Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solution to $\frac{d}{d\frac{1}{x}} x$ If I want to solve
$$\frac{d}{d\frac{1}{x}} x$$
is my approach correct?
As
$$\begin{align*} \frac{d}{d\frac{1}{x}}x&=\\
\text{with }\frac{1}{x}&=y\\
\frac{d}{dy}\frac{1}{y}&=-\frac{1}{y^2}\\
&=-\frac{1}{\left(\frac{1}{x}\right)^2}\\
&=-x^2
\end{align*}$$
Is this approach correct,... | $\frac{\mathrm{d} x}{\mathrm{d} \frac{1}{x}}=\left (\frac{\mathrm{d} \frac{1}{x}}{\mathrm{d} x} \right )^{-1}=\left ( \frac{-1}{x^{2}} \right )^{-1}=-x^{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1645747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
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Divergence of squared sum of Chebyshev Polynomials $\equiv L+R$ has empty point spectrum The Chebyshev Polynomials of the second kind $U_n$ are the solutions of the differential equation
$$(1-x^2)U_n''(x)-3xU_n'(x)+n(n+2)U_n(x)=0$$
Alternatively they are defined inductively:
$$U_0(x)=1 \qquad U_1(x)=2x \qquad U_{n+1}(x... | if $a_n = 0$ for $n < 0$, $a_0 = 1$ and $a_{n+1} = 2 x a_n - a_{n-1} $ for $n > 1$ then by multiplying by $z^{-n}$ and summing we get the Z-transform :
$$\sum_{n=0}^\infty (a_{n+1} - 2 x a_n + a_{n-1}) z^{-n} = \sum_{n=0}^\infty \delta(n) z^{-n} = 1$$
and we have also $\sum_{n=0}^\infty (a_{n+1} - 2 x a_n + a_{n-1}) z^... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Prove that one of the lines represented by $ax^2+2hxy+by^2=0$ will bisect the angle between the coordinate axes if $(a+b)^2=4h^2$. Prove that one of the lines represented by $ax^2+2hxy+by^2=0$ will bisect the angle between the coordinate axes if $(a+b)^2=4h^2$.
Solution
I calculated the two lines represented by $ax^2+... | If one of the line of
$$ax^2 +2hxy+by^2=0 \tag{1}$$
bisects the coordinate axes then the eqn. of that line might be $y=x$ OR, $y =-x$.
Combining these eqns we get $y=\pm x$ and substituting this value of $y$ in eqn. (1) we get
\begin{eqnarray}
ax^2+2hx(\pm x)+by^2=0 \\
ax^2±2hx^2+by^2=0 \\
x^2(a±2h+b)=0 \\
a±2h+b=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 5
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Prove that $a+\frac{1}{b}>2$ or $b+\frac{1}{a}>2$ for two strict positive numbers Another Olympiad Problem, let $x$ and $a$ and $b$ be strictly real positive numbers.
*
*Prove that $x$+$\frac{1}{x}$$>$$2$ (proven)
*Than conclude that $a$+$\frac{1}{b}$$>$$2$ or
$b$+$\frac{1}{a}$$>$$2$
For the second ques... | Slightly different.
If $a = b$ then $a + 1/b = b + 1/a =a + 1/a \ge 2$.
If $a > b$ then $a + 1/b > a + 1/a \ge 2$.
If $a < b$ then $b + 1/a > b + 1/b \ge 2$.
(If $a = b = 1$ then $a + 1/b = 2$. Otherwise $a + 1/b > 2$ or $b + 1/a > 2$ or both.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1649513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Solve equation $\frac{1}{x}+\frac{1}{y}=\frac{2}{101}$ in naturals My try was $$\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{2}{101}\\x+y=2k,xy=101k\\x=2k-y\\y(2k-y)=101k\\2ky-y^2=101k\\y^2-2ky+101k=0\\y=k+\sqrt{k^2-101k}\\x=k-\sqrt{k^2-101k}$$
Now $\sqrt{k^2-101k}$ has to be either integer or rational,if it's an integ... | Sneaky hint: Think about $\frac{1}{a}+\frac{1}{a(2a-1)}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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A variant of Viète's formula (the 2's replaced by 3's) I am wondering whether there exists an easy way to evaluate the following infinite product :
$\sqrt{\frac{1}{3}}.\sqrt{\frac{1}{3}+\frac{1}{3}.\sqrt{\frac{1}{3}}}.\sqrt{\frac{1}{3}+\frac{1}{3}.\sqrt{\frac{1}{3}+\frac{1}{3}.\sqrt{\frac{1}{3}}}}...$
or even better (a... | Where does this problem come from? Anyway, for $a>0$, let $u_0^{(a)} = 0$ and $u_{n+1}^{(a)} = \sqrt{\frac{1+u_n^{(a)}}{a}} = \sqrt{\frac{2}{a}}\sqrt{\frac{1+u_n^{(a)}}{2}}$. We wish to determine $\prod_{n=1}^{\infty}u_n^{(a)}$. Repeating the trigonometric substitutions in your post, we find
$$\prod_{n=1}^{\infty}u_n^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there any value for $x$ that would make the statement $(x+3)^3 = x^3+3^3$ true?
Is there any value for $x$ that would make the statement $(x+3)^3 = x^3+3^3$ true?
I understand that when factored out, you have $(x+3)^3 = x^3+9x^2+27x+27$ as opposed to the other side which is $x^3+27$, so even if you plug in $0$ the... | $(x+3)^3 = x^3+3^3$
$\Leftrightarrow x^3+9x^2+27x+27 = x^3+27$
$\Leftrightarrow 9x^2+27x = 0$
$\Leftrightarrow x^2+3x = 0$
$\Leftrightarrow x(x+3)=0$
$\Leftrightarrow x=0$ or $x=-3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\sin A+\sin B+\sin C=\cos A+\cos B+\cos C=0$,
If $$\sin A+\sin B+\sin C=\cos A+\cos B+\cos C=0,$$ prove that
$$\cos 3A+\cos 3B+\cos 3C=3\cos(A+B+C).$$
My solution:
From the given,
$$\cos^3A+\cos^3B+\cos^3C=3\cos A\cos B\cos C$$
Now,
L.H.S$$=\cos3A+\cos3B+\cos3C$$
$$=4\cos^3A-3\cos A+4\cos^3B-3\cos B+4\cos^3C... | we have $$sinA+sinB=-sinC$$ and $$cosA+cosB=-cosC$$ squaring and adding both we get
$$2+2cos(A-B)=1$$ that is
$$cos(A-B)=\frac{-1}{2}$$ $\implies$
$$A-B=\frac{2\pi}{3}$$ similarly
$$B-C=\frac{-4\pi}{3}$$ and
$$C-A=\frac{2\pi}{3}$$
Now $$cos3A=cos\left(3\left(B+\frac{2\pi}{3}\right)\right)=cos(3B+2\pi)=cos3B$$
similarly... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What is wrong with this infinite sum We know that:
https://www.youtube.com/watch?v=w-I6XTVZXww
$$S=1+2+3+4+\cdots = -\frac{1}{12}$$
So multiplying each terms in the left hand side by $2$ gives:
$$2S =2+4+6+8+\cdots = -\frac{1}{6}$$
This is the sum of the even numbers
Furthermore, we can add it to itself but shifting th... | As pointed out in section 3, page 1191 of this article, the rules for manipulating divergent series are more restrictive than those for convergent series. As pointed out in the article, to avoid problems you should work with the power series obtained by multiplying the nth term by $x^n$, instead.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1653416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Total number of $4$ digit numbers whose product of digits is $72$
Total number of $4$ digit numbers whose product of digits is $72$
$\bf{My\; Try::}$ Here the possible factor of $72 = 2^{3}\cdot 3^2$
Now here we divide $2^3\cdot 3^3$ into product of $4$ numbers
Like $1\cdot 1\cdot 8\cdot 9$
and $1\cdot 2\cdot 4\cdot ... | $1\cdot 1\cdot 8\cdot 9$
and $1\cdot 2\cdot 4\cdot 9$
and $1\cdot 2\cdot 6\cdot 6$
and $1\cdot 3\cdot 4\cdot 6$
and $1\cdot 3\cdot 3\cdot 8$
and $2\cdot 2\cdot 3\cdot 6$
and $2\cdot 2 \cdot 2\cdot 9$
and $2\cdot 3 \cdot 3\cdot 4$
So arrangement of these numbers is $$\frac{4!}{2!}+4!+\frac{4!}{2!}+4!+\frac{4!}{2!}+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1654265",
"timestamp": "2023-03-29T00:00:00",
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Third degree polynomial for $ \sin(x^2+y^2)?$ Let R be an open region containing the point $(x_0,y_0).$ Let f, g, and h be functions defined on R, except possibly at$ (x_0,y_0).$ Suppose that for every $(x,y) \in R$ not equal to $(x_0,y_0)$, we have:
$g(x,y) \leq f(x,y) \leq h(x,y)$.
If
$\lim_{(x,y)\to(x_0,y_0)} g(x,y... | Treat $x^2+y^2$ as a single variable. Recall that
$$z-\frac16z^3\le \sin(z)\le z$$
for $0<z<1$. Then, let $z=x^2+y^2$ reveals
$$(x^2+y^2)-\frac16(x^2+y^2)^3\le \sin(x^2+y^2)\le (x^2+y^2)$$
Dividing by $x^2+y^2$ yields
$$1-\frac16(x^2+y^2)^2\le \frac{\sin(x^2+y^2)}{x^2+y^2}\le 1$$
whereupon applying the squeeze theo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1656273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integration with double absolute value How to prove that?
$$
\int\left|\sqrt{|x|}\right|\ \mathrm{d}x
= \frac{1}{3} x
\left(\left(\sqrt{x} - \sqrt{-x}\right)\mathrm{sgn}(x)
+ \sqrt{x} + \sqrt{-x}\right)
$$
I cannot understand this result.
Where does the $1/3$ factor come out from?
| I find the outer $| \,|$ incomprehensible since $\sqrt{|x|}\ge0$ according to the usual definition of square root. I will answer ignoring that.
For $x\ge 0$ we have the usual result $\int x^{1/2}\,dx={2\over 3}x^{3/2}$ plus constant, which you can verify by taking the derivative of ${2\over 3}x^{3/2}$.
Suppose $x\ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1657677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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taylor expansion and limit of a series?? $f\left(x\right)=∫_0^x\tan ^{-1}tdt$
what is the taylor expansion about the origin of this function?
and how do i use this to get the limit of the series
$1-\frac{1}{2}-\frac {1}{3}+\frac {1}{4}+\frac {1}{5}-\frac{1}{6}-\frac {1}{7}.......$
i could get the limit by using concep... | The Taylor expansion can be obtained by observing
$$ f(x) = \int_0^x\arctan(t)dt=x\arctan x - \frac{1}{2}\ln(1+x^2)$$
Using the Mercator series we have
$$\ln(1+x^2) = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}x^{2k}$$
and for the first summand there is the well-know expansion
$$x\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}x^{2k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of $\frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56} + \frac{1}{72} + \frac{1}{90}$ Find the value of $p+q$, where $p$ and $q$ are two positive integers such that $p$ and $q$ have no common factor larger than $1$ and
$$\frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56} + \frac{1}{72} + \... | $1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90 = $
$1/4*5 + 1/5*6 + 1/6*7 + 1/7*8 + 1/8*9 + 1/9*10 = $
Note: $\frac{1}{n(n+1)} + \frac{1}{(n +1)(n +2)} = \frac{n+2 + n}{n(n+1)(n+2} = \frac {2n + 1}{n(n+1)(n+2)} = \frac{2}{n(n+2)}$
And $\frac{k}{n(n+k)} + \frac{1}{(n+k)(n + k + 1)} = \frac{k(n + k + 1) + n}{n(n+k)(n + k + 1)... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Value of $a^3+b^3+c^3$ when values of $a+b+c$, $abc$ and $ab+bc+ca$ are known. Is there a way to to find out what $a^3+b^3+c^3$ evaluates to, when the values of $abc$, $ab+bc+ca$ and $a+b+c$ are given?
Alternatively, is there a way to express $a^3+b^3+c^3$ in terms of the aforementioned expressions?
| Use Newton's identities:
\begin{align}
p_1 &= e_1\\
p_2 &= e_1p_1-2e_2\\
p_3 &= e_1p_2 - e_2p_1 + 3e_3 \\
\end{align}
where
\begin{align}
e_1 &= a+b+c &\qquad p_1 &= a^1+b^1+c^1\\
e_2 &= ab+bc+ca &\qquad p_2 &= a^2+b^2+c^2\\
e_3 &=abc &\qquad p_3 &= a^3+b^3+c^3\\
\end{align}
Newton's identities giv... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What's $\alpha+\beta$ if we have: $\alpha^3-6\alpha^2+13\alpha=1$ and $\beta^3-6\beta^2+13\beta=19$ ($\alpha$ and $\beta$ are Real) What's $\alpha+\beta$ if we have $\alpha^3-6\alpha^2+13\alpha=1$ and $\beta^3-6\beta^2+13\beta=19$? Here $\alpha$ and $\beta$ are real.
Firstly, I subtracted the two equations and got the... | Let $f(x)=x^3-6x^2+13x,$ Then $f'(x)=3x^2-12x+13 = 3[(x-2)^2]+1>0$
So $f(x)$ is strictly increasing function.
Now above we have given $f(\alpha)=1$ and $f(\beta) = 19\;,$ Where $\alpha<\beta$
Then $f(\alpha)<f(\beta)$, because the function is increasing.
Now \begin{align*}f(4-x)&=(4-x)^3-6(4-x)^2+13(4-x)\\&=64-x^3-48x+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to rigorously prove the following continued fraction identity? The following identity seems to be satisfied for any $z \in C$
$$\cfrac{z^2}{1+z^2-\cfrac{z^2}{1+z^2-\cfrac{z^2}{1+z^2-\cfrac{z^2}{1+z^2-\cdots}}}}=\begin{cases}1 & |z| \geq 1\\z^2 & |z| \leq 1\end{cases}$$
I checked it numerically for real and for comp... | The convergents of this continued fraction are easy to compute explicitly:$$\underbrace{\cfrac{z^2}{1+z^2-\cfrac{z^2}{1+z^2-\cfrac{\ddots}{\ddots-\cfrac{z^2}{1+z^2}}}}}_{n\text{ numerators/denominators}}=\frac{z^2(1-z^{2n})}{1-z^{2n+2}}$$ (understood in the limit sense if $z^{2n+2}=1$, and proven by induction). Now tak... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the value of $\lim _{x\to 0}\left\lfloor\frac{\tan x \sin x}{x^2}\right\rfloor$ How do I evaluate the limit
$$\lim _{x\to 0}\left\lfloor\frac{\tan x \sin x}{x^2}\right\rfloor$$
where $\lfloor\cdot\rfloor$ denotes greatest integer function.
I know that $x>\sin x$ and $x < \tan x$ but how do I use these results ... | Using Taylor–Young expansions: we know that
$$\tan(x)=x+\frac{x^3}3+o(x^4),\qquad\sin(x)=x-\frac{x^3}6+o(x^4),$$
hence
$$\tan(x)\sin(x)=x^2+\frac{x^4}6+o(x^5),$$
and hence
$$\frac{\tan(x)\sin(x)}{x^2}=1+\frac{x^2}6+o(x^3).$$
From here, we conclude that there exists a punctured neighborhood $V$ of $0$ such that
$$\foral... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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integrate $\int_0^{\frac{\pi}{4}}\frac{dx}{2+\tan x}$
$$\int^{\frac{\pi}{4}}_{0}\frac{dx}{2+\tan x}$$
$v=\tan(\frac{x}{2})$
$\tan x=\frac{2v}{1-v^2}$
$dx=\frac{2\,dv}{1+v^2}$
$$\int^{\frac{\pi}{4}}_0 \frac{dx}{2+\tan x}=\int^{\frac{\pi}{8}}_0 \frac{\frac{2\,dv}{1+v^2}}{2+\frac{2v}{1-v^2}}=\int^{\frac{\pi}{8}}_0 \frac... | You have made a mistake in simplifying the integrand. You should have
$$\int \frac{\frac{2dv}{1+v^2}}{2+\frac{2v}{1-v^2}} = \int \frac{1-v^2}{(1+v^2)(1+v-v^2)}dv.$$
Now do partial practions. You get
$$ \frac{1-v^2}{(1+v^2)(1+v-v^2)} = \frac{2v-1}{5(v^2-v-1)} -\frac{2v-2}{5(v^2+1)} .$$
The first integral is a U-substitu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
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Prove that $(x+y+z)^2(yz+zx+xy)^2 \leq 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2)$ This is Problem 6 of the 2007 Indian National Math Olympiad (INMO).
If $x, y, z$ are positive real numbers, prove that
$(x+y+z)^2(yz+zx+xy)^2 \leq 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2).$
My best idea was to expand this and simplify. Althoug... | Let I = 3(x² + xy + y²)(y² + yz + z²)(z² + zx + x²)
Then, By Holder's Inequality, it follows that
I >= 81x²y²z²
Thus, we need to prove that
81x²y²z² >= (x + y + z)²(xy + yz + zx)²
Or 9xyz >= (x + y + z)(xy + yz + zx)
Expanding R.H.S. gives
9xyz >= xy(x + y) + yz(y + z) + zx(z + x) + 3xyz
Or xy(x + y) + yz(y + z) + zx(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Algebraic proof by contradiction I have the following problem:
For the following predicate from domain of all positive real numbers
$$ \frac{}{x+1} < \frac{+1}{x+2}$$
a. Write down the predicate logic statement for the theory above
b. Negate your predicate logic statement in (a)
c. Show that negated predicate statemen... | Assume that $x$ is positive and:
$$
\frac{x}{x+1} \geq \frac{x + 1}{x + 2}
$$
This leads to:
\begin{align}
\frac{x}{x+1} \geq&\ \frac{x + 1}{x + 2}\\
x(x + 2) \geq&\ (x + 1)^2 \\
x^2 + 2x \geq&\ x^2 + 2x + 1 \\
0 \geq 1
\end{align}
This is universally false meaning that the original statement, $\frac{x}{x+1} < \frac{x ... | {
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"question_score": "1",
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$x^4 -10x^2 +1 $ is irreducible over $\mathbb Q$ I have seen the thread Show that $x^4-10x^2+1$ is irreducible over $\mathbb{Q}$ but this didn't really have a full solution.
Is it true that if it is reducible then it can be factored into a linear factor or quadratic factor in the form $x^2 - a$. Is $a$ in the rationals... | First note that it suffices to show the polynomial is irreducible over $\mathbb{Z}$. We then note that $p(x)=x^4-10x^2-1$ splits either as
$$(x-a)(x^3+bx^2+cx+d)$$
or as
$$(x^2+bx+c)(x^2+dx+e)$$
In the first case $a$ would be a root of $p$, but by the rational root theorem the polynomial has no roots in $\mathbb{Q}$ (a... | {
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"url": "https://math.stackexchange.com/questions/1665435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 4
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Calculate the limit $\lim_{x \to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$ Calculate the limit
$$\lim_{x \to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$$
I tried to factorise and to simplify, but I can't find anything good.
$$\lim_{x \to 2} \frac{\frac{x^2(x+2)-8\sqrt{x+2}}{\sqrt{x+2}}}{(4-x^2)}$$
| Since the OP does not know L'Hospital's rule (yet), this approach is probably not the best way. When you multiply top and bottom by $x^2\sqrt{x+2}+8$ you get the fraction: $\frac{x^5+2x^4-64}{(4-x^2)(x^2\sqrt{x+2}+8)}$ Both NUM and DENOM have a factor $x-2$ In case of the NUM, you could perform long division to arrive ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Solving $\int {\frac{2 \sin(x) \cos(x)}{\sin^4(x) + \cos^4(x)}}\, dx$ See this link here. I am having trouble with this one as well. Any hints or help appreciated.
$$\int{\frac{2\sin(x)\cos(x)}{\sin^4(x)+\cos^4(x)}}\,dx$$
| Hint
For the denominator
$$\sin^4(x)+\cos^4(x)=(\sin^2(x)+\cos^2(x))^2-2\sin^2(x)\cos^2(x)=1-\frac 12\sin^2(2x)=\frac 12(1+\cos^2(2x))$$
For the numerator $$2\sin(x)\cos(x)=\sin(2x)$$
So $$\int {\frac{2 \sin(x) \cos(x)}{\sin^4(x) + \cos^4(x)}}\, dx=\int \frac{2\sin(2x)}{1+\cos^2(2x)}\,dx$$
Changing variable $t=\cos(2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\angle{BEC}=\angle{DAC}$. A convex quadrilateral $\quad{ABCD}$ has $AD=CD$ and$\angle{DAB}=\angle{ABC}<90$. The line through $D$ and the midpoint of $BC$ meets line $AB$ in $E$. Prove that $\angle{BEC}=\angle{DAC}$.
I have to approaches: Either we can prove that $EB.EA=EC^2$ or we can prove that line $AD$ i... |
Let $F$ be the intersection of line $AD$ and line $BC$. Since $\angle BDA = \angle ABC$, $AF=FB$.
Let $O$ be the center of the circumcircle of $\triangle ABC$ and let $G$, $I$ be the projections of $O$ onto $AF$ and $CD$. Then
*
*$OG = OM$ because $O$ and $F$ lie on the perpendicular bisector of $AB$.
*$OG = OI$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1668438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
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Prove convergence of trignometric sum from $-\infty$ to $\infty$
Prove convergence of $$\pi^2\sum_{n=-\infty}^{\infty}\left ( \frac{1}{\cos^2\left (\pi\left (n-\frac{1}{2}\right )\tau \right )}-\frac{1}{\sin^2\left (\pi\left (n-\frac{1}{2}\right )\tau\right )}\right )$$ and $$\pi^2\sum_{n=-\infty}^{\infty}\left ( \fra... | (1) How $e^{|n|\pi\text{Im}(\tau)}$ relates to $\cos$ and $\sin$ :
For simplicity we first argue about $$\sum_{n=1}^\infty \frac{1}{\cos^2 \left (\pi\left (n-\frac{1}{2}\right )\tau \right )}.$$
Since $$\left|e^{i\pi\left(n-\frac{1}{2}\right)\tau}\right|=
e^{-\pi\left(n-\frac{1}{2}\right)\sigma}
\quad\text{and}\quad
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1668605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\frac {1 + a + a^2 + \ldots +a^{n - 1} }{n} < \frac {1 + a + a^2 + \ldots +a^{n - 1} + a^n}{n + 1}$ if $1 < a$
Prove $\frac {1 + a + a^2 + \ldots +a^{n - 1} }{n} < \frac {1 + a + a^2 + \ldots +a^{n - 1} + a^n}{n + 1}$ if $1 < a$
Tried induction. Not sure where my mistake is, but what I did doesn't seem to make... | You have a more general result:
If $\{a_n \}$ is strictly increasing then also the sequence of the avarages over the first $N$ terms is strictly increasing.
In fact we have
$$
S_N=\frac 1 N \sum_{k=1}^N a_k<\frac 1 N \sum_{k=1}^N a_N=a_N<a_{N+1}
$$
and therefore
$$
S_{N+1}=\frac {\sum_{k=1}^N a_k+ {a_{N+1}}} {N+1}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Calculating $\int_0^{\pi/2} \sqrt{\cot x} + \sqrt{\cos x} dx$ How should I solve the following integral:
$$\int_0^{\pi/2} (\sqrt{\cot x} + \sqrt{\cos x} )\,\mathrm dx$$
| I think can use
\begin{align}I&= \int_0^{\frac{\pi}{2}}\left(\sqrt{\cot (x)}+\sqrt{\cos (x)}\right)\,\mathrm dx\\&=\int _0^1\frac{1}{\cos(x)}\left(\sqrt{\cot (x)}+\sqrt{\cos (x)}\right)\, \mathrm d(\sin(x))\\&=\int _0^1\frac{1}{\sqrt{1-a^2}}\left(\sqrt{\frac{\sqrt{1-a^2}}{a}}+\sqrt{\sqrt{1-a^2}}\right)\,\mathrm da\\ &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate this integral $\int_{0}^{\infty }\frac{\ln\left ( 1+x^{3} \right )}{1+x^{2}}\mathrm{d}x$ How to evaluate this integral
$$\mathcal{I}=\int_{0}^{\infty }\frac{\ln\left ( 1+x^{3} \right )}{1+x^{2}}\mathrm{d}x$$
Mathematica gave me the answer below
$$\mathcal{I}=\frac{\pi }{4}\ln 2+\frac{2}{3}\pi \ln\left (... | Lemma 1::$$\int_{0}^{\infty}\dfrac{\ln{(x^2-x+1)}}{x^2+1}=\dfrac{2\pi}{3}\ln{(2+\sqrt{3})}-\dfrac{4}{3}G$$
Use this well known
$$\int_{0}^{+\infty}\dfrac{\ln{(x^2+2\sin{a}\cdot x+1)}}{1+x^2}dx=\pi\ln{2\cos{\dfrac{a}{2}}}+a\ln{|\tan{\dfrac{a}{2}}|}+2\sum_{k=0}^{+\infty}\dfrac{\sin{(2k+1)a}}{(2k+1)^2}$$
this indentity pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
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Can we prove $\sum_{i=1}^{k} \frac{\binom{n-2}{i}}{\binom{n}{i}} \le \frac{1}{(n-1)e}$ and arrive at an upper bound for $k$? Does this inequality hold true?
Given $\frac{n}{2} \ge k$,
$$\sum_{i=1}^{k} \frac{\binom{n-2}{i}}{\binom{n}{i}} \le \frac{1}{(n-1)e}$$
where $e = 2.71828\dots$
I have reduced this to
$$\sum_{i=1... | Without restrictions on $n$ (and/or $k$), this will not be true. In particular, for $k=n-1$ you get
$$
\sum_{i=1}^{n-1} \frac{\binom{n-2}{i}}{\binom{n}{i}} = \frac{n-2}{3}.
$$
and for $k=\frac{n}{2}$ something that behaves asymptotically as
$$
\sum_{i=1}^{n-1} \frac{\binom{n-2}{i}}{\binom{n}{i}} = \frac{7}{6} n + o(n).... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1670291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $a_n=2 a_{n-1} - a_{n-2} + 2^n$ using generating function I'm preparing to an exam and trying to solve $a_n=2 a_{n-1} - a_{n-2} + 2^n$, where $a_0=0$ and $a_1=1$.
This is my approach:
Let $A(z)=\sum_{n \geq 0} a_{n+2} z^{n+2}$, then:
$$\sum a_{n+2} z^{n+2} = 2 \sum a_{n+1} z^{n+2} - \sum a_n z^{n+2} + \sum (2z)^... | While the generating function approach works systematically, you can solve this example by hand directly without too much trouble. First I'll define $a_{-1}=1$ (so that the recurrence holds down to $n=1$) and define $b_n=a_n-a_{n-1}$. Then your recurrence becomes $b_n=b_{n-1}+2^n$ for all positive $n$, and from the fir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1672276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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How to find the limit of this hard sequence? How to find this limits
$$\lim_{n \to \infty}\sum_{i=1}^{n}\frac{n^2-(i-1)^2}{n\left\{[2n(a+b)]^2+[a(2i-1)]^2\right\}}$$
Where $a$ and $b$ are real positive constants.
My friend asked this question,but I have no idea how to do it.
| I'm sorry I post this as an answer when it is only a lower bound, but I do not have enough reputation to comment.
First note that, after the swap $i \to n-i+1$ the sum is
$$
\frac{1}{n} \sum_{i=1}^n \frac{i(2n-i)}{4n^2(a+b)^2+a^2(2n+1-2i)^2}
$$
By AM-GM, the $n$-th term is greater than or equal to
\begin{align*}
\left ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1673252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $d$ when $(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x$ There exist constants $a$, $b$, $c$, and $d$ such that
$(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x$
for all angles $x$. Find $d$.
| $$\begin{align} & \sin^7x \\ &
=\sin^6x\cdot \sin x \\ &
=\frac{1}{16}\cdot 16\sin^6x\cdot \sin x \\ &
=\frac{1}{16}\cdot (4\sin^3x)^2 \cdot \sin x \\ &
=\frac{1}{16}\cdot (3\sin x-\sin 3x)^2 \cdot \sin x \\ &
=\frac{1}{16}\cdot (9\sin^2 x-6\sin x\sin 3x+\sin^2 3x) \cdot \sin x \\ &
=\frac{1}{16}\cdot [\frac{9}{2}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Solving system of three quadratic equations $$\begin{cases}
x^2 = yz + 1 \\
y^2 = xz + 2 \\
z^2 = xy + 4
\end{cases}
$$
How to solve above system of equations in real numbers? I have multiplied all the equations by 2 and added them, then got $(x - y)^2 + (y - z)^2 + (x - z)^2 = 14$, but it leads to nowhere.
| Multiplying both sides of the first equation by $z$, of the second by $x$, and of the third by $y$, we get
$$x^2z=z^2y+z, \quad y^2x=x^2z+2x,\quad z^2y=y^2x+4y.$$
Adding up and cancelling, we get
$$2x+4y+z=0.$$
Similarly,
$$x^2y=y^2z+y, \quad y^2z=z^2x+2z, \quad z^2x=x^2y+4x,$$
giving
$$4x+y+2z=0.$$
To finish, use t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Comparing Coefficients for Partial Fractions See this simple example :
$$\frac{x+1}{(x-1)(x-2)}\equiv \frac{A}{(x-1)}+\frac{B}{(x-2)}$$
Then we can get
$x+1 \equiv A(x-2)+B(x-1)$ for $x \neq 1 ,2$
My Question :
Is it correct to put $x=1$ to find $A$ ,
Is it correct to put $x=2$ to find $B$, other than comparing coeff... | Yes, it is perfectly valid to do that. Notice that while the fractions are not defined at x= 1 or x= 2, once you have multiplied by (x- 1)(x- 2) those fractions are no longer a problem. If you set x= 1, 1+ 1= 2= A(1- 2)+ B(0) or 2= -A so A= -2. If you set x= 2, you get 2+ 1= 3= A(0)+ B(2- 1) or 3= B.
Another thing y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Maximizing $\sqrt6xy+4yz$ Let $x, y, z$ be real numbers such that $x^2+y^2+z^2=1$.
Let $A$ be maximum value of $\sqrt6xy+4yz$. Find $2A^2-4$.
An initial approach was trying out the inequality $\text{RMS-AM-GM}$. Then I tried a parametric substitution: $x=\cos\alpha, y=\sin\alpha cos\beta, z=\sin\beta \sin\alpha$.
How w... | HINT
$$1=x^2+y^2+z^2=x^2+ay^2+(1-a)y^2+z^2 \ge 2\sqrt{a}xy+2\sqrt{(1-a)}yz \quad (\because \text{AM-GM})$$
$$a=\frac{3}{11}$$
Note that $2\sqrt{a}:2\sqrt{(1-a)}=\sqrt{6}:4$.
Let $2\sqrt{a}= \sqrt{6}k$, and multiply $k$ on each side.
Can you take it from here?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $n(n+1)(n+5)$ is a multiple of $6$ I need to prove that $n(n+1)(n+5)$ is divisible by 6. where $n$ is a natural number. I have used the method of induction. But not successful
I got the expression $(k^3+6k^2+5k)+3k^2+15k+12$ when $n=k+1$.
The term inside the bracket is divisible by 6 since we have assumed t... | Suppose $n$ is a multiple of 6. Then we are done. Suppose $n$ is neither a multiple of 2 or 3. Then $n+1$ must be even, and if $n+1$ is still not a multiple of 3 then $n+5$ is. Note $n+5\equiv n+2$ (mod 3) and since neither $n$ nor $n+1$ is a multiple of 3, then $n+2$ is. If $n$ is even but not a multiple of 3, we proc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 5
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Linear Factorization of Complex Polynomials I am trying to find a linear factorization of the polynomial $$p(z) = 1 +z+z^2 +z^3 +z^4 +z^5 + z^6 +z^7 +z^8$$
I know what it means by linear factorization in the sense of non-complex polynomials, but i'm not sure where to begin for a complex polynomial of degree 8. I tried ... | The polynomial factorizes, over the reals as ($z\neq 1$)
$$p(z) = 1 +z+z^2 +z^3 +z^4 +z^5 + z^6 +z^7 +z^8=\frac{1-z^9}{1-z}=(1+z+z^2)(1+z^3+z^6)$$
and its roots are the (complex) $9$-th roots of unity (except $1$ of course):
$$-(-1)^{\frac{1}{9}},(-1)^{\frac{2}{9}},-(-1)^{\frac{3}{9}}, (-1)^{\frac{4}{9}},-(-1)^{\frac{5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Multivariable partial differentiation
$f(r,\theta)=r^5\cos \theta$,$x=r\cos \theta$, $y=r\sin \theta$
Find $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial x}$ in therms of $r$ and $\theta$
I know that $$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial r}\cdot \frac{\partial r}{\partial x}+\fra... | \begin{align}
f &= r^5 \cos \theta = r^4 \cdot r \cos \theta = x \cdot \left ( x^2 + y^2\right)^2 \implies \\
f_x &= \left ( x^2 + y^2\right)^2 + x \cdot 2(x^2 + y^2) \cdot 2x = \left ( x^2 + y^2\right) \left( x^2 + y^2 + 4x^2\right) = \\
&= \left ( x^2 + y^2\right) \left ( 5x^2 + y^2\right) = r^4 \left ( 5 \cos^2 \the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1692749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Taylor series of $\ln{\sqrt[4]{\frac{x-2}{5-x}}}$ to $o((x-x_0)^n)$ when $x_0 = 3$ Well I have tried to get it as
$$f(x) = f(x_0) + \frac{f'(x_0)(x-x_0)}{1!} + \frac{f''(x_0)(x-x_0)^2}{2!} + ... + o((x-x_0)^n)$$
and got wrong results:
First: $$f'(x) = \frac{3}{4(x-2)(5-x)}$$
Second: $$f''(x) = \frac{3(2x-7)}{4(x-2)^2(... | Answer is $$-\frac{ln2}{4} + \sum^n_{k=1}{\frac{(-1)^{k-1} + 2^{-k}(x-3)^k}{4k}} + o((x-3)^n)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the laurent series for $\frac{1}{z(z-2)^2}$ centered at z=2 and specify the region in which it converges. My attempt:
$$\frac{1}{z(z-2)^2}$$
$$\frac{1}{z(z-2)^2} = \frac{A}{z}+\frac{B}{z-2}+\frac{C}{(z-2)^2}$$
$$\frac{1}{z(z-2)^2} = \frac{(1/4)}{z}+\frac{(-1/4)}{z-2}+\frac{(1/2)}{(z-2)^2}$$
This is where I get stu... | Let $\zeta = z - 2$. We are trying to find the Laurent series for $1/(\zeta+2)\zeta^2$ centred at $\zeta = 0$, so
$$ f = \frac{1}{(\zeta+2)\zeta^2} = \zeta^{-2} \frac{1}{\zeta + 2}
$$
The latter is holomorphic in a neighbourhood of $0$, so let $(a_j)_{j=0}^\infty$ such that
$$ \frac{1}{\zeta + 2} = \sum_{j=0}^\infty a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how to find the following limit Problem: Find the following limit $${\displaystyle \lim_{n\rightarrow\infty}\left[\left(1+\frac{1}{n}\right)\sin\frac{\pi}{n^{2}}+\left(1+\frac{2}{n}\right)\sin\frac{2\pi}{n^{2}}+\ldots+\left(1+\frac{n-1}{n}\right)\sin\frac{\left(n-1\right)}{n^{2}}\pi\right]}$$.
Attempt:
Observe that
\b... | Yes, you did this perfectly except for one (inconsequential) mistake. In "note that", you should have
$$ \begin{align*}
\sum_{k=1}^n \left( 1 + \dfrac{k}{n} \right) \dfrac{1}{n^3} & = \dfrac{1}{n^2} + \sum_{k=1}^n \dfrac{k}{n^4} \\
& < \dfrac{1}{n^2} + \mathbf{ \sum_{k=1}^n \dfrac{n}{n^4} } \\
& = \dfrac{1}{n^2} + \dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Rationalize a surd $\frac{1}{1+\sqrt{2}-\sqrt{3}}$ How can I rationalize the following surd
$$\frac{1}{1+\sqrt{2}-\sqrt{3}}$$
What would be the conjugate of the denominator
| Rationalise twice, because after rationalising once , there would still remain a surd in the denominator.
First multiply and divide by $1+\sqrt{2}+ \sqrt{3}$
$$\frac{1}{1+\sqrt{2}-\sqrt{3}}\times\frac{1+\sqrt{2}+\sqrt{3}}{1+\sqrt{2}+\sqrt{3}}$$
$$= \frac{1+\sqrt{2}+\sqrt{3}}{1+2\sqrt{2}+2-3}=\frac{1+\sqrt{2}+\sqrt{3}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Real part of $(1+2i)^n$ Is it true that for all $n\in \mathbb{N}$, $n\ge 2$ we have
$$|\textrm{Re}((1+2i)^n)|>1?$$
I do know de Moivre's Theorem.
I do not know how to show that $|\sqrt{5}^n\cos(n\arccos\left ( \frac{1}{5} \right ))|>1$ because the value $\cos(n\arccos\left ( \frac{1}{5} \right ))$ can become (theoreti... | Recursion for $\boldsymbol{r_n=\mathrm{Re}\!\left((1+2i)^n\right)}$
Both $1+2i$ and $1-2i$ satisfy the equation $z^2-2z+5=0$. Therefore, the real part of $(1+2i)^n$
$$
r_n=\frac{(1+2i)^n+(1-2i)^n}2
$$
satisfies
$$
r_n=2r_{n-1}-5r_{n-2}
$$
mod $\boldsymbol{4}$
Since $r_0=r_1=1$, we have that $r_n\equiv1\pmod{4}$ for al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
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Prove that the value of $(abc)-(ab+bc+ca)+3(a+b+c)$ is $0$ If the points $\big(\frac{a^3}{a-1}, \frac{a^2-3}{a-1}),(\frac{b^3}{b-1}, \frac{b^2-3}{b-1}) ,\big(\frac{c^3}{c-1}, \frac{c^2-3}{c-1}\big)$ are collinear for three distinct values of $a,b,c$ and $a,b,c\neq1$, then prove that the value of $(abc)-(ab+bc+ca)+3(a+b... | HINT.- I do not know if my answer could be “the smart approach to this question” as you want. Anyway here I give you the following.
Equalizing slopes
$$\frac{\frac{a^3-3}{a-1}-\frac{b^3-3}{b-1}}{\frac{a^3}{a-1}-\frac{b^3}{b-1}}=1-\frac{\frac{3}{b-1}-\frac{3}{a-1}}{\frac{a^3}{a-1}-\frac{b^3}{b-1}}=1-\frac{3}{ab(a+b)-(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
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Stuck on definite integral problem due to inappropriate $\log$ I have this definite integral problem which I have solved correctly but I'm stuck in one of the steps. I have manipulated it but I think it's not feasible to solve it that way.
$$\int_0^a(a^2 + x^2)^\frac{5}{2} dx$$
I have substituted $$x = a\cot\theta$$
an... | When you want do integrate a function of $a^2+x^2$ you could let $x=$ any of $a\tan\theta, a\cot\theta, a\sinh\theta, \text{or} \,\,a\,\text{csch}\,\theta$. The substitution $x=a\sinh\theta$ has many advantages for this kind of problem. Then $a^2+x^2=a^2\cosh\theta$ and $dx=a\cosh\theta d\theta$ and so
$$\int_0^a\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1699822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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There are 3 red balls and 7 blue balls in a bag. We take 4 balls randomly from the bag without replacement. What is the probability of getting at least $2$ blue balls?
When it asks for specific quantities of balls, i.e. $2$ red balls and $2$ blue balls, I understand to do
$\dfrac{\binom{3}{2}\cdot\binom{7}{2}}{\binom{... | Let $X$ be the random variable denoting how many blue balls are selected.
The question asks us to find $Pr(X\geq 2)$
Breaking apart via cases:
$Pr(X\geq 2)=Pr(X=2)+Pr(X=3)+Pr(X=4)$
In other words, to have at least two, this is the same as asking having exactly two or exactly three or exactly four.
You say you know how ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Inequality involving circumradii Let $ABC$ be a triangle and $M$ a point on the side $BC$. Let $R_1$,$R_2$, and $R$ be the circumradii of the triangles $ABM, ACM$, and $ABC$. Show that $\max\{R_1,R_2\} \geq R\cos\frac A 2$.
| In $\Delta ABC$ we have:
$$\frac{a}{\sin A}=2R$$
$\Rightarrow 2R\cdot\sin A=a\Rightarrow2R(2\sin\frac{A}{2}\cdot\cos\frac{A}{2})=a
\Rightarrow R\cdot\cos\frac{A}{2}=\dfrac{1}{4}\cdot\dfrac{a}{\sin\frac{A}{2}}$
Now assume $AM$ divides $\angle A$ into $\angle A_1$ and $\angle A_2$, also consider
$|\alpha|\lt\dfrac{A}{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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I'm a new to modular maths and I need to find 'a' given $(x+y)^p \equiv a^b+c^d\pmod p$ I'm new to modular maths and I've been asked to do the following:
given p is prime and $(x+y)^p \equiv a^b+c^d\pmod p$ , find a=?,b=?,c=?,d=?
Can anyone help me with the same?
Or atleast point me in the right direction?
Or give me a... | Binomial coefficients
You may know these formulas:
\begin{align*}
(a+b)^2 &= a^2 + 2ab + b^2 \\
(a+b)^3 &= a^3 + 3a^2b + 3ab^2 + b^3 \\
(a+b)^4 &= a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 \\
\text{etc.}
\end{align*}
Each term in the expansion of $(a+b)^n$ clearly has the form $c \cdot a^k b^{n-k}$, where the coefficient $c$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove $P(x)=1-\frac{x^2}{2}$ is a good approximation of order $3$ for $f(x)=\cos x$ near $x=0$? Let $f$ be a function and we want to approximate $f$ using a different function $P$ near $x=0$. The error of approximation is $E(x)=f(x)-P(x)$. If the approximation is going to be any good, we want $\lim\limits_{x\ri... | Your mostly done already. The only jump in logic is the statement that
"Clearly $\lim\limits_{x\rightarrow 0}\frac{E(x)}{x^3}=0$."
This is not clear based on the previous line as the prior line if you were to plug in 0 would get you 0 over 0. You can deal with this by using L'Hospital's rule.
$\lim\limits_{x\rightarrow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1704648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Divisibility of $n^4 -n^2$ by 4 (induction proof) We have to show that $$ n^4 -n^2 $$ is divisible by 3 and 4 by mathematical induction
Proving the first case is easy however I do not know how what to do in the inductive step. Thank you.
| Doing it by the book, though you were already told there are simpler/faster ways:
For $\;n=1:\;\;1^4-1^2=0\;\color{green}\checkmark\;$
Suppose it is true for $\;n\;$ and prove for $\;n+1\;$
$$(n+1)^4-(n+1)^2=(n+1)^2\left((n+1)^2-1\right)=(n+1)^2(n+1-1)(n+1+1)=$$
$$(n^2+2n+1)(n^2+2n)=n^4+4n^3+5n^2+2n=n^4-n^2+2n(2n^2+3n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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The graph of the equation $x+y=x^3+y^3$ is the union of The graph of the equation $x+y=x^3+y^3$ is the union of
$(A)$line and an ellipse$(B)$line and a parabola$(C)$line and hyperbola$(D)$line and a point
I tried to factorize the given equation.
$x^3-x+y^3-y=0$
$x(x^2-1)+y(y^2-1)=0$
The answer given is a line and an e... | Note that
$$x+y=x^3+y^3=(x+y)(x^2-xy+y^2)$$
Setting $$x=\cos\left(\frac{\pi}{4}\right)X+\sin\left(\frac{\pi}{4}\right)Y=\frac{X+Y}{\sqrt 2}$$$$y=-\sin\left(\frac{\pi}{4}\right)X+\cos\left(\frac{\pi}{4}\right)Y=\frac{-X+Y}{\sqrt 2}$$
gives
$$x^2-xy+y^2=1\Rightarrow 3X^2+Y^2=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1706000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Induction for divisibility: $3\mid 12^n -7^n -4^n -1$ I must use mathematical induction to show that
$a_{n} = 12^n −7^n −4^n −1$ is divisible by 3 for all positive integers n.
Assume true for $n=k$
$a_{k} = 12^k -7^k -4^k -1$
Prove true for $n=k+1$
$a_{k} = 12^{k+1} -7^{k+1} -4^{k+1} -1$
$ = (12^k)(12) - (7^k)(7) - (... | First, show that this is true for $n=1$:
$12^1−7^1−4^1−1=0$
Second, assume that this is true for $n$:
$12^n−7^n−4^n−1=3k$
Third, prove that this is true for $n+1$:
$12^{n+1}−7^{n+1}−4^{n+1}−1=$
$\color\red{12^n−7^n−4^n−1}+11\cdot12^n-6\cdot7^n-3\cdot4^n=$
$\color\red{3k}+11\cdot12^n-6\cdot7^n-3\cdot4^n=$
$3k+11\cdot12\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1711256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Find the volume of the areas bounded by following:
*
*$(x^2+y^2+z^2)^2=a^2(x^2+y^2-z^2)$ with $a = const$
*$z=x^2+y^2, z^2=2(x^2+y^2), xy=a^2 , xy=2a^2,x=2y, 2x=y$ and $x > 0, y> 0$
First of all, I have never heard of this first geometric shape defined in (1)
This was given as a homework assignment but we haven't ... | 1.
Start by making the replacement $x' = x/a, y' = y/a, z' = z/a$, then the equation becomes $(x'^2+y'^2+z'^2)^2=(x'^2+y'^2-z'^2)$, and $dx\,dy\,dz = a^3\,dx'\,dy'\,dz'$, so the volume is related by $V = a^3V'$. We can assume $a=1$, and get the more general case by multiplying the volume by $a^3$.
Next, switch to cylin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\lim_{x \to 3} \sqrt{x+1} = 2$ Prove that $\displaystyle\lim_{x \to 3} \sqrt{x+1} = 2$
Attempt:
$0 < |x - 3| < \delta \Rightarrow |\sqrt{x+1} - 2| < \epsilon$
Well $|\sqrt{x+1} - 2| = |(\sqrt{x+1} - 2) \cdot \displaystyle\frac{\sqrt{x+1} + 2}{\sqrt{x+1}+2}| = |\frac{x-3}{\sqrt{x+1} + 2}| = |x-3| \cdot \fra... | Let $x = y+3$,
so that
$x \to 3$
is the same as
$y \to 0$.
$\begin{array}\\
\sqrt{x+1}-2
&=\sqrt{y+4}-2\\
&=(\sqrt{y+4}-2)\dfrac{\sqrt{y+4}+2}{\sqrt{y+4}+2}\\
&=\dfrac{(\sqrt{y+4}-2)(\sqrt{y+4}+2)}{\sqrt{y+4}+2}\\
&=\dfrac{(y+4)-4}{\sqrt{y+4}+2}\\
&=\dfrac{y}{\sqrt{y+4}+2}\\
&\to 0 \text{ as } y \to 0\\
\end{array}
$
M... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Steps to solve $\lim_{n \to \infty} (\frac{ \sqrt{n^4+1} - \sqrt{n^4-1}}{ \frac1{(2n+1)^2}} ) = 4 $ $$\lim_{n \to \infty} \left(\frac{ \sqrt{n^4+1} - \sqrt{n^4-1}}{ \frac1{(2n+1)^2}} \right) = 4 $$
I think $\sqrt{n^4+1} - \sqrt{n^4-1}$ is approaching to zero, but it is not correct. What steps can evaluate above limit t... | Set $1/n=h\implies h\to0^+, h>0$
$$\lim_{n \to \infty} \left(\frac{ \sqrt{n^4+1} - \sqrt{n^4-1}}{ \dfrac1{(2n+1)^2}} \right) =\lim_{ h\to0^+}\dfrac{(\sqrt{1+h^4}-\sqrt{1-h^4})(2+h)^2}{h^4} $$
$$=\lim_{ h\to0^+}(2+h)^2\cdot\dfrac1{\lim_{ h\to0^+}(\sqrt{1+h^4}+\sqrt{1-h^4})}\cdot\lim_{ h\to0^+}\dfrac{1+h^4-(1-h^4)}{h^4}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find the gcd of $f(x)$ and $g(x)$ Use the division algorithm to the find the gcd of $f(x)= 2x^4 + 5x^3 -5x -2$ and $g(x)= 2x^3 -3x^2 -2x$ in $\mathbb Q[x]$.
I used long division and found the following
$$f(x) = (x+4)g(x) + \left( 14x^2 + 3x -2 \right)$$
So I have found the polynomials $q(x)$ and $r(x)$ such that $$f(x)... | They key property is that if $f=gq+r$, then $$\tag{1}\gcd(f,g)=\gcd(g,r).$$
So, you already found that
$$
f(x) = (x+4)g(x) + \left( 14x^2 + 3x -2 \right).
$$
Now
$$
g(x)=\left(\frac17\,x-\frac{12}{49}\right)(14x^2+3x-2)+\left(-\frac{48}{49}x-\frac{24}{49}\right).
$$
And
$$
14x^2+3x-2=\left(-\frac{343}{24}x+\frac{49}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $A+B+C = \pi$, show that $\sum_{\text{cyclic}} \tan A \tan B \neq 0$
If $A+B+C = \pi$, and $A,B,C > 0$ show that $$\sum_{\text{cyclic}} \tan A \tan B \neq 0$$
without using the fact that if $A+B+C = \pi$, then $\tan A + \tan B + \tan C = \tan A \tan B \tan C$
My Work
I tried to show it as follows:
$$\tan(A+B+C) = ... | First
$$
\tan C=\tan(\pi-A-B)=-\tan(A+B).
$$
Hence
$$
0=\tan A\tan B+\tan A\tan C+\tan C\tan B=\tan A\tan B+\tan C(\tan A+\tan B) \\=\tan A\tan B-\frac{(\tan A+\tan B)^2}{1-\tan A\tan B}
$$
Thus
$$
(\tan A+\tan B)^2=\tan A\tan B(1-\tan A\tan B)
$$
or
$$
(\tan A)^2+(\tan B)^2+\tan A\tan B=-(\tan A)^2(\tan B)^2.
$$
Contr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1716873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Use an expression for $\frac{\sin(5\theta)}{\sin(\theta)}$ to find the roots of the equation $x^4-3x^2+1=0$ in trigonometric form
Question: Use an expression for $\frac{\sin(5\theta)}{\sin(\theta)}$ , ($\theta \neq k \pi)$ , k an integer to find the roots of the equation $x^4-3x^2+1=0$ in trigonometric form?
What ... | HINT:
Using Prosthaphaeresis Formula,
$$\sin5x-\sin x=2\sin2x\cos3x=4\sin x\cos x\cos3x$$
If $\sin x\ne0,$
$$\dfrac{\sin5x}{\sin x}-1=4\cos x\cos3x=4\cos x(4\cos^3x-3\cos x)=(4\cos^2x)^2-3(4\cos^2x)$$
OR replace $\sin^2x$ with $1-\cos^2x$ in your $$ 5\cos^4x-10\cos^2x\sin^2x + \sin^4x$$
Now if $\sin5x=0,5x=n\pi$ where ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1718300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Do there exist other tricks for trig with Cauchy's Theorem? I have noticed that $$\int_{\theta=0}^{2\pi} \frac{\mathrm{d}\theta}{a+b\cos\theta}=\frac{2\pi}{\sqrt{a^2-b^2}}$$ with $|b|<|a|$
Are there any other tricks like this for say $\int_{\theta=0}^{2\pi} \frac{\mathrm{d}\theta}{a+b\sin\theta}$?
| If one wishes to use contour integration to evaluate the integral of interest, then one simply enforces the substitution $z=e^{i\theta}$. Thus, $\cos (\theta)=\frac{z+z^{-1}}{2}$, $d\theta =\frac{1}{iz}\,dz$ and we have
$$\begin{align}
\int_0^{2\pi}\frac{1}{a+b\cos(\theta)}\,d\theta&=\oint_{|z|=1}\frac{1}{a+b\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1721078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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What's the sum of the series $\sum _{n=0}^{\infty }\left(n^2a^n\right)$ $\sum _{n=0}^{\infty }\left(n^2a^n\right)$, I have seen a similar sum with just $n$ instead of the $n^2$, yet I'm not sure how to solve this one.
| I assume $0<|a|<1$. Let $S = \sum_n a^n$, $T = \sum_n n^2 a^n$. Observe that $S(n) = \frac{1}{1-a}$ (geometric series) which implies $\frac{d}{da} S = \frac{1}{(1-a)^2}$ and $\frac{d}{da} S = \frac{-2}{(1-a)^3}$. Then
$$\frac{d^2}{da^2} S = \sum_n n(n-1)a^{n-2} = \frac{1}{a^2} \underbrace{\sum_n n^2 a^n}_{=T}- \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find solution set $\sqrt{x-5} + 5 =0$ Find solution set :
$\sqrt{x-5} + 5 =0$
Is there a solution to a set of complex numbers ?
| No. There is no solution because $\sqrt{}$ is by definition non-negative.
However $\pm\sqrt{x-5} + 5 = 0$ does have a solution.
$\pm\sqrt{x-5} + 5 = 0$
$\pm\sqrt{x-5} = -5$
$(\pm\sqrt{x-5})^2 = (-5)^2$
$x-5 = 25$
$x = 30$.
But seriously. You are NOT allowed to say $\sqrt{x-5} = -5$. You just can't. It's against the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1726541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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How can I solve this using prime factors? I'm stuck with this problem: $2^x \cdot 3^3 \cdot 26^y = 39^z$ for $x, y, z \in \mathbb{N}$.
I know that there isn't a natural solution for the equation, but I need to "prove" it using prime factors. I'm stuck here:
$$
2^x \cdot 3^3 \cdot (2\cdot 13)^y = (3\cdot 13)^z \\
\impli... | $2^x \cdot 3^3 \cdot 26^y = 39^z\iff 2^{x+y}\cdot3^3\cdot13^y=3^z\cdot13^z\Rightarrow2^{x+y}=3^{z-3}\cdot13^{z-y}$
This is possible just when you have $1=1\cdot1$ i. e. $x+y=z-3=z-y=0$. Thus $$(x,y,z)=(-3,3,3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the sum of series $\sum\limits_{n=2}^{+\infty}\frac{(-1)^n}{n^2+n-2}$. Using the power series $\sum\limits_{n=2}^{+\infty}\frac{1}{n^2+n-2}x^{n}$, the first derivative gives $\sum\limits_{n=2}^{+\infty}\frac{n}{n^2+n-2}x^{n-1}$ where the sum doesn't exist.
Integration of function $\frac{1}{n^2+n-2}x^n$ given $\fra... | $$\begin{eqnarray*}\sum_{n\geq 2}\frac{(-1)^n}{(n+2)(n-1)}&=&\int_{0}^{1}\int_{0}^{1}\sum_{n\geq 2}(-1)^n x^{n+1} y^{n-2}\,dx\,dy\\&=&\int_{0}^{1}\int_{0}^{1}\frac{x^3}{1+xy}\,dy\,dx\\&=&\int_{0}^{1}x^2\log(1+x)\,dx\\&=&\frac{\log 2}{3}-\int_{0}^{1}\frac{x^3}{3(1+x)}\,dx\end{eqnarray*}$$
The last integral is straightfo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1732188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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If $a,b,c>0\;,$ Then value of $\displaystyle \lfloor \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\rfloor $,
If $a,b,c>0\;,$ Then value of $\displaystyle \bigg \lfloor \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\bigg\rfloor $, Where $\lfloor x \rfloor $ Rep. floor function of $x$.
$\bf{My\; Try::}$ Using $b+c>a$ and $c+a>b... | $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<\frac{a+a}{b+c+a}+\frac{b+b}{c+a+b}+\frac{c+c}{a+b+c}=$$
$$=\frac{2(a+b+c)}{a+b+c}=2$$
And
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}>\frac{a}{a+b+c}+\frac{b}{c+a+b}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1$$
So
$$ \bigg\lfloor \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\bigg\rfl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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For $\triangle ABC,$ $r_1+r_3+r=r_2$, find $\sec^2A+\csc^2B-\cot^2C.$ In a triangle $ABC,$ if $r_1+r_3+r=r_2$,then find the value of $\sec^2A+\csc^2B-\cot^2C.$,where symbols have their usual meanings.
Here $r_1=\dfrac{\Delta}{s-a},r_2=\dfrac{\Delta}{s-b},r_3=\dfrac{\Delta}{s-c},r=\dfrac{\Delta}{s}$
I put these values ... | $$r_1+r_3+r-r_2=0$$
We can write $$\frac{\Delta}{s-a}+\frac{\Delta}{s-c}+\frac{\Delta}{s}-\frac{\Delta}{s-b}=0$$
to have
$$\frac{1}{b+c-a}+\frac{1}{a+b-c}+\frac{1}{a+b+c}-\frac{1}{c+a-b}=0$$
and so
$$\frac{4b(a^2+c^2-b^2)}{(b+c-a)(a+b-c)(a+b+c)(c+a-b)}=0$$
Since we have $b^2=a^2+c^2$, we know that $\triangle{ABC}$ is a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Compute $\frac{1^2 t}{1!}+\frac{2^2 t^2}{3!}+\frac{3^2 t^3}{5!}+\frac{4^2 t^4}{7!}+\ldots+\frac{n^2 t^n}{(2n-1)!}+\ldots$ I have to compute $$\frac{1^2 t}{1!}+\frac{2^2 t^2}{3!}+\frac{3^2 t^3}{5!}+\frac{4^2 t^4}{7!}+\ldots+\frac{n^2 t^n}{(2n-1)!}+\ldots$$ I know that $\sinh t$ can be represented as a series1, but for t... | Let's carry this computation out for instructional purposes.
Let
$$f(t) = \sum_{n=1}^{\infty} \frac{t^n}{(2 n)!} = \cosh{\left (\sqrt{t}\right )}-1$$
Then the sum we seek is
$$2 \sum_{n=1}^{\infty} \frac{n^3 t^n}{(2 n)!} = \left (t \frac{d}{dt} \right )^3 f(t) = 2 t f'(t) + 6 t^2 f''(t) + 2 t^3 f'''(t)$$
$$f'(t) = \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Prove $\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{1-\cos\theta}{\sin\theta}$ Prove that $\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{1-\cos\theta}{\sin\theta}$
$$\frac{2\sec\... | hint $\frac{1-\cos(\theta)}{sin(\theta)}=\tan(\theta/2)$ so now convert everything to tan by using $\sin(2x)=\frac{2tan(x)}{1+\tan^2(x)},\cos(2x)=\frac{1-\tan^2(x)}{1+\tan^2(x)}$ then just simplify it. as $1+\tan^2(x)$ cancels off from the expression by taking lcm then use $tan(x)=t$ for algebraic simplifications. I do... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Bézout's Identity of polynomials?
Let $P=X^3−7X+6$, $Q = 2X^2+ 5X − 3$ and $R = X^2 − 9 ∈\mathbb Q[X]$. What are $S$ and $T ∈\mathbb Q[X]$ such that $PS + QT = R$?
I have calculate the greatest common divisor of $P,Q,R$ are $(x+3)$,
But this can only prove that the existence of $S,T$.
So what should I do next?
| .We know that the gcd is $x+3$, as you have found. Divide both polynomials by $(x+3)$ to get: $(x^2-3x+2)$ and $2x-1$ respectively. Now the gcd of these two polynomials is $1$. We do the following "reverse of the Euclidean algorithm":
$x^2-3x+2 = 0.5x(2x-1) + (-2.5x + 2)$ and,
$2x-1 = -0.8(-2.5x+2)+0.6$. Therefore,now... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to calculate the sum: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$? How to calculate the sum: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$ ?
I know the sum converges because it is a positive sum for every $n$ and it is smaller than $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ that converges and equals $1$. I need a direction...
| You can use double integral to calculate. Let $f(x)=\sum_{n=1}^\infty\frac{1}{n(n+3)}x^n$. Then $f(1)=\sum_{n=1}^\infty\frac{1}{n(n+3)}$. Clearly
$$ f'(x)=\sum_{n=1}^\infty\frac{1}{n+3}x^{n-1}$$
and
$$ (x^4f'(x))'=\sum_{n=1}^\infty x^{n+2}=\frac{x^3}{1-x}.$$
Thus
\begin{eqnarray}
f(1)&=&\int_0^1\left(\int_0^x\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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Solve the equation $\sqrt{1-x}=2x^2-1+2x\sqrt{1-x^2}$ Solve the following equation: $\sqrt{1-x}=2x^2-1+2x\sqrt{1-x^2}$
Unfortunately I have no idea.
| Let $x=\cos2y$
WLOG $0\le2y\le180^\circ\implies\sin y\ge0$
$$\sqrt2\sin y=\cos4y+\sin4y\iff\sin\left(4y+45^\circ\right)=\sin y$$
$\implies$
either $4y+45^\circ=360^\circ n+y\iff y=120^\circ n-15^\circ\implies y=(120-15)^\circ\implies x=\cos2y=?$
or $4y+45^\circ=(2n+1)180^\circ-y\iff y=72^\circ n+27^\circ\implies y=27^\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1741924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Convergence of $\sum \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}$ I need to prove $\sum \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}$ converges. The root test is inconclusive, so I check in W.A., $\sum \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}$ converges by comparison test, but I don't know what seri... | Another possible solution:
$$
0 < a_n := \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}
= \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4(n+1) - 1}} \\
< \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4(n+1) + 1}} =: b_n
$$
and $\sum b_n$ is convergent as a telescoping series.
It follows that $\sum a_n$ is convergent and
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Lucas numbers proof I'm running through some example problems and encountered this one:
Define a sequence of integers $L_n$ by $L_1=1, L_2=3, L_{n+1}=L_n+L_{n-1}.$ Show that $L_n = a\cdot \left(\frac{1+\sqrt{5}}{2}\right)^n + b\cdot \left(\frac{1-\sqrt{5}}{2}\right)^n$ and find the values of $a$ and $b$.
I wasn't able... | The recurrence is $L_{n+1}=L_{n}+L_{n-1}$.Now I am using the characteristic
equation of this above recurrence to solve the recurrence. So the characteristic
equation of this above recurrence is as follows $\implies$
Put $L_{n}={r^n}$ and solve the equation $\implies ({r^2}-r-1)=0$
By solving this equation we get $r=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Limit $\lim_{n\to\infty}\sum_{k=0}^n\frac{n}{n^2+k}$ I came across this problem that I'm supposed to be able to solve in under 5 minutes (for a competition).
$$\lim_{n\to \infty} \sum_{k=0}^n \frac {n}{n^2+k}$$
I tried solving this for small sums, $\sum_{k=0}^2 \frac {n}{n^2+k}$, $\sum_{k=0}^3 \frac {n}{n^2+k} $ and m... | Slightly more involved solution:
$$
\frac{1}{n} \sum_{k=0}^{n} \frac{1}{1+\frac{k}{n^2}} = \frac{1}{n} \sum_{k=0}^{n} e^{- \log (1+\frac{k}{n^2}) } \sim \frac{1}{n} \sum_{k=0}^{n} e^{-\frac{k}{n^2}} = \frac{1}{n} \times \bigg(\frac{1-e^{-\frac{1}{n^2}(n+1)}}{1-e^{-\frac{1}{n^2}}} \bigg) \sim \frac{1}{n} \times \bigg( ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1748429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How to calculate the series $-\frac12-\frac14+\frac13-\frac16-\frac18+\frac15-\frac{1}{10}...$? $-\frac12-\frac14+\frac13-\frac16-\frac18+\frac15-\frac{1}{10}...$
After rearrangement the series looks like $\sum^{\infty}_{n=2}\frac{(-1)^{n+1}}{n}$.
My way of doing this is using Taylor series of $\ln{(1+x)}=\sum^{\infty}... | You cannot rearrange an infinite series which is not absolutely convergent. We can evaluate it as follows. Your infinite series is
\begin{align}
\sum_{n=1}^{\infty}\left(-\dfrac1{4n-2} - \dfrac1{4n}+\dfrac1{2n+1}\right) & = \sum_{n=1}^{\infty}\left(-\dfrac1{4n-2} - \dfrac1{4n}+ 2\cdot \dfrac1{4n+2}\right)\\
& = \sum_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1749052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculate $\sin\frac{3\pi}{14}-\sin\frac{\pi}{14}-\sin\frac{5\pi}{14}$ I have interesting trigonometric expression for professionals in mathematical science. So, here it is:
$$\sin\dfrac{3\pi}{14}-\sin\dfrac{\pi}{14}-\sin\dfrac{5\pi}{14};$$
Okay! I attempt calculate it:
\begin{gather}
\sin\dfrac{3\pi}{14}-\left(\sin\df... | For simplicity, let $x = \frac{\pi}{14}$, then we want to simplify:
$$\sin 3x-\sin x -\sin 5x$$
Multiply by $\cos x$ to get:
$$\color{blue}{\sin 3x\cos x}-\color{green}{\sin x\cos x} -\color{red}{\sin 5x\cos x} \quad (*)$$
With $\sin\alpha\cos\beta = \tfrac{1}{2}\left( \sin(\alpha+\beta)+\sin(\alpha-\beta) \right)$, yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1749340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Hint solving an integral using trig substitution I want a hint for the following integral:
$$\int \frac{\sqrt{1- a^{2}+x^{2}}}{x^{2}(a^2-x^2)} dx$$
where $a$ is a real constant.
My attempt:
$$\frac{\sqrt{1- a^{2}+x^{2}}}{x^{2}(a^2-x^2)} = \frac{1}{x^{2}\sqrt{1- a^{2}+x^{2}}}(1-\frac{1}{a^2-x^2})$$
I know from the post
... | Let be
$$
I=\int
\frac{\sqrt{1-a^2+x^2}}{x^2(a^2-x^2)}\mathrm d x
$$
with the substitution
$
\frac{\sqrt{1-a^2+x^2}}{x}=u
$ and $x^2=\frac{1-a^2}{u^2-1}$ and $\mathrm d u=\frac{a^2-1}{x^2\sqrt{1-a^2+x^2}}\mathrm d x$ and then $\frac{x^2\sqrt{1-a^2+x^2}}{a^2-1}\mathrm d u=\mathrm d x$ the integral $I$ becomes
$$
J=\int ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1749688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Suppose $P (1) + P (2) = 3P (3)$ and $P (2) + P (3) = P (1)$ for a sample space $(S,P)$ . Find the probability function. So we have the following equations:
$P(1) + P(2) = 3P(3)$
$P(2) + P(3) = P(1)$
I know that by definition, $P(1) + P(2) + P(3) = 1$
I'm not entirely sure what I'm looking for as my answer though. Are ... | Here's one way to solve this problem.
Since we know that $P(1) + P(2) + P(3) = 1$ by definition, we have the following three equations:
$P(1) + P(2) = 3P(3)$
$P(2) + P(3) = P(1)$
$P(1) + P(2) + P(3) = 1$
Plugging the first equation into the third, we get $4P(3) = 1$. So $P(3) = \frac{1}{4}$.
From the last equation, we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate this infinite product: $\prod_{n=3}^{\infty} \left(\;1-\frac{4}{n^2}\;\right)$
$$\prod_{n=3}^{\infty} \left(\;1-\frac{4}{n^2}\;\right)\;=\;\text{???}$$
I took the LCM and split the numerator as $(n+2)(n-2)$ and then took the product of the numerator and the denominator separately but I was not able to get th... | $$ \prod_{n=3}^{\infty} \left( 1 - \frac{4}{n^2} \right) = \prod_{n=3}^{\infty} \left( \frac{(n-2)(n+2)}{n^2} \right) = \prod_{n=3}^{\infty} \left( \frac{n-2}{n} \right)\prod_{n=3}^{\infty} \left( \frac{n+2}{n} \right) = \lim_{n\rightarrow \infty} \frac{2^2}{4!} \frac{(n-2)!}{n!} \frac{(n+2)!}{n!} $$
So now we pull out... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
How to evaluate this limit using Taylor expansions? I am trying to evaluate this limit:
$\lim_{x \to 0} \dfrac{(\sin x -x)(\cos x -1)}{x(e^x-1)^4}$
I know that I need to use Taylor expansions for $\sin x -x$, $\cos x -1$ and $e^x-1$. I also realise that all of these are just their regular Taylor expansions with their f... | Just in case you wanted something to check your answer off of:
\begin{align}
\lim_{x \to 0} \dfrac{(\sin x -x)(\cos x -1)}{x(e^x-1)^4} &= \lim_{x\to 0} \frac{\left(\frac{-x^3}{3!}+\frac{x^5}{5!} - \cdots\right)\left(\frac{-x^2}{2!} + \frac{x^4}{4!}- \cdots\right)}{x\left(x+\frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Showing H is a normal subgroup by calculating left and right coset If $G = \begin{pmatrix} a & b \\
0 & 1 \end{pmatrix} a,b\in (\mathbb{R}) : a \neq 0$)
and assume G is a group under matrix multpication
Prove that H = ($\begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}: t\in\mathbb{R}$) is a normal subgroup of G
I'm meant ... | First let me solve the problem the way you asked: proving that every left coset is a right coset.
Given $g=\begin{pmatrix}a & b \\ 0 & 1\end{pmatrix}$ one must prove that $gH=Hg$ (introducing an additional symbol $g'$ is wrong). Your question correctly describes the general format of matrices in the sets $gH$ and $Hg$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show convexity of a function via inequalities I am stuck with deriving the convexity of the function
$$
f(x) = \sqrt{1 + x^2}
$$
from first principles, that is I would like to show that for any $x,y \in \mathbb R$ and $\lambda \in (0,1)$ we have
$$
f(\lambda x + (1 - \lambda)y) \le \lambda f(x) + (1 - \lambda) f(y)
$... | To make the two sides of the inequalities easier to compare, put $a:=\lambda$ and $b:=(1-\lambda)$ and then square both sides (a valid operation since both sides are nonnegative and therefore $x\mapsto x^{2}$ is invertible and monotonic, thus inequality preserving).
After doing this, we have on the one hand
$$\begin{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How do I simplify $\frac{1}{1+\frac{x^2}{2}+\frac{5x^4}{48}+\frac{7x^6}{576}\dots}$ using long division? The infinite series $\frac{1}{1+\frac{x^2}{2}+\frac{5x^4}{48}+\frac{7x^6}{576}\dots}$ is supposed to simplify to $1-\frac{x^2}{2}+\frac{7x^4}{48}+\frac{19x^6}{576}\cdots$ but I don't know how this was calculated.
Di... | Actually, I got
$$1-\frac{x^2}{2}+\frac{7x^4}{48}-\frac{19x^6}{576}\cdots$$
Let
$$D = 1+\frac{x^2}{2}+\frac{5x^4}{48}+\frac{7x^6}{576}\dots,$$
your divisor.
First remainder is
$$R_1 = 1 - 1D = -\frac{x^2}{2} - \frac{5x^4}{48} - \frac{7x^6}{576} + O(x^8).$$
Second remainder (for the $-x^2/2$ term) is
$$R_2 = R_1 - \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Without using modular equivalence, show that: $\gcd(4n^2+1,24)=1$ Without using modular equivalence, show that: $\gcd(4n^2+1,24)=1$
Let $d=\gcd(4n^2+1,24)$ then we have:
$$d|24n^2+6,24n^2\ \Rightarrow\ d|6\ \Rightarrow\ d|6n^2,4n^2+1\ \Rightarrow\ d|12n^2,12n^2+3\ \Rightarrow\ d|3\ \Rightarrow\ d=1\ or\ 3$$
Using modul... | If
$n=0,1,2,3,4,5$,
$4n^2+1
=1, 5, 17, 37, 65, 101
$,
and all of these
are relatively prime to 24.
If
$n = 6m+k$
where
$0 \le k \le 5 $,
then
$\begin{array}\\
4n^2+1
&=4(6m+k)^2+1\\
&=4(36m^2+12mk+k^2)+1\\
&=4(36m^2+12mk)+4k^2+1\\
&=48(3m^2+mk)+4k^2+1\\
\end{array}
$
so
if $d$ divides both
$4n^2+1$ and $24$,
it also di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How do I show that $\sum_{i = 1}^n \frac 1{\sqrt{a_n}} \lt \frac {\sqrt 3}6$ for $a_n = 4n(4n + 1)(4n + 2)$? Let $a_n = 4n(4n + 1)(4n + 2)$, show that $$\sum_{i = 1}^n \frac 1{\sqrt{a_i}} \lt \frac {\sqrt 3}6 \quad \forall n \in \mathbb{N}^+.$$
I know I need to find an upper bound for $1/\sqrt{a_n}$ but I can't see how... | $$
\begin{align}
\frac12\sum_{k=1}^\infty\left(\frac1{\sqrt{4k-1}}-\frac1{\sqrt{4k+3}}\right)
&=\sum_{k=1}^\infty\frac2{\sqrt{4k-1}\sqrt{4k+3}\left(\sqrt{4k-1}+\sqrt{4k+3}\right)}\tag{1}\\
&\ge\sum_{k=1}^\infty\frac1{\sqrt{4k-1}\sqrt{4k+3}\sqrt{4k+1}}\tag{2}\\
&\ge\sum_{k=1}^\infty\frac1{\sqrt{4k}\sqrt{4k+2}\sqrt{4k+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1756693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Finite sum $\sum_{n=2}^N\frac{1}{n^2}\sin^2(\pi x)\csc^2(\frac{\pi x}{n})$ I was looking for a closed form but it seemed too difficult. Now I'm seeking help to simplify this sum. The 50 bounty points or more will be awarded for any meaningful simplification of this sum.
I found this function that has very interesting p... | An idea:
Note the identity
$$\sin n \theta=U_{n-1}\left( \cos \theta \right) \sin \theta ,$$
where $U_i$ is the $i^{th}$ Chebyshev orthogonal polynomial of the second kind. This gives
$\sin^2 \pi x=\sin^2 \left(n \frac{\pi x}{n}\right)=U_{n-1}^2\left( \cos \frac{\pi x}{n} \right) \sin^2 \frac{\pi x}{n},$
and then
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1759629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 1,
"answer_id": 0
} |
Finding out a limit using Taylor series. So the limit is the following:
$$\lim_{x \to 0}{\frac{x^2-\frac{x^6}{2}-x^2 \cos (x^2)}{\sin (x^{10})}}$$
Expansions for $\sin(x)$ and $\cos(x)$ are given:
$$\sin x = x-\frac{x^3}{3!} + \frac{x^5}{5!}-...+(-1)^{n-1}\frac{x^{2n-1}}{(2n-1)!} + o(x^{2n})$$
$$\cos x = 1-\frac{x^2}{... | $$\lim_{x\to0}\frac{x^2-\frac{x^6}{2}-x^2\cos(x^2)}{\sin(x^{10})} = \lim_{x\to0}\frac{x^2-\frac{x^6}{2}-x^2(1-\frac{x^4}{2}+\frac{x^8}{24}-\frac{x^{12}}{720}+\cdots)}{x^{10}-\frac{x^{30}}{6}+\cdots}$$
$$=\lim_{x\to0}\frac{x^2-\frac{x^6}{2}+(-x^2+\frac{x^6}{2}-\frac{x^{10}}{24}+\frac{x^{12}}{720}-\cdots)}{x^{10}-\frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1760199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Evaluating $\int_{0}^{3} \sqrt{1+x}\: dx$ using Limit of a Sum approach Evaluate $\int_{0}^{3} \sqrt{1+x}\: dx$ using Limit of a Sum approach.
Using the formula $$\int_{a}^{b} f(x)\:dx=(b-a) \times \lim_{n \to \infty} \frac{1}{n} \times \sum_{k=1}^{n}f\left(a+\frac{(b-a)k}{n}\right)$$ we have
$$I=\int_{0}^{3} f(x)\:dx=... | Consider
$$\int_0^3 \sqrt{1+x} \, dx = \lim_{n \to \infty} \frac{3}{n}\sum_{k=1}^n\sqrt{1 + \frac{3k}{n}} = \lim_{n \to \infty} \frac{3}{n^{3/2}}\sum_{k=1}^n\sqrt{n + 3k}.$$
Using the binomial expansion, we have
$$(n + 3k - 3)^{3/2} = (n + 3k)^{3/2} \left(1 - \frac{3}{n+3k} \right)^{3/2} \\ = (n + 3k)^{3/2}\left(1 - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1761200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Solution of functional equation $f(x+y)=f(x)+f(y)+y\sqrt{f(x)}$
If $x,y\in \mathbb{R}$ and $f(x+y)=f(x)+f(y)+y\sqrt{f(x)}$ and $f'(0)=0\;,$ Then $f(x)$ is
$\bf{My\; Try::}$ Using $$f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} = \lim_{h\rightarrow 0}\frac{f(x)+f(h)+h\sqrt{f(x)}-f(x)}{h}$$
Now Put $x=y=0$ in $$f... | $$f(x+y)=f(x)+f(y)+y\sqrt{f(x)}=f(y+x)=f(y)+f(x)+x \sqrt{f(y)}$$
Subtracting $f(x)+f(y)$ from each side and squaring , we have that $$y^2f(x)=x^2f(y) \Leftrightarrow \frac{f(x)}{x^2}=\frac{f(y)}{y^2}$$
So we have $\frac{f(x)}{x^2}$ is a constant function.
Now put $f(x)=cx^2$ in the original equation, where $c$ is a co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1762184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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$\sum_{k=1}^{n}|1+z_{k}|+\frac{1}{n-1}\sum\limits_{1\le iLet $n\ge 2$is a integer,$z_{1},z_{2},\cdots,z_{n}$ are $n$ complex numbers
Prove that
$$\color{crimson}{\sum_{k=1}^{n}|1+z_{k}|+\dfrac{1}{n-1}\sum_{1\le i<j\le n}|1+z_{i}z_{j}|\ge\sum_{k=1}^{n}|z_{k}|}$$
Proof for $n=2$:
$$
\DeclareMathOperator{\Re}{Re}
| 1+x |... | For $z, w \in \Bbb C$ we have (and this is what you already did)
$$
\DeclareMathOperator{\Re}{Re}
\bigl( \lvert 1+z \rvert + \lvert 1+w \rvert + \lvert 1+zw \rvert \bigr)^2 \\
= 1 + \lvert z \rvert ^2 + 2 \Re z + 1 + \lvert w \rvert ^2 + 2 \Re w + 1 + \lvert zw \rvert ^2 + 2 \Re (zw) \\
+ 2 \lvert 1+z \rvert \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1762753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How can we show that $ \sum_{n=0}^{\infty}\frac{2^nn[n(\pi^3+1)+\pi^2](n^2+n-1)}{(2n+1)(2n+3){2n \choose n}}=1+\pi+\pi^2+\pi^3+\pi^4 ?$ We proposed this sum, but we are lacking in knowledge of this area of maths and we would ask if any of the authors would be willing to show us step by step how to go about proving this... | A quite boring approach. We can write your series as $$\left(\pi^{3}+1\right)\sum_{n\geq0}\frac{2^{n}n^{4}}{\left(2n+1\right)\left(2n+3\right)\dbinom{2n}{n}}+\left(\pi^{3}+\pi^{2}+1\right)\sum_{n\geq0}\frac{2^{n}n^{3}}{\left(2n+1\right)\left(2n+3\right)\dbinom{2n}{n}}
$$ $$+\left(-\pi^{3}+\pi^{2}-1\right)\sum_{n\geq0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1765705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
} |
Evaluate the $\lim_{x \to \ -\infty} (x + \sqrt{x^2 + 2x})$ Evaluate : $$\lim_{x \to \ -\infty} (x + \sqrt{x^2 + 2x})$$
I've tried some basic algebraic manipulation to get it into a form where I can apply L'Hopital's Rule, but it's still going to be indeterminate form.
This is what I've done so far
\begin{align}
... | $$\lim _{ x\to -\infty } \left( \frac { -2x }{ x-\sqrt { x^{ 2 }+2x } } \right) =\lim _{ x\rightarrow -\infty }{ \left( \frac { -2x }{ x-\sqrt { { x }^{ 2 }\left( 1+\frac { 2 }{ x } \right) } } \right) =\lim _{ x\rightarrow -\infty }{ \frac { -2x }{ x-\left| x \right| \sqrt { 1+\frac { 2 }{ x } } } = } } \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1766582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Find LU decomposition of a matrix using partial pivoting I've the following matrix:
$$ A= \begin{bmatrix}
0& 7& 5& 1
\\ 4& 3& 2& 1
\\0 &0& 0& 1
\\ 0& 0& -1& -2 \end{bmatrix} $$
and I need to find the matrices $P, L$ and $U$, such that
$$PA = LU$$
what I need was to find the ma... | For $A= \begin{pmatrix}
0& 7& 5& 1
\\ 4& 3& 2& 1
\\0 &0& 0& 1
\\ 0& 0& -1& -2 \end{pmatrix} $, we have:
$$\text{ P = }\left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ \end{array} \right), \text{ L = }\left( \begin{array}{cccc} 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Decompose $V \otimes V \otimes V$ into irreducible representations of $SL_2(\mathbb{R})$ Let $V=\mathbb{C^2}$ be the standard representation of $SL_2(\mathbb{R})$
Decompose $V \otimes V \otimes V$ into irreducible representations of $SL_2(\mathbb{R})$
I will just consider $SL_2(\mathbb{C})$ since there is a 1:1 corre... | One such decomposition is
$
A=\begin{pmatrix}0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\cr -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0\cr 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\cr 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0\end{pmatri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Prove that $(2\sqrt3+4)\sin x+4\cos x$ lies between $-2(2+\sqrt5)$ and $2(2+\sqrt5)$. Prove that $(2\sqrt3+4)\sin x+4\cos x$ lies between $-2(2+\sqrt5)$ and $2(2+\sqrt5)$.
Since we know that the minimum and maximum values of $a\cos x+b\sin x$ is $-\sqrt{a^2+b^2}$ and $\sqrt{a^2+b^2}$
I applied this formula to get the ... | $$2\sqrt{11+4\sqrt{3}} \approx 8.46834180469$$
$$2(2+\sqrt(5))\approx 8.472135955$$
$$2(2+\sqrt(5)) >2\sqrt{11+4\sqrt{3}}$$
Alternatively, (ignoring the factor of 2, squaring, and then subtracting 9 from both sides):
$$4\sqrt{5} \overset{?}{>} 2+4\sqrt{3}$$
Square again
$$80 \overset{?}{>} 52 + 16\sqrt{3}$$
Subtract 52... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1769226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$(f(1 - f(x)) = 1 - x^9$, $f(1) = 0$ and $f'(1) < 0$, then where is the real number $r$ such that $f(r) = r^{99}$? If $f(1 - f(x)) = 1 - x^9$, $f$: R $\to$ R is differentiable, $f(1) = 0$ and $f'(1) < 0$, how to show there is a real number $r$ such that $$f(r) = r^{99}?$$
Edit: Taylor Theorem makes no use. I try to tak... | After almost 4 years I finally think of these. Let me know anywhere wrong.
Firstly if $f(1 - f(x)) = 1 - x^9$, then let $a \in \mathbb{R}$ so that $f(a) = 1$, hence \begin{align} f(0) = f(1 - f(a)) = 1 - a^9. (1) \end{align}
Also we rearrange the terms in expressing $f(1 - f(x))$ to be $f(x) = 1 - f^{-1}(1 - x^9)$, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1769898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
limit of the function of two variables ?? Here is the function :
\begin{equation}
f(x,y)=a^2 \left(\frac{x}{a^2-3 x^2}-\frac{y}{3 y^2-a^2}\right).
\end{equation}
We know that $y\geq x \geq 0$, $~a$ is a constant and $a^2=x^2+xy+y^2$.
I want to derive the limit of the function in the case $x\rightarrow a/\sqrt{3}=r_0$ a... | The first method makes no sense: once you decide that $x=r_0-\delta$, $y=r_0+\delta$, you have
$$
x^2+xy+y^2=(r_0+\delta)^2+(r_0-\delta)(r_0+\delta)+(r_0-\delta)^2=3r_0^2+\delta^2=a^2+\delta
$$
and your condition $x^2+xy+y^2=a^2$ is not satisfied.
In the second method you are not showing your work so I cannot comment... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$a,b,c >0$, prove $\sqrt[2]{\frac{a}{b+c}}+\sqrt[3]{\frac{b}{c+a}}+\sqrt[4]{\frac{c}{a+b}} \geqslant \frac{7}{12} \cdot2^{\frac67} \cdot 3^{\frac47}$ $a,b,c >0$, prove
$$\sqrt[2]{\frac{a}{b+c}}+\sqrt[3]{\frac{b}{c+a}}+\sqrt[4]{\frac{c}{a+b}} \geqslant \frac{7}{12} \cdot2^{\frac67} \cdot 3^{\frac47}$$
What I tried:
1) I... | A solution will be given by first showing that two of the arguments of the roots are always less than 1, and that for any fixed sum $s = a+b$ (or of any other two variables), the minimum of the LHS is obtained if one variable is taken to be zero. After that, the LHS can be minimized.
It shows (following the comment by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 1
} |
How to prove $\cos^4x - \sin^4x - \cos^2x + \sin^2x$ is always $0$? So I have a small problem here where I have to prove the following :
$$\cos^4x - \sin^4x - \cos^2x + \sin^2x = 0 $$
I know that the 2nd part is always $1$, so I need to prove that the first part also equals $1$. So how should I prove it ?
Edit : Sorr... | The edited identity is correct.
Use $a^2 - b^2 = (a+b)(a-b)$ to rewrite $\cos^4 x - \sin^4 x = (\cos^2 x - \sin^2 x)(cos^2 x + \sin^2 x)$.
The second factor is one ($\sin^2 x + \cos^2 x = 1$), reducing the original expression to:
$\cos^2x - \sin^2x - \cos^2x + \sin^2x = 0$ as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1779361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Integral involving inverse of a function If $f(x)=x^3-x^2+x$, evaluate $$\lim_{n \to \infty} \int _{n}^{2n}\frac{dx}{(f^{-1}(x))^3+f^{-1}(x)}$$ I substituted $x=f(t)$ but was unable to convert it back to $x$ and I think there would be a better approach. Some hints please. Thanks.
| For any $x>1$ we have $\left(x-\frac{1}{3}\right)^3 \leq f(x) \leq x^3 $, hence $f^{-1}(2n)-f^{-1}(n)$ is $O(\sqrt[3]{n})$ for large $n$ and $\frac{f^{-1}(2n)}{f^{-1}(n)}\to \sqrt[3]{2}$ as $n\to +\infty$
Moreover,
$$ \int \frac{3x^2-2x+1}{x^3+x}\,dx = C-2\arctan(x)+\log(x)+\log(1+x^2)\tag{1}$$
hence we may deal with
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1780861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
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Fractions in Questions and Answers
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