Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solving $\int(x^2 + 1)^7x^3{\mathrm d}x$ without expansion What is $\int(x^2 + 1)^7x^3{\mathrm d}x$?
I wasn't able to come up with a substitution so I attempted integration by parts:
$$u = x^2 + 1, u' = 2x, v' = x^3, v = \frac{x^4}{4}$$
$$(x^2+1)\frac{x^4}{4} - \frac{x^6}{12}$$
The derivative clearly shows that this is... | Quite simply, $$\int (x^2 + 1)^7 x^3 \, dx = \frac{1}{2} \int (x^2 + 1)^7 (x^2)(2x) \, dx,$$ and with the substitution $u = x^2 + 1$, $du = 2x \, dx$, we readily obtain $$\int (x^2 + 1)^7 x^3 \, dx = \frac{1}{2} \int u^7 (u-1) \, du = \frac{1}{2} \int u^8 - u^7 \, du = \frac{1}{2} \left(\frac{u^9}{9} - \frac{u^8}{8}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2625480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Simplify $\tan^{-1}\Big[\frac{\cos x-\sin x}{\cos x+\sin x}\Big]$, where $x<\pi$ Write in the simplist form
$$
\tan^{-1}\bigg[\frac{\cos x-\sin x}{\cos x+\sin x}\bigg],\quad x<\pi
$$
My Attempt:
$$
x<\pi\implies -x>-\pi\implies \frac{\pi}{4}-x>\frac{-3\pi}{4}
$$
$$
\frac{\cos x-\sin x}{\cos x+\sin x}=\frac{\cos^2 x-... | Indeed the given answer seems strange, as the usual range of the $\tan^{-1}$ function is $\left(-\frac\pi2,\frac\pi2\right)$, and if $\frac{3\pi}4 \leq x < \pi$ then $\frac\pi4-x$ is not in this range. I think you have the domain right.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2626744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Manipulating a functional equation
Consider a function $f$ such that $$f(x)f(y)=f(xy)+f\left (\frac {x}{y}\right )$$ then find $$f\left (\frac {3-2\sqrt 2}{\sqrt 2 + 1}\right )- f\left (\frac {3+2\sqrt 2}{\sqrt 2 - 1}\right )$$
And the options are as follows
A) $2f\left (\frac {3-2\sqrt 2}{\sqrt 2 + 1}\right )$
B) ... | By the properties of multiplication from this equation: $$f(x)f(y)=f(xy)+f\left (\frac {x}{y}\right )$$ we obviously have: $$f\left (\frac {x}{y}\right )=f\left (\frac {y}{x}\right )$$ which means that for any $x$: $$f\left ( x \right )=f\left (\frac {1}{x}\right )$$
Since the arguments in:
$$f\left (\frac {3-2\sqrt 2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2628351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why are we allowed to cancel fractions in limits? For example:
$$\lim_{x\to 1} \frac{x^4-1}{x-1}$$
We could expand and simplify like so:
$$\lim_{x\to 1} \frac{(x-1)(x^3 + x^2 + x + 1)}{x-1} = \lim_{x\to 1} (x^3 + x^2 + x + 1) = (1^3 + 1^2 + 1^1 + 1) = 4$$
In this case we divided out $x-1$ on top and bottom even though ... | Algebraic Limit Theorem: Let the limits exist:
$$\lim_\limits{x\to a} f(x)=L \quad \text{and} \quad \lim_\limits{x\to a} g(x)=M.$$
Then:
$$\begin{align}&1) \ \lim_\limits{x\to a} (f(x)\pm g(x))=\lim_\limits{x\to a} f(x)\pm \lim_\limits{x\to a} g(x)=L\pm M;\\
&2) \ \lim_\limits{x\to a} (f(x)\cdot g(x))=\lim_\limits{x\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2628911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
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"answer_id": 5
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Another summation identity with binomial coefficients Recently I have encountered on this page a rather nice identity:
$$
\sum_{k=0}^m4^k\frac{\binom{n}{k}\binom{m}{k}}{\binom{2n}{2k}\binom{2k}{k}}=\frac{2n+1}{2n-2m+1},
$$
which however is valid only for $n\ge m$. This motivated me to try finding out a more symmetric e... | We prove a generalisation of OPs two binomial identities. The first being
\begin{align*}
\sum_{k=0}^m4^k\frac{\binom{\color{blue}{n}}{k}\binom{m}{k}}{\binom{\color{blue}{2n}}{2k}\binom{2k}{k}}=\frac{2n+1}{2n-2m+1}\tag{1}
\end{align*}
In the second identity we exchange the variables $m$ and $n$ to get the identity more... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2631444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
$\lim\limits_{t\to 0}(\frac{2+\cos2t}{3\sin^4t}-\frac{1}{t^4})$ It's from this answer.
https://math.stackexchange.com/a/368574/481435
I cannot evaluate the last limit.
I arranged that limit to slightly clearer form.
$$\lim\limits_{t\to 0}\left(\frac{2+\cos2t}{3\sin^4t}-\frac{1}{t^4}\right)=\frac{1}{45}$$
I have no idea... | Note that
$$\frac{2+\cos2t}{3\sin^4t}-\frac{1}{t^4}=\frac{2t^4+t^4\cdot\cos2t-3\sin^4t}{3t^4\cdot\sin^4t}$$
thus we need to expand in such way to have $8^{th}$ order terms, then
*
*$\cos2t=1-\frac{4t^2}{2}+\frac{16t^4}{24}+o(t^4)=1-2t^2+\frac{2t^4}{3}+o(t^4)$
*$\sin t=t-\frac{t^3}{6}+o(t^3)$
*$\sin^4
t=\left(t-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2631581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Which of the two sums gives a better approximation to $\pi^2/6$? Which of the following two sums
$$\sum_{n=1}^{1000000}\frac1{n^2} \quad \text{or} \quad 1+\sum_{n=1}^{1000}\frac1{n^2(n+1)}$$
gives a better approximation to $\pi^2/6?$
I tested this on MATLAB and surprisingly obtained as a result the second sum. However... | Let: $$G_{10^6}=\sum_{n=1}^{1000000}\frac1{n^2} \quad \text{and} \quad H_{10^3}+1=\sum_{n=1}^{1000}\frac1{n^2(n+1)}+1.$$
We can now compare the remainders of the two and see which is the smallest:
$$\begin{aligned}
&1) \int_{10^6+1}^\infty\frac1{x^2}dx\le\underbrace{\frac{\pi^2}6-G_{10^6}}_{\text{$R_G$}}\le\int_{10^6}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Solving an exponential with three different bases $$2^x+4^x=8^x$$ Solve for $x$. I reduced the bases to $2$ and try to use logarithms but could not get passed the logarithm of an expression. When typing into online calculators they say there is no solution but graphing shows an answer.
| Let $2^x=y$.
The equation becomes $$y+y^2=y^3\\
1+y=y^2\\
y^2-y-1=0\\
y=\frac{1\pm\sqrt{5}}{2}\\
2^x=\frac{1\pm\sqrt{5}}{2}\\
x=\log_2{\frac{1\pm\sqrt{5}}{2}}$$
Because $\frac{1-\sqrt{5}}{2}<0$ the logarithm of a negative number does not exist,
$$x=\log_2{\frac{1+\sqrt{5}}{2}}\text.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Converting polar equations to cartesian equations. Where $$r=\sin(3\theta)$$ and $$y=r\sin(\theta),~x=r\cos(\theta),~r^2=x^2+y^2$$ I have started by saying that $$ r=\sin (2\theta) \cos (\theta) +\sin (\theta) \cos (2\theta) \\ r=2\sin (\theta) \cos ^2 (\theta) +\sin (\theta) (1-2\sin ^2 (\theta)) \\r=2\sin (\theta) \... | $$\begin{align}r&=\sin(3\theta)\\r&=\sin\theta\cos2\theta+\cos\theta\sin2\theta\\r^2&=r\sin\theta\cos2\theta+r\cos\theta\sin2\theta\\x^2+y^2&=y\cos2\theta+x\sin2\theta\\r^2(x^2+y^2)&=yr^2(1-2\sin^2\theta)+2xr^2\sin\theta\cos\theta\\(x^2+y^2)^2&=y(x^2+y^2)-2y^3+2x^2y\\(x^2+y^2)^2&=3x^2y-y^3\end{align}$$
I don't see a di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
continued fraction of $\sqrt{41}$
Show that $\sqrt{41} = [6;\overline {2,2,12}]$
here's my try:
$$\sqrt{36}<\sqrt{41}<\sqrt{49}\implies6<\sqrt{41}<7\implies\lfloor\sqrt{41}\rfloor=6$$
$$\sqrt{41}=6+\sqrt{41}-6=6+\frac{1}{\frac{1}{\sqrt{41}-6}}$$
$$\frac{1}{\sqrt{41}-6}=\frac{\sqrt{41}+6}{41-36}=\frac{\sqrt{41}+6}{5}=... | Method described by Prof. Lubin at Continued fraction of $\sqrt{67} - 4$
$$ \sqrt { 41} = 6 + \frac{ \sqrt {41} - 6 }{ 1 } $$
$$ \frac{ 1 }{ \sqrt {41} - 6 } = \frac{ \sqrt {41} + 6 }{5 } = 2 + \frac{ \sqrt {41} - 4 }{5 } $$
$$ \frac{ 5 }{ \sqrt {41} - 4 } = \frac{ \sqrt {41} + 4 }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2635420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the minimum value of $f(x,y,z)=\frac{x^2}{(x+y)(x+z)}+\frac{y^2}{(y+z)(y+x)}+\frac{z^2}{(z+x)(z+y)}$. Find the minimum value of $f(x,y,z)=\frac{x^2}{(x+y)(x+z)}+\frac{y^2}{(y+z)(y+x)}+\frac{z^2}{(z+x)(z+y)}$ for all notnegative value of $x,y,z$.
I think that minimum value is $\frac{3}{4}$ when $x=y=z$ but I have n... | $f(x;y;z)=\frac{(xy+yz+zx)(y-z)^2 + (xy+xz-2yz)^2}{4(x+y)(z+x)(y+z)^2} +\frac{3}{4} \geq \frac{3}{4}$
Equality holds when $x=y=z$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2637671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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prove the following algebraically $\left( \begin{array}{c} 2n \\ 2\ \end{array} \right) = 2 \left( \begin{array}{c} n \\ 2\ \end{array} \right) + n^2$ I came across the following proof in my textbook that was used as a end of chapter review. How can I prove the following algebraically?
$$\left( \begin{array}{c} 2n \\ ... | $$\binom{2n}{2} = \frac{2n(2n-1)}{2} = \frac{2n(n-1) + 2n^2}{2}
= 2 \binom{n}{2} + n^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Positive Integer Divison Proof Prove that $6|n(n + 1)(n + 2)$ for any integer $n ≥ 1$
I have attempted to do this but never seem to prove it.
| Proof by contradiction
Assumption: There are $n$ where $6$ doesn't divide $n(n+1)(n+2)$.
Let $x$ be the smallest $n$ where $6$ doesn't divide $n(n+1)(n+2)$.
Since $6|1\times 2\times 3$ and $6|2\times 3\times 4$ that $x$ has to be at least $3$.
If $6$ doesn't divide $x(x+1)(x+2)$ then it doesn't divide $x(x+1)(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2641108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 5
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Show the following inequality. Let $a,b,c \in \mathbb R^+$ and $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} =1$$
Show that $$(a^2 -3a +3)(b^2-3b+3)(c^2-3c+3) \ge 27$$
I tried using using the AM-GM inequality and some algebraic manipulation to who each of the quadratic terms cannot be smaller than $0.75$ and other little r... | For positive variables let $\frac{1}{a}=\frac{x}{3},$ $\frac{1}{b}=\frac{y}{3}$ and $\frac{1} {c}=\frac{z}{3}.$
Thus, $x+y+z=3$ and we need to prove that
$$\sum_{cyc}\left(\ln\left(\frac{9}{x^2}-\frac{9}{x}+3\right)-\ln3\right)\geq0$$ or
$$\sum_{cyc}\left(\ln(x^2-3x+3)-2\ln{x}\right)\geq0$$ or
$$\sum_{cyc}\left(\ln(x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2641810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
U-substitution of 2x in trigonometric substitution Find
$$\int^{3\sqrt{3}/2}_0\frac{x^3}{(4x^2+9)^{3/2}} \,\mathrm{d}x.$$
The text says to use substitution of $u = 2x$. How did they get $u = 2x$ and not $u = x^3$?
| Alternatively, change:
$$4x^2+9=u \Rightarrow x^2=\frac{u-9}{4}; xdx=\frac{du}{8};$$
$$\int_{9}^{36} \frac{\frac{u-9}{4}\cdot \frac{du}{8}}{u^{3/2}}=\frac{1}{32}\int_{9}^{36} \frac{1}{u^{1/2}}du-\frac{9}{32}\int_{9}^{36} \frac{1}{u^{3/2}}du=\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Removing middle fourths instead of middle thirds in Cantor Set If you removed middle fourths instead of middle thirds to form a generalized Cantor set G, then what would the Lebesgue measure $m(G)$?
The geometric series that I obtained from removing middle fourths was:
$\displaystyle\frac{1}{4}+(2\cdot\frac{1}{4}\cdot... | Yes, your geometric series is correct and your new Cantor dust has measure zero.
You may try different middle intervals to remove and get different Cantor dusts with the same measure.
Note that while these Cantor sets have the same measure, their fractal dimensions are quite different.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing that $ 1 + 2 x + 3 x^2 + 4 x^3 + \cdots + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$ I was studying a polynomial and Wolfram|Alpha had the following alternate form:
$$P(x) = 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$$
Of course, we can ... | Use the distributive property.
$$(1 + x + x^2 + x^3 + x^4 + x^5)(1 + x + x^2 + x^3 + x^4 + x^5) $$
is equal to
$$
\begin{matrix}
(1)(1 + x + x^2 + x^3 + x^4 + x^5) \\
(x)(1 + x + x^2 + x^3 + x^4 + x^5) \\
(x^2)(1 + x + x^2 + x^3 + x^4 + x^5) \\
(x^3)(1 + x + x^2 + x^3 + x^4 + x^5) \\
(x^4)(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 3
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Maxima and minima of function with limiting condition (lagrange multiplier) problem
I have function:
$$ f(x,y,z)=x+2y+\frac{z^2}{2} $$ and i would want to find it's minima and maxima with condition $x^2+y^2+z^2=5$
Attempt to solve
I can form Lagrange function with the limiting condition and the function itself.
Lag... | why don't you consider the function $$g(x,y)=x+2y+\frac{5-x^2-y^2}{2}$$ this simplifies the problem
and this is $$-\frac{1}{2}(x-1)^2-\frac{1}{2}(y-2)^2+5$$
and the minimum is given by $-5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2645664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Compute $\iint_D\frac{xy}{(1+y^2)^2}\,\mathrm{d}x\,\mathrm{d}y$ where $D=\{(x,y):x\geq 0,\ y\geq 0,\ x^2+y^2 \leq 1\}$
Compute $$I=\iint_D\frac{xy}{(1+y^2)^2}\,dx\,dy,$$ where $D = \{(x,y):x\geq
0, \ y\geq 0,\ x^2+y^2 \leq 1\}.$
In the $xy-$plane, this is just a quarter circle disk in the first quadrant. So choosing... | Try this
\begin{eqnarray}
I&=&\int_{0}^{1}\left(\int_0^{\sqrt{1-x^2}}\frac{xy}{(1+y^2)^2}dy\right)dx\\&=&\int_0^1x\left(\int_0^{\sqrt{1-x^2}}\frac{y}{(1+y^2)^2} \ dy\right)dx \\ &=& \int_0^1x\left(-\frac{1}{2(1+y^2)^2} \right)\Large|_{0}^{\sqrt{1-x^2}} dx\\
&=& \int_0^1x\left(-\frac{1}{2(2-x^2)^2}+\frac{1}{2} \right)dx... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2645797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\lim x_n$ when $\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}$
Let $x_n$ be the unique solution of the equation $$\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}$$
Find $\lim_{n \to \infty} x_n$
I think that the limit must be $\frac{1}... | Answer. $x_n\to \dfrac{1}{2}$
Explanation. Taylor series remainder
$$
\mathrm{e}=1+\frac{1}{1!}+\cdots+\frac{1}{n!}+\frac{\mathrm{e}^{\xi_n}}{(n+1)!}
$$
for some $\xi_n\in(0,1)$. Since
$$\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!},$$
Then
$$
n+x_n=\frac{\log\left(\mathrm{e}-\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2646289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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Is there any way to prove $ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $ by induction since $ \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} $ we have that for each $n\in \Bbb N$ , $ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $
my problem i... | The statement that we will prove with induction is that for every $K > k^*$,
$$\sum_{n=1}^K \frac {1}{n^2} \le \frac {\pi ^2} 6 - \frac 1K $$
where $k^* = 4091641$.
This implies that, for every $K$,
$$\sum_{n=1}^K \frac {1}{n^2} \le \frac {\pi ^2} 6 $$
The idea is to find a function $f(K) > 0$ such that
$$\sum_{n=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2650348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
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Determining $A+B$, given $\sin A + \sin B = \sqrt{\frac{3}{2}}$ and $\cos A - \cos B = \sqrt{\frac12}$. Different approaches give different answers. The question:
Determine $A + B$ if $A$ and $B$ are acute angles such that:
$$\sin A + \sin B = \sqrt{\frac{3}{2}}$$
$$\cos A - \cos B = \sqrt{\frac{1}{2}} $$
Here are th... | Your first way is right.
In the second solution $$A+\frac{\pi}{6}=B-\frac{\pi}{6},$$ which is very well or $$A+B=0,$$ which is impossible.
Another way:
$$2\sin\frac{A+B}{2}\cos\frac{A-B}{2}=\sqrt{\frac{3}{2}}$$ and
$$2\sin\frac{A+B}{2}\sin\frac{B-A}{2}=\sqrt{\frac{1}{2}}.$$
Thus,
$$\tan\frac{B-A}{2}=\frac{1}{\sqrt3}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Evaluate $\sum\limits_{i=1}^\infty a_i $ when $\sum\limits_{i=1}^na_i=\frac {n-1}{n+1}$ Question: $S_n=\frac {n-1}{n+1}$; Find $\sum_1^\infty a_i $
My answer: $a_n=S_n -S_{n-1}=\frac{2}{n^2+n}=\frac{2}{n}-\frac{2}{n+1}$
$\sum_1^\infty a_i=(\frac{2}{1}-\frac{2}{2})+(\frac{2}{2}-\frac{2}{3})+(\frac{2}{3}-\frac{2}{4})+...... | An option
$S_n:= \sum _{i=1}^{n}a_i = $
$\dfrac{n-1}{n+1}=1- 2\dfrac{1}{n+1}.$
$\lim_{ n \rightarrow \infty} S_n = 1- 2\lim_{n \rightarrow \infty}\dfrac{1}{n+1}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
} |
Given $x, y$ are acute angles such that $\sin y= 3\cos(x+y)\sin x$. Find the maximum value of $\tan y$. Given $x, y$ are acute angles such that $\sin y= 3\cos(x+y)\sin x$. Find the maximum value of $\tan y$.
Attempt at a solution:
$$(\sin y = 3(\cos x \cos y - \sin x \sin y) \sin x) \frac{1}{\cos y}$$
$$\tan y = (3 \co... | I start, like everyone else with $3\sin2x/(5-3\cos2x)$.
Consider a circle, radius 3, around 5+0i in the complex plane. The highest value for tan y is for the line through the origin, tangent to the circle. That forms a 3-4-5 right-angled triangle, so $\tan y=3/4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2655203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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Central Limit Theorem Proof using Logarithm Expansion I am trying to go over a proof of the CLT given by this site.
I understood everything up to the point where the expansion for the logarithm was used:
$$x=\frac{t^2}{2n}\ +\ \frac{t^3}{3! n^{\frac{3}{2}}}E(U_i^3).$$
This was used to expand:
$$n\ln\left(1+\frac{t^2}{2... | The logarithm of the moment generating function is
$$\ln m_u(t)=n\cdot \ln\left(1+\underbrace{\frac{t^2}{2n}\ +\ \frac{t^3}{3! n^{\frac{3}{2}}}E(U_i^3)}_{\color{red}x} +\ldots\right)$$
Now you set $x$ equal to $\frac{t^2}{2n}\ +\ \frac{t^3}{3! n^{\frac{3}{2}}}E(U_i^3) \quad (*)$
Thus you can use
$$\ln m_u(t)=n\cdot \l... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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general solution of the FODE $\frac{1}{\sin(x)} \frac{dy}{dx} = y\sec(x) -2$ What's the general solution of the first order differential equation ?
$$
\begin{align*}
\frac{1}{\sin(x)} \frac{dy}{dx} = y\sec(x)-2, \quad 0<x< \frac{\pi}{2}
\end{align*}
$$
my solution is as follows but i think it's wrong
$$
\frac{dy}{dx} ... | It seems correct to me...
$$y'-y\tan(x)=-2\sin(x)$$
$$y'-y\frac {\sin(x)}{\cos(x)}=-2\sin(x)$$
$$\frac {y'\cos(x)-y\sin(x)}{\cos(x)}=-2\sin(x)$$
$$({y}{\cos(x)})'=-\sin(2x)$$
Integrating ..
$${y(x)}{\cos(x)}=-\int \sin(2x) dx$$
$$y(x)=\frac 1 {\cos(x)}(\frac {\cos(2x)}2+K)$$
The same answer as yours.....with $\cos(2x)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2659871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the limit of $\frac{1}{n^2-\sqrt{n^4+4n^2+n}}$ I was trying to solve the following task but I stumbled across something I do not understand:
Calculate:
$$\lim_{n \to \infty}\frac{1}{n^2-\sqrt{n^4+4n^2+n}}$$
my attempt was to factorize n^2 out of the squareroot:
$$$$
$$\lim_{n \to \infty}\frac{1}{n^2-\sqrt{n^4+4n^2... | $$ (n^2 + 2)^2 < n^4 + 4 n^2 + n < \left(n^2 + 2 + \frac{1}{2n} \right)^2 $$
$$ n^2 + 2 < \sqrt{n^4 + 4 n^2 + n \;} \; < \; n^2 + 2 + \frac{1}{2n} $$
$$ - 2 > n^2 -\sqrt{n^4 + 4 n^2 + n \;} \; > \; - 2 - \frac{1}{2n} $$
$$ - \frac{1}{2} < \frac{1}{n^2 -\sqrt{n^4 + 4 n^2 + n \;} } \; < \; - \frac{1}{2 + \frac{1}{2n... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Cauchy can't be applied on this inequality for this condition of a, b, c Given three numbers $a,b,c$ satisfy $0\le a,b,c\le 2$ and $a+b+c=3$.
Prove that $a^2+b^2+c^2\le \frac{1}{ab+bc+ca}+\frac{9}{2}$.
Attempt:
By trial and error, I know that the equality does not hold for $a=b=c=1$. However, the equality holds for all... | Your way is right.
You need only to end it:
Let $a\geq b\geq c$.
Thus, $1\leq a\leq2$ and $$ab+ac+bc\geq a(b+c)=a(3-a)=2+(2-a)(a-1)\geq2.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $A$, $B$ and $C$ are the angles of a triangle then find the value of $\Delta$ I'll state the question from my book below:
If $A$, $B$ and $C$ are the angles of a triangle, then find the determinant value of
$$\Delta = \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B & \cot B & 1 \\ \sin^2C & \cot C & 1\end{vmatrix}... | By Euler's theorem $O,G,H$ are collinear. By considering their trilinear coordinates it follows that
$$ \det\begin{pmatrix}\cos(A) & \cos(B) & \cos(C) \\ \frac{1}{\sin A}&\frac{1}{\sin B}&\frac{1}{\sin C}\\ \frac{1}{\cos A}&\frac{1}{\cos B}&\frac{1}{\cos C}\end{pmatrix}=0$$
and by multiplying the first column by $\cos(... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Find the Cartesian equation of the locus described by $|z+2-7i| = 2|z-10+2i|$
Find the Cartesian equation of the locus described by $|z+2-7i| =
2|z-10+2i|$ Write your answer in the form $(x+a)^2+(y+b)^2=k$.
This was a question from my end of year exams just gone and I'm unsure as to where I have gone wrong :(. If an... | Consider $|z+a+bi|=2|z+c+di|$. Squaring gives
$$(x+a)^2+(y+b)^2=4(x+c)^2+4(y+d)^2$$
that is
$$3(x^2+y^2)+(8c-2a)x+(8d-2b)y+4c^2+4d^2-a^2-b^2=0.$$
In general this is a circle. Now put in your values of $a,\ldots,d$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $24|n^2-1$, if $(n,6)=1$ $n^2-1 = (n-1)(n+1)$
Then $24|(n-1)(n+1)$
$(n,6)=1$: $\exists a,b\in\mathbb{Z}$ that $n = 6\cdot a+b$
Investigate the residues, which arise when dividing the number n by two and three:
$\frac{6\cdot a+b}{3} = \frac{6\cdot a}{3}+\frac{b}{3} = 2\cdot a+\frac{b}{3}$
$\frac{6\cdot a+b}{2... | Note that $\gcd(n,6)=1 \implies n = 6 a + 1$ or $n=6a+5$, where $a \in \mathbb{Z}$.
For $n=6a+1$,
$$(n-1)(n+1) = 12a(3a+1).$$
Since $a(3a+1)$ is always even, i.e., $a(3a+1)=2k$, $(n-1)(n+1) = 24k$.
Similar is the case for $n = 6a+5$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Reducing a matrix to reduced row echelon form without introducing fractions I've been trying to figure out how to reduce this matrix without introducing fractions in the intermediate stages but can't figure out how to do it.
$$\begin{bmatrix}2&1&3\\
0&-2&-29\\
3&4&5\end{bmatrix}$$
| Here's one way.
$\begin{bmatrix}2&1&3\\
0&-2&-29\\
3&4&5\end{bmatrix}
\xrightarrow {R3-R1}
\begin{bmatrix}2&1&3\\
0&-2&-29\\
1&3&2
\end{bmatrix}
\xrightarrow {R1-2R3}
\begin{bmatrix}0&-5&-1\\
0&-2&-29\\
1&3&2
\end{bmatrix}
\xrightarrow {-R1}
\begin{bmatrix}0&5&1\\
0&-2&-29\\
1&3&2
\end{bmatrix}
\xrightarrow {R1+2R2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Product of equation of two lines Why does the product of equations of straight lines represents 2 straight lines?
Why does a 2nd degree equation is so linear?
| In the language of conics,
\begin{align}
0 &= (ax+by+c)(a'x+b'y+c') \\[10pt]
\iff 0 &=
\begin{pmatrix}
x & y & 1
\end{pmatrix}
\begin{pmatrix}
aa' & \frac{ab'+a'b}{2} & \frac{ac'+a'c}{2} \\
\frac{ab'+a'b}{2} & bb' & \frac{bc'+b'c}{2} \\
\frac{ac'+a'c}{2} & \frac{bc'+b'c}{2} & cc'
\end{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Is $5^{1/5} - 3\cdot i$ algebraic? I am studying the book Complex Variables with Applications written by Herb Silverman. In this book, problem number 8 in Question 1.7 is as in the following.
Is $5^{1/5} - 3\cdot i$ algebraic? (i.e, Is the $5$th-root of $5$ minus $3\cdot i$ algebraic?)
Can you help me to solve this?
| Multinomial Combiniations of Algebraic Integers
Given a set of monic polynomials defining a set of algebraic integers, how can we derive a polynomial for a multinomial combination of those algebraic integers?
For example, given $\alpha^2-3=0$ and $\beta^3-2=0$, how do we derive a polynomial for $\alpha+\beta$?
First, c... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Prove that every element in $V$ is on the form $\vec{v}$ Let $u_1=\begin{pmatrix}1\\ 2\\ 3\end{pmatrix}$, $u_2=\begin{pmatrix}2\\ 3\\ 4\end{pmatrix}$, $v_1=\begin{pmatrix}1\\ 1\\ 2\end{pmatrix}$, $v_2=\begin{pmatrix}2\\ 2\\ 3\end{pmatrix}$.
Let $U=span(\vec{u_1},\vec{u_2})$ and $V=span(\vec{v_1},\vec{v_2})$.
I've alrea... | 1) Since $V=span(\vec v_1,\vec v_2)$, there is nothing to prove, because by definition
$$
span(\vec v_1,\vec v_2)=\left\{\alpha\vec v_1+\beta\vec v_2~:~\alpha,\beta\in\mathbb R\right\}=\left\{\begin{pmatrix}\alpha+2\beta\\\alpha+2\beta\\2\alpha+3\beta\end{pmatrix}~:~\alpha,\beta\in\mathbb R\right\}.
$$
2) Write
$$
\beg... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Squaring a Binomial How to find ab if:
$$
a - b = 3
$$
$$
a^2 - b^2 = 21
$$
What I already know is the following:
To find ab for:
$$
a - b = 3
$$
$$
a^2 + b^2 = 29
$$
ab is:
$$
2ab = (a^2 + b^2) - (a - b)^2
$$
| From $a-b=3$ and $(a-b)(a+b)=21$ we get $a-b=3$ and $a+b=7$ and $a=5$, $b=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2680453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that $ \frac{(99)!!}{(100)!!} < \frac{1}{10}$ How to prove that $ \dfrac{(99)!!}{(100)!!}=\dfrac{1\cdot3\cdot5\cdot7\cdot9 \cdots99}{2\cdot4\cdot6\cdot8\cdot10\cdots100} < \dfrac{1}{10}$
Any hint to prove it?
| We know that $\dfrac{99!}{100!}=\dfrac{1}{100}$
We rewrite $\dfrac{1}{100}=\dfrac{99!}{100!}=\dfrac{\color{blue}{1}\cdot\color{red}{2}\cdot\color{blue}{3}\cdot\color{red}{4}\cdots\color{red}{98}\cdot\color{blue}{99}}{\color{blue}{1}\cdot\color{red}{2}\cdot\color{blue}{3}\cdot\color{red}{4}\cdots\color{blue}{99}\cdot\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2681690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\lim_{(x,y,z)\to (0,0,0)}\frac{1}{x^2+y^2+z^2}e^{-\frac{1}{\sqrt{x^2+y^2+z^2}}}$
Evaluate $$\lim_{(x,y,z)\to (0,0,0)}\frac{1}{x^2+y^2+z^2}e^{-\frac{1}{\sqrt{x^2+y^2+z^2}}}$$
Let $t=\sqrt{x^2+y^2+z^2}$
$$\lim_{t\to 0}\frac{1}{t^2}e^{-\frac{1}{t}}$$
Let $r=\frac{1}{t}$
$$\lim_{r\to \infty}r^2 e^{-r}=\lim_{r\t... | HINT
By spherical coordinates $\sqrt{x^2+y^2+z^2}=r\to 0$
$$\frac{1}{ x^2+y^2+z^2 }e^{-\frac{1}{\sqrt{x^2+y^2+z^2}}}=\frac1{r^2}e^{-\frac1r}$$
and for $y=\frac1r\to \infty$ since eventually $e^y\ge y^3$
$$\frac1{r^2}e^{-\frac1r}=\frac{y^2}{e^y}\le\frac{y^2}{y^3}=\frac1y\to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2682660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How to determine whether an integral is convergent or divergent? For this question, I'm not sure if I'm doing it right, can anyone please help me out?
Determine whether the following integral is convergent or divergent.
$$\int_5^6 \frac 1 {(x-3)\sqrt {x-5}} \, dx$$
$$ \frac{1}{(x-3)\sqrt {x-5}}\le \frac{1}{x-3}$$
Since... | The inequality is not valid. Rather,
\begin{align*}
0<\dfrac{1}{(x-3)\sqrt{x-5}}<\dfrac{1}{2\sqrt{x-5}},~~~~x\in(5,6],
\end{align*}
and
\begin{align*}
\int_{5}^{6}\dfrac{1}{\sqrt{x-5}}&=\lim_{\eta\rightarrow 5^{+}}\int_{\eta}^{6}\dfrac{1}{\sqrt{x-5}}dx\\
&=\lim_{\eta\rightarrow 5^{+}}2\sqrt{x-5}\bigg|_{\eta}^{6}\\
&=... | {
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"timestamp": "2023-03-29T00:00:00",
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Finding all positive integers $x,y,z$ that satisfy $2^x=3^y7^z+1$
Find all positive integers $x,y,z$ that satisfy $$2^x=3^y7^z+1$$.
I think that $(x,y,z)=(6,2,1)$ is the only solution, But how can I prove this?
| If we allow $y$ and $z$ to be zero, this can be rephrased as:
For which positive integers $x$ is $2^{x} - 1$ is not divisible by any prime other than $3$ or $7$?
Call the set of such integers $S$.
Suppose $x \in S$. If $t$ divides $x$, then $2^{t} - 1$ divides $2^{x} - 1$, so $t \in S$ as well.
$3$ divides $2^{x} - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2688972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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guess 3 of 5 repeating non-ordered numbers from 1 to 6 I'm trying to calculate the probability of guessing 3 numbers on a set of 5.
The range is $[1, 6]$, the order is not relevant and numbers can repeat.
I did a small script to calculate it by brute force, and I got something unexpected.
Choosing 3 identical numbers (... | Guessing $1,1,1$ will match the sets that have $3$ or more $1$. There is $1$ set of $5$ ones, there are $5*5=25$ sets that have $4$ ones ($5$ positions for the non-one to be in and $5$ options for the value of the non-one), there are ${5\choose 2}*5 = 50$ ways to have $3$ ones and two of another value. And ${5\choose... | {
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"timestamp": "2023-03-29T00:00:00",
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Calculating $\lim_{x \rightarrow 0} \frac{\tan x - \sin x}{x^3}$. I have a difficulty in calculating this limit:
$$\lim_{x \rightarrow 0} \frac{\tan x - \sin x}{x^3},$$
I have tried $\tan x = \frac{\sin x}{\cos x}$, then I unified the denominator of the numerator of the given limit problem finally I got $$\lim_{x \rig... | HINT:
We can write
$$\begin{align}
\frac{\tan(x)-\sin(x)}{x^3}&=\frac{\sin(x)(1-\cos(x))}{\cos(x)x^3}\\\\
&=\left(\frac{1}{\cos(x)}\right)\left(\frac{\sin(x)}{x}\right)\left(\frac{1-\cos(x)}{x^2}\right)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2690311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Proving $\sum_{n=1}^{+\infty}\frac{1}{n\sqrt{n+2}}<2$ without calculus I would like to prove the following inequality without using calculus :
$$
\sum_{n=1}^{+\infty}\frac{1}{n\sqrt{n+2}}<2
$$
Any hint?
Thank you very much!
| We have
\begin{align*}
\frac{1}{n\sqrt{n+2}}\leq\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n+1}},\qquad n\geq 2,
\end{align*}
and here is why: First note that $\sqrt{n^2-1}\leq n$ for all $n\geq 1$. Moreover, $\sqrt{n+1}-\sqrt{n-1}=\frac{2}{\sqrt{n+1}+\sqrt{n-1}}\geq \frac{1}{\sqrt{n+2}}$. From this we get
\begin{align*}
\fr... | {
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"timestamp": "2023-03-29T00:00:00",
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On finding $\sup\left\{3(-1)^n-\frac{1}{n^2+1}\right\}$ How does one find $\sup(A)$ where $A = \left\{3(-1)^n-\frac{1}{n^2+1}: n \in \mathbb{N}\right\}$?
I've tried as follows, but I'm not so sure.
$\displaystyle 3(-1)^n-\frac{1}{n^2+1} \le 3-\frac{1}{n^2+1} \le 3. $ So $3$ is an upper bound for $A$.
Let $\epsilon >0$... | It is fine. Only a technical detail.
$$3-\frac1{2N^2}=3-\frac1{4n_0^2}>3-\frac{2\epsilon}4$$
This is certainly greater than $3-\epsilon$, so your proof is correct, but it would be more elegant if you simply say that $N>1/\sqrt\epsilon$ and even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2691964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate $f(x)$ at a specific point Question:
Calculate $f(x) = \frac{49}{x^2} + x^2$ at points for which $\frac{7}{x}+x =3$
My attempt:-
I tried to find the value of $x$ and insert in $f(x)$
$$\frac{7}{x}+x =3$$
$$7+x^2 =3x$$
$$x^2 -3x + 7=0$$
$$x = \frac{3+\sqrt{9-7*4}}{2}$$
Now $x$ is coming out to be irrational an... | For $x \not =0$, $x$ real:
1) No real, negative $x$ satisfies above equation.
2) For $x >0$ :AM-GM:
$(7/x) +x \ge 2\sqrt{7} \gt 4 $.
Hence there is no real $x$ with: $(7/x) +x = 3$ .
Left to do, find a complex solution.
Note:
$(7/x+x)^2 =$
$ 49/x^2 +2(7) + x^2 =9.$
$49/x^2 +x^2 =-5.$
Hence $f(x)=-5.$
| {
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"url": "https://math.stackexchange.com/questions/2698609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Finding the the derivative of $y=\sqrt{1-\sin x}; 0A question I'm attempting is:
Find the derivative of $ y = \sqrt {1 - \sin x} ; 0 < x <\pi/2$.
I did this:
$y = \sqrt {1 - \sin x} = \sqrt {\cos^2\frac{x}{2} + \sin^2\frac{x}{2} - 2\sin \frac{x}{2}\cos \frac{x}{2}} = \sqrt { (\sin \frac{x}{2}-\cos \frac{x}{2})^2} = \... | An option:
$y^2 = 1-\sin x$; $0\lt x\lt π/2.$
Differentiate both sides with respect to x:
$2y\dfrac{dy}{dx} = -\cos x;$
Since $y \not =0:$
$\dfrac{dy}{dx} = -\dfrac{\cos x}{2\sqrt{1-\sin x}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2698798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Proving that an equation holds when we divide a square in triangles Consider the following square of side $1$ divided into right triangles:
Prove that the following equation holds:
I tried first using the Pythagorean theorem, getting 7 equations, one for each triangle, and after combining the equations, I got the fol... | First prove all the triangles are similar. This is easy as all angles are right or supplementary or congruent.
So
$AH = x$. $AB = 1$. $\frac {BI}{x} = \frac 1{\sqrt{1 + x^2}}; BI = \frac x{\sqrt{1+x^2}}$. $HI = \sqrt{ 1+x^2}- \frac x{\sqrt{1+x^2}}$.
$\frac {IG}{HI} = \frac x1; IG = x(\sqrt{ 1-x^2}- \frac x{\sqrt{1-x... | {
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How to find the sum of all elements of inverse of a matrix without finding the inverse explicitly? Problem Statement:
Find all real $x$ so that the matrix is invertible. $$\begin{bmatrix}
2+x & 2 & 2 & 2 \\ 2 & 2+x & 2 & 2 \\ 2 & 2 & 2+x & 2 \\ 2 & 2 & 2 &
2+x \end{bmatrix}$$
Assuming that $A^{-1}$ exists find the s... | As suggested here, the sum of all elements in a matrix $M$ is given by $\langle Me, e\rangle$ where $e = \pmatrix{1 \\ 1 \\ 1 \\ 1}$.
In that case, if $v$ is a vector such that $Av = e$, then we have $$\langle A^{-1}e, e\rangle = \langle v, e\rangle$$
Therefore, we have to solve the system
$$\pmatrix{2+x & 2 & 2 &2 \\ ... | {
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Some fundamental theory of calc 2 questions. I want to evaluate these integrals. I think I understand it, but I just wanted to check:
*
*$$\int_{-1}^1 x^{100} dx$$
$$\frac{1^{101}}{101} - \frac{-1^{101}}{101}$$
$$= \frac{1}{101} + \frac{1}{101} = \frac{2}{101}$$
*$$\int_{1}^9 \sqrt{x} dx$$
$$\frac{2}{3} \cdot 9^{\f... | As @littleO points out, the antiderivative of part 3 is incorrect.
$$\int_{1}^8 x^{-2/3} dx = \left[ \frac{x^{1/3}}{1/3} \right]_1^8 = [3\sqrt[3]{x}]_1^8 = 3(2-1) = 3$$
The rest is correct.
| {
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Asymptotic for the following series. $$S_N=\frac{\sum_{i=1}^{2N}\sqrt{i}+\sqrt{i-1}}{\sum_{i=1}^{N}\sqrt{i}+\sqrt{i-1}}$$
I want to find an equivalence/asymptotic for $S_N$ as $N$ become very large. I tried the following:
Edit
We know that $$\sum_{i=1}^{N}\sqrt{i}\approx\frac{2N^{3/2}}{3}$$ and so $$S_N\approx\frac{(2... | Here is a
reasonably elementary proof
for an arbitrary exponent,
not just $\frac12$.
It shows that
if the exponent is $a$
with $a > 0$,
then the limit is
$2^{a+1}$.
It also gives explicit bounds.
It is based on
this:
If $a > 0$ then
$\dfrac{n^{a+1}}{a+1}+n^a
\gt \sum_{1}^{n} k^a
\gt \dfrac{n^{a+1}}{a+1}
$.
(Similar b... | {
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$ x^{2}- x- 4+ \frac{4\left ( x- 1 \right )^{2}}{\sqrt{3+ 2x-x^{2}}+ x+ 1}> 0 $ Solve the equation
$$x^{3}+ x+ 6= 2\left ( x+ 1 \right )\sqrt{3+ 2x-x^{2}}$$
The only answer is $x= 1$ so
I have rewritten it to
$$\left ( x- 1 \right )\left ( x^{2}- x- 4 \right )= 2\left ( x+ 1 \right ) \left (\sqrt{3+ 2x-x^{2}}- x- 1 \... | Let $t=x-1$ and rewrite the equation like this
$$ t^3+3t^2+4t+8 = 2(t+2)\sqrt{4-t^2}$$
We see that $t=0$ (and so $x=1$) is a solution. Let us show it is the only one.
We also see that $t\in[-2,2]$. Now it is easy to see that $y=4t+8$ is tangent at $f(t) = t^3+3t^2+4t+8$. Let's prove that for all $t\in[-2,2]$ we have:
... | {
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Let $a$ be the real root of the equation $x^3+x+1=0$ Let $a$ be the real root of the equation $x^3+x+1=0$
Calculate $$\sqrt[\leftroot{-2}\uproot{2}3]{{(3a^{2}-2a+2)(3a^{2}+2a)}}+a^{2}$$
The correct answer should be $ 1 $. I've tried to write $a^3$ as $-a-1$ but that didn't too much, I guess there is some trick here :s
| Maybe there is a trick to it, but good ol' factoring can work here.
It is easy to check (using the identity $a^3=-a-1$) that
$$ (3a^2-2a+2)(3a^2+2a) = -5a-7a^2 $$
Let $\sqrt[3]{-5a-7a^2}+a^2 = y$. Given that $a\in \mathbb{R}$, we know that also $y\in \mathbb{R}$, and we must solve for it.
$$y-a^2 = \sqrt[3]{-5a-7a^2} \... | {
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How can I solve this inequality? $ \frac{x+14|x|-10}{|4x-6|-21}>3$ First I looked the x that doesnt belong to this function.
$$|4x -6| - 21 \neq 0$$
$$ x \neq \frac{-15}{4}$$ and $$ x \neq \frac{27}{4}$$
Then I found the roots of the x
$$x = 0$$
$$x = \frac{3}{2}$$
After I found the roots I wrote the inequality like th... | Transform the inequality:
$$\frac{x+14|x|-10-3|4x-6|+63}{|4x-6|-21}>0.$$
Consider the $3$ cases:
$$1) \ \begin{cases} x<0 \\ \frac{x-35}{4x+15}>0\end{cases} \Rightarrow \begin{cases} x<0 \\ x<-\frac{15}{4} \ \ \text{or} \ \ x>35\end{cases} \Rightarrow \color{blue}{x<-\frac{15}{4}}.$$
$$2) \ \begin{cases} 0<x<\frac32 \... | {
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Evaluate $\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$ I want to evaluate $$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$$
First,I tried to evaluate like this:
$$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx=\int_{0}^{\frac{\pi}{2}}x^2\left(\frac{1+\cos x}{\sin x}\right)\frac{dx}{1+\cos x}=\int_{0}^{\frac{\pi}{2}}x^... | $$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \frac{x^{2}}{\sin x} d x
=& \int_{0}^{\frac{\pi}{2}} x^{2} d\left[\ln \left(\tan \frac{x}{2}\right)\right] \\
=& {\left[x^{2} \ln \left(\tan \frac{x}{2}\right)\right]_{0}^{\frac{\pi}{2}}-2 \int_{0}^{\frac{\pi}{2}} x \ln \left(\tan \frac{x}{2}\right) d x } \\
=&-8 \int_{0}^{... | {
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"source": "stackexchange",
"question_score": "15",
"answer_count": 9,
"answer_id": 8
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Find $\frac{dy}{dx}$ if $y=\sin^{-1}[x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}]$, $0
Find derivative of $f(x)=\sin^{-1}[x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}]$, $0<x<1$
Let $x=\sin a$ and $\sqrt{x}=\cos b$
Then I'll get:
$$
y=\sin^{-1}[\sin a\cos b-\cos a\sin b]=\sin^{-1}[\sin(a-b)]\\
\implies\sin y=\sin(a-b)\\
\implies y=n\pi+(-1)... | Use that $$(\arcsin(x))'=\frac{1}{\sqrt{1-x^2}}$$
the whole derivative is given by $$f'(x)={\frac {1}{\sqrt {- \left( x\sqrt {1-x}+\sqrt {x}\sqrt {-{x}^{2}+1}
\right) ^{2}+1}} \left( \sqrt {1-x}-1/2\,{\frac {x}{\sqrt {1-x}}}+1/2
\,{\frac {\sqrt {-{x}^{2}+1}}{\sqrt {x}}}-{\frac {{x}^{3/2}}{\sqrt {-{
x}^{2}+1}}} \right)... | {
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"source": "stackexchange",
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Integration by Substitution of Fraction involving e
Find $\int\frac{2}{e^{2x}+4}$ using $u=e^{2x}+4$
The answer is $\frac{1}{2}x-\frac{1}{4}\ln(e^{2x}+4)+c$
I must have made a mistake somewhere as my answer is not the same. Apologies the question may be too specific, but I am teaching myself calculus.
$\int\frac{2}{e... | You wrote
$$\int\frac{1}{u^2-4u}du=\int u^{-2}-\frac{1}{4}u^{-1}du$$
Which is incorrect
Try writing it as
$$\int\frac{1}{u^2-4u}du=\frac{1}{4}\int \frac{1}{u-4}-\frac{1}{u}du$$
Or
$$\int\frac{1}{u^2-4u}du=\int\frac{1}{(u-2)^2-4}du$$
And proceed through substitution
| {
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Calculating the largest possible area of a rectangle inscribed in an ellipse So i got the equation $4x^2 + 9y^2 = 3600$
What i've done so far is:
$A= (2x)(2y) = 4xy$
Then I find the expression of $y$
$9y^2= 3600 -4x^2$
$y = \pm \sqrt{3600 -4x^2 / 9} = 2/3(\sqrt {900 - x^2} \quad 2/3(900 -x^2)^{1/2}$
Then i set
$A = 4... | Rather than using the constraint to find $x$ in terms of $y$ or vice versa, use implicit differentiation.
Differentiate the objective and set it equal to $0.$
$A = 4xy\\
\frac {dA}{dx} = 4y + 4x\frac {dy}{dx} = 0$
Differentiate the constraint.
$\frac {d}{dx} (4x^2 + 9y^2 = 3600)\\
8x + 18y\frac {dy}{dx} = 0\\
\frac {dy... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Find the $\lim\limits_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}$ The task is to find $$\lim_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}$$
What I've tried is dividing both the numerator and the denominator by $x$, but I just can't calculate it completely.
I know it should be s... | We can rewrite the fraction as follows
\begin{align*}
\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}&=\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}\times\frac{\sqrt{x^2+a^2}-x}{\sqrt{x^2+a^2}-x}\times \frac{\sqrt{x^2+b^2}-x}{\sqrt{x^2+b^2}-x}\\[4pt]
&=\frac{x^2+a^2-x^2}{x^2+b^2-x^2}\times\frac{\sqrt{x^2+b^2}-x}{\sqrt{x^2+a^2}-x... | {
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"source": "stackexchange",
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"answer_count": 4,
"answer_id": 2
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Definite integral of $y=\sqrt{(16-x^2)}$ The integral is
$$\int^3_1\sqrt{16-x^2}dx$$
I've used the trig substitution method, replacing $x$ with $4\sin\theta$:
$$x=4\sin\theta, \quad \theta=\arcsin\left(\frac x4\right), \quad dx=4\cos\theta \ d\theta$$
(I've excluded the intervals for the definite integral for now)
\be... | $\sin(2\sin^{-1}(3/4))=2\sin(\sin^{-1}(3/4))\cos(\sin^{-1}(3/4))=2\cdot(3/4)\cdot(\sqrt{7}/4)$ because of the identity that $\cos(\sin^{-1}(3/4))=\sqrt{1-\sin^{2}(\sin^{-1}(3/4))}=\sqrt{1-9/16}=\sqrt{7}/4$.
| {
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Using the Euclidean algorithm, deduce that $\gcd(x^3+2x^2+x +4;x^2+1)=1$ So, I've tried it but I keep getting to $$\frac{x^2+1} 2$$ and don't know how to proceed.
Question is Using the Euclidean algorithm, deduce that $$\gcd(x^3+2 x^2+x +4,\;x^2+1)=1.$$
| $$ \left( x^{3} + 2 x^{2} + x + 4 \right) $$
$$ \left( x^{2} + 1 \right) $$
$$ \left( x^{3} + 2 x^{2} + x + 4 \right) = \left( x^{2} + 1 \right) \cdot \color{magenta}{ \left( x + 2 \right) } + \left( 2 \right) $$
$$ \left( x^{2} + 1 \right) = \left( 2 \right) \cdot \color{ma... | {
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If sides $a$, $b$, $c$ of $\triangle ABC$ are in arithmetic progression, then $3\tan\frac{A}{2}\tan\frac {C}{2}=1$
If sides $a$, $b$, $c$ of $\triangle ABC$ (with $a$ opposite $A$, etc) are in arithmetic progression, then prove that
$$3\tan\frac{A}{2}\tan\frac{C}{2}=1$$
My attempt:
$a$, $b$, $c$ are in arithmetic ... | Expand your last line: $$2\left(\cos\frac A2\cos\frac C2 - \sin\frac A2\sin\frac C2\right)=\left(\cos\frac A2\cos\frac C2 +\sin\frac A2\sin\frac C2\right)$$
and your result is immediate after a cancellation.
| {
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"source": "stackexchange",
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Finding the area of circles in triangle In the figure there are infinitely many circles approaching the
vertices of an equilateral triangle, each circle touching other circles and sides of the triangle. If the triangle has sides of
length $1$, how can i find the total area occupied by the circles?
| Denote the circumradius by $R$ and the radii of the circles by $r_1,r_2,r_3,...$.
We will find $R$ and $r_1$:
$$r_1=\frac{2S}{a+b+c}=\frac{2\cdot\frac{\sqrt{3}}{4}}{3}=\frac{1}{2\sqrt{3}} \\
R=\frac{abc}{4S}=\frac{1}{4\cdot \frac{\sqrt{3}}{4}}=\frac{1}{\sqrt{3}}.$$
From the similarity of the triangles:
$$\begin{align... | {
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Solving $\sqrt{a +\sqrt{a-x}}+\sqrt{a-\sqrt{a+x}}=2x, \ a\in\Bbb{R}$
Given
$$\sqrt{a +\sqrt{a-x}}+\sqrt{a-\sqrt{a+x}}=2x$$
and $a\in\Bbb{R}$, express $x$ in terms of $a$.
I rationalised the above expression and then again rationalised which gave me :
$$\sqrt{a+\sqrt{a-x}}-\sqrt{a-\sqrt{a+x}}=\frac{1}{\sqrt{a+x}-\... | Squaring both sides, $$a\require{\cancel}\cancel{+\sqrt{a-x}}+a\cancel{-\sqrt{a-x}}+2\sqrt{\left(a+\sqrt{a-x}\right)\left(a-\sqrt{a-x}\right)}=2\left(a+\sqrt{a^2-a+x}\right)=4x^2.$$ since for all $m,n$ one has that $(m+n)(m-n)=m^2-n^2$. Now, dividing both sides by $2$,$$\begin{align} a+\sqrt{a^2-a+x}&=2x^2 \\ \Leftrigh... | {
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Show that $ \sum_{k=0}^{j} \binom{2j+1}{k+j+1} \ \frac{(-1)^k}{2k+1}=2^{2j-1}\ B(j+1,1/2)$. I want to prove that
$$ \sum_{k=0}^{j} \binom{2j+1}{k+j+1} \ \frac{(-1)^k}{2k+1}=2^{2j-1}\ B(j+1,1/2),$$
where $B(\cdot , \cdot)$ is the beta function.
My idea was to change it to something like
my previous question.
Edit 1.It... | $S_n=\sum\limits_{k=0}^{n} \dbinom{2n+1}{k+n+1} \ \dfrac{(-1)^k}{2k+1}$
where $j=n$ (I am sorry I used $n$ instead of $j$ during the long proving).
1,
$\dbinom{2n+1}{k+n+1}=\dfrac{(2n+1)!}{(n+k+1)!(n-k)!}\dfrac{n!}{k!}\dfrac{k!}{n!}\dfrac{(n+1)!}{(n+1)!}=$ $\dbinom{n}{k}\dfrac{\dbinom{2n+1}{n}}{\dbinom{n+k+1}{k}}$
So... | {
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What is the limit of the average value of the first $n$ terms of $(1, 2, 1, 1, 1, 2, 1, 1, 2, 1, ...)$ as $n\to\infty$? There exists a sequence $a_n$ that begins $(1, 2, 1, 1, 1, 2, 1, 1, 2, 1, ...)$. It is fully defined on the OEIS at A293630, but I will give a simple explanation here. The sequence starts $1, 2$. The ... | We establish the convergence of the limit.
Step 1. Designate $(1,2)$ as the $1$-st stage and let $L_n$ be the length of the sequence at the $n$-th stage. Also, let $S_n = \sum_{k=1}^{n} a_k$ and $T_n = S_{L_n}$. Then
$$
L_{n+1}
= \begin{cases}
2L_n - 1, & \text{if } a_{L_n} = 1 \\
3L_n - 2, & \text{if } a_{L_n} = 2
\en... | {
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Adding $k^2$ to $1^2 + 2^2 + \cdots + (k - 1)^2$. In his book, Calculus Vol. 1, Tom Apostol mentions that adding $k^2$ to the predicate
$$A(k): 1^2 + 2^2 + \cdots + (k - 1)^2 < \frac{k^3}{3}$$
gives the inequality
$$ 1^2 + 2^2 + \cdots + k^2 < \frac{k^3}{3} + k^2.$$
Why does the RHS $$1^2 + 2^2 + \cdots + (k - 1)^2 $$ ... | The reason why,$$1^2 + 2^2 + \cdots + (k-1)^2,$$ becomes, $$1^2 + 2^2 + \cdots + (k - 1)^2+ k^2,$$ is because you're adding the $k$th term squared on both sides. Thus, instead of having the sum up to the $k-1$ th term squared, now you have the sum up to the $k$th term squared.
| {
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"source": "stackexchange",
"question_score": "1",
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Mini-Challenge on a condition Hello I would like to purpose to you an enigma this is the following :
Let $a,b,c$ be positive real number find the condition on $abc$ and $ab+bc+ca$ and $a+b+c$ to have $a^9+b^9+c^9=3$
I have a solution using the following identity :
$$a^9+b^9+c^9= 3a^3b^3c^3−45abc(ab+bc+ca)(a+b+c)^4+ ... | If you set $e_1=a+b+c,$ $e_2=ab+ac+bc,$ and $e_3=abc$ and you define a recurrence relation as follows:
\begin{eqnarray*}
x_0 &=& 3 \\
x_1 &=& e_1 \\
x_2 &=& e_1^2-2e_2 \\
x_{n+3} &=& e_1x_{n+2} - e_2x_{n+1} + e_3x_n
\end{eqnarray*}
then it can be shown that $x_n = a^n+b^n+c^n.$ In order to have $a^9+b^9+c^9=3,$ there m... | {
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"question_score": "2",
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Proving that ${4a \choose 2a} - {2a \choose a}$ is divisible by 4 I'm trying to prove that ${4a \choose 2a} - {2a \choose a}$ is divisible by $4$, and I've currently tried using induction, though I have been unsuccessful. I'm not really sure where to go to for now, any help would be appreciated.
This is what I have cur... | Beware: I am going for the overkill. By Legendre's theorem exponent of the largest power of $2$ dividing $n!$ is $\sum_{k\geq 1}\left\lfloor\frac{n}{2^k}\right\rfloor$, hence
$$ \nu_2\binom{2n}{n} = \sum_{k\geq 1}\left(\left\lfloor\frac{2n}{2^k}\right\rfloor-2\left\lfloor\frac{n}{2^k}\right\rfloor\right)\tag{1}$$
and t... | {
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$\sin x-\frac{1}{2}\sin ^2x+\frac{1}{4}\sin ^3x-\frac{1}{8}\sin ^4x....\forall x \in \mathbb{R}$ is convergent? Is $\sin x-\frac{1}{2}\sin ^2x+\frac{1}{4}\sin ^3x-\frac{1}{8}\sin ^4x....\forall x \in \mathbb{R}$ convergent? If it is convergent find the sum of the series.
Gives series $$\sin x-\frac{1}{2}\sin ^2x+\frac... | Let $a:=\sin x$ then the series can be written as
$$\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\sin^nx}{2^{n-1}}=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{a^n}{2^{n-1}}=a\sum_{n=1}^{\infty}(-1)^{n-1}\Big(\frac{a}{2}\Big)^{n-1}=a\frac{1}{1-(-a/2)}=\frac{2a}{2+a}$$
by the geometric series whenever $|a|<2$ which is equivalent to $|\sin... | {
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"source": "stackexchange",
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How to find $y^{(y^2-6)}$? $$\frac{3}{1-3^{x-2}} + \frac{3}{1-3^{2-x}} = y$$ $$x≠2$$ $$y^{(y^2-6)} = ?$$What is $y^{(y^2-6)}$? Could you please explain to me how to solve this question step by step?
| Note that
$$y=\frac{3}{1-3^{x-2}} + \frac{3}{1-3^{2-x}} = \frac{3}{1-3^{x-2}} + \frac{3}{1-\frac1{3^{x-2}}}=\\= \frac{3}{1-3^{x-2}} - \frac{3^{x-1} }{1-3^{x-2}}=\frac{3-3^{x-1}}{1-3^{x-2}}=3\frac{1-3^{x-2}}{1-3^{x-2}}=3 $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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An inequality involving three consecutive primes Can you provide a proof or a counterexample to the following claim :
Let $p,q,r$ be three consecutive prime numbers such that $p\ge 11 $ and $p<q<r$ , then $\frac{1}{p^2}< \frac{1}{q^2} + \frac{1}{r^2}$ .
I have tested this claim up to $10^{10}$ .
For $p>5$ we get $\pi... | This is a comment as opposed to an answer
All primes $p_{n+1} < 2p_n$. Let $p'$ be the prime before $p$, $q'$ be the prime before $q$, and $r'$ be the prime before $r$. Then, $$p+q+r< 2(p' + q' + r')$$ So $$\frac{1}{p^2} < \frac{1}{\big(2(p' + q' + r') - q - r\big)^2}$$ Also, $$\frac{1}{p^2} < \frac{1}{q^2} + \frac{2}... | {
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"source": "stackexchange",
"question_score": "20",
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Matrix Notation Form of Roots of a Quadratic Equation We know that the quadratic equation
$$f(x)=ax^2+bx+c=0$$
has roots
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=-\frac b{2a}\pm \frac 1a\sqrt{-\left(ac-\frac {b^2}4\right)}$$
Also, $f(x)$ can be written in matrix notation as follows:
$$f(x)=
\left(\begin{matrix}x&1\\\end{m... | Let us switch to homogeneous coordinates and use $2b$ instead of $b$ for convenience.
$$ax^2+2bxy+cy^2=\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}a&b\\b&c\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\mathbf{p^TQp}=0.$$
Assuming $Q$ diagonalizable, we have (the transformation $P$ can be taken orthonormal)
$$\mathbf{... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Find the remainder when $(x+1)^n$ is divided by $(x-1)^3$. I know this question has been answered before, but I have a slightly different different question.
I saw the solution of this question in my book and the author has solved it by substituting $x-1=y$ and then equating the coefficients of $y^2$, $y^1$ and $y^0$ ... | $$
(x+1)^n = Q(x)(x-1)^3 + ax^2+b x+c
$$
$$ 2^n = a+b+c $$
$$
n(x+1)^{n-1} = Q'(x)(x-1)^3+3Q(x)(x-1)^2 + 2ax+b \Rightarrow n2^{n-1}=2a+b
$$
In the same way
$$
n(n-1)(n-2)2^{n-2} = 2a
$$
etc.
| {
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"source": "stackexchange",
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Show that the solutions from one quadratic equation are reciprocal to the solutions of another quadratic equation From Sullivan's Algebra & Trigonometry book: Chapter 1.2; Exercise 116:
Show that the real solutions of the equation $ax^2+bx+c=0$ are the
reciprocals of the real solutions of the equation $cx^2+bx+a=0$.... | If $r$ is a root of $ax^2+bx+c$, then $ar^2+br+c=0$ and\begin{align}ar^2+br+c=0&\iff \frac{ar^2+br+c}{r^2}=0\\&\iff a++b\times\frac1r+c\times\frac1{r^2}=0\\&\iff\frac1r\text{ is a root of }cx^2+bx+c=0.\end{align}
| {
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Convex quadrilateral; Area and parallel sides Consider a convex quadrilateral $ABCD$ with diagonals $AC$ and $BD$ intersecting at $O$.
Prove that the area of triangle $AOB$ = the area of the triangle $COD$ if an only if $AD$ and $BC$ are parallel.
Where should I start with this problem?
Thanks!
| Good to have a diagram.
Now, if $AD || BC$ then $w = z$ and $x = y$, so we get that triangles $AOD$ and $BOC$ are similar. Hence, we get $\frac{OB}{OD} = \frac{OC}{OA}$, and therefore $OA \cdot OB = OC \cdot OD$. We know that $\angle AOB = \angle COD$, hence from the formula for area,
$$
\mbox{Area}(AOB) = \frac 12 O... | {
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Prove that $\int_{\mathbb{R}} \frac{1}{(a^2+t^2)(b^2+t^2)} dt= \frac{\pi}{ab(a+b)}$
Prove that $\int_{\mathbb{R}} \frac{1}{(a^2+t^2)(b^2+t^2)} dt= \frac{\pi}{ab(a+b)}$
For $a,b>0$, the Fourier transform of $f(x) = e^{-a|x|}$ is $\widehat{f}(t) = \dfrac{2a}{a^2+t^2}$.
Otherwise, we have $\int_{\mathbb{R}}\dfrac{1}{(a^... | Hint:
$$\frac{1}{(a^2+t^2)(b^2+t^2)}=\frac{1}{b^2-a^2}\left(\frac{1}{a^2+t^2}-\frac{1}{b^2+t^2}\right).$$
| {
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Prove the given determinant Prove the given determinant:
$$\left|
\begin{matrix}
a&b&ax+by \\
b&c&bx+cy \\
ax+by&bx+cy&0 \\
\end{matrix}\right|=(b^2-ac)(ax^2+2bxy+cy^2)$$
I didn't get any idea. Please help me solve this.
| A bit late in the day, here is the symmetric matrix version of this, writing $P^TAP = B,$ where $B$ is also symmetric with obvious determinant. We say that $A$ and $B$ are "congruent"
$$
\left(
\begin{array}{ccc}
1&0&0 \\
0&1&0 \\
-x&-y&0 \\
\end{array}
\right)
\left(
\begin{array}{ccc}
a&b&ax+by \\
b&c&bx+cy \\
ax+by&... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Without using a calculator, is $\sqrt[8]{8!}$ or $\sqrt[9]{9!}$ greater? Which is greater between $$\sqrt[8]{8!}$$ and $$\sqrt[9]{9!}$$?
I want to know if my proof is correct...
\begin{align}
\sqrt[8]{8!} &< \sqrt[9]{9!} \\
(8!)^{(1/8)} &< (9!)^{(1/9)} \\
(8!)^{(1/8)} - (9!)^{(1/9)} &< 0 \\
(8!)^{(9/72)} - (9!)^{8... | This step doesn't look right to me:
\begin{gather}
(8!)^{(9/72)} - (9!)^{8/72} < 0 \\[6px]
(9!)^{8/72} \left(\left( \frac{8!}{9!} \right)^{(1/72)} - 1\right) < 0
\end{gather}
When you divide $(8!)^{9/72} $ by $(9!)^{8/72}$, you should get
$$
\frac{(8!)^{9/72}}{(9!)^{8/72}} =
\frac{(8!)^{9/72}}{(9\cdot8!)^{8/72}} =
\fr... | {
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"source": "stackexchange",
"question_score": "34",
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"answer_id": 7
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A limit about $\left(1+\frac{1}{n}\right)^{n}$? Here is my question:
$$\displaystyle\lim_{n\rightarrow \infty}n^2\left[\left(1+\frac{1}{1+n}\right)^{n+1}-\left(1+\frac{1}{n}\right)^{n}\right]=?$$
Any hints will be fine. Thank you!
| HINT
Note that (to be made more rigorous)
*
*$(1+\frac{1}{1+n}) ^{n+1}=e^{(n+1)\log\left(1+\frac{1}{1+n}\right)}\sim e^{(n+1)\left(\frac{1}{1+n}-\frac{1}{2(1+n)^2}\right) }=e^{1-\frac{1}{2(1+n)}}\sim e\left(1-\frac{1}{2(1+n)}\right)$
*$(1+\frac{1}{n}) ^{n}=e^{n\log\left(1+\frac{1}{n}\right)}\sim e^{n\left(\frac{1}{... | {
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Laurent series for $\frac{1}{z(z+3)(z-1)^2}$ Find the Laurent Series for
$$\frac{1}{z(z+3)(z-1)^2}$$ in $1 < |z-1| < 4$
So I did the partial fraction decomposition which yields:
$$\frac{1}{4(-1+z)^2} - \frac{5}{16(-1+z)} + \frac{1}{3z} - \frac{1}{48(3+z)}$$
Can anyone help me finish this problem?
| You are on the right track. Partial fraction expansion reveals
$$\frac{1}{z(z+3)(z-1)^2}=\frac{1/3}{z}-\frac{1/48}{z+3}-\frac{5/16}{z-1}+\frac{1/4}{(z-1)^2}\tag1$$
The last two terms on the right-hand side of $(1)$ constitute part of the Laurent expansion in the annulus $1<|z-1|<4$.
We now expand the first term, $\fra... | {
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If $a,b,c$ are the roots of $x^3+x^2-2x+1=0$ then what is the value of $\det\Delta$?
If $a,b,c$ are the roots of $ x^3+x^2-2x+1=0$, then what is the value of $\det\Delta$ where:
$$\Delta=\begin{bmatrix}
c^2 & b^2 & 2bc-a^2 \\ 2ac-b^2 & a^2 & c^2 \\ a^2 & 2ab-c^2 & b^2 \\ \end{bmatrix}$$
... | Hint: For your determinant we get $$ \left( b+a+c \right) ^{2} \left( {b}^{2}-ab-bc-ac+{a}^{2}+{c}^{2}
\right) ^{2}
$$
| {
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Simplifying the derivative of $x^{\frac{2}{3}} \cdot (6-x)^{\frac{1}{3}}$ $$x^{\frac{2}{3}} \cdot (6-x)^{\frac{1}{3}}$$
So I get:
$$-x^{\frac{2}{3}} \cdot \frac{1}{3} (6-x) ^{\frac{-2}{3}} + (6-x) ^{\frac{1}{3}} \cdot \frac{2}{3} x ^ {\frac{-1}{3}}$$
How does one go about simplifying this?
I guess I can pull out common... | We've got to preserve the terms here, and we see that the ordering is mismatched, as in
$$\frac{d}{dx}(x^{\frac{2}{3}}(6-x)^{\frac{2}{3}})=\frac{2}{3}(\frac{6-x}{x})^{\frac{1}{3}}-\frac{1}{3}(\frac{x}{6-x})^{\frac{2}{3}}$$
Then we can factor out $(\frac{6-x}{x})^{\frac{1}{3}}$ to make the second term much nicer with n... | {
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"source": "stackexchange",
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Maximum power of $2$ which divides $3^{1024}-1$ What is the maximum power of $2$ which completely divides $3^{1024}-1$?
I proceeded thus:
$\phi(2^n)=2^{n-1}$ for all $n\ge1$
$$3^{1024}=3^{2^{10}}\equiv1\pmod {2^{11}}$$
$$3^{1024}-1\equiv0\pmod {2^{11}}$$
Since $\phi(2^{11})=2^{10}$. So, maximum power of $2$ must be $1... | If $2^n$ divides $3^{1024}-1$, then $3$ has order at most $1024=2^{10}$, mod $2^n$.
Now, $3$ has order $2^{n-2}$ mod $2^n$. (*)
So, $n-2=10$ is the maximum possible, that is, $n=12$.
(*) By induction, $3^{2^{n-3}} \equiv 1+2^{n-1} \bmod 2^n$ and $3^{2^{n-2}} \equiv 1 \bmod 2^n$ for $n\ge 4$.
| {
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"question_score": "2",
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Calculate $\sum\limits_{k=0}^{20}(-1)^k\binom{k+2}2$ without calculating each term separately Is it possible to calculate $\sum_{k=0}^{20}(-1)^k\binom{k+2}2$ without calculating each term separately?
The original question was find the number of solutions to $2x+y+z=20$ which I calculated to be the coefficient of $x^{20... | There is! You can use the identity $\binom{k+2}2 = \sum_{i=0}^{k+1}i$.
Our sum is $$\sum_{k=0}^n (-1)^k \binom{k+2}{2} = \sum_{k=0}^n (-1)^k \sum_{i=0}^{k+1}i$$
For odd $n$: Letting $m=\frac {n-1}2$ and pairing up the terms we get
$$\begin{align}\sum_{k=0}^n (-1)^k \sum_{i=0}^{k+1}i &= \sum_{j=0}^m [(-1)^{2j}\sum_{i=0}... | {
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How to transform arbitrary rectangle into specific parallelogram? I have a rectangle $S$ in $(u,v)$ coordinates. $S = [a,b] \times [c,d]$.
I want to find the transformation that yields a parallelogram given by $(0,0),(-2,1),(2,4),(4,3)$. I have deduced this is equivalent to
$$\bigg\lbrace (x,y) \bigg | \text{area bou... | After trial and error, I found that
Translation:
$$T = \begin{bmatrix}
-a \\ -c \\
\end{bmatrix}$$
Rescale:
$$A_1 = \begin{bmatrix}
\frac{5}{b-a} & 0 \\
& \frac{2}{d-c}
\end{bmatrix}$$
Shear:
$$A_2 = \begin{bmatrix}
1 & -\frac{1}{2} \\
0 & 1
\end{bmatrix}$$
Rotation:
$$R = \begin{bmatrix}
4/5 & -3/5 \\
3/5 & 4/... | {
"language": "en",
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"source": "stackexchange",
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How to find all rational solutions of $\ x^2 + 3y^2 = 7 $? I knew that for $ x^2 + y^2 = 1$ the x and y can be expressed by introducing one more variable where $\ m=y/(x+1) $, then $\ x= 2m/(1+m^2) $ and $\ y= (1-m^2)/(1+m^2) $. What about $\ x^2 + 3y^2 = 7 $, should I divide the equation by 7 in order to get the 1 at ... | Using the method of pg 7 of this paper on this related equation
$$x^2+3y^2=7z^2 \quad \text{with initial solution} \quad (x,y,z)=(2,1,1)$$
A line $y=t(x-2)+1$, which will cut through the ellipse $x^2 + 3y^2 = 7$ at rational points if $t$ is rational.... when substituted into the ellipse yields:
$$\begin{align}
x^2+3\le... | {
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Inhomogeneous Wave Equation Energy Method Let $u$ be a solution to the following equation:
$$
u_{tt}=u_{xx}−u^3
$$
Assume that $u(x,0) =u_t(x,0) = 0$ for all $x\in[a, b]$. Prove that $u(x, t) = 0$ if $a+t < x < b−t$.
I initially thought of using Duhamel's principle on this but realized that the cube in the integrand w... | Note that, thanks to the Leibniz integral rule,
\begin{align}
&\frac{\rm d}{{\rm d}t}\int_{a+t}^{b-t}\left(\frac{1}{2}u_t^2+\frac{1}{2}u_x^2+\frac{1}{4}u^4\right){\rm d}x\\
&=\int_{a+t}^{b-t}\left(u_tu_{tt}+u_xu_{xt}+u^3u_t\right){\rm d}x-\frac{1}{2}\left(u_t^2+u_x^2+\frac{1}{2}u^4\right)\Bigg|_{x=b-t}-\frac{1}{2}\left... | {
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"source": "stackexchange",
"question_score": "2",
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Values of $p$ such that the roots of $x^2+px+p$ are greater than $1$ I found this question in a book, the answer is given as $(-\infty,-5)$.
I know that for both roots of a quadratic polynomial to be greater than 1, $D \ge 0$, which implies that $p^2-4p \ge 0$, which means $p \ge 0$ and $p \ge 4$, or $p < 0$ and $p < ... | Let $x=a+1$, then the equation is equivalent to
$(a+1)^2+p(a+1)+p=0$
$\Leftrightarrow a^2+2a+1+ap+p+p=0$
$\Leftrightarrow a^2+(p+2)a+2p+1=0$
We need to find $p$ so that $x>1$, or $a+1>1$, or $a>0$, which means we need to find $a$ so that the equation $a^2+(p+2)a+2p+1=0$ has two roots and both of them are positive roots... | {
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"source": "stackexchange",
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Showing $\lim_{x \to2^-}\frac{1}{x-2} = -\infty$ using the definition of the limit
Show
$$\lim_{x \to2^-}\frac{1}{x-2} = -\infty $$
My answer:
For all $B<0, \exists \delta >0 $ such that $\frac{1}{x-2} < B$ always that $-\delta < x-2 < \delta$, "in this part I don't understand because write $-\delta < x-2 < 0$, ... | Remember we are assuming that $x < 2$ as $x\to 2^-$.
$-\delta < x - 2 < \delta;$ and $x < 2 \iff $
$-\delta < x-2 < 0 \iff$
$0 < 2-x < \delta \iff$
$\frac 1{x-2} > \frac 1{\delta} \iff$
$\frac 1{2-x} < -\frac 1{\delta}$
So if we set $-\frac 1{\delta} = B$ or in other words if we set $\delta = -\frac 1B$ we have our pro... | {
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Evaluating $\lim_{n\rightarrow\infty} \int_{0}^{\pi} \frac {\sin x}{1+ \cos^2 (nx)} dx$ Greetings I want to evaluate $\displaystyle\lim_{n\rightarrow\infty} \int_{0}^{\pi} \frac {\sin x}{1+ \cos^2 (nx)} dx$.
Here is my try: We have that $x\in[0,\pi]$ so $$\cos(n\pi)\le \cos(nx) \le 1.$$ Here I am not sure, but if it's ... | One approach can be as follows:
\begin{align}
\int_0^{\pi}\frac{\sin x}{1+\cos^2nx}\,dx&\stackrel{(1)}{=}
\frac{1}{n}\int_0^{n\pi}\frac{\sin(y/n)}{1+\cos^2y}\,dy=
\frac{1}{n}\sum_{k=0}^{n-1}\int_{k\pi}^{(k+1)\pi}\frac{\sin(y/n)}{1+\cos^2y}\,dy\stackrel{(2)}{=}\\
&\stackrel{(2)}{=}\frac{1}{n}\sum_{k=0}^{n-1}\int_{0}^{\p... | {
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"source": "stackexchange",
"question_score": "4",
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Limit of $\frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n$ Compute the limit
$$
\lim_{n \to \infty} \frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n
$$
How can this be done? The best I could do was rewrite the limit as
$$
\lim_{n \to \infty} \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \r... | If you know Sterling's Approximation:
$$
n! \sim \left(\frac{n}{e}\right)^n \sqrt{2\pi n},
$$
then you could approach it as follows:
$$
\lim_{n \to \infty} \frac{\sqrt[n]{(n + 1)(n + 2)\cdots (2n)}}{n}
= \lim_{n \to \infty} \frac{1}{n} \sqrt[n]{\frac{(2n)!}{n!}}
= \lim_{n \to \infty} \frac{1}{n} \sqrt[n]{\frac{... | {
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Probability of flipping a coin an exact number of times A biased coin flips heads with probability 4/9 and tails with probability 5/9. The coin is flipped 90 times. What is the probability that heads is flipped exactly 40 times?
I believe this question requires the utilisation of the binomial distribution in which:
$Pr... | The standard deviation for this is $$\sqrt{np(1-p)} = \sqrt{90\cdot \frac{4}{9}\cdot\frac{5}{9}} = 4.714$$
Roughly $2/3$ of the outcomes for the number of heads would be between about $35$ and $45$. That's $11$ outcomes each with a probability of less than $0.09$ to ensure the total is less than $1$. So your answer of ... | {
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"source": "stackexchange",
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Concurrency (I Think Using Menelaus' Theorem) Let $ABC$ be a triangle with incenter $I$, and let $B'$ and $C'$ be points on $BC$ such that $\angle{BIB'} = \angle{CIC'} = 90^\circ$. Let $AB'$ meet $CI$ at $P$, and let $AC'$ meet $BI$ at $Q$. Prove that $PQ$, $BC$, and the tangent to $\odot(BIC)$ at $I$ are concurrent.
S... | Can be done with barycentric coordinates. Wlog, let $BC + AC + AB = a + b + c = 1$. The condition for the orthogonality of $P_1P_2$ and $P_3P_4$ is that the permanent of the matrix with the rows $(a^2, b^2, c^2), P_2 - P_1, P_4 - P_3$ is zero. For $IB$ and $IB'$, this becomes
$$\operatorname{perm} \begin{pmatrix}
a^2 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2783834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How to apply the Chinese remainder theorem to $x^{3}+2x+1 \equiv 0 \bmod 15$? I want to apply the Chinese remainder theorem to the polynomial equation $ x^{3}+2x+1 \equiv 0 \bmod 15 $ which I have split into two equations $x^{3}+2x+1 \equiv 0 \bmod 3 $ and $x^{3}+2x+1 \equiv 0 \bmod 5$.
Looking at the Chinese remainde... | Yes, you may use the chinese remainder theorem.
$x^3+ 2x + 1 \equiv 0 \mod 3$
By Fermat $x^3 \equiv x \mod 3$ and we get the impossible $x^3 + 2x + 1 \equiv x + 2x + 1 \equiv 3x + 1 \equiv 1 \equiv 0\mod 3$. (Or we could have tried them each individually-- there's only three of them.)
so the chinese remainder theorem s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2787929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Definite integral concerning the greatest integer function Evaluate the integral $$I=\int_{0}^{x} \lfloor t+1 \rfloor^3 dt$$where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$.
The answer is given as $$\Bigg[\frac{\lfloor x \rfloor(\lfloor x \rfloor +1)}{2}\Bigg]^2+(\lfloor x \rfloor+1)^3 \... |
According to OP's stated answer we have
\begin{align*}
\color{blue}{\int_{0}^{x} \lfloor t+1 \rfloor^3\,dt=\left(\frac{\lfloor x \rfloor(\lfloor x \rfloor +1)}{2}\right)^2+(\lfloor x \rfloor+1)^3 \{x\}}\tag{1}
\end{align*}
This can be shown as follows
\begin{align*}
\int_{0}^{x}\lfloor t+1\rfloor ^3\,dt&=\int_{1}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2788856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Verification of Computation of Galois group Let $f = x^8-10x^4 + 1$. We want the Galois group over $\mathbb{Q}$
The roots are $\pm \sqrt{ \pm \sqrt{5 \pm 2\sqrt6}}$. I'm not sure about the splitting field,but I think that it would just be $\mathbb{Q}(i, \sqrt[4]{5 + 2\sqrt6})$, but I can't figure out if this would co... | Note that the Galois group is non-abelian as complex conjugation will not commute with any automorphism that sends $\sqrt[4]{5 + 2\sqrt6} \to i\sqrt[4]{5 + 2\sqrt6}$. Now $\sqrt[4]{5 + 2\sqrt6} = \sqrt{\sqrt{2}+\sqrt{3}}=\alpha$, say, so we see that the splitting field contains at least 7 quadratic subfields, $\Bbb{Q}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Indefinite integral of $\sqrt{x^2-x}$ i was trying to compute the indefinite integral:
$$
\int\sqrt{x^2-x}dx
$$
but i got stuck: after a few (unsuccessful) attempts for some $u$-substitution, i tried integration by parts:
$$
\int\sqrt{x^2-x} \ dx=\int(x)'\sqrt{x^2-x} \ dx= \\ x\sqrt{x^2-x}-\int x(\sqrt{x^2-x})'dx= \\
=... | $$\int\sqrt{x^2-x}\; dx=\int\sqrt{x}\; \sqrt{x-1}\; dx\qquad\qquad x\rightarrow\; \cosh^2\theta\quad dx\rightarrow2\cosh\theta\sinh\theta\; d\theta$$
$$=2\int\cosh^2\theta\sinh^2\theta\; d\theta\ =2\int\cosh^2\theta(\cosh^2\theta-1)\; d\theta\ = 2\int\cosh^4\theta\; d\theta -2\int\cosh^2\theta\; d\theta\ $$
$$=\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2790831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Finding the Jordan canonical form of A and Choose the correct option Let $$ A = \begin{pmatrix}
0&0&0&-4 \\ 1&0&0&0 \\ 0&1&0&5 \\ 0&0&1&0
\end{pmatrix}$$
Then a Jordan canonical form of A is
Choose the correct option
$a) \begin{pmatrix}
-1&0&0&0 \\ 0&1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
\end{pmatrix}$
$b) \begin{pmatri... | Another answer that requires only a very minimal amount of computations.
The trace of $A$ is preserved under similarity transformations. $\operatorname{Tr}(A) = 0+0+0+0 = \lambda_1+\lambda_2+\lambda_3+\lambda_4$ is enough to exclude (c) and (d). As others have noted, (b) isn't even a Jordan canonical form (and by the w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Find $\int{\arctan x}\,\mathrm dx$ without substitution 2018-08-15: I'm still looking for an answer that does not rely on
$$\int{f\left[g(x)\right]g'(x)}\,\mathrm dx = F\left[g(x)\right]$$
I'm refreshing my old calculus skills, and the textbook (Kalkulus by Tom Lindstrøm, 3rd edition, a Norwegian book) asks me to fin... | Observe that
$$I=\int{x\cdot \frac{1}{x^2+1}}\,\mathrm dx$$
can be rewritten as
$$I=\frac{1}{2}\int{\frac{\mathrm d(x^2+1)}{x^2+1}}=\ln(x^2+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2799115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Prove using calculus or otherwise the following inequality Prove that$$ \frac {2}{\pi -2}\ \le\ \frac {\sin x-x\cos x}{x-\sin x}\ <\ 2$$where $$x\in\left(0,\frac{\pi}{2}\right]$$
My Attempt:$$ f'(x)=\frac{ x\sin x(x-\sin x)-(1-\cos x)(\sin x-x\cos x)}{(x-\sin x)^2}=\frac{ x^2\sin x-(1-\cos x)x-\sin x(1-\cos x)}{(x-\sin... | If we can show that $f$ is decreasing on $(0,\frac{\pi}{2}]$, then it is easy to finish by checking the endpoints.
To show that $f$ is decreasing on this interval, it is enough to show that $f'$ is negative on this interval. You computed
$$f'(x)=\frac{ x^2\sin x-(1-\cos x)x-\sin x(1-\cos x)}{(x-\sin x)^2}$$
which is n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.