Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
High dimensional generalizations of $\int_{\sqrt{2} }^{\sqrt{3} } \frac{\arctan(y)}{(y^2-1)\sqrt{y^2-2} } dy =\frac{5\pi^2}{96}$ Let $Q=n_1+n_2+n_3+1$,$\mathbf{s}=(n_1,n_2,n_3)$.
Define
$$
A(\mathbf{s})
=\int_{D}\prod_{n=1}^{Q-1}\frac{1}{1+x_n^2}\int_{1}^{\infty}\left(Q+\sum_{n=1}^{Q-1}x_n^2 +y^2\right)^{-1}\mathrm{d}y
\text{d}x_i.
$$
Where $D=[0,\infty]^{n_1}\times[0,1]^{n_2}
\times[1,\infty]^{n_3}\subset\mathbb{R}^{Q-1}$,
$\mathrm{d}x_i
=\prod_{n=1}^{Q-1} \text{d}x_n$.
*
*Special case.
For $\mathbf{s}=(0,0,n_3)$, we have
$$
A(\mathbf{s})
=\frac{1}{Q}\left ( \frac{\pi}{4} \right )^Q.
$$
*Question 1
Prove
$$
\pi^{-Q}A(\mathbf{s})\in\mathbb{Q}.
$$
My ultimate goal is to evaluate $A(\mathbf{s})$. Numerical calculations suggest that
$$A(1,0,0)=\frac{\pi^2}{12} \quad A(0,1,0)=\frac{5\pi^2}{96}\quad A(0,0,1)=\frac{\pi^2}{32}$$
$$A(2,0,0)=\frac{\pi^3}{32} \quad A(1,1,0)=\frac{\pi^3}{80}$$
$$A(0,3,0)=\frac{93\pi^4}{35840} \qquad A(0,4,0)=\frac{193\pi^5}{322560}$$
Actually, if we explicit calculate the multiple integrals, it yields
$$
\begin{aligned}
&\int_{\sqrt{2} }^{\sqrt{3} }
\frac{\arctan(y)}{(y^2-1)\sqrt{y^2-2} } \text{d}y
=\frac{5\pi^2}{96},\\
&\frac{\pi}{6} \int_{\sqrt{3} }^{\sqrt{5} }
\frac{\arctan(y)}{(y^2-1)\sqrt{y^2-2} } \text{d}y
-\int_{2}^{\sqrt{5} }
\frac{\displaystyle{\arctan (y)\arctan \sqrt{\frac{y^2-4}{y^2-2} } } }{(y^2-1)\sqrt{y^2-2} }
\text{d}y=\frac{11\pi^3}{5760},\\
&\frac{\pi}{6} \int_{\sqrt{3} }^{\sqrt{5} }
\frac{\arctan\left(y\sqrt{2+y^2} \right)}{(y^2-1)\sqrt{y^2-2} } \text{d}y
-\int_{2}^{\sqrt{5} }
\frac{\displaystyle{\arctan\left(y\sqrt{2+y^2}\right)\arctan \sqrt{\frac{y^2-4}{y^2-2} } } }{(y^2-1)\sqrt{y^2-2} }
\text{d}y=\frac{\pi^3}{420}.
\end{aligned}
$$
Are there any other simple results? For $Q=4$, we may meet some 'troubles', such as this one:
$$
\int_{0}^{1} \int_{1}^{\sqrt{2} }
\frac{u\left(\pi-2\arctan\sqrt{u^4-1}-2\arctan \sqrt{\frac{u^2-1}{u^2+1} }
\right) \arctan \sqrt{4+u^2+v^2} }
{(1+v^2)\sqrt{1+u^2}(2+u^2) \sqrt{4+u^2+v^2} } \text{d}u\text{d}v.
$$
I can hardly convert into a 'simple' form.
*
*Question 2.
Can we evaluate a more general family of this kind of integrals?
$$A(\alpha,\mathbf{s})
=\int_{D}\prod_{n=1}^{Q-1}\frac{1}{\alpha^2+x_n^2}\int_{1}^{\infty}\left(\alpha^2 Q+\sum_{n=1}^{Q-1}x_n^2 +y^2\right)^{-1}\mathrm{d}y
\text{d}x_i.$$
| The claim is non-trivial. Let $$g(n_1,n_2) = \int_1^\infty\int_{{{[0,\pi /4]}^{{n_1}}}{{\times [\pi /4,\pi /2]}^{n_2}}} {\frac{d{x_i}}{{{1+a^2 + {{\sec }^2}{x_1} + ... + {{\sec }^2}{x_n}}}}} da$$
We need to prove $$\tag{*}g(n_1,n_2)\in \pi^{1+n_1+n_2} \mathbb{Q}$$
The solution is largely in parallel to the one I wrote here. One should read that answer first. For any (measurable) set $A\subset \mathbb{R}^{2n}$, let
$$\nu(A) = \int_1^\infty \int_{A} \frac{dx_i}{(1+a^2+x_1^2+\cdots+x_{2n}^2)^{n+1}} da$$
With $T,R$ as in the linked answer, mutatis mutandis, we have (with $n=n_1+n_2$), $$\tag{1}\nu({T^{{n_1}}} \times {R^{{n_2}}}) = \sum\limits_{k = 0}^{{n_2}} {\binom{n_2}{k}\frac{{{{( - 1)}^k}}}{{{2^{{n_1} + k}}}}\nu({{[0,1]}^{2{n_1} + {n_2} + k}} \times {{[0,\infty ]}^{{n_2} - k}})} $$
The following lemma proved there
(Lemma) Let $n_1,n_2$ be nonnegative integers, $n=n_1+n_2$, $m,r>0$. If $mr=n+1$, then $$\int_{{{[0,1]}^{{n_1}}}{{\times[0,\infty ]}^{{n_2}}}} {\frac{1}{{{{(1 + {x_1}^r + ... + {x_n}^r)}^m}}}d{x_i}} = \frac{r}{{\Gamma (m)}}\frac{{\Gamma {{(1 + \frac{1}{r})}^{n + 1}}}}{{{n_1} + 1}} $$
implies $$\tag{2}\nu({{[0,1]}^{2{n_1} + {n_2} + k}} \times {{[0,\infty ]}^{{n_2} - k}}) = \frac{\pi^{n+1}}{2^{2n+1}n!} \frac{1}{(2n_1+n_2+k+2)(2n_1+n_2+k+1)}$$
On the other hand, polar coordinates on each pairs $(x_1,x_2), (x_3,x_4),\cdots$ gives,
$$\begin{aligned}\nu({T^{{n_1}}} \times {R^{{n_2}}}) &= \int_1^\infty\int_{{{[0,\sec {\theta _i}]}^n} \times {{[0,\pi /4]}^{{n_1}}} \times {{[\pi/4,\pi /2]}^{{n_2}}}} {\frac{{{r_1}...{r_n}d{r_i}d{\theta _i}}}{{{{(1 + {r_1}^2 + ... + {r_n}^2)}^{n+1}}}}} da\\
&=\frac{1}{2^nn!} {\sum\limits_{i,j \ge 0} {{{(\frac{\pi }{4})}^{{n_1} + {n_2} - i - j}}{{( - 1)}^{i + j}}\binom{n_1}{i}\binom{n_2}{j}g(i,j)} } \end{aligned}$$
Compare this with $(1),(2)$ will give a recurrence of $g(i,j)$, which allows one to calculate all $g(i,j)$ starting from inital value $g(0,0) = \frac{\pi}{4}$, proving $(*)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4303148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Combinatoric question - arrange balls in cells [Probability] We want to arrange $10$ (different) balls in $4$ cells, what is the chance that $5$ will be entered to cell number $1$, and the rest of them will enter an even cell ($2$ or $4$)?
My attempt:
Note that we have a symmetric space, in which we have $4^{10}$ options to arrange the balls.
Let us choose $5$ balls to enter cell $1$ - $\binom{10}{5}$.
Then we have to arrange the next $5$ balls in an even cell, so each one of them has $2$ options, to enter cell number $2$ or cell number $4$, which is: $\binom{5}{5} \cdot 2^5$.
So in total we have: $$\frac{\binom{10}{5} + \binom{5}{5} \cdot 2^5}{4^{10}}=0.027 ~\% $$
Is that correct?
Thanks!
| Probability for a ball to end up in any of the $4$ bins is uniform $(=0.25)$ so
$$P=\sum_{k=0}^5\underbrace{\frac{10!}{5!(5-k)!0!k!}(0.25)^{10}}_{\textrm{Probability of 5 balls in B1, 5-k in B2, 0 in B3, k in B4}}$$
so
$$\frac{10!\cdot 0.25^{10}\cdot 2}{5!}\bigg(\frac{1}{5!}+\frac{1}{4!}+\frac{1}{3!2!}\bigg)\approx 0.0077$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4304153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find coefficient in this expression I want to find the coefficient of $x^{21}$ in this expression: $(x^3+x^4+x^5+\ldots +x^{10})^4$. The first thing I did was $(x^3+x^4+x^5+\ldots +x^{10})^4=(x^3)^4(1+x+x^2+\ldots+x^7)^4$. So the problem is reduced to finding the coefficient of $x^9$ in the expression $(1+x+x^2+\ldots+x^7)^4$. However, I am not able to find that coefficient :/
| $\mathbf{\text{HINT}}:$Remember that $$(1+x+x^2 +...+x^7)^4= \bigg(\frac{1-x^8}{1-x}\bigg)^4$$ , so you have binomial expressions.
So, $(1-x^8)^4 (1-x)^{-4}$ , then $(1+(-x^8))^4 (1-x)^{-4}$ ,
you will just find coefficients of binomial expressions such that
For $(1+(-x^8))$ : $\binom{4}{m}(-x^8)^m$
For $(\frac{1}{1-x})^4$ : $\binom{n+4-1}{n}x^n$ , where $8m+n=9$
Find all $m,n$ values satisfying the equation such as $(0,9),(1,1)$ , then put them into formulas.
Then , $$1 \times\bigg( \binom{9+4-1}{9}\bigg)+ (-4)\times \bigg(\binom{1+4-1}{1}\bigg)=204$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4304489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\lim_{x\rightarrow \infty} \frac{1}{\ln(x+1)-\ln(x)}-x$ I'm supposed to compute
$$\lim_{x\rightarrow \infty}\left( \frac{1}{\ln(x+1)-\ln(x)}-x\right).$$
However, I keep getting the wrong answer, so I'll present my solution for you, and I hope you can give me any tips on how to solve it.
Rewriting using logarithm laws, we have
$$\lim_{x\rightarrow \infty} \left(\dfrac{1}{\ln\frac{x+1}{x}}-x\right).$$
Simplyfing further, we have:
$$\lim_{x\rightarrow \infty} \dfrac{1-x\ln\frac{x+1}{x}}{\ln\dfrac{x+1}{x}}= \lim_{x\rightarrow \infty} \frac{1-\ln(1+\frac{1}{x})^x}{\ln(1+\frac{1}{x})} = \frac{1-e}{0} \rightarrow -\infty.$$
However, the answer sheet tells me that it's $1/2$, and I don't really see where I did something wrong in the solution. Thanks.
| $$\begin{aligned}\frac{1}{\ln(x+1)-\ln(x)}-x
&= \frac{1}{\ln\left(1 + \frac{1}{x}\right)}-x \\
&= \frac{1}{\frac{1}{x} - \frac{1}{2x^2}} -x + o(1) \\
&= x \left(\frac{1}{1-\frac{1}{2x}}-1 \right) + o(1) \\
&= x \left(\frac{1}{1-\frac{1}{2x}}-1 \right) + o(1) \\
&= x \left(\frac{1}{1-\frac{1}{2x}}-1 \right) + o(1) \\
&= \frac{1}{2} + o(1) \\
\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4308661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Proving that $3^{(3^4)}>4^{(4^3)}$ without a calculator Is there a slick elementary way of proving that $3^{(3^4)}>4^{(4^3)}$ without using a calculator?
Here is what I was thinking:
$$4^4=256>243=3^5,$$
hence
$$4^{4^3}=4^{64}=(4^4)^{16}=(3^5)^{16}\cdot\left(\dfrac{256}{243}\right)^{16}=3^{80}\cdot\left(\dfrac{256}{243}\right)^{16}<3^{81}=3^{3^4}\,.$$
It is possible to prove that $\left(\dfrac{256}{243}\right)^{16}<3$ without a calculator by making a comparison such as $\dfrac{256}{243}=1+\dfrac{13}{243}<1+\dfrac{15}{240}=\dfrac{17}{16}$ and $\left(1+\dfrac{1}{16}\right)^{16}<e$ so $\left(\dfrac{256}{243}\right)^{16}<\left(1+\dfrac{1}{16}\right)^{16}<e<3$.
Is there a more elegant elementary proof? Ideally I'm looking for a proof that doesn't rely on calculus.
Source of problem: I made up this question but it is inspired by a similar question that appeared in the British Mathematical Olympiad round 1 in 2014.
| $3^4-4^3=17$ therefore we can rewrite the inequality as:
$$3^{17} > \left(\frac{4}{3}\right)^{4^3}$$
we have $\left(\frac{4}{3}\right)^8 < 10$, hence it suffices to show:
$$3^{17} > 10^8$$
or
$$3 \cdot 81^4 > 10^8$$
which is satisfied if:
$$3 \cdot 80^4 > 10^8$$
or
$$3 \cdot 2^{12} \cdot 10^4 > 2^{4} \cdot 5^4 \cdot 10^4$$
$$3 \cdot 2^8 > 5^4$$
i.e.
$$768 > 625$$
ADDENDUM
$4^8=4^4 \cdot 4^4 = 256 \cdot 256$ and $3^8=3^4 \cdot 3^4 = 81 \cdot 81$, it is not too difficult to make the two multiplications and then the division.
Also, if you are a programmer, you know already that $4^8 = 2^{16} = 65536$, so you just need to make $81 \times 81$ and then the division. Rather, you don't need any division because you see immediately $81 \cdot 81 = 6561 \gt 65536 / 10$.
Anyway, I admit, I have used the calculator :-)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4310803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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Find all $f(x)$ such that $x(f(x+1)-f(x))=f(x)$ The problem
Find all $f(x): \mathbb{R} \to \mathbb{R} $ such that $x(f(x+1)-f(x))=f(x)$ and $|f(x) -f(y)| \le |x-y|, \forall x,y \in \mathbb{R}$
My approach
Obviously $f(x)=x$ is one solution, I suspect it is the only one. Now I substitute $x=0$ in the first condition, I get $f(0)=0$.
Thus substitute $y=0$ in the second condition I get $|f(x) | \le | x | \tag{1}$
Another thing I got from the first condition:
\begin{align}
\frac{f(x+1)}{x+1}=\frac{f(x)}{x}
\end{align}
So I denote $c_x= x -\lfloor{x}\rfloor$, and assign a random function $f(x)=k_xx, \forall x\in [0,1], k_x$ a constant chosen independently for $c_x$, then $f(x)=k_xx, \forall x\in\mathbb{R}$.
Now what I can conclude is that $k_x$ all are smaller than $1$, because of $(1)$. But how do I proceed? Should I plug it into the second condition of the problem and try to deduce something?
Any help is appreciated!
EDITS
Earlier I mistakenly assumed that $f(x)$ is a polynomial. It's not. It is a function
| I will show that the only such functions satisfying the conditions in your question are:
$$
f(x) = ax, \;\text{where $-1 \leq a \leq 1$}
$$
Let $f$ be a $1$-Lipschitz function satisfying $x(f(x+1) - f(x)) = f(x)$. Let $f(1) = a$.
Claim. For any $x \in [0,1]$, we have $f(x) = ax$.
Proof. Suppose $f(c) = b$ for some $c \in (0,1]$. We observe that:
$$
f(c + 1) = \frac{c + 1}{c}f(c) = \frac{c+1}{c}b \\
f(c + 2) = \frac{c + 2}{c + 1}f(c + 1) = \frac{c+2}{c}b \\
f(c + 3) = \frac{c + 3}{c + 2}f(c + 2) = \frac{c+3}{c}b \\
\vdots
$$
Thus, one can prove by induction that $f(c + n) = \frac{c + n}{c}b$ (in particular, $f(n) = na$). Now if $c \in (0,1)$ and $b \neq ac$, then we have:
\begin{align*}
|f(c + n) - f(n)| &= \left|\frac{c + n}{c}b - an\right| \\
&= \left|b + \frac{n}{c}(b - ac)\right|
\end{align*}
Thus, as $n$ gets arbitrarily large, $|f(c + n) - f(n)|$ gets arbitrarily large, yet $|(c + n) - n| = c$ remains constant, which contradicts the $1$-Lipschitzness of $f$. $\blacksquare$
For any $x > 0$, we may write $x = n + c$ for some $n \in \mathbb{Z}_{\geq 0}$ and $c \in [0,1)$. If $c = 0$, then as mentioned above we have $f(n) = an$. If $c > 0$, then:
$$
f(x) = f(n + c) = \frac{n + c}{c}f(c) = \frac{n + c}{c}ac = a(n + c) = ax
$$
Therefore, $f(x) = ax$ for $x \geq 0$. The case for $x < 0$ can be done similarly: Observe that we have:
$$
\frac{f(x - 1)}{x - 1} = \frac{f(x)}{x}
$$
Therefore, we may again prove by induction that for $c \in (0,1]$:
$$
f(c - n) = \frac{c - n}{c}f(c)
$$
Any $x < 0$ may be written as $x = c - n$ for some $c \in (0,1)$ and $n \in \mathbb{Z}^+$, so:
$$
f(x) = f(c - n) = \frac{c - n}{c}f(c) = \frac{c - n}{c}(ac) = a(c - n) = ax
$$
so $f(x) = ax$ for $x < 0$. Finally, to see that we must have $-1 \leq a \leq 1$, we simply observe that:
$$
|a| = |a||1 - 0| = |a - 0| = |f(1) - f(0)| \leq |1 - 0| = 1
$$
by the $1$-Lipschitzness of $f$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4313746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find the sum $\sum_{k=1}^{\lfloor n/2\rfloor}\frac{2^{n-2k}\binom{n-2}{2k-2}\binom{2k-2}{k-1}}{k}$ Let $n$ be positive integer, find the value
$$f(n)=\sum_{k=1}^{\lfloor n/2 \rfloor}\dfrac{2^{n-2k}\binom{n-2}{2k-2}\binom{2k-2}{k-1}}{k}. $$
I have found
$$ f(2)=1, \quad f(3)=2, \quad f(4)=5, \quad f(5)=14, \quad f(6)=42, \quad f(7)=132. $$
It seem OEIS (A000108):
$$ f(n)=\dfrac{1}{n}\binom{2n-2}{n-1},$$
but how to prove it?
| In seeking to evaluate where the sum is zero when $n\lt 2$
$$\sum_{k=1}^{\lfloor n/2\rfloor}
2^{n-2k} {n-2\choose n-2k} \frac{1}{k} {2k-2\choose k-1}$$
we recognize the Catalan number and obtain
$$\sum_{k=1}^{\lfloor n/2\rfloor}
2^{n-2k} {n-2\choose n-2k}
[z^{k}] \frac{1-\sqrt{1-4z}}{2}
\\ = [w^n] (1+w)^{n-2}
\sum_{k=1}^{\lfloor n/2\rfloor}
2^{n-2k} w^{2k}
[z^{k}] \frac{1-\sqrt{1-4z}}{2}.$$
We get a zero contribution when $k=0$ from the coefficient extractor in
$z$ as well as when $2k\gt n$ from the coefficient extractor in $w$ so we
may extend the sum to
$$2^n [w^n] (1+w)^{n-2}
\sum_{k\ge 0}
2^{-2k} w^{2k}
[z^{k}] \frac{1-\sqrt{1-4z}}{2}
\\ = 2^n [w^n] (1+w)^{n-2}
\frac{1-\sqrt{1-w^2}}{2}.$$
This is
$$2^n \;\underset{w}{\mathrm{res}}\;
\frac{1}{w^{n+1}} (1+w)^{n-2}
\frac{1-\sqrt{1-w^2}}{2}.$$
Now we put $w/(1+w) = v$ so that $w=v/(1-v)$ and $dw = 1/(1-v)^2 \; dv$
to get
$$2^n \;\underset{v}{\mathrm{res}}\;
\frac{1}{v^{n+1}} (1-v)^3
\frac{1-\sqrt{1-v^2/(1-v)^2}}{2}
\frac{1}{(1-v)^2}
\\ = 2^n [v^n] \frac{1-v-\sqrt{1-2v}}{2}
= [v^n] \frac{1-2v-\sqrt{1-4v}}{2}
\\ = [v^{n-1}] \frac{1-2v-\sqrt{1-4v}}{2v}
= [v^{n-1}] (-1+\frac{1-\sqrt{1-4v}}{2v}).$$
This is the Catalan number $C_{n-1}$ when $n\ge 2$ and when $n=1$
we get $-1+C_0 = 0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4317353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
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What is $ \lim_{x \rightarrow 0}\left(\frac{1}{\ln\cos (x)}+\frac{2}{\sin ^{2}(x)}\right) $? The answer of the following limit:
$$
\lim_{x \rightarrow 0}\left(\frac{1}{\ln \cos (x)}+\frac{2}{\sin ^{2}(x)}\right)
$$
is 1 by Wolfram Alpha.
But I tried to find it and I got $2/3$ :
My approach :
$1)$
$
\ln(\cos x)=\ln\left(1-\frac{x^{2}}{2}+o\left(x^{3}\right)\right)=-\frac{x^{2}}{2}+o\left(x^{3}\right)
$
$2)$
$
\sin ^{2}(x)=\left(x-\frac{x^{3}}{3!}+o\left(x^{3}\right)\right)^{2}=x^{2}-\frac{x^{4}}{3}+o\left(x^{4}\right)
$
$3)$
$\begin{aligned} \frac{1}{-\frac{x^{2}}{2}+o\left(x^{3}\right)}+\frac{2}{x^{2}-\frac{x^{4}}{3}+o\left(x^{4}\right)}=\frac{-x^{2}+x^{2}-\frac{x^{4}}{3}+o\left(x^{3}\right)}{-\frac{x^{4}}{2}+o\left(x^{5}\right)}=\frac{-\frac{1}{3}+o\left(x^{3}\right)}{-\frac{1}{2}+o\left(x^{5}\right)} \end{aligned}$
$4)$
$\lim _{x \rightarrow 0} \frac{-\frac{1}{3}+o\left(x^{3}\right)}{-\frac{1}{2}+o\left(x^{5}\right)}=\frac{-\frac{1}{3}}{-\frac{1}{2}}=\frac{2}{3}$
So where is the mistake in my approach?
Note: $o$ denotes the little-o notation
Edit : I've understood where's my mistake is, but another question popped up reading the answers which is : does $o(1/x)$ tends to zero as x tends to zero?
| Hint:
$$\dfrac1{\ln\cos x}=\dfrac2{\ln(1-\sin^2x)}$$
Now writing $\sin^2x=y,$
$$\lim_{y\to0^+}2\cdot\dfrac{y+\ln(1-y)}{y\ln(1-y)}=2\lim_y\dfrac{y+\left(-y-y^2/2-y^3/3-\cdots\right)}{-y^2\cdot\dfrac{\ln(1-y)}{-y}}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4324786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 5
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Solve the ordinary differential equation.
Solve the equation: $$x(2y+1)dx=y(x^2-3x+2)dy$$
To solve this, I first move $y(x^2-3x+2)dy$ to the LHS, then divide both sides by $(2 y + 1)(x^2 - 3 x + 2)$, the equation becomes:
$$\dfrac{x}{x^2-3x+2}dx-\dfrac{y}{2y+1}dy=0$$
After that, I integrate both sides: $$\int\dfrac{x}{x^2-3x+2}dx-\int\dfrac{y}{2y+1}dy=C$$
Which gives: $$2\ln|x-2|-\ln|x-1|+\dfrac{\ln|2y+1|-2y-1}{4}=C$$
Did I do this correctly?
| There is a mistake when you divide both sides by $(2y+1)(x^2-3x+2)$.The correct form is
$$
\frac{x}{x^2-3x+2}\mathrm{d}x=\frac{y}{2y+1}\mathrm{d}y
$$
so integrating both sides gives
$$
2\ln |x-2|-\ln |x-1|=\frac{1}{4}\left( \left( 2y+1 \right) -\ln |2y+1|+C^\prime \right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4327237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Are there any two real nonzero numbers $a$ and $b$ such that $a^2+ab+b^2 = 0$ Why or why not? Are there any two real nonzero numbers $a$ and $b$ such that $a^2+ab+b^2 = 0$ Why or why not?
$$a^2+ab+b^2=(a+b)^2-ab=0$$ iff $$(a+b)^2=ab \tag{1}$$
but $(a+b)^2 = a^2+2ab+b^2 $ so equation 1 couldn't possibly be true.
Also, when $a=b\ne 0$, $(a^2+ab+b^2)(a-b) = a^3-b^3 =0$.
| $$a^2 + ab + b^2 = \left( a + \frac{b}{2}\right)^2 + \frac{3b^2}{4}=0$$
Since the two terms are non-negative, thus
$$\left( a + \frac{b}{2}\right)^2= \frac{3b^2}{4}=0$$
Therefore
$$a=0,\quad b=0$$
| {
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"url": "https://math.stackexchange.com/questions/4327817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve $(x - 1)(x + 7) < 0$ I tried this:
\begin{align*}
(x - 1)(x + 7) < 0
&\iff x-1<0 \vee x+7<0\\
&\iff x < 1 \vee x < -7\iff x\in(-7,1).
\end{align*}
The thing though, is that I'm not sure if $ab<0 \implies a<0$ or $b<0$.
| A product $ab < 0$ if $a$ and $b$ have opposite signs. When $a = x - 1$ and $b = x + 7$
$$(x - 1 > 0 \wedge x + 7 < 0) \vee (x - 1 < 0 \wedge x + 7 > 0)$$
Since $x - 1 > 0 \implies x > 1$ and $x + 7 < 0 \implies x < -7$, the statement $$x - 1 > 0 \wedge x + 7 < 0$$ is never true since it is not possible for a real number to be both greater than $1$ and less than $-7$.
Since $x - 1 < 0 \implies x < 1$ and $x + 7 > 0 \implies x > -7$, the statement $$x - 1 < 0 \wedge x + 7 > 0$$ is true when $-7 < x < 1$.
Hence, the solution set is $S = (-7, 1)$.
Another way to do the problem is to perform a line analysis. Since $(x + 7)(x - 1) = 0$ when $x = -7$ or $x = 1$ and $(x + 7)(x - 1)$ is continuous, the sign of the product can change at the points $x = -7$ or $x = 1$. We draw a number line with the points $x = -7$ and $x = 1$ marked as zeros. We wish to determine the sign of the product $(x + 7)(x - 1)$ in the three intervals $(-\infty, -7), (-7, 1), (1, \infty)$. To do so, we determine the signs of $x + 7$ and $x - 1$ in each of these intervals. The term $x + 7$ is negative when $x < -7$, zero at $x = -7$, and positive when $x > -7$. The term $x - 1$ is negative when $x < 1$, zero at $x = 1$, and positive when $x > 1$. The sign of $(x + 7)(x - 1)$ in each interval is found by multiplying the signs of the factors $x + 7$ and $x - 1$ in that interval.
Since we wish to determine where $(x + 7)(x - 1) < 0$, our solution set occurs where $(x + 7)(x - 1)$ is negative, which is the interval $(-7, 1)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the volume between the sphere $x ^2 + y^2 + z^2 = 4$ and the plane $z = 1$ Suppose $y \geq 3$. I want to compute the volume between the sphere $x ^2 + (y − 2)^2 + z^2 = 4$
and the plane $y = 3$.
So I move left the sphere and and the plan, and rotate it counterclockwise. I got the new sphere and the new plan:
Suppose $z \geq 1$. Then compute the volume between $x ^2 + y^2 + z^2 = 4$ and the plan $z = 1$.
Here is my attempt using spherical coordinates:
$$\int_{0}^{2 \pi} \int_{0}^{\frac{\pi}{6}} \int_1^2 \rho^2 \sin \phi dp d \phi d\theta = 2\pi (1- \frac{\sqrt{3}}{2})\frac{1}{3}$$ I am supposed to get $\frac{5 \pi}{3}$. Where am I wrong?
| OP's attempt can be rectified after correction the following two mistakes:
*
*At different $\phi$ values, the $\rho$ does not always vary from 1 to 2. By drawing a diagram, we can check that at angle $\phi$, the range of $\rho$ is from $\frac{1}{\cos \phi}$ to 2.
*$\phi$ is supposed to range from 0 to $\pi/3$ (calculated from $\cos^{-1} (\frac{1}{2})$; again a diagram will help)
Thus, the correct integral to compute is
$$\int\limits_{0}^{2\pi} d\theta \int\limits_{0}^{\pi/3}\int\limits_{\frac{1}{\cos\phi}}^{2} \rho^2\sin\phi \ d\rho\ d\phi$$
$$ = 2\pi \int\limits_{0}^{\pi/3} \left.\left(\frac{1}{3}\rho^3\sin\phi\right)\right\rvert_{\frac{1}{\cos\phi}}^{2} \ d\phi$$
$$= 2\pi \int\limits_{0}^{\pi/3} \frac{8}{3}\sin\phi-\frac{1}{3}\tan\phi\sec^2\phi\ d\phi$$
$$= 2\pi \left.\left(-\frac{8}{3}\cos\phi-\frac{1}{6}\tan^2\phi\right)\right\rvert_{0}^{\pi/3}$$
$$ = \frac{5\pi}{3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine the value of the expression $\frac {5}{\sin^2A} + \frac{7}{\cos^2B}$ if $\cos A = \frac{-2}{3}$ and $\sin B=\frac{3}{4}$
Determine the value of the expression $\frac {5}{\sin^2A} + \frac{7}{\cos^2B}$ if $\cos A = \frac{-2}{3}$ and $\sin B=\frac{3}{4}$.
I applied $$\cos^2B\;=\;1-\sin^2B,$$
and $$\tan^2A\;=\;\frac{\sin^2A}{\cos^2A}.$$ But I am not sure what to do from there.
| Using this trigononometric identity
$$ \cos^{2}\theta+\sin^{2}\theta =1 $$
we get that
$$ \cos^{2}B = 1-\sin^{2}B $$
similary we can find $\sin^{2}A$, and then we just replace
$$ \frac{5}{1-\cos^{2}A} + \frac{7}{1-\sin^{2}B}=\frac{5}{1-\left(\frac {-2}{3}\right)^{2}} + \frac{7}{1-\left(\frac {3}{4}\right)^{2}} $$
there you go...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing $f(kx,ky)\geq f(x,y)$ We have a function $$f(x,y)={y\choose x}$$ for $1<x<y$. We are trying to show that for a positive integer $k$, $f(kx,ky)\geq f(x,y)$.
So what I did so far was to write
$$\frac{ky(ky-1)\cdots(ky-kx+1)(ky-kx)!}{(ky-kx)!(kx)!} \geq \frac{y(y-1)\cdots(y-x+1)(y-x)!}{(y-x)!x!}$$
So we get
$$\frac{ky(ky-1)\cdots(ky-kx+1)}{(kx)!} \geq \frac{y(y-1)\cdots(y-x+1)}{x!}$$
Then I got stuck because every term on the numerator of the left hand side is greater than or equal to that of its right hand side but same goes for the denominator!
Could you anyone please give me a hint on what to do next? Is this even the right direction?
| We will use the well-known absorption identity of binomial coefficients:
\begin{align*}
\binom{y}{x} = \frac{y}{x} \binom{y-1}{x-1}
\end{align*}
Hence
\begin{align*}
\binom{ky}{kx} = \binom{ky - (k-1)x}{x} \prod_{n=x+1}^{kx} \frac{ky - kx + n}{n} \ge \binom{ky - (k-1)x}{x}
\end{align*}
Since $y > x$ and therefore $\frac{ky - kx + n}{n} \ge \frac{n}{n} = 1$. Pascal's identity also tells us
\begin{align*}
\binom{y}{x} = \binom{y-1}{x} + \binom{y-1}{x-1} \ge \binom{y-1}{x}
\end{align*}
In other words, $f(y) = \binom{y}{x}$ is an increasing function for any fixed $x$. Therefore, to conclude
\begin{align*}
\binom{ky - (k-1)x}{x} \ge \binom{y}{x}
\end{align*}
We need to show $ky - (k-1)x \ge y \iff y \ge x$, which is given.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove: $\frac{\frac{1}{a}+b}{\sqrt{\frac{1}{a}+a}}+\frac{\frac{1}{b}+c}{\sqrt{\frac{1}{b}+b}}+\frac{\frac{1}{c}+a}{\sqrt{\frac{1}{c}+c}}\ge3\sqrt{2}$
Let $a,b,c>0$. Prove that: $$\frac{\frac{1}{a}+b}{\sqrt{\frac{1}{a}+a}}+\frac{\frac{1}{b}+c}{\sqrt{\frac{1}{b}+b}}+\frac{\frac{1}{c}+a}{\sqrt{\frac{1}{c}+c}}\ge3\sqrt{2}$$
Anyone help me a hint to solve above problem?
I tried by AM-GM without success: $$LHS\ge3\sqrt[3]{\frac{\left(\frac{1}{a}+b\right)\left(\frac{1}{b}+c\right)\left(\frac{1}{c}+a\right)}{\sqrt{\left(\frac{1}{a}+a\right)\left(\frac{1}{b}+b\right)\left(\frac{1}{c}+c\right)}}}$$
The rest is proving: $$\left(\frac{1}{a}+b\right)\left(\frac{1}{b}+c\right)\left(\frac{1}{c}+a\right)\ge2\sqrt{2\left(\frac{1}{a}+a\right)\left(\frac{1}{b}+b\right)\left(\frac{1}{c}+c\right)}$$ which is not true by a=60,b=0.02 and c=0.9
Thank you for your help!
| Lemma: From this problem that the OP posed (with terms rearranged),for $ a_i > 0$,
$$ \frac{ \frac{1}{ a_i } + a_{i+1} } { \sqrt{ \frac{1}{ a_i } + a_{i} } } \geq \sqrt{2} ( \sqrt{a_{i+1}} - \sqrt{a_i} + 1 ). $$
Corollary: This inequality follows by summing up over $a_i$, and the RHS simplifies to $ 3 \sqrt{2} $.
Notes
*
*Furthermore, we get the obvious generalization to $n$ variables, $ \sum \frac{ \frac{1}{ a_i } + a_{i+1} } { \sqrt{ \frac{1}{ a_i } + a_{i} } } \geq n \sqrt{2}$.
*Guessing this lemma isn't obvious to me. Proving it also requires some manipulation.
| {
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Why am I getting the correct value for $\sin\left(2\tan^{-1}\frac{4}{3}\right)$ even though the usage of the formula is incorrect? The expression:
$$\sin\left(2\tan^{-1}\left(\frac{4}{3}\right)\right)$$
Way 1:
If I punch the above expression in my calculator, I get $\frac{24}{25}$.
Way 2:
$$\sin\left(2\tan^{-1}\left(\frac{4}{3}\right)\right)$$
$$\sin\left(\sin^{-1}\left(\frac{2\times\frac{4}{3}}{1+(\frac{4}{3})^{2}}\right)\right)$$
$$[\text{Using $2\tan^{-1}x=\sin^{-1}\frac{2x}{1+x^2}$}]$$
$$\frac{2\times\frac{4}{3}}{1+(\frac{4}{3})^{2}}$$
$$\frac{24}{25}$$
My comments:
One of the conditions of $2\tan^{-1}x=\sin^{-1}\frac{2x}{1+x^2}$ is that $|x|\leq1$. In this case $|\frac{4}{3}|\nleq1$, but still I'm getting the correct answer using the formula. Why is that?
Related:
*
*Why am I getting a different value for $\sin\left(2\tan^{-1}\frac{4}{3}\right)$ than my calculator?
*Why does the equation with $2 \arctan(x)$ and other Inverse Trigonometric functions have weird conditions?
| When $x > 1,$ as in this case, $\arctan x > \frac\pi4,$
and therefore $2 \arctan x > \frac\pi2.$
That is, $2 \arctan x$ is in the second quadrant.
But $\arcsin$ can only produce angles in the first and fourth quadrants.
There is a similar difficulty when $x < -1.$
That's why the formula $2\tan^{-1}x=\sin^{-1}\frac{2x}{1+x^2}$
is good only for $\lvert x\rvert \leq 1.$
But when $x > 1,$ you need a different formula. One correct formula is
$$2\tan^{-1}x = \pi - \sin^{-1}\frac{2x}{1+x^2}.$$
And $\sin(\pi - \theta) = \sin \theta,$ so you get the same result in the end.
Postscript:
The following is a kind of extended comment on the question. You could also view it as an answer to a question that was not literally asked.
Note that for any real number $x$ (without restriction) we have
\begin{align}
\sin\left(\tan^{-1} x\right) &= \frac{x}{\sqrt{1+x^2}}, \\
\cos\left(\tan^{-1} x\right) &= \frac{1}{\sqrt{1+x^2}}.
\end{align}
So if $\tan^{-1} x = \alpha$ then
\begin{align}
\sin\left(2\tan^{-1} x\right) = \sin(2\alpha)
&= 2\sin\alpha \cos\alpha \\
&= 2\left(\frac{x}{\sqrt{1+x^2}}\right)\left(\frac{1}{\sqrt{1+x^2}}\right)
\\[1ex]
&= \frac{2x}{1+x^2}.
\end{align}
So this is in fact a general formula that does not require any restrictions or special cases in order to deal with the quadrants in which the trig functions fall.
| {
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"timestamp": "2023-03-29T00:00:00",
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Minimum value of $a^2 \cot (10^ \circ)+b^2\cot (70^ \circ)+c^2\cot (130^ \circ)$ Let $a,b,c$ be real numbers such that $a+b+c=3$, then find minimum value of
$$a^2 \cot (10^ \circ)+b^2\cot (70^ \circ)+c^2\cot (130^ \circ)$$
I have solved many such question using A.M.-G.M. inequality but since $\cot (130^ \circ)$ is negative, hence it cannot be applied here. I also have a result in mind which is
$$\cot (60 ^\circ-\theta)\cot (\theta)\cot (60 ^\circ+\theta)=\cot (3\theta)$$ and in this question if we write $
\cot(130^\circ)=-\cot (50^\circ)$, then above mentioned result can be applied on $\cot (10^ \circ),\cot (70^ \circ),\cot (50^ \circ)$ but I am not able to put everything together and reach the final answer which is given as $\sqrt{27}$. Any help or hint would be appreciated.
| Since $\cot (10^ \circ)$ and $\cot (70^ \circ)$ are both positive and $a^2$ and $b^2$ are positive, because they are real, you want to minimize their contribution. Set $a=b=0$, then $c=3$ and the minimum value is $9\cot (130^ \circ) < 0$ so the answer is wrong unless we take the negative root.
$$\sqrt{27} = -3 \sqrt{3} $$ but we know that
$$\cot (120^ \circ)=\frac{-\sqrt{3}}{3}$$ so $$9 \cot (120^ \circ)=-3 \sqrt{3}=\sqrt{27}$$ so I think your book has a misprint.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the volume of the solid bounded by $z^2=xy$; $x+y=a$; $x+y=b$ $(0Find the volume of the solid bounded by $z^2=xy$; $x+y=a$; $x+y=b$ $(0<a<b)$ by applying variable substitution. $xy>0$ and $x,y$ can't be negative at the same time so $x>0;y>0.$ Now, $a - x < y < b - x$ and we get that $a < x < b$. So the integral to compute is $\int_{a}^{b}\int_{a-x}^{b-x}{\sqrt{xy}\,dxdy}$. Now this integral, as requested by the problem, should be solved by applying variable substitution. But the answer in the textbook is $\dfrac{\pi(b^3-a^3)}{12}$. I don't get where this $\pi$ comes from. The only way I know to obtain such a result is by changing to polar coordinate system, but I don't see any way to obtain relations between $a,b$, $a-x$, $b-x$ and the new coordinate system especially because in none of these equations are $x$ or $y$ squared. Am I doing some mistake or how does this $\pi$ appear in the answer? Thanks in advance.
| If you do $X=x+y$, $Y=x-y$, and $Z=z$, then $z^2=xy$ becomes $Z^2=\frac14(X^2-Y^2)$, $x+y=a$ becomes $X=a$, and $x+y=b$ becomes $X=b$. Since $Z^2\geqslant0$, $Y^2\leqslant X^2$, which means that $-X\leqslant Y\leqslant X$. Finally, you have $(x,y,z)=\left(\frac{X+Y}2,\frac{X-Y}2,Z\right)$, and the absolute value of the Jacobian of$$(X,Y,Z)\mapsto\left(\frac{X+Y}2,\frac{X-Y}2,Z\right)$$is $\frac12$. So, the volume that you're after is\begin{align}\int_a^b\int_{-X}^X\int_{-\frac12\sqrt{X^2-Y^2}}^{\frac12\sqrt{X^2-Y^2}}\frac12\,\mathrm dZ\,\mathrm dY\,\mathrm dX&=\frac12\int_a^b\int_{-X}^X\sqrt{X^2-Y^2}\,\mathrm dY\,\mathrm dX\\&=\int_a^b\frac14\pi X^2\,\mathrm dX\\&=\frac{\pi(b^3-a^3)}{12}.\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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How many ways to deal with the integral $\int \frac{d x}{1-\sin x \cos x}$? Multiplying both numerator and denominator of the integrand by $\sec^2 x$ yields
\begin{aligned}
& \int \frac{d x}{1-\sin x \cos x} \\
=& \int \frac{\sec ^{2} x}{\sec ^{2} x-\tan x} d x \\
=& \int \frac{d(\tan x)}{\tan ^{2} x-\tan x+1} \\
=& 2 \int \frac{d(2 \tan x-1)}{(2 \tan x-1)^{2}+(\sqrt{3})^{2}} \\
=& \frac{2}{\sqrt{3}} \arctan\left(\frac{2 \tan x-1}{\sqrt{3}}\right)+C .
\end{aligned}
Is there any simpler solution? Let me know if you have any, thank you for your attention.
Wish you enjoy the solution!
| We could use the double-angle formula for sine,
$$\sin 2x = 2 \sin x \cos x$$
and then get
$$\newcommand{\II}{\mathcal{I}}
\newcommand{\dd}{\mathrm{d}}
\II := \int \frac{\dd x}{1 - \sin x \cos x} = \int \frac{\dd x}{1 - \frac 1 2 \sin 2x} =2 \int \frac{\dd x}{2 - \sin 2x}$$
Then let $u = 2x$ to get
$$\II = \int \frac{\dd u}{2 - \sin u}$$
Weierstrass substitution ($t = \tan(u/2)$) then gives
$$\II = \int \frac{1}{2 - \frac{2t}{1+t^2}} \frac{2}{1+t^2} \, \dd t = \int \frac{\dd t}{t^2 - t + 1}$$
Integral $(13)$ from this PDF then gives the answer
$$\II
= \frac{2}{\sqrt 3} \arctan \left( \frac{2t - 1}{\sqrt 3} \right) + C
= \frac{2}{\sqrt 3} \arctan \left( \frac{2 \tan(x) - 1}{\sqrt 3} \right) + C $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $x^5=\frac{133x-78}{133-78x}$ Solve $$x^5=\dfrac{133x-78}{133-78x}$$ We have $D:133-78x\ne0,x\ne\dfrac{133}{78}$. In $x\in D$ the given equation is equivalent to $$x^5(133-78x)=133x-78\\78x^6-133x^5+133x-78=0$$ As $x=0$ is not a solution, we can divide both sides by $x^3\ne0$ to get $$78x^3-133x^2+\dfrac{133}{x^2}-\dfrac{78}{x^3}=0\\78\left(x^3-\dfrac{1}{x^3}\right)-133\left(x^2-\dfrac{1}{x^2}\right)=0$$ I am not able to come up with a subtle substitution. It would have been easier if we had $x^2\color{red}{+}\dfrac{1}{x^2}.$
$x=1$ is an "obvious" solution. What else? Thank you!
| Multiplying to a common denominator gives
$$
0=78x^6 - 133x^5 + 133x - 78=(13x^2 + 6x + 13)(3x - 2)(2x - 3)(x + 1)(x - 1).
$$
The factorisation arises by the Rational Root Theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral of $\int \frac{\sqrt{x^2-1}}{x}dx$
Evaluate the integral of $$\int \frac{\sqrt{x^2-1}}{x}dx$$
Attempt: I've tried taking $x=\sec y$,
$$ \int \frac{\sqrt{\sec^2y-1}}{\sec y}dx $$
How to proceed further?
| $$\int \:\frac{\sqrt{x^2-1}}{x}dx$$
Let $u=\:\sqrt{x^2-1}$ then
$$\frac{du}{dx}=\frac{x}{\sqrt{x^2-1}}\iff dx=\frac{\sqrt{x^2-1}}{x}\:du$$
Thus
\begin{align*}
\int \:\frac{\sqrt{x^2-1}}{x}dx& =\int \:\frac{u^2}{u^2+1}du\\
&=\int \:\frac{u^2 + 1 - 1}{u^2+1 + 1 - 1}du\\
&=\int \:\left(\frac{u^2+1}{u^2+1}-\frac{1}{u^2+1}\right)du\\
&=\int \:\left(1-\frac{1}{u^2+1}\right)du\\
&=u-\arctan \left(u\right)+C\\
&=\:\sqrt{x^2-1}-\arctan \left(\:\sqrt{x^2-1}\right)+C
\end{align*}
| {
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How to solve $\lfloor x \rfloor + \lfloor \frac{1}{x} \rfloor = 1$? I am stuck with this equation. All I could do is this:
$\lfloor x \rfloor$ = $\lfloor n + m \rfloor$ such that $n \in N$ and $m<1$.
We get:
$\lfloor x \rfloor + \lfloor \frac{1}{x} \rfloor = 1$
$\lfloor n + m \rfloor + \lfloor \frac{1}{n+m} \rfloor = 1$
$n + \lfloor m \rfloor + \lfloor \frac{1}{n+m} \rfloor = 1$
$n + 0 + \lfloor \frac{1}{n+m} \rfloor = 1$
$n + \lfloor \frac{1}{n+m} \rfloor = 1$
From here on I have no idea what to do!
Edit: It is easy to see that any value $1<x<2$ satisfies the equation, but can I find all the solutions?
| First, we can notice that for $
x>1,$ $\lfloor \frac{1}{x} \rfloor=0$
Then x solution and $ x>1$ if and only if $ x$ strictly between 1 and 2
for $0<x<1 : \lfloor x \rfloor=0$ , then $x$ solution if and only if $\lfloor\frac{1}{x} \rfloor=1$ which can easily be solved using the definition of the floor function , then
$x$ solution if and only if $x$ strictly between $0.5 $ and $1$
Clearly if $x$ negative x is not a solution .
| {
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(Dis)prove the inequality $2^{\frac{1}{2}}\cdot\left(f\left(xa\right)+f\left(\frac{1}{xa}\right)\right)^{\frac{1}{2}}\leq g(x)$ for $a,x>0$ Problem :
Let $a,x>0$ then define :
$$f\left(x\right)=x^{\frac{x}{x^{2}+1}}$$
And :
$$g\left(x\right)=\sqrt{\frac{x^{a}}{a}}+\sqrt{\frac{a^{x}}{x}}$$
Then prove or disprove that :
$$2^{\frac{1}{2}}\cdot\left(f\left(xa\right)+f\left(\frac{1}{xa}\right)\right)^{\frac{1}{2}}\leq g(x)\tag{E}$$
My attempt
From here New bound for Am-Gm of 2 variables we have :
$$h(x)=\left(-\frac{1}{x}+\frac{1}{x^{2}}+1\right)^{-\frac{x}{2}}\geq x^{\frac{x}{x^{2}+1}}$$
So we need to show :
$$\sqrt{2}\sqrt{h\left(xa\right)+h\left(\frac{1}{xa}\right)}\leq g(x)$$
Now using Binomial theorem (second order at $x=1$) for $1\leq x\leq 2$ and $0.5\leq a\leq 1$ we need to show :
$$\sqrt{2}\sqrt{r\left(xa\right)+r\left(\frac{1}{xa}\right)}\leq g(x)$$
Where :
$$r(x)=\left(1+\left(-\frac{1}{x}+\frac{1}{x^{2}}\right)x+\frac{1}{2}\left(-\frac{1}{x}+\frac{1}{x^{2}}\right)^{2}\cdot x\cdot\left(x-1\right)\right)^{-\frac{1}{2}}$$
I have not tried but here show this inequality $\sqrt{\frac{a^b}{b}}+\sqrt{\frac{b^a}{a}}\ge 2$ user RiverLi's provide some lower bound again I haven't checked but (perhaps?) it works with this inequality .If it doesn't work we need a higher order in the Pade approximation .
Edit 06/01/2022 :
Using the nice solution due to user RiverLi it seems we have for $0.7\leq a \leq 1$ and $1\leq x\leq 1.5$ :
$$\frac{1}{a}\cdot\frac{1+x+(x-1)a^{2}}{1+x-(x-1)a^{2}}+\frac{1}{x}\cdot\frac{1+a+(a-1)x^{2}}{1+a-(a-1)x^{2}}\geq \sqrt{2}\sqrt{r\left(a^{2}x^{2}\right)+r\left(\frac{1}{a^{2}x^{2}}\right)}$$
If true and proved it provides a partial solution .
Edit 07/01/2022:
Define :
$$t\left(x\right)=\left(\ln\left(\left(-\frac{1}{x}+\frac{1}{x^{2}}+1\right)^{\frac{x}{2}}+1\right)\right)^{-1}$$
As accurate inequality we have for $x\geq 1$ :
$$\left(-\frac{1}{x}+\frac{1}{x^{2}}+1\right)^{-\frac{x}{2}}\leq \left(\ln\left(\left(-\frac{1}{x}+\frac{1}{x^{2}}+1\right)^{\frac{x}{2}}+1\right)\right)^{-1}+h(1)-t(1)$$
Again it seems we have for $0<x\leq 1$ :
$$\left(\ln\left(\left(-\frac{1}{x}+\frac{1}{x^{2}}+1\right)^{\frac{x}{2}}+1\right)\right)^{-\frac{96}{100}}+h(1)-(t(1))^{\frac{96}{100}}\geq \left(-\frac{1}{x}+\frac{1}{x^{2}}+1\right)^{-\frac{x}{2}}$$
If true we can use the power series of $\ln(e^x+1)$ around zero and hope .
Last edit 08/01/2022:
I found a simpler one it seems we have firstly :
On $(0,1]$ :
$$\left(1+\frac{1}{x^{2}}-\frac{1}{x}\right)^{-\frac{x}{2}}-\frac{x^{2}}{x^{2}+1}-\left(1-\frac{0.5\cdot1.25x}{x+0.25}\right)\leq 0$$
And on $[1,8]$ :
$$\left(1+\frac{1}{x^{2}}-\frac{1}{x}\right)^{-\frac{x}{2}}-\frac{x^{2}}{x^{2}+1}-\frac{0.5\cdot1.25x}{x+0.25}\leq 0$$
Question :
How to show or disprove $(E)$ ?
Thanks really .
| Just preliminary remarks (hoping that I shall be able to go further.
$$ f(a x)\times f\left(\frac{1}{a x}\right)=1$$
The maximum value of
$$2^{\frac 12} \sqrt{f(a x)+ f\left(\frac{1}{a x}\right)}$$ is attained for $a=2$ and $x=\frac 53$ and, for this couple,
$$2^{\frac 12} \sqrt{\left(\frac{3}{10}\right)^{30/109}+\left(\frac{10}{3}\right)^{30/109}}=2.05466 \qquad > 2$$ while the minimum value of $g(x)$ is $2$ (obtained for $x=a=1$).
The equality is precisely obtained for $x=a=1$.
At the point where the equality occurs $(x=a=1)$, we have for the function
$$h(x,a)=g(x)-\sqrt{f(a x)+ f\left(\frac{1}{a x}\right)}$$
$$\frac{\partial h(x,a)}{\partial x}=\frac{\partial h(x,a)}{\partial a}=0$$
$$\frac{\partial^2 h(x,a)}{\partial x^2}=\frac{\partial^2 h(x,a)}{\partial a^2}=\frac{\partial^2 h(x,a)}{\partial x\,\partial a}=\frac 14$$ which seems to be a good sign (at least locally).
To be continued (I hope)
| {
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The Golden Ratio shows up in the cosine function. Do other related numbers do so as well? I just learned that
*
*$\cos\left(\frac{π}{5}\right)$ = $\frac{φ}{2}$,
*$\cos\left(\frac{2π}{5}\right)$ = $\frac{1}{2φ}$ = $-\frac{ψ}{2}$,
*$\cos\left(\frac{3π}{5}\right)$ = $-\frac{1}{2φ}$ = $\frac{ψ}{2}$, and
*$\cos\left(\frac{4π}{5}\right)$ = $-\frac{φ}{2}$,
where $φ = \frac{1 + \sqrt{5}}{2}$ is the Golden Ratio and $ψ = \frac{1 - \sqrt{5}}{2}$ is the Golden Ratio Conjugate.
It seems suggestive to me that the Golden Ratio involves the square root of five and also appears in the output of the cosine function for inputs proportional to the inverse of five. So I did some digging.
The Golden Ratio is a member of the group of numbers which are solutions to the cubic equation $x^3 - a x - 1 = 0$, with $a = 1$ giving the Plastic Ratio $ρ = \sqrt[3]{\frac{9 + \sqrt{69}}{18}} + \sqrt[3]{\frac{9 - \sqrt{69}}{18}}$ and two complex roots, and $a = 2$ giving the Golden Ratio $φ$ plus the Golden Ratio Conjugate $ψ$ and $-1$. The value for $a = 3$ does not seem to have a name that I can find, but it does also show up in the cosine function.
If we denote the real-valued roots as $R_n(a)$, going from most positive at $n = 1$ to most negative at $n = 3$, then
*
*$\cos\left(\frac{π}{9}\right) = \frac{R_1(3)}{2} ≈ \frac{1.879385}{2}$,
*$\cos\left(\frac{2π}{9}\right) = -\frac{R_3(3)}{2} ≈ -\frac{-1.532089}{2}$,
*$\cos\left(\frac{4π}{9}\right) = -\frac{R_2(3)}{2} ≈ -\frac{-0.347296}{2}$,
*$\cos\left(\frac{5π}{9}\right) = \frac{R_2(3)}{2} ≈ \frac{-0.347296}{2}$,
*$\cos\left(\frac{7π}{9}\right) = \frac{R_3(3)}{2} ≈ \frac{-1.532089}{2}$, and
*$\cos\left(\frac{8π}{9}\right) = \frac{-R_1(3)}{2} ≈ -\frac{1.879385}{2}$.
Given that these two sets of roots for $a = 2$ and $a = 3$ show up, it makes me wonder if more do so as well. However, I haven't been able to find an way to fit the Plastic Ratio into the cosine function for $a = 1$, or the unnamed roots for $a = 4$.
Do the roots for other values of $a$ appear in the output of the cosine function? Is the appearance of these two sets of roots purely a coincidence, or is there some deeper mathematical connection?
| Substitute $x=m \cos t$ to get $$m^3 \cos^3 t - a m \cos t -1=0$$
Let $$m=\sqrt{\frac{4a}{3}}$$ then it rearranges (assuming $a,m\ne 0$) to
$$4 \cos^3 t - 3 \cos t = \sqrt{\frac{27}{4a^3}}$$
Using trigonometric identity
$$\cos (3t) = \sqrt{\frac{27}{4a^3}}$$
So
$$x = \sqrt{\frac{4a}{3}} \cos\left(\frac{1}{3} \cos^{-1}\left(\sqrt{\frac{27}{4a^3}}\right)\right)$$
When $a=1$ the R.H.S. of the penultimate equation is bigger than $1$ so $\cos^{-1}$ is problematic. Note that $\cos^{-1}/3$ is multi-valued.
| {
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Notice that $\sqrt{51}\approx 7+\frac{\sqrt{2}}{10}$ I was doing some other stuff and noticed that:
$$\sqrt{51} - \left(7 + \dfrac{\sqrt{2}}{10}\right) = 0.0000070723\,\, (*)$$
and this immediately made me think of their respective continued fractions, which are,
$$\sqrt{51} = 7 + \dfrac{1}{7+\dfrac{1}{14+\dfrac{1}{7+\dfrac{1}{14+\dots}}}} = [7;7,14]$$
and
$$\sqrt{2} = 1+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dots}}} = [1;2].$$
But I do not an immediate way to derive the above approximation based on these continued fractions and so I was wondering if there is an easy trick here.
For what it's worth, when you repeatedly square $(*)$ to get rid of the square roots, it amounts to:
$$4999^2-51\cdot 700^2 = 1.$$
But they are not the smallest solutions to the above Pell's equation since
$$50^2 - 51*7^2 = 1.$$
So I guess my question is can we derive $(*)$ from some known identity/approximations? Also, I would be curious to see if someone know of some other near-identites involving small numbers like $(*).$
| To approach with continued fractions:
$$\begin{align}a_n=\sqrt{n^2+2}&=[n,n,2n,n,2n,\dots]\\\sqrt{n^2+1}&=[n,2n,2n,\dots]\\
b_n=n+\frac{1}{\sqrt{n^2+1}}&=[n,n,2n,2n,2n,\dots]
\end{align}$$
So the third continued fraction for both $a_n$ and $b_n$ are $$c_3=\frac{2n^3+3n}{2n^2+1}.$$ The fourth fraction for $a_n$ is: $$c_4=\frac{2n^4+4n^2+1}{2n^3+2n}.$$ We know $c_3<b_n<a_n<c_4,$ so $$0<a_n-b_n<c_4-c_3=\frac{1}{(2n^3+2n)(2n^2+1)}<\frac{1}{4n^5}$$
Your case is $n=7,$ because $\frac{\sqrt2}{10}=\frac1{\sqrt{7^2+1}}.$
If you have solutions of $n^2-2m^2=-1,$ you get the approximation: $$\sqrt{n^2+2}\approx n+\frac{\sqrt{2}}{2m}$$ There are infinitely many integer solutions $(n,m)$ computed as $n+m\sqrt2=(1+\sqrt2)(3+\sqrt2)^k.$
For example, $(n.m)=(7,5)$ or $(n,m)=(41,29).$ The latter gives:
$$\sqrt{1683}-41-\frac{\sqrt2}{58}=0.000000001077321\approx\frac12\cdot\frac{1}{4\cdot 41^5}$$
That $1/2$ is probably because the fourth coefficients of $a_n$ and $b_n$ are $n$ and $2n,$ respectively.
A little more work shows that $$a_n-b_n\sim\frac1{8n^5}$$
Specifically, I think you get: $$0<\frac{1}{2n(n^2+1)(4n^2+3)}-\left(a_n-b_n\right)<\frac{1}{32n^7}$$
| {
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Find the area of a regular pentagon as a function of its diagonal For reference:
Calculate the area of a regular pentagon
as a function of its diagonal of length $a$. (Answer:$\frac{a^2}{4}\sqrt\frac{25-5\sqrt5}{2}$)
My progress:
$R$ = radius inscribed circle
$S = \frac{5R^2}{8}(\sqrt{10+2\sqrt5})$
$L=\frac{R}{2}(\sqrt{10-2\sqrt5})\implies R^2 = \frac{4L^2}{10-2\sqrt5}\tag{I}$
$\cos36^\circ=\frac{a}{2L} \implies a = L(\frac{1+\sqrt5)}{2}\implies L= \frac{2a}{1+\sqrt5}$
$\implies L^2 = \frac{a^2}{4}(6-2\sqrt5)\tag{II}$
$\text{From (I)}: S = \frac{20L^2}{8(10-2\sqrt5)}.(\sqrt{10+2\sqrt5})$
$\implies S =\frac{5L^2}{2(10-2\sqrt5)}.(\sqrt{10+2\sqrt5})$
$\text{From (II)}: S = \frac{5a^2(6-2\sqrt5)}{8.(10-2\sqrt5)}\cdot(\sqrt{10+2\sqrt5})$
$S = \frac{a^2(5-\sqrt5)}{16}(\sqrt{10+2\sqrt5})$
$\boxed{S = \frac{a^2}{4} \sqrt{\frac{25-5\sqrt5}{2} }}$
| While you are starting with a formula for the pentagon in terms of $R$, you can derive directly as follows -
Given $a$ is diagonal,
$ \displaystyle S_{AED} = S_{BCD} = \frac 12 \cdot a \cdot \frac {a \tan36^0}{2}$
$ \displaystyle S_{ADB} = \frac 12 \cdot a^2 \sin36^\circ$
So, $~ \displaystyle S = \frac{a^2}{2} \sin 36^\circ \cdot \frac{1 + \cos36^\circ}{\cos36^\circ}$
As $\cos 36^\circ = \dfrac{1 + \sqrt5}{4}$
$ \displaystyle \sin 36^\circ = \sqrt{1 - \frac{6+2 \sqrt5}{16}} = \frac 12 \cdot \sqrt{\frac{5-\sqrt5}{2}}$
Also, $\dfrac{1 + \cos36^\circ}{\cos36^\circ} = \dfrac{5 + \sqrt5}{1 + \sqrt5} = \sqrt5$
So, $ \displaystyle S = \frac{a^2}{4} \sqrt{\frac{25-5\sqrt5}{2}}$
| {
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Confusion about the use of principal roots and their use in expressions I'm struggling to derive the quadratic equation using the principal root $(\sqrt{x^2} = |x|).$
Taking $ax^2 + bx + c = 0$ and solving for $x$, I got $$\left(x+\frac{b}{2a}\right)^2 = \frac{b^2-4ac}{4a^2}.$$
Taking the principal root both sides gave me $$\left|x+\frac{b}{2a}\right| = \frac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}\\x+\frac{b}{2a} = \frac{±\sqrt{b^2-4ac}}{2|a|}.$$
And there's where I struggle. While for $a \geq 0, |a|=a,$ and it goes just well.
But that's not true if $a<0$, I can´t get it right with $a<0.$
Edit : Changed a typing error into the expression of the discriminant
| You are making a mistake; you should have $b^2\color{red}-4ac$ instead of $b^2\color{red}+4ac$. Besides, you are using the fact that $a^2=b^2\iff|a|=|b|$, which is correct. However, it is better here to use the fact that $a^2=b^2\iff a=\pm b$. So\begin{align}\left(x+\frac b{2a}\right)^2=\frac{b^2-4ac}{4a^2}\left(=\left(\frac{\sqrt{b^2-4ac}}{2a}\right)^2\right)&\iff x+\frac b{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\\&\iff x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.\end{align}
| {
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Find the sum (Includes binomial coefficients) I came across this question under summation:
Find the sum: $$\sum_{r=0}^n \left[\frac{r}{n} - \alpha \right]^2 {n \choose r}x^r(1-x)^{n-r} $$
To start with this, I wrote it as $\sum_{r=0}^n \left[\frac{r-n\alpha}{n} \right]^2 {n \choose r}\left(\frac{x}{1-x}\right)^r(1-x)^{n}$
= $(1-x)^{n}\sum_{r=0}^n \left[\frac{r^2+n^2\alpha^2-2n\alpha r}{n^2} \right] {n \choose r}\left(\frac{x}{1-x}\right)^r$
= $(1-x)^{n}\left[\sum_{r=0}^n \frac{r^2}{n^2} {n \choose r}\left(\frac{x}{1-x}\right)^r + \sum_{r=0}^n \alpha ^2 {n \choose r}\left(\frac{x}{1-x}\right)^r - \sum_{r=0}^n \frac{2\alpha r}{n} {n \choose r}\left(\frac{x}{1-x}\right)^r\right]$
Here, I found the values of individual sums by differentiating the binomial expansion of $\left(1+\frac{x}{1-x}\right)^n$.
I got :
$\sum_{r=0}^n \frac{r^2}{n^2} {n \choose r}\left(\frac{x}{1-x}\right)^r = \frac{nx+n(n-1)x^2}{(1-x)^nn^2}$
$\sum_{r=0}^n \alpha ^2 {n \choose r}\left(\frac{x}{1-x}\right)^r$ = $\alpha^2 \left(\frac{1}{1-x}\right)^n$
$\sum_{r=0}^n \frac{2\alpha r}{n} {n \choose r}\left(\frac{x}{1-x}\right)^r$ = $\frac{2\alpha}{n}\left(\frac{nx}{(1-x)^n}\right)$
Substituting the values, finally I got the sum to be $\frac{x(1-x)}{n}+(x-\alpha)^2$.
Can anyone please verify if the answer is correct?
| I can also verify that your approach and results are correct.
| {
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Number of regular polygons possible
Thirty six equally spaced points are plotted on a circle, and some of
these points are joined successively to form a polygon. How many
distinct such regular polygons are possible.
What I thought that the answer should be $^{36}C_3 + ^{36}C_4 + ^{36}C_5 + ^{36}C_6 + ^{36}C_7 + \cdots + ^{36}C_{36}$ but this is not the correct answer that has been provided. What am I doing wrong? Please help !!!
Thanks in advance !!!
| Since the $k$ vertices of a regular polygon inscribed in a circle must be equally spaced, to form a regular polygon using the $36$ equally spaced points on the circle, the number of sides must be a divisor of $36$. Hence, the regular polygon must have either $3, 4, 6, 9, 12, 18,$ or $36$ sides. There are $36/k$ distinguishable regular polygons with $k$ sides since there are $36$ possible points at which you could start drawing the regular polygon, but doing so counts each such regular polygon $k$ times, once for each of the $k$ vertices of the polygon where you could start your drawing. Hence, the number of distinct regular polygons which could be drawn using $36$ equally spaced points on the circle is
$$\frac{36}{3} + \frac{36}{4} + \frac{36}{6} + \frac{36}{9} + \frac{36}{12} + \frac{36}{18} + \frac{36}{36} = 12 + 9 + 6 + 4 + 3 + 2 + 1 = 37$$
| {
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An asymptotic estimate for $\binom{n}{k}$ when $n$ is much larger than $k$ According to Wikipedia, if $n,k \to \infty$ and $k / n \to 0$, then
$$\binom{n}{k} \approx \left( \frac{en}{k} \right)^k \cdot (2 \pi k)^{-1/2} \cdot \exp \left( -\frac{k^2}{2n} (1+o(1)) \right).$$
Stirling's approximation gives that
$$\binom{n}{k} \approx \frac{ \sqrt{2 \pi n } (n/e)^n}{\sqrt{2 \pi k} (k/e)^k \sqrt{2 \pi (n-k)}((n-k)/e)^{n-k} } \approx \frac{n^n}{\sqrt{2 \pi k} \cdot k^k \cdot (n-k)^{n-k}},$$
since $k/n \to 0$, but I still don't see how to get from here to there. In particular, I don't see where the smaller order term $$\exp \left( -\frac{k^2}{2n} (1+o(1)) \right)$$
comes from.
| We have
$$
\frac{{n^n }}{{\sqrt {2\pi k} k^k (n - k)^{n - k} }} = \left( {\frac{{en}}{k}} \right)^k (2\pi k)^{ - 1/2} \left( {1 - \frac{k}{n}} \right)^{k - n} e^{ - k} .
$$
Now
\begin{align*}
\left( {1 - \frac{k}{n}} \right)^{k - n} e^{ - k} & = \exp \left( {(k - n)\log \left( {1 - \frac{k}{n}} \right) - k} \right)
\\ &
= \exp \left( {(k - n)\left( { - \frac{k}{n} - \frac{{k^2 }}{{2n^2 }} + \mathcal{O}\!\left( {\frac{{k^3 }}{{n^3 }}} \right)} \right) - k} \right)
\\ &
= \exp \left( { - \frac{{k^2 }}{{2n}} - \frac{{k^3 }}{{2n^2 }} + \mathcal{O}\!\left( {\frac{{k^4 }}{{n^3 }}} \right) + \mathcal{O}\!\left( {\frac{{k^3 }}{{n^3 }}} \right)} \right)
\\ &
= \exp \left( { - \frac{{k^2 }}{{2n}}\left( {1 + \frac{k}{n} + \mathcal{O}\!\left( {\frac{{k^3 }}{{n^2 }}} \right)} \right)} \right)
\\ &
= \exp \left( { - \frac{{k^2 }}{{2n}}\left( {1 + o(1)} \right)} \right),
\end{align*}
provided $k = o(n^{2/3} )$.
Addendum. We can derive the approximation under the weaker assumption $k=o(n)$ as follows. We write
$$
\binom{n}{k} = \frac{{n^k }}{{k!}}\prod\limits_{m = 1}^{k - 1} {\left( {1 - \frac{m}{n}} \right)} \sim \left( {\frac{{en}}{k}} \right)^k (2\pi k)^{ - 1/2} \prod\limits_{m = 1}^{k - 1} {\left( {1 - \frac{m}{n}} \right)}
$$
for large $n$ and $k$. Then
\begin{align*}
\prod\limits_{m = 1}^{k - 1} {\left( {1 - \frac{m}{n}} \right)} & = \exp \left( {\sum\limits_{m = 1}^{k - 1} {\log \left( {1 - \frac{m}{n}} \right)} } \right) = \exp \left( {\sum\limits_{m = 1}^{k - 1} {\left( { - \frac{m}{n} + \mathcal{O}\!\left( {\frac{{m^2 }}{{n^2 }}} \right)} \right)} } \right)
\\ & = \exp \left( { - \frac{1}{n}\sum\limits_{m = 1}^{k - 1} m + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)\sum\limits_{m = 1}^{k - 1} {m^2 } } \right) = \exp \left( { - \frac{{k^2 }}{{2n}} + o(1) + \mathcal{O}\!\left( {\frac{{k^3 }}{{n^2 }}} \right)} \right)
\\ & = \exp \left( { - \frac{{k^2 }}{{2n}}\left( {1 + \mathcal{O}\!\left( {\frac{k}{n}} \right)} \right)} \right) = \exp \left( { - \frac{{k^2 }}{{2n}}(1 + o(1))} \right),
\end{align*}
provided $k=o(n)$.
| {
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Maximum value of $ (1- \tan A)(1- \tan B)(1- \tan C)$ Assume that we have a triangle with angles A, B, C. so $A+B+C = \pi$.
What is the maximum value of $(1- \tan A)(1- \tan B)(1- \tan C)$?
My try:
I know that given $A+B+C = \pi$ we have that
$\tan A + \tan B + \tan C = \tan A \tan B \tan C$ and $\tan C = - \tan(A+B)$
Using the equalities above, I came to the result
$(1- \tan A)(1- \tan B)(1- \tan C) = 1 - 2(\tan A+ \tan B +\tan C )+(\tan A \tan B + \tan A \tan C +\tan B \tan C)$
and this is where I'm stuck, using the second equality I didn't get anything helpful.
I'm trying to solve this without using differentiation. any hints would be appreciated.
| The maximum value is infinity. Take
$$A = \frac{\pi}{2} - \epsilon, B = \frac{\pi}{3} + \epsilon, C = \frac{\pi}{6}$$
for $\epsilon > 0$ arbitrarily small.
The minimum value is also infinity. Take
$$A = \frac{\pi}{2} + \epsilon, B = \frac{\pi}{3} - \epsilon, C = \frac{\pi}{6}$$
for $\epsilon > 0$ arbitrarily small.
| {
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Number of Ways of Ordering Numbers How many ways you can order the numbers $0, 1, 2, 3,..., 12$ using each number exactly once, such that the sum of two adjacent numbers are not greater than $13$? (This is a first round's question of the four rounds of Bangladesh Mathematical Olympiad for class $11$–$12$.)
For example, these are some orderings which satisfy the condition:
*
*$0, 12, 1, 11, 2, 10, 3, 9, 4, 8, 5, 7, 6$
*$12, 1, 11, 2, 10, 3, 9, 4, 8, 5, 7, 6, 0$
*$1, 12, 0, 11, 2, 10, 3, 9, 4, 8, 5, 6, 7$
*$11, 2, 10, 1, 12, 0, 3, 9, 4, 8, 5, 6, 7$
*$6, 7, 2, 11, 0, 12, 1, 10, 3, 5, 8, 4, 9$
| Denote the number by $P(12)$.
There are six different arrangements for the placing of the $12$. It can be at one end and adjacent to $0$ or $1$ or it can be adjacent to both $0$ and $1$ in some order.
Consider a line starting $12,0$. The number of possibilities is then the number of arrangements of $1$ to $11$ with sum not greater than $13$ but (subtracting $1$ from each number) this is the same as $P(10)$.
The line starting $12,1$ is the same since neither $0$ nor $1$ add to greater than $13$ with a remaining number.
In the case of $0,12,1$ occurring in a line, replace this sequence by $1$ and then we again require the number of arrangements of $1$ to $11$ with sum not greater than $13$.
Thus $P(12)=6P(10)=36P(8)= ... =6^5.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4370524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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} |
Prove that $\displaystyle\int_{0}^{\frac{1}{\sqrt{3}}}\frac{\arctan\left(\frac{1}{\sqrt{1-2x^2}}\right)}{1+x^2} \mathrm{dx}=\frac{13\pi^2}{288}$ I need to prove that,
$\displaystyle \tag*{} \int_{0}^{\frac{1}{\sqrt{3}}}\frac{\arctan\left(\frac{1}{\sqrt{1-2x^2}}\right)}{1+x^2} \mathrm{dx}=\frac{13\pi^2}{288}$
Here is what I tried:
I tried to solve the integral by Integrating by parts.
We have:
$\displaystyle \tag{1} \int f(x) g'(x) \mathrm{ dx} = f(x)g(x) - \int f'(x) g(x) \mathrm{ dx}$
I noticed ,
$\displaystyle \tag*{} \dfrac{1}{1+x^2} = \arctan '(x)$
$\displaystyle \tag*{} g'(x) = \dfrac{1}{1+x^2} \Leftrightarrow g(x) = \arctan(x)$
Now, I defined $f(x)$
$\displaystyle \tag*{} f(x) = \arctan \left( \dfrac{1}{\sqrt{1-2x^2}}\right )$
and
$\displaystyle \tag*{} f'(x) = \arctan'\left(\dfrac{1}{\sqrt{1-2x^2}} \right ) = \text{arccot}'(\sqrt{1-2x^2}) = \dfrac{x}{\sqrt{1-2x^2}(1-x^2)}$
Now, using $(1)$ and substituting the values of $f(x)$ and $g(x)$, My indefinite integral becomes:
$\displaystyle \tag*{} \arctan(x) \arctan \left (\dfrac{1}{\sqrt{1-2x^2}} \right ) - \int \dfrac{x \arctan(x)}{\sqrt{1-2x^2}(1-x^2)} \mathrm{ dx}$
Now, we want to evaluate:
$\displaystyle \tag*{} \int \dfrac{x \arctan(x)}{\sqrt{1-2x^2}(1-x^2)} \mathrm{ dx}$
Now this is integral I am having trouble solving. Any hints or different methods would be greatly appreciated. Thank you.
| A large portion of this answer is taken directly from this answer!
Enforce the substitution
$$y=\sqrt{1-2x^2}\qquad\qquad x^2=\frac {1-y^2}{2}\qquad\qquad\mathrm dx=-\frac {y}{\sqrt{2(1-y^2)}}\,\mathrm dy$$
The integral now becomes
$$\begin{align*}\int\limits_0^{1/\sqrt3}\frac {\arctan\left(\frac 1{\sqrt{1-2x^2}}\right)}{1+x^2}\,\mathrm dx & =\sqrt{2}\int\limits_{1/\sqrt{3}}^1\frac {y\left(\frac {\pi}2-\arctan y\right)}{\sqrt{1-y^2}(3-y^2)}\,\mathrm dy\\ & =\frac {\pi}{\sqrt{2}}\int\limits_{1/\sqrt{3}}^1\frac {y\,\mathrm dy}{\sqrt{1-y^2}(3-y^2)}-\sqrt{2}\int\limits_{1/\sqrt3}^1\frac {y\arctan y}{\sqrt{1-y^2}(3-y^2)}\,\mathrm dy\end{align*}$$
Splitting up the integrand, the first integral can be evaluated with the substitution $y\mapsto\sqrt{1-y^2}$, giving
$$\frac {\pi}{\sqrt{2}}\int\limits_{1/\sqrt{3}}^1\frac {y}{\sqrt{1-y^2}(3-y^2)}\,\mathrm dy=\frac {\pi^2}{12}$$
And enforcing the substitution $t=\sqrt{y}$ on the second integral, we have that
$$\int\limits_0^{1/\sqrt3}\frac {\arctan\left(\frac 1{\sqrt{1-2x^2}}\right)}{1+x^2}\,\mathrm dx=\frac {\pi^2}{12}-\frac 1{\sqrt{2}}\int\limits_{1/3}^1\frac {\arctan\sqrt{t}}{(3-t)\sqrt{1-t}}\,\mathrm dt$$
The last integral is difficult, but using the same methodology and formulas as the answer I have linked above, we first rewrite the integral such that the lower limit is zero.
$$\int\limits_{1/3}^1\frac {\arctan\sqrt t}{(3-t)\sqrt{1-t}}\,\mathrm dt=\underbrace{\int\limits_0^1\frac {\arctan\sqrt{t}}{(3-t)\sqrt{1-t}}\,\mathrm dt}_{J}-\underbrace{\int\limits_0^{1/3}\frac {\arctan\sqrt{t}}{(3-t)\sqrt{1-t}}\,\mathrm dt}_{K}$$
Next, we use the following formula to evaluate the right-hand side
$$\int\limits_0^x\frac {\arctan\sqrt t}{(a-t)\sqrt{b-t}}\,\mathrm dt=\frac 1{\sqrt{a-b}}S\left(\arctan\sqrt{\frac {b-x}{a-b}},\arctan\sqrt{\frac {b+1}{a-b}},\arctan\frac 1{\sqrt{a}}\right)$$
There are two important observations that will help us evaluate the two integrals, namely
*
*$S(0,\beta,\gamma)=\pi(\beta-\gamma)$
*When $\sin^2\alpha+\sin^2\gamma=\sin^2\beta$, then $S(\alpha,\beta,\gamma)=-\alpha^2+\beta^2-\gamma^2$
Substituting $a=3$ and $b=1$, then
$$\begin{align*}J & =\frac 1{\sqrt2}S\left(0,\frac {\pi}4,\frac {\pi}6\right)=\frac {\pi^2}{12\sqrt2}\\K & =\frac 1{\sqrt2}S\left(\frac {\pi}6,\frac {\pi}4,\frac {\pi}6\right)=\frac {\pi^2}{144\sqrt2}\end{align*}$$
Where I have used the first observation to evaluate $J$ and the second observation to evaluate $K$. Taking the difference $J-K$, then
$$\int\limits_{1/3}^1\frac {\arctan x}{(3-x)\sqrt{1-x}}\,\mathrm dx=\frac {11\pi^2}{144\sqrt2}$$
Putting everything together, we get that
$$\int\limits_0^{1/\sqrt{3}}\frac {\arctan\left(\frac 1{\sqrt{1-2x^2}}\right)}{1+x^2}\,\mathrm dx=\frac {\pi^2}{12}-\frac {11\pi^2}{288}\color{blue}{=\frac {13\pi^2}{288}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4372882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Solve differential equation $(x^2-y^2+2x-y)dx+(x^2-y^2+x-2y)dy=0$ Solve the differential equation: $(x^2-y^2+2x-y)dx+(x^2-y^2+x-2y)dy=0$.
My attempt :
I factored and did some calculation to get $ dx+dy+\dfrac{xdx-ydy}{(x-y)(x+y+1)}=0.$
I am stuck after it . It could have been a lot easier if there were $xdy-ydx$. Will partial fraction decomposition work here ?
| Let
\begin{equation}
\begin{split}
x+y =z\\
x-y =\widetilde{z}
\end{split}
\end{equation}
Using this substitution, the differential equation you posted can be written as
\begin{equation}
\begin{split}
\left(z \widetilde{z}+ \widetilde{z} \right) \mathrm{d} z = -\frac{1}{2}\left( z \mathrm{d}\widetilde{z}+\widetilde{z} \mathrm{d}z\right)
\end{split}
\end{equation}
By just adding $-x\mathrm{d} x+ y\mathrm{d}y = -\frac{1}{2}\mathrm{d}(x^2-y^2)$ on both sides.
Now, this resulting equation can be easily solved by simplifying it as
\begin{equation}
\begin{split}
\frac{2}{z}\left(z+\frac{3}{2}\right) \mathrm{d}z = -\frac{\mathrm{d}\widetilde{z}}{\widetilde{z}}
\end{split}
\end{equation}
Giving
\begin{equation}
e^{2 z} z^3 \widetilde{z} = k
\end{equation}
Where $k$ is the integration constant.
Or,
The solution to the differential equation you posed is the solution to equation
$$e^{2(x+y)} (x+y)^3 (x-y) =k$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4374002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find floor of $ 1+\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} + ...+\frac{1}{\sqrt{2017}}$? How to find floor of $S=1+\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} + ...+\frac{1}{\sqrt{2017}}$?
With
$$\frac{1}{\sqrt{i}} > 2(\sqrt{i+1} -\sqrt{i})$$
and
$$\frac{1}{\sqrt{i}} < 2(\sqrt{i} -\sqrt{i-1}),$$
I'm able to get to
$$2 \sqrt{n+1}-2 < \sum_{i=1}^n \frac{1}{\sqrt{i}} < 2 \sqrt{n}-1.$$
Given $n=2017$, the result is $88.37282...$ (using Wolfram), so the floor should be $88$.
But I just get to $87.844<S<88.822$, and therefore I don't get the exact floor.
With ALGEBRA ONLY (no integral, limit, Harmonic Series), could you please help me to solve this problem?
Thank you!
| Thanks to @Gary's help, I'm able to solve this problem. Thank you!
With $k≥1$, we have:
$$ \frac{2}{\sqrt{k} + \sqrt{k+1}} < \frac{2}{\sqrt{k} + \sqrt{k}} < \frac{2}{\sqrt{k} + \sqrt{k-1}} $$
So
$$ \frac{2}{\sqrt{k} + \sqrt{k+1}} < \frac{1}{\sqrt{k}} < \frac{2}{\sqrt{k} + \sqrt{k-1}} $$
and so
$$ 2(\sqrt{k+1} - \sqrt{k}) < \frac{1}{\sqrt{k}} < 2(\sqrt{k} - \sqrt{k-1})$$
*
*With $k=2$: $ 2(\sqrt{3} - \sqrt{2}) < \frac{1}{\sqrt{2}} < 2(\sqrt{2} - \sqrt{1})$
*With $k=3$: $ 2(\sqrt{4} - \sqrt{3}) < \frac{1}{\sqrt{3}} < 2(\sqrt{3} - \sqrt{2})$
...
*With $k=2017$: $ 2(\sqrt{2018} - \sqrt{2017}) < \frac{1}{\sqrt{2017}} < 2(\sqrt{2017} - \sqrt{2016})$
So we have:
$1 + 2(\sqrt{2018} - \sqrt{2}) < S < 1 + 2(\sqrt{2017} - 1)$
And therefore $88.016<S<88.822$ which implies that $$\lfloor S\rfloor = 88.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4374424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve for $x$ in $\frac{\sin(30)\cdot\sin(60)\cdot\sin(x+10)}{\sin(80)\cdot\sin(40)\cdot\sin(x)}=1$ I have reduced a variant of the $80-80-20$ triangle problem to this equation-
$$\frac{\sin\left(30\right)\cdot\sin\left(60\right)}{\sin\left(80\right)\cdot\sin\left(40\right)}\cdot\frac{\sin\left(x+10\right)}{\sin\left(x\right)}=1$$ I'm looking for a nice transformation of the equation to get $x$. I already solved the problem by reflection but want an alternative solution with sine rule. Please give hints or another better equation if you can find one. I found the original problem solved on AoPS using law of sines but could not apply the similar methods here.
| Let us use the identity $$\sin\theta\cdot\sin(60^\circ-\theta)\cdot\sin(60^\circ+\theta)=\frac{\sin3\theta}4$$
to get $$\frac{\sin40^\circ\sin80^\circ}{\sin60^\circ}=\frac1{4\sin20^\circ}.$$
In your equation, leave the terms containing $x$ in LHS and move the others to RHS, then use the derived relation.
$$\frac{\sin(x+10^\circ)}{\sin x}=\frac{\sin40^\circ\sin80^\circ}{\sin30^\circ\sin60^\circ}=\frac1{2\sin20^\circ}=\frac{\sin30^\circ}{\sin20^\circ}$$
From here, considering the nature of the original problem, $x=20^\circ$ is an obvious solution. To confirm further, expand $\sin$ to get,
$$\frac{\sin(x+10^\circ)}{\sin x}=\frac{\sin(20^\circ+10^\circ)}{\sin20^\circ}$$
$$\cos10^\circ+\cot x\sin10^\circ=\cos10^\circ+\cot20^\circ\sin10^\circ$$
$$\cot x=\cot 20^\circ.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4375873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate the limit: $ \lim\limits_{(x,y)\to(0,0)}{\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}}$
Calculate the limit:
$$\lim\limits_{(x,y)\to(0,0)}{\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}}
$$
I tried to change to polar coordinates like that:
\begin{align*}
\lim\limits_{(x,y)\to(0,0)}{\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}} &=
\lim\limits_{(x,y)\to(0,0)}{\frac{e^{-\frac{1}{x^2+y^2}}}{(x^2+y^2)^2-2x^2y^2}}\\
&=\lim\limits_{r\to0}{\frac{e^{-\frac{1}{r^2}}}{r^4(1-2\cos^2\theta \sin^2\theta)}}\\
\end{align*}
and I'm not sure how to continue from this point.
Thank you!
| Since $1-2\cos^2 \theta \sin^2 \theta \ne 0$ the limit in polar coordinates becomes
$$
\frac{1}{1-2\cos^2 \theta \sin^2 \theta}\cdot \lim_{r\to 0}\dfrac{e^{-1/r^2}}{r^4}=0, \quad \forall \theta \in[0,2\pi)
$$
This shows that the original limit is zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4377737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Simplify $\sin[\arctan(4x)]$ for $ - \frac 14 \le x \le \frac 14 $ I am coming up with the wrong answer.
The problem is $\sin[\arctan(4x)]$
Here are the reciprocal functions I am using $\sin\theta= \frac {1}{\csc\theta}$ and $\cot\theta = \frac{1}{\tan\theta}$
Here is the Pythagorean identity I am using $\csc^2 \theta = 1+\cot^2 \theta$
I am trying to rewrite the problem $\sin[\arctan(4x)]$ in terms of the Pythagorean Identity using the reciprocal identities. $$ \frac{1}{\csc^2 \theta} = \frac {1}{1 + \frac{1}{1+\tan^2 \theta}}$$ This seems to be a stretch, I do not think I am expressing the Pythagorean Identity right, and I am not sure how to either. I did start to try to plug in the numbers.
$$ \frac{1}{\csc^2 \theta} = \frac {1}{1 + \frac {1}{(4x^2)}}$$
$$ \frac{1}{\csc \theta} = \sqrt{1+4x} $$
$$\sin\theta = \frac{1}{\sqrt{1+4x}} $$
I could rationalize the denominator and multiply the by its conjugate to get; $$\frac{\sqrt{1-4x}}{\sqrt{(1+4x)(1-4x)}} $$ But, at this point I know I am wrong. The answer is $$ \frac{4x}{\sqrt{16x^2+1}}$$
| As egreg said, the condition is redundant. On the other words, for any real $x$,$$\sin(\arctan(4x))=\frac{4x}{\sqrt{16x^2+1}}.$$
I like to prove it by 2 cases. Let $y=\arctan 4x$.
A. When $x\geq 0,$
$$\sin ^{2} y=\tan ^{2} y \cos ^{2} y=16 x^{2}\left(1-\sin ^{2} y\right) \Rightarrow \sin ^{2} y=\frac{16 x^{2}}{16 x^{2}+1} \Rightarrow \sin y=\frac{4 x}{\sqrt{16 x^{2}+1}} $$
Hence $$
\sin (\arctan (4 x))=\frac{4 x}{\sqrt{16 x^{2}+1}}.
$$
B. When $x< 0,$ let $z=-x>0,$ then using the result in case A yields
$$
\sin (\arctan 4 z)=\frac{4 z}{\sqrt{16 z^{2}+1}}.
$$
Now putting $z=-x$ yields $$
\sin (\arctan 4(-x))=\frac{-4 x}{\sqrt{16 (-x)^{2}+1}}.
$$
Using the facts that $\sin x$ and $\arctan x $are odd functions, we have$$
-\sin (\arctan (4x))=-\frac{4 x}{\sqrt{16 x^{2}+1}}.
$$
$$
\sin (\arctan (4x))=\frac{4 x}{\sqrt{16 x^{2}+1}}.
$$
Therefore we can conclude that for any real $x$, $$
\boxed{\sin (\arctan (4x))=\frac{4 x}{\sqrt{16 x^{2}+1}}}.
$$
In general, for any real $x$, $$\boxed{\sin (\arctan (x))=\frac{x}{\sqrt{x^{2}+1}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4379873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
How to solve derivative/limit of $f(x)=x\sqrt{4-x^2}$ I'm trying to differentiate $x\sqrt{4-x^2}$ using the definition of derivative.
So it would be something like
$$\underset{h\to 0}{\text{lim}}\frac{(h+x) \sqrt{4-\left(h^2+2 h x+x^2\right)}-x \sqrt{4-x^2}}{h}$$
I was trying to solve and I just can end up with something like
$$\underset{h\to 0}{\text{lim}}\frac{(x+h)\sqrt{4-x^2-2xh-h^2}-x\sqrt{4-x^2}}h \cdot \frac{\sqrt{4-x^2-2xh-h^2}+\sqrt{4-x^2}}{\sqrt{4-x^2-2xh-h^2}+\sqrt{4-x^2}}$$
$$\underset{h\to 0}{\text{lim}}\frac{-3x^2h-3xh^2+4h-h^3+\sqrt{4-x^2}-\sqrt{4-x^2-2xh+h^2}}{h\sqrt{4-x^2-2xh-h^2}+\sqrt{4-x^2}}$$
Now if I group on h, I will have some tricky 3 instead of 2.
The idea is I should have something like $h(2x^2+4)$ that would cancel up.
I'm quite stuck can I ask a little of help? I know wolframalpha exists but it refuses to create the step by step solution with the error "Ops we don't have a step by step solution for this query".
The final result shall be
$$-\frac{2 \left(x^2-2\right)}{\sqrt{4-x^2}}$$
| Clearly, when $h\to 0$:
$$\frac{h\cdot \sqrt{4-\left(h^2+2 h x+x^2\right)}}{h}\to \sqrt{4-x^2}$$
So, the limit simplify to:
$$\lim_{h\to 0}\frac{x\sqrt{4-\left(h^2+2 h x+x^2\right)}-x \sqrt{4-x^2}}{h}=\sqrt{4-x^2}+\lim_{h\to 0}\frac{x\left[\sqrt{4-\left(h^2+2 h x+x^2\right)}-\sqrt{4-x^2}\right]}{h}=\sqrt{4-x^2}+\lim_{h\to 0}\frac{x\cdot \sqrt{4-x^2}\left[\sqrt{1-\frac{h^2+2hx}{4-x^2}}-1\right]}{h}$$
Using the fact that:
$$\sqrt{1+x}\,\, \sim\,\, \frac{1}{2}x\,\, x \to 0$$
We obtain:
$$\lim_{h\to 0}\frac{x\cdot \sqrt{4-x^2}\left[\sqrt{1-\frac{h^2+2hx}{4-x^2}}-1\right]}{h}\,\, \sim\,\, \lim_{h\to 0}\frac{x\cdot \sqrt{4-x^2}\cdot\left(-\frac{1}{2}\cdot\frac{2hx}{4-x^2}\right)}{h}=\lim_{h\to 0}\frac{-\frac{hx^2}{\sqrt{4-x^2}}}{h}=-\frac{x^2}{\sqrt{4-x^2}}$$
So, the limit is:
$$\sqrt{4-x^2}-\frac{x^2}{\sqrt{4-x^2}}=\frac{4-x^2-x^2}{\sqrt{4-x^2}}=2\cdot\frac{2-x^2}{\sqrt{4-x^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove by induction that $4^n+5^n+6^n$ is divisible by $15$ for all odd $n\in \mathbb{N}$.
Prove by induction that $4^n+5^n+6^n$ is divisible by $15$ for all odd $n\in \mathbb{N}$.
My proof :
I'm straight going to the induction hypothesis part.
Let for some odd $m(\gt2)\in \mathbb{N}, 15 | 4^m+5^m+6^m$.
Then, $$ 4^m+5^m+6^m = 15q$$ for some $q\in \mathbb{Z}$.
Now, $$4^m = 15q-5^m-6^m$$
Since $m$ is odd then the next odd number is $m+2$.
Now, $$ 4^{m+2} + 5^{m+2} + 6^{m+2} \\
= 4^m \cdot 16 + 5^m \cdot 25 + 6^m \cdot 36 \\
= (15q-5^m-6^m) \cdot 16 + 5^m \cdot 25 + 6^m \cdot 36 \\
= 15 \cdot 16q + 9 \cdot 5^m + 20 \cdot 6^m \\
= 15 \cdot 16q + 9 \cdot 5 \cdot 5^{m-1} + 20 \cdot 6 \cdot 6^{m-1} \\
= 15\big( 16q + 3 \cdot 5^{m-1} + 8 \cdot 6^{m-1} \big) \\
\\
\implies 15 | 4^{m+2} + 5^{m+2} + 6^{m+2} $$
Thus from induction the statement is true for every odd $n\in \mathbb{N}$
Is my proof correct?
| If you know about linear recurrence: Let $f(n)=4^n+5^n+6^n,$ then you get the recurrence $$\begin{align}
f(n+3)&=(4+5+6)f(n+2)-(4\cdot 5+4\cdot 6+5\cdot 6)f(n+1)+4\cdot 5\cdot 6 f(n)\\
&=15f(n+2)-74f(n+1)+240f(n)\\&\equiv f(n+1)\pmod{15}
\end{align} $$ So not only is $$f(2k+1)\equiv f(1)=15\equiv 0\pmod {15},$$ but $$f(2k)\equiv f(2)=77\equiv 2\pmod{15},$$ for $k\geq 1.$
If $a,b,c$ are integers, and $g(n)=a4^n+b5^n+c6^n,$ you will get the same congruence, $g(n+3)\equiv g(n+1)\pmod {15}.$
Even more generally, if $a,b,c,d$ are integers and $3\not\mid d,$ and $h(n)=a(d-1)^n+bd^n+c(d+1)^n,$ you also get:
$$h(n+3)\equiv h(n+1)\pmod {3d}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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'Higher order' complex numbers I recently learned that complex numbers can also be represented as matrices of the form: $$\begin{pmatrix} x & -y \\ y & x\end{pmatrix}$$ where complex multiplication corresponds to matrix multiplication and conjugation corresponds to transposition. I thought it was interesting how this representation corresponds to the $2 \times 2$ unit rotation matrix:
$$\begin{pmatrix} \cos \theta & -\sin\theta \\ \sin \theta & \cos \theta \end{pmatrix}$$
I was wondering if, then, there is a higher order version of the complex numbers? e.g. numbers that correspond to the $3 \times 3$ unit rotation matrix:
$$\begin{pmatrix} \sin \theta \cos \phi & \sin \theta \sin \phi & \cos \theta \\ \cos \theta \cos \phi & \cos \theta \sin \phi & - \sin \theta \\ -\sin \phi & \cos \phi & 0\end{pmatrix}$$ etc...
| Usually with even number of dimensions you have better results than with odd. Thus, you do not have anything useful or interesting in 3 dimensions. But in 4 dimensions you have multiple interesting algebras, some of which are more or less numbers-like.
Eevee Trainer has already mentioned quaternions, which are a division algebra but not commutative, but there are also other approaches which give algebras with different properties in 4D.
*
*First of all, possibly the simplest approach is to consider the $2\times2$ real matrices themselves, which is a 4D space. Matrix operations on $2\times2$ matrices are isomorphic to so-called split-quaternions:
$\begin{align}
\boldsymbol{1} =\begin{pmatrix}1&0\\0&1\end{pmatrix},\qquad&\boldsymbol{i} =\begin{pmatrix}0&1\\-1&0\end{pmatrix},\\
\boldsymbol{j} =\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad&\boldsymbol{k} =\begin{pmatrix}1&0\\0&-1\end{pmatrix}.\end{align}$
They are neither commutative, nor a division-algebra, but allow to apply any functions that can be generalized to square matrices. They have zero divisors, nilpotents and idempotents.
*Dual complex numbers. Formed by adding the dual unity $\varepsilon$ such that $\varepsilon^2=0$ to complex numbers. They can be represented by matrices of the form
$\left(
\begin{array}{cc}
u & w \\
0 & v \\
\end{array}
\right)$, where $u,v,w$ are complex numbers or matrices of the form $\left(
\begin{array}{cccc}
a & b & c & d \\
-b & a & -d & c \\
0 & 0 & a & b \\
0 & 0 & -b & a \\
\end{array}
\right)$ or $\left(
\begin{array}{cccc}
a & c & b & d \\
0 & a & 0 & b \\
-b & -d & a & c \\
0 & -b & 0 & a \\
\end{array}
\right)$ where $a,b,c,d$ are real numbers ($a$ is real part, $b$ is imaginary part, $c$ is dual part, $d$ is imaginary dual part). This system is commutative but has zero divisors and nilpotents.
*Tessarines. Formed by adding split-complex unity $j$ such as $j^2=1$ to complex numbers. Can be represented as matrices of the form $\left(
\begin{array}{cc}
u & w \\
w & u \\
\end{array}
\right)$, where $u,w$ are complex numbers or matrices of the form $\left(
\begin{array}{cccc}
a & b & c & d \\
-b & a & -d & c \\
c & d & a & b \\
-d & c & -b & a \\
\end{array}
\right)$ or $\left(
\begin{array}{cccc}
a & c & b & d \\
c & a & d & b \\
-b & -d & a & c \\
-d & -b & c & a \\
\end{array}
\right)$ where $a,b,c,d$ are real numbers ($a$ is real part, $b$ is imaginary part, $c$ is split-imaginary part, $d$ is the coefficient of $ij$ or $-ij=k$). This system is commutative algebra, but it has zero divisors and idempotents.
If one adds another complex unity $i_1$ such that $i_1^2=-1$ instead of split-complex unity $j$, one will get bicomplex numbers, which are isomorphic to this system (the split-complex unity will arise automatically as $j=-i i_1$).
*Dual split-complex numbers. They are formed by adding split-complex unity $j$ and dual unity $\varepsilon$ to real numbers. They can be represented as real matrices of the form $\left(
\begin{array}{cccc}
a & b & c & d \\
b & a & d & c \\
0 & 0 & a & b \\
0 & 0 & b & a \\
\end{array}
\right)$ or $\left(
\begin{array}{cccc}
a & c & b & d \\
0 & a & 0 & b \\
b & d & a & c \\
0 & b & 0 & a \\
\end{array}
\right)$ where $a,b,c,d$ are real numbers ($a$ is real part, $b$ is split-imaginary part, $c$ is dual part, $d$ is split-imaginary dual part).
This is a commutative algebra but with zero divisors, nilpotents and idempotents.
*4D split-complex numbers. These are formed by adding two split-complex unities $j$ and $j_1$ such that $j^2=j_1^2=1$. They are isomorphic to the matrices of the form $\left(
\begin{array}{cccc}
a & b & c & d \\
b & a & d & c \\
c & d & a & b \\
d & c & b & a \\
\end{array}
\right)$ or $\left(
\begin{array}{cccc}
a & c & b & d \\
c & a & d & b \\
b & d & a & c \\
d & b & c & a \\
\end{array}
\right).$ This is a commutative system, with zero divisors and idempotents.
*4D dual numbers. These are formed by adding two dual unities to real numbers: $\varepsilon$ and $\varepsilon_1$, such that $\varepsilon^2=\varepsilon_1^2=\varepsilon\varepsilon_1=0$. They are isomorphic to the real matrices of the following form:
$\left(
\begin{array}{cccc}
a & b & c & d \\
0 & a & 0 & c \\
0 & 0 & a & b \\
0 & 0 & 0 & a \\
\end{array}
\right)$. This system is commutative but has zero divisors and nilpotents.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Rectangle in a circle of radius a that maximizes x^n+ y^n
Consider a rectangle with sides $2x$ and $2y$ inscribed in a given fixed circle $$x^2 + y^2 = a^2,$$ and let $n$ be a positive number. We wish to find the rectangle that maximizes the quantity $$z = x^n + y^n.$$ If $n = 2$, it is clear that $z$ has the constant value $a^2$ for all rectangles.
*
*If $n < 2$, show that the square maximizes $z$; and
*if $n > 2$, show that $z$ is maximized by a degenerate rectangle in which $x$ or $y$ is $0$.
My approach.
$$x^2 + y^2 = a^2$$
$$\therefore\, y = (a^2 - x^2)^{1\over 2}$$
\begin{align}
z &= x^n + y^n \\
{d\over dx}z & = nx^{n-1} - nx(a^2-x^2)^{n-2\over 2}
\end{align}
Equating with $0$ gives:
$$nx^{n-1} = nx(a^2-x^2)^{n-2\over 2}$$
$$x^{n-2} = (a^2 - x^2)^{n-2\over 2}$$
Squaring on both sides gives:
$$(x^2)^{n-2} = (a^2 - x^2)^{n-2}$$
Taking $\ln$ on both sides gives
$$(n-2) \ln(x^2) = (n-2)\ln(a^2 - x^2)$$33
Equating $x^2 = a^2 - x^2$ , we get $x = {a\over \sqrt{2}}$ and $y = {a\over\sqrt{2}}$. In particular $x=y$ implies that the optimal rectange is a square.
My computation appears to be independent of $n$, so I am puzzled about the statement in the question for $n>2$ and $n<2$.
| ${d\over dx}z$ =
$nx^{n-1}$ - $nx(a^2-x^2)^{n-2\over 2}$
Equating with $0$ gives
$x= {a\over \sqrt {2}}$ so, $y = {a\over \sqrt {2}}$
and $x= 0$ so, $y = a$
Now on double differentiating z
${d^2\over dx^2}z$ = $n(n-1)x^{n-2} - n(a^2 - x^2)^{n-2\over 2} + n(n-2)x^2(a^2 - x^2)^{n-4\over 2}$
At $x= {a\over \sqrt {2}}$,
${d^2\over dx^2}z$ simplifies to be $2n(n-2)({a^2\over 2})^{n-2\over 2}$
So, if $n<2$,
${d^2\over dx^2}z$ At $x= {a\over \sqrt {2}}$ is < $0$
So, at $x= {a\over \sqrt {2}}$, Z = maximum when $n<2$ and minimum when $n>2$
When $ n < 2$
At $x =0$ ${d^2\over dx^2}z$ is undefined due to the term $n(n-1)x^{n - 2}$ = $n(n-1)$$({1\over 0})^{2 - n}$
When n>2
At $x = 0$
${d^2\over dx^2}z$ simplifies to be $-n(a^2)^{n-2\over 2}$ which is $< 0$ as $n > 0$
So $z$ is maximum at $x = 0$
This shows that when $n<2$ maximum value of $z$ occurs when $x = y = {a\over \sqrt {2}}$ thus a square
When $n= 2$, $z$ simply equates to $a^2$
When n > 2 $z$ is maximum when $x = 0$ , $y = a$ and since $x$ and $y$ are symmetric so they are interchangeable this means z is also maximum when $x = a$ , $y = 0$
and thus it is a degenerate rectangle which gives maximum value of $z$ when $n>2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to get $\int_0^∞\frac{dx}{(x^2+1)^n}=\frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2}\overset{?}=\frac{\sqrt{\pi}}{2}\frac{Γ(n-\frac{1}{2})}{Γ(n)}$? I recently solved the following integral using a recursive formula and integration by parts.
$$\int_0^\infty\frac{dx}{(x^2+1)^n}=\frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2}$$
Where $(n)!!$ represents the double factorial, not to be confused with $(n!)!$.
But when I plug this same integral into WolframAlpha I get this:
$$\int_0^\infty\frac{dx}{(x^2+1)^n}=\frac{\sqrt{\pi}}{2}\frac{\Gamma(n-\frac{1}{2})}{\Gamma(n)}$$
How do I prove that these two results are equivalent?
| Recursion gives
$$\newcommand{\g}[1]{\Gamma \left( #1 \right)}
\newcommand{\para}[1]{\left( #1 \right)}
\begin{align*}
\g{ n - \frac 1 2 } &= \para{ n -1 - \frac 1 2 }\cdot \g{(n-1) - \frac 1 2}\\
\g{ n - \frac 1 2 } &= \para{ n -1 - \frac 1 2 }\para{ n -2 - \frac 1 2 }\cdot \g{(n-2) - \frac 1 2}\\
&= \vdots \\
&=\para{ n -1 - \frac 1 2 }\para{ n -2 - \frac 1 2 }\cdot \cdots \cdot \para{ \frac 3 2 } \para{ \frac 1 2 }\cdot \g{ \frac 1 2 } \\
&=\para{ \frac{2n-3}{2} }\para{ \frac{2n-5}{2} }\para{ \frac{2n-7}{2} } \cdots\para{ \frac{1}{2} } \sqrt \pi \\
&= \frac{(2n-3)!!}{2^{n-1}} \sqrt \pi
\end{align*}$$
Of course,
$$\g{n} = (n-1)!$$
Then their ratio is
$$\frac{\g{n-1/2}}{\g{n}} = \frac{\sqrt \pi}{2^{n-1}} \frac{(2n-3)!!}{(n-1)!}$$
By a property of the double factorial,
$$2^{n-1}(n-1)! = (2n-2)!!$$
so
$$\frac{\g{n-1/2}}{\g{n}} = \sqrt{\pi} \frac{(2n-3)!!}{(2n-2)!!}$$
The desired result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4383273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Area of shaded region in a square
My approach
The area of shaded region = 8×the area of
Given that the points of the shaded region are closer to the center than the boundary of the square.
Let's talk about the boundary of the shaded region
The boundary of the shaded region therefore must be the locus of all the points whose distance from the center of the square = distance from the boundary.
Let's find the locus of the boundary of the shaded region
From the second figure
${\sqrt {h^2+k^2} } = {\sqrt {(h-h)^2 + (k-{a\over 2})^2 }}$
This simplifies to be:
$k = {a^2 - 4h^2\over 4a}$
$y = {a^2 - 4x^2\over 4a}$
Also the curve intersects the line( the hypotenuse of the triangle) y= x
For point of intersection :
$ {a^2 - 4x^2\over 4a} = x $
$4x^2 +4ax - a^2 = 0$
$x = {-4a ± \sqrt{16a^2 - 4×4×(-a^2)}\over 8}$
$x = a{(\sqrt{2} ± 1)\over 2}$
Solution 1: $x = a{(\sqrt{2} + 1)\over 2}$
Solution 2: $x = a{(\sqrt{2} - 1)\over 2}$
Solution 1 can be discarded as $ x = a{(\sqrt{2} + 1)\over 2} ≈ 1.207106 a > {a\over 2} $
Solution 2: $ a{(\sqrt{2} - 1)\over 2} ≈ 0.2071 a < {a\over 2} $
from 0 to $a{(\sqrt{2} - 1)\over 2}$ :
The curve (boundary of shaded region) lies above the line $ y=x$
so $dA = ({a^2\over 4a} - {4x^2\over 4a} - x)dx $
$dA = ({a\over 4} - {x^2\over a} - x)dx $
$\int_{0}^{A} dA = \int_{0}^{a{(\sqrt{2} - 1)\over 2}} ({a^2\over 4a} - {4x^2\over 4a} - x)dx $
A = $({a\over 4}x - {x^3\over 3a} - {x^2\over 2})]_{0}^{{a{(\sqrt{2} - 1)\over 2}}}$
$ A = {a^2\over 8}(\sqrt{2}-1) - {a^2\over 24}(\sqrt{2}-1)^3 - {a^2\over 8}
(\sqrt{2}-1)^2 $
$ = {a^2\over 8}(\sqrt{2}-1) \Biggl( 1 - {(\sqrt{2}-1)^2\over 3} - (\sqrt{2} -1)\Biggr)$
This simplifies to be equal to
${a^2\over 8}(\sqrt{2}-1)({3+5\sqrt{2}\over 3})$
$= {(7- 2\sqrt{2})\over 8×3}a^2$
Area of shaded figure = 8A
$A_{total}$ = ${(7- 2\sqrt{2})\over 3}a^2$
But the answer is :
${4\sqrt{2}-5\over 3}a^2 $
I don't know where I got it wrong, and also I have re calculated this and the result is same.
Did I miss something important or calculated wrongly
Any help of hint or suggestion or worked out solution would be appreciated.
| The problem was here
A = ${a^2\over 8}(\sqrt{2}-1) \Biggl( 1 - {(\sqrt{2}-1)^2\over 3} - (\sqrt{2} -1)\Biggr)$
This equals to $a^2(\sqrt{2} -1){(3-\sqrt{2})\over 8×3}$
∴ $A_{total}$ = $8A = a^2{(4\sqrt{2}-5)\over 3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4385102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Alternative way to solve a limit problem $$
\lim _{n \rightarrow \infty} \frac{1}{1+n^{2}}+\frac{2}{2+n^{2}}+\cdots+\frac{n}{n+n^{2}}
$$
I want to find the limit of this infinite series which I found in a book. The answer is $1/2$.
The solution to this limit was given by Sandwich/Squeeze Theorem, which was basically that the above function lies between:
$$
\frac{1}{n+n^{2}}+\frac{2}{n+n^{2}}+\frac{3}{n+n^{2}}+\cdots+\frac{n}{n+n^{2}}
$$
And,
$$
\frac{1}{1+n^{2}}+\frac{2}{1+n^{2}}+\cdots+\frac{n}{1+n^{2}}
$$
series and the limit of both of these series tend to $1/2$ as $n \to \infty$.
I fully understood the solution, but I find that this isn't something that naturally/intuitively comes to your mind. I mean we need to find two different series by trial and error, both of which need to converge to a single number.
Is there any different solution to this limit problem, like dividing by powers of n, or maybe telescoping sums?
| $$S_n=\sum_{r=1}^n \frac r{r+n^2}=\sum_{r=1}^n \frac{\frac r{n^2}}{ 1+\frac{r}{n^2}}=\sum_{r=1}^n \frac{1+\frac r{n^2}-1}{ 1+\frac{r}{n^2}}=\sum_{r=1}^n 1-\sum_{r=1}^n \frac{1}{ 1+\frac{r}{n^2}}=n-\sum_{r=1}^n \frac{1}{ 1+\frac{r}{n^2}}$$
$$\sum_{r=1}^n \frac{1}{ 1+\frac{r}{n^2}}=n^2\Bigg[\sum_{r=1}^{n^2+n}\frac 1r -\sum_{r=1}^{n^2}\frac 1r\Bigg]$$ So, using harmonic numbers
$$S_n=n+n^2 \left(H_{n^2}-H_{n^2+n}\right)$$ Using the asymptotics
$$H_p=\log (p)+\gamma +\frac{1}{2 p}-\frac{1}{12 p^2}+\frac{1}{120
p^4}+O\left(\frac{1}{p^{6}}\right)$$ apply it twice and continue with Taylor series to simplify.
$$H_{n^2}-H_{n^2+n}=-\frac{1}{n}+\frac{1}{2 n^2}+\frac{1}{6 n^3}-\frac{1}{4 n^4}+\frac{2}{15 n^5}+O\left(\frac{1}{n^{6}}\right)$$
$$S_n=\frac{1}{2}+\frac{1}{6 n}-\frac{1}{4 n^2}+\frac{2}{15 n^3}+O\left(\frac{1}{n^{4}}\right)$$ which is a quite good approximation. For example,
$$S_{10}=\frac{3039003639041255}{5909102214621606}=0.514292$$ while the truncated series gives exactly $0.5143$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4385416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 2
} |
What are reciprocal vectors ? I did not undertand the definition in this question
Two groups of vectors a, b, c, and a', b', c'
are called of reciprocal if a.a' = b.b' = c.c'
= 1 and whathever other mixed scalar product like a.b' = 0. Show that:
*
*c' = (a x b)/Q
*a' = (b x c)/Q
*b' = (c x a)/Q
Q = a . (b x c)
I understood that:
*
*a $\perp$ b $\perp$ c
and:
*
*a // a'
*b // b'
*c // c'
In this way, in three dimensions, we have that b x c = a, so Q = 1. And it makes no sense, because this way would be a = a', b = b' and c = c'.
How to solve that ?
| One way to do this is to show that
*
*The given formulas for $\mathbf{a}'$, $\mathbf{b}'$, $\mathbf{c}'$ satisfy the definition, and
*Show these reciprocal vectors are unique, i.e. if $\mathbf{a}''$, $\mathbf{b}''$, $\mathbf{c}''$ are an alternate set of reciprocal vectors to $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$, then $\mathbf{a}' = \mathbf{a}''$, $\mathbf{b}' = \mathbf{b}''$, $\mathbf{c}' = \mathbf{c}''$.
The second step is important from a logical standpoint, but I have a feeling that whoever asked you the question was not really expecting you to complete it.
To complete the first step, recall the scalar triple product:
$$[\mathbf{p}, \mathbf{q}, \mathbf{r}] = (\mathbf{p} \times \mathbf{q}) \cdot \mathbf{r}.$$
We will need the following properties:
*
*$\displaystyle{[\mathbf{p}, \mathbf{q}, \mathbf{r}] = [\mathbf{q}, \mathbf{r}, \mathbf{p}] = [\mathbf{r}, \mathbf{p}, \mathbf{q}]}$
*If any two of $\mathbf{p}, \mathbf{q}, \mathbf{r}$ are the same, then $\displaystyle{[\mathbf{p}, \mathbf{q}, \mathbf{r}] = 0}$.
Then,
$$Q = [\mathbf{a}, \mathbf{b}, \mathbf{c}] = [\mathbf{b}, \mathbf{c}, \mathbf{a}] = [\mathbf{c}, \mathbf{a}, \mathbf{b}].$$
This means, for example,
$$\mathbf{c}' \cdot \mathbf{c} = \frac{\mathbf{a} \times \mathbf{b}}{Q} \cdot \mathbf{c} = \frac{[\mathbf{a}, \mathbf{b}, \mathbf{c}]}{Q} = 1,$$
and
$$\mathbf{a}' \cdot \mathbf{b} = \frac{\mathbf{b} \times \mathbf{c}}{Q} \cdot \mathbf{c} = \frac{[\mathbf{b}, \mathbf{c}, \mathbf{c}]}{Q} = 0.$$
The other $4$ calculations follow similarly.
To show uniqueness, we first show that $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ form a basis for $\Bbb{R}^3$. Since there are three such vectors, it suffices to show linear independence.
Suppose $d\mathbf{a} + e\mathbf{b} + f\mathbf{c} = \mathbf{0}$. Taking the inner product of both sides with, say, $\mathbf{a}'$, we get
\begin{align*}
&d(\mathbf{a} \cdot \mathbf{a}') + e(\mathbf{b} \cdot \mathbf{a}') + f(\mathbf{c} \cdot \mathbf{a}') = \mathbf{0} \cdot \mathbf{a}' \\
\implies &d \times 1 + e \times 0 + f \times 0 = 0 \\
&\implies d = 0.
\end{align*}
Taking dot products with $\mathbf{b}$ and $\mathbf{c}$ imply $e = 0$ and $f = 0$ respectively. Thus, $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ are linearly independent, and hence form a basis (in particular, they are spanning). One nice property is that, if a vector $\mathbf{v}$ is perpendicular to each of $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$, then it must be $\mathbf{0}$.
Suppose $\mathbf{a}''$, $\mathbf{b}''$, $\mathbf{c}''$ and $\mathbf{a}'$, $\mathbf{b}'$, $\mathbf{c}'$ are both reciprocal to $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$. Consider, for example, $\mathbf{v} = \mathbf{a}' - \mathbf{a}''$. Note that it is perpendicular to $\mathbf{a}$, as
$$\mathbf{a} \cdot \mathbf{v} = \mathbf{a} \cdot \mathbf{a}' - \mathbf{a} \cdot \mathbf{a}'' = 1 - 1 = 0,$$
and it is perpendicular to $\mathbf{b}$ (or similarly $\mathbf{c}$) as
$$\mathbf{b} \cdot \mathbf{v} = \mathbf{b} \cdot \mathbf{a}' - \mathbf{b} \cdot \mathbf{a}'' = 0 - 0 = 0.$$
Therefore, $\mathbf{a}' - \mathbf{a}'' = 0$, i.e. $\mathbf{a}' = \mathbf{a}''$. Similar arguments show $\mathbf{b}' = \mathbf{b}''$ and $\mathbf{c}' = \mathbf{c}''$ as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4387115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $AT \cdot XY = AX \cdot BY$ In figure below $T,X,Y$ are tangency points and the big circles are tangent to each other. And $TB$ is parallel to that common tangent below.
So, I made more calculations than I should. Let $R$ be the radius of the blue $\Omega$ circle and $r$ be the radius of the green circle.
I managed, through heavy calculation, to come to:
$R = r + \frac{AB^2}{16r}$
I guess this is a good start, there should be an easy way to prove it.
By defining $TA = t, AX = a, XY = 2k, BY = b$
I also came to the following relation:
$(4t^2 + 4t(a+b+2k) + (a+b+2k)^2)\frac1{\sqrt{8t^2 + 8t(a+b+2k) + (a+b+2k)^2}} k = k^2 + (a+k)(b+k)$
which is nice, has the same variables that the question asks, can anyone solve this one?
My calculations:
$M =$ midpoint of $\overline{AB}$
$C = \Omega \cap$ the common tangent, $CA=CB$
$D$ is such that $TMCD$ is a rectangle and $E$ is the tangency point between $\Omega$ and green circle. $TE = x \implies EC = \frac{R}{r} x$ (homothety centered at $E$)
$TC^2 = TD^2 + DC^2 = (2r)^2 + (2\sqrt{Rr})^2 = (x\frac{R+r}r)^2$
so $x = \frac{2r}{R+r} \sqrt{r(r+R)}$ RIGHT?
$x \cdot TC = TA \cdot TB \implies 4r^2 = TA \cdot TB$ RIGHT?
but $TB = 2\sqrt{Rr} + \frac{AB}2$
So now:
$TA = \frac{\sqrt{16^2-AB^2}-AB}2$ oops, I think I see my problem.... I tried to isolate $R$ in $TA \cdot (2\sqrt{Rr} + \frac {AB}2) = 4r^2$ and I changed a signal.... welp if someone wants to continue from here.
| The answer is written before the edit to your question...
Let $R$ be the radius of $\Omega$, $p$ be the distance of the center of $\Omega$ to $TB$. The circle tangent at $X$ and $Y$ have radius $r_1$ and $r_2$, and the centers of whom are $U,V$. The center of $\Omega$ is $W$.
So we have $UW=R-r_1$, the vertical distance of $U$ and $W$ is $r_1+p$, so $$AX=\sqrt{R^2-p^2}-\sqrt{(R-r_1)^2-(p+r4_1)^2}=\sqrt{R+p}(\sqrt{R-p}-\sqrt{R-p-2r_1}).$$ Similarly, we have $$YB=\sqrt{R+p}(\sqrt{R-p}-\sqrt{R-p4-2r_2}).$$ Also, $XY$ is the sum of two horizontal distances, that is, $$\sqrt{(R-r_1)^2-(p+r_1)^2}+\sqrt{(R-r_2)^2-(p+r_2)^2}=\sqrt{R+p}(\sqrt{R-p-2r_1}+\sqrt{R-p-2r_2}).$$ Also we know that this $XY$ also equals to $\sqrt{4r_1r_2}$. So we have the relation
$$\sqrt{R+p}(\sqrt{R-p-2r_1}+\sqrt{R-p-2r_2})=\sqrt{4r_1r_2}$$
Also, the radius of the green circle is $\frac{R+p}{2}$. So the common tangent of the green circle and $\Omega$ is $\sqrt{2R(R+p)}$, and thus $AT=\sqrt{2R(R+p)}-\sqrt{R^2-p^2}=\sqrt{R+p}(\sqrt{2R}-\sqrt{R-p})$. So we have
$$AT\times XY=AX\times BY$$
is equivalent to
$$(\sqrt{2R}-\sqrt{R-p})(\sqrt{R-p-2r_1}+\sqrt{R-p-2r_2})=(\sqrt{R-p}-\sqrt{R-p-2r_1})(\sqrt{R-p}-\sqrt{R-p-2r_2})$$
That is, we have
$$\sqrt{2R}(\sqrt{R-p-2r_1}+\sqrt{R-p-2r_2})=R-p+\sqrt{R-p-2r_1}\sqrt{R-p-2r_2}$$
Let $x=\sqrt{R-p-2r_1}$, $y=\sqrt{R-p-2r_2}$, we want to prove that
$$\sqrt{2R}(x+y)=R-p+xy$$
Since we have the relation
$$\sqrt{R+p}(\sqrt{R-p-2r_1}+\sqrt{R-p-2r_2})=\sqrt{4r_1r_2}$$
That is,
$$\sqrt{R+p}(x+y)=\sqrt{(R-p-x^2)(R-p-y^2)}$$
Square it, we have
$${(R+p)}(x+y)^2=(R-p-x^2)(R-p-y^2)$$
Or,
$${(R+p)}(x+y)^2=(R-p)^2-(R-p)(x+y)^2+2(R-p)(xy)+(xy)^2$$
Move the term, we have
$${2R}(x+y)^2=(R-p)^2+2(R-p)(xy)+(xy)^2$$
Or,
$${2R}(x+y)^2=(R-p+xy)^2$$
take the square root, we have
$$\sqrt{2R}(x+y)=R-p+xy$$
Which yields the desired result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How to show that $\left(\sqrt{\binom{n}{1}}+\sqrt{\binom{n}{2}}+...+\sqrt{\binom{n}{n}}\right)^2≤n(2^n-1)$? I want to show that $\left(\sqrt{\binom{n}{1}}+\sqrt{\binom{n}{2}}+...+\sqrt{\binom{n}{n}}\right)^2≤n(2^n-1)$.
My attempt:
We know that $\sqrt{ab}≤\frac{a+b}{2}$ because $(\sqrt{a}-\sqrt{b})^2≥0$ and that $\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+...+\binom{n}{n}=2^n$ by expanding $(1+x)^n$ and substituting $x=1$.
Next, I re-wrote $\left(\sqrt{\binom{n}{1}}+\sqrt{\binom{n}{2}}+...+\sqrt{\binom{n}{n}}\right)^2$ as $\binom{n}{1}+\binom{n}{2}+...+\binom{n}{1}+\sum_{0<i<j}\sqrt{\binom{n}{i}\binom{n}{j}}$
$≤2^n-1+\sum_{0<i<j}\frac{\binom{n}{i}+\binom{n}{j}}{2}$ using the two term $A_m G_m$ inequality.
$=2^n-1+\frac{1}{2}\left(\binom{n}{1}+\binom{n}{2}+\binom{n}{1}+\binom{n}{3}+...+\binom{n}{1}+\binom{n}{n}+\binom{n}{2}+\binom{n}{3}+\binom{n}{2}+\binom{n}{4}+...+\binom{n}{n-1}+\binom{n}{n}\right)$
$=2^n-1+\frac{n}{2}\left(\binom{n}{1}+\binom{n}{2}+\binom{n}{3}+...+\binom{n}{n}\right)$
$=2^n-1+\frac{n}{2}(2^n-1)$
$=\frac{n+2}{2}(2^n-1)$, which is clearly not the result.
Could someone point out the mistake in my working and help me finish the proof?
| There is an error in your calculation. You should have
$$
\left(\sqrt{\binom{n}{1}}+\sqrt{\binom{n}{2}}+...+\sqrt{\binom{n}{n}}\right)^2 = \sum_{i=1}^n \binom ni + 2 \sum_{0<i<j}\sqrt{\binom{n}{i}\binom{n}{j}} \\
\le
\sum_{i=1}^n \binom ni + \sum_{0<i<j} \left(\binom{n}{i} + \binom{n}{j}\right)\, ,
$$
note the factor $2$ which is missing in your equation. The right-hand side is equal to
$$
\frac 12 \sum_{i=1}^n\sum_{j=1}^n \left(\binom{n}{i} + \binom{n}{j}\right) = n \sum_{i=1}^n \binom{n}{i} = n (2^n-1) \, .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4388240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve \begin{cases} x'=2x+4y+\cos t\\ y'=-x-2y+\sin t \end{cases} Solve
\begin{cases}
x'=2x+4y+\cos t\\
y'=-x-2y+\sin t
\end{cases}
My solution:
First I solve :
\begin{cases}
x'=2x+4y\\
y'=-x-2y
\end{cases}
$$\begin{pmatrix} {x'}\\ {y'}
\end{pmatrix}=\begin{pmatrix} {2} & 4\\ {-1} & -2
\end{pmatrix} \begin{pmatrix} {x}\\ {y}
\end{pmatrix}$$
The eigenvalue is $0$.
The eigevector is \begin{pmatrix} {-2}\\ {1}
\end{pmatrix}
The generalized eigenvector is \begin{pmatrix} {0}\\ -{1\over 2}
\end{pmatrix}
Then, the solution is:
$y=c_1\begin{pmatrix} {0}\\ {-1\over 2}
\end{pmatrix} + c_2 \bigg [t\begin{pmatrix} {-2}\\ {1}
\end{pmatrix}+\begin{pmatrix} {0}\\ {-1\over 2}
\end{pmatrix}\bigg]$
Using variation of parameters method:
$$\begin{pmatrix} {0} & -2t\\ {-1\over 2} & t- {1\over 2}
\end{pmatrix} \begin{pmatrix} {c'_1}\\ {c'_2}
\end{pmatrix}=\begin{pmatrix} {\cos t}\\ {\sin t}
\end{pmatrix}$$
\begin{cases}
-2tc'_2=\cos t\\
-\frac{1}{2} c'_1+ (t- {1\over 2})c'_2=\sin t
\end{cases}
$c'_2=\frac{cos t}{ -2t} \implies c_2=\int \frac{\cos t}{-2t}$
I don't know how to solve it
Where am I wrong?
Thanks !
| The DE-system is given by
\begin{align*}
\underbrace{\frac{{\rm d}}{{\rm d}t}\begin{bmatrix} x(t)\\ y(t)\end{bmatrix}}_{X'(t)}=\underbrace{\begin{bmatrix} 2 & 4\\ -1 & -2\end{bmatrix}}_{A}\underbrace{\begin{bmatrix} x(t)\\ y(t) \end{bmatrix}}_{X(t)} +\underbrace{\begin{bmatrix} \cos t\\ \sin t\end{bmatrix}}_{F(t)}
\end{align*}
*
*Solve $X'(t)=AX(t)$:
Since
$$\det(A-\lambda I)=0 \implies \lambda=0$$
So we have $\lambda=0$ is an eigenvalue with algebraic multiplicity equals $2$, then
$$(A-0\cdot I)v=0 \implies v=\begin{bmatrix} -2\\ 1\end{bmatrix} \longleftarrow \quad \text{eigenvector}$$
and
$$(A-0\cdot I)u=v \implies u=\begin{bmatrix} -1\\ 0 \end{bmatrix} \longleftarrow \quad \text{ generalized eigenvector} $$
Hence
$$X_{1}(t)=\begin{bmatrix} -2\\ 1\end{bmatrix}e^{0\cdot t}=\color{red}{\begin{bmatrix} -2\\ 1\end{bmatrix}},\quad X_{2}(t)=\begin{bmatrix} -2\\ 1\end{bmatrix}te^{0\cdot t}+\begin{bmatrix} -1\\ 0\end{bmatrix}e^{0\cdot t}=\color{red}{\begin{bmatrix} -2\\ 1\end{bmatrix}t+\begin{bmatrix} -1\\ 0\end{bmatrix}}.$$
We get,
$$X(t)=c_{1}X_{1}(t)+c_{2}X_{2}(t)$$
$$\color{green}{X_{c}(t)=c_{1}\begin{bmatrix} -2\\ 1\end{bmatrix}+c_{2}\left(\begin{bmatrix} -2\\ 1\end{bmatrix}t+\begin{bmatrix} -1\\ 0\end{bmatrix} \right)}.$$
*
*Solve $X'(t)=AX(t)+F(t)$:
By variation of paramters,
$$\color{blue}{X_{p}(t)=\Phi(t)\int \Phi^{-1}(t)F(t){\rm d}t},$$
where
$$\Phi(t)=\begin{bmatrix}-2 & -2t-1\\ 1 & t \end{bmatrix},\quad \Phi^{-1}(t)=\begin{bmatrix} \frac{t}{2} & t+1\\ -\frac{1}{2} & -1\end{bmatrix}.$$
Hence,
\begin{align*}
X_{p}(t)&=\begin{bmatrix}-2 & -2t-1\\ 1 & t \end{bmatrix}\int \begin{bmatrix} \frac{t}{2} & t+1\\ -\frac{1}{2} & -1\end{bmatrix}\begin{bmatrix} \cos t\\ \sin t\end{bmatrix}{\rm d}t,\\
&=\begin{bmatrix}-2 & -2t-1\\ 1 & t \end{bmatrix}\int \begin{bmatrix}\frac{t}{2}\cos t+(t+1)\sin t\\ -\frac{1}{2}\sin t-\sin t \end{bmatrix} {\rm d}t\\
&=\begin{bmatrix}-2 & -2t-1\\ 1 & t \end{bmatrix}\begin{bmatrix}\left(\frac{2+t}{2}\right)\sin t-(1+2t)\cos t\\ \frac{3}{2}\cos t \end{bmatrix}\\
&=\begin{bmatrix}-2\left[\left(\frac{2+t}{2}\right)\sin t-(1+2t)\cos t \right] +(-2t-1)\left[\frac{3}{2}\cos t \right]\\ 1\left[\left(\frac{2+t}{2}\right)\sin t-(1+2t)\cos t \right]+t\left[\frac{3}{2}\cos t \right]\end{bmatrix}
\end{align*}
Hence,
$$\color{green}{X_{p}(t)=\begin{bmatrix}-2\left[\left(\frac{2+t}{2}\right)\sin t-(1+2t)\cos t \right] +(-2t-1)\left[\frac{3}{2}\cos t \right]\\ 1\left[\left(\frac{2+t}{2}\right)\sin t-(1+2t)\cos t \right]+t\left[\frac{3}{2}\cos t \right]\end{bmatrix}}$$
*
*Solution
$$\color{green}{X(t)=X_{c}(t)+X_{p}(t),\quad -\infty<t<+\infty}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4389523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Minimizing problem. The real numbers $x_1,$ $x_2,$ $\dots,$ $x_{10}$ satisfy
$$\sqrt{x_1 - 1^2} + 2 \sqrt{x_2 - 2^2} + 3 \sqrt{x_3 - 3^2} + \dots + 10 \sqrt{x_{10} - 10^2} = \frac{x_1 + x_2 + \dots + x_{10}}{2}.$$Find the minimum value of $x_1 + x_2 + x_3 + \dots + x_{10}.$
I tried using Cauchy to get $$\left(1^2(x_1-1^2)+2^2(x_2-2^2)+\cdots+10^2(x_{10}-10^2)\right)(1^2+\cdots+1^2)\geq \left(\sqrt{x_1 - 1^2} + 2 \sqrt{x_2 - 2^2} + 3 \sqrt{x_3 - 3^2} + \dots + 10 \sqrt{x_{10} - 10^2}\right)^2,$$ but this doesn't really do much. I'm not really sure what else to do... It would be great if anyone could help!
Thanks in advance!!
| \begin{align}
& \text{Since $\sqrt{x_i-i^2}$ is real number, $x_i \geq i^2. $} \\
\text{let } & \sum_{x=1}^{10} x_i = S. \\
\therefore \; & S \geq \sum_{i=1}^{10} i^2 = 385. \\
& \Bigg( \sum_{i=1}^{10} x_i-i^2\Bigg)\Bigg(\sum_{i=1}^{10} i^2 \Bigg) \geq \Bigg( \sum_{i=1}^{10} i\sqrt{x_i-i^2} \Bigg). & \text{(C.S. Inequality.)} \\
\therefore \; & (S-385)(385) \geq \Bigg( \dfrac {S}{2} \Bigg)^2 \\
\Rightarrow \; & 385S-385^2 \geq \dfrac {S^2}{4}. \\
\therefore \; & S^2-4\cdot385\cdot S + 4 \cdot385^2 \leq 0. \\
\ \\
\therefore \; & (S-770)^2 \leq 0, S = 770.
\end{align}
Is it just me or the answer is this...? I thought I could answer this...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4393840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimum swaps necessary for every person to meet every other person at every location The setting:
*
*There are 9 people.
*There are 3 locations.
*At any time, there are 3 people at each location.
*People can only swap locations at the same time.
The goal:
*
*Every person visits each location once.
*Every person meets each person at least once.
*At each swap, one person stays at the location.
Example:
The numbers in the table below represent which person is at which location. e.g. in iteration 1, persons 1, 2, and 3 are at location 1. All goals are met in this example, but can it be done in fewer iterations?
Iteration
Location 1
Location 2
Location 3
1
1, 2, 3
4, 5, 6
7, 8, 9
2
4, 2, 9
1, 5, 7
3, 8, 6
3
4, 1, 8
2, 6, 7
3, 9, 5
4
4, 3, 7
2, 5, 8
1, 9, 6
5
5, 6, 7
3, 8, 9
1, 2, 4
| No, it is not possible to do it in fewer iterations. There are $\binom 92=36$ pairs of people, and each round allows at most $9$ of the pairs to meet, so you need at least four rounds. However, we can show four rounds are impossible.
Without loss of generality, the first round is $123\mid 456\mid 789$. Every location needs someone to stick around for the second round, WLOG the people who remain are $1,4,$ and $7$. Since $1$ needs to visit every location, $1$ needs to visit locations $2$ and $3$ during rounds $3$ and $4$. WLOG, $1$ visits location $2$ first.
So far, we have the following partial schedule:
Iteration
Location 1
Location 2
Location 3
1
1, 2, 3
4, 5, 6
7, 8, 9
2
1,
4,
7,
3
1,
4
1,
At some point, $1$ and $4$ need to meet. They cannot meet in round $3$, since then $4$ would only have time to visit two locations. So they must meet in round $4$. This means that the group that meets at round $4$ and location $3$ is $\{1,4,x\}$ for some $x$.
Since every round causes at most $9$ new pairs of people to meet, and there are $36$ pairs of people who need to meet, there cannot be any repetitions of pairs of people meeting, else the schedule would not be efficient enough for all pairs to meet. Since each pair of people can only meet once, $x$ cannot be $2$ or $3$ (since $1$ met $2$ and $3$ in round $1$), and $x$ cannot be $5$ or $6$ (since $4$ met $5$ and $6$ in round $1$), so $x$ must be $7$, $8$ or $9$. However, if $x$ was $7$, then $7$ would visit at most $2$ locations, so $x$ must be $8$ or $9$. WLOG, $x=8$. Let us update our partial schedule:
Iteration
Location 1
Location 2
Location 3
1
1, 2, 3
4, 5, 6
7, 8, 9
2
1,
4,
7,
3
1,
4
1, 4, 8
At last, we get our contradiction. Consider the location of $8$ in round $2$.
*
*If $8$ is in location $1$ in round $2$, then $1$ and $8$ would meet twice.
*If $8$ is in location $2$ in round $2$, then $4$ and $8$ would meet twice.
*If $8$ is in location $3$ in round $2$, then $7$ and $8$ would meet twice.
There is no possible place for $8$ in round $2$, so no four-round schedule is possible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4394545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Calculating the Area when inverse function is given For all the real number, $f\left( x \right)$ is an increasing function that is differentiable while satisfying the following conditions?
(A) $f\left( 1 \right) = 1,\int\limits_1^2 {f\left( x \right)dx} = \frac{5}{4}$
(B) Say g(x) is an inverse function of then, for all real number of x as $x \ge 1$, $g\left( {2x} \right) = 2f\left( x \right)$
If $\int\limits_1^8 {xf'\left( x \right)dx} = \frac{p}{q}$, then what is $p+q=$________?
My solution is as follow
It is given that $x \ge 1$ and $f\left( 1 \right) = 1,\int\limits_1^2 {f\left( x \right)dx} = \frac{5}{4}$
${A_2} = 2 \times 2 - \left( {1 \times 1 + \frac{5}{4}} \right) = 4 - \frac{9}{4} = \frac{7}{4}$
$g\left( 1 \right) = g\left( {f\left( 1 \right)} \right) = 1$
$g\left( 2 \right) = 2f\left( 1 \right) = 2$
$g\left( 4 \right) = 2f\left( 2 \right)$
Also $g\left( {2x} \right) = 2f\left( x \right)$ , we need to find the value of $\int\limits_1^8 {xf'\left( x \right)dx} = \frac{p}{q}$
$T = \int\limits_1^8 {xf'\left( x \right)dx} = \left[ {xf\left( x \right)} \right]_1^8 - \int\limits_1^8 {\left( {\left( {\frac{d}{{dx}}x} \right)\int {f'\left( x \right)dx} } \right)dx} $
$T = \int\limits_1^8 {xf'\left( x \right)dx} = \left[ {xf\left( x \right)} \right]_1^8 - \int\limits_1^8 {\left( {\left( {\frac{d}{{dx}}x} \right)\int {f'\left( x \right)dx} } \right)dx} $
$T = \left[ {8f\left( 8 \right) - f\left( 1 \right)} \right] - \int\limits_1^8 {f\left( x \right)dx} \Rightarrow T = 8f\left( 8 \right) - 1 - \int\limits_1^8 {f\left( x \right)dx} $
I have elaborated my steps, how do I proceed
| Here is a really big hint.
First, show using induction that $f(2^{n-1})=2^{n-1}$ for all $n\geq 1$.
Next, for each $n\geq 1$, set $$a_n=\int_{2^{n-1}}^{2^n}f(x)\mathrm{d}x$$ Enforce the substitution $x \mapsto \frac{f(x)}{2}$ on the integral for $a_{n+1}$ then follow up with integration by parts to get a nice recurrence relation: $$\begin{eqnarray*}a_{n+1}&=&\int_{2^n}^{2^{n+1}}f(x)\mathrm{d}x \\ &=& \int _{2^{n-1}}^{2^n}2x\cdot \frac{2\mathrm{d}x}{f'\left(f^{-1}(2x)\right)} \\ &=& \int_{2^{n-1}}^{2^n}2x\frac{\mathrm{d}}{\mathrm{d}x}\left(f^{-1}(2x)\right)\mathrm{d}x \\ &=& \int_{2^{n-1}}^{2^n}2x\frac{\mathrm{d}}{\mathrm{d}x}\left(2f(x)\right)\mathrm{d}x \\ &=& \int_{2^{n-1}}^{2^n}4xf'(x)\mathrm{d}x \\ &=& 4xf(x)\Bigg|_{x=2^{n-1}}^{x=2^n} - 4\int_{2^{n-1}}^{2^n}f(x)\mathrm{d}x \\ &=& 4^{n+1}-4^{n}-4a_n\end{eqnarray*}$$ Express $\int_1^8$ as $\int_1^2+\int_2^4+\int_4^8$ then use the fact that $a_1=5/4$ to get your answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4395179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof of Pythagorean quadruple Given $x^2 + y^2 + z^2 = m^2$ with integers $x, y, z$ and $m > 0$, z odd and $(x,y,z)=1$.
Set $x_{1}=\frac{1}{2}x$ and $y_{1}=\frac{1}{2}y$.
Then $x_{1}^2+y_{1}^2=\frac{m+z}{2}\frac{m-z}{2}$.
Set $f=(x_{1},y_{1})$, $f_{1}=(f,\frac{1}{2}(m+z))$, $f_{2}=(f,\frac{1}{2}(m-z))$.
The prof says, that with a simple argument you can see, that $(f_{1},f_{2})=1$.
Can you help me with this argument? I tried to understand it, but I failed.
Thanks for your help!
| Suppose, for a contradiction, that $(f_{1}, f_{2}) = d > 1.$ Since $f_{1} = (f, \frac{1}{2}(m+z))$, this implies that $d$ divides both $f$ and $\frac{1}{2}(m+z).$ Similarly, by the definition of $f_{2},$ it follows that $d$ divides $\frac{1}{2}(m-z).$
Since $d$ divides $\frac{1}{2}(m+z)$ and $\frac{1}{2}(m-z),$ $d$ divides $z = \frac{1}{2}(m+z) - \frac{1}{2}(m-z).$
Then, since $d$ divides $f = (x_{1}, y_{1}),$ it follows that $d$ divides $x_{1} = \frac{1}{2}x.$ So, $d$ must also divide $x.$ By the same argument, $d$ divides $y.$
We see that $d > 1$ divides $x, y$ and $z,$ contradicting the assumption that $(x, y, z) = 1.$
So, our initial assumption was wrong, and $(f_{1}, f_{2})$ must be $1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4397874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the area of a triangle $\triangle MNT$ I've found this geometry problem in a math competition:
Let $\ \triangle ABC$ be a triangle, with $AB=52$, $BC=56$, $AC=60$, $\ \Gamma$ be the circumscribed circle of $\ \triangle ABC$, $\ M$ and $N$ be the midpoints of $AB$ and $AC$, respectively. The point $T$ lies on the minor arc $\overset{\frown}{BC}$, so that the circumscribed circle of the triangle $\ \triangle MNT$ is internally tangent to $\Gamma$. What's the area of $MNT$?
The answer to this problem is $525$.
I have drawn this figure of the situation described in the problem.
Applying the intercept theorem, I know that $MN=\dfrac{BC}{2}=28$.
Applying the law of cosines, I've also found the cosine of the three angles of the triangle $ABC$, $\cos\hat{A}=\dfrac{33}{65}$, $\cos \hat{B}=\dfrac{5}{13}$, $\cos \hat{C}=\dfrac{3}{5}$.
Now, how can I use the condition of tangency in order to resolve the problem?
|
As $a = 56, b = 60, c = 52$, $s = 84$
Using Heron's formula, $\Delta = 1344$
Circumradius $ \displaystyle R = \frac{abc}{4 \Delta} = \frac{65}{2}$
If altitude from $A$ to $BC$ is $h$, $1344 = \frac 12 \cdot a \cdot h$
$~~~~\implies h = 48$
If the perp from $A$ to $BC$ meets $BC$ at $H$ and midpoint of $BC$ is $G$,
$BH^2 = c^2 - h^2 \implies BH = 20$, $GH = 8$
That shows that the horizontal distance between midpoints of $MN$ and $BC$ or in other words, $OI = 4$
$ \displaystyle OG^2 + (a/2)^2 = R^2 \implies OG = \frac{39}{2}$. So, $IJ = \displaystyle 24 - \frac{39}{2} = \frac 92 $
If $r$ is circumradius of $\triangle MNT$ and $OO' = x$,
$ \displaystyle O'M = r = \frac{65}{2} - x$
$ \displaystyle \left(\frac{65}{2} - x\right)^2 = 14^2 + \left(\frac 92 + \sqrt{x^2 - 16}\right)^2$
Solve for $x$,
Then altitude from $T$ to $MN$ is,
$\displaystyle h_1 = \frac{9}{2} + \frac{\sqrt{x^2-16}}{x} \cdot \frac {65}{2}$
Finally, $\Delta_{MNT} = \frac 12 \cdot 28 \cdot h_1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4399706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that for $a,b,c \in \mathbb{R}^+$ with $abc = 1$, $\frac{1}{a^3(b + c)} + \frac{1}{b^3(a + c)} + \frac{1}{c^3(a + b)} \ge \frac{3}{2}$ I would like confirmation that I did this proof correctly. If I did, it would be a milestone in my mathematical journey as it would be my first IMO problem.
By Cauchy, we have that $$\left( \frac{1}{a^3(b + c)} + \frac{1}{b^3(a + c)} + \frac{1}{c^3(a + b)} \right) ( (b + c) + (a + c) + (a + b) ) \ge \left( \frac{1}{a^{3/2}} + \frac{1}{b^{3/2}} + \frac{1}{c^{3/2}} \right)^2.$$
By AM-GM, we have that $$\frac{1}{a^{3/2}} + \frac{1}{b^{3/2}} + \frac{1}{c^{3/2}} \ge 3\sqrt[3]{\frac{1}{a^{3/2}b^{3/2}c^{3/2}}} = 3\sqrt[3]{\frac{1}{(abc)^{3/2}}} = 3.$$
Then, again by AM-GM, we have that $a + b + c \ge 3\sqrt[3]{abc} = 3$, and so $(b + c) + (a + c) + (a + b) = 2(a + b + c) \ge 2\cdot3 = 6.$
Thus, we have that $$\left( \frac{1}{a^3(b + c)} + \frac{1}{b^3(a + c)} + \frac{1}{c^3(a + b)} \right) ( (b + c) + (a + c) + (a + b) ) \ge \left( \frac{1}{a^3(b + c)} + \frac{1}{b^3(a + c)} + \frac{1}{c^3(a + b)} \right)6 \ge 3^2 = 9,$$ and so $$\frac{1}{a^3(b + c)} + \frac{1}{b^3(a + c)} + \frac{1}{c^3(a + b)} \ge \frac 96 = \frac 32,$$ and we are done.
| Your solution is incorrect (see my comment.)
One solution is indeed by Cauchy Schwartz:
$$\left( \frac{1}{a^3(b + c)} + \frac{1}{b^3(a + c)} + \frac{1}{c^3(a + b)} \right) ( a(b + c) + b(a + c) + c(a + b) )\ge(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2=\frac{(ab+bc+ca)^2}{(abc)^2}=(ab+bc+ca)^2$$
Thus, we have
$$\left( \frac{1}{a^3(b + c)} + \frac{1}{b^3(a + c)} + \frac{1}{c^3(a + b)} \right)\ge \frac{(ab+bc+ca)^2}{ ( a(b + c) + b(a + c) + c(a + b) )}=\frac{(ab+bc+ca)^2}{2(ab+bc+ca)}=\frac{ab+bc+ca}{2}$$
Since $abc=1$ so $ab+bc+ca\ge3\sqrt[3]{ab\times bc\times ca}=3\sqrt[3]{(abc)^2}=3$, and this can yield the result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find all the integer pairs $(x,y)$ satisfying the equation $x^{17}+6 x^{14}+2 x^{5}=y !+2 $ We are going to solve the equation
$$x^{17}+6 x^{14}+2 x^{5}=y !+2 \tag*{(*)}$$
by the Fermat Little Theorem:
$$x^3\equiv x \quad \pmod 3 \quad \forall x\in Z.$$
Assume that $y \geqslant 3$.
Applying the Fermat Little Theorem to $(*)$ in modulo $3$ yields
$$x^{5} \equiv x^3\cdot x ^2\equiv x^3 \equiv x$$
$$x^{17} \equiv\left(x^{5}\right)^{3} \cdot x^2 \equiv x^{3}\cdot x^2\equiv x\\y !=0 $$
Expressing the equation $(*)$ in modulo $3$ yields
$$ x+0+ 2x\equiv 2 \quad(\bmod 3) \\ 0\equiv 2 \quad(\bmod 3) $$
which is a contradiction.
$$\boxed{ \textrm{There is no such integer pairs for }y\geqslant 3.}$$
Now let’s check for the integer pairs
$(x,y)$ for $y=0,1,2$.
When $y=2$,
$$0 \equiv x+0+2x\equiv 2!+2 \equiv 1 \pmod 3, $$
which is a contradiction.
$\therefore$ there is no such integer solution.
When $y=0,1$, $(*)$ becomes
$$f(x)=x^{17}+6 x^{14}+2 x^{5}-3=0$$
because $f(-1)=0$ but $f(1) \neq 0$ and $f(\pm 3) \neq 0$
$\therefore x=-1$ is the ONLY integer solution.
Now we can conclude that the integer pairs $(x, y)$ satisfying $(*)$ are only $(-1,0)$ and $(-1,1)$.
My Question
Is there simpler solutions?
| You could also note that $y!+2 \pmod 8 \in \{2,4\}$ for any $y \ge 2$, which is even, so $x$ must be even. But then the RHS would be a multiple of $x^5$ and thus as $2|x^5$, a multiple of $32$. So then the RHS $x^{17}+6x^{14}+2^{x^5}$ must be $0 \pmod{8}$ $\not \in \{2,4\}$....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4404447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Triangle-geometry problem Here is the question:
$\cos(A-B) = \frac{7}{8}$, $\cos(C) = ?$
By the Law of Cosines, I get:
$AB^2 = 41-40\cos(C)$
I also tried to expand $\cos(A-B)$ by the compound angle formula, getting:
*
*$\cos(A)\cos(B) + \sin(A)\sin(B)$
Which by the Law of Sines becomes:
*
*$\cos(A)\cos(B) + \frac{5}{4} \sin(B)^2 = \frac{7}{8}$
That's where I have been able to get so far. One thing though that has been bothering me is whether $AB =3$. I am tempted to go down that way because of the Pythagorean triple $3^2 + 4^2 = 5^2$. However, they have not specified that $\angle{A} = \frac{\pi}{2}$, so I am worried about wrongly assuming it. Any assistance would be greatly appreciated.
|
Draw $\angle BAD = \angle B$. Then $\angle CAD = \angle A - \angle B$. Say $\angle A - \angle B = \alpha$.
If $BD = x, AD = x, CD = 5 - x$.
Applying law of cosines in $\triangle CAD$,
$(5-x)^2 = x^2 + 4^2 - 2 \cdot 4 \cdot x \cdot \cos \alpha$
$25 + x^2 - 10 x = 16 + x^2 - 7x \implies x = 3$
We now know the sides of $\triangle CAD$. Applying law of cosines again, we find $ ~ \cos \angle C = \dfrac{11}{16}$.
| {
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"url": "https://math.stackexchange.com/questions/4408222",
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"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
} |
Finding the equation of an ellipse given the foci I just wanted someone to check my solutions for this problem:
Find the equation of the ellipse with Foci (2,3) and (-1,1) where the distances from any point on the ellipse to the focus sums to 10. Write your answer in the form $Ax^2+Bxy+Cy^2=D$. Sketch the graph of the ellipse.
Solution:
$d_1$ = distance between point $P(x,y)$ and $(-1,1)$
$d_2$ = distance between point $P(x,y)$ and $(2,3)$
$d_1 = \sqrt{(x+1)^2+(y-1)^2}$
$d_2 = \sqrt{(x-2)^2+(y-3)^2}$
$d_1+d_2=10$
$d_1=10-d_2$
$\sqrt{(x+1)^2+(y-1)^2} = 10 - \sqrt{(x-2)^2+(y-3)^2}$
$(x+1)^2+(y-1)^2=100-20\sqrt{(x-2)^2+(y-3)^2}+(x-2)^2+(y-3)^2$
$x^2+2x+1+y^2-2y+1=x^2-4x+4+y^2-6y+9+100-20\sqrt{(x-2)^2+(y-3)^2}$
$6x+4y-11=100\sqrt{(x-2)^2+(y-3)^2}$
Squaring both sides:
$36x^2+16y^2+48xy-132x-88y+121=100^2[(x-2)^2+(y-3)^2]$
Is this the correct answer and is it in the correct form? Or would I have to continue expanding terms out and simplifying?
Also, how would I then end up graphing it? I know that the Foci are (2,3) and (-1,1). I also know that $d_1+d_2=10$ so $2a=10$ so $a=5$.
| HINT
Suppose you are given an ellipse as follows:
\begin{align*}
\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1
\end{align*}
Then we can rotate and translate it in order to obtain the general equation of an ellipse in $\mathbb{R}^{2}$.
Let us start with rotating it first:
\begin{align*}
\frac{(\cos(\alpha)x - \sin(\alpha)y)^{2}}{a^{2}} + \frac{(\sin(\alpha)x + \cos(\alpha)y)^{2}}{b^{2}} = 1
\end{align*}
Now we can translate it with respect to the point $(x_{0},y_{0})$:
\begin{align*}
\frac{(\cos(\alpha)(x - x_{0}) - \sin(\alpha)(y - y_{0}))^{2}}{a^{2}} + \frac{(\sin(\alpha)(x - x_{0}) + \cos(\alpha)(y - y_{0}))^{2}}{b^{2}} = 1
\end{align*}
At the given example, $(x_{0},y_{0}) = (0.5,2)$ and $\tan(\alpha) = 2/3$.
Can you determine $a$ and $b$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4408821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Finding roots of the equation $8x^3+4x^2-4x-1=0$ which are in form of cosine of angles. So, I start this by reducing it to a cubic with 0 coefficient of the 2nd degree term which gives me :
$$\left(x + \frac{1}{6}\right)^3 - \frac{7}{12}\left(x + \frac{1}{6}\right)-\frac{7}{216}=0$$
Replacing $\left(x + \frac{1}{6}\right)$ by $y$:
$$y^3-\frac{7}{12}y -\frac{7}{216}=0$$
Then replacing $y$ by $r \cos(\alpha)$ and comparing the equation with :
$$\cos^3 (\alpha) - \frac{3}{4}\cos(\alpha)-\frac{1}{4}\cos(3\alpha)=0$$
I get:
$$r = \sqrt{\frac{7}{3}}, \alpha = \frac{1}{3}\cos^{-1}\left(\frac{1}{2\sqrt{7}}\right)$$
which finally give:
$$\theta = \cos^{-1}\left(\frac{\sqrt{7}}{9}\cos^{-1} \left(\frac{1}{2\sqrt{7}}\right)-\frac{1}{6}\right)$$
which is approximately 1.329 on putting it in the calculator.
But if I consider the common problem:
$8x^3-4x^2-4x+1=0$ having roots $\cos(\pi/7), \cos(3\pi/7), \cos(5\pi/7)$.
If I replace $x$ by $-x$ in this equation, we get:
$$8x^3+4x^2-4x-1=0$$
which is same the expression in the actual problem. So its roots must be $-\cos(\pi/7), -\cos(3\pi/7), -\cos(5\pi/7)$. This clearly doesn't matches with the answer I got.
Where I go wrong? Also can someone confirm while comparing coefficients in $cos(\alpha)$ variable cubic equation, we treat $\cos(3\alpha)$ as a constant. My book give a special section in which it shows this method of comparing in this way. So I guess its probably right? But I am not sure how is it right, isn't $\cos(3\alpha)$ also dependent on the main variable $\cos(\alpha)$.
| I am deriving answer to this question in some other way
We know $$ 8\cos\frac{\pi}{7}\cdot\cos\frac{2\pi}{7}\cdot\cos\frac{4\pi}{7} = -1$$
By applying transformation formula
$$ 4\cos\frac{\pi}{7}\cdot2(\cos\frac{2\pi}{7}\cdot\cos\frac{4\pi}{7}) = -1$$
$$ 4\cos\frac{\pi}{7}\cdot(\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}) = -1$$
$$ 4\cos\frac{\pi}{7}\cdot(-\cos\frac{\pi}{7}+\cos\frac{2\pi}{7}) = -1$$
$$ 4\cos\frac{\pi}{7}\cdot(-\cos\frac{\pi}{7}+2\cos^2\frac{\pi}{7}-1) = -1$$
Putting $ \cos\frac{\pi}{7} = x $
We will get the equation $$8x^3-4x^2-4x+1 =0 $$
Similarly other angles like $\cos\frac{2\pi}{7}$ and $\cos\frac{3\pi}{7}$ may be derived in terms of cubic equations
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4409527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Computing $I=\iint_D\sqrt{\frac{1-x^2-y^2}{1+x^2+y^2}}dxdy$
Compute $$I=\iint_D\sqrt{\frac{1-x^2-y^2}{1+x^2+y^2}}dxdy,$$ where $D=\{(x,y)\in\Bbb R^2\mid x^2+y^2\le 1, x\ge 0, y\ge 0\}.$
Source: Berman, task 3537
My attempt:
I switched to polar coordinates $$\begin{cases}x=r\cos\varphi\\y=r\sin\varphi\end{cases}, 0\le\varphi\le \pi/2,0\le r\le 1$$
$$\begin{aligned}I&=\int_0^{\pi/2}\int_0^1\sqrt{\frac{1-r^2}{1+r^2}}rdrd\varphi\\&=\frac\pi2\int_0^1\sqrt{\frac{1-r^2}{1+r^2}}rdr\\&=\begin{bmatrix}t&=r^2\\dt&=2rdr\end{bmatrix}\\&=\frac\pi4\int_0^1\sqrt{\frac{1-t}{1+t}}dt\\&=\begin{bmatrix}s=\sqrt{\frac{1-t}{1+t}}\implies(1+t)s^2=1-t\implies s^2+ts^2=1-t\implies(1+s^2)t=1-s^2\implies t=\frac{1-s^2}{1+s^2}\\\implies dt=-\frac{4sds}{(1+s^2)^2}\end{bmatrix}\\&=-\frac\pi4\int_1^0\frac{4s}{(1+s^2)^2}ds\\&=\frac\pi2\int_0^1\frac{2sds}{{(1+s^2)}^2}\\&=\begin{bmatrix}v=1+s^2\\dv=2sds\end{bmatrix}\\&=\frac\pi2\int_1^2\frac{dv}{v^2}\\&=-\frac\pi2\frac1v\Big|_1^2\\&=\frac\pi4\end{aligned}$$
The solution in the book is $\frac{\pi(\pi-2)}8.$ What did I do wrong?
| Under the change of variables
$$t \mapsto \frac{1 - s^2}{1 + s^2} \quad \left(\textrm{that is, } s = \sqrt{\frac{1 - t}{1 + t}}\right), \qquad dt = -\frac{4 s\,ds}{(1 + s^2)^2},$$
the integral transforms as
$$\int_0^1 \sqrt{\frac{1 - t}{1 + t}} \,dt = \int_1^0 s \cdot -\frac{4 s\,ds}{(1 + s^2)^2} = 4 \int_0^1 \frac{s^2 \,ds}{(1 + s^2)^2}.$$
In your solution, the numerator in the integral in $s$ only contains one factor of $s$. The correctly transformed integral does lead to the text's solution, $\frac{(\pi - 2) \pi}{8}$.
Remark The correctly transformed integral in $s$ can be evaluated efficiently via a standard trigonometric substitution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Examples where $(a+b+\cdots)^2 = (a^2+b^2+\cdots)$ Consider the two infinite series
$$
\frac{\pi}{\sqrt{8}} = 1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \cdots
$$
and
$$
\frac{\pi^2}{8} = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{9^2} + \frac{1}{11^2} + \cdots
$$
(Notice that the first series has signs that go two-by-two rather than every-other.)
Squaring the first equality also gives $\pi^2/8$ and so these two, when put together, satisfy the 'highschooler's dream' for squaring a sum: just square each term and sum,
$$
(a + b + c + \cdots)^2 = (a^2 + b^2 + c^2 + \cdots)
$$
with nothing like $2ab + 2ac + 2bc + \cdots$ needed.
A trivial example of this would be
$$
(a + 0)^2 = a^2 + 2a0 + 0^2 = a^2 + 0^2
$$
but it only succeeds because one addend is zero.
My questions are
*
*Are there any other simple nontrivial examples? I believe any other nontrivial example must be an infinite sum. edit: John Omielan provides the simple finite example $(1+1-\frac{1}{2})^2 = 1^2 + 1^2 + \frac{1}{2^2}$.
*Is there an "obvious" demonstration that the above sum (other than the direct evaluation) satisfies the highschooler's dream? Put another way, is there a simple demonstration that the infinite sum of "cross terms" vanishes?
| My answer is the same as Ivan's. However I give an alternative proof of his statement, which is too long for a comment.
You can take almost arbitrary first $k-1$ numbers $a_1$, $a_2$, ... ,
$a_{k-1}$ and always exists $a_k$ such that
$$(a_1+a_2+...+a_{k-1}+a_k)^2={a_1}^2+{a_2}^2+...+{a_{k-1}}^2+{a_k}^2$$
My Proof:
We have:
$$ (a_1+a_2+...+a_{k-1})^2=a_1^2+a_2^2+...+{a_{k-1}}^2 \iff \sum_{1\ \leq\ i\ <\ j\ \leq\ k-1} a_i a_j = 0. $$
Now let
$$a_k = \frac{ - \left(\displaystyle\sum_{1\ \leq\ i\ <\ j\ \leq\ k-1} a_i a_j\right) } { \displaystyle\sum_{i=1}^{k-1} a_i}.$$
This implies:
$$ \sum_{1\ \leq\ i\ <\ j\ \leq\ k} a_i a_j = \sum_{1\ \leq\ i\ <\ j\ \leq\ k-1} a_i a_j + a_k \sum_{i=1}^{k-1} a_i = 0, $$
with the definition of $a_k$ being possible if and only if $\displaystyle\sum_{i=1}^{k} a_i \neq 0.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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Investigate whether the sequence $\frac{nx^n + 3\sin{2n\pi x}}{n}$ is uniformly convergent Let $(f_n)$ be a sequence of functions defined by
\begin{equation*}
f_n(x) = \frac{nx^n + 3\sin{(2n\pi x)}}{n}
\end{equation*}
for all $x \in [0,1]$ and $n \in \Bbb N$.
Show that $f_n$ is converges pointwise to a function $f:[0,1]\to \Bbb R$ on $[0,1]$.
Investigate whether $f_n$ is uniformly convergent to $f$ on $[0,1]$ or not.
Attempt:
*
*For $x=0$: We have $f_n(0)=0 \to 0$.
*For $x=1$: We have $f_n(1)=1 \to 1$.
*For $x \in (0,1)$: I claimed that $f_n \to 0$. The proof is goes as follows:
Let $\varepsilon > 0$ and $x \in (0,1)$ be arbitrary. Choose $k \in \Bbb n$ such that
$k > \max\left\{\frac{2x}{\varepsilon - \varepsilon x}, \frac{6}{\varepsilon} \right\}$. Then, for any $n \in \Bbb n$ with $n \ge k$, we have
\begin{align*}
\left|f_n(x)-f(x) \right| &= \left|\frac{nx^n + 3\sin{(2n\pi x)}}{n}-0 \right| \\
&\le x^n + \frac{3}{n} \\
&= \frac{1}{\left(1+\left(\frac{1}{x}-1\right)\right)^n} + \frac{3}{n} \\
&\le \frac{1}{1+n\left(\frac{1}{x} -1 \right)} + \frac{3}{n} \\
&< \frac{1}{n\left(\frac{1}{x} -1 \right)} + \frac{3}{n} \\
&\le \frac{1}{k\left(\frac{1}{x} -1 \right)} + \frac{3}{k} \\
&< \frac{\frac{\varepsilon}{2}\left(\frac{1}{x} -1 \right)}{\frac{1}{x}-1} + \frac{\varepsilon}{2} \\
&= \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.
\end{align*}
Hence, $f_n \to f$ where
\begin{equation*}
f(x) =
\begin{cases}
0, \qquad \text{if} \ 0 \le x < 1 \\
1, \qquad \text{if} \ x = 1
\end{cases}.
\end{equation*}
Now, my intuition is said that $f_n$ is not uniformly convergent to $f$ on $[0,1]$. But, I didn't be able yet to find the subsequence such that the sequence did not uniformly convergent. On the other hand, I tried using the uniform norm
\begin{align*}
||f_n - f||_{[0,1]} =
\sup \left\{\begin{array}{lr}
\frac{nx^n + 3\sin{(2n\pi x)}}{n}, & \text{for} \ 0 \le x < 1\\
0, & \text{for} \ x=1
\end{array}\right\} = \frac{13}{4} \ne 0.
\end{align*}
But, is this correct? Any ideas? Thanks in advanced.
| If $x\in[0,1)$, then$$\lim_{n\to\infty}f_n(x)=\lim_{n\to\infty}x^n+\frac{3\sin(2\pi x)}n=0$$and you always have $f_n(1)=1$. Therefore, your sequence converges pointwise to$$\begin{array}{rccc}f\colon&[0,1]&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}0&\text{ if }x<1\\1&\text{ if }x=1,\end{cases}\end{array}$$which is a discontinuous function. Since each $f_n$ is continuous, the convergence is not uniform.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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pdf of $\frac{(X_1-X_2)^2}{2}$ Let $X_1$ and $X_2$ be independent normally distributed random variables.
$$X_1 \sim N(0,1) \\ X_2 \sim N(0,1)$$
Find the pdf of $\frac{(X_1-X_2)^2}{2}$
We have that
By example we know that when $Y = X^2$ we have that
$$\begin{align}f_Y(y) &= \frac{e^{-\frac{(-\sqrt{y})^2}{2}}}{\sqrt{2\pi}}\left|-\frac{1}{2\sqrt{y}}\right|+\frac{e^{-\frac{(\sqrt{y})^2}{2}}}{\sqrt{2\pi}}\left|\frac{1}{2\sqrt{y}}\right| \\ &= \frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{y}}e^{-\frac{y}{2}} \end{align}$$
Which follows achi-square distribution with 1 degree of freedom, therefore
$$\frac{(X_1-X_2)^2}{2} \sim \chi_1^2$$
However, how can I do the calculation for this using multivariate transformations, I cannot get the distribution?
| Let $U=\frac{(X_1-X_2)}{\sqrt{2}}$ and $M_{x_i(t)}$ be moment generating function of $X_i$.
$M_{\frac{X_1-X_2}{\sqrt{2}}}(t)=E(e^{(\frac{X_1-X_2}{\sqrt{2}})t})=E(e^{\frac{X_1}{\sqrt{2}}t}) E(e^{\frac{X_1}{\sqrt{2}}t})=M_{X_1}(\frac{t}{\sqrt{2}})M_{X_2}(\frac{-t}{\sqrt{2}})=e^{\frac{1}{2}(\frac{t}{\sqrt{2}})^2}e^{\frac{1}{2}(\frac{-t}{\sqrt{2}})^2}=e^{\frac{t^2}{2}}$ So $U=\frac{(X_1-X_2)}{\sqrt{2}} \sim N(0,1)$ Therefore $U^2=(\frac{X_1-X_2}{\sqrt{2}})^2 \sim \chi^2_1$
| {
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Showing $\sin^s\theta+\cos^s\theta\le1-K_s\cos^2\theta\sin^2\theta$, for $s>2$ and $K_s=\min\{\frac14s(s-2),2\}$ I was reading through an article (X. Lu, B. Wennberg, Solutions with increasing energy for the spatially homogeneous Boltzmann equation (2002)) and this elementary inequality is used in a proof:
For $s>2$ and $\theta\in [0,\pi/2]$,
$$\sin^s(\theta) + \cos^s(\theta) \le 1- K_s \cos^2(\theta)\sin^2(\theta)$$
where $K_s=\min\left\{\frac{1}{4}s(s-2), 2 \right\}$.
I'm trying to show it but I'm not succeeding.
I tried to use another another elementary inequality found in the same proof: for $s>2$ and $a,b \ge 0$
$$a^s+b^s \le (a+b)^s \le a^s + b^s + 2^s(a^{s-1}b+ab^{s-1}),$$
but I didn't succeed as well. Do you have any suggestions? Thank you in advance
| $\newcommand{\d}{\mathrm{d}}$
Let $x=\cos^2 \theta$. Over $0\leq x \leq 1$, we want to uniformly bound
$$\begin{split}
\frac{1 - \cos^s \theta - \sin^s\theta}{\cos^2 \theta \sin^2 \theta}
&= \frac{1 - x^{s/2} - (1-x)^{s/2}}{x(1-x)} \\
&= \frac{1 - x^{s/2 -1}}{1 - x} + \frac{1 - (1-x)^{s/2-1}}{x} \\
&= (\tfrac{s}{2}-1)\int_0^1\left((1-tx)^{s/2-2}+(1-t(1-x))^{s/2-2}\right)\d t
\end{split}$$
If $2<s\leq 4$ or $6\leq s$, then $u^{s/2 -2}$ is convex, so that by Jensen's inequality we have
$$\begin{split}
(\tfrac{s}{2}-1)\int_0^1\left((1-tx)^{s/2-2}+(1-t(1-x))^{s/2-2}\right)\d t
&\geq 2(\tfrac{s}{2}-1)\int_0^1 (1-\tfrac{t}{2})^{s/2 - 2} \d t \\
&= 4 - 2^{3-s/2}\text{.}
\end{split}$$
In this case, the inequality is an equality at $x=\tfrac{1}{2}$.
If $4\leq s \leq 6$, then $u^{s/2 -2}$ is concave, so it is bounded below by the secant from $1$ to $u$:
$$\begin{split}
(\tfrac{s}{2}-1)\int_0^1\left((1-tx)^{s/2-2}+(1-t(1-x))^{s/2-2}\right)\d t
&\geq (\tfrac{s}{2}-1)\int_0^1 (1 + (1-t)^{s/2 - 2}) \d t \\
&= \tfrac{s}{2}\text{.}
\end{split}$$
In this case, the inequality is an equality at $x\in\{0,1\}$.
We can summarize the established minimum as
$$\begin{split}\min\frac{1-\cos^s\theta -\sin^s\theta}{\cos^2 \theta \sin^2\theta} &=
\begin{cases}
4-2^{3-s/2} & 2 < s\leq 4 \\
\tfrac{s}{2} & 4 \leq s \leq 6 \\
4 - 2^{3-s/2} & 6\leq s
\end{cases}\\
&=\min\{4-2^{3-s/2},\tfrac{s}{2}\}\text{,}\end{split}$$
and one can show that $\min\{\tfrac{s(s-2)}{4},2\}\leq \min\{4-2^{3-s/2},\tfrac{s}{2}\}$.
| {
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Prove that $\frac{a-c}{k}=\frac{d+b}{n}$, where $N=a^2+b^2=c^2+d^2$ and $k=\gcd(a-c,d-b)$, $n=\gcd(a+c, d+b)$ Let $N$ be odd and $N = a^2 + b^2 = c^2 + d^2$, where $a, b, c, d \in \mathbb{N}$ and WLOG let $a, c$ be odd, $b, d$ be even, $a > c$, and $b < d$. Prove that $\frac{a-c}{k}=\frac{d+b}{n}$.
I first began the proof my showing that $k$ and $n$ are positive integers.
Let $k= \gcd(a -c, d-b)$ and $n= \gcd(a + c, d + b)$. If $a$ and $c$ are both odd, then $a\pm c$ is even. Similarly, if $d$ and $b$ are both even, then $b\pm d$ is also even. Thus, $k$ and $n$ are also both even.
Now I wish to show that $\frac{a-c}{k}=\frac{d+b}{n}$.
I first tried to use the fact that $N = a^2 + b^2 = c^2 + d^2$.
$$
\begin{align*}
a^2 + b^2 &= c^2 + d^2\\
a^2 -c^2 &= d^2-b^2\\
(a+c)(a-c)&=(d+b)(d-b)\\
\end{align*}
$$
This seems like I'm getting close but I can't quite see how to finish it. Any hints are appreciated.
| We have that
$$k = \gcd(a - c, d - b) \; \; \to \; \; a - c = ke, \; d - b = kf, \; \gcd(e, f) = 1 \tag{1}\label{eq1A}$$
$$n = \gcd(a + c, d + b) \; \; \to \; \; a + c = ng, \; d + b = nh, \; \gcd(g, h) = 1 \tag{2}\label{eq2A}$$
Using \eqref{eq1A} and \eqref{eq2A} with your factorization gives
$$\begin{equation}\begin{aligned}
(a + c)(a - c) & = (d + b)(d - b) \\
(ng)(ke) & = (nh)(kf) \\
ge & = hf
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
With \eqref{eq1A} giving $\gcd(e,f) = 1$, then \eqref{eq3A} shows that
$$f \mid g \; \; \to \; \; g = sf, \; s \ge 1 \tag{4}\label{eq4A}$$
Similarly, since $\gcd(g,h) = 1$ from \eqref{eq2A}, then
$$h \mid e \; \to \; e = th, \; t \ge 1 \tag{5}\label{eq5A}$$
Thus, \eqref{eq3A} becomes
$$(sf)(th) = hf \; \; \to \; \; st = 1 \; \; \to \; \; s = t = 1 \tag{6}\label{eq6A}$$
Therefore, using \eqref{eq1A}, \eqref{eq2A}, \eqref{eq5A} and \eqref{eq6A}, we have that
$$e = h \; \; \to \; \; \frac{a - c}{k} = \frac{d + b}{n} \tag{7}\label{eq7A}$$
| {
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Evaluating $\int_{1/\sqrt{2}}^1\frac{1}{(3y^2-1)\sqrt{2y^2-1}}\,\mathrm{d}y$; how do you avoid using a complex substitution? $\newcommand{\d}{\mathrm{d}}$The given exam question - I provide the beginning for context:
Let: $$I=\int\frac{1}{(b^2-y^2)\sqrt{c^2-y^2}}\d y$$Where $b,c\gt0$, and employ the substitution $y=\frac{cx}{\sqrt{x^2+1}}$ to show that: $$I=\int\frac{1}{b^2+(b^2-c^2)x^2}\d x$$And hence evaluate: $$\int_1^{\sqrt{2}}\frac{1}{(3-y^2)\sqrt{2-y^2}}\d y$$And: $$\int_{1/\sqrt{2}}^1\frac{y}{(3y^2-1)\sqrt{2y^2-1}}\d y$$
This I managed straightforwardly. The next part:
By means of a suitable substitution, evaluate: $$\int_{1/\sqrt{2}}^1\frac{1}{(3y^2-1)\sqrt{2y^2-1}}\d y$$
I ordinarily wouldn't have a clue about evaluating that integral, as I am not very well practised in integration (which is why I am seeking out such exercises) so I assumed they wanted us to use the same ideas. My attempt is based on the context of the question at the start, but I shouldn't have needed a complex variable substitution.
$$\begin{align}I:=&\int_{1/\sqrt{2}}^1\frac{1}{(3y^2-1)\sqrt{2y^2-1}}\d y\\\overset{y\mapsto1/y}{=}&\int_1^{\sqrt{2}}\frac{y}{(3-y^2)\sqrt{2-y^2}}\d y\end{align}$$I optimistically employ the same substitution $y=\frac{\sqrt{2}x}{\sqrt{x^2+1}}$ to get: $$\begin{align}I=&\,\sqrt{2}\int_1^\infty\frac{x}{(x^2+3)\sqrt{x^2+1}}\d x\\\overset{x\mapsto1/x}{=}&\sqrt{2}\int_0^1\frac{1}{(1+3x^2)\sqrt{1+x^2}}\d x\end{align}$$
Now, I decided that instead of trying for perhaps a very long time to come up with a similar substitution to the one at the beginning (the problem being that we have $b^2+y^2$ instead of $b^2-y^2$) I decided to enforce negative signs by introducing a complex variable substitution - which will cause some weird problem.
Let $x=\frac{iz}{\sqrt{z^2+1}}$. Then $x^2(z^2+1)=-z^2,\,z^2(x^2+1)=-x^2,\,z=\frac{-ix}{\sqrt{x^2+1}}$ (the minus sign comes from pluggin $z$ back in and seeing which of $\pm i$ works) and $\frac{\d x}{\d z}=\frac{i}{(z^2+1)\sqrt{z^2+1}}$ and we get: $$\begin{align}I&=i\sqrt{2}\int_0^{-i/\sqrt{2}}\frac{1}{1-2z^2}\d z\\&\overset{z=-\frac{iu}{\sqrt{2}}}{=}\int_0^{1}\frac{1}{1+u^2}\d u\\&=\frac{\pi}{4}\end{align}$$
However, although this isn't really complex integration, I am extremely certain we were not supposed to do it this way (since the syllabus is for pre-university mathematics) so how were we supposed to do it?
Many thanks.
| Here is one way to carry out the integration. Substitute $\sqrt2 y= \cosh t$
\begin{align}
&\int_{1/\sqrt{2}}^1\frac{1}{(3y^2-1)\sqrt{2y^2-1}}\,{d}y\\
=&\>\frac1{\sqrt2}\int_0^{\cosh^{-1}\sqrt2}
\frac1{\frac32\cosh^2t-1}dt
= \frac1{\sqrt2}\int_0^{\tanh^{-1}\frac1{\sqrt2}}
\frac{d(\tanh t)}{\frac12+ \tanh^2t}dt\\
=&\>\tan^{-1}\left(\sqrt2\tanh t\right)\bigg|_0^{\tanh t=\frac1{\sqrt2}}=\tan^{-1}(1)=\frac\pi4
\end{align}
| {
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How to prove that $g(x)=x^2$ is integrable on $[2,5]$ using regular partitions? So I've been trying to prove that $g(x)=x^2$ is integrable on the interval $[2,5]$ using regular partitions and the theorem that a function is integrable if
$$\lim_{n\to\infty}(U(f,P_n)-L(f,P_n)) = 0.$$
This is what I have so far:
Let
$$g:[2,5]\to \mathbb{R},\quad g(x)=x^2,$$
let $P_n=\{x_0, x_1, x_2,...,x_n\}$ be defined by $\Delta x=\frac{b-a}{n}$, $x_k=x_0+k\Delta x$. Set
$$m_k=\inf\{g(x)\,|\,x\in [x_{k-1}, x_k]\}=x_{k-1}^2$$
and
$$M_k=\sup\{g(x)\,|\,x\in [x_{k-1}, x_k]\}=x_k^2.$$
Then
\begin{align*}
L(g, P_n)
&=\sum_{n=1}^{k} m_k(x_k-x_{k-1}) \\
&=\sum_{n=1}^{k} x_{k-1}^2 \Delta x \\
&= \Delta x \sum_{n=1}^{k} x_{k-1}^2 \\
&= \Delta x \sum_{n=1}^{k} (x_0+(k-1)\Delta x)^2 \\
&= \Delta x\sum_{n=1}^{k} (x_0^2+2x_0\Delta x (k-1) + ((k-1)\Delta x)^2) \\
&= \Delta x\left(nx_0^2+2nx_0\Delta x \sum_{n=1}^{k}(k-1) + \left(n\Delta x\sum_{n=1}^{k}(k-1)\right)^2\right) \\
&= nx_0^2 \frac{3}{n}+2nx_0\frac{9}{n^2}(0+1+2+...+(n-1)) + n\frac{27}{n^3}(0+1+2+...+(n-1))^2 \\
&=3x_0^2+2x_0\frac{9}{n}\frac{n(n-1)}{2} + \frac{27}{n^2}\frac{(n(n-1))^2}{4} \\
&=3x_0^2+x_0{9}(n-1) + \frac{27(n-1)^2}{4} \\
&= 12 + 18(n-1) + \frac{27(n-1)^2}{4}.
\end{align*}
Similarly
$$U(f,P_n)=\sum_{n=1}^{k} M_k(x_k-x_{k-1})=12+18(n+1)+\frac{27(n+1)^2}{4}.$$
Then
$$\lim_{n\to\infty}(U(f,P_n)-L(f,P_n)) = \lim_{n\to\infty}\left(12 + 18(n-1) + \frac{27(n-1)^2}{4}-\left(12+18(n+1)+\frac{27(n+1)^2}{4}\right)\right)$$
And this is where I run into issues because then $\lim_{n\to\infty}(U(f,P_n)-L(f,P_n))$ does not equal zero like it is supposed to. I believe I am running into a computational error somewhere, but I can't seem to find it.
Thanks in advance.
| The first error is that you have confused $n$ for $k$ at various points: when you say $\sum_{n=0}^k$, I think you mean $\sum_{k=0}^n$.
The second error, possibly related, is that you have introduced a factor of $n$ (the sum index) in places where the sum remains e.g. $\sum_{n=1}^k c(k-1)=^?nc\sum_{n=1}^k (k-1)$ for some constant $c$.
The third error is that you have confused $\sum_{i=1}^m i^2=\frac{m(m+1)(2m+1)}{6}$ for $\sum_{i=1}^m i^3=\frac{(m(m+1))^2}{4}$.
I believe it should read:
$$
\begin{align*}
L(g, P_n)
&=\cdots \\
&= \Delta x\sum_{k=1}^{n} (x_0^2+2x_0\Delta x (k-1) + ((k-1)\Delta x)^2) \\
&= \Delta x\left(nx_0^2+2x_0\Delta x \sum_{k=1}^{n}(k-1) + \left(\Delta x\sum_{k=1}^{n}(k-1)\right)^2\right) \\
&= nx_0^2 \frac{3}{n}+2x_0\frac{9}{n^2}(0+1+2+...+(n-1)) + \frac{27}{n^3}(0+1+2+...+(n-1))^2 \\
&=3x_0^2+2x_0\frac{9}{n^2}\frac{n(n-1)}{2} + \frac{27}{n^3}\frac{(n-1)\times n\times 2n}{6} \\
&=3x_0^2+x_0{9}\frac{n-1}{n} + 9\frac{n-1}{n} \\
&= 12 + 27\frac{n-1}{n}.
\end{align*}
$$
Indeed, as $n\to\infty$, $L(g,P_n)\to 39$ and $\int_2^5 x^2 dx=39$.
You will need to calculate $U(g,P_n)$ again with this in mind.
| {
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Computing $\int_{0}^{1} \frac{\sin ^{-1} (x) \ln (1+x)}{x^{2}} d x$ Applying integration by parts splits the integral into 3 integrals,
$\displaystyle \begin{aligned}I&=\int_{0}^{1} \frac{\sin ^{-1} x \ln (1+x)}{x^{2}} d x\\&=-\int_{0}^{1} \sin ^{-1} x \ln (1+x) d\left(\frac{1}{x}\right) \\&=-\left[\frac{\sin ^{-1} x \ln (1+x)}{x}\right]_{0}^{1}+\underbrace{\int_{0}^{1} \frac{\ln (1+x)}{x \sqrt{1-x^{2}}}}_{K} +\underbrace{\int_{0}^{1}\frac{\sin ^{-1} x}{x}}_{L} d x-\underbrace{\int_{0}^{1} \frac{\sin ^{-1} x}{1+x}}_{M} d x \end{aligned} \tag*{} $
Letting $x= \cos \theta$ for $K$ and $\sin^{-1}x \mapsto x$ for $L$ and $M$, yields
$\displaystyle I=-\frac{\pi}{2} \ln 2 +\underbrace{\int_{0}^{\frac{\pi}{2}} \frac{\ln (1+\cos \theta)}{\cos \theta} d \theta}_{K}+\underbrace{\int_{0}^{\frac{\pi}{2}} \frac{x\cos x }{\sin x} d x}_{L}-\underbrace{\int_{0}^{\frac{\pi}{2}} \frac{x\cos x }{1+\sin x} d x }_{M}\tag*{} $
For the integral $ K,$putting $ a=1$ in my post yields
$\displaystyle \boxed{K=\frac{\pi^{2}}{8}}\tag*{} $
For the integral $ L,$ integration by parts yields
$\displaystyle \begin{aligned}L &=\int_{0}^{\frac{\pi}{2}} x d \ln (\sin x) \\&=[x \ln (\sin x)]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} \ln (\sin x) d x \\&=\boxed{\frac{\pi}{2} \ln 2}\end{aligned}\tag*{} $
For the integral $ M,$ integration by parts yields
$\displaystyle \begin{aligned}M &=\int_{0}^{\frac{\pi}{2}} x d \ln (1+\sin x)\\&=[x \ln (1+\sin x)]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} \ln (1+\sin x) d x \\&=\frac{\pi}{2} \ln 2-\underbrace{\int_0^{\frac{\pi}{2} }\ln (1+\sin x) d x}_{N}\end{aligned}\tag*{} $
For the integral $ N,$ using my post in the second last step yields
$\displaystyle \begin{aligned}N \stackrel{x\mapsto\frac{\pi}{2}-x}{=} &\int_{0}^{\frac{\pi}{2}} \ln (1+\cos x) d x \\=&\int_{0}^{\frac{\pi}{2}} \ln \left(2 \cos ^{2} \frac{x}{2}\right) d x \\=&\frac{\pi}{2} \ln 2+2 \int_{0}^{\frac{\pi}{2}} \ln \left(\cos \frac{x}{2}\right) d x \\=&\frac{\pi}{2} \ln 2+4 \int_{0}^{\frac{\pi}{4}} \ln (\cos x) d x \\=&\frac{\pi}{2} \ln 2+4\left(\frac{1}{4}(2 G-\pi \ln 2)\right) \\=&\boxed{-\frac{\pi}{2} \ln 2+2 G}\end{aligned}\tag*{} $
where $ G$ is the Catalan’s Constant.
Putting them together yields
$\displaystyle \boxed{I=-\pi \ln 2+\frac{\pi^{2}}{8}+2G} \tag*{} $
Question: Is there any shorter solution?
| A self-contained solution
\begin{align}
I=&\int_{0}^{1} \frac{\sin ^{-1} x \ln (1+x)}{x^{2}} d x
=\int_{0}^{1} \sin ^{-1}x \>d\left( \ln x -\frac{1+x}x \ln (1+x)\right)\\
\overset{ibp} =&\> -\pi \ln 2-{\int_{0}^{1} \frac{\ln \frac x{1+x}-\frac1x \ln(1+x)}{\sqrt{1-x^{2}}}}\>
\overset{x=\frac{2t}{1+t^2}}{dx}\\
=&\>-\pi\ln2 + \int_0^1 \underset{=K}{\frac{\ln\frac{(1+t)^2}{1+t^2}}{t}}dt -2\int_0^1 \underset{=J}{\frac{\ln\frac{2t}{(1+t)^2}}{1+t^2}}dt
\end{align}
with
\begin{align}
K=&\>\int_0^1 \frac{\ln (1+t)^2}{t}dt
-\int_0^1 \frac{\ln (1+t^2)}{t}\overset{t^2\to t}{dt}
=\frac32 \int_0^1 \frac{\ln (1+t)}{t}dt =\frac{\pi^2}8\\
J=&
\int_0^1 \frac{\ln t}{1+t^2}dt
+\int_0^1 \frac{\ln \frac2{(1+t)^2}}{1+t^2}\overset{t\to\frac{1-t}{1+t}}{dt}= \int_0^1 \frac{\ln t}{1+t^2}dt=-G\\
\end{align}
Thus
$$I=-\pi \ln 2+\frac{\pi^{2}}{8}+2G$$
| {
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Show $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{1999}-\frac{1}{2000} =\frac{1}{1001}+\frac{1}{1002}+\ldots+\frac{1}{1999}+\frac{1}{2000}$ I am trying to show that
$$
1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{1999}-\frac{1}{2000}
=\frac{1}{1001}+\frac{1}{1002}+\ldots+\frac{1}{1999}+\frac{1}{2000}.
$$
It seems there is some sort of generalizable pattern here, so I will verify it for smaller numbers:
$$
\begin{align*}
\text{Say, }n=4 \hspace{35pt} 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}&=\frac{1}{3}+\frac{1}{4}\\
\frac{12}{12}-\frac{6}{12}+\frac{4}{12}-\frac{3}{12}&=\frac{4}{12}+\frac{3}{12}\\
\frac{7}{12}&=\frac{7}{12}
\end{align*}
$$
So, my guess on the general formula is
$$
1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{n-1}-\frac{1}{n}=\frac{1}{n/2+1}+\ldots+\frac{1}{n-1}+\frac{1}{n}.
$$
This really seems like I am getting somewhere, but how can I finish off the proof? Is induction viable?
| Proceed with induction. We seek to show, in essence, for all $n$ sufficiently large,
$$\sum_{k=1}^{2n} \frac{(-1)^{k-1}}{k} = \sum_{k=n+1}^{2n} \frac{1}{k}$$
We have a base case. So let's try to show that - assuming the above holds for $n$ - that it holds for $n+1$ too. We have
$$\begin{align*}
\sum_{k=1}^{2(n+1)} \frac{(-1)^{k-1}}{k}
&= \frac{(-1)^{2n}}{2n+1} + \frac{(-1)^{2n+1}}{2n+2} + \sum_{k=1}^{2n} \frac{(-1)^k}{k} \\
&= \frac{1}{2n+1} + \frac{-1}{2n+2} + \sum_{k=1}^{2n} \frac{(-1)^{k-1}}{k} \\
&= \frac{1}{2n+1} + \frac{-1}{2n+2} + \sum_{k=n+1}^{2n} \frac{1}{k} \\
\sum_{k=(n+1)+1}^{2(n+1)} \frac{1}{k}
&= \sum_{k=n+2}^{2n+2} \frac{1}{k} \\
&= \frac{1}{2n+2} + \frac{1}{2n+1}+ \sum_{k=n+2}^{2n} \frac{1}{k} \\
&= \frac{1}{2n+2} + \frac{1}{2n+1} - \frac{1}{n+1} + \sum_{k=n+1}^{2n} \frac{1}{k} \\
\end{align*}$$
It is trivial to show that
$$\frac{1}{2n+1} + \frac{-1}{2n+2} = \frac{1}{2n+2} + \frac{1}{2n+1} - \frac{1}{n+1}$$
so the induction follows and the original equation holds for $n \ge 4$. In particular, $n=1000$ gives you your desired result.
| {
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Inequality help: $\frac{a+b+c}{3}\geq\sqrt{\frac{ab+bc+ca}{3}}\geq\sqrt[3]{abc}$ I'm working through "Calculus of One Variable" by Joseph Kitchen, and this inequality (problem 3 from section 1.3) is causing me quite a bit of pain to prove:
$$\frac{a+b+c}{3}\geq\sqrt{\frac{ab+bc+ca}{3}}\geq\sqrt[3]{abc},$$
where $a,b,c\geq 0.$
The main issue I'm having is that I cannot use the fact that $0\leq a < b$ if and only if $a^{2} < b^{2}$, as this fact is proven in a later exercise.
If you want to take a look at what is proven and what isn't, the first 80 pages of the book is available on a Google Books preview.
Although a hint for this problem is provided in the back of the book, it has not been useful to me. The hint states: "Use Example 2 of the text. Define $A = (x+y+z)/3$ and $H^{-1} = (x^{-1}+y^{-1}+z^{-1})/3$." Note that $A$ and $H$ refer to the arithmetic and harmonic means, respectively.
Example 2 proved that if $a$, $b$, and $c$ are positive numbers which are not all equal, then $(a+b+c)(bc+ca+ab) > 9abc$. This was proved by noting that $(a+b+c)(bc+ca+ab)-9abc = a(b-c)^{2} + b(c-a)^{2} + c(a-b)^{2}$. With little work, it is not hard to see that when $a$, $b$, and $c$ are non-negative, then $(a+b+c)(bc+ca+ab) \geq 9abc$. However, I'm having a difficult time seeing how this fact is useful for my current problem. Using the hint to define $A = (x+y+z)/3$ and $H^{-1} = (x^{-1}+y^{-1}+z^{-1})/3$, I proved that $A\geq H$:
$$A = \frac{x+y+z}{3}\geq^{(1)} \frac{9xyz}{yz+zx+xy} = \frac{9}{x^{-1}+y^{-1}+z^{-1}}\geq H,$$
where the first inequality (1) holds by Example 2. However, I do not see how this is at all helpful. Here's some of the things that I've tried (that do not involve the hint).
Define $x = \sqrt[3]{a}$, $y = \sqrt[3]{b}$, and $z = \sqrt[3]{c}$. Then, observe that
$$\frac{a+b+c}{3} - \sqrt[3]{abc} \,=\, \frac{x^{3} + y^{3} + z^{3} - 3xyz}{3} = \frac{(x+y+z)\left((x-y)^{2} + (y-z)^{2} + (z-x)^{2}\right)}{6}\geq 0.$$
Thus, $(a+b+c)/3 \geq \sqrt[3]{abc}$. I was also able to prove that $(a+b+c)/3\geq\sqrt{(ab+bc+ca)/3}$, however, I used the fact that $0\leq a < b$ if and only if $a^{2} < b^{2}$, which I'm not supposed to use. I only include this proof since it may give insight into the problem. (Some of these algebraic "tricks" were hard to figure out!)
Since $\frac{1}{2}\left((a-b)^{2} + (b-c)^{2} + (c-a)^{2}\right) = (a+b+c)^{2} - 3(ab + bc + ca)\geq 0$, it follows that $3(a+b+c)^{2}\geq 9(ab+bc+ca)$. And therefore, $(a+b+c)^{2}/9 \geq (ab+bc+ca)/3 \geq 0$. Thus, it follows that $(a+b+c)/3\geq\sqrt{(ab+bc+ca)/3}$.
So... to recap, using the appropriate methods, I have prove that $(a+b+c)/3 \geq \sqrt[3]{abc}$ and that $(a+b+c)/3 \geq3/(a^{-1}+b^{-1}+c^{-1})$. However, the rest is giving me a headache. Perhaps I've missed something obvious, but I'm stuck...
| Assuming that you can use $$\frac{x+y+z}{3} \ge \sqrt[3]{xyz}$$ for $x,y,z \ge 0$ Substitute
for $x=ab$, $y=bc$, and $z=ac$.
You will get $$\frac{ab + bc + ac}{3} \ge \sqrt[3]{(abc)^2}$$ and after taking root of both sides you will get the right hand side inequality.
| {
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An inequality $16(ab + ac + ad + bc + bd + cd) \le 5(a + b + c + d) + 16(abc + abd + acd + bcd)$ Let $a, b, c, d \ge 0$ be nonnegative real numbers such that $a + b + c + d \le 1$.
Show that $16(ab + ac + ad + bc + bd + cd) \le 5(a + b + c + d) + 16(abc + abd + acd + bcd)$.
If I set the power means as $A := (a + b + c + d)/4$ and $Q := \sqrt{(a^2 + b^2 + c^2 + d^2)/4}$ and $C := \sqrt[3]{(a^3 + b^3 + c^3 + d^3)/4}$, then
$$RHS - LHS = \frac{4}{3}(128A^3 - 96A^2 - 96QA + 15A + 16C^3 + 24Q^2),$$
where $0 \le A \le 1/4$. The power mean inquality states that $A \le Q \le C$. What can I do with the $-96QA$ term?
| Another way which works for several such inequalities, consider minimising $RHS - LHS$ (note there will exist a minimum for this function in this domain). At the point of minimum, if possible let there be two unequal variables, WLOG say $a \neq b$. Then consider
$$RHS - LHS = 16\left(ab(c+d)+cd(a+b)\right)+5(a+b+c+d) - 16\left(ab+(a+b)(c+d)+cd \right) \\= -16(1-c-d)\color{red}{ab} + (\color{red}{a+b})\left(cd+5-16(c+d)\right)+5(c+d)-16cd$$
Now it can be seen that replacing both $a, b$ with $\frac12(a+b)$ will change only the first term, and in fact it will reduce $RHS-LHS$ as $ab$ will increase. Hence at the minimum, we must have $a=b=c=d$, so it is enough to check the inequality for this case, which is a lot easier.
$16(6a^2) \leqslant 5(4a) + 16(4a^3) \iff 4a(1-4a)(5-4a) \geqslant 0$ which is true for $a \in [0, \frac14]$.
| {
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Find number of positive integer solutions to the equation $a+b+c=15$ where a,b are even and c is odd This is equivalent to:
Coefficient of $x^{15}$ in $(1+x^2+x^4+\cdots+x^{14})^2(x+x^3+\cdots+x^{15})$
Or
Coefficient of $x^{14}$ in $(1+x^2+x^4+\cdots+x^{14})^3$
Is this approach correct?
| Why not just solving the problem with a direct approach? You know that $a$ and $b$ are even and $c$ is odd, so $a = 2A, b = 2B$, and $c = 2C+1$, with $A,B$ positive integers and $C \geq 0$. Now the problem amounts to solving
$$2(A+B+C)+1 = 15,$$
so you want $A,B > 0$ and $C \geq 0$ integers such that
$$A+B+C = 7.$$
You can just do this by trying case by case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4439082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the number of non-negative integer solutions to $x+y+z=11$ How do I find the number of nonnegative integer solutions to $x+y+z=11$ provided that $x\leq 3, y\leq 4, z\leq 6$ using the sum rule (counting)?
I know the answer is 6, but I'm having difficulty understanding why.
| One way is to expand $$(1+x+x^2+x^3)(1+x+x^2+x^3+x^4)(1+x+x^2+x^3+x^4+x^5+x^6)$$ and get the coefficient of $x^{11}.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Which way do I prove that $36\le 4\sum_{cyc} a^3-\sum_{cyc}a^4 \le 48$? It is given that $a+b+c+d=6$ and $a^2+b^2+c^2+d^2=12$
| Note that $(a-1)^4=a^4-4a^3+6a^2-4a+1$. On cyclically summing and using the data given, we have that $\sum_{cyc} (a-1)^4=\sum_{cyc}a^4-4\sum_{cyc}a^3+52$. Hence we have to prove that $$36\le 52- \sum_{cyc}(a-1)^4\le 48 \iff 4\le \sum_{cyc} (a-1)^4 \le 16$$ Now, we have that $\sum_{cyc} (a-1)^2=(\sum_{cyc} a^2)- 2(\sum_{cyc}a)+4=4$. Hence, the right inequality is clear (just note that $16=4^2=(\sum_{cyc} (a-1)^2)^2$ will have some other non negative terms other than the desired ones.) The left inequality follows from Power-mean/QM-AM/cauchy-schwartz: $$\sum_{cyc}(a-1)^4 \ge \frac{\left(\sum_{cyc} (a-1)^2\right)^2}{4}=4$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Determine rank of $ A = \begin{bmatrix} 2 & 1 & -2 & 1 \\ 4 & 1 & -2 & -3 \\ 1 & -1 & 2 & -3 \\ 2 & 2 & -4 & -5 \\ 3 & 1 & -2 & 2 \end{bmatrix}$ Could you give me your feedback ? I've verified with https://matrix.reshish.com/rankCalculation.php but maybe there are things that could be done differently
Determine the rank of the following matrices
$$ A = \begin{bmatrix} 2 & 1 & -2 & 1 \\
4 & 1 & -2 & -3 \\
1 & -1 & 2 & -3 \\
2 & 2 & -4 & -5 \\
3 & 1 & -2 & 2
\end{bmatrix}, B = \begin{bmatrix} 1 & 2 & 1 \\
2 & 1 & 1 \\
1 & 2 & 2 \\
2 & 1 & 0 \\
\end{bmatrix}, C = \begin{bmatrix} 1 & 1 & 0 \\
-1 & 2 & 2 \\
2 & -13 & -10 \\
2 & -1 & -2 \\
\end{bmatrix}$$
For $A$, add 2 first row to second row, 1/2 first row to third row, 1 first row to fourth row, 3/2 first row to fifth row. Then add -3/2 second row to third row, add 1 second row to fourth row, add -1/2 second row to fifth row. That will zero out the third column too below the second row. Lastly, for the fourth column, add 11/4 third row to fourth row, add -3/4 third row to fifth row. We get
$$\begin{bmatrix} 2 & 1 & -2 & 1 \\
0 & -1 & 2 & -5 \\
0 & 0 & 0 & 4 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}$$ which has rank 3.
For $B$, add -2 first row to second row, -1 first row to third row, -2 first row to fourth row. Then add -1 second row to fourth row, add 1 third row to fourth row. We get
$$\begin{bmatrix} 1 & 2 & 1 \\
0 & -3 & -1 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{bmatrix}$$ which has rank 3.
For $C$, add -1 first row to second row, add -2 first row to third row, add -2 first row to fourth row. Then add 5 second row to third row, 1 second row to fourth row. We get
$$\begin{bmatrix} 1 & 1 & 0 \\
0 & 3 & 2 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{bmatrix}$$ whihc has rank 2.
Is this correct ? Could you give me your feedback ?
Thanks for your help
| You should write the result each row reduction, this helps avoid mistakes and it's a good practice for this calculation.
Now, returning the question
*
*$A$ is similar by row reduction to $\begin{bmatrix}\boxed{1}&0&0&0\\0&\boxed{1}&-2&0\\0&0&0&\boxed{1}\\0&0&0&0\\0&0&0&0\end{bmatrix}$ so $\rm rank (A)=3$.
*$B$ is similar by row reduction to $\begin{bmatrix}\boxed{1}&0&0\\0&\boxed{1}&0\\0&0&\boxed{1}\\0&0&0\end{bmatrix}$ so $\rm rank(B)=3$.
*$C$ is similar by row reduction to $\begin{bmatrix}\boxed{1}&0&-2/3\\ 0&\boxed{1}&2/3\\0&0&0\\0&0&0\end{bmatrix}$ so $\rm rank( C)= 2$.
Here, we're pivoting with $1$ but we can pivote with a number different of $0$ and $1$ as we can see in your approach. So your conclusion they're correct.
| {
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"source": "stackexchange",
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Knowing that $\int_0^\pi\frac{x\sin^{2022}x}{\sin^{2022}x+\cos^{2022}x}\, \mathrm dx=\frac{\pi^a}{b} \ (a, b \in \mathbb N)$, calculate $2a^2 + 3b^3$.
Knowing that $\displaystyle \int_0^\pi\dfrac{x\sin^{2022}x}{\sin^{2022}x + \cos^{2022}x}\, \mathrm dx = \dfrac{\pi^a}{b}$, where $a$ and $b$ are positive integers, calculate the value of $2a^2 + 3b^3$.
[For context, this question is taken from an exam whose format consists of 50 multiple-choice questions with a time limit of 90 minutes. Calculators are the only electronic device allowed in the testing room. (You know those scientific calculators sold at stationery stores and sometimes bookstores? They are the goods.) I need a solution that works within these constraints. Thanks for your cooperation, as always. (Do I need to sound this professional?)
By the way, if the wording of the problem sounds rough, sorry for that. I'm not an expert at translating documents.]
Let's pretend I know what's going on.
As $f(0) = f(\pi)$ where $f(x) = \dfrac{x\sin^{2022}x}{\sin^{2022}x + \cos^{2022}x}$, it can be proven that $$\displaystyle \int_0^\pi\dfrac{x\sin^{2022}x}{\sin^{2022}x + \cos^{2022}x}\, \mathrm dx = \dfrac{\pi}{2}\int_0^\pi\dfrac{\sin^{2022}x}{\sin^{2022}x + \cos^{2022}x}\, \mathrm dx$$
You know what's the deal with years being used in problems? They're interchangeable.
Consider sequence $(u_n)$ where $\displaystyle u_n = \int_0^\pi\dfrac{\sin^{2n}x}{\sin^{2n}x + \cos^{2n}x}\, \mathrm dx, \forall x \in \mathbb N$.
(Why shouldn't it be $\displaystyle u_n = \int_0^\pi\dfrac{\sin^{n}x}{\sin^{n}x + \cos^{n}x}\, \mathrm dx, \forall x \in \mathbb N$? Well, $x = -\dfrac{\pi}{4} + k\pi, k \in \mathbb Z$ is a root for all equations $\sin^{2n + 1}x + \cos^{2n + 1}x = 0, n \in \mathbb N$, which means the integral does not converge at $x = \dfrac{3\pi}{4}$.)
We have that $$\begin{aligned} u_{n + 1} - u_n &= \int_0^\pi\dfrac{\sin^{2n + 2}x}{\sin^{2n + 2}x + \cos^{2n + 2}x}\, \mathrm dx - \int_0^\pi\dfrac{\sin^{2n}x}{\sin^{2n}x + \cos^{2n}x}\, \mathrm dx\\ &= \int_0^\pi\dfrac{\sin^{2n}2x(2\sin^2x - 1)\, \mathrm dx}{2^n\sin^2x(\sin^{2n + 2}x + \cos^{2n + 2}x)(\sin^{2n}x + \cos^{2n}x)}, \forall n \in \mathbb N \end{aligned}$$...I think.
There's also $$\displaystyle \int_0^\pi\dfrac{\sin^{2022}x}{\sin^{2022}x + \cos^{2022}x}\, \mathrm dx + \displaystyle \int_0^\pi\dfrac{\cos^{2022}x}{\sin^{2022}x + \cos^{2022}x}\, \mathrm dx = \displaystyle \int_0^\pi\, \mathrm dx = \pi$$
Perhaps a step in the right direction would be solving $\displaystyle \int_0^\pi\dfrac{\sin^{2022}x - \cos^{2022}x}{\sin^{2022}x + \cos^{2022}x}\, \mathrm dx$.
It's also interesting to note that $\displaystyle \int\dfrac{\sin x - \cos x}{(\sin x + \cos x)^2}\, \mathrm dx = \dfrac{1}{\sin x + \cos x}$ and $\displaystyle \int\dfrac{2}{(\sin x + \cos x)^2} = \dfrac{\sin x - \cos x}{\sin x + \cos x}$.
But that's all for now, this was a mess. As always, thanks for reading, (and even more so if you could help), and have a great tomorrow~
By the way, the choices were $32, 194, 200$ and $100$. WolframAlpha gives the result of $\dfrac{\pi^2}{4}$, so the answer is $2 \times 2^2 + 3 \times 4^3 = 200$.
| Continue with
\begin{align}
&\int_0^\pi\dfrac{x\sin^{2022}x}{\sin^{2022}x + \cos^{2022}x}\,dx
= \dfrac{\pi}{2}\int_0^\pi\dfrac{\sin^{2022}x}{\sin^{2022}x + \cos^{2022}x}\, dx\\
= &\ \dfrac{\pi}{2}\int_0^\frac\pi2\dfrac{\sin^{2022}x}{\sin^{2022}x + \cos^{2022}x}\, dx
+ \frac{\pi}{2}\int^\pi_\frac\pi2\dfrac{\sin^{2022}x}{\sin^{2022}x + \cos^{2022}x}\, \overset{x\to \frac\pi2+x}{dx}\\
= &\ \dfrac{\pi}{2}\int_0^\frac\pi2\dfrac{\sin^{2022}x}{\sin^{2022}x + \cos^{2022}x}\, dx
+\frac{\pi}{2}\int_0^\frac\pi2\dfrac{\cos^{2022}x}{\sin^{2022}x + \cos^{2022}x}\, dx\\
= &\>\frac\pi2\int_0^\frac\pi2 dx=\frac{\pi^2}4
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4447417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all integer values of $m$ such that the equations $x+y+\sqrt{4-x^2}+\sqrt{4-y^2}=4$ and $x\sqrt{4-x^2}+y\sqrt{4-y^2}=m+2$ have a solution.
Consider the system of equation $\left\{ \begin{aligned} x + y + \sqrt{4 - x^2} + \sqrt{4 - y^2} &= 4\\ x\sqrt{4 - x^2} + y\sqrt{4 - y^2} &= m + 2 \end{aligned} \right. (x; y \in \mathbb R)$. How many integer values of $m$ are there such that there exists a solution to the above system?
[For context, this question is taken from an exam whose format consists of 50 multiple-choice questions with a time limit of 90 minutes. Calculators are the only electronic device allowed in the testing room. (You know those scientific calculators sold at stationery stores and sometimes bookstores? They are the goods.) I need a solution that works within these constraints. Thanks for your cooperation, as always. (Do I need to sound this professional?)
By the way, if the wording of the problem sounds rough, sorry for that. I'm not an expert at translating documents.]
Of course, it's definitely not $$x\sqrt{4 - x^2} + y\sqrt{4 - y^2} \le \dfrac{(x + \sqrt{4 - x})^2 + (y + \sqrt{4 - y^2})^2}{2}$$, nor is it $$\left(x\sqrt{4 - x^2} + y\sqrt{4 - y^2}\right)^2 \le (x^2 + y^2)[(4 - x^2) + (4 - y^2)]$$
Let's just try to solve this the typical way.
Consider the function $f(z) = z + \sqrt{4 - z^2}$. The domain and codomain of the function are respectively $[-2; 2]$ and $[-2; 2\sqrt 2]$.
(This is an unnecessary step, but I'm going to include it anyway.
Since $f(x) + f(y) = 4$, this means $$f(x), f(y) \in [4 - 2\sqrt 2; 2\sqrt 2] \iff x, y \in \left[2 - \sqrt 2 - 2\sqrt{\sqrt 2 - 1}; 2\right]$$ This is just to further strengthen the domains of $f(x)$ and $f(y)$ specifically. Strengthen's definitely not the right word.)
Consider function $g(z) = z\sqrt{4 - z^2}$. For $\displaystyle x, y \in \left[2 - \sqrt 2 - 2\sqrt{\sqrt 2 - 1}; 2\right]$, the codomain of $g(x)$ and $g(y)$ are $[10 - 8\sqrt{2}; 2]$. Therefore, $g(x) + g(y) \in [20 - 16\sqrt{2}; 4]$, right?
If only it was that easy.
How about this? Let $\left\{ \begin{aligned} x = 2\cos a, \sqrt{4 - x^2} &= 2\sin a\\ y = 2\cos b, \sqrt{4 - y^2} &= 2\sin b \end{aligned} \right.$, (since $x^2 + (\sqrt{4 - x^2})^2 = y^2 + (\sqrt{4 - y^2})^2 = 4$). Afterwards, the system of equation could be rewritten as $$\left\{ \begin{aligned} \sin\left(\dfrac{a + b}{2} + \dfrac{\pi}{4}\right)\cos\left(\dfrac{a - b}{2}\right) &= \dfrac{\sqrt 2}{2} &&(1)\\ \sin(a + b)\cos(a - b) &= \dfrac{m + 2}{4} &&(2) \end{aligned} \right.$$
Combining $(1)$ and $(2)$, we have $\sin(a + b)\left[\dfrac{1}{\sin^2\left(\dfrac{a + b}{2} + \dfrac{\pi}{4}\right)} - 1\right] = \dfrac{m + 2}{4}$.
Let $\dfrac{a + b}{2} = z$ and function $$f(z) = 2\sin z\cos z\left[\dfrac{2}{(\sin z + \cos z)^2} - 1\right] \implies f'(z) = 2 \times \left[\dfrac{2(\cos z - \sin z)}{(\cos z + \sin z)^3} - \cos 2z\right]$$
How would one go about actually solving $f'(z) = 0$?
Hmmm~ well, it's obvious that $\cos z - \sin z = 0 \iff z = n\pi + \dfrac{\pi}{4}, \forall n \in \mathbb Z$ is one of the subset of roots.
Actually, let's graph the function.
(seeing that the $y$-intercept extends to $-\infty$) How about we don't graph the function?
Aside from the apparent fact that $z\sqrt{4 - z^2} = \dfrac{(z + \sqrt{4 - z^2})^2 - 4}{2}$, I don't know what to do next. Also, it's past midnight already. So as always, thanks for reading, (and even so if you could help), and have a great tomorrow, everyone~
By the way, the options were $1, 2, 3$ and $4$, and there are seven integers between $20 - 16\sqrt{2}$ and $4$, so my first attempt would have led me to a dead end.
| If you solve the system as if $\sqrt{4-x^2}$ and $\sqrt{4-y^2}$ were the variables in a system of two linear equations, then you get:
$$\sqrt{4-x^2} = \frac{-4y + xy + y^2 + m + 2}{x - y}$$
$$\sqrt{4-y^2} = \frac{4x - x^2 - xy - m - 2}{x - y}$$
Note that $\sqrt{4-x^2} + \sqrt{4-y^2}$ simplifies to just $x + y + 4$, so the first equation gives $2x + 2y + 4 = 4$, or $y = -x$. Substituting this into the second equation gives:
$$x \sqrt{4-x^2} - x \sqrt{4-x^2} = m + 2$$
$$0 = m + 2$$
$$m = -2$$
So, we now have a unique solution for $m$. We just need to verify that there exists at least one real solution to $x \sqrt{4-x²} + y \sqrt{4-y²} = 0$. And clearly, anything with $y = -x$, $|x| \le 2$, and $|y| \le 2$ will work.
Therefore, a solution exists with $m = -2$.
I don't know what you did all that trig for ;-)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $ \int \frac{dx}{(x^2 + a^2)^3} $ I would like to find the anti-derivative
$$\displaystyle \int \dfrac{dx}{(x^2 + a^2)^3 }$$
My attempt:
By substitution: $ x = a \tan(\theta) \Rightarrow dx = a \sec^2(\theta) d\theta$
Then the integral becomes
$$\displaystyle \int \dfrac{ a \sec^2(\theta) d\theta }{a^6 \sec^6(\theta) } $$
And this reduces to
$$\dfrac{1}{a^5} \displaystyle \int \cos^4(\theta) d\theta $$
Now, $\cos^4(\theta) = ( \cos^2(\theta) )^2 = \dfrac{1}{4} (1 + \cos(2 \theta) )^2 = \dfrac{1}{4} ( 1 + 2 \cos(2 \theta) + \cos^2(2 \theta) ) $
So this reduces to
$$ \cos^4(\theta) = \dfrac{1}{4} + \dfrac{1}{2} \cos(2 \theta) + \dfrac{1}{8} (1 + \cos(4 \theta) ) $$
and these are integrable easily, but I want to relate their integrals to the original variable $x$.
So, now I have the anti-derivative as
$$ \dfrac{3}{8} \theta + \dfrac{1}{4} \sin(2 \theta) + \dfrac{1}{32} \sin(4 \theta) $$
Since $ x = a \tan(\theta) $, then $ \cos(\theta) = \dfrac{a}{\sqrt{x^2 + a^2}}$, and $\sin(\theta) = \dfrac{x}{\sqrt{x^2 + a^2}} $
Hence
$$\sin(2 \theta) = \dfrac{ 2 a x }{x^2 + a^2} $$
$$\cos(2 \theta) = \dfrac{a^2 - x^2}{x^2 + a^2} $$
And finally,
$$\sin(4 \theta) = \dfrac{4 a x (a^2 - x^2) }{ (x^2 + a^2)^2} $$
Hence, my integral becomes
$$ \displaystyle \int \dfrac{dx}{(x^2 + a^2)^3} = \dfrac{1}{a^5}\left( \dfrac{3}{8} \tan^{-1}\left(\dfrac{x}{a}\right) + \dfrac{1}{2} \dfrac{a x }{x^2 + a^2} + \dfrac{1}{8} \dfrac{ a x (a^2 - x^2) }{ (x^2 + a^2)^2 } \right) + C $$
where $C$ is an arbitrary constant of integration.
And my question is: Are my approach and derivation correct?
| Furthermore, I am going to find the integral in general
$$
I_{n}=\int \frac{1}{\left(x^{2}+a^{2}\right)^{n}} d x
$$
by a reduction formula by noting that
$$
\begin{aligned}
\frac{d}{d x}\left[\frac{x}{\left(x^{2}+a^{2}\right)^{n}}\right] &=\frac{a^{2}+(1-2 n) x^{2}}{\left(x^{2}+a^{2}\right)^{n+1}} \\
&=\frac{1}{\left(x^{2}+a^{2}\right)^{n}}-2 n \cdot \frac{x^{2}+a^{2}-a^{2}}{\left(x^{2}+a^{2}\right)^{n+1}} \\
&=\frac{1}{\left(x^{2}+a^{2}\right)^{n}}-2 n\left[\frac{1}{\left(x^{2}+a^{2}\right)^{n}}-\frac{a^{2}}{\left(x^{2}+a^{2}\right)^{n+1}}\right]
\end{aligned}
$$
Integrating both sides w.r.t. $x$ yields $$
\boxed{I_{n+1}=\frac{x}{2 n a^{2}\left(x^{2}+a^{2}\right)^{n}}+\frac{2 n-1}{2 n a^{2}} I_{n}}
$$
In particular, when $n=2$,
$$
\begin{aligned}
I_{3} &=\frac{x}{4 a^{2}\left(x^{2}+a^{2}\right)^{2}}+\frac{3}{4 a^{2}} I_{2} \\
&=\frac{x}{4 a^{2}\left(x^{2}+a^{2}\right)^{2}}+\frac{3}{4 a^{2}}\left[\frac{x}{2 a^{2}\left(x^{2}+a^{2}\right)}+\frac{1}{2 a^{2}} I_{1}\right] \\
&=\boxed{\frac{x}{4 a^{2}\left(x^{2}+a^{2}\right)^{2}}+\frac{3 x}{8 a^{4}\left(x^{2}+a^{2}\right)}+\frac{3}{8 a^{5}} \tan ^{-1} \frac{x}{a}+C}
\end{aligned}
$$
| {
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"url": "https://math.stackexchange.com/questions/4452736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $\sqrt{x-1} + \sqrt{y-1} \le xy, x \ge 1, y \ge 1$ Let $x$ and $y$ be real numbers, such that $x \ge 1$ and $y \ge 1$.
Prove that this inequality is true: $\sqrt{x-1} + \sqrt{y-1} \le xy$
Can someone show me steps to solve it.
PS:I need to give steps on how to solve it.
| Using Cauchy Schwarz and AM-GM
$$\sqrt{x-1}+\sqrt{y-1} \le \sqrt{2(x+y-2)} \le \frac{2(x+y-2)+1}{2} \le x+y-1 $$
Now it suffices to prove that
$$x+y-1 \le xy$$
Wich is equivalent to $$(x-1)(y-1) \ge 0$$
So we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4455028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Expand [a+b*log(1-x)]^{-1} as power seires in x Consider the function $[a+b\ln(1-x)]^{-1}$ near $x=0$, where $a$ and $b$ are constants.
As well-known, the Taylor series of $\ln(1-x)$ is $\ln(1-x)=-x-\frac{1}{2}x^2-\frac{1}{3}x^3-\frac{1}{4}x^4+...$. To expand the above function in Taylor series, can we just plug the expansion of $\ln(1-x)$ to get:
$$\left[a+b(-x-\frac{1}{2}x^2-\frac{1}{3}x^3+...)\right]^{-1}=a^{-1}\left[1+\frac{b}{a}(-x-\frac{1}{2}x^2-\frac{1}{3}x^3+...)\right]^{-1},$$
and then use the Taylor series of $(1+x)^{-1}=1-x+x^2-x^3+...$?
That is,
$$\left[a+b(-x-\frac{1}{2}x^2-\frac{1}{3}x^3+...)\right]^{-1}=a^{-1}\left[1+\frac{b}{a}(-x-\frac{1}{2}x^2-\frac{1}{3}x^3+...)\right]^{-1}=a^{-1}\left[1-\frac{b}{a}(-x-\frac{1}{2}x^2-\frac{1}{3}x^3+...)+(\frac{b}{a})^2(-x-\frac{1}{2}x^2-\frac{1}{3}x^3+...)^2-...\right].$$
To keep terms up to $x^2$, it produces $a^{-1}\left[1-\frac{b}{a}(-x-\frac{1}{2}x^2)+(\frac{b}{a})^2x^2-...\right]=a^{-1}\left[1+\frac{b}{a}x+(\frac{b}{2a}+\frac{b^2}{a^2})x^2+O(x^3)\right]$. Is it correct?
Thank you in advance?
| Looks good to me. Since we can let $r = b/a$ and write $$f(x;a,b) = (a + b \log (1-x))^{-1} = \frac{1}{a} (1 + r \log (1-x))^{-1},$$ the series expansion can be written more compactly in terms of $r$; e.g.,
$$f(x;a,b) = \frac{1}{a} \sum_{k=0}^\infty \frac{c_k}{k!} x^k,$$ where $$\begin{array}{c|c}
k & c_k \\
\hline
0 & 1 \\
1 & r \\
2 & r(1 + 2r) \\
3 & 2r(1 + 3r + 3r^2) \\
4 & 2r(3 + 11r + 18r^2 + 12r^3) \\
5 & 2r(12 + 50r + 105r^2 + 120r^3 + 60r^4) \\
6 & 2r(60 + 274r + 675r^2 + 1020r^3 + 900r^5 + 360r^6) \\
\vdots & \vdots
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4455301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral of $\frac1{|x|^3}$ on a circular segment How can I evaluate
$$\int \int _C |(x,y)|^{-3} dx dy$$
Where $C$ is the part between the chord $AB$ and the arc $AB$, and $|(x,y)| = \sqrt{x^2 + y^2}$? The radius of the circle is $R$.
I tried using polar coordinates, but I can’t find a way to represent $C$ with polar coordinates.
| The function over which you are integrating is rotationally symmetric around the origin $(0,0)$, thus we can rotate the coordinate system in such a ways that $A$, $B$ are at $\pm15^\circ$ from an axis and, say, $x=\mathrm{const}$ for points on $\gamma$. One such rotation is such that
$$\begin{align}
A &= (\cos 15^\circ, \sin 15^\circ) = (\cos(\pi/12), \sin(\pi/12)) \\
&= \tfrac14 (\sqrt6+\sqrt2, \sqrt6-\sqrt2) =:(c,s)\\
B &= (\cos 15^\circ, -\sin 15^\circ) = (c,-s)
\end{align}$$
and all points in the chord have $x=\cos15^\circ = c$. Then
$$\begin{align}
\iint_C |(x,y)|^{-3}dxdy
&= \int_{+\sin(\pi/12)}^{-\sin(\pi/12)} \frac{dy}{(y^2+\cos^2(\pi/12))^{3/2}} \\
&= \int_{+s}^{-s} \frac{dy}{(y^2+c^2)^{3/2}} \\
&= \frac{y}{c^2(y^2+c^2)^{1/2}} \Big|_{y=s}^{y=-s}\\
&= \frac{-2s}{c^2(s^2+c^2)^{1/2}} = -2\frac{s}{c^2} \\
&= -2 \frac{\sqrt6-\sqrt2}{2+\sqrt3}
\end{align}$$
where in the second-last line we use $s^2+c^2=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4459059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Evaluating $\int_{0}^{\frac{\pi}{2}} x^{2} \ln (\cos x) d x$ Latest Edit
As suggested by Mr Claude Leibovici, I go further to investigate the integral in general,
$$I_{m}:=\int_{0}^{\frac{\pi}{2}} x^{m} \ln (\cos x) d x$$
By the Fourier Series of ln(cos x)
,
$ \displaystyle \ln (\cos x)=-\ln 2+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \cos (2 n x) \tag*{(*)} $
Multiplying (*) by $ x^m$ and then Integrating both sides from $0 $ to $\frac{\pi}{2}$ yields
$$\begin{aligned}
I_m&=-\int_{0}^{\frac{\pi}{2}} x^{m} \ln 2 d x+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n !} \int_{0}^{\frac{\pi}{2}} x^{m} \cos (2 n x) d x\\&= -\frac{\pi^{m+1}\ln 2}{(m+1) 2^{m+1}}+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} J(m, n)\end{aligned}
$$
Next we need a reduction formula on $J(m,n)$.
Applying integration by parts twice, we obtain
$$
\begin{aligned}J(m, n)&=\frac{1}{2 n} \int_{0}^{\frac{\pi}{2}} x^{m} d(\sin 2 n x)\\ &= -\frac{m}{2 n} \int_{0}^{\frac{\pi}{2}} x^{m-1} \sin 2 n x d x\\&=\frac{m}{4 n^{2}} \int_{0}^{\frac{\pi}{2}} x^{m-1} d(\cos 2 n x)\\&=\frac{(-1)^{n} \pi^{m-1} m}{2^{m+1} n^{2}} -\frac{m(m-1)}{4 n^{2}} J (m-2, n)\end{aligned}
$$
By the reduction formula of $J(m,n)$, we can find $I_m$ by plugging $J(m,n)$.
By the Fourier Series of ln(cos x)
,
$ \displaystyle \ln (\cos x)=-\ln 2+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \cos (2 n x) \tag*{(*)} $
Multiplying (*) by $ x^2$ and then Integrating both sides from $0 $ to $\frac{\pi}{2}$ yields
$\displaystyle I=-\underbrace{\int_{0}^{\frac{\pi}{2}} x^{2} \ln 2 d x}_{\frac{\pi^{3} \ln 2}{24}}+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \underbrace{ \int_{0}^{\frac{\pi}{2}} x^{2} \cos (2 n x) d x}_{J_n}\tag*{} $
Integrating by parts twice yields
$\displaystyle \begin{aligned}J_{n} &=\frac{1}{2 n} \int_{0}^{\frac{\pi}{2}} x^{2} d(\sin 2 n x) \\&=\frac{1}{2 n}\left[x^{2} \sin 2 n x\right]_{0}^{\frac{\pi}{2}}-\frac{1}{n} \int_{0}^{\frac{\pi}{2}} x \sin 2 n x d x \\&=\frac{1}{2 n^{2}} \int_{0}^{\frac{\pi}{2}} x d(\cos 2 n x) \\&=\left[\frac{1}{2 n^{2}} x \cos 2 n x\right]_{0}^{\frac{\pi}{2}}-\frac{1}{2 n^{2}} \int_{0}^{\frac{\pi}{2}} \cos 2 n x d x \\&=\frac{\pi}{4 n^{2}} \cos n \pi-\frac{1}{2 n^{2}}\left[\frac{\sin 2 n x}{2 n}\right]_{0}^{\frac{\pi}{2}} \\&=\frac{\pi}{4 n^{2}} \cos n \pi\end{aligned}\tag*{} $
We can now conclude that
$\displaystyle \begin{aligned}I &=-\frac{\pi^{3} \ln 2}{24}+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \cdot \frac{\pi}{4 n^{2}} \cos n \pi \\&=-\frac{\pi^{3} \ln 2}{24}+\frac{\pi}{4} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(-1)^{n}}{n^{3}} \\&=-\frac{\pi^{3} \ln 2}{24}-\frac{\pi}{4} \sum_{n=1}^{\infty} \frac{1}{n^{3}} \\&=\boxed{-\frac{\pi^{3} \ln 2}{24}-\frac{\pi}{4} \zeta(3)}\end{aligned}\tag*{} $
Is there any other method to evaluate $I$?
| Hint for an alternative approach:
First, reduce the integral
$$
\int_0^{\pi/2} x^2\log(2\cos x)\,dx
= \frac{4\pi}7 \int_0^{\pi/2} x\log(2\cos x)\,dx
$$
and, then, show that
\begin{align}
&\int_0^{\pi/2} x\log(2\cos x)\,dx
= -\frac12\int_0^\infty \frac{\ln t\tan^{-1}t}{1+t^2} dt
=- \frac 7{16}\zeta(3)
\end{align}
| {
"language": "en",
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"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $y'$ for $y - x^{2}y^{2} - \cos(xy) = 4$ I want to evaluate the title's implicit differentiation. Here are the steps I've taken:
1. Derive both sides of the equation
$$\frac {d}{dx}(y - x^2y^2 - cos(xy)) = \frac {d}{dx}(4)$$
2. Note that the derivative of a constant is 0
$$\frac {d}{dx}(y - x^2y^2 - cos(xy)) = 0$$
3. Apply rule of difference
$$\frac {d}{dx}(y) - \frac {d}{dx}(x^2y^2) - \frac {d}{dx}(cos(xy)) = 0$$
4. Apply chain rule
$$\frac {d}{dx}(y) - \frac {d}{dx}(x^2y^2) - \left(\frac {d}{dx}cos(xy)*\frac {d}{dx}(xy)\right) = 0$$
5. Apply product rule
$$\frac {d}{dx}(y) - \left(\frac {d}{dx}(x^2)y^2 + x^2\frac {d}{dx}(y^2)\right) - \left(\frac {d}{dx}cos(xy)*\left(\frac {d}{dx}(x)y + x\frac {d}{dx} (y)\right)\right) = 0$$
6. Solve derivatives
$$\frac {d}{dx}(y) - \left(2xy^2 + 2x^2y\frac {dy}{dx}\right) - \left(-sin(xy)*\left(1y + x\frac {d}{dx} (y)\right)\right) = 0$$
7. Factorize and isolate the derivative of y wrt x
$$\frac {dy}{dx}(1-2x^2y+xsin(xy)) = 2xy^2 - ysin(xy) \Rightarrow$$
$$\frac {dy}{dx} = \frac {2xy^2 - ysin(xy)}{1+2x^2y+xsin(xy)}$$
Is this correct? And if so, is this the best approach? Also, how can I improve my mathematical dissertation?
| Hey I see your solution is almost correct, but you got the sign wrong. Here is the similar method.
*
*First, we derive the following equation via implicit differentiation.
\begin{equation*} \frac{d}{dx} \left( y-x^{2}y^{2} -\cos(xy) = 4 \right) = \frac{d}{dx}(4) \end{equation*}
*Second, we sum up the term differentiate and manager the constant factor.
\begin{equation*} \left( - \left( \frac{d}{dx} \cos(xy) \right) + \frac{d}{dx}(y)-\frac{d}{dx}(x^{2}y^{2}) \right) =\frac{d}{dx}(4) \end{equation*}
*We apply the chain rule, $\frac{d}{dx}(\cos(xy))=\frac{d\cos(u)}{du}\frac{du}{dx}$, where $u=xy$ and $\frac{d}{dx}(\cos(u))=-sin(u) $
, we get
\begin{align*} \frac{d}{dx}(y)-\frac{d}{dx}(x^{2}y^{2})+\frac{d}{dx}(xy)\sin(xy) &= \frac{d}{dx}(4) \\
-\left(\frac{d}{dx}(x^{2}y^{2} \right)+\sin(xy)\left(x\left(\frac{d}{dx}(y) \right) + \left( \frac{d}{dx}(x) \right) y \right)+y'(x) &= \frac{d}{dx}(4) \\
\end{align*}
*Use product rule, $\frac{d}{dx}(uv) = v\frac{du}{dx}+u\frac{dv}{dx}$, where $u=x^{2}$ and $v=y^{2}$:
\begin{align*} -\left( x^{2} \frac{d}{dx}(y^{2}) + y^{2}\frac{d}{dx}(x^{2})\right) + \sin(xy)\left(x\frac{d}{dx}(y)+y\frac{d}{dx}(x) \right) y'(x) &= \frac{d}{dx}(4) \end{align*}.
*After the whole substitution, we arrived at
\begin{equation*} (1+x\sin(xy)-2x^{2}y)y'(x) =-sin(xy)y+2xy^{2} \end{equation*}
*Divide both side by $-2x^{2}y+xsin(xy)+1 $
\begin{equation*} y'(x) = \frac{-sin(xy)y+2xy^{2}}{1+xsin(xy)-2x^{2}y} \end{equation*}
Well, just carefully with the sign!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4460899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Mertens' theorem with the numerator other than $1$ The Mertens' theorem is of the form
$$\prod_{p\le X}\left(1-\frac{1}{p}\right)=\frac{e^{-\gamma}}{\log X}\left\{1+o\left(\frac{1}{\log X}\right)\right\}.$$
I want to replace $1$ with others, for example $2$. At the first glance I think that (and I believe I have seen it somewhere)
$$\prod_{2\neq p\le X}\left(1-\frac{2}{p}\right)\sim\prod_{p\le X}\left(1-\frac{1}{p}\right)^2.$$
However, the following arguments (that I tried) give
$$\prod_{2\neq p\le X}\left(1-\dfrac{2}{p}\right)=\left(\dfrac{2e^{-\gamma}}{\log X}\right)^2\prod_{p>2}\left(1-\dfrac{1}{(p-1)^2}\right)\left\{1+o\left(\dfrac{1}{\log X}\right)\right\},$$
This suggest another constant in the middle above that I'm not sure I did it right. Here are the arguments I mentioned.
We first have for $p\neq 2$,
$$\log\left(1-\frac{2}{p}\right)=\log\left(1-\frac{1}{p}\right)^2-\log\left(1+\frac{1}{p(p-2)}\right).$$
Thus, by Mertens' theorem (with P.N.T. form) that
$$\sum_{p\le X}\log\left(1-\frac{1}{p}\right)^{-1}=\log\log X+\gamma+o\left(\frac{1}{\log X}\right),$$
\begin{align*}
\sum_{2< p\le X}\log\left(1-\frac{2}{p}\right) &= \sum_{2< p\le X}2\log\left(1-\frac{1}{p}\right)-\sum_{2< p\le X}\log\left(1+\frac{1}{p(p-2)}\right)\\
&= -2\log\log X-2\gamma+\log 4+o\left(\frac{1}{\log X}\right)+\sum_{2< p\le X}\log\left(1-\frac{1}{(p-1)^2}\right).
\end{align*}
Also, $$\sum_{2< p\le X}\log\left(1-\frac{1}{(p-1)^2}\right)=c+O\left(\sum_{p>X}\log\left(1-\frac{1}{(p-1)^2}\right)\right)=c+O(1/X),$$
where $c=\sum_{p>2}\log\Big(1-1/(p-1)^2\Big)$.
Therefore, we obtain
$$\prod_{2<p\le X}\left(1-\frac{2}{p}\right)=\left(\frac{2e^{-\gamma}}{\log X}\right)^2\prod_{p>2}\left(1-\frac{1}{(p-1)^2}\right)\left\{1+o\left(\frac{1}{\log X}\right)\right\},$$
as desired.
| That first asymptotic relation is incorrect.
For $p \geq 3$,
$$
\frac{(1-1/p)^2}{1-2/p} = \frac{1-2/p+ 1/p^2}{1-2/p} = 1 + \frac{1}{p(p-2)}
$$
and $\prod_{p\geq 3} (1 + \frac{1}{p(p-2)})$ converges by an argument similar to convergence of $\prod_{p} (1 + \frac{1}{p^2})$. So
$$
\prod_{3 \leq p \leq X} \left(1 - \frac{1}{p}\right)^2 \sim
c\prod_{3 \leq p \leq X}\left(1 - \frac{2}{p}\right)
$$
for the constant $c = \prod_{p \geq 3} (1 + \frac{1}{p(p-2)}) > 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4464594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Proof $ \lim_{x \to 3} \frac{x - 3}{x^{2} - 5x + 6} = 1 $ using $ \epsilon-\delta $ definition I'm trying to prove that $ \lim_{x \to 3} \frac{x - 3}{x^{2} - 5x + 6} = 1 $ using $ \epsilon-\delta $ definition. Here is my proof so far:
My effort: Let $ \epsilon \gt 0 $, Choose $ \delta = min\{1, \}.$ Suppose $ 0 \lt \left|x - 3\right| < \delta.$
Check: $ \left|\frac{x-3}{x^2 - 5x + 6} - 1\right| = \left|\frac{x - 3}{\left(x - 2\right)\left(x - 3\right)} - 1\right| = \left|\frac{1}{x-2} - 1\right| = \left|\frac{1 - x + 2}{x - 2}\right| = \left|\frac{-x + 3}{x - 2}\right| = \frac{\left|x-3\right|}{\left|x-2\right|}.$
I want to find a lower bound $ M \gt 0$ such that $ \left|x - 2\right| \ge M $ so that I can write $ \frac{\left|x - 3\right|}{\left|x - 2\right|} \le \frac{\left|x -3\right|}{M}$.
I know that $ -1 \lt x-3 \lt 1 $ and therefore $ 0 \lt x - 2 \lt 2 $, but I can't choose $ 0 $ to be the lower bound because it needs to be positive. What can I do?
| Consider ${|x-3|}$ is at least less than ${\frac{1}{2}}$. Then we have the set ${|x-3|<\frac{1}{2}}$ and will further restrict it for smaller desired epsilons.
But if ${|x-3|<\frac{1}{2}}$, then we know that ${|x-2| > |x-3|}$. Hence we know that ${\frac{1}{|x-2|}<\frac{1}{|x-3|}<2}$.
As you've written the reduced equation is ${\frac{|x-3|}{|x-2|}=\frac{1}{|x-2|}|x-3|}$. Since ${\frac{1}{|x-2|}<2}$ we have ${\frac{1}{|x-2|}|x-3|<2|x-3|}$.
So if we can make ${2|x-3|<\epsilon}$ then we certainly have ${\frac{|x-3|}{|x-2|}<\epsilon}$. Hence, we need ${|x-3|<\frac{1}{2}\epsilon}$ to also be true.
Finally, taking the intersection of this with our initial condition, we have ${|x-3|<\frac{1}{2}}$ and ${|x-3|<\frac{1}{2}\epsilon}$ ${\implies |x-3|<min(\frac{1}{2}, \frac{1}{2}\epsilon)}$.
Let's go back to the original definition of the epsilon-delta limit. For any desired ${\epsilon > 0}$, we need to provide a bound on ${|x-3|}$ that causes the equation to be less than ${\epsilon}$ away from the limit of 1. This function would be as follows:
${\delta(\epsilon) = \begin{cases}
\frac{1}{2}\epsilon & 0\lt \epsilon \lt 1\\
\frac{1}{2} & \epsilon \geq 1 \\
\end{cases}
}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $\sqrt{x+3}+\sqrt{5-x}-2\sqrt{15+2x-x^2}=-4$
Solve the following equation: $\sqrt{x+3}+\sqrt{5-x}-2\sqrt{15+2x-x^2}=-4$
Since real solutions are to be found, the domain of $x$ is $[-3; 5]$.
I immediately found that $15+2x-x^2$ can be factored to $(5-x)(x+3)$, this gave:
$\sqrt{x+3}+\sqrt{5-x}-2\sqrt{(5-x)(x+3)}=-4$
Squaring both sides was too complicated so I tried to substitute $a=\sqrt{5-x}$ for $a \in [0; \sqrt{8}]$ and $b=\sqrt{x+3}$ for $b \in [0; \sqrt{8}]$, this gave the new multivariable equation: $a+b-2ab=-4$
Another thing I noticed was $a^2+b^2=8$, together with the above equation, I got this system of equations: $\left\{ \begin{array}{l}
a + b - 2ab = - 4\\
{a^2} + {b^2} = 8
\end{array} \right.$
. Solving this by elimination is quite difficult for me.
This problem needs to be solved using algebra so I wonder how do I continue with this or are there any better way to solve this algebraically?
| Squaring both sides works out not so bad. Move the big radical to the other side and square:
$$\sqrt{x+3}+\sqrt{5-x} = 2\sqrt{15+2x-x^2}-4$$
$$8 +2\sqrt{15+2x-x^2} = 4(15+2x-x^2)-16\sqrt{15+2x-x^2} +16$$
Let $k=\sqrt{15+2x-x^2}$ to get
$$8+2k = 4k^2-16k+16$$
$$2k^2-9k+4=0$$
$$(2k-1)(k-4)=0.$$
That gives you two quadratics to solve $k=4$ yields $x=1$. $2k=1$ yields two solutions which are extraneous.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4470136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Looking for a basis of a vector subspace Let $n \geq 4$ and $\mathbb{R_n}$[x] a vector space of all polynomials of degree n at most.
There is given a set:
$
U = \{p \in \mathbb{R_n}[x]; p(1) = p(-1), p''(0) = 2p(1)\}
$
Find some basis of a vector subspace U and deterime $dimU$
My attempted solution:
$p(x) = a_nx^n + \dots + a_1x + a_0$
$p'(x) = na_nx^{n-1} + \dots + 2a_2x +a_1$
$p''(x) = n(n-1)a_nx^{n-2} + (n-1)(n-2)a_{n-1}x^{n-3} + \dots + 6a_3x +2a_2$
$p(1) = a_n + \dots + a_0$, if $n = 2k, k \in \mathbb{N}$: $p(-1) = a_n - a_{n-1} + \dots -a_1
+a_0$
$\rightarrow 2(a_{n-1} + a_{n-3} + \dots a_1) = 0$
$p''(0) = 2a_2 = 2p(1) = 2(a_n + \dots a_0) \rightarrow a_n + \dots + a_3 + a_1 + a_0 =0$
$$
a_1= -a_{n-1}-a_{n-3}- \dots a_3
\\ a_0 = -a_n - \dots - a_3 -a_1
$$
$p(x) = a_nx^n + \dots + a_2x^2 - x(a_{n-1} + a_{n-3} + \dots a_3) - a_n - \dots - a_3 - a_1 =
\\=a_n(x^n-1) + a_{n-1}(x^{n-1}-x-1) + a_{n-2}(x^{n-2}-1)+ \dots + a_3(x^3-x-1) + a_2x^2 - a_1$
Not sure if the process is correct, because the polynomials by the coefficients, which should form a basis, seem different from the solution.
| You have not computed any basis. You have tried to eliminate the coefficients using the equations but only have done so for $a_{0}$ and not $a_{1}$. Note that you can actually eliminate both $a_{1}$ and $a_{0}$ and write your $p(x)$ in terms of $a_{n},...a_{2}$ .
You should proceed like this:-
Let $n$ be even . Let $n=2m$ , $m\geq 2 $
Then $p(1)=p(-1)\implies a_{0}+a_{1}+...+a_{n}=a_{0}-a_{1}+...-a_{n-1}+a_{n}$
Which gives $a_{1}+a_{3}+...+a_{2m-1}=0$
$p''(0)=2\cdot(\text{coefficient of}\, x^{2})=2a_{2}$ .
So $2a_{2}=2(a_{0}+a_{1}+...+a_{2m})\implies a_{2}=a_{0}-(a_{3}+a_{5}+...+a_{2m-1})+a_{2}+...+a_{2m}\\=a_{0}+a_{2}+...+a_{2m}$ .
So $a_{0}+a_{4}+a_{6}+...+a_{2m}=0$ . So $a_{0}=-(a_{4}+a_{6}+...+a_{2m})$
This means $p(x)=-(a_{4}+...+a_{2m})-(a_{3}+a_{5}+...+a_{2m-1})x+a_{2}x^{2}+a_{3}x^{3}+...+x^{2m}a_{2m}$
So $p(x)=a_{2}x^{2}+a_{3}(x^{3}-1)+a_{4}(x^{4}-1)+...+(x^{2m}-1)a_{2m}$.
So $U=\{ a_{2}x^{2}+a_{3}(x^{3}-x)+a_{4}(x^{4}-1)+a_{5}(x^{5}-x)+...+(x^{2m}-1)a_{2m}\,,a_{i}\in\Bbb{R}\}$
That is $U=\text{span}\{x^{2},x^{3}-x,x^{4}-1,...,x^{2m-1}-x,x^{2m}-1\}$
Now you need to extract a linearly independent set out of the above. Matrices and Row Reduced Echelon Form is the best way to go.
These above vectors represented as row vectors of an $(2m-1)\times (2m+1)$ matrix gives us the following :-
$$\begin{bmatrix} 1 & 0 & 0 &\cdots & 0 & 0 & -1 \\ 0 & 1 & 0 &\cdots & 0 & -1 &0 \\ \vdots &\vdots&\vdots &\cdots&\vdots&\vdots&\vdots\\ 0 & 0 & 0 &\cdots & 0 & -1 & 0 \\ 0 & 0 & 0&\cdots & 1 & 0 & 0 \end{bmatrix}$$ .
Now you can use induction to prove that this matrix has full rank of $2m-1$ (as is evident from the Row reduced form of the matrix which it is already in ) and this would imply that $\{x^{2m}-1,x^{2m-1}-x,...,x^{3}-x,x^{2}\}$ is a basis for $U$.
For the case $m=1$ we have that $p(1)=p(-1)\implies a_{0}+a_{1}+a_{2}=a_{0}-a_{1}+a_{2}\implies a_{1}=0 $ and $2a_{2}=2(a_{0}+a_{1}+a_{2})\implies a_{0}=0 $ and $U=\{a_{2}x^{2}\,,a_{2}\in\Bbb{R}\}=\text{span}\{x^{2}\}$ .
Now use the same method to conclude when $n$ is an odd number . First solve the case for $n=1$ and then for $n=2m-1\,m\geq 2 $.
| {
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"source": "stackexchange",
"question_score": "4",
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Why can we simplify $(x^2-4)/(x-2)$ to get $x+2,$ but what happens at $x=2$ when $y=(x^2-4)/(x-2)?$ When teaching algebra I have always taught that we can simplify $\dfrac{x^2-4}{x-2}$ by factorising the top line and "cancelling" the $(x-2)$ in the top and bottom because they are a factor of $1.$
But what about if $x=2\,?$ I have recently been sketching functions and was asked to sketch $\dfrac{f(x)}{g(x)}$ where $f(x)=x^2-4$ and $g(x)=x-2$ and the answer in the book has the line $y=x+2$ with the point where $x=2$ removed.
Are these questions subtly different or is the simplification technically incorrect? Does it have an implication that $x$ is not equal to $2?$
|
we can simplify $\dfrac{x^2-4}{x-2}$ by "cancelling" the $(x-2)$
Does it have an implication that $x$ is not equal to $2?$
Indeed, this simplification is valid only if $x$ doesn't equal $2:$
\begin{align}\frac{(x+2)(x-2)}{x-2}=x+2&\implies x\ne2.\end{align}
is the simplification technically incorrect?
It depends on the context. While $$\frac{(x+2)(x-2)}{x-2}\not\equiv x+2,$$ the simplification is correct when, for example, we are dealing with only negative values: \begin{align}x<0 &\implies \frac{(x+2)(x-2)}{x-2}=x+2.\end{align}
In general, $$y=\frac{h(x) \times g(x)}{g(x)} \\\iff \\y=h(x)\quad\text{and}\quad g(x) \ne0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $x^2 + 3x +1 = 0$. Is $x^{2048} + \dfrac{1}{x^{2048}}$ divisible by 3? Let $x^2 + 3x +1 = 0$. Solve for $x^{2048} + \dfrac{1}{x^{2048}}$. Is it divisible by 3?
$x^2 + 1 = -3x \Rightarrow x+ \dfrac{1}{x} = -3$
$x^2 + \dfrac{1}{x^2} = (x + \dfrac{1}{x})^2 - 2 = 9 -2 = 7$
$x^4 + \dfrac{1}{x^4} = (x^2 + \dfrac{1}{x^2})^2 - 2 = 49 - 2 = 47$
seeing the pattern, let $s_n = x^{2^n} + \dfrac{1}{x^{2^n}}$
I need $s_{11}$
$s_1 = -3$
$s_2 = 7$
$s_3 = 47$
$s_4 = 47^2 - 2 = 2207$
$\vdots$
$((2207^2 -2)^2 - 2)^2 -2)^2 ... - 2)$
quite big
But I realized I didn't have to simplify it, I just have to check if
$((((((2207^2 - 2)^2 -2)^2 -2)^2 -2)^2 -2)^2 - 2)^2 -2$ is divisible by 3
| Let $a_n = x^n + \dfrac{1}{x^n}\implies a_1 = 3, a_2 = 7, a_{n+1}= x^{n+1}+\dfrac{1}{x^{n+1}}= \left(x^n+\dfrac{1}{x^n}\right)\left(x+\dfrac{1}{x}\right) - \left(x^{n-1}+\dfrac{1}{x^{n-1}}\right)= 3a_n-a_{n-1}\implies a_{n+1} - 3a_n + a_{n-1} = 0$. From here you can use characteristic equation for difference equation and find the general term $a_n$ and from it you can find a closed form for $a_{2048}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
complex integral $\int_0^{2\pi} (\cos x)^n \;dx$ I tried to solve following integral $$\int_0^{2\pi} (\cos x)^n \;dx$$
by saying $z = e^{ix}$ we get $dx=\frac{-i \;dz}{z}$ and $\cos x = \frac{1}{2}\left( z + \frac{1}{z}\right)$
$$\int_0^{2\pi} (\cos x)^n \;dx = \frac{-i}{2^n}\oint_{|z|=1} \frac{1}{z}\left( z + \frac{1}{z}\right)^n \; dz = 2\pi i \left( \frac{-i}{2^n} \right) \cdot Res(f(z) = \frac{z^2 + 1} {z^{n+1}}; z_0 = 0)$$
$$= \frac{\pi}{2^{n-1} n!} \left[ z^2 + 1\right]^{(n)}(0)$$
if $n=0$ we get $\frac{\pi}{2^{-1}} = 2\pi$, if $n=1$ we get $0$, if $n=2$ we get $\frac{\pi}{2!} = \frac{\pi}{2}$ and for $n = 3$ and $n > 3$ we get $0$.
But if I use integral solver for $n=4$ the result is $\frac{3\pi}{4}$ and for $n=2$ we get $\pi$. Why residue theorem doesn't work here? Or have I done some stupid mistake somewhere and I can't spot it?
| You've just made a simple algebra mistake: notice that
$$\left(z + \frac1z\right)^n = \left(\frac{z^2 + 1}{z}\right)^n = \frac{(z^2 + 1)^n}{z^n}$$
so
$$\frac{-i}{2^n} \oint_{|z| = 1} \frac{1}{z}\left(z + \frac1z\right)^n dz = \frac{\pi}{2^{n-1}} \text{Res}_{z = 0}\left[\frac{(z^2 + 1)^n}{z^{n+1}}\right]$$
whereas your solution dropped the power of $n$ in the numerator.
Now, in order to efficiently find the necessary residue I would suggest considering a Laurent series: we can expand the power in the numerator using the binomial theorem:
$$(z^2 + 1)^n = \sum_{k = 0}^n \binom{n}{k} z^{2k}$$
so dividing by the denominator gives us our Laurent series:
$$\frac{(z^2 + 1)^n}{z^{n+1}} = \sum_{k = 0}^n \binom{n}{k} z^{2k-n-1}$$
and we just need to find the coefficient of $z^{-1}.$ This would require $2k = n,$ so for even $n$ we get a residue of $\binom{n}{n/2}$ and for odd $n$ we get no residue because there is no $z^{-1}$ term.
So, we get that the value of the desired integral is
$$\int_0^{2\pi} \cos^n(x) dx = \begin{cases}\frac{\pi\binom{n}{n/2}}{2^{n-1}} & n = 2k \\ 0 & n = 2k+1\end{cases}$$
for integers $n$ and $k.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4474046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $\mathbb{E}\left[X|Y=\frac{1}{4}\right]$. Let $X,Y$ be random variables with joint density given by
$$f(x,y)=\begin{cases}
\frac{3}{8}\left ( x+y^{2} \right ) & \text{ if } 0<x<2,\text{ }0<y<1 \\
0 & \text{ otherwise.}
\end{cases}$$
Find $\mathbb{E}\left[X|Y=\frac{1}{4}\right]$.
$\textbf{My attempt:}$
$$
\begin{align*}
\mathbb{E}\left[X|Y=\frac{1}{4}\right]
&=\int_{0}^{2}xf\left(x,\frac{1}{4}\right)dx\\
&=\frac{3}{8}\int_{0}^{2}x\left(x+\frac{1}{16}\right)dx\\
&=\frac{67}{64}.
\end{align*}$$
Why is this wrong?
| Basically, you forgot the definition of $p(x|y)$ and didn't divide by $p(y=\frac14)=\frac{51}{64}$. So the final answer should have been $\frac{67}{51}$.
Instead of $f(x,y)$, I will use the notation
$$
p(x,y)=\frac38 (x+y^2)
\mathbf{1}_{x\in(0,2)}
\mathbf{1}_{y\in (0,1)}.
$$
We have
$$
\begin{align}
p(y)
&=\int_{-\infty}^\infty p(x,y)dx\\
&=\int_{-\infty}^\infty
\frac38 (x+y^2)
\mathbf{1}_{x\in(0,2)}
\mathbf{1}_{y\in (0,1)}
dx\\
&=\int_0^2
\frac38 (x+y^2)
dx
\cdot \mathbf{1}_{y\in (0,1)}
\\
&=\frac34 (1+y^2)\mathbf{1}_{y\in (0,1)}\\
&>0.
\end{align}
$$
Recall the definitions $\mathbb{E}(X|Y=y)=\int xp(x|y)dx$, and $p(x,y)=p(x|y)p(y)$. The case where $p(y)>0$ enables us to write $p(x|y)=\frac{p(x,y)}{p(y)}$.
So we calculate $$
\begin{align}
\mathbb{E}\left(X|Y=\frac14\right)
&=\left.\int_{-\infty}^\infty xp(x|y)dx \right|_{y=1/4}\\
&=\left.\int_{-\infty}^\infty x\frac{p(x,y)}{p(y)}dx\right|_{y=1/4}\\
&=\left. \int_{-\infty}^\infty x \cdot \frac{
\frac38 (x+y^2)
\mathbf{1}_{x\in(0,2)}
\mathbf{1}_{y\in (0,1)}
}
{
\frac34 (1+y^2)
\mathbf{1}_{y\in (0,1)}
}
dx
\right|_{y=1/4}\\
&=\int_0^2 x\cdot \frac12 \frac{x+(1/4)^2}{1+(1/4)^2}dx \\
&=\frac{67}{51}.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4474965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the power series in powers of $x$ for the function Find the power series in powers of $x$ for the function
$$f(x) = \int_0^x \frac{t^2}{1-t^4} \,dt$$
Solution Verification Requesteed
$$f(x) = \int_0^x \frac{t^2}{1-t^4} \,dt \implies f'(x) = \frac{x^2}{1-x^4}$$
Now, we proceed with partial fraction decomposition. We obtain
$$f'(x) = \frac{x^2}{1-x^4} = - \frac{x^2}{x^4-1} = -x^2 \cdot \frac{1}{(x^2+1)(x+1)(-1)} = \frac{Ax+B}{x^2+1} + \frac{C}{x+1} + \frac{D}{x-1}$$
We do partial fractions and obtain $A=0, B= -\frac{1}{2}, C = \frac{1}{4}, D = - \frac{1}{4}$. Now, we see that
$$-x^2 \cdot \frac{1}{x^4-1} = - \frac{1}{2(x^2+1)} + \frac{1}{4(x+1)} - \frac{1}{4(x-1)}$$
$$f'(x) = \frac{x^2}{x^4-1} = \frac{1}{2(x^2+1)} - \frac{1}{4(x+1)} + \frac{1}{4(x-1)}$$
$$f'(x)= \frac{1}{2} \cdot \frac{1}{1-(-x^2)} - \frac{1}{4} \cdot \frac{1}{1-(-x)} + \frac{1}{4} \cdot \frac{1}{-(1-x)}$$
$$f'(x)= \frac{1}{2} \sum_{n=0}^\infty (-1)^nx^{2n} - \frac{1}{4} \sum_{n=0}^\infty (-1)^n x^n - \frac{1}{4} \sum_{n=0}^\infty x^n$$
We can integrate each sum term-by-term to obtain
$$f(x) = \frac{1}{2} \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1} - \frac{1}{4} \sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1} - \frac{1}{4} \sum_{n=0}^\infty \frac{x^{n+1}}{n+1} $$
$$f(x) = \frac{1}{2} \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1} - \frac{1}{4} \Big[ \sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1} + \sum_{n=0}^\infty \frac{x^{n+1}}{n+1}\Big]$$
$$f(x) = \frac{1}{2} \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1} + \Big[(x- \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots) + (x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4}) \Big]$$
$$f(x) = \frac{1}{2} \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1} + 2 \sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1}$$
I'd simplify it further into one series but that dang $(-1)^n$ in the first sum keeps things from being nice and intuitive.
UPDATE
$$f'(x) = x^2 \cdot \frac{1}{1-(x^4)} = x^2 \sum_{n=0}^\infty x^{4n} = \sum_{n=0}^\infty x^{4n+2}$$
$$\implies f(x) = \sum_{n=0}^\infty \frac{x^{4n+3}}{4n+3}$$ as Jack D'Aurizio has concluded.
| You can also proceed by identification:
$\begin{align}(1-x^4)f'(x)
&=(1-x^4)\sum\limits_{n=1}^\infty na_nx^{n-1}\\
&=\sum\limits_{n=1}^\infty na_nx^{n-1}-\sum\limits_{n=1}^\infty na_nx^{n+3}\\
&=\sum\limits_{n=0}^\infty (n+1)a_{n+1}x^n-\sum\limits_{n=4}^\infty (n-3)a_{n-3}x^n\\
&=a_1+2a_2x+3a_3x^2+4a_4x^3+\sum\limits_{n=4}^\infty \Big((n+1)a_{n+1}-(n-3)a_{n-3}\Big)x^n\\
&=x^2
\end{align}$
Leads to the system:
$\begin{cases}
a_1=0\\
a_2=0\\
a_3=\frac 13\\
a_4=0\\
na_n=(n-4)a_{n-4}&\forall n\ge 5\end{cases}$
Also $a_0=f(0)=0$
Therefore all coefficients are zero except for $i=4n+3$ which verifies the equation $i\times a_i=1$
(i.e. $3a_3=1$ propagates to $7a_7=(7-4)a_3=3a_3=1$ and so on).
$$f(x)=\sum\limits_{n=0}^\infty \frac{x^{4n+3}}{4n+3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4475334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
problem canceling in multiplication of fractions now that i have a grasp on formatting mathematical fractions and whatnot i present you with my latest confusion
$$3\frac15 \times 1\frac23 \times 2\frac 34$$
converted into improper fractions
$$\frac{16}5 \times \frac 53 \times \frac{11}4$$
this much i understand. the following bit is whats driving me insane
$$\frac41 \times \frac13 \times \frac{11}1$$
i simply dont get how it goes from the improper fractions to the next set of fractions. apparently its called canceling but i cant see any sense in it
| First, $\frac{16}5 \times \frac 53 \times \frac{11}4$ is the same as $\frac{16}1 \times \frac 13 \times \frac{11}4$, since we can cancel the $5$ from both numerator and denominator. And in turn, $\frac{16}1 \times \frac 13 \times \frac{11}4$ is the same as $\frac41 \times \frac13 \times \frac{11}1$, since we can cancel a factor of $4$ from both numerator and denominator.
It might be able to see the validity of these cancellations more easily if, instead of writing $\frac{16}5 \times \frac 53 \times \frac{11}4 = \frac{16}1 \times \frac 13 \times \frac{11}4 = \frac41 \times \frac13 \times \frac{11}1$, we instead write the equivalent $\frac{16\times5\times11}{5\times3\times4} = \frac{16\times1\times11}{1\times3\times4} = \frac{4\times1\times11}{1\times3\times1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Optimize $xyz$ where $x+y+z=1$ and $x^2+y^2+z^2=1$? Im trying to optimize
$f(x,y,z)=xyz$ restricted to $g(x,y,z)=x+y+z=1$ and $h(x,y,z)=x^2+y^2+z^2=1$.
$∇f=(yz,xz,xy)$, $∇g=(1,1,1)$ and $∇h=(2x,2y,2z)$.
I tried using the determinant $det(∇f,∇g,∇h)=yz(2z-2y)-xz(2z-2x)+xy(2y-2x)=0$ which I dont know what to do with and I cant simplify the determinant in a good way with row operations.
I also tried solving $z$ from $g=1$. $z=1-x-y$.
$f(x,y,1-x-y)=xy-x^2y-xy^2$ restricted to $h(x,y)=2x^2+2y^2-2x-2y+2xy+1$ with Lagrange multiplier but I made no progress there either as the partial derivatives got too messy.
| The intersection locus of the sphere $x^2 + y^2 + z^2 = 1$ and the plane $x + y + z = 1$ is a circle with center at $(\dfrac{1}{3}, \dfrac{1}{3}, \dfrac{1}{3})$ and radius $\sqrt{\dfrac{2}{3}}$ spanned by the orthogonal unit vectors
$u_1 = \dfrac{1}{\sqrt{2}} (1, -1, 0) $
and
$ u_2 = \dfrac{1}{\sqrt{6}} (1, 1, -2 ) $
Therefore, the circle parametrically is given by $q(t) = (x(t), y(t), z(t))$ where
$ x(t) = \dfrac{1}{3} + \dfrac{1}{\sqrt{3}} \cos(t) + \dfrac{1}{3} \sin(t) $
$ y(t) = \dfrac{1}{3} - \dfrac{1}{\sqrt{3}} \cos(t) + \dfrac{1}{3} \sin(t) $
$z(t) = \dfrac{1}{3} - \dfrac{2}{3} \sin(t) $
It follows that
$ x(t) y(t) = \bigg( \dfrac{1}{3} (1 + \sin(t) ) \bigg)^2 - \dfrac{1}{3} \cos^2(t) $
and this simplifies to
$ x(t) y(t) = - \dfrac{2}{9} + \dfrac{2}{9} \sin(t) + \dfrac{4}{9} \sin^2(t) $
Multiplying by $z(t)$ gives
$ f(t) = x(t) y(t) z(t) = \dfrac{2}{27} (-1 + \sin(t) + 2 \sin^2(t) ) ( 1 - 2 \sin(t) ) \\
= \dfrac{2}{27} (2 \sin(t) - 1)(\sin(t) + 1 )(1 - 2 \sin(t) ) $
Let $g(r) = (2 r - 1)(r + 1)(1 - 2r) = -(r + 1)(4 r^2 - 4 r + 1 ) = - (4 r^3 - 3 r + 1 ) $
$g'(r) = - (12 r^2 - 3 ) = -3 ( 4 r^2 - 1) $
Hence the the critical points are $ r = \pm \dfrac{1}{2} $. And we have a local minimum at $r = - \dfrac{1}{2} $ where $g(- \dfrac{1}{2}) = -2 $ and a local maximum at $r = \dfrac{1}{2} $ where $g(\dfrac{1}{2}) = 0 $. And we also have $g(-1) = 0 $ and $g(1) = -2 $
Based upon all that, we deduce that
The maximum is $0$ and the minimum is $ \dfrac{-4}{27} $
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Show that $(1 + x^2 y^2)^{-\frac{1}{x^2 + y^2}}$ is $1$ as $(x,y) \to(0,0)$ I want to show that:
$$\lim_{(x,y)\to(0,0)} (1 + x^2 y^2)^{-\frac{1}{x^2 + y^2}} = 1$$
Direct substitution doesn't work, because of the denominator of $\frac{1}{x^2 + y^2}$.
Usually these problems are solved by bounding the following expression:
$$\left|(1 + x^2 y^2)^{-\frac{1}{x^2 + y^2}} -1 \right| = \left| \frac{1 - (1 + x^2 y^2)^{\frac{1}{x^2 + y^2}} }{ (1 + x^2 y^2)^{\frac{1}{x^2 + y^2}}} \right|$$
I don't know how to continue.
| An answer without polar coordinates, using bounds as OP was interested in. The Bernoulli inequality yields $(1+x^2y^2)^{(x^2+y^2)^{-1}}\ge1+\frac{x^2y^2}{x^2+y^2}$ so we can straightaway bound: $$\left|\frac{1-(1+x^2y^2)^{(x^2+y^2)^{-1}}}{(1+x^2y^2)^{(x^2+y^2)^{-1}}}\right|\le\frac{|1-(1+x^2y^2)^{(x^2+y^2)^{-1}}|}{1+\frac{x^2y^2}{x^2+y^2}}$$
Similarly using a variant of the Bernoulli inequality ($(1+x)^r\le e^{xr}$ for $r\gt0$, $x\gt0$) and choosing $(x,y)$ in a neighbourhood of the origin so that the ratio $\frac{x^2y^2}{x^2+y^2}\lt1$: $$0\le\frac{(1+x^2y^2)^{(x^2+y^2)^{-1}}-1}{1+\frac{x^2y^2}{x^2+y^2}}\le\frac{e^{\frac{x^2y^2}{x^2+y^2}}-1}{1+\frac{x^2y^2}{x^2+y^2}}\le\frac{\frac{x^2y^2}{x^2+y^2}(e-1)}{1+\frac{x^2y^2}{x^2+y^2}}$$So we are now interested in showing that: $$\frac{x^2y^2}{x^2+y^2+x^2y^2}\overset{(x,y)\to0}{\longrightarrow}0$$
On any path of approach with $|x|\le|y|$ we have: $$\frac{x^2y^2}{x^2+y^2+x^2y^2}\le\frac{y^2}{1+1+y^2}\to0$$Symmetrically we get the same when approaching $|y|\le|x|$. The "sequence of approach" criterion of limits in this space concludes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4481514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Proving irreducibility for $\frac{a^2+b^2}{ac+bd}$
For positive integers $a,b,c,d$ satisfying $ad-bc=1,$ prove that $\frac{a^2+b^2}{ac+bd}$ is irreducible. That is, $\gcd(a^2+b^2, ac+bd)=1.$
I know how to prove $\frac{a+b}{c+d}$ is irreducible if $ad-bc$ with bezouts theorem, but am not sure how to prove $\frac{a^2+b^2}{ac+bd}$ is irreducible. I tried using proof by contradiction, writing $km=a^2+b^2$ and $kn=ac+bd$ and $\gcd(m,n)=1$ for $k>1.$ Then from the first equation I got $b=\sqrt{km-a^2}.$ Then plugging it into the second equation I got $kn=ac+(\sqrt{km-a^2})d.$ Then, $c=\tfrac{kn-d\sqrt{km-a^2}}{a}.$ So then $$ad-bc=ad-(\sqrt{km-a^2})\left(\frac{kn-d\sqrt{km-a^2}}{a}\right)=ad-\left(\frac{kn\sqrt{km-a^2}}{a}-\frac{dkm}{a}+ad\right).$$ Thus, $$ad-bc=\frac{dkm}{a}-\frac{kn\sqrt{km-a^2}}{a}=1.$$ This doesn't seem to lead anywhere useful, though. There are two many variables and there seems like no good way to prove that $k=1.$ Is there a better way then this? Possibly a way to prove irreducibility like proving $\frac{a+b}{c+d}$ is irreducible using bezouts theorem? Thanks in advance.
|
Prove that if $ad-bc=1$, $(a^2+b^2, ac+bd)=1.$
First, we can recall the Brahmagupta-Fibonacci identity.
$$ (ad-bc)^2+(ac+bd)^2=(a^2+b^2)(c^2+d^2). $$
Then, it's easy to find that:
$$ (ac+bd)^2+1=(a^2+b^2)(c^2+d^2). $$
So,
$$ \Big((ac+bd)^2, a^2+b^2\Big)=1. $$
Therefore,
$$ (a^2+b^2, ac+bd)=1. \blacksquare$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4484541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Conditional probability with 4 coins The question: A box contains four coins, two of which are fair, one double-headed (i.e., heads on both sides), and the third is biased in such a way that it comes up heads with probability 1/4. A coin is drawn at random from the box and flipped twice. If both flips result in heads, what is the probability that the coin drawn was double-headed?
From what I understand, there are 2 fair coins with a 1/2 chance to get heads, 1 coin that has 100% chance of getting heads, and one coin with a 1/4 chance to get heads.
Does this mean, for example, that the chance of getting heads in both flips from a fair coin is 1/8? Since there's a 2/4 (1/2) chance to pick one? I just cannot understand the formulation of the question but I assume we have to use Baye's theorem here? I would appreciate any help here! Thanks
| Probability of choosing double-hedaded is $\frac{1}{4}\times 1$
Probability of choosing normal one is $\frac{2}{4}\times \frac{1}{2}$
Probability of choosing biased, heads with probability 1/4 is $\frac{1}{4}\times \frac{1}{4}$
If the result is H
$$P=\frac{\frac{1}{4}\times 1}{\frac{1}{4}\times 1 + \frac{2}{4}\times \frac{1}{2} + \frac{1}{4}\times \frac{1}{4}}=\frac{4}{9}$$
If the result is HH
$$P=\frac{\frac{1}{4}\times 1 \times 1}{\frac{1}{4}\times 1 \times 1 + \frac{2}{4}\times \frac{1}{2}\times \frac{1}{2} + \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}}=\frac{16}{25}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to obtain $\sqrt{5}$ (without a calculator) using complex numbers & roots of unity? For context, here is the entire question:
15. (a) Find the fifth roots of unity in exponential form.$\hspace{14.5pt}$(b) Let $\alpha$ be the complex fifth root of unity with the smallest positive argument, and suppose that $u = \alpha + \alpha^4$ and $v = \alpha^2 + \alpha^3$,$\hspace{30pt}$(i) Find the values of $u + v$ and $u - v$.
Particularly, I am struggling with the second part of (b)(i) where the answer is $\sqrt{5}$ for $u-v$. However, I cannot fathom how this answer is obtained without using a calculator.
What I have done so far: $u-v = \alpha + \alpha^4 - \alpha^2 - \alpha^3$, then through substitution of $\alpha = e^{i(2\pi/5)}$, I eventually obtained $u - v = 2\cos(2\pi/5) - 2\cos(4\pi/5)$.
How do I go further from here to calculate √5, or is it a distinct approach to the question entirely?
Any help greatly appreciated!
| $z^5 - 1 = 0\\
(z-1)(z^4+z^3+z^2 + z+1)=0$
$z^4+z^3+z^2 +z+ 1$ might not be immediately obvious how to factor, but it is a symmetric polynomial so we can do this.
$z^2(z^2 + \frac 1{z^2} + z+ \frac {1}{z} + 1)$
let $s = z+ \frac 1z$
$s^2 = z^2 + 2 + \frac {1}{z^2}$
$s^2 + s - 1 = 0$
$s = \frac {-1\pm\sqrt{5}}{2}$
There is a shortcut from here to the end, and the next few lines can be bypassed.
$z + \frac 1z = \frac {-1\pm\sqrt{5}}{2} $
$z^2 + \frac {1 \pm \sqrt 5}{2}z + 1 = 0$
$z = \frac {-1\pm\sqrt{5}}{4} \pm \sqrt{\frac {5 \pm \sqrt 5}{8}}i$
or $z = \cos(\frac {2n\pi}{5})+i\sin(\frac {2n\pi}{5})$
The root with the smallest argument is $a = \frac {-1 + \sqrt{5}}{4} + \sqrt{\frac {5 + \sqrt 5}{8}}i = \cos(\frac {2\pi}{5})+i\sin(\frac {2\pi}{5}) = e^{\frac {2\pi}5 i} $
$u = a+a^4 = e^{\frac {2\pi}5 i}+ e^{\frac {8\pi}5 i} = e^{\frac {2\pi}5 i}+e^{\frac {-2\pi}5 i} = 2\text{ Re} (a) = \frac{-1+\sqrt 5}{2}$
$v= a^2+a^3 = e^{\frac {4\pi}5 i}+ e^{\frac {6\pi}5 i} = e^{\frac {4\pi}5 i}+e^{\frac {-4\pi}5 i} = 2\text{ Re} (a^2) = \frac {-1-\sqrt 5}{2}$
Since we only needed the real part of $a$ we could save ourselves a trip through the quadratic formula to find the imaginary part of $a$
$u+v = -1$
$u-v = \sqrt 5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4490293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
If a particle travels 30 meters in any 3 seconds, does it necessarily travel 20 meters in any 2 seconds? Is it possible if its motion could've been discontinuous?
I'm trying to understand if there exists a function that has this property, but I chose to say it in terms of motion because it's easy to explain.
| It isn't true, not even for continuous motion.
For example, let $V(t)=10+\cos\left(\dfrac{2\pi}{3}t\right)$ in meters/sec.
Then
\begin{eqnarray}
D&=&\int_0^3V(t)\,dt\\
&=&\left[10t+\frac{3}{2\pi}\sin\left(\dfrac{2\pi}{3}t\right)\right]_0^3\\
&=& 30
\end{eqnarray}
but
\begin{eqnarray}
D&=&\int_0^2V(t)\,dt\\
&=&\left[10t+\frac{3}{2\pi}\sin\left(\dfrac{2\pi}{3}t\right)\right]_0^2\\
&=& 20+\frac{3}{2\pi}\sin\left(\dfrac{4\pi}{3}\right)\\
&=&20-\frac{3\sqrt{3}}{4\pi}
\end{eqnarray}
For any three second time interval the distance traveled will be 30 meters because of the period of the sinusoidal portion of the velocity.
ADDENDUM: In response to a comment from @TheRubberDuck
\begin{eqnarray}
D&=&\int_x^{x+3}V(t)\,dt\\
&=&\left[10t+\frac{3}{2\pi}\sin\left(\dfrac{2\pi}{3}t\right)\right]_x^{x+3}\\
&=& 10(x+3)+\frac{3}{2\pi}\sin\left(\dfrac{2\pi}{3}(x+3)\right)-
\left(10x+\frac{3}{2\pi}\sin\left(\dfrac{2\pi}{3}x\right)\right)\\
&=&30+\frac{3}{2\pi}\left[ \sin\left(\frac{2\pi}{3}x+2\pi \right)-\sin\left(\dfrac{2\pi}{3}x\right) \right]\\
&=&30
\end{eqnarray}
However, integrating between $x$ and $x+2$ gives a factor
\begin{eqnarray}
\sin\left(\frac{2\pi}{3}x+\frac{4\pi}{3} \right)-\sin\left(\dfrac{2\pi}{3}x\right)
\end{eqnarray}
which is not identically $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4493208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 1
} |
(Absolute) convergence of $\sum_{k=1}^\infty \frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k+1}}$ We want to check if the following series converges (absolutely).
$$\sum_{k=1}^\infty \frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k+1}}$$
This is what I have so far:
\begin{align}
\sum_{k=1}^\infty\frac{\sqrt{k+1}-\sqrt k}{\sqrt{k+1}}
&=\sum_{k=1}^\infty\frac{(\sqrt{k+1}-\sqrt k)(\sqrt{k+1}+\sqrt k)}{\sqrt{k+1}(\sqrt{k+1}+\sqrt k)}\\
&=\sum_{k=1}^\infty\frac1{\sqrt{k+1}(\sqrt{k+1}+\sqrt k)}\\
&= \sum_{k=1}^\infty \frac{1}{k+1+\sqrt{k^2+k}}
\end{align}
But how do we continue from here?
| Note that
$$
\forall k \in \Bbb N,\quad \sqrt{k+1}-\sqrt{k} = \sqrt{k+1}\left(1-\sqrt{1-\frac{1}{k+1}}\right)
$$
and since $\sqrt{1+x} \underset{x\to 0}{=} 1 +\frac{1}{2}x + o(x)$, it follows that the general term in the summation satisfies
$$
\frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k+1}} \underset{k\to\infty}{=} \frac{1}{2(k+1)} + o\left(\frac{1}{k+1}\right).
$$
The general term in the summation is thus equivalent to $\frac{1}{2(k+1)}$, from which we deduce that the sum is divergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
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