Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Every positive integer can be written as form of $ab-cd$?
Show that: every positive integer can be written as form of $ab-cd$,where $a-b=2(c-d),a,b,c,d\in Z$
for example
(1):$a=1,b=1,c=d=0$ then we have
$\color{red}1=1\cdot 1-0\cdot 0,$ such $a-b=2(c-d)$
(2):$a=4,b=2,c=3,d=2$ then we have
$\color{red}2=4\cdot 2-3\cdot 2,a-b=2(c-d)$
(3) $a=7,b=1,c=4,d=1$
$\color{red}3=7\cdot 1-4\cdot 1,a-b=2(c-d)$
form this by some manipulation and by comparison we can obtain the first implication i don't know how to proceed further。
| For every $n \in \Bbb N$ you can simply choose $a = 2n+1$, $c=n+1$
and $b = d = 1$.
Then $ab - cd = (2n+1) - (n+1) = n$ and $a-b = 2n = 2(c-d)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
solution for complex number equation with power i have the following equation $$z^3 = 2 + 2i$$
and I need to find the solutions of this equation in trigonometric form.
how do I find multiple solutions of such an equation?
| $$z^3=2+2i$$
$$z^3=|2+2i|e^{\arg(2+2i)i}$$
$$z^3=2\sqrt{2}e^{\tan^{-1}\left(\frac{2}{2}\right)i}$$
$$z^3=2\sqrt{2}e^{\tan^{-1}(1)i}$$
$$z^3=2\sqrt{2}e^{\frac{\pi}{4}i}$$
$$z=\left(2\sqrt{2}e^{\left(\frac{\pi}{4}+2\pi k\right)i}\right)^{\frac{1}{3}}$$
$$z=\sqrt{2}e^{\left(\frac{1}{12}\pi +\frac{2}{3}\pi k\right) i}$$
And k goes from 0-2 -> $k=0-2$ and k is the element of Z. So:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1280977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Simplify the expression (combination and factorial) Simplify the following expression:
$\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!}$
My attempt:
$\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n+1-3)!} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n-2)!} * \frac{(n-1)! + (n-2)!}{(n+1)!}$
and this is where I get stuck... How to continue?
When I put $\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!}$ into Wolfram Alpha it simplifies it into: $\frac{n}{6}$
When I put $\frac{(n+1)!}{3!(n+1-3)!} * \frac{(n-1)! + (n-2)!}{(n+1)!}$ into Wolfram Alpha it simplifies it into: $\frac{1}{6} * (n^{3} - n +1)$
| You must have made a typo with your second entrance into WA. To confirm your first answer:
\begin{align}\require{cancel}
\binom{n+1}{3}\cdot\frac{(n-1)!+(n-2)!}{(n+1)!}&= \frac{\cancel{(n+1)!}}{3!(n-2)!}\cdot\frac{(n-1)!+(n-2)!}{\cancel{(n+1)!}}\\[1em]
&= \frac{(n-1)!+(n-2)!}{3!(n-2)!}\\[1em]
&= \frac{\cancel{(n-2)!}\cdot[(n-1)+1]}{3!\cancel{(n-2)!}}\\[1em]
&= \frac{n}{3!}\\[1em]
&= \frac{n}{6}
\end{align}
This verifies your first simplification by WA. There must have been a typo in the second one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1283618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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A limit involving nested trigonometric functions and logarithms
Evaulate
$$ L = \lim_{x \to 0} \frac{1-\cos(\sin x)+\ln(\cos x)}{x^4}. $$
I can solve it using Maclaurin series, but I'm trying to figure out a way to the solution without using it. L'Hopital would probably work but needs to be applied 4 times, and the given function in numerator is too complicated for things to work out nicely on paper.
I think we can somehow use $\displaystyle\lim_{x\to0}\frac{1-\cos(x)}{x^2}=\frac12$ and $\displaystyle\lim_{x\to0}\frac{\ln(1+x)}{x}=1$.
With these, I can lower the degree of denominator to $2$, like this:
$$ L = \lim_{x \to 0}\frac{\frac{1 - \cos(\sin x)}{\sin^2 x}\frac{\sin^2 x}{x^2}+\frac{\ln(1+(\cos x-1))}{\cos x - 1}\frac{\cos x - 1}{x^2}}{x^2}. $$
This seems closer because left and right terms in the numerator are made of known limits, but the problem is that the whole expression is still $(1/2 - 1/2)/0 = 0/0$ so I can not break down $L$ into two limits and work things out.
I'm looking only for solutions involving manipulating limits in this way. No Maclaurin and no L'Hopital if it's gonna get too messy on paper.
| Instead of using the usual limits
$$
\lim_{y\rightarrow 0}\frac{1-\cos y}{y^{2}}=\frac{1}{2},\ \ \ \ \ \ \ \
\lim_{y\rightarrow 0}\frac{\log (1+y)}{y}=1.
$$
which lead to
$$
L=\lim_{x\rightarrow 0}\frac{\frac{1-\cos (\sin x)}{\sin ^{2}x}\frac{\sin
^{2}x}{x^{2}}+\frac{\ln (1+(\cos x-1))}{\cos x-1}\frac{\cos x-1}{x^{2}}}{%
x^{2}}.
$$
I will push one more order and use the (maybe less usual but easy to obtain) following limits
$$
\lim_{y\rightarrow 0}\frac{1-\frac{y^{2}}{2}-\cos y}{y^{4}}=-\frac{1}{24},\
\ \ \ \ and \ \ \ \ \ \ \
\lim_{y\rightarrow 0}\frac{\log (1-y)+y}{y^{2}}=-\frac{1}{2}.
$$
Each one of them maybe obtained by making use of L'Hospital's rule.
\begin{eqnarray*}
L &=&\lim_{x\rightarrow 0}\frac{1-\cos \left( \sin x\right) +\ln \left( \cos
x\right) }{x^{4}} \\
&=&\lim_{x\rightarrow 0}\frac{1-\frac{\sin ^{2}\left( x\right) }{2}-\cos
\left( \sin x\right) +\frac{\sin ^{2}\left( x\right) }{2}+\ln \left( \cos
x\right) }{x^{4}} \\
&=&\lim_{x\rightarrow 0}\left( \frac{1-\frac{\sin ^{2}\left( x\right) }{2}%
-\cos \left( \sin x\right) }{\sin ^{4}x}\right) \left( \frac{\sin x}{x}%
\right) ^{4}+\frac{\frac{\sin ^{2}\left( x\right) }{2}+\ln \left( \sqrt{%
1-\sin ^{2}x}\right) }{x^{4}} \\
&=&\lim_{x\rightarrow 0}\left( \frac{1-\frac{\sin ^{2}\left( x\right) }{2}%
-\cos \left( \sin x\right) }{\sin ^{4}x}\right) \left( \frac{\sin x}{x}%
\right) ^{4}+\frac{1}{2}\frac{\sin ^{2}x+\ln \left( 1-\sin ^{2}x\right) }{%
\left( \sin ^{2}x\right) ^{2}}\left( \frac{\sin x}{x}\right) ^{4} \\
&=&\lim_{y\rightarrow 0}\left( \frac{1-\frac{y}{2}-\cos \left( y\right) }{%
y^{4}}\right) \cdot \left( 1\right) ^{4}+\frac{1}{2}\lim_{y\rightarrow 0}%
\frac{y+\ln \left( 1-y\right) }{y^{2}}\cdot \left( 1\right) ^{4} \\
&=&-\frac{1}{24}+\frac{1}{2}\cdot \frac{-1}{2}=-\frac{7}{24}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1284141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $\sum_{k = 0}^{4} (1+x)^k = \sum_{k=1}^5 \binom{5}{k}x^{k-1}$ Question:
Show that:
$$\sum_{k = 0}^{4} (1+x)^k = \sum_{k=1}^5 \binom{5}{k}x^{k-1}$$
then go on to prove the general case that:
$$\sum_{k = 0}^{n-1} (1+x)^k = \sum_{k=1}^n \binom{n}{k}x^{k-1}$$
Attempted solution:
It might be doable to first prove the general case and then say that it is true for the specific case, but for the specific case I decided to write out the term and show that they were identical since they were so few.
For the LHS
$$\sum_{k = 0}^{4} (1+x)^k = (1+x)^0 + (1+x)^1 + (1+x)^2 + (1+x)^3 + (1+x)^4$$
$$ = (1) + (1 + x) + (x^2 +2x + 1) + (x^3 + 3x^2 + 3x + 1) + (x^4+4 x^3+6 x^2+4 x+1)$$
$$=5 + 10x + 10x^2 + 5x^3 + x^4$$
For the RHS:
$$\sum_{k=1}^5 \binom{5}{k}x^{k-1} = \binom{5}{1}x^{1-1} + \binom{5}{2}x^{2-1} + \binom{5}{3}x^{3-1} + \binom{5}{4}x^{4-1} + \binom{5}{5}x^{5-1}$$
$$= \frac{5!}{1!4!} x^0 + \frac{5!}{2!3!} x^1 + \frac{5!}{3!2!} x^2 + \frac{5!}{4!1!} x^3 + \frac{5!}{5!0!} x^4$$
$$ = 5 + 10x + 10x^2 + 5x^3 + x^4$$
That completes the first step of the question. So far so good.
For the general case, I started by using the binomial theorem for the binom and then writing out the inner-most sum:
$$\sum_{k = 0}^{n-1} (1+x)^k = \sum_{k = 0}^{n-1} \sum_{i = 0}^{k} \binom{k}{i}x^i = \sum_{k = 0}^{n-1} \left( \binom{k}{0}x^0 + \binom{k}{1}x^1 + ... + \binom{k}{k}x^{k}\right)$$
I can imagine that each step in the sum will decide the coefficients for the various powers of x and thus be identical to the RHS of the general case.
However, I run of out steam here and do not at the moment see any obvious way forward. What are some productive approaches for the general case? Am I doing things too complicated?
| For $\sum_{k = 0}^{n-1} (1+x)^k$ simplify with the geometric summation formula
For $\sum_{k=1}^n \binom{n}{k}x^{k-1}$ use the binomial theorem
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
solving $a = \sqrt{b + x} + \sqrt{c + x}$ for $x$ I'm trying to solve a very simple looking square root equation but nothing seems to work. The equation has this form (solve for $x$):
$$
a = \sqrt{b + x} + \sqrt{c + x}
$$
Squaring both sides obviously doesn't help since it will still give me a square root. Rearranging and then squaring doesn't help either.
The problem looks very simple to me but I have no idea on how to approach this.
| HINT
If we agree that $\alpha = \sqrt{b + x}$ and $\beta = \sqrt{c + x}$ we obtain that $\alpha + \beta = a$ and $\alpha^{2} -\beta^{2} = b - c$. Hence it can be claimed $(\alpha + \beta)(\alpha - \beta) = a(\alpha - \beta) = b-c$ which implies the following system of equations
\begin{align*}
\begin{cases}
\alpha - \beta = \displaystyle\frac{b-c}{a}\\
\alpha + \beta = a
\end{cases} \Rightarrow 2\alpha = a + \frac{b-c}{a} = \frac{a^{2} + b - c}{a}\Leftrightarrow \boxed{\alpha = \sqrt{b+x} = \frac{a^{2} + b - c}{2a}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
Formal power series coefficient problem Find the coefficient of:
$[x^{33}](x+x^3)(1+5x^6)^{-13}(1-8x^9)^{-37}$
I have figured out that I need to use this identity:
$(1-x)^{-k} = \sum\limits_{i>=0} \binom {n+k-1} {k-1} x^n $
But I have no clue how to proceed with this, I have been stuck with this for hours please help.
| You know that
$$(1+5x^6)^{-13}=\sum_{n\ge 0}\binom{n+12}{12}\left(-5x^6\right)^n=\sum_{n\ge 0}\binom{n+12}{12}(-1)^n5^nx^{6n}\tag{1}$$
and
$$(1-8x^9)^{-37}=\sum_{n\ge 0}\binom{n+36}{36}\left(8x^9\right)^n=\sum_{n\ge 0}\binom{n+36}{36}8^nx^{9n}\;.\tag{2}$$
The coefficient of $x^{33}$ in $(x+x^3)(1+5x^6)^{-13}(1-8x^9)^{-37}$ is the sum of the coefficients of $x^{33}$ in $x(1+5x^6)^{-13}(1-8x^9)^{-37}$ and $x^3(1+5x^6)^{-13}(1-8x^9)^{-37}$. Clearly
$$[x^{33}]x(1+5x^6)^{-13}(1-8x^9)^{-37}=[x^{32}](1+5x^6)^{-13}(1-8x^9)^{-37}$$
and
$$[x^{33}]x^3(1+5x^6)^{-13}(1-8x^9)^{-37}=[x^{30}](1+5x^6)^{-13}(1-8x^9)^{-37}\;,$$
so you need to find the coefficients of $x^{30}$ and $x^{32}$ in $(1+5x^6)^{-13}(1-8x^9)^{-37}$. If for convenience we write
$$(1+5x^6)^{-13}=\sum_{n\ge 0}a_nx^n\qquad\text{and}\qquad(1-8x^9)^{-37}=\sum_{n\ge 0}b_nx^n\;,$$
then we know that the coefficients of $30$ and $32$ in $(1+5x^6)^{-13}(1-8x^9)^{-37}$ are
$$\sum_{k=0}^{30}a_kb_{30-k}\qquad\text{and}\qquad\sum_{k=0}^{32}a_kb_{32-k}\;,\tag{3}$$
respectively. Now use $(1)$ and $(2)$ to evaluate these coefficients. Note that the only powers of $x$ with non-zero coefficients in $(1)$ are those whose exponents are multiples of $6$, and the only powers of $x$ with non-zero coefficients in $(2)$ are those whose exponents are multiples of $9$, so most of the terms in $(3)$ will be $0$. (In fact, with a little thought you can determine that one of the coefficients is $0$ without actually doing any arithmetic.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1289014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Convergence of $\sum_{n=0}^\infty \frac{\ln(1+2^n)}{n^2+x^{2n}}$ Let $x \in \mathbb{R}$. Define the series:
$$\sum_{n=0}^\infty \frac{\ln(1+2^n)}{n^2+x^{2n}}.$$
For what $x$ does it converge?
It clearly has positive terms. The ratio and root tests seem ineffective, and I am personally not able to study the confrontation with the harmonic series, that is, the limit: $$\lim_n n^\alpha \frac{\ln(1+2^n)}{n^2+x^{2n}}.$$ Can you give me a hand?
| For $|x|\le 1$ we have
$$
\frac{\ln(1+2^n)}{n^2+x^{2n}}>\frac{\ln2^n}{n^2+n^2}=\frac{\ln2}{2}\cdot\frac{1}{n}\quad \forall n\ge 1.
$$
Since the Harmonic series $\sum_{n=1}^\infty\frac{1}{n}$ is divergent, so is the series $\sum_{n=0}^\infty\frac{\ln(1+2^n)}{n^2+x^{2n}}$.
For $|x|>1$ we have:
$$
\frac{\ln(1+2^{n+1})}{(n+1)^2+x^{2n+2}}\cdot\frac{n^2+x^{2n}}{\ln(1+2^n)}=\frac{(n+1)\ln2+\ln(1+2^{-n-1})}{n\ln2+\ln(1+2^{-n})}\cdot\frac{n^2+x^{2n}}{(n+1)^2+x^{2n+2}},
$$
and therefore
$$
\lim_{n\to\infty}\frac{\ln(1+2^{n+1})}{(n+1)^2+x^{2n+2}}\cdot\frac{n^2+x^{2n}}{\ln(1+2^n)}=
x^{-2} \mbox{ if } |x|>1
$$
Thanks to the ratio test the series $\sum_{n=0}^\infty\frac{\ln(1+2^n)}{n^2+x^{2n}}$ converges provided $x^{-2}<1$, i.e. if $|x|>1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\alpha$ lies between $0$ and $4$.
Let $a,b,c$ be the length of the sides of the triangle $ABC$ .
Given $(a+b+c)(b+c-a)=\alpha bc$.Then Prove that the value of $\alpha$ lies in between $0$ and $4$.
$\begin{align}(a+b+c)(b+c-a)&=\alpha bc\\
\implies \alpha&=\dfrac{b^2+c^2-a^2}{bc}+2\\
\alpha&=2\cos A+2\\
\end{align}$
I also know a relation like $a+b>c\\b+c>a\\a+c>b$
I have studied maths up to $12th$ grade.
| We have
$$(b+c)^2-a^2 = \alpha bc \implies b^2+c^2-a^2 = bc(\alpha-2) \implies \alpha = 2 + \dfrac{b^2+c^2-a^2}{bc} = 2+2\cos(A)$$
Note that $\cos(A) \in [-1,1]$. Hence, we have
$$\alpha \in \left[2-2,2+2\right] = \left[0,4\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1291780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Need help with taylor series.
Evaluate the limit
$$\lim\limits_{x \to 1} \frac{1-x + \ln x}{1+ \cos πx}$$
The limit im trying to get is $-\frac{1}{π^2}$ as I've solved from l'Hopitals rule.
Now I need to solve the limit by using Taylor Series and this is what i did so far
$$\begin{align*}
f(x) &= 1-x + \ln x = 1 -x + (x-1) + \frac{1}{2} (x-1)^2 + \frac{1}{3} (x-1)^3 - \frac{1}{4} (x-1)^4 + \ldots \\
g(x) &= 1+\cos πx = 1+\left[ 1+\frac{1}{2!} (πx)^2 + \frac{1}{4!} (πx)^4 - \frac{1}{6!} (πx)^6 +\ldots \right] \\
\frac {f(x)}{g(x)} & = \frac{\frac{1}{2} (x-1)^2 + \frac{1}{3} (x-1)^3 - \frac{1}{4}(x-1)^4 + \ldots} {2-\frac{1}{2!} (πx)^2 + \frac{1}{4!} (πx)^4 - \frac{1}{6!} (πx)^6+\ldots}
\end{align*}$$
I have no idea where to go to solve for $-\frac{1}{π^2}$ now. Please help
| First, let $h=x-1$ then $\cos \pi x=\cos (\pi +\pi h)=-\cos (\pi h).$
Next re-write the original fraction as the following product
\begin{equation*}
\frac{1-x+\ln x}{1+\cos \pi x}=\frac{-h+\ln (1+h)}{1-\cos (\pi h)}=\frac{%
-h+\ln (1+h)}{h^{2}}\times \frac{(\pi h)^{2}}{(1-\cos (\pi h))}\times \frac{1%
}{\pi ^{2}}.
\end{equation*}
By Taylor expansion we can note that
\begin{align}
\cos \pi h& =1-\frac{(\pi h)^{2}}{2}+o(h^{3}) \notag \\
\log (1+h)& =h-\frac{h^{2}}{2}+o(h^{3}). \notag
\end{align}
then
\begin{equation*}
\lim_{h\rightarrow 0}\frac{-h+\log (1+h)}{h^{2}}=\lim_{h\rightarrow 0}(-%
\frac{1}{2}+o(h))=-\frac{1}{2}.
\end{equation*}
\begin{equation}
\lim_{h\rightarrow 0}\frac{1-\cos (\pi h)}{(\pi h)^{2}}=\lim_{h\rightarrow 0}%
\frac{\frac{(\pi h)^{2}}{2}+o(h^{3})}{(\pi h)^{2}}=\lim_{h\rightarrow 0}%
\frac{1}{2}+\frac{1}{\pi ^{2}}o(h)=\frac{1}{2}.
\end{equation}
Therefore
\begin{eqnarray*}
\lim_{x\rightarrow 1}\frac{1-x+\ln x}{1+\cos \pi x} &=&\lim_{h\rightarrow 0}%
\frac{-h+\ln (1+h)}{h^{2}}\cdot \frac{(\pi h)^{2}}{(1-\cos (\pi h))}\cdot
\frac{1}{\pi ^{2}}. \\
&=&\lim_{h\rightarrow 0}\frac{-h+\ln (1+h)}{h^{2}}\cdot \lim_{h\rightarrow 0}%
\frac{(\pi h)^{2}}{(1-\cos (\pi h))}\cdot \lim_{h\rightarrow 0}\frac{1}{\pi
^{2}}. \\
&=&-\frac{1}{2}\cdot \frac{2}{1}\cdot \frac{1}{\pi ^{2}} \\
&=&-\frac{1}{\pi ^{2}}.
\end{eqnarray*}
$\bf{EDIT:}$ You stack because you tried to deal with the whole expression at ONCE. My computations were done TWO times but on A SMALL piece at a time: Divide and conquer!
| {
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"url": "https://math.stackexchange.com/questions/1292372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Creative way to find this area Let's say We have a circle with center at $(0,0)$ with radius $r$
and we have the line $y=a$ where $0 \leq a \leq r$.
the question is what is the area that between the circle and the line $y=a$(the area that above the line).
illustration for $r=5$ and $a=4$: http://www.wolframalpha.com/input/?i=x%5E2%2By%5E2%3D25+and+y%3D4
We will end with the area as a function of $a$. So far I have tried to do integration to get the area ... it is not pretty at all.
| Using, geometry we have the equation of the circle centered at the origin as:
$$x^2+y^2=r^2$$
Now, substitute $y=a$ in the above equation
$$x^2+a^2=r^2 \implies x=\pm \sqrt{r^2-a^2}$$
Now, join the points of intersection $(\sqrt{r^2-a^2}, a),\,(-\sqrt{r^2-a^2}, a)$ to the origin, to get a sector with radius $r$ and an isosceles triangle with sides $r,\,r,\,2\sqrt{r^2-a^2}$
The aperture angle $\alpha$ of the sector is calculated as
$$\alpha=2\cos^{-1}\left(\frac{a}{r}\right)$$
Now, the area between the circle and the line is
$$
\begin{align}
&\frac{1}{2}\cdot\alpha r^2-\frac{1}{2}(a)(2\sqrt{r^2-a^2})\\
=\:&\frac{1}{2}\left(2\cos^{-1}\left(\frac{a}{r}\right)\right)r^2-\frac{1}{2}a\left(2\sqrt{r^2-a^2}\right)\\
=\:&r^2\cos^{-1}\left(\frac{a}{r}\right)-a\sqrt{r^2-a^2}
\end{align}
$$
Where, $0\leq a\leq r$
Hence, as you mentioned for $r=5,\,a=4$. substituting these values in the expression, we get the area:
$$
\begin{align}
&5^2\cos^{-1}\left(\frac{4}{5}\right)-4\sqrt{5^2-4^2}\\
=\:&25\cos^{-1}\left(\frac{4}{5}\right)-12\\
\approx\:&4.08752772 \space \text{unit}^2
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding $\int\frac{1}{x^{11}+4x^6}dx$ I wanted to find out if there is an easy way to evaluate $$\displaystyle\int\frac{1}{x^{11}+4x^6}dx.$$
I substituted $u=x^5$ and then used partial fractions, but maybe there is a simpler way to find this.
| $$\int\frac{1}{x^{11}+4x^6}dx=$$
$$\int \left(\frac{1}{4x^6}+\frac{x^4}{16(x^5+4)}-\frac{1}{16x}\right)dx=$$
$$\frac{1}{16}\int \frac{x^4}{x^5+4}dx-\frac{1}{16}\int \frac{1}{x}dx+\frac{1}{4}\int \frac{1}{x^6}dx=$$
$$\frac{1}{80}\int \frac{1}{u}du-\frac{1}{16}\int\frac{1}{x}dx+\frac{1}{4}\int\frac{1}{x^6}dx=$$
$$\frac{\ln(u)}{80}-\frac{1}{16}\int\frac{1}{x}dx+\frac{1}{4}\int\frac{1}{x^6}dx=$$
$$\frac{\ln(u)}{80}-\frac{\ln(x)}{16}+\frac{1}{4}\int\frac{1}{x^6}dx=$$
$$\frac{\ln(u)}{80}-\frac{1}{20x^5}-\frac{\ln(x)}{16}+C=$$
$$\frac{1}{80}\left(-\frac{4}{x^5}+\ln(x^5+4)-5\ln(x)\right)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Method to solve equation to find the value of each variable Ok lets say I am given a question:
There are 17 males students and 19 female students.
out of the 17 male students 11 of them do hip hop while the other 6 do drag racing
while in the female side 11 of them do hip hop while the other 8 drag racing.
out of the people who do hip hop 13 of them like to eat and 9 of them like to sleep
out of the people who do drag racing 5 of them like to eat and 9 of them like to sleep
so what is the number of males that like to hip hop and eat
below are the equation i have come up with
$$mhe+fhe+mhs+fhs+mde+fde+mds+fds = 36$$
$$mhe+mhs+mde+mds = 17$$
$$fhe+fhs+fde+fds = 19$$
$$fhe+fhs+mhe+mhs = 22$$
$$mde+mds+fde+fds = 14$$
$$mhe + fhe = 13$$
$$mhs + fhs = 9$$
$$mde +fde =5$$
$$mds+ fds = 9$$
| Assuming e.g. mhe represents a single variable (i.e., not $m \times h \times e$), we can perform Gauss-Jordan Elimination to obtain the reduced row echelon form:
$$\left(\begin{array}{cccccccc|c} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 36 \\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 17 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 19 \\ 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 22 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 14 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 13 \\ 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 9 \\ 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 5 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 9 \\ \end{array}\right) \xrightarrow{\text{Gauss-Jordan}} \left(\begin{array}{cccccccc|c} 1 & 0 & 0 & -1 & 0 & -1 & 0 & -1 & -6 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 19 \\ 0 & 0 & 1 &
1 & 0 & 0 & 0 & 0 & 9 \\ 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 5 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 9 \\ 0 & 0 &
0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 &
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array}\right)$$
So there are infinitely many solutions, given by
$$
(-6+\alpha+\beta+\gamma,19-\alpha-\beta-\gamma,9-\alpha,\alpha,5-\beta,\beta,9-\gamma,\gamma)
$$
where $\alpha,\beta,\gamma$ are arbitrary.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A not so hard basic calculus problem? But it appears to be very lengthy Find the coordinates of the two points on the curve $y=4-x^2$ whose tangents pass through the point $(-1,7)$.
My work:
Let the two points be $(a,b)$ and $(c,d)$. And $\frac{dy}{dx}=-2x$, so the gradients of the tangents are $-2a$ and $-2c$. We get $\frac{7-d}{-1-c}=-2c\implies 7-d=2c+2c^2$ and $\frac{7-b}{-1-a}=-2a\implies 7-b=2a+2a^2$. Now we have 2 equations and 4 unknowns. So we need two more equations.
We have the equations of the tangents are $y=-2cx+2c^2+d$ and $y=-2ax+2a^2+b$. The solutions of these two equations are $(-1,7)$, putting $(-1,7)$ in we get $7(a-c)=-2ac(a-c)+ad-bc$ and $2(a-c)=-2(a-c)(a+c)+d-b$.
We can also equate $4-x^2=-2cx+2c^2+d$ and $4-x^2=-2ax+2a^2+b$, then we get $a^2-c^2=d-b$. From here, we still have to substitute and do some algebra to get $(a,b)$ and $(c,d)$.
Is my working correct? But I feel like the working is too lengthy. Is there any faster way?
Many thanks!
| You made it complicated. Let the one point be $(a,\ 4 - a^2)$ and $\frac{\text{d}y}{\text{d}x} = −2x$, so the gradient of the tangent is $−2a$; we get $\frac{7 - (4 - a^2)}{(-1-a)} = -2a$, so you can get the answer easily: $(1,3)$, $(-3,-5)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove this is an isosceles triangle In a $\triangle ABC$, $$\qquad \sin B\cdot\sin C=\cos^2\left(\frac{A}{2}\right)$$
Prove that this is an isosceles triangle.
Can anyone guide me to prove this? Thanks!
| Note: in a triangle ABC, we have $A+B+C=\pi=180^o$
Given $$\sin B\sin C=\cos^2\frac{A}{2} \implies 2\sin B\sin C=2\cos^2\frac{A}{2} $$$$\implies \cos(B-C)-\cos(B+C)=2\cos^2\frac{A}{2} $$$$\implies \cos(B-C)-\cos(180^o-A)=2\cos^2\frac{A}{2} $$ $$\implies \cos(B-C)+\cos A=2\cos^2\frac{A}{2} $$$$\implies \cos(B-C)+2\cos^2\frac{A}{2}-1=2\cos^2\frac{A}{2}$$ $$\cos(B-C)=1 \implies B-C=0 \quad \text{or} \quad B=C$$ Hence, the triangle ABC is an isosceles triangle
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Question in regard to solving for inverse laplace transform I am having some confusion when it comes to solving for the inverse laplace transform. ( We are allowed the tables with the common values by the way).
Il give an example.
Take,
$$y''+4y'+8y=-6e^{-2t}cos(3t)$$ with $y(0)=-2$ and $y'(0)=-5$
Now, by invoking the linearity of the Laplace transform and using the tables, and letting $Y(s)=L[y]$
I get to the form,
$$(s^2+4s+8)Y(s)+2s+13=-6\frac{s+2}{(s+2)^2+3^2}$$
Now here is where I am having more trouble because I don't know the least messy way to get the RHS in such a form that I can use the tables to find the inverse transforms, using linearity of course.
I know I could get it in the form $$Y(s)=\frac{-6\frac{s+2}{(s+2)^2+3^2}-2s-13}{s^2+4s+8}$$ but that doesn't seem to be of any help.
I know also that we can usually use partial fractions, but how could I apply it in such a case when we have a rational in the numerator?
Does anyone have any hints and tricks to this type of thing?
Thanks!
| Your work is correct. Gather everything under common denominator:
$$\frac{-2 s^3-21 s^2-84 s-181}{\left(s^2+4 s+8\right) \left(s^2+4 s+13\right)}=\frac{A s+B}{s^2+4 s+8}+\frac{C s+D}{s^2+4 s+13}$$
This will be super messy, I will only calculate $A$ and $B$. Multiply everything by ${s^2+4 s+8}$
$$\frac{-2 s^3-21 s^2-84 s-181}{s^2+4 s+13}=(A s+B)+\frac{(s^2+4 s+8)(C s+D)}{s^2+4 s+13}$$
As usual, we want to set $s^2+4 s+8=0$. This makes the second term in $ RHS =0$. Inserting $s^2=-4s-8$, LHS becomes:
$$\frac{1}{5} \left(8 s^2+16 s-13\right)$$
Insert $s^2=-4s-8$ once more:
$$\frac{1}{5} (-16 s-77)=A\,s+B$$
You do the same for $s^2+4 s+13=0$ and you'll get:
$$\frac{6 s+12}{5 \left(s^2+4 s+13\right)}+\frac{-16 s-77}{5 \left(s^2+4 s+8\right)}=\frac{6 s+12}{5 \left((s+2)^2+3^2\right)}+\frac{-16 s-77}{5 \left((s+2)^2+2^2\right)}$$
we get:
$$\frac 6 5 \frac{s+2}{(s+2)^2+3^3}\rightarrow\frac 6 5 e^{-2t}\cos (3t)$$
and:
$$\frac{-16 s-77}{5 \left((s+2)^2+2^2\right)}=-\frac{16}{5}\frac{s+2}{(s+2)^2+4}-\frac{45}{10}\frac{2}{(s+2)^2+4}\\ \rightarrow-\frac{16}{5}e^{-2t}\cos (2t)-\frac{45}{10}e^{-2t}\sin (2t)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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taking the limit $\lim\limits_{n\rightarrow \infty} {\frac{(3^{n+1} + 4)(7^n-47)}{(7^{n+1}-47)(3^n +4)} }$ I need help with a guide on how i will deal with this kind of problem.. This a part of my solution in series convergence. I find it hard taking the limit of this: $$\lim_{n\rightarrow \infty} {\frac{(3^{n+1} + 4)(7^n-47)}{(7^{n+1}-47)(3^n +4)} }$$
| You can factor out the exponentials to get:
$$
\frac{3^{n + 1}7^n\left(1 + \frac{4}{3^{n + 1}}\right)\left(1 - \frac{47}{7^n}\right)}{7^{n+1}3^n\left(1 - \frac{47}{7^{n + 1}}\right)\left(1 + \frac{4}{3^n}\right)}
$$
Which leaves us with:
$$
\frac{3}{7}\cdot\frac{\left(1 + \frac{4}{3^{n + 1}}\right)\left(1 - \frac{47}{7^n}\right)}{\left(1 - \frac{47}{7^{n + 1}}\right)\left(1 + \frac{4}{3^n}\right)}
$$
Now when you take the limit as $n$ goes to $\infty$, you simply end up with $\frac{3}{7}\cdot\frac{1\cdot1}{1\cdot1} = \frac{3}{7}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Simple trigonometry equation The previous class we were doing trigonometry exercises. Before the class finished, our teacher wrote exercises on the table. I am stuck with the following one:
$$
\cos(2x) + 1 + 3\sin x = 0
$$
I have come up with this:
$$
1= \sin^2 x + \cos^2 x
$$
$$
\cos(2x) = \cos^2 x - \sin^2 x
$$
When we substitute we get
$$
2\cos^2 x + 3\sin x = 0
$$
Need to find $x\ldots$
| $$\cos(2x)+1+3\sin x=0$$
now let's substitute $\cos (2x)=1-2\sin^2x$
$2-2\sin^2x+3\sin x=0$
And then, if we denote $y=\sin x$ we get a quadratic equation
$$2-2y^2+3y=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1301550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Inequality and Trigonometric Substitution Prove that for all positive real $a,b,c$, we have
$$(a^2+2)(b^2+2)(c^2+2) \geq 9(ab+bc+ca).$$
Because of the term $a^2+2$, this motiveates me to substitute $a=\sqrt{2}\tan A, b=\sqrt{2}\tan B, c=\sqrt{2}\tan C$ and the fact that $1+\tan^2x=\sec^2x$, then the desired inequality becomes
$$2^2\sec^2 A\sec^2 B \sec^2 C\geq 3^2(\tan A\tan C + \tan B\tan C + \tan C \tan A).$$
But then I was stuck and couldn't move further, please helps.
| Ok,if you use trigonometry to solve it
as you
$$\Longleftrightarrow \cos{A}\cos{B}\cos{C}(\cos{A}\sin{B}\sin{C}+\sin{A}\cos{B}\sin{C}+\sin{A}\sin{B}\cos{C})\le\dfrac{4}{9}$$
$$\cos{A}\cos{B}\cos{C}(\cos{A}\cos{B}\cos{C}-\cos{(A+B+C)})\le\dfrac{4}{9}$$
By Jenson inequality we have
$$\cos{A}\cos{B}\cos{C}\le\cos^3{\dfrac{A+B+C}{3}}$$
so let $$\cos{\dfrac{A+B+C}{3}}=x,\Longrightarrow \cos{(A+B+C)}=4x^3-3x$$
$$\Longleftrightarrow x^4(1-x^2)\le \dfrac{4}{27},0<x<1$$
Use AM-GM inequality we have
$$x^4(1-x^2)=\dfrac{1}{2}x^2\cdot x^2\cdot (2-2x^2)\le\dfrac{4}{27}$$
PS:you can prove this stronger $$(a^2+2)(b^2+2)(c^2+2)\ge 3(a+b+c)^2\ge 9(ab+bc+ac)$$
see Showing $~\prod (a^{2}+2)\geq 9\sum ab$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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For which values of $a$ does the matrix can be diagnolized? Given $$A=\begin{pmatrix}
2 & 0 & 0\\
a & 2& 0\\
a+3 & a &-1
\end{pmatrix}$$
For which values of $a$ can $A$ be diagonal?
I found that $p_A(x)=(x-2)^2(x+1)$ and tried to find the eigen subspace of 2, to see if the geomtric multiplicity of the eigenvalue $2$ is $2$.
I got a set of equations:$$2x=2x ; ax+2y=2y ; (a+3)x+ay-z=2z$$
But I could not understand how to extract the relevant information from it.
| You'll get for $a \neq 0$, that the nullspace of $A-2I$ is less than $2$. To see why, you have that
\begin{align*} A - 2I = \left(\begin{matrix} 0 & 0 & 0 \\ a & 0 & 0 \\ a+3 & a & - 3 \end{matrix}\right) \end{align*}
is row equivalent to
\begin{align*}\left(\begin{matrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ a+3 & a & - 3 \end{matrix}\right) \end{align*}
since $a \neq 0$. This is row equivalent to
\begin{align*}\left(\begin{matrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & a & -3 \end{matrix}\right)\end{align*}
regardless if $a = -3$ or not. So that this is row equivalent to
\begin{align*}\left(\begin{matrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & -\frac{3}{a} \end{matrix}\right)\end{align*}
since $a \neq 0$. I think you can extract the relevent information from there.
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I show this inequality: $-2 \le \cos \theta (\sin \theta +\sqrt{\sin ^2 \theta +3})\le 2$
Show that $$-2 \le \cos \theta ~ (\sin \theta +\sqrt{\sin ^2 \theta +3})\le 2$$ for all value of $\theta$.
Trial: I know that $0\le \sin^2 \theta \le1 $. So, I have $\sqrt3 \le \sqrt{\sin ^2 \theta +3} \le 2 $. After that I am unable to solve the problem.
| Let $\cos \theta ~ (\sin \theta +\sqrt{\sin ^2 \theta +3})=y$
$\iff \sin \theta +\sqrt{\sin ^2 \theta +3}=y\sec\theta$
$\iff \sqrt{\sin ^2 \theta +3}=y\sec\theta-\sin \theta$
Squaring we get $\sin ^2 \theta +3=y^2(1+\tan^2\theta)+\sin^2\theta-2y\tan\theta$
$\iff y^2(\tan^2\theta)-2y(\tan\theta)+y^2-3=0$
As $\tan\theta$ is real, the discriminant must be $\ge0\implies(y-2)(y+2)\le0\iff\cdots$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is the argument I used to evaluate the convergence of the series $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{n+a}{(n+b)(n+c)}$ right?
If $a,b,c$ be real constants, analyze the convergence of $$\sum_{n=1}^{\infty} (-1)^{n-1}\frac{n+a}{(n+b)(n+c)}$$
What I tried to:
I compared the general term of my series to $\frac{1}{n}$: $$\lim \limits_{n \to \infty} \frac{\frac{n+a}{(n+b)(n+c)}}{\frac{1}{n}}= $$ $$\lim\limits_{n \to \infty} \frac{n^2 (1 + \frac{a}{n})}{n^2 \left ( 1 + \frac{b+c}{n} + \frac{bc}{n^2} \right ) } = 1 $$
So, as $\frac{n+a}{(n+b)(n+c)} \sim \frac{1}{n}$ when $n \to \infty$, the series (conditionally) converges because the alternating harmonic series converges.
| As Julian Aguirre pointed, we are not allowed to use a comparison with an alternating series. However, Dirichlet's test is enough to ensure convergence, since the partial sums of $(-1)^n$ are bounded and $\frac{(n+a)}{(n+b)(n+c)}$ is eventually decreasing to zero as $n\to +\infty$.
We may notice that:
$$\frac{(n+a)}{(n+b)(n+c)}=\frac{b-a}{b-c}\cdot\frac{1}{(n+b)}+\frac{c-a}{c-b}\cdot\frac{1}{(n+c)}$$
from which it follows that:
$$\begin{eqnarray*} \sum_{n\geq 1}(-1)^{n-1}\frac{(n+a)}{(n+b)(n+c)}&=&\int_{0}^{1}\sum_{n\geq 1}(-1)^{n-1}\left(\frac{b-a}{b-c}x^{n+b-1}+\frac{c-a}{c-b}x^{n+c-1}\right)\,dx\\&=&\frac{1}{b-c}\int_{0}^{1}\frac{(b-a)x^b+(a-c)x^c}{1+x}\,dx.\end{eqnarray*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the Matrix of T with respect to basis B. A linear transformation $T : P_2 \rightarrow P_2$ is given by
$T(a+bx+cx^2) = (a−b+c)+(b+c)x+(2b−a)x^2$.
It is given that the set $B=\{1+x+x^2, x+x^2, x^2\}$
is a basis for $P_2$.
(a) Find the matrix of $T$ with respect to the basis $B$ in both the domain and the
codomain.
I am really not sure how to answer this question, please help me.
I know the $T(x+x^2)=-(x+x^2)+2(x+x^2)=x+x^2$ therefore $0(1+x+x^2)+2(x+x^2)+0(x^2)$
so $\begin{pmatrix}0 & 2 & 0\end{pmatrix}$ but the answer for $T(1+x+x^2)$ and $T(x^2)$ is different on my mark scheme.
Thank you.
| The matrix $T$, with respect to the basis $v_1= 1 + x + x^2$, $v_2= x + x^2$, and $v_3=x^2$ can be computed directly as follows
$T(v_1)=T(1+x+x^2)=(1-1+1) + (1+1)x + (2-1)x^2=1 + 2x + x^2= v_1+ v_2 - v_3$,
$T(v_2)=T(x+x^2)= 2x + 2x^2 = 2v_2$,
$T(v_3)=T(x^2)= 1 + x = v_1-v_3$
Therefore, the matrix $T=(a_{ij})$ is the matrix which solves the equation
$\begin{pmatrix}
v_1 & v_2 & v_3
\end{pmatrix} T = \begin{pmatrix} v_1 + v_2 - v_3 & 2v_2 & v_1-v_3 \end{pmatrix}$. I'll let you solve explicitly for the values of $T$.
| {
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Simple argument for $\frac{(x+y)^2}{x^2+xy+y^2}\le 4/3$ I would like to show that $\forall x,y\in\mathbb R^+:\frac{(x+y)^2}{x^2+xy+y^2}\le 4/3$.
The inequality is indeed true as the maximum of $\frac{(x+y)^2}{x^2+xy+y^2}$ is reached for $x=y$ and its value is $4/3$.
Except for the standard way of computing partial derivatives and finding the maximum, is there a simple argument that imply this inequality (perhaps using symmetry somehow?).
Thanks !
| Since $x^2+xy+y^2=\left(x+\frac y2\right)^2+\frac{3y^2}{4}\gt 0$, one has$$\begin{align}\frac{(x+y)^2}{x^2+xy+y^2}\le\frac 43&\iff (x+y)^2\le\frac{4}{3}(x^2+xy+y^2)\\&\iff \frac 13x^2-\frac 23xy+\frac 13y^2\ge0\\&\iff \frac 13(x-y)^2\ge 0\end{align}$$
This is true.
| {
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How to distribute $x \times \sqrt{x^2 + 1}$ How do I distribute the x in this problem?
How do I "gain access" so to say. Does it become $x^\frac{1}{2}(x^2 + 1)^\frac{1}{2}$ or $(x^3+x)^\frac{1}{2}$ ?
Or do I need to even do that in order to integrate $\int x(x^2+1)^\frac{1}{2}$ ?
| You don't need to expand in order to integrate. Just use a $u$ substitution with $u = x^2 + 1$ and so $du = 2x \space dx$ and so $dx = \frac{du}{2x}$ and so
your integral becomes $$\large{\int \color{red}{x}u^\frac{1}{2} \times \frac{du}{2\color{red}{x}}}$$ and so you cancel out the $x$ to end up with $$ \large{\int u^\frac{1}{2} du = \frac{u^\frac{3}{2}}{ \frac{3}{2}}} =\frac {2u^\frac{3}{2}}{3}=\frac{2(x^2+1)^\frac{3}{2}}{3} + C$$
| {
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Evaluate $\lim\limits_{x\to+\infty}\frac{\left(x+\sqrt{x^2-1}\right)^n+\left(x-\sqrt{x^2-1}\right)^n}{x^n},n\in \mathbb{N}$ If we use the following
$$a^n-b^n=\left(a-b\right)\left(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}\right)=u\times t$$
$$u=x+\sqrt{x^2-1}-x+\sqrt{x^2-1}=2\sqrt{x^2-1}=2x\sqrt{1-\frac{1}{x}}$$
Now, the limit is
$$2\lim\limits_{x\to+\infty}\frac{x\sqrt{1-\frac{1}{x}}\times t}{x^n}$$
What to do next, if this is a good approach?
| Let $a= x+\sqrt{x^2-1}$, then we have $x-\sqrt{x^2-1}=\frac{1}{a}$ and $a+ \frac{1}{a}= 2x$. Next, we can write $$\lim\limits_{x\to+\infty}\frac{(x+\sqrt{x^2-1})^n+(x-\sqrt{x^2-1})^n}{x^n}
=\lim\limits_{a\to+\infty}\frac{a^n+\frac{1}{a^n}}{\frac{1}{2^n}(a+\frac{1}{a})^n}=\lim\limits_{a\to+\infty}\frac{a^n}{\frac{1}{2^n}(a)^n}= 2^{n}$$
| {
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What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:
Theorem: $\sqrt{2}$ is irrational.
Proof:
$3^2-2\cdot 2^2 = 1$.
(That's it)
That is a corollary of
this result:
Theorem:
If $n$ is a positive integer
and there are positive integers
$x$ and $y$ such that
$x^2-ny^2 = 1$,
then
$\sqrt{n}$ is irrational.
The proof is in two parts,
each of which
has a one line proof.
Part 1:
Lemma: If
$x^2-ny^2 = 1$,
then there are arbitrarily large integers
$u$ and $v$ such that
$u^2-nv^2 = 1$.
Proof of part 1:
Apply the identity
$(x^2+ny^2)^2-n(2xy)^2
=(x^2-ny^2)^2
$
as many times as needed.
Part 2:
Lemma: If
$x^2-ny^2 = 1$
and
$\sqrt{n} = \frac{a}{b}$
then
$x < b$.
Proof of part 2:
$1
= x^2-ny^2
= x^2-\frac{a^2}{b^2}y^2
= \frac{x^2b^2-y^2a^2}{b^2}
$
or
$b^2
= x^2b^2-y^2a^2
= (xb-ya)(xb+ya)
\ge xb+ya
> xb
$
so
$x < b$.
These two parts
are contradictory,
so
$\sqrt{n}$
must be irrational.
Two things to note about
this proof.
First,
this does not need
Lagrange's theorem
that for every
non-square positive integer $n$
there are
positive integers $x$ and $y$
such that
$x^2-ny^2 = 1$.
Second,
the key property of
positive integers needed
is that
if $n > 0$
then
$n \ge 1$.
| The irrationality of $\sqrt{2}$ can be deduced from the following
Theorem (Fermat, 1640): The number $1$ is not congruent.
Reasoning: If $\sqrt{2}$ were rational then $\sqrt{2},\sqrt{2}$,and $2$ would be the sides of a rational right triangle with area $1$. This is a contradiction of $1$ not being a congruent number.
A positive rational number $n$ is called a congruent number if there is a rational right triangle with area $n$: there are rational $a,b,c>0$ such that
$$a^2+b^2=c^2\qquad\text{ and }\qquad\frac{1}{2}ab=n$$
A proof of this theorem based upon Fermat's method of descent is given in The congruent number problem Theorem 2.1 by Keith Conrad.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "114",
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Using an inverse operator to find a particular solution to a differential equation. I am just learning about inverse operators in solving a differential equation, but I don't understand exactly how they work. For example, find a particular solution to
$$4y''-3y'+9y=5x^2$$
using inverse operators.
The above equation is equivalent to
$$(4D^2-3D+9)y=5x^2$$
Now the way to solve this would be to use the inverse operator as follows:
$$y_p=(4D^2-3D+9)^{-1}5x^2$$
or
$$y_p=\frac {1}{4D^2-3D+9}5x^2$$
The book I am reading uses "simple division" and arrives at a result, but I don't quite understand how that works.
What is a step-by-step method to solve the above problem?
| Notice that
$$4D^2 - 3D + 9 = \left [D - \left ( \frac{3}{8} (1 - i \sqrt{15} )\right)\right ] \left [D - \left ( \frac{3}{8} (1 + i \sqrt{15} )\right)\right ] $$
Also note that
$$ \frac{1}{-a( 1 - D /a )} = -\frac{1}{a} \left ( 1 + \frac{D}{a} + \frac{D^2}{a^2} + \ldots \right) $$
Thus to find $y_p$, we simply have to compute ( calling the roots $\lambda_\pm = 3/8 ( 1 \pm i \sqrt{15})$)
$$ y_p = \frac{1}{4D^2 - 3D + 9} 5x^2 = \frac{1}{ | \lambda_\pm|^2} \frac{1}{1- D/\lambda_+} \frac{ 1}{1 - D / \lambda_-} 5x^2= \frac{1}{|\lambda_\pm|^2}\left ( 1 + \frac{D}{\lambda_+} + \frac{D^2}{\lambda_+^2} \right ) \left ( 1 + \frac{D}{\lambda_-} + \frac{D^2}{\lambda_-^2} \right) 5x^2 = \frac{5}{|\lambda_\pm|^2} \left ( 1+ \frac{ \lambda_+ + \lambda_-}{|\lambda_\pm|^2} D + \frac{(\lambda_- + \lambda_+)^2} {|\lambda_\pm|^4 } D^2 \right ) x^2$$
Now you just have to evaluate the derivatives and you're done. (where $|\lambda_\pm|^2 = \lambda_+ \lambda_-$)
Note that I stopped at the second differential operator since $D^3 ( 5x^2) = 0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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An asymptotic formula for the bounded sum of primes. How to prove the following asymptotic formula?:
$$
\sum\limits_{p\leq x} p \sim \frac{x^2}{2 \log x}
$$
I'm stuck and I don't know where to start. I've been suggested the use of $\pi (x) = \frac{x}{\log x} + O\left( \frac{x}{\log ^2 x} \right) $. But I'm unsure about how to procede.
| If $\displaystyle S(n) = \sum\limits_{p\leq n} p$, (sum of primes not exceeding $n$),
then by Abel's summation formula, we may write:
$$S(n) = n\pi(n) - \sum\limits_{j=2}^{n-1} \pi(j)$$
So, using the estimate $\displaystyle \pi (x) = \frac{x}{\log x} + O\left( \frac{x}{\log ^2 x} \right)$,
$\begin{align}S(n) &= \frac{n^2}{\log n} + O\left(\frac{n^2}{\log^2 n}\right) - \sum\limits_{j=2}^{n-1} \frac{j}{\log j}\\&= \frac{n^2}{\log n} + O\left(\frac{n^2}{\log^2 n}\right) - \left(\int_{2}^{n} \frac{x}{\log x}\,dx + O\left(\frac{n}{\log n}\right)\right) \,\textrm{(Since, $\frac{x}{\log x}$ is monotone)} \\ &= \frac{n^2}{\log n} - \left(\left.\frac{x^2}{2\log x}\right\vert_{2}^{n} + \int_2^{n} \frac{x}{2\log^2 x}\,dx\right)+ O\left(\frac{n^2}{\log^2 n}\right) \\ &= \frac{n^2}{\log n} - \left(\frac{n^2}{2\log n} + O\left(\frac{n^2}{\log^2 n}\right)\right)+ O\left(\frac{n^2}{\log^2 n}\right)\\ &= \frac{n^2}{2\log n} + O\left(\frac{n^2}{\log^2 n}\right)\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Homework question: Represent function as power series Represent the following function as a power series of powers of x-2.
$$f(x)=\frac{x-5}{3x^2+5x-2}$$
Now, the Taylor series got me thinking. Knowing my professor, she wouldn't have us solving huge derivatives like this one. So I'm wondering, is there any other way to solve this?
| Yes, there is another way of approaching this.
First, it will make things a bit easier to let $y=x-2$.
Then, write
$$\begin{align}
3x^2+5x-2&=(3x-1)(x+2)\\\\
&=(3(x-2)+5)((x-2)+4)\\\\
&=(3y+5)(y+4)
\end{align}$$
Next, using partial fraction expansion reveals that
$$\begin{align}
\frac{x-5}{3x^2+5x-2}&=\frac{y-3}{(3y+5)(y+4)}\\\\
&=\frac{1}{y+4}+\frac{-2}{3y+5} \tag 1
\end{align}$$
Now, we can expand each term of the right-hand side of $(1)$ separately. For the first term, we have
$$\begin{align}
\frac{1}{y+4}&=\frac{1/4}{1+(y/4)}\\\\
&=\frac14\sum_{n=0}^{\infty}(-y/4)^n \tag 2
\end{align}$$
which is valid for $|y|<4$.
For the second term, we have
$$\begin{align}
\frac{-2}{3y+5}&=-\frac{2/5}{1+(3y/5)}\\\\
&=-\frac25 \sum_{n=0}^{\infty}(-3y/5)^n \tag 3
\end{align}$$
which is valid for $|y|<5/3$.
Now, adding $(2)$ and $(3)$ yields
$$\begin{align}
\frac{y-3}{(3y+5)(y+4)}=\sum_{n=0}^{\infty}\left(\frac{1}{4^{n+1}}-\frac{2(3)^n}{5^{n+1}}\right)(-1)^ny^n
\end{align}$$
whereupon substituting back $y=x-2$ yields
$$\begin{align}
\frac{x-5}{3x^2+5x-2}=\sum_{n=0}^{\infty}(-1)^n\left(\left(\frac{1}{4}\right)^{n+1}-\frac23\left(\frac{3}{5}\right)^{n+1}\right)(x-2)^n
\end{align}$$
where the series converges for $|x-2|<5/3$.
NOTE:
We can obtain a series representation for $|x-2|>5/3$ in terms of inverse powers of $x-2$. That exercise is left to the reader.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For what $k$ does $\lim_{x \to-\infty} \frac{5^{kx}-1}{5^{-2x} + 1}$ exist? For what values of $k$ does this limit exist?
$$\lim_{x \to-\infty} \frac{5^{kx}-1}{5^{-2x} + 1}$$
Progress: I worked it by dividing everything by $5^{-2x}$ and now have $5^{x(k+2)}$ since the bottom half ended up being $1$ and the top ends up $5^{x(k+2)} - 0$. But I just don't know what to make of it from here. Could $k$ be any number except $2$?
| One more hint:
$$\frac{5^{kx}-1}{5^{-2x} + 1} = \frac{5^{kx}}{5^{-2x} + 1} - \frac{1}{5^{-2x} + 1} = \frac{1}{5^{-(2+k)x} + 5^{-kx}} - \frac{1}{5^{-2x} + 1} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Taylor expansion of $f(x) = \frac{x}{x+3}\frac{1}{x-2}$ near $x=2$. I am trying to Taylor expand the function $$f(x) = \frac{x}{x+3}\frac{1}{x-2}$$ around the point $x_0 = 2$. Clearly, the last factor explodes around this point, so I will try and expand that term. However, terms on the form $\frac{1}{x-2}$ are a geometric series. Is that simply what I have to use?
| I think here we may rather talk about a Laurent series expansion around $x=2$.
We have, as $x \to 2$,
$$
\begin{align}
f(x) &= \frac{1}{x-2}\frac{x}{x+3}\\\\
&= \frac{1}{5(x-2)}\frac{2+(x-2)}{\left(1+\dfrac{(x-2)}5\right)}\\\\
&= \frac{2+(x-2)}{5(x-2)}\sum_{n\geq0}\dfrac{(-1)^n}{5^n}\left(x-2\right)^n,\quad 0<|x-2|<5,\\\\
&= \left(\frac15+\frac{2}{5(x-2)}\right)\sum_{n\geq0}\dfrac{(-1)^n}{5^n}\left(x-2\right)^n\\\\
&=\frac25\frac{1}{(x-2)}+\frac{3}{25}-\frac{3(x-2)}{125}+\frac{1}{625} (x-2)^2-\frac{(x-2)^3}{3125}+\cdots\\\\
&=\frac25\frac{1}{(x-2)}+3\sum_{n\geq0}\dfrac{(-1)^n}{5^{n+2}}\left(x-2\right)^n.
\end{align}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Asymptotic expansion of integral Find the first two terms in asymptotic expansion (as $x \rightarrow \infty$) of the following integral
$$\int_{0}^{\frac{\pi}{2}}e^{-x \sin^{2} t}dt.$$
| As Lucian pointed out in a comment $$f(x)=\int_{0}^{\frac{\pi}{2}}e^{-x \sin^{2} t}dt=\frac{ \pi}{2} e^{-x/2}\, I_0\left(\frac{x}{2}\right)$$ At this point, one could use the asymptotic expansion given by Hankel for the modified Bessel function of the first kind $$I_{\alpha}(z)=\frac{e^z}{\sqrt{2\pi z}}\Big(1-\frac {4\alpha^2-1}{8z}+\frac {(4\alpha^2-1)(4\alpha^2-9)}{2! (8z)^2}-\frac {(4\alpha^2-1)(4\alpha^2-9)(4\alpha^2-25)}{3! (8z)^3}+\cdots\Big)$$ Applied to $\alpha=0$ and $z=\frac x2$, this gives $$f(x)\approx\frac{\sqrt{\pi } }{2
\sqrt{x}}\left(1+\frac{1}{4 x}+\frac{9}{32 x^2}+\frac{75}{128 x^3}+\cdots\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1318546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\int_0^{\pi/2} \ln(\sin(x)) dx$ using the Beta, Gamma and Digamma Functions. Evaluate
$$I(\theta)=\int_0^{\pi/2} \ln(\sin(\theta))d\theta$$
$$$$
My incorrect attempt:$$$$
Consider $$J(a,0)=\int_0^{\pi/2} \sin^a(\theta)cos^0(\theta) d\theta$$
$\dfrac{d}{da}J(0,0)=I(\theta)$$$$$
Now, $J(a,0)=\beta\bigg (\dfrac{a+1}{2},\dfrac{1}{2}\bigg ) = \dfrac{\Gamma(\frac{a+1}{2})\Gamma(\frac{1}{2})}{\Gamma(\frac{a+2}{2})}$
$$\Longrightarrow I(\theta)= \dfrac{d}{da}J(0,0) = \dfrac{\partial}{\partial a}\dfrac{\Gamma(\frac{a+1}{2})\Gamma(\frac{1}{2})}{\Gamma(\frac{a+2}{2})}\bigg |^{a=0}$$
$$$$
$$\dfrac{d}{da}J(0,0) = \dfrac{\Gamma(\frac{1}{2})}{\bigg (\Gamma(\frac{a+2}{2})\bigg )^2} \bigg (\frac{1}{2}\Gamma'(\frac{a+1}{2})\Gamma(\frac{a+2}{2}) -\frac{1}{2}\Gamma ' (\frac{a+2}{2})\Gamma(\frac{a+2}{2})\bigg ) $$
$$$$
$$= \dfrac{\Gamma(\frac{1}{2})}{\Gamma(\frac{a+2}{2})} \bigg ( \frac{1}{2}\psi(\frac{a+1}{2})\Gamma(\frac{a+2}{2})-\frac{1}{2}\psi(\frac{a+2}{2})\Gamma(\frac{a+1}{2})\bigg ) $$
$$$$
$$= \dfrac{\frac{1}{2}\Gamma(\frac{a+1}{2})\Gamma(\frac{1}{2})}{\Gamma(\frac{a+2}{2})}\bigg (\psi(\frac{a+1}{2}) -\psi(\frac{a+2}{2})\bigg ) $$
$$$$
$$\Longrightarrow \dfrac{d}{da}J(0,0) = \dfrac{1}{2}\Gamma(\frac{1}{2})\Gamma(\frac{1}{2})\bigg (\psi(\frac{1}{2}) -\psi(1)\bigg ) =I(\theta) $$
$$$$I was bitterly disappointed to discover that this does not give the correct answer. This is the first time I have used these Special Functions on my own and I was really hoping I would get the correct answer. $$$$I would indeed be grateful if somebody would please be so kind as to indicate the errors I have made. Many thanks in advance.
| It does give the correct answer, indeed. I think you only need the powerful identity:
$$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}\tag{1}$$
that follows from the series definition of the digamma function. That gives, for instance:
$$\log 2 = \sum_{n\geq 1}\frac{(-1)^{n+1}}{n} = \sum_{n\geq 0}\frac{1}{(2n+1)(2n+2)} = \frac{1}{2}\left(\psi(1)-\psi\left(\frac{1}{2}\right)\right).\tag{2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove Alternating Series Approximation Prove if $S=\sum_{n=1}^{\infty}a_{n}$ is an alternating series with $\left | a_{n+1}\right | < \left | a_{n} \right |$, and $\lim_{n\to\infty}a_{n}=0$, then $\left |S-(a_{1}+a_{2}+\cdots+a_{n}) \right | \leq \left | a_{n+1} \right | $ I'm supposed to group the terms in the error as $(a_{n+1}+a_{n+2})+(a_{n+3}+a_{n+4})$ to show that the error has the same sign as $a_{n+1}$ but I don't understand what they mean by the terms in the error.
| It is I think best to look at a particular example, or two, or three. Consider the alternating series
$$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23}+\cdots.\tag{1}$$
This is the famous Leibniz series. It converges to $\frac{\pi}{4}$. Here $a_n=\frac{(-1)^{n-1}}{4n-3}$.
Suppose that we truncate this series just after the term $-\frac{1}{7}$. Then the truncation error (in the language of your question, the sum of the terms in the error) is the rest of the stuff, which is
$$\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23}+\cdots.\tag{2}$$
The expression (2) may be rewritten as
$$\left(\frac{1}{9}-\frac{1}{11}\right)+\left(\frac{1}{13}-\frac{1}{15}\right)+\left(\frac{1}{17}-\frac{1}{19}\right)+\left(\frac{1}{21}-\frac{1}{23}\right)+\cdots.\tag{3}$$
It is also useful to group a little differently, as
$$\frac{1}{9}+\left(-\frac{1}{11}+\frac{1}{13}\right)+\left(-\frac{1}{15}+\frac{1}{17}\right)+\left(-\frac{1}{19}+\frac{1}{21}\right)+\cdots.\tag{3'}$$
Note that the sum of any two grouped terms in (3) is positive. So the sum $S$ of the full series (2) is positive. The estimate $1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}$ is an underestimate of $S$.
Now look at the version (3') of the truncation error (2). The sum of any two grouped terms is negative, so the full sum is less than $\frac{1}{9}$.
We conclude that the error when we truncate just after the $-\frac{1}{7}$ term is positive but less than $\frac{1}{9}$. The absolute value of the error is less than the absolute value of the first neglected term.
I suggest you make a similar analysis of the truncation error if we truncate just after the $\frac{1}{9}$ term. Grouping in the two ways done above will show that the sum of the neglected terms is negative, but greater than $-\frac{1}{11}$. Again the truncation error is less in absolute value than the absolute value of the first neglected term.
We can make an essentially identical analysis for general series $a_1+a_2+a_3+a_4+\cdots$, where the terms alternate in sign, decrease in absolute value, and have limit $0%
The truncation error is always less in absolute value than the absolute value of the first neglected term. If the first neglected term is positive, then our estimate of the full infinite sum is an underestimate. If the first neglected term is negative, our estimate is an overestimate.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show an $\arctan$ and $\arcsin$ function is constant Show that for every $x\geq1$ the following is true:
$2\arctan x + \arcsin \frac{2x}{1+x^2} = \pi$
One way (mentioned in the link at the bottom) would be to calculate the derivative of the left side, show that it is always $0$ then show that for $x=1$ the equation is true.
I'm trying for some time to find a cleaner way to prove the equality, without so much algebra. Does anyone have any idea where to start?
Someone already mentioned the same problem here.
|
\begin{align} 2\arctan(x)+\arcsin\left(\frac{2x}{1+x^2}\right)
&=\pi
\end{align}
Let's $\chi=\arctan(x)\ge\pi/4$ since $x\ge1$.
Consider a unit circle
Then in a $\triangle OAB$ we have $|OA|=|OB|=1$,
$\angle OAB=\angle OBA=\chi$,
$\angle AOB=\alpha=\pi-2\chi$,
$|AH|=\sin(\alpha)=a$.
In the $\triangle ABH$
\begin{align}
|AH|&=|BH|\tan(\chi)
\\
\sin(\alpha)
&=
(1-\cos(\alpha))\tan(\chi)
\\
\sin(\alpha)
&=
\left(1-\sqrt{1-\sin^2(\alpha)}\right)\tan(\chi)
\\
a&=
\left(1-\sqrt{1-a^2}\right)x
\\
\left(1-\frac{a}{x}\right)^2&=
1-a^2
\\
\frac{a^2}{x^2}+a^2 &= 2\frac{a}{x}
\\
a\left(\frac{1}{x^2}+1\right)
&= \frac{2}{x}
\\
a&=\sin(\alpha)=
\frac{2x}{1+x^2}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1321351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
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Is $y^2=x^3+7$ unsolvable modulo some $n$? The equation $y^2=4x^3+7$ has no integral solution since $y^2\equiv4x^3+7\pmod4$ has no solution (i.e. has no solution in $\Bbb{Z}/4\Bbb{Z}$).
It is well known that $y^2=x^3+7$ has no integral solution, but is there an integer $n (n>1)$ such that $y^2\equiv x^3+7\pmod n$ has no solution? If not, how can I prove it? Thanks in advance.
| The discriminant of $x^3+7$ is $-27\cdot49$. Given a prime power $q$, this discriminant is non-zero in the field $F_q$ unless $q$ is $3,9,27,7,$ or $49$. For other values of $q$, $y^2=x^3+7$ is an elliptic curve over $F_q$, and we can use Hasse's bound
$$|N-(q+1)| \le 2\sqrt q$$
where $N$ is the number of points on the curve. So there are at least $q+1-2\sqrt q$ points on the curve. One of these is the point at infinity; subtracting this, we have at least $q-2\sqrt q$ solutions of $y^2=x^3+7$. This is $> 0$ (and therefore $\ge 1$) for all $q \ge 5$.
So you need only check the cases $q=2,3,4,7,9,27,$ and $49$ to prove that $y^2=x^3+7$ has a solution modulo every prime power. And as Greg Martin points out in a comment, this is enough to establish it for all $n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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The algebra generated by the set $\{1,x^2\}$ is dense in $C\left[0,1\right]$ with the supremum norm but fails to be dense in $C\left[-1,1\right]$. Show that the algebra generated by the set $\{1,x^2\}$ is dense in $C\left[0,1\right]$ with the supremum norm but fails to be dense in $C\left[-1,1\right]$.
I have know that for each $f\in$$C\left[0,1\right]$ and $\epsilon>0$, there is a polynomial $p$ such that $||f-p||_\infty<\epsilon$.
| To show that the algebra generated by $\{1,x^2\}$ is not dense in $C[-1,1]$ consider $x\in C[-1,1]$. Let $f$ be in the algebra, since $1$ and $x^2$ are both even functions, $f$ is also an even function.
Consider the cases where $x=\frac{1}{2}$ and $x=-\frac{1}{2}$. At these points, $f(-\frac{1}{2})=f(\frac{1}{2})$. Moreover, consider
$$
\left|f\left(\frac{1}{2}\right)-\frac{1}{2}\right|\qquad \left|f\left(-\frac{1}{2}\right)-\left(-\frac{1}{2}\right)\right|=\left|f\left(\frac{1}{2}\right)+\frac{1}{2}\right|.
$$
Observe, by the triangle inequality:
$$
1=\left|\frac{1}{2}-\left(-\frac{1}{2}\right)\right|=\left|\frac{1}{2}-f\left(\frac{1}{2}\right)-\left(-\frac{1}{2}\right)+f\left(\frac{1}{2}\right)\right|\leq\left|f\left(\frac{1}{2}\right)-\frac{1}{2}\right|+\left|f\left(\frac{1}{2}\right)+\frac{1}{2}\right|=\left|f\left(\frac{1}{2}\right)-\frac{1}{2}\right|+\left|f\left(-\frac{1}{2}\right)-\left(-\frac{1}{2}\right)\right|.
$$
At least one of thee two differences must be at least $\frac{1}{2}$, so the $\infty$ norm cannot be less than $\frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1322536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
On representation of even integers as $a^x+b^y$ Does there exist some $k \in \mathbb N$ such that for every even integer $n \ge k $ we can find positive integers $a,b,x,y$ such that $n=a^x+b^y$ , where $\gcd (a,b)=1 ; a,b>1$ and at least one of $x,y$ is more than $1$ ? I know that for every even integer $n>6$ there are integers $a,b>1$ such that $\gcd(a,b)=1$ and $n=a+b$ , so here $x=y=1$ for every $n$ , I want to see if at least one power can be bigger.
| Let $n \ge 6$. Choose a prime $p$ such $p^2 \le n-2$ and $p$ does not divide $n$. Then $n=p^x+(n-p^2)^y$ with $x=2$ and $y=1$, and $p$ and $n-p^2$ are greater than $1$ and relatively prime. This fails only if there is no such prime, which happens only if $n$ is divisible by every prime between $2$ and $\sqrt{n-2}$. But in that case $n \ge \prod_{p \le \sqrt{n-2}}p$, which can't happen for very large $n$, since clearly the right-hand side grows much faster than the left. Indeed,
$$
\prod_{p \le 7=\sqrt{51-2}} p= 2\cdot 3\cdot 5\cdot 7 = 210 \gg 51,
$$
so for $n\ge 51$ we can write $n=a^2+b$ where $\gcd(a,b)=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1322720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$x^2-y^2=196$, can we find the value of $x^2+y^2$? $x$ and $y$ are positive integers.
If $x^2-y^2=196$, can we know what the value of $x^2+y^2$ is?
Can anyone explain this to me? Thanks in advance.
| Rewrite the equation to get $x^2=y^2+196$. Now write $x=y+k$, then $$x^2=(y+k)^2=y^2+2yk+k^2=y^2+196$$
Therefore $2yk+k^2=196$
Now note that $k \mid 196$ and $k$ must be even.
This leaves the possibilities $2,4,14,28,98,196$.
$k=2$ gives $y=48$.
$k=4$ gives $y=22.5$, which is not an integer.
$k=14$ gives $y=0$, but that is not positive.
Similiar, for larger $k$, it won't be positive.
So $y=48$, therefore $$x^2+y^2=(x^2-y^2)+2y^2=196+2\cdot48^2=4804$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to integrate $\int x\sqrt { \frac {a^2-x^2}{a^2+x^2} } \, dx$? How to integrate $\displaystyle\int x\sqrt {\frac{a^2 - x^2}{a^2+x^2}} \, dx$ ?
I tried this for quite some time, but not being able to solve. Help!
| Hint:
$$x\sqrt{\frac{a^2-x^2}{a^2+x^2}}=x\sqrt{\frac{1-(\frac{x}{a})^2}{1+(\frac{x}{a})^2}}$$
and use the trigonometric identity
$$\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=|\tan\left(\frac{\theta}{2}\right)|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Theorem about prime numbers which are one more than a multiple of $8$ Assume the following:
A prime $p>2$ can be written as $p=m^2+n^2$ for integral $m,n$ iff $p$ is $1$ more than a multiple of $4$.
Now, prove that every prime which is one more than a multiple of $8$ be written as $x^2+16y^2$ where $x$ and $y$ are integers.
I was able to show that if $p$ is one more than a multiple of $8$, it is one more than a multiple of $4$, and since $p$ is odd, one of $m$ or $n$ is odd. But I am unable to go beyond this point. Could someone please help?
| Let $p\equiv 1\pmod{\! 8}$. Since $p\equiv 1\pmod{\! 4}$, exist $x,y\in\Bbb Z$ giving you $p=x^2+y^2$.
Exactly one of $x,y$ is even, wlog $y$ is even.
$x^2\equiv 1\pmod{\! 8}$, $y^2\equiv \{0,4\}\pmod{\! 8}$ and $p\equiv 1\pmod{\! 8}$.
So we must have $y^2\equiv 0\pmod{\! 8}$.
$8\mid y^2\iff 4\mid y$. Let $y=4m$. Then $p=x^2+16m^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Determine when value will be greater than x I want every time to add '$(b-x)$' to the value of $a$ and i want to know how many times i will repeat this operation to be the value of $a$ greater or equal to $x$.
$a = b+a - x$
I know that $b > x$ and $ a < x $
Any ideas?
| So you have an arithmetic sequence
$t_n = a + n(b - x)$, (Starting at $n = 0$).
where $a < x < b$, and you want to know the first time that $t_n > x$
\begin{align}
t_n & > x\\
a + n(b - x) & > x\\
n(b - x) & > x - a\\
n & > \dfrac{x - a}{b - x}\\
n & = \left \lceil \dfrac{x - a}{b - x} \right \rceil
\end{align}
If $\dfrac{x - a}{b - x}$ is not an integer, then
$n = \left \lfloor 1 + \dfrac{x - a}{b - x} \right \rfloor
= \left \lfloor \dfrac{b - a}{b - x} \right \rfloor$
But if $\dfrac{x - a}{b - x}$ is an integer, then the answer is still
$\left \lceil \dfrac{x - a}{b - x} \right \rceil$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1325063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that the sum of the Lagrange (interpolation) coefficients is equal to 1 Prove that the sum of the Lagrange (interpolation) coefficients is equal to 1.
Please suggest me a book-reference or give a solution for me. Thanks a lot in advance.
If $f = \sum_{i=0}^nf(x_i)L_i(x)$ then one has to prove $\sum_{i=0}^nL_i(x)=1$ where
$L_i(x)=\frac{(x-x_0)...(x-x_{i-1})(x-x_{i+1})...(x-x_n)}{(x_i-x_0)...(x_i-x_{i-1})(x_i-x_{i+1})...(x_i-x_n)}$, $i=0 \cdots n$
| Let me propose one more method of proof.
"Beginner's guide to mapping simplexes affinely", section "Lagrange interpolation", describes a determinant form of Lagrange polynomial that interpolates $(a_0;b_0)$, $\dots$, $(a_n;b_n)$
$$
P(x) = (-1)
\frac{
\det
\begin{pmatrix}
0 & b_0 & b_1 & \cdots & b_n \\
x^n & a_0^n & a_1^n & \cdots & a_n^n \\
x^{n-1} & a_0^{n-1} & a_1^{n-1} & \cdots & a_n^{n-1} \\
\cdots & \cdots & \cdots & \cdots & \cdots \\
1 & 1 & 1 & \cdots & 1 \\
\end{pmatrix}
}{
\det
\begin{pmatrix}
a_0^n & a_1^n & \cdots & a_n^n \\
a_0^{n-1} & a_1^{n-1} & \cdots & a_n^{n-1} \\
\cdots & \cdots & \cdots & \cdots \\
1 & 1 & \cdots & 1 \\
\end{pmatrix}
}.
$$
Obviously, setting all $b_i = 1$ we'll get the sum of Lagrange polynomials $L_i$, because
$$
P(x) = \sum_{i=0}^n\, \underbrace{b_i}_{=1} L_i(x) = \sum_{i=0}^n L_i(x).
$$
Now just simplify everything
$$
P(x) = (-1)
\frac{
\det
\begin{pmatrix}
0 & 1 & 1 & \cdots & 1 \\
x^n & a_0^n & a_1^n & \cdots & a_n^n \\
x^{n-1} & a_0^{n-1} & a_1^{n-1} & \cdots & a_n^{n-1} \\
\cdots & \cdots & \cdots & \cdots & \cdots \\
1 & 1 & 1 & \cdots & 1 \\
\end{pmatrix}
}{
\det
\begin{pmatrix}
a_0^n & a_1^n & \cdots & a_n^n \\
a_0^{n-1} & a_1^{n-1} & \cdots & a_n^{n-1} \\
\cdots & \cdots & \cdots & \cdots \\
1 & 1 & \cdots & 1 \\
\end{pmatrix}
} =
(-1)
\frac{
\det
\begin{pmatrix}
-1 & 0 & 0 & \cdots & 0 \\
x^n & a_0^n & a_1^n & \cdots & a_n^n \\
x^{n-1} & a_0^{n-1} & a_1^{n-1} & \cdots & a_n^{n-1} \\
\cdots & \cdots & \cdots & \cdots & \cdots \\
1 & 1 & 1 & \cdots & 1 \\
\end{pmatrix}
}{
\det
\begin{pmatrix}
a_0^n & a_1^n & \cdots & a_n^n \\
a_0^{n-1} & a_1^{n-1} & \cdots & a_n^{n-1} \\
\cdots & \cdots & \cdots & \cdots \\
1 & 1 & \cdots & 1 \\
\end{pmatrix}
}.
$$
Laplace expansion along the first row leads to
$$
P(x) =
\frac{
\det
\begin{pmatrix}
a_0^n & a_1^n & \cdots & a_n^n \\
a_0^{n-1} & a_1^{n-1} & \cdots & a_n^{n-1} \\
\cdots & \cdots & \cdots & \cdots \\
1 & 1 & \cdots & 1 \\
\end{pmatrix}
}{
\det
\begin{pmatrix}
a_0^n & a_1^n & \cdots & a_n^n \\
a_0^{n-1} & a_1^{n-1} & \cdots & a_n^{n-1} \\
\cdots & \cdots & \cdots & \cdots \\
1 & 1 & \cdots & 1 \\
\end{pmatrix}
} = 1.
$$
For practical example you may want to check "Workbook on mapping simplexes affinely", section "Lagrange interpolation".
NOTE. Using Laplace expansion along the first column in the numerator you can get expressions for coefficients at $x^i$, thus determinant form of Lagrange interpolation can be quite convenient in some cases.
| {
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"url": "https://math.stackexchange.com/questions/1325292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
converting a differential equation to polar coordinates I have the following family of autonomous systems, I'm having trouble with part b):
$$x'=x(1-\sqrt {x^2+y^2})-y-\epsilon y$$
$$y'=y(1-\sqrt {x^2+y^2})+x+\epsilon(x+x^2+y^2)$$
a) Convert the system to polar coordinates $(r(t;r_0,\theta_0,\epsilon),\theta(t;r_0,\theta_0,\epsilon))$:
$$x=r\cos\theta \qquad y=r\sin\theta \qquad r^2=x^2+y^2$$
$$x'=r'\cos\theta - r\sin\theta\theta' \quad y'=r'\sin\theta + r\cos\theta\theta'$$
$$\left ( \begin{array}\\ \cos{\theta} & -r \sin{\theta} \\ \sin{\theta} & r \cos{\theta} \end{array} \right ) \left ( \begin{array}\\ r' \\ \theta' \end{array} \right ) = \left ( \begin{array}\\ r\cos\theta(1-r)-r\sin\theta-\epsilon r \sin\theta\\r\sin\theta(1-r) + r\cos\epsilon + \epsilon(r\cos\theta+r^2) \end{array} \right )$$
I found the inverse matrix:
$$\left ( \begin{array}\\ \cos{\theta} & -r \sin{\theta} \\ \sin{\theta} & r \cos{\theta} \end{array} \right )^{-1}=\left ( \begin{array}\\ \cos{\theta} & \sin{\theta} \\ \frac{-1}{r}\sin{\theta} & \frac{1}{r} \cos{\theta} \end{array} \right )$$
Multiplied on the left side, simplified and obtained:
$$\left ( \begin{array}\\ r' \\ \theta' \end{array} \right )=\left ( \begin{array}\\ r(1+\epsilon) \\1 + \epsilon r \end{array} \right )$$
b)See that for $\epsilon=0$ there is a periodic circular orbit with radius 1, period $2\pi$, and is attractive. In particular $(r(t;1,0,0),\theta(t;1,0,0))=(1,t)$.
For $\epsilon=0$ what I get is:
$$\left ( \begin{array}\\ r' \\ \theta' \end{array} \right )=\left ( \begin{array}\\ r \\ 1 \end{array} \right )$$
$$r'(t)=r(t) \quad \theta'(t)=1$$
Therefore $\theta (t)=t$ matches the initial value problem $\theta(t;1,0,0)=t$. However $r(t)=r'(t)$ can't match the initial value problem $r(t;1,0,0)=(1)$ because:
$$r(t)=1 \Rightarrow r'(t)=0 \Rightarrow r(t) \neq r'(t)$$
Where did I go wrong?
| $$\left ( \begin{array}\\ \cos{\theta} & -r \sin{\theta} \\ \sin{\theta} & r \cos{\theta} \end{array} \right )^{-1}= \left ( \begin{array}\\ \cos{\theta} & \sin{\theta} \\ \frac{-1}{r}\sin{\theta} & \frac{1}{r} \cos{\theta} \end{array} \right )$$
Multiplied on the left side, simplified and obtained:
$$\left ( \begin{array}\\ \cos{\theta} & \sin{\theta} \\ \frac{-1}{r}\sin{\theta} & \frac{1}{r} \cos{\theta} \end{array} \right ) * \left ( \begin{array}\\ r\cos\theta(1-r)-r\sin\theta-\epsilon r \sin\theta\\r\sin\theta(1-r) + r\cos\theta + \epsilon(r\cos\theta+r^2) \end{array} \right )$$
$$\left ( \begin{array}
*r\cos^2\theta (1-r)-r\sin\theta\cos\theta-\epsilon r \sin\theta\cos\theta+r\sin^2\theta(1-r) + r\cos\theta\sin\theta + \epsilon\sin\theta (r\cos\theta+r^2)
\\ -\sin\theta \cos\theta(1-r)+\sin^2\theta+\epsilon \sin^2 \theta+\sin\theta\cos\theta(1-r)+\cos^2\theta+\epsilon(\cos^2 \theta+r\cos\theta)
\end{array} \right )$$
Cancel out:
$$\left ( \begin{array} +r(1-r) + \epsilon\sin\theta r^2
\\ +1+\epsilon+\epsilon r\cos\theta
\end{array} \right )$$
Doesn't equal:
$$\left ( \begin{array}\\ r' \\ \theta' \end{array} \right )=\left ( \begin{array} \\ r(1+\epsilon) \\1 + \epsilon r \end{array} \right )$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "9",
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Integers as a sum of $\frac{1}{n}$ Say $\sum_{i \in I} \frac{1}{n_i} = 2$, where $(n_i)_{i \in I}$ is a finite sequence of positive integers (not necessarily distinct). Is there a subsequence $(n_i)_{i \in J}$ of $(n_i)_{i \in I}$ such that $\sum_{i \in J} \frac{1}{n_i} = 1$?
| Here is a counterexample even when all fractions are required to be different!
$\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{11}+\frac{1}{13}+\frac{1}{16}+\frac{1}{17}+\frac{1}{21}+\frac{1}{1105}+\frac{1}{55692}+\frac{1}{1361360}$
This came from:
$\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{11}+\frac{1}{13}+\frac{1}{16}+\frac{1}{17}$ [only prime powers]
$+\frac{1}{3 \cdot 7}+\frac{1}{5 \cdot 13 \cdot 17}+\frac{1}{2^2 \cdot 3^2 \cdot 7 \cdot 13 \cdot 17}+\frac{1}{2^4 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17}$ [plus extra terms to get to $2$]
To prove it, first multiply throughout by $2^4 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17$.
Now consider any subset whose sum is $1$. We can assume that it does not include the last term. Modulo $13$ we see immediately that it cannot include any term with denominator divisible by $13$. Now modulo $7$ we see that it cannot include $\frac{1}{21}$. We can repeat this reasoning all the way, but now actually it is obvious that it is impossible because the remaining terms allowed are all prime powers and hence the term in the subset with largest denominator $p^k$ will create a contradiction modulo $p$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find area bounded by $(\frac{x}{3}+\frac{y}{6})^3-9xy=0$ $C = \{(x,y) \mid (\frac{x}{3}+\frac{y}{6})^3-9xy=0\}$.
I want to find the area bounded by $C$ in first quadrant.
Can you tell me how to solve exercises like this?
I have no idea what integral I need to solve, because my curve is implicitly defined.
| Let's use (slightly modified) polar coordinates. Set $x = 3r\cos\theta$ and $y = 6r\sin\theta$.
The Jacobian of this transformation is $\left|\begin{matrix}\tfrac{\partial x}{\partial r} & \tfrac{\partial y}{\partial r} \\ \tfrac{\partial x}{\partial \theta} & \tfrac{\partial y}{\partial \theta}\end{matrix}\right| = \left|\begin{matrix}3\cos\theta & 6\sin\theta \\ -3r\sin\theta & 6r\cos\theta\end{matrix}\right| = 18r$.
The curve $C$ in $(r,\theta)$ coordinates is given by:
$\left(\dfrac{3r\cos\theta}{3}+\dfrac{6r\sin\theta}{6}\right)^3-9(3r\cos\theta)(6r\sin\theta) = 0$
$r^3(\cos\theta+\sin\theta)^3 - 162r^2\cos\theta\sin\theta = 0$
$r = \dfrac{162\cos\theta\sin\theta}{(\cos\theta+\sin\theta)^3} =: R(\theta)$
Thus, the area of the region bounded by $C$ is:
$\displaystyle\iint\limits_{C}\,dx\,dy$ $=\displaystyle\int_{0}^{\pi/2}\int_{0}^{R(\theta)}18r\,dr\,d\theta$ $=\displaystyle\int_{0}^{\pi/2}9R(\theta)^2\,d\theta$ $=\displaystyle\int_{0}^{\pi/2}\dfrac{9 \cdot 162^2\cos^2\theta\sin^2\theta}{(\cos\theta+\sin\theta)^6}\,d\theta$ $=\displaystyle9 \cdot 162^2\int_{0}^{\pi/2}\dfrac{\tan^2\theta\sec^2\theta}{(1+\tan\theta)^6}\,d\theta$ $= 9 \cdot 162^2\displaystyle\int_{0}^{\infty}\dfrac{u^2}{(1+u)^6}\,du$ $= 9 \cdot 162^2\displaystyle\int_{1}^{\infty}\dfrac{(v-1)^2}{v^6}\,dv$ $= 9 \cdot 162^2\displaystyle\int_{1}^{\infty}\left(\dfrac{1}{v^4}-\dfrac{2}{v^5}+\dfrac{1}{v^6}\right)\,dv$ $= 9 \cdot 162^2 \left[-\dfrac{1}{3u^3}+\dfrac{1}{2u^4}-\dfrac{2}{5u^5}\right]_{1}^{\infty}$ $= 7873.2$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the half range cosine fourier series expansion for $f(x)=(x-1)^2,\quad 0Find the half range cosine fourier series expansion for
$$f(x)=(x-1)^2,\quad 0<x<1$$
and hence deduce that
$$\pi^2=8\left(\frac 1 {1^2}+\frac 1 {3^2}+\frac 1 {5^2}+\ldots\right)\tag{1}$$
My work
I have derived that the expansion is
$$f(x)=\frac 1 3+\sum\limits_{n=1}^\infty \frac 4 {n^2\pi^2} \cos n\pi x$$
But I could not deduce that $(1)$.
| Define $f$ as a periodic function of $1$ on $\mathbb{R}$ as
$$
f(x)=(x−1)^2, \hspace{5 mm} 0\le x<1
$$
Set $x=0$, then
$$
f(0)=1=\frac 1 3+\sum\limits_{n=1}^\infty \frac 4 {n^2\pi^2}
$$
So we have
$$
\dfrac{2}{3}=\sum\limits_{n=1}^\infty \frac 4 {n^2\pi^2} \hspace{4 mm} \text{and} \hspace{4 mm} \dfrac{\pi^2}{6}=\sum\limits_{n=1}^\infty \dfrac1{n^2}
$$
And set $x\to1^{-}$, then
$$
\lim\limits_{x\to1^{-}}f(x)=0=\frac 1 3+\sum\limits_{n=1}^\infty \frac 4 {n^2\pi^2} \cos n\pi
$$
Thus
\begin{align}
\dfrac{\pi^2}{12}&=\sum\limits_{n=1}^\infty \dfrac1 {(2n-1)^2}-\sum\limits_{n=1}^\infty \dfrac1 {(2n)^2}
\\
&=\sum\limits_{n=1}^\infty \dfrac1 {(2n-1)^2}-\frac1{4}\sum\limits_{n=1}^\infty \dfrac1 {n^2}
\\
&=\sum\limits_{n=1}^\infty\dfrac1 {(2n-1)^2}-\dfrac{\pi^2}{24}
\end{align}
So
$$
\sum\limits_{n=1}^\infty\dfrac1 {(2n-1)^2}=\dfrac{\pi^2}{8}
$$
| {
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Variation on Pythagoras: If $a^2 + b^2 = c^2$, then $a + b \leq c\sqrt{2}$ This is a very interesting word problem that I came across in an old textbook of mine. So I know its got something to do with derivative of the Pythagorean Theorem using calculus, trigonometry, geometry, or plain old algebra, which yields the shortest, simplest proofs, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:
Prove that for $a, b, c > 0$,
if $a^2 + b^2 = c^2$, then $a + b \leq c\sqrt{2}$
| We solve this derivative of the Pythagorean Theorem using calculus, trigonometry,
geometry, or plain old algebra, which yields the shortest, simplest proofs!
Calculus:
The heavy hand of calculus does not provide a short
or elegant proof, but I always like seeing more than one approach.
$a^2 + b^2 = c^2$
$a^2 = c^2 - b^2 $
$a = \sqrt{c^2 - b^2} $
$a + b = \sqrt{c^2 - b^2} + b $
To maximize $a + b$, we take its derivative and set it to $0$.
Derivative is $$\frac{(\sqrt{c^2 - b^2} - b)}{\sqrt{c^2 - b^2}} $$
Setting that to $0$, we have
$\sqrt{c^2-b^2} = b$
$c^2 - b^2 = b^2 $
$c^2 = 2b^2$
We can also reach this point by seeing
that the derivative is also equal to $\frac{a - b}{a}
Setting $a - b = 0$, we have $a = b$,
$b = a = \frac{c}{\sqrt2} = \sqrt2 \times \frac c2$
At the maximum, $a + b = \sqrt2 \times c$,
therefore $a + b \leq \sqrt2 c$.
Trigonometry:
$a = c$ $\cos t$ and $b = c \sin t$
So $a + b = c \times ( \cos t + \sin t)$
As $c$ is constant, we need to maximise $\cos t + \sin{t}$
This is of the form $a \cos t + b \sin t$
$a = 1$ and $b = 1$
let $a = 1 = r \cos a$
B = 1 = r sin a
square and add: r = sqrt(2) and tan a = 1
so 1 cos t + 1 sin t
$= r \cos a \cos t + r \sin a \sin t $
$= r \cos(t-a)$
$= \sqrt2 \cos \times\ (t- \frac{\pi}{4})$
$a+b = \sqrt2 \cos (t-\frac{\pi}{4})\times c$
so $\leq \sqrt2 \times C$ as $\cos(t- \frac{\pi}{4}) \leq 1$
And the simplest proofs follow.
Geometry:
$a,b,c$ are the sides of a right triangle.
Since $a, b$, and $c$ are positive real numbers:
$a+b > \sqrt2 \times c \geq (a+b)^2 > 2c^2 \geq 2ab > c^2$.
(since $a^2 + b^2 = c^2$)
The triangle $ABC$ is a right triangle, so it can be
inscribed in a circle with diameter $c$.
The height of this triangle, $h, \leq \frac c2$ (radius of circle).
The area of triangle = $\frac{ch}{2} = \frac{ab}{2}$.
$h \leq \frac c2$ (radius of circle).
$h² \leq \frac{ch}{2} \leq (\frac c2)^2 = \frac{c^2}{4} $
So $\frac{ab}{2} \leq \frac{c^2}{4}$
and $2ab \leq c^2$
Algebra:
We obviously have that: $0 \leq (a - b)^2$
$0 \leq a^2 - 2ab + b^2$
$a^2 + 2ab + b^2 \leq 2a^2 + 2b^2 = 2c^2 $
$(a + b)^2 \leq 2c^2$
EDIT: I should have just written:
$(a + b)^2 \leq (a - b)^2 + (a + b)^2 = 2c^2$
It would have been a nice one liner.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 9,
"answer_id": 4
} |
How to evaluate $\int \frac { \sin x+\cos x }{ \sin^4 x+\cos^4x}\, dx$? How can one find $$\int \frac { \sin x+\cos x }{ \sin^4 x+\cos^4x}\, dx?$$
| Split the integral as
$$
\int\frac {\cos x}{\sin^4 x + (1 - \sin^2 x)^2} dx
+ \int \frac{\sin x}{(1 - \cos^2 x)^2 + \cos^4 x} dx
$$
Compute the integrals with the substitutions $u = \sin x$ and $u = \cos x$ respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Nesbitt's Inequality for 4 Variables I'm reading Pham Kim Hung's 'Secrets in Inequalities - Volume 1', and I have to say from the first few examples, that it is not a very good book. Definitely not beginner friendly.
Anyway, it is proven by the author, that for four variables $a, b, c$, and $d$, each being a non-negative real number, the following inequality holds:
$$\frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b}\ge 2$$
I have no idea how the author proves this. It comes under the very first section, AM-GM. I get the original Nesbitt's inequality in 3 variables that the author proves (which is also cryptic, but I was able to decipher it).
My effort: I understood how the author defines the variables $M, N$ and $S$.
$$S = \frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b}$$
$$M = \frac{b}{b+c} + \frac{c}{c+d} + \frac{d}{d+a} + \frac{a}{a+b}$$
$$N = \frac{c}{b+c} + \frac{d}{c+d} + \frac{a}{d+a} + \frac{b}{a+b}$$
$M + N = 4$, pretty straightforward. The numerators and denominators cross out to give four 1s.
Then the author, without any expansion/explanation, says
$$M + S = \frac{a+b}{b+c} + \frac{b+c}{c+d} + \frac{c+d}{d+a} + \frac{d+a}{a+b}\ge 4$$
Which is also true, since the AM-GM inequality says
$$\frac{M+S}{4}\ge \left(\frac{a+b}{b+c}\cdot\frac{b+c}{c+d}\cdot\frac{c+d}{d+a}\cdot\frac{d+a}{a+b}\right)^{1/4}$$
The RHS above evaluates to $1^{1/4}$ since all the numerators and denominators cancel out.
The next part is the crux of my question.
The author claims,
$$N + S =\frac{a+c}{b+c}+\frac{a+c}{a+d}+\frac{b+d}{c+d} + \frac{b+d}{a+b}\ge\frac{4(a+c)}{a+b+c+d}+\frac{4(b+d)}{a+b+c+d}$$
This is completely bizarre for me! Where did the author manage to get a sum of $(a+b+c+d)$??
As a side note, I'd definitely not recommend this book for any beginner in basic algebraic inequalities (even though the title of the book promotes that it's a treatment of basic inequalities). The author takes certain 'leaps of faith', just assuming that the student reading the book would be able to follow.
| The simplest is to use the AM-HM inequality; which is a consequence of the AM-GM inequality, but can be proved independently.
Remember the harmonic mean of $x$ and $y$ is the number whose inverse is the arithmetic mean of the inverses of $x$ and $y$. Explicitly:
$$\frac1H=\frac12\Bigl(\frac1x+\frac1y\Bigr)\ge \frac1A=\frac2{x+y}$$
Thus $\,\dfrac1x+\dfrac1y\ge\dfrac4{x+y}$.
Apply this inequality twice in
$$(a+c)\Bigl(\frac1{b+c}+\frac1{a+d}\Bigl)+(b+d)\Bigl(\frac1{c+d}+\frac1{a+b}\Bigl). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Complex integration with trigonometric and logarithm
Show that $\int_0^{2\pi}\log\sin^22\theta dx=4\int_0^\pi\log\sin \theta d\theta=-4\pi
\log2$
I did $$\int_0^{2\pi}\log\sin^22\theta d\theta=4\int_0^{\frac{\pi}{4}}\log\sin^22\theta d\theta$$
taking $u=2\theta$
$$2\int_0^\frac{\pi}{2}\log\sin^2\theta d\theta$$
but I need to get $\sin \theta$ and $\sin^2 \theta=\frac{1}{2}(1-\cos2\theta)$ does not help much. I tried some trigonometric manipulation but could not get anything
| Consider what has already been demonstrated
\begin{align}
\int_{0}^{2\pi}\log(\sin^{2}(2\theta)) \, d\theta &= \frac{1}{2} \, \int_{0}^{\pi} \log(\sin^{2}(\theta)) \, d\theta = \frac{I}{2}
\end{align}
Now let $2 \sin^{2}(\theta) = 1 - \cos(2\theta)$ to obtain
\begin{align}
I &= - \ln(2) \, \int_{0}^{\pi} d\theta + \int_{0}^{\pi} \ln(1-\cos(2\theta)) \, d\theta \\
&= - \pi \, \ln 2 - \sum_{n=1}^{\infty} \frac{1}{n} \, \int_{0}^{\pi} \cos^{n}(2\theta) \, d\theta \\
&= - \pi \, \ln 2 - \sum_{n=1}^{\infty} \frac{1}{2 n} \, \int_{0}^{2\pi} \cos^{n}(x) \, dx \hspace{5mm} x = 2\theta \\
&= - \pi \, \ln 2 - \sum_{n=1}^{\infty} \frac{1}{2 n} \, \left( \int_{0}^{\pi} \cos^{n}(x) \, dx + \int_{\pi}^{2\pi} \cos^{n}(x) \, dx \right) \\
&= - \pi \, \ln 2 - \sum_{n=1}^{\infty} \frac{1+(-1)^{n}}{2 n} \, \int_{0}^{\pi} \cos^{n}(x) \, dx \\
&= - \pi \, \ln 2 - \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{2n} \cdot \frac{(2)^{2} \, \Gamma\left(\frac{1}{2} \right) \, \Gamma\left(\frac{2n+3}{2}\right)}{(2n+1) \, \Gamma\left(n + 1\right)} \\
I &= - \pi \, \ln 2 - \pi \, \ln 2 = - 2 \pi \, \ln 2.
\end{align}
With this result it is evident that
\begin{align}
\int_{0}^{2\pi} \ln(\sin^{2}(2\theta)) \, d\theta = - \pi \, \ln 2.
\end{align}
This result may also be obtained by the proper changes and the link to a former question presented by Chapper's comments.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1335164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Why does $\cos(x) + \cos(y) - \cos(x + y) = 0$ look like an ellipse? The solution set of $\cos(x) + \cos(y) - \cos(x + y) = 0$ looks like an ellipse. Is it actually an ellipse, and if so, is there a way of writing down its equation (without any trig functions)?
What motivates this is the following example. The solution set of $\cos(x) - \cos(3x + 2y) = 0$ looks like two straight lines, and indeed we can determine the equations of those lines.
$$
\begin{align}
\cos(x) &= \cos(3x + 2y) \\
\implies x &= \pm (3x + 2y) \\
\implies x + y &= 0 \text{ or } 2x + y = 0
\end{align}
$$
Can we do a similar thing for the first equation?
| we have $$\begin{align} 0 &= \cos x +\cos y -\cos(x + y)\\
&=\cos x +\cos y -\cos x \cos y+ \sin x \sin y\\
&=(1-\cos x)\cos y+\sin x \sin y+\cos x\\
&=2\sin^2 (x/2)\cos y+2\sin(x/2)\cos(x/2)\sin y + \cos x\\
&=2\sin(x/2)\left(\sin (x/2)\cos y+\cos(x/2)\sin y)\right)+\cos x\\
&=2\sin(x/2)\sin (x/2 + y)+\cos x\\
\end{align} $$
therefore $$y=-\frac x2-\sin^{-1}\left(\frac{\cos x}{2\sin (x/2)}\right)+2k\pi,-\frac x2 +\sin^{-1} \left(\frac{\cos x}{2\sin (x/2)}\right)+2k\pi + \pi .\tag 1 $$
when i graphed $(1),$ what i see kind of ellipses all running parallel to the line $y = -x/2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1336698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
"answer_count": 9,
"answer_id": 3
} |
Two different trigonometric identities giving two different solutions Using two different sum-difference trigonometric identities gives two different results in a task where the choice of identity seemed unimportant. The task goes as following:
Given $\cos 2x =-\frac {63}{65} $ , $\cos y=\frac {7} {\sqrt{130}}$, under the condition $0<x,y<\frac {\pi}{2}$, calculate $x+y$.
The first couple of steps are the same: finding $\sin$ and $\cos$ values for both $x$ and $y$.
From $\cos 2x$ we have:
\begin{align*}
\cos 2x &=-\frac {63}{65} \\
\cos2x &= \cos^2x-\sin^2x = \cos^2x - (1-\cos^2x)=2\cos^2x -1 \\
2\cos^2x -1 &= -\frac {63}{65} \\
2\cos^2x &=\frac {-63+65}{65} \\
\cos^2x &=\frac {1}{65} \\
\cos x &=\frac {1} {\sqrt{65}}
\end{align*}
(taking only the positive value of $\cos x$ because $\cos x$ is always positive under the given domain)
\begin{align*}
\sin^2x &=1-\frac {1}{65} \\
\sin^2x &=\frac {64}{65} \\
\sin x &=\frac {8} {\sqrt{65}}
\end{align*}
(again, only positive value)
From $\cos y$ we have:
\begin{align*}
\cos y &=\frac {7} {\sqrt{130}} \\
\cos^2y &=\frac {49} {130} \\
\sin^2y &=1-\frac {49} {130} \\
\sin^2y &=\frac {81} {130} \\
\sin y &=\frac {9} {\sqrt{130}}
\end{align*}
Now that we've gathered necessary information, we proceed to calculate value of some trigonometric function of $x+y$, hoping we will get some basic angle:
sin(x+y):
\begin{align*}
\sin(x+y) &=\sin x \cos y + \sin y \cos x =\frac {8} {\sqrt{65}} \frac {7} {\sqrt{130}} + \frac {9} {\sqrt{130}}\frac {1} {\sqrt{65}} \\
\sin(x+y) &=\frac {65} {\sqrt{65}\sqrt{130}} \\
\sin(x+y) &=\frac {\sqrt{2}}{2}
\end{align*}
Thus,
$x+y =\frac {\pi}{4}+2k{\pi}$ OR $x+y =\frac {3\pi}{4}+2k{\pi}$
Since $x$ and $y$ are in the first quadrant, their sum must lie in first or second quadrant.
Solutions are:
$x+y= \{\frac {\pi}{4}, \frac {3\pi}{4} \}$
cos(x+y):
\begin{align*}
\cos(x+y) &= \cos x \cos y - \sin x \sin y =\frac {1} {\sqrt{65}} \frac {7} {\sqrt{130}} - \frac {8} {\sqrt{65}}\frac {9} {\sqrt{130}} \\
\cos(x+y) &=-\frac {65} {\sqrt{65}\sqrt{130}} \\
\cos(x+y) &=-\frac {\sqrt{2}}{2}
\end{align*}
Thus,
$x+y =\frac {3\pi}{4}+2k{\pi} $ OR $x+y =\frac {5\pi}{4}+2k{\pi}$
Now we can only have one solution: $x+y=\{\frac {3\pi}{4}\}$
Similar happens with $\cos(x-y)$.
My question is: why do these two formulas give two different solutions? General insight would be great, since I found a lot of examples with similar problems. Thank you in advance.
| In your calculation of $x+y$ via $\sin(x+y)$, you wrote that $x+y=\frac{\pi}{4}$ is a possibility. It is not, since already $\sin y=\frac{9}{\sqrt{130}}\gt\frac{1}{\sqrt{2}}$, so $y\gt \frac{\pi}{4}$.
Remark: In this case, computing $\cos(x+y)$ is the better strategy, since the cosine here unambiguously identifies $x+y$. Computing $\sin(x+y)$ also works, but requires an extra step to disambiguate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
How to solve this system of equations for $x^2+y^2+z^2$? For the complex numbers $x,y,z$, the system of equations
$x^2-yz=i~~~~~
y^2-zx=i~~~~~
z^2-xy=i$
It is not easy for me to get $x^2+y^2+z^2$ from the above. I don't need the values of $x,y,z$
I'm stuck in what to do at first. Any advice would be helpful.
My attempt : we can get $x^2+y^2+z^2-xy-yz-zx=3i$,
so $(x-y)^2+(y-z)^2+(z-x)^2=6i$
I actually don't know whether this is a right way to solve the above.
| Subtracting the equations in pairs we get
$$(x-y)(x+y+z) = 0$$
$$(x-z)(x+y+z) = 0$$
$$(y-z)(x+y+z) = 0$$
If $x+y+z \not = 0$ then $x=y=z$, but then $x^2-yz = 0 \not = i$ so we must have $x+y+z = 0$.
Next we can use the identity
$$x^2 + y^2 + z^2 = (x+y+z)^2 - 2(xy+yz+xz)$$
togeather with the equation system to arrive at
$$x^2 + y^2 + z^2 = 2 i + \frac{(x+y+z)^2}{3} = 2i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Expand $\binom{xy}{n}$ in terms of $\binom{x}{k}$'s and $\binom{y}{k}$'s Motivated by this question, I want to find a complete set of relations for the ring of integer-valued polynomials, where the generators are the polynomials $\binom{x}{n}$ for $n\in \mathbb{N}$. The best way to do this is would be to describe how to decompose $\binom{x+y}{n}$ and $\binom{xy}{n}$ as a sum of products of $\binom{x}{0},\dots, \binom{x}{n}$ and $\binom{y}{0},\dots,\binom{y}{n}$. This can be done in principle by peeling off the binomials starting with the highest degree and working one's way down. Playing around with Sage, one soon guesses that
$\binom{x+y}{n} = \sum_{k=0}^n \binom{x}{k}\binom{y}{n-k}$
and in fact, I think this has a combinatorial proof which straightforwardly generalizes that of the identity $\binom{m}{n} = \binom{m}{n-1} + \binom{m-1}{n-1}$.
But $\binom{xy}{n}$ seems to be not so straightforward. The first few expansions are:
$\binom{xy}{2} = 2\binom{x}{2}\binom{y}{2} + x\binom{y}{2} + y \binom{x}{2}$
$\binom{xy}{3} = 6\binom{x}{3} \binom{y}{3} + $
$\qquad ~ ~ 6 \binom{x}{3}\binom{y}{2} + 6\binom{x}{2}\binom{y}{3} + $
$\qquad ~ ~ x \binom{y}{3} + 4 \binom{x}{2} \binom{y}{2} + y \binom{x}{3} $
$\binom{xy}{4} = 24\binom{x}{4}\binom{y}{4} + $
$\qquad ~ ~ 36\binom{x}{3}\binom{y}{4} + 36 \binom{x}{4}\binom{y}{3} + $
$ \qquad ~ ~ 14 \binom{x}{2}\binom{y}{4} + 45\binom{x}{3}\binom{y}{3} + 14 \binom{x}{4}\binom{y}{2} + $
$\qquad ~ ~ 12 \binom{x}{2}\binom{y}{3} + 12 \binom{x}{3}\binom{y}{2} + $
$\qquad ~ ~ \binom{x}{2}\binom{y}{2}$
and it is not so easy to discern a pattern.
This must be well-known: what is a closed-form expression for the expansion of $\binom{xy}{n}$?
| A beautiful combinatorial answer was found by Gjergji Zajmi. For an expanded version, see the solution to Exercise 3.9 in my Notes on the combinatorial fundamentals of algebra. (At least, it is Exercise 3.9 in the version of 10 January 2019. In future versions, the numbering can shift.)
Unlike Martin's answer, this one directly gives a formula for $\dbinom{xy}{n}$ as a nonnegative linear combination of products $\dbinom{x}{k}\dbinom{y}{\ell}$, as opposed to a polynomial in the $\dbinom{x}{k}$ and the $\dbinom{y}{\ell}$. This, of course, does not work for arbitrary $\lambda$-rings.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Finding $\int_{0}^{\frac{\pi}{2}}\frac{1}{\cos (x-\frac{\pi}{3}).\cos (x-\frac{\pi}{6})}\mathrm{d}x$ How can I find $$\int_{0}^{\frac{\pi}{2}}\frac{1}{\cos (x-\frac{\pi}{3}).\cos (x-\frac{\pi}{6})}\mathrm{d}x$$ ? I suspect this has something simple to do with the basic definite integral properties; but can't find a way through.
| Let $$\displaystyle y = \frac{x-\frac{\pi}{3}+x-\frac{\pi}{6}}{2} = x-\frac{\pi}{4}\Rightarrow x= \left(y+\frac{\pi}{4}\right)$$ and $dx = dy$ and changing limit
Put into $$\displaystyle I = \int_{0}^{\frac{\pi}{2}}\frac{1}{\cos \left(x-\frac{\pi}{3}\right)\cdot \cos \left(x-\frac{\pi}{6}\right)}dx = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{1}{\cos \left(y-\frac{\pi}{12}\right)\cdot \cos \left(y+\frac{\pi}{12}\right)}dy$$
Now Using the formula $$\bullet \; \cos(A+B)\cdot \cos(A-B) = \cos^2A-\sin^2 B.$$
So Integral $$\displaystyle I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{1}{\cos^2 y-\sin^2 \left(\frac{\pi}{12}\right)}dy = 2\int_{0}^{\frac{\pi}{4}}\frac{1}{\cos^2 y -a^2}dy = 2\int_{0}^{\frac{\pi}{4}}\frac{\sec^2 y}{1 -a^2(1+\tan^2 y)}dy$$
Where $\displaystyle a= \sin \frac{\pi}{12}$
Now Let $\tan y = t\;,$ Then $\sec^2 ydy = dt$ and Changing Limit, We get
$$\displaystyle I = 2\int_{0}^{1}\frac{1}{1-a^2(1+t^2)}dt = \frac{1}{a^2}\int_{0}^{1}\frac{1}{\left(\frac{\sqrt{1-a^2}}{a}\right)^2-t^2}dt = \frac{2}{a^2}\cdot \frac{a}{2\sqrt{1-a^2}}\ln\left|\frac{\sqrt{1-a^2}+at}{\sqrt{1-a^2}-at}\right|_{0}^{1}$$
So $$\displaystyle I = \frac{1}{a\sqrt{1-a^2}}\ln \left|\frac{\sqrt{1-a^2}+a}{\sqrt{1-a^2}-a}\right| = 2\cdot \ln \left|\frac{\cos \frac{\pi}{12}+\sin \frac{\pi}{12}}{\cos \frac{\pi}{12}-\sin \frac{\pi}{12}}\right|^2=2\ln (3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to solve the integral $\int\tan^{3}x \sec^{3/2}x\; dx$? How to solve the following indefinite integral
$$\int \tan^{3}x \sec^{3/2}x \; dx$$
to get the solution in the form of
$$\large\frac{2}{7}\sec^{7/2}x - \frac{2}{3}\sec^{3/2}x +c$$
I tried taking
$$u = \sec^{2}x \implies du = \tan x \; dx$$
$$\large \int \tan^{3} x \sec^{\frac{3}{2}}x\;dx = \int (u-1) u ^{\frac{3}{4}}du = \int u^{\frac{7}{4}}- u^{\frac{3}{4}}du = \frac{4}{11} (\sec x)^{\frac{11}{2}} \frac{4}{7} (\sec x)^{\frac{7}{2}} +c$$
Where did I go wrong?
| Your $du$ is wrong. $(\tan x)'=\sec^2 x$ (not the other way around).
Instead, write the integrand as $\sec x \tan x \tan^2 x \, (\color{maroon}{\sec x})^{1/2}$ and let $u=\color{maroon}{\sec x}$. Write $\tan^2 x$ in terms of $\color{maroon}{\sec x}$ and note $du$ then is $\sec x\tan x\, dx$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1347207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The Laurent series around $z=0$ of the function $f(z) = \frac{z}{(z-i)(z-2)}$ in the annulus $A(0,1,2)$ What I got so far:
$$
\frac{z}{(z-i)(z-2)} = \frac{z}{(2-i)(z-i)} + \frac{z}{(2-i)(z-2)}
$$
which is equal to
$$
\frac{z}{(2-i)(z-i)} + \frac{z}{(2-i)(z-2)} = \frac{z}{(2-i)z + 1-2i} + \frac{z}{(2-i)z + 2i - 4}
$$
and
$$
\frac{z}{(2-i)z + 1-2i} + \frac{z}{(2-i)z + 2i - 4} = \frac{1}{z(2-i)}\bigg(\frac{1}{1 + \frac{1 - 2i}{z(2-i)}} - \frac{1}{1 + \frac{2i - 4}{z(2-i)}}\bigg)
$$
Now if we just look at
$$
\frac{1}{1 + \frac{2i - 4}{z(2-i)}} = \frac{1}{1 - \frac{2 - i}{z}}
$$
and this reminds me of a geometric progression. But I don't know how to continu.. Could anyone finish this for me or give me tips? Thanks.
| (z/((z-i)(z-2)))=(z/((2-i)(z-i)))+(z/((2-i)(z-2))),1<∣z∣<2. The expansion in powers of z is
(z/((2-i)z(1-i/z)))-(z/((2-i)2(1-z/2)))
=(1/(2-i))∑₀((i/z))ⁿ-(1/(4-2i))∑₀((z/2))ⁿ.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral of $\frac{x^2+1}{(1-x^2)\sqrt{1+x^4}}$ So we have to evaluate $\int\frac{x^2+1}{(1-x^2)\sqrt{1+x^4}}dx$.
My work-
We can write the integrand as $\frac{(x+1)^2-2x}{(1-x)(1+x)\sqrt{1+x^4}}dx$.
So we wish to deduce $\int\frac{(x+1)}{(1-x)\sqrt{1+x^4}}dx-\int\frac{2x}{(1-x^2)\sqrt{1+x^4}}dx$
So lets write it as $I_1+I_2$.
I tried and I can't evaluate $I_1$.
For $I_2$, I took $x^2=t;\ 2xdx=dt$.
The integral becomes $\int\frac{dt}{(1-t)\sqrt{1+t^2}}dt$.
Okay, now we do $t=\tan\theta; \ dt=\sec^2\theta d\theta.$
$\int\frac{\sec\theta}{1-\tan\theta} d\theta$
Now what to do? Please guide me.?
| If we replace $x$ by $\sqrt{u}$ we have:
$$ \int\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}\,dx=\int\frac{(1+u)}{2(1-u)\sqrt{u(1+u^2)}}\,du$$
Then, if we replace $\frac{1+u}{1-u}$ by $v$ we have:
$$ \int\frac{(1+u)}{2(1-u)\sqrt{u(1+u^2)}}\,du =\frac{1}{2\sqrt{2}}\int\frac{v}{\sqrt{v^4-1}}\,dv=\frac{1}{2\sqrt{2}}\,\log\left(v^2+\sqrt{v^4-1}\right)$$
so:
$$ \int\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}\,dx= \frac{1}{2\sqrt{2}}\log\left(\left(\frac{x^2+1}{x^2-1}\right)^2+\sqrt{\left(\frac{x^2+1}{x^2-1}\right)^4-1}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
A limit problem: $\lim\limits_{n\to\infty}\frac{1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n} }{1+\frac{1}{3}+\frac{1}{9}+\cdots+\frac{1}{3^n} }$ I need help in solving the limit below:
$$\lim_{n\to\infty}\frac{1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n} }{1+\frac{1}{3}+\frac{1}{9}+\cdots+\frac{1}{3^n} }$$
What I've done is to simplify the upper part to: $$\frac{2^{n+1}-1}{2^n}$$
Any hints or solutions will be greatly appreciated.
| If you multiply the top by $2$, you get $2x = 2+x$, whence $x = 2$
Likewise, the bottom multiplied by $3$ gives $3y = 3+y$, gives $y = \frac{3}{2}$.
So you then find $\left(\frac{2}{1.5}\right) = \frac{4}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
If det $A = 0$ and $\det B \neq 0$ then show that $abc = -1$ This has been hurting my head for a while now....
If
$$
\det\begin{bmatrix}a&a^2&1+a^3\\b&b^2&1+b^3\\c&c^2&1+c^3\end{bmatrix}=0
$$
And
$$
\det\begin{bmatrix}a&a^2&1\\b&b^2&1\\c&c^2&1\end{bmatrix} ≠0
$$
Then show that $abc=-1$.
| The hard part is to compute the determinants and factor them nicely:
$
\det\begin{bmatrix}a&a^2&1\\b&b^2&1\\c&c^2&1\end{bmatrix} = (ab^2+bc^2+ca^2)- (b^2c+c^2a+a^2b) = (a-b)(b-c)(c-a)
$
and
$\det\begin{bmatrix}a&a^2&1+a^3\\b&b^2&1+b^3\\c&c^2&1+c^3\end{bmatrix}$
$= (ab^2(1+c^3)+bc^2(1+a^3)+ca^2(1+b^3)) - ((1+a^3)b^2c+(1+b^3)c^2a+(1+c^3)a^2b)$
$= (ab^2+bc^2+ca^2) - (b^2c+c^2a+a^2b)$ $+ (ab^2c^3+bc^2a^3+ca^2b^3) - (a^3b^2c+b^3c^2a+c^3a^2b)$
$= \left[(ab^2+bc^2+ca^2) - (b^2c+c^2a+a^2b)\right] + abc\left[(bc^2+ca^2+ab^2) - (a^2b+b^2c+c^2a)\right]$
$= \left[(ab^2+bc^2+ca^2) - (b^2c+c^2a+a^2b)\right](1+abc)$
$= (a-b)(b-c)(c-a)(1+abc)$
The rest is easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Find the cubic equation of roots $α, β, γ$. Taken from Fitzpatrick $4$ unit course textbook. The question says:
If the cubic equation $\ ax^3+bx^2+cx+d$ has roots $α, β, γ$. Find the cubic equation who's roots are $α^2, β^2, γ^2$
I keep getting a $±$ sign that I can't get rid of. The answer in the back is $x(ax+c)^2=(bx+d)^2$
Thanks for any help.
| $$ax^3+bx^2+cx+d=a^3(x-\alpha)(x-\beta)(x-\gamma)$$
Set $x=y$ to get $$ay^3+by^2+cy+d=a^3(y-\alpha)(y-\beta)(y-\gamma)\ \ \ \ (1)$$
Set $x=-y$ to get $$a(-y)^3+b(-y)^2+c(-y)+d=a^3(-y-\alpha)(-y-\beta)(-y-\gamma)$$
$$\iff ay^3-by^2+cy-d=a^3(y+\alpha)(y+\beta)(y+\gamma)\ \ \ \ (2)$$
Multiply $(1),(2)$ to get $$a^6(y^2-\alpha^2)(y^2-\beta^2)(y^2-\gamma^2)=(ay^3+cy)^2-(by^2+d)^2$$
Set $y^2=u$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Straight line is tangent to the curve. The straight line $y=mx+1$ is tangent to the curve $x^2+y^2-2x+4y=0$. Find the possible values of $m$.
My attempt
Substitute the $y=mx+1$ into the equation $x^2+y^2-2x+4y=0$. $$x^2+(mx+1)^2-2x+4(mx+1)=0$$ $$x^2+m^2x^2+2mx+1-2x+4mx+4=0$$ $$(1+m^2)x^2+6mx-2x+5=0$$ $$(1+m^2)x^2+(6m-2)x+5=0$$
I think what I did is wrong as I don't know how to continue from my steps. Can anyone explains it? Thanks
| Hint: Differentiating implicitly yields $$2x + 2y \frac{\mathrm{d}y}{\mathrm{d}x} - 2 + 4\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \iff \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1-x}{y+2}$$
What can you say about gradient of tangent lines and derivatives of functions?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Uniform Convergence of $\sum_{n=0}^\infty (1-x^2)^2x^n$ on $[0,1]$; subsequent integral Let $a_n = (1-x^2)^2 x^n$. Show that $\sum_{n=0}^\infty a_n$ converges uniformly on $[0,1]$ and deduce that $\int_0^1 \frac{(1-x^2)^2}{1-x} dx = \sum_{n=1}^\infty \frac{8}{n(n+2)(n+4)}$.
Attempt: Denoting partial sums by $S_n$, we have
\begin{eqnarray*}
S_n &=& \sum_{k=0}^n (1-x^2)^2 x^k \\
&=&(1-x^2)^2 \frac{1-x^{n+1}}{1-x} \\
&=& (1+x)(1-x^2)(1-x^{n+1}).
\end{eqnarray*}
I tried to show uniform convergence by showing convergence of $\sup_{x \in [0,1]} S_n(x)$, but I seem unable to solve the resulting first order condition to find the maximizing x. Is there a simpler way to do this?
If I could show uniform convergence, I could integrate the infinite series term by term yielding
\begin{eqnarray*}
\int \frac{(1-x^2)^2}{1-x} dx &=& \sum_{n=0}^\infty \int_0^1 (1-x^2)^2 x^n dx \\ &=&\sum_{n=0}^\infty \int_0^1 x^n - 2x^{2+n} + x^{4+n} dx
\\ &=& \dots
\\ &=&\sum_{n=0}^\infty \frac{8}{(n+1)(n+3)(n+5)}
\\ &=&\sum_{n=1}^\infty \frac{8}{n(n+2)(n+4)}
\end{eqnarray*}
and I would be done.
| To show uniform convergence of this series we can use the Weierstrass M-test. How big can $f_n(x)=(1-x^2)x^n$ be on $[0,1]?$ Set $f_n'(x) = 0.$ We get $x=\sqrt {n/(4+n)}.$ At this $x$ we have
$$ f_n(\sqrt {n/(4+n)}) = [1-n/(4+n)]^2\cdot[\sqrt {n/(4+n)}\,]^n.$$
The first factor on the right is $[4/(4+n)]^2.$ The second one is $\le1$ for all $n.$ Thus $0\le f_n(x) \le [4/(4+n)]^2, x\in [0,1].$ Since $\sum [4/(4+n)]^2<\infty,\sum f_n$ converges uniformly on $[0,1]$ by Weierstrass M.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1355046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Generate all multisets of length k for n symbols I am trying to generate a list of all multisets of length $k$ in a set with $n$ symbols. For example, if I had the set
$S = {A, B, C}$
I would expect the following output for $k = 2$ and $n = 3$:
$O = {(A,A),(A,B),(A,C),(B,B),(B,C),(C,C)}$
What is the proper way to go about generating this list?
| I presume you want to do this algorithmically. In Mathematica:
Union[Sort /@ Tuples[{A, B, C}, 2]]
{{A, A}, {A, B}, {A, C}, {B, B}, {B, C}, {C, C}}
More generally:
myMultiSet[n_Integer, k_Integer] := Union[Sort /@ Tuples[Range[n], k]]
So for instance myMultiSet[5,3] yields
{{1, 1, 1}, {1, 1, 2}, {1, 1, 3}, {1, 1, 4}, {1, 1, 5}, {1, 2, 2}, {1,
2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 3}, {1, 3, 4}, {1, 3, 5}, {1,
4, 4}, {1, 4, 5}, {1, 5, 5}, {2, 2, 2}, {2, 2, 3}, {2, 2, 4}, {2, 2,
5}, {2, 3, 3}, {2, 3, 4}, {2, 3, 5}, {2, 4, 4}, {2, 4, 5}, {2, 5,
5}, {3, 3, 3}, {3, 3, 4}, {3, 3, 5}, {3, 4, 4}, {3, 4, 5}, {3, 5,
5}, {4, 4, 4}, {4, 4, 5}, {4, 5, 5}, {5, 5, 5}}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Evaluate $ \lim_{(x,y)\to(0,0)} \frac {e^{x+y^2}-1-\sin \left ( x + \frac{y^2}{2} \right )}{x^2+y^2} $
$$ \lim_{(x,y)\to(0,0)} \frac {e^{x+y^2}-1-\sin \left ( x +
\frac{y^2}{2} \right )}{x^2+y^2} $$
I've a few doubts about this limit. I mean, if I take polar coordinates, I get that the limit doesn't exist. And Wolfram agrees with me. Even though, I've found a solution of this problem that doesn't say the same thing; which I transcribe next:
· The Taylor's second order polynomial of $ e^{x+y^2} $ in $ (0,0) $ is $ 1 + x + \frac{1}{2}x^2 + y^2 $.
· The Taylor's second order polynomial of $ \sin \left ( x + \frac{y^2}{2} \right ) $ in $ (0,0) $ is $ x + \frac{1}{2}y^2 $.
Then:
$$ \lim_{(x,y)\to(0,0)} \frac {e^{x+y^2}-1-\sin \left ( x + \frac{y^2}{2} \right )}{x^2+y^2} = \lim_{(x,y)\to(0,0)} \frac {1 + x + \frac{1}{2}x^2 + y^2-1-(x + \frac{1}{2}y^2)}{x^2+y^2} = \frac {1}{2} $$
Where's the mistake in this?
What I've tried so far:
$$ \lim_{(x,y)\to(0,0)} \frac {e^{x+y^2}-1-\sin \left ( x + \frac{y^2}{2} \right )}{x^2+y^2} $$
Let be $ \rho \geq 0 $ and $ \varphi \in [0,2\pi) $, so:
$$ \left\{\begin{matrix}
x = \rho\cos(\varphi)\\
y = \rho\sin(\varphi)
\end{matrix}\right. $$
Then:
$$ \begin{align*}
\lim_{\rho\to 0} \frac {e^{\rho\cos(\varphi)+(\rho\sin(\varphi))^2}-1-\sin \left ( \rho\cos(\varphi) + \frac{(\rho\sin(\varphi))^2}{2} \right )}{(\rho\cos(\varphi))^2+(\rho\sin(\varphi))^2} &= \lim_{\rho\to 0} \frac {\rho\cos(\varphi)+(\rho\sin(\varphi))^2-\left ( \rho\cos(\varphi) + \frac{(\rho\sin(\varphi))^2}{2} \right )}{(\rho\cos(\varphi))^2+(\rho\sin(\varphi))^2} \\
&= \lim_{\rho\to 0} \frac {\rho\cos(\varphi)+(\rho\sin(\varphi))^2-\left ( \rho\cos(\varphi) + \frac{(\rho\sin(\varphi))^2}{2} \right )}{\rho^2} \\
&= \lim_{\rho\to 0} \frac {(\rho\sin(\varphi))^2- \frac{(\rho\sin(\varphi))^2}{2}}{\rho^2} \\
&= \lim_{\rho\to 0} \frac {1}{2}\sin^2(\varphi)
\end{align*} $$
Which limit doesn't exit, because the result depends of $ \varphi $ which varies in $ [0,2\pi) $.
And here you can see that Wolfram agrees.
According to a comment below, I've calculated a third order Taylor's polynomial (respecto to $ \rho$) of $ e^{\rho\cos(\varphi)+(\rho\sin(\varphi))^2} $. I found that any Taylor's polynomial (in $ \rho = 0 $) of order grater than 2 is exactly $ 1+\rho\cos(\varphi)+(\rho\sin(\varphi))^2 $.
| For $(x,y)$ near $(0,0),$ we have
$$e^{x+y^2} = 1 + (x+y^2) +(x+y^2)^2/2 + O((x+y^2)^3), \sin (x+y^2/2) = (x+y^2/2) +O((x+y^2/2)^3).$$
So the numerator of our expression is
$$(y^2/2 + x^2/2) + xy^2 + y^4/2 + O((x+y^2)^3) + O((x+y^2/2)^3)= (y^2/2 + x^2/2) + r(x,y).$$
Dividing by $x^2 + y^2$ we get $1/2 + r(x,y)/(x^2+y^2).$ Now $r(x,y)/(x^2+y^2)\to 0.$ I'll leave this to you for now. Thus the desired limit is $1/2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1357224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Proving $\sin 20^\circ \, \sin 40^\circ \, \sin 60^\circ \, \sin 80^\circ =\frac{3}{16}$ The problem is to prove that
$$\sin 20^\circ \, \sin 40^\circ \, \sin 60^\circ \, \sin 80^\circ =\frac{3}{16}$$
All my attempts were to get them in $\sin (2A)$ form after eliminating $\sin 60^\circ$ in both sides. Unfortunately, all these attempts were futile.
Any hints are welcomed.
| Denote $Q = \sin 20 \sin 40 \sin 60 \sin 80$. Observe first that
\begin{align}
\sin 60 & = \sin (20+40) \\
& = \sin 20 \cos 40 + \cos 20 \sin 40 \\
& = \sin 20 (2\cos^2 20 - 1) + \cos 20 (2 \sin 20 \cos 20) \\
& = (4\cos^2 20 - 1) \sin 20
\end{align}
We can now write
\begin{align}
Q & = \sin 20 \sin (60-20) \sin 60 \sin (60+20) \\
& = \sin 20 (\sin 60 \cos 20 - \cos 60 \sin 20)
\sin 60 (\sin 60 \cos 20 + \cos 60 \sin 20) \\
& = \sin 20 \sin 60 (\sin^2 60 \cos^2 20 - \cos^2 60 \sin^2 20) \\
& = \sin 20 \sin 60 \frac{3\cos^2 20 - \sin^2 20}{4} \\
& = \sin 20 \sin 60 \frac{4\cos^2 20 - 1}{4} \\
& = \sin 60 \frac{(4\cos^2 20 - 1)\sin 20}{4} \\
& = \sin 60 \frac{\sin 60}{4} \\
& = \frac{\sin^2 60}{4} = \frac{3}{16}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1357880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Evaluate this limit at infinity $\lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2}$ Problem: Find the limit of \begin{align*} \lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2}
\end{align*}
Attempt at solution. The back of my textbook gives the answer as $-\frac{1}{4} \sqrt{2}$. Here's what I did: \begin{align*} \lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2} \\ = \lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{x^2 (4-6/x+7/x^2)} \\ \\ = \lim_{x \to \infty} \frac{\sqrt{x^2(1/x+1/x^2)}(1-\sqrt{x^2(2/x+3/x^2)}}{x(4-6/x+7/x^2)} \\ \\ \lim_{x \to \infty} \frac{\sqrt{1/x+1/x^2}(1-x \sqrt{2/x+3/x^2})}{4-6/x+7/x^2}
\end{align*} If I now evaluate this limit, everything in the numerator goes to zero except $1$. And the denominator leaves me with $4$. So I thought the answer should be $1/4$?
| The $x\sqrt{2/x+\cdots}$ term does not go to zero, but to infinity. Even when you multiply out the numerator and get a $\bigl(\sqrt{1/x+\cdots}\bigr)x\sqrt{2/x+\cdots}$ term, this goes to $\sqrt2$ rather than to $0$.
The term that arises from the $1$ when you multiply out does go to zero.
What you should probably do instead is rewrite the original numerator as
$$ x\sqrt{x+1}(1-\sqrt{2x+3}) = x^2\sqrt{1+1/x}\left(\sqrt{1/x}-\sqrt{2+3/x}\right) $$
where you can then reduce by $x^2$ and see the remaining factors go towards $1$ and $-\sqrt 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Find the Sum using bijection Find the sum of $S=\displaystyle\sum_{i,j,k \ge 0, i+j+k=17} ijk$.
I am looking for a solution that uses some bijection.
I couldn't find any bijection.
I am able to do the problem by other method by observing that,
$S$ is the coefficient of $x^{17}$ in,
$(x+2x^2+3x^3+4x^4........)^3$ and then sum the last formal power series and then find the coefficient of $x^{17}$ in that sum.
But the hint says find a bijection.
so, please help.
| For this you'll require the sums up to fourth powers:
$\begin{align}
\sum_{0 \le k \le n} k
&= \frac{n (n + 1)}{2} \\
\sum_{0 \le k \le n} k^2
&= \frac{n (n + 1) (2 n + 1)}{6} \\
\sum_{0 \le k \le n} k^3
&= \frac{n^2 (n + 1)^2}{4} \\
\sum_{0 \le k \le n} n^4
&= \frac{n^2 (6 n^3 + 15 n^2 + 10 n - 1)}{30}
\end{align}$
So you want:
$\begin{align}
\sum_{\substack{i + j + k = n \\
i, j, k \ge 0}}
i j k
&= \sum_{0 \le i \le n}
i \sum_{0 \le j \le n - i}
j (n - i - j) \\
&= \sum_{0 \le i \le n}
i \sum_{0 \le j \le n - i} (n j - i j - j^2) \\
&= \sum_{0 \le i \le n}
i \left(
n \frac{(n - i) (n - i + 1)}{2}
- i \frac{(n - i) (n - i + 1)}{2}
- \frac{(n - i) (n - i + 1) (2 n - 2 i + 1}{6}
\right) \\
&= \sum_{0 \le i \le n}
\frac{i (n^3 - n) - i^2 (3 n^2 - 1) + 3 i^3 n - i^4}{6} \\
&= \frac{(n - 1) n (3 n^3 + 3 n^2 - 12 n - 10)}{360}
\end{align}$
Phew!
Thanks to maxima for routine algebra help. Any transcription errors are mine alone.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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"answer_id": 1
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Fast way to get a position of combination (without repetitions) This question has an inverse: (Fast way to) Get a combination given its position in (reverse-)lexicographic order
What would be the most efficient way to translate a combination of $k$-tuple into its positions within the $\left(\!\!\binom{n}{k}\!\!\right)$ combinations?
I need this to be fast for combinations of $\left(\!\!\binom{70}{7}\!\!\right)$ order of magnitude - very large, but not exceeding 2 billion (fits into int32 maximum value).
Below is an example of $\left(\!\!\binom{6}{3}\!\!\right)$, where the aim is to quickly translate (a, d, f) tuple to value 9, while in the real problem $k$ ranges between 5 and 8.
$$\begin{array}
{cccccc|c}
a&b&c&d&e&f&^{combination}/_{sort\_order}&
\\\hline
x&x&x& & & &1\\
x&x& &x& & &2\\
x&x& & &x& &3\\
x&x& & & &x&4\\
x& &x&x& & &5\\
x& &x& &x& &6\\
x& &x& & &x&7\\
x& & &x&x& &8\\
x& & &x& &x&9\\
x& & & &x&x&10\\
.&.&.&.&.&.&.\\
& & &x&x&x&20\\
\end{array}$$
I know that I could pre-calculate all the combinations and reverse the lookup dictionary. However, such dictionary would not be efficient in terms of memory usage. Therefore I am looking for either calculation-based approach, or a more efficient data structure to perform this mapping.
| Let us denote your tuple [a b c] as [1 1 1 0 0 0] and so on.
Define $\binom{n}{r}=0$ for $n<r$
For your tuple: $[a d f] = [1 0 0 1 0 1]$
$$P = 1\cdot \binom{0}{1}+0\cdot \binom{1}{1}+0\cdot \binom{2}{1}+1\cdot \binom{3}{2}+0\cdot\binom{4}{2}+1\cdot\binom{5}{3} + 1$$
$$P=0 + 0 +0 +3+0+10+0+1 = 14$$
General Algorithm:
*
*Calculate the position value of each binary digit using $\binom{n}{r}$
*Take $n$ as position of the digit from left, for leftmost digit $n=0$.
*Write $r$ for each position as the number of 'ONES' counted from left, including the one at current position.
Example-1: [a b c] = [1 1 1 0 0 0]
Calculate the position of the tuple as sum:
$$P = 1\cdot \binom{0}{1}+1\cdot \binom{1}{2}+1\cdot \binom{2}{3}+0\cdot \binom{3}{3}+0\cdot\binom{4}{3}+0\cdot\binom{5}{3} + 1$$
$$P=0 + 0 +0 +0+0+0+0+1 = 1$$
Example-2: [d e f] = [0 0 0 1 1 1]
$$P = 0\cdot \binom{0}{0}+0\cdot \binom{1}{0}+0\cdot \binom{2}{0}+1\cdot \binom{3}{1}+1\cdot\binom{4}{2}+1\cdot\binom{5}{3} + 1$$
$$S=0+0+0+3+6+10+1=20$$
The lone ONE is added because you are not starting at zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 0
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Solve this exponential equation: $3^{2x}+\left(\frac{1}{2}\right)^{-x} \cdot 3^{x+1}-2^{2x+2}=0$ I tried solving this equation $$3^{2x}+\left(\frac{1}{2}\right)^{-x} \cdot 3^{x+1}-2^{2x+2}=0$$ by taking the log of both sides, but with no results, what do I do? Sorry if this equation is very easy, I couldn't solve it...
| we have $$(3^x)^2+2^x\cdot3^x\cdot 3-(2^x)^2\cdot2^2=0$$ divided by $(2^x)^2$ we have $$\left(\frac{3^x}{2^x}\right)^2+3\cdot \frac{3^x}{2^x}-4=0$$
setting $$u=\frac{3^x}{2^x}$$ we have a quadratic equation in $$u$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How find all complex numbers such that: $|\,1 - z\,| < k\ (1 - |\,z\,|\, )$? Let $k > 1$ be a real number. How may one find all complex numbers such that:
$|\,1 - z\,| < k\ (1 - |\,z\,|\, )$?
................................................................................................................................................................
Will it form a circle, with radius $\frac{k}{\sqrt{1+k^2}}$ and with its center at $\left( \frac{1}{\sqrt{1+k^2}}, 0 \right)$?
| This never can be a circle. However :
$$ z= x +iy $$
$$ |1-z| =\sqrt{ (1-x)^2 +y^2 } $$
$$ 1- |z|=1-\sqrt{x^2 +y^2} $$
$$ (1-x)^2 +y^2 <(K-K\sqrt{x^2 +y^2})^2$$
$$ (1-x)^2 +y^2 -K^2 -K^2(x^2+y^2)<-2K^2\sqrt{x^2 +y^2}$$
$$ 1+x^2-2x+y^2-K^2-K^2x^2-K^2y^2<-2K^2\sqrt{x^2 +y^2}$$
$$1-K^2+x^2(1-K^2) + y^2(1-K^2)-2x<-2K^2\sqrt{x^2 +y^2}$$
$$1-\frac{2}{(1-K^2)}x+x^2 + y^2<\frac{-2K^2}{(1-K^2)}\sqrt{x^2 +y^2}$$
$$\frac{(1-\frac{2}{(1-K^2)}x+x^2 + y^2)^2}{x^2 +y^2}<\frac{4K^4}{(1-K^2)^2}$$
$$\frac{(1-\frac{2}{(1-K^2)}x+x^2 + y^2)^2}{x^2 +y^2}<\frac{4K^4}{(1-K^2)^2}$$
$$ x=rcos\theta $$
$$ r=rsin\theta$$
$$\frac{(1+r^2-\frac{2rcos\theta}{(1-K^2)})^2}{r^2}<\frac{4K^4}{(1-K^2)^2}$$
you can plot this expression to investigate what it really is.
(The polar is optional)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
The converges of $ \sqrt 2 +\sqrt { 2-\sqrt 2} +\sqrt { 2-\sqrt { 2+\sqrt 2} } + \cdots =$ I would like to know wheather this series converge or diverge?
$\sqrt 2 +\sqrt { 2-\sqrt 2} +\sqrt { 2-\sqrt { 2+\sqrt 2} } +\sqrt { 2-\sqrt { 2+\sqrt { 2+\sqrt 2} } } +\cdots$
My work:
multiplyIing both demominator and numerator to $\sqrt { 2+\sqrt { 2 } } $,$\sqrt { 2+\sqrt { 2+\sqrt { 2 } } } $,$\sqrt { 2+\sqrt { 2+\sqrt { 2+\sqrt { 2 } } } } $ respectively and so on..
$$\sqrt { 2 } +\frac { \sqrt { 2 } }{ \sqrt { 2+\sqrt { 2 } } } +\frac { \sqrt { 2-\sqrt { 2 } } }{ \sqrt { 2+\sqrt { 2+\sqrt { 2 } } } } +\frac { \sqrt { 2-\sqrt { 2+\sqrt { 2 } } } }{ \sqrt { 2+\sqrt { 2+\sqrt { 2+\sqrt { 2 } } } } } +...=\\ =\sqrt { 2 } +\frac { \sqrt { 2 } }{ \sqrt { 2+\sqrt { 2 } } } +\frac { \sqrt { 2 } }{ \left( \sqrt { 2+\sqrt { 2 } } \right) \left( \sqrt { 2+\sqrt { 2+\sqrt { 2 } } } \right) } +\frac { \sqrt { 2 } }{ \left( \sqrt { 2+\sqrt { 2 } } \right) \left( \sqrt { 2+\sqrt { 2+\sqrt { 2 } } } \right) \left( \sqrt { 2+\sqrt { 2+\sqrt { 2+\sqrt { 2 } } } } \right) } +...=\\ =\sqrt { 2 } \left[ 1+\frac { 1 }{ \sqrt { 2+\sqrt { 2 } } } +\frac { 1 }{ \left( \sqrt { 2+\sqrt { 2 } } \right) \left( \sqrt { 2+\sqrt { 2+\sqrt { 2 } } } \right) } +\frac { 1 }{ \left( \sqrt { 2+\sqrt { 2 } } \right) \left( \sqrt { 2+\sqrt { 2+\sqrt { 2 } } } \right) \left( \sqrt { 2+\sqrt { 2+\sqrt { 2+\sqrt { 2 } } } } \right) } +... \right] $$
we can write every term as :${ a }_{ n+1 }=\frac { 1 }{ \sqrt { 2+{ a }_{ n } } } ,{ a }_{ 0 }=\sqrt { 2 } $
it is obvious that the sequences are monotone decreasing so ${ a }_{ n }>{ a }_{ n+1 }$ and I wrote sum as:$$S_{ \infty }=\sqrt { 2 } \sum _{ i=0 }^{ \infty }{ \frac { 1 }{ { a }_{ i+1 }\sqrt { 2+{ a }_{ i } } } } $$(i am not sure about it)
I am stuck here,I suspect series converges but how to show,which convergence tests can i use i don't know?Any hints,help will be appriceated?
P.S.I apologize for my english
| Let:
$$ c_1=\sqrt{2},\quad c_2=\sqrt{2+\sqrt{2}},\quad c_{n+1}=\sqrt{2+c_n}$$
and $d_n=c_n/2$. Then $d_1=\cos\frac{\pi}{4}$ and $d_{n+1}=\sqrt{\frac{1+d_n}{2}}$, by recognizing the cosine duplication formula, give:
$$ c_n = 2 \cos\frac{\pi}{2^{n+1}} $$
so:
$$ \sqrt{2} = 2\sin\frac{\pi}{4},\quad \sqrt{2-\sqrt{2}}=2\sin\frac{\pi}{8},\qquad \sqrt{2-c_n} = 2\sin\frac{\pi}{2^{n+2}}$$
and we are asking if:
$$ \sum_{n\geq 0}2\sin\frac{\pi}{2^{n+2}} $$
is convergent, but that is trivial since $0<\sin\frac{\pi}{2^{n+2}}<\frac{\pi}{2^{n+2}}$. We may also compute such a sum:
$$ 2\sum_{n\geq 0}\sin\frac{\pi}{2^{n+2}}=2\sum_{m\geq 0}\frac{(-1)^m \pi^{2m+1}}{(2m+1)!}\sum_{n\geq 0}\frac{1}{2^{(n+2)(2m+1)}}=2\sum_{m\geq 0}\frac{(-1)^m \pi^{2m+1}}{(2m+1)!\left(2^{4m+2}-2^{2m+1}\right)}.$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that $(\sqrt2 − 1)^n, \forall n \in \mathbb{Z^+}$ can be represented as $\sqrt{m} − \sqrt{m−1}$ for some $m \in \mathbb{Z^+}$ (no induction). From the 1994 Canada National Olympiad:
Prove that $(\sqrt2 − 1)^n, \forall n \in \mathbb{Z^+}$ can be represented as $\sqrt{m} − \sqrt{m−1}$ for some $m \in \mathbb{Z^+}$.
I think one solution method is fairly straightforward:
*
*Prove the dual claim that $(\sqrt2 + 1)^n, \forall n \in \mathbb{Z^+}$ can be represented as $\sqrt{m} + \sqrt{m−1}$ for some $m \in \mathbb{Z^+}$, for then $(\sqrt2 − 1)^n = \dfrac{1}{(\sqrt2 + 1)^n} = \dfrac{1}{\sqrt{m} + \sqrt{m−1}} = \sqrt{m} - \sqrt{m−1}$.
*Express $(\sqrt2 + 1)^n = a_n\sqrt2 + b_n$, and formulate a suitable Induction Hypothesis: $IH: (\sqrt2 + 1)^n = a_n\sqrt2 + b_n\text{ with }{b_n}^2-2{a_n}^2 = (-1)^n, n \in \mathbb{Z^+}$.
*Verify $IH$ for $n=1$, and prove the induction step.
Does anyone know of an elegant non-inductive alternative to this method?
Closed form $a_n, b_n$ are fine as long they are not conjured out of thin air (OEIS etc.).
| Step 1
Let us define
$$
\psi(p,q) = p \sqrt{2} + q.
$$
Let us define
$$
\Psi = \big\{ \psi(p,q) | p, q \in \mathbb{Z} \big\}.
$$
We have
$$
\big(p \sqrt{2} + q \big) \big(r \sqrt{2} + s \big)
= \big( p s + q r \big) \sqrt{2} + \big( 2 p r + q s \big).
$$
Therefore
$$
\forall \psi_1, \psi_2 \in \Psi : \psi_1 \psi_2 \in \Psi.
$$
But also
$$
\forall \psi \in \Psi, \forall n \in \mathbb{N} : \psi^n \in \Psi.
$$
Whence for $\psi = \sqrt{2} + 1 \in \Psi$, we get
$$
\forall n \in \mathbb{N} : \big( \sqrt{2} + 1 \big)^n \in \Psi\\
\Downarrow
$$
$$
\exists p, q \mathbb{Z} : \big( \sqrt{2} + 1 \big)^n = p \sqrt{2} + q.
$$
Step 2
Let us define $p_n \in \mathbb{Z}$ and $q_n \in \mathbb{Z}$ such that
$$
\big( \sqrt{2} + 1 \big)^n = p_n \sqrt{2} + q_n.
$$
Thus
$$
p_{n+1} \sqrt{2} + q_{n+1}
= \big( \sqrt{2} + 1 \big) \big( p_n \sqrt{2} + q_n \big)
= \big( p_n + q_n \big) \sqrt{2} + \big( 2 p_n + q_n \big).
$$
So we obtain the recursion relation
$$
\left[
\begin{array}{rcl}
p_0 &=& 0\\\\
q_0 &=& 1\\\\
p_{n+1} &=& p_n + q_n\\\\
q_{n+1} &=& 2 p_n + q_n
\end{array}
\right.
$$
Step 3
Whence
$$
p_1 = p_0 + q_0 = 1
$$
and
$$
p_{n+2} = p_{n+1} + q_{n+1}
= p_{n+1} + 2 p_n + q_n
= p_{n+1} + 2 p_n + p_{n+1} - p_{n}
= 2 p_{n+1} + p_{n}.
$$
Therefore
$$
\left[
\begin{array}{rcl}
p_0 &=& 0\\\\
p_1 &=& 1\\\\
p_{n+2} &=& 2 p_{n+1} + p_{n}\\\\
\hline\\
q_{n} &=& p_{n+1} - p_{n}
\end{array}
\right.
$$
Step 4
The recursion
$$
p_{n+2} = 2 p_{n+1} + p_{n}
$$
is a brother of Fibonacci, as Fibonacci is given by
$F_{n+2} = F_{n+1} + F_{n}$.
We can write
$$
p_{n+2} = 2 p_{n+1} + p_{n}\\
\Downarrow\\
p_{n+2} + \big( \phi - 2 \big) p_{n+1}
= \phi p_{n+1} + p_{n}\\
\Downarrow\\
\phi p_{n+2} + \big( \phi^2 - 2 \phi \big) p_{n+1}
= \phi \big( \phi p_{n+1} + p_{n} \big).
$$
The case
$$
\phi^2 - 2 \phi = 1
$$
yields
$$
\phi p_{n+2} + p_{n+1} = \phi \big( \phi p_{n+1} + p_{n} \big).
$$
So
$$
\phi p_{n+1} + p_{n} = \phi^n \big( \phi p_{1} + p_{0} \big).
$$
As
$$
\phi^2 - 2 \phi = 1 \Rightarrow \phi_\pm = 1 \pm \sqrt{2}
$$
we obtain
$$
\begin{array}{rclc}
\phi_+ \phi_- p_{n+1} + \phi_+ p_{n} &=&
\phi_+ \phi_-^n \big( \phi_- p_{1} + p_{0} \big)\\
\phi_+ \phi_- p_{n+1} + \phi_- p_{n} &=&
\phi_- \phi_+^n \big( \phi_+ p_{1} + p_{0} \big)\\
&&&-\\
\hline\\
\big( \phi_+ - \phi_- \big) p_{n} &=&
\phi_+ \phi_-^n \big( \phi_- p_{1} + p_{0} \big)
- \phi_- \phi_+^n \big( \phi_+ p_{1} + p_{0} \big)
\end{array}
$$
Whence
$$
p_{n} = - \phi_+ \phi_- \frac{ \phi_+^n - \phi_-^n }{ \phi_+ - \phi_- } p_1
+ \phi_+ \phi_- \frac{ \phi_+^{n-1} - \phi_-^{n-1} }{ \phi_+ - \phi_- } p_0.
$$
As $p_0=0$, $p_1=1$ and $\phi_+ \phi_- = -1$, we obtain
$$
p_{n} = \frac{ \phi_+^n - \phi_-^n }{ \phi_+ - \phi_- } \in \mathbb{Z}.
$$
As
$$
q_n = p_{n+1} - p_n,
$$
we get
$$
q_n = \frac{ \phi_+^{n+1} - \phi_-^{n+1} }{ \phi_+ - \phi_- }
- \frac{ \phi_+^n - \phi_-^n }{ \phi_+ - \phi_- }
= \frac{ [\phi_+ - 1 ]\phi_+^n - [ \phi_- - 1] \phi_-^n }{ \phi_+ - \phi_- }
= \frac{ \phi_+^n + \phi_-^n }{ \phi_+ - \phi_- } \sqrt{2},
$$
so we obtain
$$
q_{n} = \frac{ \phi_+^n + \phi_-^n }{ \phi_+ - \phi_- } \sqrt{2} \in \mathbb{Z}.
$$
Step 5
Eventually we obtain
$$
( \sqrt{2} + 1 )^n = \frac{ \phi_+^n - \phi_-^n }{ \phi_+ - \phi_- } \sqrt{2}
+ \frac{ \phi_+^n + \phi_-^n }{ \phi_+ - \phi_- } \sqrt{2}.
$$
So
$$
( \sqrt{2} + 1 )^n =
\sqrt{ 2 \left( \frac{ \phi_+^n - \phi_-^n }{ \phi_+ - \phi_- } \right)^2 }
+ \sqrt{ 2 \left( \frac{ \phi_+^n + \phi_-^n }{ \phi_+ - \phi_- } \right)^2 }
$$
Note that
$$
p_{n} = \frac{ \phi_+^n - \phi_-^n }{ \phi_+ - \phi_- }
\in \mathbb{Z} \Rightarrow
2 \left( \frac{ \phi_+^n - \phi_-^n }{ \phi_+ - \phi_- } \right)^2
\in \mathbb{Z},
$$
Now comes the fun part:
$$
2 \left( \frac{ \phi_+^n + \phi_-^n }{ \phi_+ - \phi_- } \right)^2
= 2 \left( \frac{ \phi_+^n - \phi_-^n }{ \phi_+ - \phi_- } \right)^2
+ 8 \frac{ \big( \phi_+ \phi_- \big)^n }{ \big( \phi_+ - \phi_- \big)^2 },
$$
and as $\phi_+ \phi_- = -1$ and $\phi_+ - \phi_- = 2 \sqrt{2}$,
we get
$$
2 \left( \frac{ \phi_+^n + \phi_-^n }{ \phi_+ - \phi_- } \right)^2
= 2 \left( \frac{ \phi_+^n - \phi_-^n }{ \phi_+ - \phi_- } \right)^2
+ (-1)^n.
$$
Let
$$
m = 2 \left( \frac{ \phi_+^n - \phi_-^n }{ \phi_+ - \phi_- } \right)^2
+ \frac{1 + (-1)^n}{2}.
$$
Whence
$$
( \sqrt{2} + 1 )^n = \sqrt{m} + \sqrt{m-1}.
$$
Step 6
Note that
$$
( \sqrt{2} + 1 ) ( \sqrt{2} - 1 ) = 1
$$
and
$$
( \sqrt{m} + \sqrt{m-1} ) ( \sqrt{m} - \sqrt{m-1} ) = 1,
$$
then
$$
( \sqrt{2} - 1 )^n
= \frac{1}{ ( \sqrt{2} + 1 )^n }
= \frac{1}{ \sqrt{m} + \sqrt{m-1} }
= \sqrt{m} - \sqrt{m-1}.
$$
Whence
$$
( \sqrt{2} - 1 )^n = \sqrt{m} - \sqrt{m-1}.
$$
Conclusion
$$
( \sqrt{2} \pm 1 )^n = \sqrt{m} \pm \sqrt{m-1},
$$
where
$$
m = 2 \left( \frac{ \phi_+^n - \phi_-^n }{ \phi_+ - \phi_- } \right)^2
+ \frac{1 + (-1)^n}{2},
$$
and
$$
\phi_\pm = 1 \pm \sqrt{2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
simplify and evaluate $\frac{\tan80^\circ-\tan20^\circ}{1+\tan80^\circ\tan20^\circ}$ How do you simplify and evaluate $\dfrac{\tan80^\circ-\tan20^\circ}{1+\tan80^\circ\tan20^\circ}$?
What is the problem asking?
| Hint:
Try to reduce this problem to one that makes use of the more familiar $\sin$ and $\cos$ sum of angle identities.
$$\begin{array}{lll}
\displaystyle\frac{\tan 80^\circ - \tan 20^\circ}{1+\tan80^\circ\tan20^\circ} &=&\displaystyle\frac{\tan 80^\circ + \tan (-20^\circ)}{1-\tan80^\circ\tan(-20^\circ)}\\
&=&\displaystyle\frac{\tan 80^\circ + \tan (-20^\circ)}{1-\tan80^\circ\tan(-20^\circ)}\cdot\frac{\cos 80^\circ\cos(-20^\circ)}{\cos 80^\circ\cos(-20^\circ)}\\
&=&\dots
\end{array}$$
Hint 2: $\tan 80^\circ\cos 80^\circ = ?$
You can take it from here I think.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Help needed to solve combinatorics problem. I have been revisiting my old probability courses and I found a problem, which I can't figure out how to solve or at least what I get differs from the answer in the book.
The problem reads as follows:
Given 8 balls i.e. 5 white 1 black, 1 blue and 1 red, find how many sequences of 6 objects can be formed (ordering matters!).
Hint1:
Only permutations and permutations with repetition might be used. This is taken from a German textbook (I have translated it to my best), and the question comes at the end of the chapter on permutations. Notice that variations $\frac{n!}{(n-k)!}$ and combinations $\frac{n!}{(n-k)!k!}$ are not covered yet.
Hint2:
The answer according to the textbook is 178.
Please note, that all the answers given below use the combination formula to get the answer and they all differ from 178, which is supposed to be the right answer.
| Use exponential generating functions. The white balls give:
$$
1 + \frac{z}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \frac{z^4}{4!} + \frac{z^5}{5!}
$$
Each of the other colors gives:
$$
1 + \frac{z}{1!}
$$
In all, we want:
$$
6! [z^6] \left( 1 + \frac{z}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \frac{z^4}{4!} + \frac{z^5}{5!} \right) \cdot ( 1 + z)^3
$$
My tame symbolic algebra system (maxima) tells me this is 228.
Alternatively, it can be computed by hand (ellipses are terms that don't affect the result):
$\begin{align}
6! [z^6] \left( e^z - \frac{z^6}{6!} - \dots \right) \cdot (1 + z)^3
&= 6! \left( [z^6] e^z (1 + 3 z + 3 z^2 + z^3) - [z^6] \frac{z^6}{6!} \cdot (1 + \dots) \right) \\
&= 6! \left(\frac{1}{6!}
+ 3 \cdot \frac{1}{5!}
+ 3 \cdot \frac{1}{4!}
+ \frac{1}{3!}
- \frac{1}{6!}
\right) \\
&= 228
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
How to solve $z^3 + \overline z = 0$ I need to solve this:
$$z^3 + \overline z = 0$$
how should I manage the 0?
I know that a complex number is in this form: z = a + ib so:
$$z^3 = \rho^3\lbrace \cos(3\theta) + i \sin (3\theta)\rbrace$$
$$\overline z = \rho\lbrace \cos(-\theta) + i \sin (-\theta)\rbrace$$
but how about the 0?
EDIT:
ok, following some of your comments/answers this is what I have done:
$$z^3 = - \overline z$$
$$\rho^3\lbrace \cos(3\theta) + i \sin (3\theta)\rbrace = \rho\lbrace \cos(-\theta) + i \sin (-\theta)\rbrace$$
So
$$
\begin{Bmatrix}
\rho^3 = \rho\\
3\theta = -\theta + 2k\pi
\end{Bmatrix}$$
$$
\begin{Bmatrix}
\rho^3 = \rho\\
2\theta = 2k\pi
\end{Bmatrix}$$
$$
\begin{Bmatrix}
\rho = 0 or \rho = 1\\
\theta = k\frac{\pi}{2}
\end{Bmatrix}$$
is this the right way?
| One solution is $z=0$. When $z\ne0$, multiplying by $z$ will not alter the solutions of the equations; so we can look at $z^4+|z|^2=0$. If we write $z=re^{it}$, the equation becomes
$$
r^4e^{4it}+r^2=0.
$$
As we are assuming $r\ne0$, this reduces to $-r^2e^{4it}=1$. So $r=1$ and $e^{4it}=-1$, that is (since $-1=e^{i\pi}$), $4t=\pi+2k\pi$. Thus
$$
t=\frac14\,\left(\pi+2k\pi\right)=\frac{\pi}4+\frac{k\pi}2,\ \ k=0,1,2,3.
$$
So we get four solutions, from $z=\cos t+ i\sin t$, which are
$$
\frac{\sqrt2}2+i\,\frac{\sqrt2}2,\ \ -\frac{\sqrt2}2+i\,\frac{\sqrt2}2,\ \ \frac{\sqrt2}2-i\,\frac{\sqrt2}2,\ \ \text{ and }-\frac{\sqrt2}2-i\,\frac{\sqrt2}2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 7
} |
Third point of intersection is also a point of inflection? Let $C \subset \mathbb{P}_2$ be a nonsingular cubic. If $L$ is a line through two distinct points of inflection on $C$, how do I show that the third point of intersection is also a point of inflection?
| This is Problem 2 in section 5.7, "Elliptic Curves," of I. Niven, H. S. Zuckerman, H. L. Montgomery, An Introduction to the Theory of Numbers, 5th ed., Wiley (New York), 1991. I use the nomenclature from that book.
After I worked out this answer, I realized that it is the same as Lubin's.
$C$ is elliptic. Let $\mathbf{A}$ and $\mathbf{B}$ be the two distinct inflection points and $\mathbf{AB}$ the third point. Choose $\mathbf{0}$ to be an inflection point. We use the following facts. The proof of the main result is farther down.
*
*Fact 1: $\mathbf{AA} = \mathbf{A}$, $\mathbf{BB} = \mathbf{B}$, and $\mathbf{00} = \mathbf{0}$.
*Converse of Fact 1: If $\mathbf{PP} = \mathbf{P}$, then $\mathbf{P}$ is an inflection point.
*Fact 2: $\mathbf{P}(\mathbf{0P}) = \mathbf{0}$. Indeed, $\mathbf{0P}$ is the third point on the line through $\mathbf{0}$ and $\mathbf{P}$. The line through $\mathbf{P}$ and $\mathbf{0P}$ is the same line, so the third point on the line is the original point $\mathbf{0}$.
*Fact 3: $\mathbf{A} + \mathbf{A} + \mathbf{A} = \mathbf{0}$, proved as follows:
\begin{align}
\mathbf{A} + \mathbf{A} + \mathbf{A} & = \mathbf{A} + \mathbf{0(AA)} &&\text{definition of +}\\
& = \mathbf{A} + \mathbf{0(A)} &&\text{Fact 1}\\
& = \mathbf{0(A(0A))} &&\text{definition of +}\\
& = \mathbf{00} &&\text{Fact 2}\\
& = \mathbf{0} &&\text{Fact 1}
\end{align}
Likewise, $\mathbf{B} + \mathbf{B} + \mathbf{B} = \mathbf{0}$.
*Converse of Fact 3: If $\mathbf{P} + \mathbf{P} + \mathbf{P} = \mathbf{0}$, then $\mathbf{P}$ is an inflection point. To prove that, we have
\begin{align}
\mathbf{P} + \mathbf{P} + \mathbf{P} & = \mathbf{0}\\
\mathbf{0(P(0(PP)))} & = \mathbf{00} &&\text{definition of + on left; Fact 1 on right}\\
\mathbf{P(0(PP))} & = \mathbf{0} &&\text{by uniqueness}
\end{align}
We also know that $\mathbf{P(0P)} = \mathbf{0}$ by Fact 2. Comparing these last two equations, we see that $\mathbf{PP} = \mathbf{P}$ by uniqueness. Hence, $\mathbf{P}$ is an inflection point by the converse of Fact 1.
*Fact 4: $\mathbf{A} + \mathbf{B} + \mathbf{AB} = \mathbf{0}$ is proved at The sum of three colinear rational points is equal to $O$. (That proof applies to general points, not necessarily rational.)
We are now ready for the proof of the main result:
\begin{align}
\mathbf{0} & = \mathbf{A} + \mathbf{A} + \mathbf{A} + \mathbf{B} + \mathbf{B} + \mathbf{B} &&\text{Fact 3}\\
& = \mathbf{A} + \mathbf{B} + \mathbf{A} + \mathbf{B} + \mathbf{A} + \mathbf{B} &&\text{+ is commutative}\\
& = -\mathbf{AB} - \mathbf{AB} - \mathbf{AB} &&\text{Fact 4}
\end{align}
Because $-\mathbf{0} = \mathbf{0}$, we have that $\mathbf{AB} + \mathbf{AB} + \mathbf{AB} = \mathbf{0}$. Thus, by the converse of Fact 3, $\mathbf{AB}$ is an inflection point.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding Linear Combination of Polynomials I am stuck on a question involving finding the greatest common divisor of polynomials and then solving to find the linear combination of them yielding the greatest common divisor. My work thus far is as follows:
Define: $f(x) = x^3-3x+3$, $g(x) = x^2 - 4$.
Execute the Euclidean Algorithm for Polynomials as follows:
$x^3-3x+3 = x(x^2 -4)+(x+3)$
$x^2-4 = (x-3)(x+3) +5$
$x+3 = \frac x5(5) +3$
$5 = 1(3) +2$
$3 = 1(2) +1$
$2 = 1(2) +0$
So we know that $\gcd(f(x),g(x)) = 1$. Now, my objective is to find polynomials $s(x), t(x)$ such that $f(x)s(x) + g(x)t(x) = 1$.
I have tried using back substitution, a method involving matrix multiplication of coefficients, and a method using continuous fractions. None of these have worked- all returning incorrect answers, two of which were the same. Does anyone have any suggestions as to where I am going wrong, or a simpler way to finish this computation?
Note: it is not specified in my book which field these polynomials are defined over. I am assuming the reals.
| We chose polynomials $f(x)$ and $g(x)$ such that $$f(x)(x^2-4) + g(x)(x^3-3x+3)=1$$ and for this it is necessary the equality of degrees.
So $d(f)+2=d(g)+3\iff d(f)=d(g)+1$.
Chose the simplest possibilities
$f(x)=ax^2+bx+c$ and $g(x)=dx+e$ and calculate the coefficients.It follows
$(a+d)x^4+(b+e)x^3+(-4a+c-3d)x^2+(-4b+3d-3e)x+(-4c+3e)=1$
Solving the resulting five equations equalizing coefficients we get
$f(x)=\frac {x^2-3x+1}{5}$ and $g(x)=\frac {-x+3}{5}$ which gives the identity
$$(x^2-3x+1)(x^2-4)+ (-x+3)(x^3-3x+3)=5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
First principle derivative of a square root and conjugates I'm trying to find the derivative of the equation: $$g(x)=\sqrt {x+2}-3x^2$$.
I can find the solution just fine using the power rule but am finding trouble with First Principles.
Essentially, I understand getting as far as $$\displaystyle\lim_{h\to 0}\frac{\sqrt {x+h+2}-3(x+h)^2 -\sqrt {x+2}+3x^2}{h}.$$ From here I can expand out to $$\lim_{h\to 0}\frac{\sqrt {x+h+2}-3x^2-6xh-3h^2 -\sqrt {x+2}+3x^2}{h}.$$ But then I get stuck.
I'm not sure if I should use the conjugate rule now (but then how would I even apply that?) or if I'm supposed to try and simplify.
The answer is $\dfrac{1}{2\sqrt {x+2}}-6x$ that I got using the power rule.
Any help and guidance is appreciated.
| $$\lim_{h\to0}{\frac{g(x+h)-g(x)}{h}}$$
$$\lim_{h\to0}{\frac{\sqrt{x+h+2}+3(x+h)^2-\sqrt{x+2}-3x^2}{h}=\frac{\sqrt{x+h+2}-\sqrt{x+2}-3x^2-6xh-3h^2+3x^2}{h}}$$
$3x^2$ is eliminated, and $3h^2$ is negligible:
$$=\frac{\sqrt{x+h+2}-\sqrt{x+2}-6xh}{h}=\frac{\sqrt{x+h+2}-\sqrt{x+2}}{h}-\frac{6xh}{h}$$
now using the conjugate rule
$$=\frac{(\sqrt{x+h+2}-\sqrt{x+2})(\sqrt{x+h+2}+\sqrt{x+2})}{h(\sqrt{x+h+2}+\sqrt{x+2})}-6x=\frac{(x+h+2)-(x+2)}{h(\sqrt{x+h+2}+\sqrt{x+2})}-6x=\frac{h}{h(\sqrt{x+h+2}+\sqrt{x+2})}-6x$$
$$=\frac{1}{(\sqrt{x+h+2}+\sqrt{x+2})}-6x$$
and since $h\to0$ we have:
$$\frac{1}{2\sqrt{x+2}}-6x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Given the differential equation, how to solve the y function with x as the independent variable? $y\frac{dy}{dx} = x(y^4 + 2y^2 + 1)$
$y = 1$ when $x = 4$
I tired to integrate by substitution, but it doesn't seem to work out.
| Notice, we have $$y\frac{dy}{dx}=x(y^4+2y^2+1)$$ $$\frac{y}{y^4+2y^2+1}dy=xdx$$ $$\frac{y}{(y^2+1)^2}dy=xdx$$ Now, integrating both the sides we have $$\int \frac{y}{(y^2+1)^2}dy=\int xdx$$ $$\frac{1}{2}\frac{(y^2+1)^{-1}}{-1}=\frac{x^2}{2}+c$$ $$-\frac{1}{2(y^2+1)}=\frac{x^2}{2}+c$$ Substituting $y=1$ & $x=4$, we get
$$\frac{-1}{2((1)^2+1)}=\frac{(4)^2}{2}+c\implies c=\frac{-31}{4}$$ Hence, the solution is $$-\frac{1}{2(y^2+1)}=\frac{x^2}{2}-\frac{31}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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How to integrate a function of form f (x)/g (x) $\frac {(ax^n+b) dx}{cx^m+d}$ How to find $\int \frac {(ax^n+b) dx}{cx^m+d}$ (where m>n and m,n are natural numbers,and a,b,c,d are integers) ?
| The general approach will involve:
*
*If $n\geq m$, use long division to get a polynomial plus a fraction of two polynomials where the degree of the numerator is smaller than $m$.
*Factoring $cx^m+d$ in factors with linear and quadratic terms.
*Applying partial fraction decomposition.
*Seperating all partial fractions in individual integrals.
*Applying either the ln method or the arctan method.
$m=1$, $n=1$. Then \begin{align*} \int \frac{ax+b}{cx+d} dx &= \int \frac{a}{c} + \frac{b-\frac{da}{c}}{cx+d} dx \\ &= \int \frac{a}{c} + \frac{bc-da}{c^2x+cd} dx \\ &= \int \frac{a}{c}dx + \int \frac{bc-da}{c^2x+cd} dx \\ &= \frac{a}{c}x + \frac{bc-da}{c}\int \frac{1}{cx+d} dx\end{align*}
Substitution $u=cx+d$.
\begin{align*}\\ \frac{a}{c}x + \frac{bc-da}{c}\int \frac{1}{cu} du &= \frac{a}{c}x + \frac{bc-da}{c^2}\int \frac{1}{u} du \\ &= \frac{a}{c}x + \frac{bc-da}{c^2}\ln(u) \\ &= \frac{a}{c}x + \frac{bc-da}{c^2}\ln(cx+d) \end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
New Idea to prove $1+2x+3x^2+\cdots=(1-x)^{-2}$
Given $|x|<1 $ prove that $\\1+2x+3x^2+4x^3+5x^4+...=\frac{1}{(1-x)^2}$.
1st Proof: Let $s$ be defined as
$$
s=1+2x+3x^2+4x^3+5x^4+\cdots
$$
Then we have
$$
\begin{align}
xs&=x+2x^2+3x^3+4x^4+5x^5+\cdots\\
s-xs&=1+(2x-x)+(3x^2-2x^2)+\cdots\\
s-xs&=1+x+x^2+x^3+\cdots\\
s-xs&=\frac{1}{1-x}\\
s(1-x)&=\frac{1}{1-x}\\
s&= \frac{1}{(1-x)^2}
\end{align}
$$
2nd proof:
$$
\begin{align}
s&=1+2x+3x^2+4x^3+5x^4+\cdots\\
&=\left(1+x+x^2+x^3+\cdots\right)'\\
&=\left(\frac{1}{1-x}\right)'\\
&=\frac{0-(-1)}{(1-x)^2}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
3rd Proof:
$$
\begin{align}
s=&1+2x+3x^2+4x^3+5x^4+\cdots\\
=&1+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+0+x^3+x^4+x^5+\cdots\\
&+\cdots
\end{align}
$$
$$
\begin{align}
s&=\frac{1}{1-x}+\frac{x}{1-x}+\frac{x^2}{1-x}+\frac{x^3}{1-x}+\cdots\\
&=\frac{1+x+x^2+x^3+x^4+x^5+...}{1-x}\\
&=\frac{\frac{1}{1-x}}{1-x}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
These are my three proofs to date. I'm looking for more ways to prove the statement.
| Let $S=1+2x+3x^2+4x^3+\dotsb$.
\begin{align}
\phantom{-x^2}S&=1+2x+3x^2+4x^3+\dotsb\\
\phantom{^2}-xS&=\phantom1-\phantom2x-2x^2-3x^3-\dotsb\\
\phantom{^2}-xS&=\phantom1-\phantom2x-2x^2-3x^3-\dotsb\\
\phantom{-}x^2S&=\phantom{1+2x+2}x^2+2x^3+\dotsb
\end{align}
Adding them together:
\begin{align}
(1-2x&+x^2)S\\
&=1+0x+0x^2+0x^3+\dotsb\\
&=1\\
S&=\frac1{1-2x+x^2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 15,
"answer_id": 14
} |
Trigonometric equation cos sin and power The problem is $2\cos t - 3\sin^2t +2 = 0$.
I get to $2\cos t -3\sin^2t =-2$
I think that I need to use a trigonometric identity like $\cos(x+y)$ and to divide $2\cos t -3\sin^2t$ with the $\sqrt{2^2+3^2}$
Do you know how to solve this? It should be $\sqrt{2^2 + 3^2}$
| $$2 \cos t - 3 \sin^2t +2 = 0\\.$$ Write $$\sin^2 t=1-\cos^2 t, $$ then you have a quadratic equation. Solve for $\cos t$ $$ 2\cos t -3(1-\cos^2 t)+2=0\\3\cos ^2 t+2\cos t-3+2=0\\\cos t=-1,\;\frac{1}{3}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Trigonometric equation with sine and cosine So the equation is $3\cos ^2t + 5\sin t = 1$
Now I have simplified this to $$3(1-\sin ^2t) + 5\sin t -1 = 0$$
which leads to $$-3\sin ^2t + 5\sin t + 2 = 0$$
Then I get $$-3t^2 + 5 t +2 = 0$$
Is this the correct way to go with this equation then use $t = t/2 \pm \sqrt {(t/2)^ 2 + y}$ where $y$ in this case will be $2/3$ ?
| $$3\cos ^2t + 5\sin t = 1$$
$$3(1-\sin ^2t) + 5\sin t -1 = 0$$
$$-3\sin ^2t + 5\sin t + 2 = 0$$
$$3\sin ^2t - 5\sin t - 2 = 0$$
Now , Let $\sin t =x$, then equation reduces to,
$$3x^2-5x-2=0$$
$$(3x+1)(x-2)=0$$, Now can you finish from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Solve $2\sin^3x + \sin3x +3\sin^2x \cos x + \cos^3x=0$ $2\sin^3x + \sin3x +3\sin^2x\cos x + \cos^3x=0$
My try:
$$2\sin^3x +3\sin x - 4\sin^3x +\cos x(3\sin^2x+\cos^2x)=0 $$
$$ \cos x(2\sin^2x+1) - 2\sin^3x+3\sin x=0.$$
And then i have no idea.
| Notice, $$2\sin^3x + \sin3x +3\sin^2x\cos x + \cos^3x=0$$
$$2\sin^3x +3\sin x - 4\sin^3x +3\cos x(1-\cos^2x)+cos^3x=0 $$
$$3\sin x - 2\sin^3x +3\cos x-3\cos^3x+cos^3x=0 $$
$$ 3(\cos x+\sin x)-2(\cos^3x+\sin^3x)=0$$ $$ 3(\cos x+\sin x)-2(\cos x+\sin x)(\cos^2x+\sin^2x-\cos x\sin x)=0$$ $$(\cos x+\sin x)(3-2(1-\cos x\sin x))=0$$$$\implies \cos x+\sin x=0\implies \frac{1}{\sqrt{2}}\cos x+\frac{1}{\sqrt{2}}\sin x=0$$ $$\implies \cos\left(x-\frac{\pi}{4}\right)=0 \implies x-\frac{\pi}{4}=(2n+1)\frac{\pi}{2}$$ $$\implies x=(2n+1)\frac{\pi}{2}+\frac{\pi}{4}$$$$\implies \color{blue}{x=\frac{(4n+3)\pi}{4}}$$ or $$\implies 3-2(1-\cos x\sin x)=0\implies 2\sin x\cos x=-1\implies 2x=2n\pi-\frac{\pi}{2}$$ $$\implies x=n\pi-\frac{\pi}{4}=\color{blue}{\frac{(4n-1)\pi}{4}}$$ Where $\color{blue}{n}$ is any integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Finding the horizontal and vertical tangents of a parametric equation.
Find the points at which the polar curve $r=2+2\sin{(\theta)}$ has a
horizontal or vertical tangent line.
Translate the parametric equation to Cartesian coordinates:
$$
r^2=2r+2r\sin{(\theta)} \\
x^2+y^2=2\sqrt{x^2+y^2}+2y \\
x^2+y^2-2y+1=2\sqrt{x^2+y^2}+1 \\
x^2+(y-1)^2=2\sqrt{x^2+y^2}+1 \\
$$
Then differentiate with respect to $x$:
$$
\frac{\mathrm{d}}{\mathrm{d}x}\left[x^2+(y-1)^2\right]=\frac{\mathrm{d}}{\mathrm{d}x}\left[2\sqrt{x^2+y^2}+1\right] \\
2x+2(y-1)y'=\frac{2(x+y)}{\sqrt{x^2+y^2}} \\
x+(y-1)y'=\frac{x+y}{\sqrt{x^2+y^2}}y' \\
y'=\frac{x}{\frac{x+y}{\sqrt{x^2+y^2}}-(y-1)} \\
y'=\frac{x}{\frac{x+y-\sqrt{x^2+y^2}(y-1)}{\sqrt{x^2+y^2}}} \\
y'=\frac{x\sqrt{x^2+y^2}}{x+y-\sqrt{x^2+y^2}(y-1)} \\
$$
So, by now it seems I'm not making any real progress in finding the vertical and horizontal tangents of this equation.
Am I going about this wrong?
| Mistake in taking derivatives:
$$
\frac{\mathrm{d}}{\mathrm{d}x}\left[x^2+(y-1)^2\right]
=\frac{\mathrm{d}}{\mathrm{d}x}\left[2\sqrt{x^2+y^2}+1\right] \\
2x+2(y-1)y'=\frac{2x+2yy'}{\sqrt{x^2+y^2}}
$$
and you dropped the $y'$ on the RHS...
UPDATE
Another approach is to let $x=x(\theta)$ and $y = y(\theta)$ and compute
$$
\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}
$$
In our case, $r=2+2\sin \theta$ so $$x = r\cos \theta = \cos \theta (2+2\sin \theta)$$ and similarly $$y = r\sin \theta = \sin \theta (2+2\sin \theta)$$, so computing $y'(\theta)$ is straight-forward arithmetic.
Setting it to $0$ and $\pm \infty$ should give you the relevant angles...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why does $\sqrt{6} + \sqrt{10} + \sqrt{15}$ have four conjugates? I am having trouble understanding how algebraic number $\sqrt{6} + \sqrt{10} + \sqrt{15}$ has four conjugates.
Minimal polynomial is $x^4-62 x^2-240 x-239$ according to Wolfram Alpha.
Factorized:
$$\left(x-2\sqrt{15 (4-\sqrt{15})}-8\sqrt{4-\sqrt{15}}-\sqrt{15}\right)\cdot
\left(x-2\sqrt{4-\sqrt{15}}+\sqrt{15}\right) \\
\cdot \left(x+2\sqrt{4-\sqrt{15}}+\sqrt{15}\right)
\cdot\left(x+2\sqrt{15 (4-\sqrt{15})}+8\sqrt{4-\sqrt{15}}-\sqrt{15}\right)$$
| $$\sqrt{6}+\sqrt{10}+\sqrt{15}=\sqrt{2\cdot 3}+\sqrt{2\cdot 5}+\sqrt{3\cdot 5} = \frac{1}{2}\left(\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2-(2+3+5)\right)$$
where $\sqrt{2}+\sqrt{3}+\sqrt{5}$ is an algebraic number of degree $8$ over $\mathbb{Q}$, having conjugates $\pm\sqrt{2}\pm\sqrt{3}\pm\sqrt{5}$, whose minimal polynomial is an even function. It follows that $\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2$ is an algebraic number of degree $4$ over $\mathbb{Q}$, and the conjugates of $\sqrt{6}+\sqrt{10}+\sqrt{15}$ are given by:
$$ \frac{1}{2}\left(\left(\sqrt{2}\pm\sqrt{3}\pm\sqrt{5}\right)^2-(2+3+5)\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why is $\frac{1}{4/3} - \frac{1}{3/2}$ not the same as $\bigl(\frac{4}{3} - \frac{3}{2}\bigr)^{-1}$ If you have the problem:$$\frac{1}{\frac{4}{3}} - \frac{1}{\frac{3}{2}} =?$$
Why can't you change the problem to $(\frac{4}{3} - \frac{3}{2})^{-1}$ and get the same answer?
In the first scenario, the answer is $1/12$
In the second scenario, the answer is $-6$
Why doesn't this work?
| It looks like based on your post that you want to know why $\frac{1}{\frac{4}{3}} - \frac{1}{\frac{3}{2}}$ is not equal to $\frac{1}{\frac{4}{3} - \frac{3}{2}}$.
In general, you can only say a property holds like, in your case, $(a + b)^{-1} = a^{-1} + b^{-1}$ if you can show it holds for every $a$ and $b$. But as you pointed out in your question, if $a = \frac{4}{3}$ and $b = \frac{3}{2}$, this property doesn't hold. Therefore, it's not true that $(a + b)^{-1} = a^{-1} + b^{-1}$ holds. The property is false, as your example perfectly showed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380482",
"timestamp": "2023-03-29T00:00:00",
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Help in finding the integral function. Can somebody provide a hint in finding the following integral? $$\displaystyle \int \dfrac{1}{(x^3+1)^3} \text{ d}x$$ I thought of using partial fractions but that isn't making any sense.
| This following trick, which is essentially the same as @Daniel Fischer's suggestion, reduces your burden of calculating partial fraction decomposition. First, write
$$ \int \frac{dx}{(1+x^3)^3} = \int \frac{1 + x^3}{(1+x^3)^3} \, dx - \int x \cdot \frac{x^2}{(1+x^3)^3} \, dx. $$
Then by integration by parts, we have
$$ \int x \cdot \frac{x^2}{(1+x^3)^3} \, dx = - \frac{1}{6} \frac{x}{(1+x^3)^2} + \frac{1}{6}\int \frac{dx}{(1+x^3)^2}. $$
Plugging this back to the first identity, we get
$$ \int \frac{dx}{(1+x^3)^3} = \frac{1}{6} \frac{x}{(1+x^3)^2} + \frac{5}{6} \int \frac{dx}{(1+x^3)^2}. $$
Similarly, we have
$$ \int \frac{dx}{(1+x^3)^2} = \frac{1}{3} \frac{x}{(1+x^3)^2} + \frac{2}{3} \int \frac{dx}{1+x^3}. $$
Finally, the last integral can be calculated easily by partial fraction decomposition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve $2+ \cos{\frac{3x}{2}} + \sqrt{3} \sin{\frac{3x}{2}} = 4\sin^2{\frac{x}{4}}$ $$2+ \cos{\frac{3x}{2}} + \sqrt{3} \sin{\frac{3x}{2}} = 4\sin^2{\frac{x}{4}}$$
My try:
$$ \cos{\frac{3x}{2}} + \sqrt{3} \sin{\frac{3x}{2}} = \sqrt{4}\left(\frac{\sqrt{3}}{2} \sin{\frac{3x}{2}} + \frac{1}{2}\cos{\frac{3x}{2}}\right) = 2\sin\left({\frac{3x}{2} + \frac{\pi}{6}}\right)
$$
$$
2+ 2\sin\left({\frac{3x}{2} + \frac{\pi}{6}}\right) = 4\sin^2{\frac{x}{4}}
$$
$$
2\sin\left({\frac{3x}{2} + \frac{\pi}{6}}\right) + \cos{\frac{x}{2}}=0
$$
And then...
| $2+2\sin(\frac{3x}{2}+\frac{\pi}{6})=4 \sin^2(\frac{x}{4})$ simplifies to $2\sin(\frac{3x}{2}+\frac{\pi}{6})+2 cos(\frac{x}{2})=0$
$\sin(\frac{3x}{2}+\frac{\pi}{6})=-cos(\frac{x}{2})$
$\sin(\frac{3x}{2}+\frac{\pi}{6})=sin(\frac{3\pi}{2}-\frac{x}{2})$
I hope you can solve further.
| {
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New idea to solve this equation I was teaching $\left \lfloor x \right \rfloor$ function properties and equation . I solved this equation in my class . My works are show below. Some students ask me for new Idea...,and now I am looking for various method to solve this (like this) equation .$$\left \lfloor x\right \rfloor+2x=1$$
1st: $$x=n+p \\n \in\mathbb{Z} , 0 \leq p<1 \to \left \lfloor x \right \rfloor=n ,p=x-n\\ $$
and
$$ \left \lfloor x\right \rfloor+2x=1\\n+2(n+p)=1 \to 3n+2p=1 \\3n=0 ,\pm3 ,\pm 6,\pm9,... $$in this case $$3n=0 \to 2p=1 \\n=0 , p=\frac{1}{2} \to x=n+p=0+\frac{1}{2} $$
2nd:
$$\left \lfloor x\right \rfloor+2x=1 \to \left \lfloor x\right \rfloor=1-2x$$ like $f(x)=g(x)$ by drawing both of them obtain the answer
With respect to the picture ,it suffice to solve $0=1-2x$s o the answer is $x=\frac{1}{2}$
3rd :
we know $$\left \lfloor x\right \rfloor =k \in \mathbb{Z} \to k \leq x <k+1 $$ so $$\left \lfloor x\right \rfloor=1-2x \to 1-2x=k \in \mathbb{Z}\\x=\frac{1-k}{2} \to \left \lfloor \frac{1-k}{2}\right \rfloor =k$$ so we have
$$k \leq \frac{1-k}{2} <k+1 \to \\\left\{\begin{matrix}
k\leq \frac{1-k}{2} \to & 2k \leq 1-k \to & k \leq \frac{1}{3} \to k=\left \{ 0,-1,-2,-3,... \right \}\\
\frac{1-k}{2}<k \to & 1-k<2k \to &k> -\frac{1}{3} \to k=\left \{ 0,1,2,3,... \right \}
\end{matrix}\right.\\
\left \{ ...,-3,-2,-1,0 \right \}\bigcap \left \{ 0,1,2,3,... \right \}=\left \{ 0 \right \}\rightarrow k=0 \\\rightarrow x=\frac{1-k}{2}=\frac{1}{2}$$
| $x-1<\lfloor x\rfloor \leq x$, so $3x-1<\lfloor x\rfloor +2x=1\leq 3x$, this implies $x\in [1/3, 2/3)$, so $\lfloor x\rfloor=0$, then $2x=1$, $x=1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
ODE $x'' + 2x' +5x = \sin3t$, $x(0)=1,\ x'(0)=-1$, Solve using Laplace Transform While solving the differential equation
$$x'' + 2 x' + 5 x = \sin3t, \quad x(0) = 1, \quad x'(0) = -1$$
by use of Laplace transform I got to
$$X(s^2 +2s+5)=\frac{(3)}{s^2 +9} +s +1$$
$$X=\frac{\left(\:s^3+s^2+9s+12\right)}{\left(s^2+9\right)\left(s^2-2s+5^{\:}\right)}$$
Is it correct, if yes how do i find the inverse of it?
X=f(s)
| Notes:
*
*The first part of this solution is for an equation that contained an error. It remains here as a demonstration. The corrcted equation is the second half of this solution.
*The process of the proposer is correct in process but does need some guidance in finding a nice result.
Consider the differential equation
$$x'' + 2 x' + 5 x = 0, \quad x(0) = 1, \quad x'(0) = -1$$
and make use of
\begin{align}
\mathcal{L}\{x''(t)\} &= s^{2} f(s) - s f(0) - f'(0) = s^{2} f(s) - s + 1 \\
\mathcal{L}\{x'(t)\} &= s f(s) - f(0) = s f(s) - 1
\end{align}
then
\begin{align}
(s^2 + 2s + 5) f(s) = s + 1
\end{align}
such that
\begin{align}
f(s) = \frac{s+1}{(s+1)^{2} + 2^{2}}.
\end{align}
Using the known transform
\begin{align}
\mathcal{L}\{ e^{-at} \, \cos(b t)\} = \frac{s+a}{(s+a)^{2} + b^{2}}
\end{align}
then the solution to the differential equation is $x(t) = e^{-t} \, \cos(2 t)$.
Amended equation
$$x'' + 2 x' + 5 x = \sin(3 t)$$
becomes, after the Laplace transform,
\begin{align}
((s+1)^{2} + 4) \, f(s) = s+1 + \frac{3}{s^{2} + 3^{2}}
\end{align}
for which
$$f(s) = \frac{s+1}{(s+1)^{2} + 2^{2}} + \frac{1}{2} \, \frac{3}{s^{2} + 3^{2}} \cdot \frac{2}{(s+1)^{2} + 2^{2}}.$$
By using
\begin{align}
\mathcal{L}\{ e^{-at} \, \sin(bt)\} &= \frac{b}{(s+a)^{2} + b^{2}} \\
\mathcal{L}\left\{\int_{0}^{t} F(t-u) \, G(u) \, du \right\} &= f(s) \cdot g(s)
\end{align}
then
\begin{align}
x(t) &= e^{-t} \, \cos(2t) + \frac{1}{2} \, \int_{0}^{t} e^{-u} \, \sin(2u) \, \sin(3t-3u) \, du \\
&= \frac{1}{52} \, e^{-t} \, \left( 9 \, \sin(2t) + 58 \, \cos(2t) \right) - \frac{1}{26} \, \left( 2 \sin(3t) + 3 \cos(3t) \right).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Substitution to solve an initial value problem By using the substitution $y(x) = v(x)x$, how can I solve the initial value problem
$$
\frac{dy}{dx} = \frac{x^2+y^2}{xy - x^2},\quad y(1)=1
$$
And also keep my answer in the form $g(x,y)= 4e^{-1} xe^\frac{y}{x}$
| $$\frac { dy }{ dx } =\frac { { x }^{ 2 }+{ y }^{ 2 } }{ xy-{ x }^{ 2 } } =\frac { 1+\frac { { y }^{ 2 } }{ { x }^{ 2 } } }{ \frac { y }{ x } -1 } \\ \frac { y }{ x } =t\Rightarrow y=xt\Rightarrow dy=t+xdt\\ t+xdt=\frac { 1+{ t }^{ 2 } }{ t-1 } \\ xdt=\frac { 1+{ t }^{ 2 } }{ t-1 } -t=\frac { 1+{ t }^{ 2 }-{ t }^{ 2 }+t }{ t-1 } =\frac { t+1 }{ t-1 } \\ \int { \frac { t-1 }{ t+1 } dt } =\int { \frac { dx }{ x } } \\ \int { \left( 1-\frac { 2 }{ t+1 } \right) } dt=\int { \frac { dx }{ x } } \\ t-2\ln { \left| t+1 \right| =\ln { \left| x \right| +C } } \\ \frac { y }{ x } -2\ln { \left| \frac { y }{ x } +1 \right| =\ln { \left| x \right| +C } } \\ y\left( 1 \right) =1\\ 1-\ln { 4=C } \\ \frac { y }{ x } -2\ln { \left| \frac { y }{ x } +1 \right| =\ln { \left| x \right| +1- } \ln { 4 } } \\ \frac { y }{ x } =\ln { \left( { \frac { e }{ 4 } x\left( \frac { x+y }{ x } \right) }^{ 2 } \right) } \\ \\ { \frac { e }{ 4 } x\left( \frac { x+y }{ x } \right) }^{ 2 }={ e }^{ \frac { y }{ x } }\\ { \left( x+y \right) }^{ 2 }=4{ e }^{ \frac { y }{ x } -1 }\\ \\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$A+B+C=2149$, Find $A$ In the following form of odd numbers
If the numbers
taken from the form where $A+B+C=2149$
Find $A$
any help will be appreciate it, thanks.
| If $A$ were on the $n^{th}$ row, then $B=A+2n$ and $C=A+2(n+1)$.
Out of laziness, we will try to figure out what row this would need to be to be in the right ballpark through estimation. $\frac{2149}{3}\approx 716$, so $A<716<B<C$. Which row would $715$ fit into?
By changing your figure some by first adding one to each entry and then dividing by two, we will get the chart $\left[\begin{array}{}\color{red}{1}\\2&\color{red}{3}\\4&5&\color{red}{6}\\7&8&9&\color{red}{10}\\\vdots\end{array}\right]$, and these numbers are very familiar to us. The red numbers are the triangle numbers, $T(n)=\binom{n+1}{2}=\frac{n^2+n}{2}$.
So, $T(26)=351<\frac{715+1}{2}=358<378=T(27)$, so we suppose that $715$ occurs on the $26^{th}$ row, implying that either $A$ is on the $25^{th}$, $26^{th}$, or $27^{th}$ row (since we have been estimating up to this point).
So, we try to solve now, $A+B+C=2149=A+(A+2n)+(A+2n+2)=3A+4n+2$
In the case that $n=25$, this would be $3A+100+2=2149\Rightarrow 3A=2047\Rightarrow A=\frac{2047}{3}\not\in\mathbb{Z}$, so we know that $n=25$ was not possible.
In the case that $n=26$, this would be $3A+104+2=2149\Rightarrow 3A=2043\Rightarrow A=681$
In the case that $n=27$, this would be $3A+108+2=2149\Rightarrow 3A=2039\Rightarrow A=\frac{2039}{3}\not\in\mathbb{Z}$, so we know that $n=27$ was not possible.
We can similarly show that if $n=24$ that leads to a contradiction as well.
We believe then that our answer is $A=681, B=733, C=735$. All that remains to check is that $681$ is indeed on the $26^{th}$ row to confirm our calculations by checking that $T(25)$ is less than $\frac{A+1}{2}$ and $T(26)$ is greater than $\frac{A+1}{2}$.
Indeed, $T(25)=325<\frac{681+1}{2}=341<351=T(26)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 0
} |
$\frac{CE}{DE}=\frac{(a+b)^2}{kc^2}$ If bisector of angle C of an acute triangle ABC cuts the side AB in D and the circumcircle of triangle ABC in E,then $\frac{CE}{DE}=\frac{(a+b)^2}{kc^2}$.Then what is value of k?
Since angle bisector of C,bisects the arc BC.And circumcenter O will fall on line CE.And $\angle AOB=2\angle C$.These concepts i know but i caanot relate them to get answer.maybe i am lacking somewhere.Any suggestions or guidance is welcome.
|
$\frac{CE}{DE}=\frac{(a+b)^2}{kc^2}$. What is value of $k$?
Note that
\begin{align}
\angle EAB&=\angle EBA=\tfrac\gamma2
\end{align}
Assuming that
$|AB|=c$, $|BC|=a$, $|CA|=b$,
$\angle BAC=\alpha$,
$\angle ABA=\beta$,
$\angle ACB=\gamma$,
we can find $|CD|$ and $|CE|$ from the area relations:
\begin{align}
S_{\triangle ABC}&=
S_{\triangle ADC}+
S_{\triangle BDC}
\\
S_{CAEB}&=
S_{\triangle ABC}+S_{\triangle AEB}
=
S_{\triangle ACE}+S_{\triangle BCE}
\\
S_{\triangle AEB}&=
\tfrac14 c^2\tan\tfrac\gamma2
\end{align}
which gives
\begin{align}
|CD|
&=
\frac{ab\sin\gamma}{(a+b)\sin\tfrac\gamma2}
\\
|CE|
&=
\frac{ab\sin\gamma+\tfrac12 c^2\tan\tfrac\gamma2}{(a+b)\sin\tfrac\gamma2}
\\
|DE|&=|CE|-|CD|
=
\frac{\tfrac12 c^2\tan\tfrac\gamma2}{(a+b)\sin\tfrac\gamma2}
\\
\frac{|CE|}{|DE|}
&=
\frac{2ab\sin\gamma}{c^2\tan\tfrac\gamma2}+1
=
\frac{2ab(\cos\gamma+1)}{c^2}+1
\\
&=
\frac{a^2+b^2-c^2+2ab +c^2}{c^2}
=
\frac{(a+b)^2}{c^2}.
\end{align}
Hence $k=1$.
Edit:
Note that $|CE|$ is a common side of $\triangle BCE$ and $\triangle ACE$:
\begin{align}
S_{BCE}+S_{ACE}&=S_{ABC}+S_{AEB}
\\
\tfrac12|CE|a\sin\tfrac\gamma2
+
\tfrac12|CE|b\sin\tfrac\gamma2
&=
\tfrac12ab\sin\gamma
+
\tfrac14c^2\tan\tfrac\gamma2
\\
|CE|\sin\tfrac\gamma2(a+b)
&=
ab\sin\gamma
+
\tfrac12c^2\tan\tfrac\gamma2
\\
|CE|&=
\frac{ab\sin\gamma
+
\tfrac12c^2\tan\tfrac\gamma2}{\sin\tfrac\gamma2(a+b)}.
\end{align}
Edit 2:
In $\triangle AEB$ we have $\angle EAB=\angle EBA=\tfrac\gamma2$,
thus it's isosceles and $|AE|=|BE|$.
Its area $S_{AEB}=\tfrac12 h |AB|$, where
$h$ is the altitude, which splits
isosceles $\triangle AEB$
into two congruent right triangles
with the base $\tfrac12c$, so
$h=\tfrac12|AB|\tan\angle EAB=
\tfrac12c \tan\tfrac\gamma2$,
hence $S_{AEB}=\tfrac12 h c=\tfrac14c^2 \tan\tfrac\gamma2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $ (a+b) \cos(x) = a-b$, if $ \sin(x) + \tan(x) = \frac{4a\sqrt{ab}}{a^2-b^2},$ $ \tan(x) - \sin(x) = \frac{4a\sqrt{ab}}{a^2-b^2}.$ Prove that $ (a+b) \cos(x) = a-b$, if $$ \sin(x) + \tan(x) = \frac{4a\sqrt{ab}}{a^2-b^2},$$
$$ \tan(x) - \sin(x) = \frac{4a\sqrt{ab}}{a^2-b^2}.$$
I tried solving it with system, but with no result.
| grand_chat has already pointed out that you may want
$$\tan x-\sin x=\frac{4\color{red}{b}\sqrt{ab}}{a^2-b^2}\tag 1$$
With
$$\tan x+\sin x=\frac{4a\sqrt{ab}}{a^2-b^2}\tag 2$$
now $(1)+(2)$ gives
$$2\tan x=\frac{4(a+b)\sqrt{ab}}{(a+b)(a-b)}\Rightarrow \tan x=\frac{2\sqrt{ab}}{a-b}$$
Also, $(2)-(1)$ gives
$$2\sin x=\frac{4(a-b)\sqrt{ab}}{(a-b)(a+b)}\Rightarrow \sin x=\frac{2\sqrt{ab}}{a+b}$$
Here, use
$$\tan x=\frac{\sin x}{\cos x}\Rightarrow \cos x=\frac{\sin x}{\tan x}=\left(\frac{2\sqrt{ab}}{a+b}\right)/\left(\frac{2\sqrt{ab}}{a-b}\right)=\frac{a-b}{a+b}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Closed form for an infinite product The following fascinating formula appears in the paper "On gamma quotients and infinite products" by M.Chamberland and A.Straub (see page 9):
$$\prod_{n=2}^\infty\left(1-\frac{1}{n(n-1)}\right)=-\frac1\pi\cos\left(\frac{\sqrt5}{2}\pi\right).$$
Unfortunately, it's given without any reference or explanation. Could anyone give one?
| We will be only dealing with the following identities:
$$
\Gamma (z) = \frac{ e^{ -\gamma z }}{ z } \, \prod_{ k=1 }^{ \infty } \left\{ \left( 1+\frac{ z }{ k } \right)^{ -1 } \, e^{ \frac { z }{ k } } \right\}
\hspace{10mm}
\Gamma (-z)\Gamma (z)\quad =\quad \frac { -\pi }{ z \sin { (\pi z) } }
$$
Let's start then ^^:
\begin{align}
\prod_{ n=2 }^{ \infty }{ \left( 1-\frac { 1 }{ n(n-1) } \right) } &= \prod_{ n=2 }^{ \infty }{ \left( \frac { { n }^{ 2 }-n-1 }{ n(n-1) } \right) } \\
&= \prod_{ n=2 }^{ \infty }{ \left( \frac { (n-{ \phi }_{ + }) }{ n } \right) } \prod _{ n=2 }^{ \infty }{ \left( \frac { (n-{ \phi }_{ - }) }{ (n-1) } \right) } \\
&= \prod_{ n=2 }^{ \infty }{ \left( 1-\frac { { \phi }_{ + } }{ n } \right) } \prod _{ k=1 }^{ \infty }{ \left( \frac { k+1-{ \phi }_{ - } }{ k } \right) } \\
&= \frac { \prod _{ k=1 }^{ \infty }{ \left( 1-\frac { { \emptyset }_{ + } }{ k } \right) } }{ 1-{ \phi }_{ + } } \prod _{ k=1 }^{ \infty }{ \left( 1+\frac { 1-{ \phi }_{ - } }{ k } \right) }
\end{align}
Now let's calculate each product using our identity:
\begin{align}
\prod_{ k=1 }^{ \infty }{ \left( 1+\frac { 1-{ \phi }_{ - } }{ k } \right) { e }^{ -\frac { 1-{ \phi }_{ - } }{ k } } }
&= \frac { { e }^{ -\gamma (1-{ \phi }_{ - }) } }{ (1-{ \phi }_{ - })\Gamma (1-{ \phi }_{ - }) } \\
\prod_{ k=1 }^{ \infty }{ \left( 1-\frac { { \phi }_{ + } }{ k } \right) } { e }^{ \frac { { \phi }_{ + } }{ k } }
&= \frac { { e }^{ \gamma { \phi }_{ + } } }{ -{ \phi }_{ + }\Gamma (-{ \phi }_{ + }) }
\end{align}
Now we will multiply both the products(observe that ${ \phi }_{ + }+{ \phi }_{ - } = 1$):
\begin{align}
\prod_{ k=1 }^{ \infty }{ \left( 1+\frac { 1-{ \phi }_{ - } }{ k } \right) } { e }^{ \frac { { \phi }_{ - }-1 }{ k } }\prod _{ k=1 }^{ \infty }{ \left( 1-\frac { { \phi }_{ + } }{ k } \right) } { e }^{ \frac { { \phi }_{ + } }{ k } }
&= \prod _{ k=1 }^{ \infty }{ \left( 1+\frac { 1-{ \phi }_{ - } }{ k } \right) \left( 1-\frac { {\phi }_{ + } }{ k } \right) } \\
&= \frac { 1 }{ -\Gamma (-{ \phi }_{ + })\Gamma (1-{ \phi }_{ - }){ \phi }_{ + }(1-{ \phi }_{ - }) }
\end{align}
Observe that:
$$
\Gamma (-{ \phi }_{ + })\Gamma (1-{ \phi }_{ - }) = \Gamma (-{ \phi }_{ + })\Gamma ({ \phi }_{ + }) = \frac { -\pi \csc { (\pi { \phi }_{ + }) } }{ { \phi }_{ + } }
$$
so:
$$
-\Gamma (-{ \phi }_{ + })\Gamma (1-{ \phi }_{ - }){ { \left( { \phi }_{ + } \right) }^{ 2 } } = \pi \csc { (\pi { \phi }_{ + }){ \phi }_{ + } }
$$
We get:
\begin{align}
\prod_{ n=1 }^{ \infty }{ \left( 1-\frac { 1 }{ n(n-1) } \right) }
&= \frac { \prod _{ k=1 }^{ \infty }{ \left( 1-\frac { { \phi }_{ + } }{ k } \right) } \prod _{ k=1 }^{ \infty }{ \left( 1+\frac { 1-{ \phi }_{ - } }{ k } \right) } }{ 1-{ \phi }_{ + } } \\
&= \frac { \sin { (\pi { \phi }_{ + }) } }{ \pi { \phi }_{ + }(1-{ \phi }_{ + }) } \\
&= -\frac { \sin { (\frac { \pi }{ 2 } +\frac { \sqrt { 5 } }{ 2 } \pi ) } }{ \pi } \\
&= -\frac { \cos { \left( \frac { \sqrt { 5 } \pi }{ 2 } \right) } }{ \pi }
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 3
} |
prove that $\int_0^\frac{\pi}{4} \frac{1-\cos x}{x} \,dx < \frac{\pi^2}{64}$ prove that $\int_0^\frac{\pi}{4} \frac{1-\cos x}{x} \,dx < \frac{\pi^2}{64}$
I showed that in
$$ \forall x \in [0,\frac{\pi}{4}] \quad \int_0^\frac{\pi}{4} \frac{1-\cos x}{x} \,dx \le \int_0^\frac{\pi}{4} \frac{1-\cos(\frac{\pi}{4})}{\frac{\pi}{4}} \,dx = 1 -\frac{\sqrt2}{2} $$
but it's not good enough
| Since:
$$\frac{\sin x}{x}=\prod_{n\geq 1}\left(1-\frac{x^2}{n^2\pi^2}\right) $$
over the interval $\left[0,\frac{\pi}{4}\right]$ we have:
$$ \frac{1-\cos x}{x}=\left(\frac{\sin(x/2)}{x/2}\right)^2\cdot\frac{x}{2}\leq\left(1-\frac{x^2}{4\pi^2}\right)^2\frac{x}{2} $$
so:
$$ \int_{0}^{\frac{\pi}{4}}\frac{1-\cos x}{x}\,dx \color{red}{\leq} \int_{0}^{\pi/4}\left(1-\frac{x^2}{4\pi^2}\right)^2\frac{x}{2}\,dx = \frac{\pi^2}{64}-\frac{191\,\pi^2}{3\cdot 2^{18}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.