \right) \left( \sqrt { 2x+5 } +\sqrt { x+7 } \right) }{ \left( x-2 \right) \left( \sqrt { 2x+5 } +\sqrt { x+7 } \right) } =\\ =\frac { 1 }{ \sqrt { 2x+5 } +\sqrt { x+7 } } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1920906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
${\sqrt{a+b+c} + \sqrt{a}\over b+c} + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c}+ \sqrt{c}\over b+a} \ge {9-3\sqrt{3}\over2\sqrt{a+b+c}}$
$${\sqrt{a+b+c} + \sqrt{a}\over b+c} + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c} + \sqrt{c}\over b+a} \ge {9+3\sqrt{3}\over2\sqrt{a+b+c}}$$
I tried AM-GM, which onl... | Hint:
If you multiply $a$, $b$ and $C$ by a same positive constant, $k^2$, then both LHS and RHS get multiplied by $k^{-1}$ (i.e. the inequality is homogeneous). Therefore you can assume WLOG that $a+b+c= 1$.
Simplify the inequality and you get:
$$\dfrac{1}{1-\sqrt{a}} + \dfrac{1}{1-\sqrt{b}} + \dfrac{1}{1-\sqrt{c}} \g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1921833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Orthogonal change of variables - elimination of cross terms
Eliminate cross terms of $Q$ and express $Q$ in terms of the new variables.
$$\textit{Q}=2x^2+5y^2+5z^2+4xy-4xz-8yz$$
$$\begin{pmatrix}
2 &2 &-2 \\
2 &5 &-4 \\
-2 &-4 &5
\end{pmatrix}$$
$$-\lambda^3+12\lambda^2-21\lambda +10$$
$$\lambda _1=1, v_1=\be... |
Updated:
Note that $\lambda_{1}=\lambda_{2}$, $\boldsymbol{v}_{1} \perp \boldsymbol{v}_{3}$ and $\boldsymbol{v}_{2} \perp \boldsymbol{v}_{3}$, $\boldsymbol{v}_{1}$ and $\boldsymbol{v}_{2}$ can be resolved into linear combination of $\boldsymbol{e}_{1}$ and $\boldsymbol{e}_{2}$.
Let $\boldsymbol{e}_{1}=\displaystyle \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Proving this identity: $\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\theta \csc^2\theta - 2$ I have tried solving this trig. identity, but I get stuck when it comes to the $-2$ part. Any suggestions?
$$\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\thet... | $$\frac{s^2}{c^2}+\frac{c^2}{s^2}=\frac{s^4+c^4}{s^2c^2}=\frac{(s^2+c^2)^2-2s^2c^2}{s^2c^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
$x^2 dy/dx + 2xy=y^{-3}$, find $y(x)$ The question states to start by substituting $u=y^{-2}$, however I am not sure how to deal with this as-as far as I can see there is no easy substitution.
Are you meant to find that $y=(1/u)^{1/2}$ and then work with that(seems a little messy)?
Thanks,
| Rewrite it as follows
$$
(2xy-y^{-3})dx+x^2dy=0
$$
Multiply it by $x^6y^3$, then
$$
(2x^7y^4-x^6)dx+x^8y^3dy=0
$$
Which is full differential. Hence:
$$
F=\int{x^8y^3dy}=\frac{x^8y^4}{4}+f(x)\\
\frac{dF}{dx}=2x^7y^4+\frac{df}{dx}=2x^7y^4-x^6\to\\
f=-\frac{x^7}{7}\to F=\frac{x^8y^4}{4}-\frac{x^7}{7}=C^*
$$
Which represen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1925305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Simplify the subtraction of two summations. $$\sum_{i=1}^n (3i^2 +4) - \sum_{j=2}^{n+1} (3j^2 +1) $$
I thought I would use the distributive rule but the question stipulates that my answer should not include a summation sign...
| Method 1:
$$\begin{align}
\sum_{i=1}^n (3i^2 +4) - \sum_{j=2}^{n+1} (3j^2 +1) & =\sum_{i=1}^n (3i^2 +4) - \sum_{j=1}^n (3(j+1)^2 +1) \\
& = \sum_{i=1}^n (3i^2 +4) - \sum_{j=2}^{n+1} (3j^2+6j+4) \\
& = \sum_{i=1}^n \left[(3i^2 +4) - (3i^2+6i+4)\right] \\
& = \sum_{i=1}^n -6i \\
& = -6\cdot\frac{n(n+1)}2 \\
\sum_{i=1}^n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1925889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
General term of recurrence relation Find the general term of the following recurrence relation:
$$a_{1} = 2$$
$$a_{n+1} = \frac{2a_{n} - 1}{3}$$
I've tried to find the first few terms:
$$a_{1} = 2$$
$$a_{2} = \frac{2 \cdot 2 - 1}{3} = 1$$
$$a_{3} = \frac{2 \cdot 1 - 1}{3} = \frac{1}{3}$$
$$a_{4} = -\frac{1}{9}$$
$$a_{... | We have
$$ 3a_{n+1}=2a_n-1 \tag{1}$$
and we may try to get rid of the inhomogeneous term $-1$ by a suitable substitution.
For instance, by setting $a_n=b_n-1$, we get $b_1=3$ and
$$ 3b_{n+1}=2b_n \tag{2} $$
from which $b_n = 3\cdot\frac{2^{n-1}}{3^{n-1}}$ readily follows. The general term of the original sequence is so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1927862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to sum up these fractions? Found this question in a competitive math test for elementary students. The long way is to add all the decimal values but is there a pattern/trick to solve this question (or these types)? I don't know how to solve this except by the long method of adding all the decimal equivalents.
The A... | Not sure if this is what you are looking for but this is what I did:
$$S = \frac17 + \frac18 + \frac19 + \frac1{10} + \frac1{11} + \frac1{12} + \frac1{14} + \frac1{15} + \frac1{18} + \frac1{22} + \frac1{24} + \frac1{28} + \frac1{33}$$
I observed the following groupings:
*
*7, 14, 28
*8, 12, 24
*11, 22, 33
*I did ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1928224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$ax^3+8x^2+bx+6$ is exactly divisible by $x^2-2x-3$, find the values of $a$ and $b$ Find the values of $a$ and $b$ for which the polynomial $ax^3+8x^2+bx+6$ is divisible by $x^2-2x-3$.
| Let $f(x)=ax^3+8x^2+bx+6$, and let $g(x)=x^2-2x-3$. Since the question says that $g(x)|f(x)$, then write $f(x)=g(x)(px+q)$. Note the linear polynomial, we did that because dividing $f(x)$ by $g(x)$ will result in degree $1$. Now, expand the LHS and RHS, and you get something like this:
$$ax^3+8x^2+bx+6=(x^2-2x-3)(px+q)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 3
} |
Finding a Power Series representation for the function $f(x) = \frac{2}{3-x}$ Let's say I want to find a Power Series representation of the function $f(x) = \frac{2}{3-x}$
Now I know we can write this as a geometric series
$$\sum_{n=0}^{\infty}ar^n = \frac{a}{1-r}$$
But I see two possible ways two write it as a geometr... | Both methods are quite ok and valid and there are even many more possibilities to expand the function into a power series.
In order to better see what's going on, we should at first look somewhat closer at the function $f$. We consider the full specification of $f$ and choose as domain and codomain
\begin{align*}
&f:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Dice roll game probability Suppose we play a 1 vs 1 game. I roll a 6 sided fair die, if I get a 1 I win, if not you roll. If you get a 6 you win, otherwise I go again.... What are my chances of winning and what are yours?
| The probability of you winning immediately is $\frac{1}{6}$. The probability of you winning on the third roll (after your opponent rolls for the first time) is $\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}$. This represents the probability that you do not win on the first roll, your opponent does not win on the second... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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A different way to prove that $\int_{1+\sqrt{2}}^{\infty} \frac{\ln(1+x)}{1+x^{2}} \, dx = \frac{3 G}{4} + \frac{\pi}{16} \, \ln 2$ From the fact that $$\int_{0}^{1} \int_{0}^{1} \frac{1}{2-x^{2}-y^{2}} \, dx \, dy $$ is an integral representation of Catalan's constant ($G$), I was able to deduce that $$\int_{1+\sqrt{2... | Perform the change of variable $y=\arctan x$,
$\displaystyle I=\int_{1+\sqrt{2}}^{\infty} \frac{\ln(1+x)}{1+x^{2}} \, dx=\int_{\tfrac{3\pi}{8}}^{\tfrac{\pi}{2}}\ln(1+\tan x)dx=\int_{\tfrac{3\pi}{8}}^{\tfrac{\pi}{2}} \ln\left(\sin x+\cos x\right)dx-\int_{\tfrac{3\pi}{8}}^{\tfrac{\pi}{2}}\ln\left(\cos x\right)dx$
$\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1931724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 0
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Do the Taylor series of $\sin x$ and $\cos x$ depend on the identity $\sin^2 x + \cos^2 x =1$? I had this crazy idea trying to prove the Pythagorean trigonometric identity;$$\sin^2x+\cos^2x=1$$by squaring the infinite Taylor series of $\sin x$ and $\cos x$. But it came out quite beautiful, involving also a combinatoric... | No, they do not depend on the Pythagoren identity. Using the differential equations that define $\sin$ and $\cos$
*
*$\cos(0) = 1$
*$\sin(0) = 0$
*$\cos' = -\sin$
*$\sin' = \cos$
it's very easy to show that $\forall x \in \mathbb{R}\cos(x) = \sum_{i = 0}^\infty \frac{(-1)^ix^{2i}}{(2x)!}$ and $\forall x \in \math... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 3,
"answer_id": 1
} |
If $\cos A=\tan B$, $\cos B=\tan C$ … If $\cos A=\tan B$, $\cos B=\tan C$ and $\cos C=\tan A$, prove that $\sin A=\sin B=\sin C$.
My Attempt.
Let us consider $x$, $y$ and $z$ as:.
$$x = \tan^2A$$
$$y = \tan^2B$$
$$z = \tan^2C$$
$$\cos^2A = \tan^2B$$
$$\frac {1}{\sec^2A}= \tan^2B$$
$$\frac {1}{1 + \tan^2A} = \tan^2B$$
$... | From
$\cos A=\tan B,
\cos B=\tan C,
\cos C=\tan A
$,
$\begin{array}\\
\cos A
&= \sin B/\cos B\\
&= \sin B/\tan C\\
&= \sin B/(\sin C/\cos C)\\
&= \sin B \cos C/\sin C\\
&= \sin B (\sin A/\cos A)/\sin C\\
\end{array}
$
or
$\cos^2 A = \sin A\sin B/\sin C
$.
Similarly,
$\begin{array}\\
\cos B
&= \sin C/\cos C\\
&= \sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1935885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$ab(a+b) + bc(b+c) + ac(a+c) \geq \frac{2}{3}(a^{2}+b^{2}+c^{2})+ 4abc$ for $\frac1a+\frac1b+\frac1c=3$ and $a,b,c>0$
Let $a$, $b$, and $c$ be positive real numbers with $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$. Prove that:
$$
ab(a+b) + bc(b+c) + ac(a+c) \geq \frac{2}{3}(a^{2}+b^{2}+c^{2})+ 4abc.
$$
... | We need to prove that
$$\sum\limits_{cyc}(a^2b+a^2c)\geq\frac{2(a^2+b^2+c^2)}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}+4abc$$ or
$$\sum\limits_{cyc}(a^2b+a^2c)\geq\frac{2abc(a^2+b^2+c^2)}{ab+ac+bc}+4abc$$ or
$$\sum\limits_{cyc}ab\sum\limits_{cyc}(a^2b+a^2c)\geq2abc(a^2+b^2+c^2)+4abc(ab+ac+bc)$$ or
$$\sum_{cyc}(a^3b^2+a^3c^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1944026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Finding solutions to the diophantine equation $7^a=3^b+100$
Find the positive integer solutions of the diophantine equation $$7^a-3^b=100.$$
So far, I only found this group $7^3-3^5=100$.
| note: this answer has been found to be -at least- incomplete, see comments of Gottfried Helms and piquito
$100$ has no primitive root therefore $7^m$ and $3^n$ are congruent with $1$ modulo $100$ for some integers $m,n$ smaller than $100$ and these numbers should divide $100$.
We have $7^4\equiv 3^{20}\equiv 1\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1946621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 0
} |
Inverse of a $2 \times 2$ block matrix Let
$$S := \pmatrix{A&B\\C&D}$$
If $A^{-1}$ or $D^{-1}$ exist, we know that matrix $S$ can be inverted.
$$S^{-1} = \pmatrix{A^{-1}+A^{-1}B(D-CA^{-1}B)^{-1}CA^{-1}&-A^{-1}B(D-CA^{-1}B)^{-1}\\-(D-CA^{-1}B)^{-1}CA^{-1}&(D-CA^{-1}B)^{-1}}$$
But, what if $A^{-1}$ and $D^{-1}$ do not ex... | One good reference is the Handbook Matrix Mathematics Theory, Facts and Formulas by Dennis S. Bernstein. Let $\mathbb{F}$ equal to $\mathbb{R}$ or $\mathbb{C}$.
Proposition 3.9.7. Let $A \in \mathbb{F}^{n \times n}, B \in \mathbb{F}^{n \times m}, C \in \mathbb{F}^{m \times n}$, and $D \in \mathbb{P}^{m \times m}$. If... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1946713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
} |
If $abc=1$ so $\sum\limits_{cyc}\sqrt{\frac{a}{b+c}}\geq\frac{9}{\sqrt{a+b+c+15}}$ Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\geq\frac{9}{\sqrt{a+b+c+15}}$$
It seems nice enough.
I proved this inequality by Holder, but it quit... | Put the following substitution :
$\frac{a}{b+c}=x$$\quad$$\frac{b}{a+c}=y$$\quad$$\frac{c}{b+a}=z$
Remark that :
$$a+b+c=((\frac{1}{x}+1)(\frac{1}{y}+1)(\frac{1}{z}+1))^{\frac{1}{3}}$$
And :
$$abc=1 \iff -2=-\frac{1}{xyz}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$
So we get the inequality related to this Post that you hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1947672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Find the set of $n\in\Bbb Z^+$ with $M=\{n,n+1,n+2,n+3,n+4,n+5\}$ partitionable into two sets
Find the set of all positive integers $n$ with the property that the set $M=\{n, n + 1,n + 2,n + 3,n + 4,n + 5\}$ can be partitioned into two sets such that the product of the numbers in one set equals the product of the numb... | Partition $M$ into $A$ and $B.$ For brevity let $A^*=\prod_{x\in A}x$ and $B^*=\prod_{y\in B}y.$ Assume that $A^*=B^*$.
We may assume $n+5\in B.$
Observe that $5|n$ otherwise exactly one member of $M$ is divisible by $5,$ implying that $5$ divides exactly one of $A^*, B^*.$ So let $M=\{5m,5m+1,5m+2,5m+3,5m+4,5m+5\}.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1948239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Proving $a^n - b^n \le na^{n - 1}(a-b)$ with induction The question is to prove $a^n - b^n \le na^{n - 1}(a-b)$ with induction where $0 < b < a$ and $ \forall n \ge 1$, $n \in \mathbb{Z}$
I finished the different parts of the proof, including the predicate, base case, and inductive step, but I want to make sure that I'... | \begin{align}
&a^{n+1} - b^{n+1} \\
=\ &a^{n+1} - a^nb + a^nb - b^{n+1} \tag{your step $1$} \\
=\ &a^n(a - b) + \color{red}{b(a^n - b^n)} \\
\leq\ &a^n(a - b) + \color{red}{a(a^n - b^n)} \\
\leq\ &a^n (a - b) + \color{red}{a \cdot na^{n-1}(a-b)} \\
\leq \ &a^n(a - b) + \color{red}{n\cdot a^n(a-b)} \tag{your step $4$}\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1952084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find all continuous functions $f(x+1)=f(x)+1$ and $f\left(\frac1x\right)=\frac1{x^2}f(x)$
Find all continuous functions $f:\mathbb R \rightarrow \mathbb R$ such that
1) $\forall x \in \mathbb R$
$$f(-x)=-f(x)$$
2) $\forall x \in \mathbb R$
$$f(x+1)=f(x)+1$$
3$\forall x \in \mathbb R/ \{0\}$
$$f\left(\frac1x\righ... | Get $y = f(x)$ , then i have
$$f(x+1)=y+1,f(\frac{1}{x+1})=\frac{y+1}{(x+1)^2},f(\frac{-1}{x+1})=\frac{-(y+1)}{(x+1)^2}$$
$$f(\frac{x}{x+1})=f(-\frac{1}{x+1}+1)=\frac{-(y+1)}{(x+1)^2}+1=\frac{x^{2}+2x-y}{(x+1)^2}$$
$$f(\frac{x}{x+1})=f(\frac{1}{\frac{x+1}{x}})=\frac{f(\frac{x+1}{x})}{(\frac{x+1}{x})^2}=\frac{\frac{x^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1956628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Show that as $n\rightarrow \infty,$ $\frac{n^4}{3^n}\rightarrow 0$ linearly with rate $\frac{1}{3}$ Show that as $n\rightarrow \infty,$ $\frac{n^4}{3^n}\rightarrow 0$ linearly with rate $\frac{1}{3}$
linearly means take $a_n=\frac{n^4}{3^n}$
$\lim_{n\rightarrow \infty} (a_n)^\frac{1}{n}$
$\lim_{n\rightarrow\infty } (\f... | Since
$$\lim_{n\to\infty}\left(\frac{n^4}{3^n}\right)^\frac{1}{n} = \lim_{n\to\infty}\frac{\left(n^4\right)^\frac{1}{n}}{\left(3^n\right)^\frac{1}{n}}=\lim_{n\to\infty}\frac{n^\frac{4}{n}}{3},$$
it will suffice to show that
$$\lim_{n\to\infty}n^\frac{4}{n} = \left(\lim_{n\to\infty} n^\frac{1}{n}\right)^4 = 1.$$
Now let... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Sum of the series $\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........$
The sum of the series $$\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........$$
$\bf{My\; Try::}$ We can write it as $\displaystyle \binom{n}{0} = $ Coefficient of $x^0$ in $(1+x)^n$
Similarly $\displaystyle \bino... | Hint:
$$\binom{n}{0}= \mbox{ coefficient of } x^n \mbox{ in } (1+x)^n \\
\binom{n-1}{1}= \mbox{ coefficient of } x^n \mbox{ in } x^2(1+x)^{n-1} \\
\binom{n-2}{2}= \mbox{ coefficient of } x^n \mbox{ in } x^4(1+x)^{n-2} \\
...$$
Hint 2:
$$(1+x)^n-x^2(1+x)^{n-1}+x^4(1+x)^{n-2}-..=(1+x)^n \left(1-(\frac{x^2}{1+x})+ (\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1960944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solving an ordinary differential equation nonlinear I have the following
$$ y' = \frac{ xy }{x^2 + y^2} $$
My approach would be to rewrite this in terms of $x$
$$ x' = \frac{x}{y} + \frac{y}{x} $$
And then let $u = \frac{y}{x} \implies u' = \frac{x + y x'}{x^2} = \frac{1}{x} + \frac{u}{x} x' = \frac{1}{x} + \frac{u}{x}... | This belongs to the class of homogeneous differential equations of order 0, that is, it satisfies $f(\lambda x, \lambda y) = f(x,y)$. One way to solve such a equation is to write the differential equation in the form: $y' = f(\frac{y}{x})$ and make the substitution $v=\frac{y}{x}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $a,b,c$ where $(a-1)(b-1)(c-1)$ is a divisor of $(abc-1)$ Find integers $a,b,c$ such that $1<a<b<c$ and $(a-1)(b-1)(c-1)$ is a divisor of $(abc-1)$ .I tried it solve using elementary number theory but I can't proceed . Somebody help me.
| Denote $\bullet = \dfrac{abc-1}{(a-1)(b-1)(c-1)}$
Note that $$
\bullet = \dfrac{a}{a-1}\cdot \dfrac{bc-1}{(b-1)(c-1)} + \dfrac{1}{(b-1)(c-1)} \\
=\dfrac{a}{a-1} \cdot \left(1 + \dfrac{1}{b-1} + \dfrac{1}{c-1}\right) + \dfrac{1}{(b-1)(c-1)}.\tag{1}
$$
$(1)\Rightarrow$ (lower bound)
$$
\dfrac{a}{a-1}<\bullet.\tag{2}
$$
... | {
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"url": "https://math.stackexchange.com/questions/1963717",
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"source": "stackexchange",
"question_score": "2",
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Find the sum $\sum_{n=1}^{\infty} \frac{4n}{n^4+2n^2+9}$ Find the sum
$$\sum_{n=1}^{\infty} \dfrac{4n}{n^4+2n^2+9}.$$
By calculator, we can predict that its sum is equal to $\dfrac{5}{6}$ so I think we should use inequalities to prove it. And I found that
$\dfrac{5}{6(n^4+n^2)} < \dfrac{4n}{n^4+2n^2+9}< \dfrac{5}{6(n^2... | $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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\newcommand{\ic}{\... | {
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Find Min of function $x+\frac{2}{x}$ for x>0 with following solution? I want to find Minimum of function $x+\frac{2}{x}$ for $x>0$ with following solution. My problem is how can we say we found the answer with following solution.
Solution:
Let Minimum be something like c. We have:
$x+\frac{2}{x} \ge c \to \frac{x^2+2}{... | Using $AM-GM$ inequality, we have(since $x>0$)
$$ x + \frac{2}{x} \geq 2 \sqrt{ x \cdot \frac{2}{x} } = 2 \sqrt{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964011",
"timestamp": "2023-03-29T00:00:00",
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Find the minimum value of $\sqrt{x+y}+\sqrt{(1-x)+y}+\sqrt{x+(1-y)}+\sqrt{(1-x)+(1-y)}$ Given that $x,y\in[0,1]$, find the minimum of the following
$$f(x,y)=\sqrt{x+y}+\sqrt{(1-x)+y}+\sqrt{x+(1-y)}+\sqrt{(1-x)+(1-y)}$$
I have found $$f(0,0)=f(0,1)=f(1,0)=f(1,1)=2+\sqrt{2}, f(\dfrac{1}{2},\dfrac{1}{2})=2+\sqrt{6}$$
so i... | $\frac{\partial^2f}{\partial x^2}=\frac{\partial^2f}{\partial y^2}=-\frac{1}{4}\left((x+y)^{-\frac{3}{2}}+(1-x+y)^{-\frac{3}{2}}+(1-y+x)^{-\frac{3}{2}}+(2-x-y)^{-\frac{3}{2}}\right)<0$,
which says that $f$ is a concave function of $x$ and of $y$.
Id est, $\min\limits_{\{x,y\}\subset[0,1]}f=\min\limits_{\{x,y\}\subset\{... | {
"language": "en",
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$q^{3}\leq aq^{2}+bq+c$ finite amount of q problem Im a bit confused why this is true it's simple but i seem to have some problems with it $q^{3}\leq aq^{2}+bq+c$ is only true for a finite amount of $\forall q \epsilon \mathbb{N}$
| First, you can assume that $a, b, c \ge 0$, for if they're negative, then $q^3 \le aq^2 + bq + c \le |a|q^2 + |b|q | c$, and if there are at most finitely many solutions the the latter, then there are to the former as well. For the same reason, we can assume $c \ge 1$, so that $a+b+c \ge 1$.
Claim: For $q \ge a+b+c$, t... | {
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$x^2$ is congruent to $-1 \bmod p$ In general, how do you solve $x^2$ is congruent to $-1 \bmod p$, where $p$ is an odd prime and $x$ is an integer.
Specifically, I need to solve $x^2\equiv-1 \pmod{29}$
| Since you can figure out
$$ 29 = 25 + 4 = 5^2 + 2^2 $$
in your head, you see
$$ 5^2 + 2^2 \equiv 0 \pmod {29}, $$
$$ 5^2 \equiv - 2^2 \pmod {29}. $$
Bothe 2 and 5 are relatively prime to $29,$ either one has a multiplicative inverse mod 29.
$$ \frac{5^2}{2^2} \equiv -1 \pmod {29}, $$
$$ \left( \frac{5}{2} \right)^2 \e... | {
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"source": "stackexchange",
"question_score": "1",
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Solving indefinite integral $\int \frac{dx}{(x^4-1)^3}$ I'm trying to solve next integral, but I can't start. WolframAlpha gives me really terrible answer and can't give any step-by-step instructions, so I simply does not know how to start.
$$\int \frac{dx}{(x^4-1)^3}$$
Please give any hint or start point of solving th... | Let
$$I:=\int\frac{dx}{x^4-1}, J:=\int\frac{dx}{(x^4-1)^2}, K:=\int\frac{dx}{(x^4-1)^3}.$$
By parts,
$$I=\frac x{x^4-1}+4\int\frac{x^4\,dx}{(x^4-1)^2}=\frac x{x^4-1}+4(I+J)$$
and
$$J=\frac x{(x^4-1)^2}+8\int\frac{x^4\,dx}{(x^4-1)^3}=\frac x{(x^4-1)^2}+8(J+K).$$
On another hand,
$$I=\int\frac{dx}{x^4-1}=\frac12\int\fra... | {
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How do I prove that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\dotsb+xy^{n-2}+y^{n-1})$ I am unsuccessfully attempting a problem from Spivak's popular book 'Calculus' 3rd edition. The problem requires proof for the following equation:
$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}+\dotsb+xy^{n-2}+y^{n-1})$$
The solution to the problem, contai... |
The step from equation 1 to equation 2 uses both the distributive law as well as the commutative law.
In order to better see what's going on, we look at small special cases $n=2,3$. We also use somewhat more detailed transformations.
n=2:
\begin{align*}
(x-y)\left(x^{1}+y^1\right)&=x\left(x^{1}+y^1\right)-y\left(x... | {
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"source": "stackexchange",
"question_score": "2",
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Closed form of a sum $ \sum_{i=1}^{n}\frac{1}{((i-1)x)^2+y^2}$ Consider a sum:
$$ \sum_{i=1}^{n}\frac{1}{((i-1)x)^2+y^2}$$
with $x$ and $y$ being (non-zero) constants. Is it possible to obtain a nice closed form of this expression?
| As already wrote, there is no nice closed form. A way to see an approximation is the following. Using Abel's summation we have $$S_{N}\left(z\right)=\sum_{n\leq N}\frac{1}{1+n^{2}z^{2}}=\frac{N}{1+N^{2}z^{2}}+2z^{2}\int_{1}^{N}\frac{\left\lfloor t\right\rfloor t}{\left(1+t^{2}z^{2}\right)^{2}}dt
$$ where $\left\lfloor... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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How to show $1+3x$ is a unit in $\mathbb{Z_6}[[x]]$ I am new to the subject of rings of formal power series. The problem is to show that $1+3x$ is a unit in $\mathbb{Z_6}[[x]]$, where $\mathbb{Z_6}[[x]]$ is the ring of formal power series with coefficients in $\mathbb{Z}_6$, also known as $ \mathbb{Z }/6\mathbb{Z}$.
M... | You need to show that
$$(1 + 3x + 3x^2 + 3x^3 + \cdots)(1 + 3x) = 1$$
using the addition and multiplication rules in $\Bbb Z_6[[x]]$. The element $1 + 3x + 3x^2 + 3x^3 + \cdots$ is $\sum\limits_{j = 0}^\infty 3x^j$, so over $\Bbb Z_6[[x]]$ we write
$$\left(\sum_{j = 0}^\infty 3^jx^j\right)(1 + 3x) = \sum_{j = 0}^\inft... | {
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"source": "stackexchange",
"question_score": "2",
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How should I calculate the determinant? $\left|\begin{array}{cccc}1&a&b&c+d\\1&b&c&a+d\\1&c&d&a+b\\1&d&a&b+c\end{array}\right|=
\left|\begin{array}{cccc}1&a&b&c\\1&b&c&a\\1&c&d&a\\1&d&a&b\end{array}\right|+
\left|\begin{array}{cccc}1&a&b&d\\1&b&c&d\\1&c&d&b\\1&d&a&c\end{array}\right|$
I tried to calculate the determina... | Compute
$$
\begin{bmatrix}
1 & a & b & c+d \\
1 & b & c & d+a \\
1 & c & d & a+b \\
1 & d & a & b+c
\end{bmatrix}
\begin{bmatrix}
a+b+c+d \\ -1 \\ -1 \\ -1
\end{bmatrix}
$$
| {
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"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Find the maximum value of the expression $ \frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}$ Find the maximum value of the expression $ \frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}$ where $x, y, z$ are real numbers satisfying the condition $x + y + z = 1 $.
My Attempt
$$
\begin{align}
\frac{x}{1+x^2}+\frac{y}{1+y^2}... | You've already been given a hint to try Jensens inequality, which works fine in this case. Alternatively by AM-GM
$$9\frac{x}{9+(3x)^2}\leqslant 9\frac{x}{10(3x)^{2/10}}=\frac3{10}(3x)^{4/5}$$
Sum that up and observe by power means:
$$\sum (3x)^{4/5} \leqslant 3\left(\frac{3x+3y+3z}3\right)^{4/5}=3$$
Equality is possi... | {
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"question_score": "3",
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Show that $\sum_{k=1}^{\infty}\frac{k}{2^k}=2$
Show that $\sum_{k=1}^{\infty}\frac{k}{2^k}=2$
I have no idea to calculate this sum. I try shift index since $k=0$ gives 0, tis won't change the sum. But I don't know how to keep going. Can someone give me a hint or suggestion to calculate this sum? Thanks
| Here's another solution.
Let
\begin{align}
S=\sum^\infty_{k=1}\frac{k}{2^k} =&\ \sum^\infty_{k=1} \frac{k-1+1}{2^k} = \frac{1}{2}\sum^\infty_{k=1} \frac{k-1}{2^{k-1}} + \sum^\infty_{k=1} \frac{1}{2^k}\\
=&\ \frac{1}{2}\sum^\infty_{k=2} \frac{k-1}{2^{k-1}}+\frac{1}{2} = \frac{1}{2}\sum^\infty_{k=1}\frac{k}{2^k}+1 \\
... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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What is the largest perfect square that divides $2014^3-2013^3+2012^3-2011^3+\ldots+2^3-1^3$ I've tried this but didn't get the answer :
Let $S=2014^3-2013^3+2012^3-2011^3+\ldots+2^3-1^3$
Using $n^3-(n-1)^3 = 3n^2-3n+1$,
\begin{align}
S &= 3(2014^2)-3(2014)+1+3(2012^2)-3(2012)+1+\ldots+3(2^2)-3(2)+1
\\&= 3\left ( 2014(... | In general you obtain, for $n^3-(n-1)^3+\cdots + 2^3-1^3$ and $n$ even the formula
$$
3\left(\sum_{k=1}^{n/2} 2k(2k-1)\right)+\frac{n}{2}.
$$
Now it is easier to see that this is divisible by $(\frac{n}{2})^2$; and you can test this first for $n=2,4,6,\ldots$ before dealing with $n=2014$. In fact, your computation is c... | {
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maximum value of $\int^{\frac{3\pi}{2}}_{-\frac{\pi}{2}}\sin xf(x)dx$ subjected to the condition$|f(x)|\leq 5$ maximum value of $\displaystyle \int^{\frac{3\pi}{2}}_{-\frac{\pi}{2}}\sin xf(x)dx$ subjected to the condition$|f(x)|\leq 5$
could some help me with this, thanks
| If $|f(x)|\le 5$ for all $x\in \left[-\frac{\pi}{2},\frac{3\pi}{2}\right]$, we have $\sin(x) f(x)\le|\sin(x) f(x)|\le 5|\sin(x)|$ for all $x\in\left[-\frac{\pi}{2},\frac{3\pi}{2}\right]$.
We thus have $$\int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}}\sin(x)f(x)\mathrm{dx}\le\int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}}5|\sin(x)|\math... | {
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"source": "stackexchange",
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How to calculate the limit of $\frac{a_n}{n^2}$ for the sequence $a_{n+1}=a_n+\frac{2 a_{n-1}}{n+1}$? Assume $\{a_n\}$is a sequence which satisfies the following recursion formula:
$$a_{n+1}=a_n+\frac{2a_{n-1}}{n+1}~(n\geqslant 1).$$
and $a_0=\pi,a_1=\pi^2 $.
How to compute $\displaystyle\lim_{n\to \infty}\frac{a_n}{n... | Let we assume that
$$ f(x) = \pi+\pi^2 x +a_2 x^2+\ldots = \sum_{n\geq 0}a_n x^n \tag{1}$$
The recursion turns into the differential equation
$$\frac{f(x)-\pi-\pi^2 x}{x}=f(x)-\pi+\frac{2}{x}\int_{0}^{x}t\,f(t)\,dt \tag{2} $$
and assuming $G'(x)=x\,f(x)$, that leads to
$$ f(x) = \frac{\pi(9-5\pi)}{4}\cdot\frac{e^{-2x}... | {
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"source": "stackexchange",
"question_score": "4",
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How to prove $\sum_{k=1}^\infty\frac{k^k}{k!}x^k=\frac{1}{2}$ where $x=\frac{1}{3}e^{-1/3}$ How to prove that
$$
\sum_{k=1}^\infty\frac{k^k}{k!}x^k=\frac{1}{2}, ~\text{where}~~ x=\frac{1}{3}e^{-1/3}~?
$$
I found this sum in my notes, but I don't remember where I got it. Any hints or references would be nice.
| Let's consider $\sum_{k=1}^\infty\frac{k^k}{k!}\big(\frac{1}{ae^{1/a}}\big)^k=\frac{1}{a-1}$ for $a > 1$ (as in the question).
Then $\frac{1}{a-1} = - 1+ \frac{1}{1-1/a} = \sum_{k=1}^\infty (\frac1a)^k$ by the geometric series. So one has to show
$\sum_{k=1}^\infty\Big[-1 + \frac{k^k}{k!}{e^{-k/a}}\Big](\frac{1}{a}\... | {
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"source": "stackexchange",
"question_score": "8",
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Prove the following Inequality. Given that $\alpha,\beta$ and $\gamma\in (0,\pi)$ and $\alpha+\gamma+\beta=\pi$, show that $$\cos\alpha+\cos\gamma+\sin\beta\leq\frac{3\sqrt3}{2}.$$
Now I am aware of the following Inequality: $\cos\alpha+\cos\gamma+\cos\beta\leq\frac{3}{2}$, but notice that the term $\cos\beta$ has been... | $\cos(\alpha)+\cos(\gamma)+\sin(\beta)=\sin(\frac{\pi}{2}-\alpha)+\sin(\frac{\pi}{2}-\gamma)+\sin(\beta)$
1) Suppose $\beta \leq \frac{\pi}{2}$
Using Jensen inequality we get: $\frac{\sin(\frac{\pi}{2}-\alpha)+\sin(\frac{\pi}{2}-\gamma)+\sin(\beta)}{3} \leq \sin(\frac{\frac{\pi}{2}-\alpha+\frac{\pi}{2}-\gamma+\beta}{3}... | {
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"source": "stackexchange",
"question_score": "2",
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To find the equation of curve. We have given a differential equation we have to find the equation of curve. I tried it a lot but not able to proceed. Can someone give me some hints.
$(1-xy+x^2y^2)~dx=x^2~dy$
I tried to substitute $xy=t$ but got no result. Please help.
| $(1-xy+x^2y^2)~dx=x^2~dy$
$\dfrac{dy}{dx}=\dfrac{1}{x^2}-\dfrac{y}{x}+y^2$
Let $y=-\dfrac{1}{u}$ ,
Then $\dfrac{dy}{dx}=\dfrac{1}{u^2}\dfrac{du}{dx}$
$\therefore\dfrac{1}{u^2}\dfrac{du}{dx}=\dfrac{1}{x^2}+\dfrac{1}{xu}+\dfrac{1}{u^2}$
$\dfrac{du}{dx}=\dfrac{u^2}{x^2}+\dfrac{u}{x}+1$
Let $v=\dfrac{u}{x}$ ,
Then $u=xv$
$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Let $\alpha$, $\beta$, and $\gamma$ be acute angles such that $\alpha$ + $\beta$ = $\gamma$. Show that Let $\alpha$, $\beta$, and $\gamma$ be acute angles such that $\alpha$ + $\beta$ = $\gamma$. Show that
cos$\alpha$ + cos$\beta$ + cos$\gamma$ - 1 $\geq$ $2\sqrt{\cos\alpha\times\cos\beta\times\cos\gamma}$.
Here's what... | Let $\alpha\geq\beta$
Since $\cos$ is a decreasing function on $\left(0,\frac{\pi}{2}\right)$, $\frac{\alpha+\beta}{2}<\frac{\pi}{4}$ and $\frac{\alpha-\beta}{2}<\frac{\pi}{4}$, we obtain:
$$\cos\alpha+\cos\beta+\cos(\alpha+\beta)-1=2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}+2\cos^2\frac{\alpha+\beta}{2}-2... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Is this valid? Show that $(x-2)\mid x^{3} - 4x$ Is this valid? Here $\textsf{R}(x)$ stands for the remainder.
Show that $(x-2)\mid x^{3} - 4x$
I know that I can simply do this: Let $p(x) = x^{3} - 4x$ then if $(x-2)\mid x^{3} - 4x$ it follows that $p(2) = 0$. So we have $2^{3} - 4(2) = 0$. However I am wondering ab... | $x^3-4x=x(x^2-4)=x(x+2)(x-2)$. So, you can clearly see that $x-2$ is a factor.
Or you can do it this way,
Let $f(x)=x^3-4x$. See that $f(2)=0$. Hence, by factor theorem, $x-2|f(x).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1987467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Summation of Central Binomial Coefficients divided by even powers of $2$ Whilst working out this problem the following summation emerged:
$$\sum_{m=0}^n\frac 1{2^{2m}}\binom {2m}m$$
The is equivalent to
$$\begin{align}
\sum_{m=0}^n \frac {(2m-1)!!}{2m!!}&=\frac 12+\frac {1\cdot3}{2\cdot 4}+\frac{1\cdot 3\cdot 5}{2\cdo... | Another approach. By Euler/De Moivre's formula we have
$$A_n=\frac{1}{4^n}\binom{2n}{n} = \frac{1}{2\pi}\int_{-\pi}^{\pi}\cos^{2n}(x)\,dx \tag{1}$$
since $\cos(x)^{2n} = \frac{1}{4^n}\sum_{j=0}^{2n}\binom{2n}{j}e^{(2n-j)ix}e^{-jix}$ and $\int_{-\pi}^{\pi}e^{kix}\,dx =2\pi \delta(k)$, hence only the contribute given by ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1989966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
Singapore math olympiad Trigonometry question: If $\sqrt{9-8\sin 50^\circ} = a+b\csc 50^\circ$, then $ab=$?
$$\text{If}\; \sqrt{9-8\sin 50^\circ} = a+b\csc 50^\circ\text{, then}\; ab=\text{?}$$
$\bf{My\; Try::}$ We can write above question as $$\sin 50^\circ\sqrt{9-8\sin 50^\circ} = a\sin 50^\circ+b$$
Now for Left si... | There is not a unique solution; we will be able to provide a solution as a function of $a$. Note that if $a = 0$, then $ab=0$. Henceforth, assume $a \neq 0$.
Let $x = \sin(50^\circ) \neq 0$ and $b = c/a$ so that we wish to solve for $c$. Then the given equation is
$$ \sqrt{9 - 8 x} = a + \frac{c}{a x} \text{.} $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Finding maximum value from a statement/equation The statement has values of $x$ and $y$ as positive integers: $$\sqrt{x} - \sqrt{11} = \sqrt{y}$$
I have to find the maximum possible values of $\frac{x}{y}$, this what I have done so far:
$$x = (\sqrt{y} + \sqrt{11})^2$$
$$y = (\sqrt{x} - \sqrt{11})^2$$
therefore: $$\fra... | Hint: $\frac{x}{y}=\frac{(\sqrt{y}+\sqrt{11})^2}{y}=\frac{y+2\sqrt{y}\sqrt{11}+11}{y}=1+\frac{2\sqrt{y}\sqrt{11}+11}{y}=f(y)$. Find zeros of $f'(y)$ and determine if they are minima or maxima by the second derivative test.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Circle incribed within a triangle percentage
I have worked out the areas as $\pi/3$ for the circle and $2/\sqrt3$ for the triangle but don't know how to convert into a percentage without a calculator.
| $$
\begin{gathered}
side_{\,tr} = 1\quad \quad h_{\,tr} = \frac{{\sqrt 3 }}
{2}\quad \quad r_{\,circ} = \frac{{\sqrt 3 }}
{6} \hfill \\
A_{\,circ} = \pi \frac{3}
{{36}} = \frac{\pi }
{{12}}\quad \quad A_{\,tr} = \frac{{\sqrt 3 }}
{4} \hfill \\
\rho = \frac{{A_{\,circ} }}
{{A_{\,tr} }} = \frac{\pi }
{{12}}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Surface area of sphere within a cylinder I have to
Compute the surface area of that portion of the sphere $x^2+y^2+z^2=a^2$ lying within the cylinder $\Bbb{T}:=\ \ x^2+y^2=by.$
My work:
I start with only the $\Bbb{S}:=\ \ z=\sqrt{a^2-x^2-y^2}$ part and will later multiply it by $2$.
$${\delta z\over \delta x}={-x\o... | Your derivatives are wrong by a factor of $2$. This causes your integrand to be
$$
\sqrt{1-{x^2+y^2\over a^2-x^2-y^2}}=\frac a{\sqrt{a^2-x^2-y^2}}
$$
If we represent the cylinder in polar coordinates, we have $$r^2=b r \sin\theta.$$After cancelling $r$, we get $r=b\sin\theta$. So the interior of the cylinder is the r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1993279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Prove: $\log_{2}{3} < \log_{3}{6}$ How should I prove that $\log_{2}{3} < \log_{3}{6}$?
I tried something like this: $2^{\log_{2}{3}}< 2^{\log_{3}{6}}$, $3<6^{\log_{3}{2}}$, $\log_{6}{3}<\log_{6}{6^{\log_{3}{2}}}=\log_{3}{2}$, $\frac{1}{\log_{3}{6}}< \log_{3}{2}$. $\frac{1}{1+\log_{3}{2}}<\log_{3}{2}$, but still nothi... | Here is mine...
$$\log_2 3=\frac{\log_2 243}5<\frac{\log_2 256}5=8/5$$
$$\log_3 2=\frac{\log_3 32}5>\frac{\log_3 27}5=3/5$$
$$\log_3 6=\log_3 2+\log_3 3>1+3/5=8/5>\log_2 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1994320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
If $A$ is a real $n \times n$ matrix satisfying $A^3 = I$ then Trace of $A$ is always If $A$ is a real $3 \times 3$ matrix satisfying $A^3$ = I such that $ A \neq I $ .Then, Trace of A is always
*
*$0$
*$1$
*$-1$
*$3$
I proceed as follows: from given,
$\min(x)=x-1$ or $x^2+x+1$ or $(x-1)(x^2+x+1)$
$\min(x)\... | $$A^3=I\iff A^2=A^{-1}$$ If the question is valid for all matrix then with a single example one can answer. Consider the matrix
$$A = \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1 \\ 1 & 0 & 0\end{pmatrix}$$
Calculation gives $$A^{-1}=\begin{pmatrix} 0 & 0 & 1\\ 1 & 0 & 0 \\ 0 & 1 & 0\end{pmatrix}\text { and } A^2=\begin{pmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1994406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Finding the subinterval for which $f_{n}(x) = \frac{n\sin(x)}{1+n^2\sin^2(x)}$ is Uniformly Convergent in Let $f_{n}(x) = \frac{n\sin(x)}{1+n^2\sin^2(x)}$ for $x \in [0,\pi]$. For $\delta > 0$, I want to find $E_{\delta} \subset [0, \pi]$ where $f_{n}(x)$ converges uniformly and $\mu([0,\pi] \setminus E_{\delta}) < \de... | Let's look on the interval $[0,\pi/2]$ and choose an integer $m > 2$.
Then for $n > 2m^2$ and $x > \frac{1}{m}$, we have
\begin{eqnarray}
2m^2 < n &\iff& 2m < \frac{n}{m}\\
&\implies& 2m < nx \tag{since $x> \tfrac{1}{m}$}\\
&\implies& \frac{2}{nx} < \frac{1}{m}
\end{eqnarray}
Now because $\sin (x) > \frac{x}{2}$ in $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1995151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Mathematical Induction for Recurrence Relation I have solved the following recurrence relationship:
$T(1) = 1$
$T(n) = T(n-1) + n + 2$
so
$T(n) = \frac{1}{2}n^2+\frac{5}{2}n -2$
I am now trying to perform mathematical induction to prove this.
$Basis:$
$T(1)=1=3-2=\frac{1}{2} + \frac{5}{2} - 2 $
$Induction:$
$T(k+1) = T... | HINT: Try to express the last formula in terms of $(k+1) $, rather than a function of $k $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1995278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Show that $f(x)=\pi-2\arctan(\sqrt{x-1})$
Given, for every $x>1$,
$$f(x)=4\arctan\frac{1}{\sqrt{x-1}+\sqrt{x}}$$
Show that $f(x)=\pi-2\arctan(\sqrt{x-1})$
I have tried to use the fact that $\arctan(x)+\arctan(1/x)=\frac{\pi}{2
}$
So I obtain: $f(x)=4(\frac{\pi}{2}-\arctan(\sqrt{x-1}+\sqrt{x})$
I am stuck here !
| Your statement is equivalent to proving that, for
$$
g(x)=2\arctan\frac{1}{\sqrt{x}+\sqrt{x-1}}
$$
we also have
$$
g(x)=\frac{\pi}{2}-\arctan\sqrt{x-1}
$$
Note that
$$
\frac{1}{\sqrt{x}+\sqrt{x-1}}=\sqrt{x}-\sqrt{x-1}
$$
Set $h(x)=\sqrt{x}-\sqrt{x-1}$ and prove that, for $x>1$, $0<h(x)<1$.
It follows that $0<g(x)<\pi/2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1996649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
How many five-digit numbers can be formed using the digits 1-9 which have at least three identical digits? How many five-digit nos. can be formed using the digits 1-9 having at least three identical digits?
My attempt:
Total no. of possible nos. with no restrictions $=9^5$
No. of numbers having two identical digits$=9... | First approach: Pick which three spots host the identical digits; there are $\binom53$ ways to do this. Fill those three spots in one of $9$ ways, and then fill the other two spots in any of $9$ ways. Thus, $\binom53\times 9^3$
Problem with this approach: the number $12222$ has been counted $4$ times, once as $1\color{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1997857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
find the range of the $b$, if $a^2+b^2+c^2=21~~~~2b=a+c$ Let $\Delta ABC$ such $|AC|=b,|BC|=a,|AB|=c$,and such $$\begin{cases}
2b=a+c\\
a^2+b^2+c^2=21
\end{cases}$$
find the range of the $b$
since
$$(a+c)^2-2ac+b^2=21$$
so we have
$$2ac=5b^2-21$$
then $a,c$ is a equation
$$x^2-2bx+\dfrac{5b^2-21}{2}=0$$ roots.so we ha... | For the lower bound we have: $b > |a-c| \implies b^2 > (a-c)^2, (2b)^2 = (a+c)^2\implies 5b^2 > (a-c)^2 + (a+c)^2 = 2(a^2+c^2) = 2(21- b^2) \implies 7b^2 > 42 \implies b^2 > 6 \implies b > \sqrt{6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Help with tricky double integral
Consider the region $R$ bounded by the circles
$$x^{2}+y^{2}=Ax$$$$x^{2}+y^{2}=Bx$$$$x^{2}+y^{2}=Cy$$$$x^{2}+y^{2}=Dy$$
where $B>A$ and $D>C$.
Use the change of variables
\begin{cases} u=\frac{x}{x^2+y^2}\\
v=\frac{y}{x^2+y^2} \end{cases}
to evaluate the integral $$\int\int_R ... | *
*You pretty much got it:
$$
R(u,v)=\left\{(u,v)\;|\; \frac{1}{B}\le u\le\frac{1}{A} , \frac{1}{D}\le u\le \frac{1}{C}\right\}
$$
The Figures below show the case where $A=C=1$, $B=D=2$:
So the region we are interested in is
*We are given \begin{cases} u=\frac{x}{x^2+y^2}\\
v=\frac{y}{x^2+y^2} \end{cases}
Solvi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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A proof related to ellipse. A tangent is drawn to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ to cut the ellipse $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$ at the points $P$ and $Q$. If tangents at $P$ and $Q$ to the ellipse $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$ intersect at right angle, then prove that $\frac{a^2}{c^2}+\fr... | 1) The first ellipse $E_1$ has equation $$F(x,y)=b^2x^2+a^2y^2-a^2b^2=0$$ and its derivative at the point $(x,y)$ is equal to $$-\dfrac{\dfrac{\partial F}{\partial x}}{\dfrac{\partial F}{\partial y}}=\frac{-b^2x}{a^2y}$$
2) Clearly for an ellipse of equation $\dfrac{X^2}{A^2}+\dfrac{Y^2}{B^2}=1$ the vertical $X=A$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding a simpler representation for the set $\{z\in\mathbb{C}|\|z+i\|+\|z-i\|=4\}$
Sketch the following set: $\{z\in\mathbb{C}|\|z+i\|+\|z-i\|=4\}$
I tried thinking about it geometrically, where we are interested in all triangles of the form (x,y-1), (x,y+1), (0,0) and we want the sum of the sides next to (0,0) be 2... | This is the "two nails and fixed cord" construction of an ellipse.
Computing the minor axis:
For $z = a + 0\cdot i$ we have
$$
\lVert z+i \rVert + \lVert z-i \rVert
= \sqrt{a^2 + 1^2} + \sqrt{a^2 + (-1)^2}
= 2 \sqrt{1 + a^2} = 4 \iff \\
1 + a^2 = 4 \iff \\
a = \sqrt{3}
$$
For $z = 0 + b\cdot i$ we have
$$
\lVert z+i \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Where's this 1/2 coming from squaring the derivative for arc length? I am currently studying arc length in calculus 2 and I am having trouble understanding one of the solutions. I understand the concept and how the procedure works, but it seems like in a lot of problems you're adding on a 1/2 and bringing it in front o... | You've forgotten the middle term in $(a-b)^2 = a^2 - 2 a b + b^2$ and that $y^3 \cdot y^{-3} = 1 \neq 0$. Distibuting twice, then collecting...
\begin{align*}
&\left( \left( \frac{1}{2} \right) y^3 - \left( \frac{1}{2} \right) y^{-3} \right)^2 \\
&\qquad = \left( \left( \frac{1}{2} \right) y^3 - \left( \frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Locus of focus of variable parabola
A variable parabola is drawn to pass through A & B, the ends of a
diameter of a given circle with centre at the origin and radius c & to
have as directrix a tangent to a concentric circle of radius 'a' (a>c) ; the axes being AB & a perpendicular diameter, prove that the locus of... | We don't need to know the equation of the parabola.
We may assume that $A(c,0),B(-c,0)$.
Let $F(X,Y)$ be the coordinates of the focus of the parabola.
The equation of the directrix is given by $mx+ny=a^2$ where $P(m,n)$ is the tangent point such that $m^2+n^2=a^2$.
Now, by the definition of the parabola, the distanc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2005235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The sum of 50 terms of the series $\frac{3}{1} $+ $\frac{5}{1+2^2} $+ $\frac{7}{1+2^2+3^2}$+ $\frac{11}{1+2^2+3^2+4^2}$... The numerator is increasing by 2.
The number of terms in denominator is increasing, and they are squares of natural numbers (i.e. $1^2, 2^2, 3^2, 4^2...$)
Options are:
(a)$\frac {100}{17}$
(b)$\fra... | Start by simplifying the denominator of the $n^{th}$ term.
$$U_n=\frac{2n+1}{\sum_{i=1}^ni^2}$$
$$=\frac{6(2n+1)}{n(n+1)(2n+1)}$$
$$=\frac{6}{n(n+1)}$$
$$=\frac{6}{n}-\frac{6}{n+1}$$
So the sum of the first 50 terms will be:
$$S_{50}=U_1+U_2+U_3+\cdots+U_{48}+U_{49}+U_{50}$$
$$=\frac{6}{1}-\frac{6}{2}+\frac{6}{2}-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2006796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to find condition on the constant term of a cubic equation with no square term such that the cubic equation has atleast two integer roots? The number of integers $k$ for which the equation $x^3-27x+k=0$ has at least two distict integer roots is ?
a. 1 b. 2 c. 3 d. 4
My attempt is to differentiate thi... | In another way,
you already found that
$$
\begin{gathered}
x_{\,1} ,\,x_{\,2} ,\,x_{\,3} ,k\;\text{integers} \hfill \\
x_{\,1} \leqslant - 3,\quad - 3 \leqslant x_{\,2} \, \leqslant 3,\quad 3 \leqslant x_{\,3} \hfill \\
\end{gathered}
$$
where we include the $=$ sign since the roots may coincide at the absci... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2013854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the integral $\int \frac{dx}{1+x^5}$
$$\int \frac{dx}{1+x^5}$$
I have tried to add a $x^5$ and subtract $x^5$, but got nothing.
| Hint. One may start with
$$
x^5+1=(x+1)\left(x^2-\frac{\sqrt{5}+1}{2} x+1\right)\left(x^2+\frac{\sqrt{5}-1}{2} x+1\right)
$$ then one may obtain a partial fraction decomposition,
$$
\frac1{x^5+1}=\frac{a_0}{x+1}+\frac{a_1x+b_1}{x^2-\frac{\sqrt{5}+1}{2} x+1}+\frac{a_2x+b_2}{x^2+\frac{\sqrt{5}-1}{2} x+1}
$$ and integrate... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2014388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
solve $x^2 - 25 xy + y^2 = 1$ does it have a solution? Usually the Pell equation is written $x^2 - dy^2 = 1$ but here I am looking for solutions to an equation of the type:
$$ x^2 - k xy + y^2 = 1 $$
and In particular, $k$ is a perfect square. So I am picking $k = 25$ an example.
If we complete the square then $25/2$ ... | Your discriminant, for $x^2 - k x y + y^2,$ is
$$ \Delta = k^2 - 4. $$ The smallest solution, in positive integers, to
$$ \tau^2 - \Delta \sigma^2 = 4 $$
is $\tau = k, \sigma = 1.$ You have the obvious $-1$ automorphism given by interchanging $x,y.$ The matrix generating the oriented automorphisms is
$$
\left(
\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2015221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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If $\alpha = 2\arctan(2\sqrt{2}-1)$ and $\beta = 3\arcsin\left(\frac{1}{3}\right)+\arcsin\left(\frac{3}{5}\right)\;,$ Then prove that $\alpha>\beta$
If $\alpha = 2\arctan(2\sqrt{2}-1)$ and $\displaystyle \beta = 3\arcsin\left(\frac{1}{3}\right)+\arcsin\left(\frac{3}{5}\right)\;,$ Then prove that $\alpha>\beta$
$\bf{M... | \begin{align}
\beta&=\sqrt{10}\left(\frac{3}{\sqrt{10}}\arcsin\frac13+\frac{1}{\sqrt{10}}\arcsin\frac35\right)\\
&<\sqrt{10}\arcsin\left(\frac{4\sqrt{10}}{25}\right)
\end{align}
where I use the convexity of $\arcsin$. On the other hand using the definition of $\arctan$ we obtain $$\arcsin\left(\frac{4\sqrt{10}}{25}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2016329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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factoring $(x^6 - y^6)$: what is going on here? I apologize if this question already exists, but it was quite difficult to word.
In my math class today, we learned how to factor a difference of two perfect cubes. One of our practice questions was:
factor $x^6 - 64$
almost everyone other than me (including my teacher, a... | It's a very common mistake (*) to think that, since $t^2+4t+16$ is irreducible over the reals, also the polynomial obtained with $t=x^2$ is irreducible.
Over the reals, every polynomial of degree $>2$ is reducible, so also $x^4+4x^2+16$ is reducible.
The trick here is to push in a difference of squares: if you add and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2017003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Expansion formula for $\tan$ We know that we can easily write the expansion of $\sin$ and $\cos$. But writing the expansion of $\tan$ becomes very difficult as coefficients of the powers of $x$ can't be found out easily. Is there a simple way of finding these coefficients, say a general formula for writing up to any nu... | If you are only concerned about the answer:
$$tanx=x+\frac{1}{3}x^3+\frac{2}{15}x^5+\frac{17}{315}x^7+\frac{62}{2835}x^9+... $$
As to the method of deriving using the expansions of $sin(x)$ and $cos(x)$, use $$cos(x) \times tan(x) = sin(x)$$
$$(1-\frac{x^2}{2!}+\frac{x^4}{4!}...) \times (a+bx+cx^2+dx^3+ex^4+fx^5...) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2017126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$f(z)=\frac{(iz+2)}{(4z+i)}$ maps the real axis in the $\mathbb{C}$-plane into a circle Find the center and radius of the circle. Also find the points on the complex plane which is mapped onto the center of the circle.
| For a real $x$, we have
$$f(x) = \frac{9x}{16x^2+1} + \frac{2(2x^2-1)}{16x^2+1}i $$
When $x=0$, we have $f(0) = -2i$ and when $x \rightarrow \infty$, $f(x) \rightarrow \frac{1}{4}i$. Clearly, changing $x$ to $-x$ does not change the imaginary part of $f(x)$ and hence $f(\mathbb{R})$ is symmetrical about the imaginary ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2018173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving this sequence ( probably using induction? ) Question:
For $n>0$ and $\theta\in\Re$, Let $$t_n = \sum_{k=1}^n (-1)^k 3^{k-1} cos^3{\frac{\theta}{3^k}}$$
Using the identity $cos^3\phi=\frac{1}{4}cos3\phi+\frac{3}{4}cos\phi$,
Show that $$t_n = \frac{(-1)^n 3^n}{4}cos\frac{\theta}{3^n} - \frac{1}{4}cos\theta$$ and... | Hint. One may observe that, for $k=1,2,3\cdots,$ we have (why?)
$$
\frac{(-1)^k 3^k}{4}\cdot\cos\frac{\theta}{3^k}-\frac{(-1)^{k-1} 3^{k-1}}{4}\cdot\cos\frac{\theta}{3^{k-1}}=(-1)^k 3^{k-1} \cdot\cos^3{\frac{\theta}{3^k}}
$$ then summing from $k=1$ to $k=n$ one may recognize a telescoping sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2018249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Linear differential equations doubt There is solution below with the question, I have doubt in the last step.
Solve $$\frac{d^2 y}{dx^2} + 2y = x^2 e^{3x} - \cos 2x + $e^x$$
Solution: Auxillary equation is $D^2 + 2 = 0$
$$D=\pm i\sqrt{2}$$
Complementary function is $e^{0x}(C_1 \cos \sqrt{2} x + C_2 \sin \sqrt{2}x)$
Pa... | $$
u=(D^2+2)^{-1}(x^2·e^{ 3x})
$$
is just a fancy way to say that $u$ is a solution of
$$
u''+2u=x^2·e^{ 3x}
$$
which does not tell you much. You could then use a trial solution with undetermined coefficients $$u=(A+Bx+Cx^2)e^{3x}$$ to get
$$
u''+2u=e^{3x}(9(A+Bx+Cx^2)+6(B+2Cx)+2C+2(A+Bx+Cx^2))\\
=(11A+6B+2C+(11B+12C)x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2018588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Minimal perimeter The problem is: Find the angle to OX axis, at
which a line should be drawn through the point A
(a,b) (a>0, b>0), so, that triangle, formed by this
line and positive coordinate semi-axes had the
minimal perimeter. I.e. the triangle vertices are (0,0), and two intersection points of line passing through... | I got the same function $f(\varphi)$ with $0<\varphi<\frac{\pi}{2}$, so I tried to derivate and to solve the equation:
\begin{equation}
f'(\varphi)=\left(a\frac{\sin\varphi}{\cos^2\varphi}-b\frac{\cos\varphi}{\sin^2\varphi}\right)(1+\cos\varphi+\sin\varphi)+\left(\frac{a}{\cos\varphi}+\frac{b}{\sin\varphi}\right)(-\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Showing $64\le\left(1+\frac1x\right)\left(1+\frac1y\right)\left(1+\frac1z\right)$
Showing $64\le\left(1+\frac1x\right)\left(1+\frac1y\right)\left(1+\frac1z\right)$ subject to $x,y,z>0$ and $x+y+z=1$
This inequality is equivalent to;
$\left(\frac{x+1}{4}\cdot\frac1x\right)\left(\frac{y+1}{4}\cdot\frac1y\right)\left(\f... | By Holder and AM-GM $\prod\limits_{cyc}\left(1+\frac{1}{x}\right)\geq\left(1+\frac{1}{\sqrt[3]{xyz}}\right)^3\geq\left(1+\frac{1}{\frac{1}{3}}\right)^3=64$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Please help me with this limit $$
\lim_{h\to 0} {\frac {\sqrt[3]{x+h}-\sqrt[3]{x}} h}
$$
I tried with the conjugation in the top and under but i still have the determiantion the result must be 1/3 i did it with Matlab, you CAN'T use Lhopital
| Just multiply both side to $ \left( \sqrt [ 3 ]{ { \left( x+h \right) }^{ 2 } } +\sqrt [ 3 ]{ x\left( x+h \right) } +\sqrt [ 3 ]{ { x }^{ 2 } } \right) $ to get announced result $$\lim _{ h\to 0 }{ \frac { \sqrt [ 3 ]{ x+h } -\sqrt [ 3 ]{ x } }{ h } } =\lim _{ h\to 0 }{ \frac { \left( \sqrt [ 3 ]{ x+h } -\sqrt [ 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2024652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the $n$ derivative of $y= e^{2x}\sin^2 x$ We have
\begin{align*}
y&= e^{2x}\sin^2 x\\
&= e^{2x}\left(\frac{1-\cos 2x}{2}\right)\\
&= \frac{e^{2x}}{2} - \frac{e^{2x}\cos 2x}{2}
\end{align*}
Then
\begin{align*}
y^{(n)} &= \left(\frac{e^{2x}}{2}\right)^{(n)} - \left(\frac{e^{2x}\cos 2x}{2}\right)^{(n)}\\
&= 2^{n-1}e... | Calculate the fist few derivatives and see if a recognizable pattern shows up.
$f(x) = \frac {e^{2x}}{2} + \frac {e^{2x}\cos2x}{2}\\
f'(x) = e^{2x} + e^{2x} \cos 2x - e^{2x}\sin 2x\\
f''(x) = 2e^{2x} - 2e^{2x}\cos 2x - 2e^{2x}\sin 2x - 2e^{2x}\sin 2x - 2e^{2x} \cos 2x = 2e^{2x} - 4e^{2x}\sin 2x\\
f'''(x) = 4e^{2x} - 8 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2024756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Taylor expansion for $\frac{1}{x^2 + 2x + 2}$ I'm trying to find the Taylor expansion of the function:
$$ f(x) = \frac{1}{x^2 + 2x + 2} $$
about the point $ x = 0 $. I have worked out the terms up to the fourth derivative, which was very tedious. I found:
$$ f(x) = \frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 + 0 x^3 ... | You can write
$$
f(x) = \frac{1}{2}\cdot\frac{1}{1+(x+\frac{x^2}{2})}.
$$
Setting $\stackrel{\rm def}{=} x+\frac{x^2}{2}\xrightarrow[x\to0]{}0$, you can use the Taylor expansion of $\frac{1}{1+u} = \sum_{n=0}^{\infty} (-1)^n u^n$ around $0$ to get
$$\begin{align}
f(x) &= \frac{1}{2}\left(1-u+u^2-u^3+u^4+o(u^4)\right)\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 2
} |
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