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Prove that $\frac{1}{n+1} + \frac{1}{n+3}+\cdots+\frac{1}{3n-1}>\frac{1}{2}$ Without using Mathematical Induction, prove that $$\frac{1}{n+1} + \frac{1}{n+3}+\cdots+\frac{1}{3n-1}>\frac{1}{2}$$
I am unable to solve this problem and don't know where to start. Please help me to solve this problem using the laws of inequality. It is a problem of Inequality.
Edit: $n$ is a positive integer such that $n>1$.
| The sum can be written as
\begin{align}
\frac{1}{n+1} + \frac{1}{n+3} + \ldots + \frac{1}{3n - 1} & = \sum_{i=1}^n \frac{1}{n + 2i - 1}.
\end{align}
Now recall the AM-HM inequality:
$$
\frac 1n\sum_{i=1}^n(n + 2i - 1) > \frac{n}{\sum_{i=1}^n \frac{1}{n + 2i - 1}}.
$$
(The requirement that $n > 1$ guarantees that the inequality is strict.)
Rearrange to get
\begin{align}
\sum_{i=1}^n \frac{1}{n + 2i - 1} & > \frac{n^2}{\sum_{i=1}^n(n + 2i - 1)} = \frac 12.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Proving $\tan A=\frac{1-\cos B}{\sin B} \;\implies\; \tan 2A=\tan B$
If $\tan A=\dfrac{1-\cos B}{\sin B}$, prove that $\tan 2A=\tan B$.
My effort:
Here
$$\tan A=\frac{1-\cos B}{\sin B}$$
Now
$$\begin{align}\text{L.H.S.} &=\tan 2A \\[4pt]
&=\frac{2\tan A}{1-\tan ^2A} \\[6pt]
&=\frac{(2-2\cos B)\over\sin B}{1-\frac{(1-\cos B)^2}{\sin^2 B}}
\end{align}$$
On simplification from here, I could not get the required R.H.S.
| You are on the right road.... Continuing calculation,
\begin{align}
\frac{\frac{2-2\cos B}{\sin B}}{1-\frac{(1-\cos B)^2}{\sin^2 B}}&=\frac{\frac{2-2\cos B}{\sin B}}{\frac{\sin^2 B -1+2\cos B-\cos^2 B }{\sin^2 B}}\\
&=\frac{\frac{2-2\cos B}{\sin B}}{\frac{\sin^2 B -\sin^2 B -\cos^2 B+2\cos B-\cos^2 B }{\sin^2 B}}\\
&=\frac{\frac{2-2\cos B}{\sin B}}{\frac{2\cos B(1-\cos B)}{\sin^2 B}}\\
&=\frac{2\sin^2 B(1-\cos B)}{2\cos B\sin B(1-\cos B)}\\
&=\frac{\sin B}{\cos B}\\
&=\tan B
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1644391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to derive the equation of tangent to an arbitrarily point on a ellipse?
Show that the equation of a tangent in a point $P\left(x_0, y_0\right)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, could be written as: $$\frac{xx_0}{a^2} + \frac{yy_0}{b^2} = 1$$
I've tried implicit differentiation $\to \frac{2x\frac{d}{dx}}{a^2}+\frac{2y\frac{d}{dy}}{b^2} = 0$, but not sure where to go from here. Substituting $P$ doesn't seem to help me much, and solving for $y$ from original equation seems to cause me more trouble than help. Please don't give me the solution, rather just give a slight hint or two:)
Thanks in advance!
Edit:
Solving for $\frac{dy}{dx}$ and putting it into point slope gives me:
$\frac{d}{dx} \left[\frac{x^2}{a^2} + \frac{y^2}{b^2}\right] = \frac{d}{dx}\left[1\right] \to \frac{2x}{a^2} + \frac{y}{b^2}\frac{dy}{dx}= 0\to \frac{dy}{dx}=-\frac{xb^2}{ya^2} \to \frac{dy}{dx}(P)=-\frac{x_0b^2}{y_0a^2}$
Then we get:
$y-y_0=\frac{dy}{dx}\left(x-x_0\right)\to y = -\frac{x_0b^2}{y_0a^2}(x-x_0)+y_0\to y = \frac{x_0^2b^2}{y_0a^2}-\frac{xx_0b^2}{y_0a^2} + y_0\to /:b^2,*y_0\to \frac{yy_0}{b^2}+\frac{xx_0}{a^2}=\frac{y_0^2}{b^2}$
Which looks close, but not exactly the expression i wanted. Where is the error?
| To show that the given line is tangent to the ellipse, one can show that the system of equations
\begin{align}
\frac{x^2}{a^2} + \frac{y^2}{b^2} &= 1 \tag{I} \\
\frac{xx_0}{a^2} + \frac{yy_0}{b^2} &= \tag{II} 1
\end{align}
with $x$, $y$ variables has exactly one solution—namely $(x_0, y_0)$. We further assume that $(x_0, y_0)$ lies on the ellipse, i.e.,
$$ \frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} = 1. \tag{III} $$
Taking $(\mathrm I) - 2(\mathrm{II}) + (\mathrm{III})$ produces
$$ \frac{x^2 - 2xx_0 + x_0^2}{a^2} + \frac{y^2 - 2yy_0 + y_0^2}{b^2} = 0 $$
or
$$ \left(\frac{x-x_0}{a}\right)^2 + \left(\frac{y-y_0}{b}\right)^2 = 0. $$
The only way for the sum of two squares to be zero is that $x=x_0$ and $y=y_0$, hence this is indeed the only solution as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Numerical evaluation of first and second derivative We start with the following function $g: (0,\infty)\rightarrow [0,\infty)$,
$$ g(x)=x+2x^{-\frac{1}{2}}-3.$$ From this function we need a 'smooth' square-root. Thus, we check $g(1)=0$,
$$g'(x)=1-x^{-\frac{3}{2}},$$ $g'(1)=0$ and
$$g''(x)=\frac{3}{2}x^{-\frac{5}{2}}\geq 0.$$
Therefore, we can define the function $f: (0,\infty)\rightarrow \mathbb{R}$,
$$f(x)=\operatorname{sign}(x-1)\sqrt{g(x)}.$$
We can compute the first derivatives as
$$
f'(x)=\frac{1}{2} \frac{g'(x)}{f(x)}=\frac{1}{2}\frac{1-x^{-\frac{3}{2}}}{\operatorname{sign}(x-1)\sqrt{x+2x^{-\frac{1}{2}}-3}}
$$
and second derivatives as
$$
f''(x)=\frac{1}{2} \frac{g''(x)f(x)-g'(x)f'(x)}{\bigl(f(x)\bigr)^2}\\
=\frac{1}{2}\frac{g''(x)-\frac{\bigl(g'(x)\bigr)^2}{2g(x)}}{f(x)}\\
=\frac{1}{2}\frac{\frac{3}{2}x^{-\frac{5}{2}}-\frac{\bigl(1-x^{-\frac{3}{2}}\bigr)^2}{2\bigl(x+2x^{-\frac{1}{2}}-3\bigr)}}{\operatorname{sign}(x-1)\sqrt{x+2x^{-\frac{1}{2}}-3}}
$$
where we used $f'(x)$ from above, reduced by $f(x)$ and used $f^2(x)=g(x)$.
For $x=1$ the first and the second derivatives of $f$ are of the type $\frac{0}{0}$.
I need a numerical stable evaluation of this derivatives.
But we have numerical cancellation especially in the nominators.
A linear Taylor-polynomial at $x=1$ is possible for $f'(x)$ but the computation of $f'''(1)$ by hand is time-consuming.
*
*Is there a better formulation for $f(x)$?
*Is there an easy way to compute the coefficients of the Taylor-polynomials at $x=1$?
*(Edit 2:) How can I evaluate $f'(x)$ and $f''(x)$ numerically stable for $x\in (0,\infty)$?
Edit 1:
The Taylor-polynomial of degree 1 for $f'(x)$ at $x=1$ is
$$
f'(x)=\frac{\sqrt{3}}{2} - \frac{5}{4\sqrt{3}}(x-1)+\mathcal{O}\bigl(\lvert x-1\rvert^2\bigr)
$$
| For the first question, since $g(x)=\frac{(\sqrt{x}-1)^2(\sqrt{x}+2)}{\sqrt{x}}$, then
$$f(x)=sign(x-1)\sqrt{\frac{(\sqrt{x}-1)^2(\sqrt{x}+2)}{\sqrt{x}}}$$
The $sign(x-1)$ part is necessary for $f$ to be smooth, as $\lim_{x\rightarrow1^-}f'(x)\neq\lim_{x\rightarrow1^+}f'(x)$.
Indeed, if we write
$$f'(x)=\frac{1}{2}\frac{1-\frac{1}{x\sqrt{x}}}{sign(x-1)\sqrt{\frac{(\sqrt{x}-1)^2(\sqrt{x}+2)}{\sqrt{x}}}}=\frac{\sqrt[4]{x}}{2\sqrt{(\sqrt{x}+2)}}\frac{1-\frac{1}{x\sqrt{x}}}{(\sqrt{x}-1)}$$
(ignoring the $sign$ part for manipulation), we see that
$$\lim_{x->1^+}f'(x)=\frac{1}{2\sqrt{3}}\frac{0}{0}$$
The part that gives the $\frac{0}{0}$ indetermination can be solved first by multiplying by the conjugate of the denominator...
$$\lim_{x->1^+}\frac{1-\frac{1}{x\sqrt{x}}}{(\sqrt{x}-1)}=\lim_{x->1^+}\frac{(1-\frac{1}{x\sqrt{x}})(\sqrt{x}+1)}{x-1}=\lim_{x->1^+}\frac{\sqrt{x}+1-\frac{1}{x}-\frac{1}{x\sqrt{x}}}{x-1}=\frac{0}{0}\dots$$
...and applying L'Hôpital's rule:
$$\dots\lim_{x->1^+}\frac{1}{2\sqrt{x}}+\frac{1}{x^2}+\frac{3}{2x^{5/2}}=3$$
Knowing this, we get that
$$\lim_{x->1^+}f'(x)=\frac{3}{2\sqrt{3}}=\frac{\sqrt{3}}{2}$$
Analysing the function $f$ taking now into account the $sign(x-1)$, we obtain that
$$\lim_{x->1^-}f'(x)=-\frac{\sqrt{3}}{2}$$
as can be seen plotting $f(x)$ and $-f(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find all functions $f:\mathbb R \rightarrow \mathbb R$ such that $f (x+xy+f(y) )= (f(x)+ \frac 12 )\ (f(y)+ \frac 12 \ ).$ Find all functions $f:\mathbb R \rightarrow \mathbb R$ such that
$$f\left (x+xy+f(y) \right )=\left (f(x)+ \frac 12 \right )\left (f(y)+ \frac 12 \right ).$$
for every $x,y \in \mathbb R$.
My work so far:
1) $y=-1$:
$$f(f(-1))=\left(f(x)+\frac 12 \right ) \cdot \left(f(-1)+\frac 12 \right ).$$
So, if $f(-1) \not = -\frac 12$ that $f=const$ ($c=(c+1/2)^2$ - contradiction)
So, $f(-1)=-\frac 12$.
2) $x=0$:
$$f(f(y))=\left(f(0)+\frac 12 \right ) \cdot \left(f(y)+\frac 12 \right ).$$
$y=0$
$$f(x+f(0))=\left(f(0)+\frac 12 \right ) \cdot \left(f(x)+\frac 12 \right )$$
| From $f(-1) = -\frac{1}{2}$ we obtain
$$ f \left( - \frac{1}{2} \right) = f (f (-1)) = 0.$$
So by choosing $y= - \frac{1}{2}$ we obtain
$$f\left( \frac{1}{2} x \right) = f\left(x - \frac{1}{2} x + f\left( - \frac{1}{2} \right) \right) = \frac{1}{2} f(x) + \frac{1}{4}.$$
So $f(0) = \frac{1}{2}$.
Choosing $y=0$ gives
$$f\left(x + \frac{1}{2}\right)=f(x) + \frac{1}{2}.$$
Combining this gives for $x= \frac{z}{2^n}$ with $z \in \mathbb{Z}$ and $n \in \mathbb{N}$ that $f(x) = x + \frac{1}{2}$.
Now, if $f$ is continuous, we have $f(x) =x + \frac{1}{2}$ as the Dyadic numbers are dense on the real line.
However, $f$ does not need to be continuous, but for $x= \frac{m}{n}$ with $m,n \neq 0$ and $y=n-1$ we have
$$m+n = f \left( x+xy + f(y) \right) = n f(x) + \frac{n}{2}$$
and so
$$f \left( \frac{m}{n} \right) = \frac{2m+n}{2n} = \frac{m}{n} + \frac{1}{2}.$$
So for all $q \in \mathbb{Q}$ we have $f(q) = q + \frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1649185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Complex Addition: $(i+\sqrt3/2)^{200} + (i-\sqrt3/2)^{200}$ Solving a question, I need to find the value of following in between the solution.
$$\left(\frac{i+\sqrt3}{2}\right)^{200} + \left(\frac{i-\sqrt3}{2}\right)^{200}$$
The only useful thing I got was
$$\left(\frac{i+\sqrt3}{2}\right)^{100}\left(\frac{i-\sqrt3}{2}\right)^{100} = 1$$
Which might be useful to complete the square.
| If $a=\dfrac{i+\sqrt3}2, b=\dfrac{\sqrt3-i}2$
$a+b=\sqrt3,ab=1\implies a^2+b^2=(a+b)^2-2ab=1$
So, $a,b$ are the roots of $t^2-\sqrt3t+1=0$
$\implies(t^2+1)^2=(\sqrt3t)^2$
$\iff t^4-t^2+1=0\implies(t^2+1)(t^4-t^2+1)=0\iff t^6=-1$
$\implies t^{2+6(2n+1)}=(t^6)^{2n+1}\cdot t^2=-t^2$
$\implies a^{12n+8}+b^{12n+8}=-(a^2+b^2)=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Let A be a square matrix such that $A^3 = 2I$ Let $A$ be a square matrix such that $A^3 = 2I$
i) Prove that $A - I$ is invertible and find its inverse
ii) Prove that $A + 2I$ is invertible and find its inverse
iii) Using (i) and (ii) or otherwise, prove that $A^2 - 2A + 2I$ is invertible and find its inverse as a polynomial in $A$
$I$ refers to identity matrix.
Am already stucked at part i). Was going along the line of showing that $(A-I)([...]) = I$ by manipulating the equation to $A^3 - I = I$ and I got stuck... :(
| For parts (i) and (ii):
\begin{align}
(A-I)^{-1} & = A^2+A+I \\
(A+2 I)^{-1} & = \frac{1}{10}(A^2-2A+4).
\end{align}
For (iii):
$$
A^2-2A+2I=A^2-2A+A^3=A(A+2I)(A-I)
$$
The inverse of $A$ is $\frac{1}{2}A^2$. So,
\begin{align}
(A^2-2A+2I)^{-1}&=(A-I)^{-1}(A+2I)^{-1}A^{-1}\\
&=\frac{1}{20}(A^2+A+I)(A^2-2A+4)A^2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Strange behavior of infinite products $\prod^{\infty}_{n=1} \ln (1+ \frac{1}{n} )^n$ and $\prod^{\infty}_{n=1} \ln (1+ \frac{1}{n} )^{n+1}$ There are two expressions marking the lower and upper bounds for number $e$:
$$\left(1+\frac{1}{n} \right)^n \leq e \leq \left(1+\frac{1}{n} \right)^{n+1}$$
Naturally, I wanted to know if infinite products of their logarithms converge to the same value. I was greatly surprised to find that not only do they not converge to the same value, but one of them converges to zero and the other to infinity:
$$\prod^{\infty}_{n=1} \ln \left(1+ \frac{1}{n} \right)^n=0$$
$$\prod^{\infty}_{n=1} \ln \left(1+ \frac{1}{n} \right)^{n+1} \rightarrow + \infty$$
On the other hand their product (or equally, the infinite product of their geometric means) converges, but not to $1$:
$$\prod^{\infty}_{n=1} \ln \left(1+ \frac{1}{n} \right)^{n} \ln \left(1+ \frac{1}{n} \right)^{n+1}=\prod^{\infty}_{n=1} n(n+1) \ln^2 \left(1+ \frac{1}{n} \right) \rightarrow P$$
Mathematica gives the following values (since $P_n$ is decreasing, it's certainly less than $1$):
$$P(14999)=0.921971686261$$
$$P(15000)=0.921971685920$$
The convergence (or divergence) can be proved using the corresponding series and the integral test:
$$\sum^{\infty}_{n=1} \ln \ln \left(1+ \frac{1}{n} \right)^n $$
$$\int^{\infty}_{1} \ln \left( x \ln \left(1+ \frac{1}{x} \right)\right) dx =\int^{1}_{0} \frac{1}{y^2} \ln \left( \frac{\ln \left(1+ y \right)}{y} \right) dy\rightarrow - \infty $$
This integral does not converge (according to Wolframalpha)
$$\sum^{\infty}_{n=1} \ln \ln \left(1+ \frac{1}{n} \right)^{n+1} $$
$$\int^{\infty}_{1} \ln \left( (x+1) \ln \left(1+ \frac{1}{x} \right)\right) dx =\int^{1}_{0} \frac{1}{y^2} \ln \left( \left(1+ \frac{1}{y} \right) \ln \left(1+ y \right) \right) dy\rightarrow + \infty $$
This integral also does not converge (according to Wolframalpha)
Finally, the 'mean' infinite product gives (see Wolframalpha):
$$\int^{1}_{0} \frac{1}{y^2} \ln \left( \frac{1}{y} \left(1+ \frac{1}{y} \right) \ln^2 \left(1+ y \right) \right) dy=-0.0569274$$
So, this infinite product converges, but not to $1$ according to Mathematica.
Is there any explanation for all this? Is it connected to the special
properties of $e$?
| Note $$\log\left(1+\frac 1n\right)=\frac{1}{n}-\frac{1}{2n^2}+o\left(\frac1{n^2}\right)$$
So $$\log\left(1+\frac 1n\right)^n=n\log\left(1+\frac 1n\right)= 1-\frac{1}{2n}+o\left(\frac1n\right).$$
Now, since $\prod \left(1-\frac{1}{2n}\right)$ does not converge to a positive value, neither does the left side product.
You can similarly deduce:
$$(n+1)\log\left(1+\frac 1n\right) =1+\frac{1}{2n}+o\left(\frac1n\right).$$
So this is not so strange.
For the product, you need more terms. You can show:
$$\log\left(1+\frac1n\right)^2 = \frac1{n^2} -\frac{1}{n^3} + \frac{11}{12}\frac{1}{n^4}+O\left(\frac1{n^5}\right)$$
So $$n(n+1)\log\left(1+\frac1n\right)^2 =1-\frac{1}{12n^2}+O\left(\frac{1}{n^3}\right)$$
You might be able to put tighter bounds on this to get that all the terms are less than $1$. This only shows that all but finitely many are less than $1$.
| {
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Prove $\binom{3n}{n,n,n}=\frac{(3n)!}{n!n!n!}$ is always divisible by $6$ when $n$ is an integer.
Prove $$\binom{3n}{n,n,n}=\frac{(3n)!}{n!n!n!}$$ is always divisible by $6$ when $n$ is an integer.
I have done a similar proof that $\binom{2n}{n}$ is divisible by $2$ by showing that $$\binom{2n}{n}=\binom{2n-1}{n-1}+\binom{2n-1}{n}=2\binom{2n-1}{n-1}$$ but I am at a loss for how to translate this to divisible by $6$. Another way to do this proof would be to show that when you shoot an $n$-element subset from $2n$ you can always match it with another subset (namely the $n$-elements that were not chosen). Again, no idea how to translate this to $6!$.
| $$\binom{3n}{n,n,n}=\frac{(3n)!}{n!n!n!} = \binom{3n}{n}\binom{2n}{n}$$
$$\binom{3n}{n} = \binom{3n -1}{n-1} +\binom{3n-1}{n} = \frac{(3n - 1)!}{(n-1)!(3n -1 -(n-1))!} + \frac{(3n - 1)!}{n! ((3n -1 -n)!}$$
$$=\frac{(3n-1)!)}{(n-1)!(2n-1)!}(\frac{1}{2n} + \frac{1}{n})$$
$$=\frac{(3n-1)!)}{(n-1)!(2n-1)!} \frac{3}{2n}$$
$$=\frac{(3n-1)!)}{(n-1)!(2n)!} * 3$$
$$=3\binom{3n -1}{n-1}$$
It has been already proved that
$$\binom{2n}{n}=\binom{2n-1}{n-1}+\binom{2n-1}{n}=2\binom{2n-1}{n-1}$$
Combining both
$$\binom{3n}{n,n,n} = \binom{3n}{n}\binom{2n}{n} = 6\binom{3n -1}{n-1}\binom{2n-1}{n-1}$$
| {
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"url": "https://math.stackexchange.com/questions/1652748",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "11",
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Let $C$ be the set of all complex numbers of the form $a+ b \sqrt {5}i$, where $a$ and $b$ are integers... Let $C$ be the set of all complex numbers of the form $a+ b \sqrt {5}i$, where $a$ and $b$ are integers. Prove that $7$, $1 + 2\sqrt {5}i$, and $1 - 2\sqrt {5} i$ are all prime in $C$.
-I am really lost in this question, the closest looking thing I could muster up from our lecture notes was the section covering the fundamental theorem of arithmetic, and more specifically covering uniqueness. Really not sure how to approach this problem, any help and jump start is appreciated.
| Hint: Use the norm: $N(a+b\sqrt 5i)= a^2+5b^2$.
Suppose $7=ab$. This implies $N(u)N(v) = N(7)=49$.
$N(u)=1, N(v)=49$, write $u=a+i\sqrt5 b$, $N(u) =a^2+5b^2=1$ implies $b=0, a^2=1$.
$N(u)=7=N(v)$. Write $u=a+ i\sqrt5 b$, $N(u) =a^2+5b^2=7$ we deduce that $\mid b\mid \leq 1$. Suppose that $\mid b\mid =1$, $a^2 +5=7$, $a^2=2$ impossible.
Suppose that $b=0$, $a^2=7$ impossible.
Write $1+2i\sqrt 5 =uv$, $N(u)N(v) =21$.
$N(u)=1, N(v)=21$ implies as above $u=1$.
$N(u)=3, N(v)=7$. The argument above shows you can't have $N(v)=7$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Trouble proving through induction after establishing a basecase What is the following sum?
$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + ... + \frac{1}{(n-1)n}$
Experiment, conjecture the value, and then prove it by induction.
I found the sum to be $\frac{n-1}{n}$ and established the base of n=2 being equal to $\frac 12$, but I don't know what to do next.
| By induction assume that:$\frac 1 2 +\frac 1 6 +\frac {1}{12}+\dots \frac{1}{n(n-1)}=\frac{n-1}{n}$
For $n+1:$
$\underbrace{\frac 1 2 +\frac 1 6 +\frac {1}{12}+\dots \frac{1}{n(n-1)}}_{=\frac{n-1}{n}}+\frac{1}{n(n+1)}\stackrel{\color{red}?}{=}\frac{n}{n+1}$
$$\frac{1}{n(n+1)}+\frac{n-1}{n}=\frac{1+(n-1)(n-1)}{n(n+1)}=\frac{n^2}{n(n+1)}=\boxed{\frac{n}{n+1}}$$
As expected
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the sum of all $abc$
Let $T$ be the set of all triplets $(a,b,c)$ of integers such that $1 \leq a < b < c \leq 6$. For each triplet $(a,b,c)$ in $T$ , take number $a \cdot b \cdot c$. Add all these numbers corresponding to all the triplets in $T$. Prove that the answer is divisible by $7$.
Attempt
Let $S = \displaystyle \sum_{b < c} bc $. Then with no restriction on $a$ we have $M = (1+2+3+4+5+6)S = 21S$. This is the sum of the cases where $a < b, a = b, $ and $a > b$. Now how can I use this to find the case where $a<b$?
| For any such triplet $(a, b, c)$ with $a<b<c$,
we have a corresponding triplet $(7-c, 7-b, 7-a)$ with $7-c<7-b<7-a$.
Since we can pair up all the triplets in this manner,
and $abc+(7-c)(7-b)(7-a)\equiv 0\pmod{7}$,
the sum of the products of each of the triplets is divisible by 7.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $3 = -x^2+4x$ by factoring I have $3 = -x^2 + 4x$ and I need to solve it by factoring. According to wolframalpha the solution is $x_1 = 1, x_2 = 3$.
\begin{align*}
3 & = -x^2 + 4x\\
x^2-4x+3 & = 0
\end{align*}
According to wolframalpha $(x-3) (x-1) = 0$ is the equation factored, which allows me to solve it, but how do I get to this step?
| $$ 3 = -x^2 +4x \implies \ x^2 -4x+3 = 0$$
$$ x^2 - 3x - x +3 = 0$$
$$ x(x-3)-1(x-3)$$
$$ (x-3)(x-1) = 0$$
$$x = 1, x = 3$$
For more on the methods, visit this link.
Alternatively, you could use the quadratic formula:
$$x = \frac{-b \pm\sqrt{b^2 - 4ac}}{2a}$$
This was obtained from the quadratic equation:
$$ax^2 + bx +c = 0$$
where $a = 1, b = -4, c = 3$
| {
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If $x^2+y^2-xy-x-y+1=0$ ($x,y$ real) then calculate $x+y$ If $x^2+y^2-xy-x-y+1=0$ ($x,y$ real) then calculate $x+y$
Ideas for solution include factorizing the expression into a multiple of $x+y$ and expressing the left hand side as a sum of some perfect square expressions.
| Let $s=x+y$ and $d=x-y$. Then $x=(s+d)/2$ and $y=(s-d)/2$. Making this substitution, we find that
$$
x^2+y^2-xy-x-y+1=\frac{3d^2}{4}+\frac{(s-2)^2}{4}.
$$
Hence, being a sum of squares, if the original expression is zero, $d=0$ and $s-2=0$. This means that $s=x+y=2$, with $x=y$.
| {
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Proving for all integer $n \ge 2$, $\sqrt n < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt n}$ Prove the following statement by mathematical induction:
For all integer $n \ge 2$, $$\sqrt n < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt n}$$
My attempt: Let the given statement be $p(n)$ .
1.\begin{align*} \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} & =\frac{\sqrt 2 +1}{\sqrt 2} \\
2 &< \sqrt 2 +1 \\
\sqrt 2 &< \frac{\sqrt 2 +1}{\sqrt 2}=\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} \end{align*}
Hence, $p(2)$ is true.
2.For an arbitrary integer $k \ge 2$, suppose $p(k)$ is true.
That is, $$\sqrt k < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt k}$$
Then we must show that $p(k+1)$ is true.
We're going to show that $$\sqrt {k+1} < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt k}+\frac{1}{\sqrt {k+1}}$$
I'm stuck on this step. I can't develop it further. How can I complete this proof?
| For the inductive step, it suffices to show $\sqrt{n+1} - \sqrt{n} < \frac {1}{\sqrt{n+1}} = \frac{\sqrt{n+1}}{n+1}$ since that implies the inequality gets stronger as $n$ increases. This is clear since
$$\sqrt{n+1} - \sqrt{n} = \frac{\sqrt{n+1}}{n+1} + \frac{(-\sqrt{n}\sqrt{n+1})(\sqrt{n+1} - \sqrt{n})}{n+1}$$
The right addend on the right equation is clearly negative.
| {
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Prove $\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2$ if $a+b+c=0$ Found this lovely identity the other day, and thought it was fun enough to share as a problem:
If $a+b+c=0$ then show $$\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2.$$
There are, of course, brute force techniques for showing this, but I'm hoping for something elegant.
| $$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=-2(ab+bc+ca)$$
The key here is the identity (for all $n\in\mathbb Z_{\ge 3}$):
$$a^n+b^n+c^n=(a+b+c)\left(a^{n-1}+b^{n-1}+c^{n-1}\right)-$$
$$-(ab+bc+ca)\left(a^{n-2}+b^{n-2}+c^{n-2}\right)+abc\left(a^{n-3}+b^{n-3}+c^{n-3}\right)$$
Therefore: $$a^3+b^3+c^3=3abc\\ a^4+b^4+c^4=2(ab+bc+ca)^2\\ a^5+b^5+c^5=-5abc(ab+bc+ca)\\a^7+b^7+c^7=7abc(ab+bc+ca)^2$$
Therefore:
$$\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2\\\frac{a^2+b^2+c^2}{2}\frac{a^3+b^3+c^3}{3}=\frac{a^5+b^5+c^5}{5}\\ \frac{a^2+b^2+c^2}{2}\frac{a^5+b^5+c^5}{5}=\frac{a^7+b^7+c^7}{7}\\$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate $I= \int \frac{x^2 -1}{x\sqrt{1+ x^4}}\,\mathrm d x$
$$I= \int \frac{x^2 -1}{x\sqrt{1+ x^4}}\,\mathrm d x$$
My Endeavour :
\begin{align}I&= \int \frac{x^2 -1}{x\sqrt{1+ x^4}}\,\mathrm d x\\ &= \int \frac{x}{\sqrt{1+ x^4}}\,\mathrm d x - \int \frac{1}{x\sqrt{1+ x^4}}\,\mathrm d x\end{align}
\begin{align}\textrm{Now,}\;\;\int \frac{x}{\sqrt{1+ x^4}}\,\mathrm d x &= \frac{1}{2}\int \frac{2x^3}{x^2\sqrt{1+ x^4}}\,\mathrm dx\\ \textrm{Taking}\,\,(1+ x^4)= z^2\,\,\textrm{and}\,\, 4x^3\,\mathrm dx= 2z\,\mathrm dz\,\, \textrm{we get} \\ &= \frac{1}{2}\int \frac{z\,\mathrm dz}{\sqrt{z^2-1}\, z}\\ &= \frac{1}{2}\int \frac{\mathrm dz}{\sqrt{z^2-1}}\\ &= \frac{1}{2}\ln|z+ \sqrt{z^2 -1}|\\ &= \frac{1}{2}\ln|\sqrt{1+x^4}+ x^2|\\ \textrm{Now, with the same substitution, we get in the second integral}\\ \int \frac{1}{x\sqrt{1+ x^4}}\,\mathrm d x &= \frac{1}{2}\int \frac{2x^3}{x^4\sqrt{1+ x^4}}\,\mathrm dx\\ &= \frac{1}{2}\int \frac{z\,\mathrm dz}{( z^2 -1)\;z} \\ &=\frac{1}{2}\int \frac{\mathrm dz}{ z^2 -1} \\ &=\frac{1}{2^2}\, \ln \left|\frac{ z+1}{z-1}\right| \\ &= \frac{1}{2^2}\, \ln \left|\frac{ \sqrt{1+ x^4}+1}{\sqrt{1+ x^4}-1}\right|\;.\end{align}
So, \begin{align}I&=\int \frac{x^2 -1}{x\sqrt{1+ x^4}}\,\mathrm d x \\ &=\frac{1}{2}\, \ln|\sqrt{1+x^4}+ x^2|- \frac{1}{2^2}\, \ln \left|\frac{ \sqrt{1+ x^4}+1}{\sqrt{1+ x^4}-1}\right| + \mathrm C\;.\end{align}
Book's solution:
\begin{align}I&=\int \frac{x^2 -1}{x\sqrt{1+ x^4}}\,\mathrm d x \\ &= \ln\left\{\frac{1+x^2 + \sqrt{1+x^4}}{x}\right\} + \mathrm C\;.\end{align}
And my hardwork's result is nowhere to the book's answer :(
Can anyone tell me where I made the blunder?
| Here is another approach.
Let $x^2=\tan(\phi)$
$$
\begin{align}
\int\frac{x^2-1}{x\sqrt{1+x^4}}\,\mathrm{d}x
&=\frac12\int\frac{x^2-1}{x^2\sqrt{1+x^4}}\,\mathrm{d}x^2\\
&=\frac12\int\frac{\tan(\phi)-1}{\tan(\phi)\sec(\phi)}\,\mathrm{d}\tan(\phi)\\
&=\frac12\int(\sec(\phi)-\csc(\phi))\,\mathrm{d}\phi\\
&=\frac12\log(\sec(\phi)+\tan(\phi))+\frac12\log(\csc(\phi)+\cot(\phi))+C\\
&=\frac12\log\left(\sqrt{1+x^4}+x^2\right)+\frac12\log\left(\sqrt{1+\frac1{x^4}}+\frac1{x^2}\right)+C\\
&=\log\left(\sqrt{1+x^4}+x^2\right)-\log(x)+C
\end{align}
$$
| {
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Where does the normal to the graph of $y = \sqrt x$ at the point $(a, \sqrt a)$ intersect the $x$-axis?
Where does the normal to the graph of $y = \sqrt x$ at the point $(a, \sqrt a)$ intersect the $x$-axis?
I know how to find equation with numbers, but got really confused with this one. If anybody could break it down it would be greatly appreciated.
Here is what I've done: Took derivative $= .5x^{-.5}$, then plugged in $x$ value $a$ to get $1/2a^{-1/2}$, so the slope of tangent is $1/2a^{-1/2}$? so the slope of normal is $2a^{-1/2}$? So $y-\sqrt{a} = 2a^{-1/2}(x-a)$?
Thank you in advance
| The function $y = \sqrt{x} = x^{\frac{1}{2}}$ has derivative
$$y' = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}$$
as you found. The tangent line to the graph of $y = \sqrt{x}$ at the point $(a, \sqrt{a})$ has slope
$$y'(a) = \frac{1}{2\sqrt{a}}$$
which is defined for $a > 0$.
You made a mistake when you calculated the slope of the normal line. Since the normal line is perpendicular to the tangent line, its slope is the negative reciprocal of the slope of the tangent line. Thus, the slope of the normal line is $-2\sqrt{a}$. Hence, the equation of the normal line at the point $(a, \sqrt{a})$ is
$$y - \sqrt{a} = -2\sqrt{a}(x - a)$$
To solve for the $x$-intercept of the normal line, we set $y = 0$ and solve for $x$.
\begin{align*}
0 - \sqrt{a} & = -2\sqrt{a}(x - a)\\
-\sqrt{a} & = -2\sqrt{a}x + 2a\sqrt{a}\\
2\sqrt{a}x & = (2a + 1)\sqrt{a}\\
x & = a + \frac{1}{2}
\end{align*}
Thus, for $a > 0$, the normal line to the graph of $y = \sqrt{x}$ at the point $(a, \sqrt{a})$ intersects the $x$-axis at the point $(a + \frac{1}{2}, 0)$.
Since
$$\lim_{x \to 0^{+}} \frac{1}{2\sqrt{x}} = \infty$$
the tangent line to the graph of $y = \sqrt{x}$ is vertical, so the normal line is horizontal. Since the tangent line to the point $(0, \sqrt{0}) = (0, 0)$ is the $y$-axis, the normal line is the $x$-axis.
| {
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Evaluation of $\lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
Evaluation of $\displaystyle \lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
$\bf{My\; Try::}$ Here $(x+1)\;,(x+2)\;,(x+3)\;,(x+4)\;,(x+5)>0\;,$ when $x\rightarrow \infty$
So Using $\bf{A.M\geq G.M}\;,$ We get $$\frac{x+1+x+2+x+3+x+4+x+5}{5}\geq \left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}$$
So $$x+3\geq \left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}$$
So $$\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\leq 3$$
and equality hold when $x+1=x+2=x+3=x+4=x+5\;,$ Where $x\rightarrow \infty$
So $$\lim_{x\rightarrow 0}\left[\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right]=3$$
Can we solve the above limit in that way, If not then how can we calculate it
and also plz explain me where i have done wrong in above method
Thanks
| Let us shift the variable by $3$ and get
$$\lim_{x\to\infty}\sqrt[5]{(x-2)(x-1)x(x+1)(x+2)}-x+3=\lim_{x\to\infty}x\left(\sqrt[5]{1-\frac5{x^2}+\frac4{x^4}}-1\right)+3.$$
Then by L'Hospital,
$$\lim_{t\to0}\frac{\sqrt[5]{1-5t^2+4t^4}-1}t=\lim_{t\to0}\frac{-10t+16t^3}{5\sqrt[5]{1-5t^2+4t^4}}=0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Different constants of integration on different intervals I tried to prove that on the interval $(-1,\infty)$ that: $$\arctan\left(\frac{x-1}{x+1}\right)=\arctan(x)-\frac{\pi}{4}$$
So I defined: $$f(x)=\arctan\left(\frac{x-1}{x+1}\right)-\arctan(x)$$
Then apparently: $$f'(x)=0$$
Therefore $f(x)$ is constant thus:
$$f(x)=\arctan\left(\frac{x-1}{x+1}\right)-\arctan(x)=C$$
$$\arctan\left(\frac{x-1}{x+1}\right)=\arctan(x)+C$$
Taking $x=0$ gives $C=-\frac{\pi}{4}$
Thus is proven: $$\arctan\left(\frac{x-1}{x+1}\right)=\arctan(x)-\frac{\pi}{4}$$
QUESTION: Why does $C$ have a different value for $x<-1$? Why does the above reasoning only apply to the interval $(-1,\infty)$ How to find $C$ for $x<-1$?
| From the Article $240,$ Ex$-5$ of Plane Trigonometry(by Loney),
$$\arctan x+\arctan y=\begin{cases} \arctan\frac{x+y}{1-xy} &\mbox{if } xy<1\\ \pi+\arctan\frac{x+y}{1-xy} & \mbox{if } xy>1 \\ \text{sgn}(x)\dfrac\pi2 & \mbox{if } xy=1 \end{cases} $$
As $\arctan(-u)=-\arctan u,$
Setting $y=-1,$
$$\arctan x-\dfrac\pi4=\begin{cases} \arctan\frac{x-1}{1+x} &\mbox{if } -x<1\iff x>-1\\ \pi+\arctan\frac{x-1}{1+x} & \mbox{if } -x>1\iff x<-1 \\ -\dfrac\pi2 & \mbox{if } -x=1\iff x=-1 \end{cases} $$
Do you notice tthe different constants?
| {
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Geometric problem about triangle [Edited]
Let $|BC|=|AD|=1, \quad\angle BCA=\frac{\pi}{2}$ and $\angle ACD=\frac{\pi}{6}$. Find $|AB|$.
Can anyone show how to solve it?
| Put $|BC|=:a$, $|AB|=:c$, $|AC|=:b$, and denote the angle at $D$ by $\beta$. Two applications of the sine law then give
$${a+c\over\sin 120^\circ}={a\over\sin\beta},\qquad {a\over\sin30^\circ}={b\over\sin\beta}\ .$$
Dividing the first of these by the second we obtain
$${a+c\over a}\cdot{1/2\over\sqrt{3}/2}={a\over b}={a\over\sqrt{c^2-a^2}}\ ,$$
and squaring leads to the equation
$$(a+c)^2(c^2-a^2)=3a^4\ .$$
When $a=1$ we obtain the equation
$$0=(1+c)^2(c^2-1)-3=(c+2)(c^3-2)\ ,$$
which has only one usable solution, namely $c=\root3\of 2$.
| {
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Number of $4\times3$ matrices of rank 3 over a field with 3 elements. I am finding number of $4\times3$ matrices of rank 3 over a field with 3 elements. If i count it as number of linearly independent columns i.e $3$ then its answer is $(3^{4}-1)(3^{4}-3)(3^{4}-3^{2}).$ But when i like to obtain the same formula as number of linearly independent rows my answer does not match. Please suggest me how to find the same formula as we look at number of linearly independent rows i.e. $3.$ Column wise already solved number of matrices of rank 3?. Thanks.
| Start with the empty set and successively add the rows. The dimension of the space spanned by the set must increase from $0$ to $3$, and it can increase by at most $1$ in each step, so there must be exactly $3$ steps where it's incremented and one step where it stagnates. The incrementing steps always have the same counts of options, $3^3-3^0$, $3^3-3^1$ and $3^3-3^2$, whereas the stagnating step has $3^k$ options if it occurs after $k$ incrementing steps. Thus the total count is
$$
(3^3-3^0)(3^3-3^1)(3^3-3^2)(3^3+3^2+3^1+3^0)=(3^4-3^0)(3^4-3^1)(3^4-3^2)\;,
$$
in agreement with the column result. (For the equality, shift one factor of $3$ from $3^3-3^2$ to each of $3^3-3^0$ and $3^3-3^1$ and use $(3^1-3^0)(3^3+3^2+3^1+3^0)=3^4-3^0$.)
| {
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Find the absolute value and argument for $(3+4i)^{-1}$ When I have to find the argument and absolute value of $z=(3+4i)^{-1}$ I thought that I had to find the inverse of $z$ by applying the rule $z^{-1}=\left(\frac{a}{a^2+b^2},\frac{-b}{a^2+b^2}\right)$ but the answer suggests that $\lvert(3+4i)^{-1}\rvert$=$\frac{1}{\lvert3+4i\rvert}$ and that $(Arg)z^{-1}=-Argz$
Why does my first approach give different solutions? Is it that the inverse definition of a complex number is not the same with the annotation $z^{-1}$?
| The inverse of a complex number $z=a+bi$ is given by:
$$z^{-1} = \frac{1}{a+bi} = \frac{a-bi}{(a+bi)(a-bi)} = \frac{a-bi}{a^2+b^2}.$$
Thus, its absolute value is
$$|z^{-1}| = \sqrt{\frac{a^2}{(a^2+b^2)^2} + \frac{b^2}{(a^2+b^2)^2}} = \frac{\sqrt{a^2+b^2}}{a^2+b^2} = \frac{1}{\sqrt{a^2+b^2}} = \frac{1}{|z|}.$$
Using the exponential, you can represent $z$ as
$$z=|z|e^{i\arg(z)}.$$
Thus
$$z^{-1}=|z^{-1}|e^{i\arg(z^{-1})} = \frac{1}{|z|} e^{i\arg(z^{-1})},$$
so
$$1 = z\,z^{-1} = |z|e^{i\arg(z)} \frac{1}{|z|} e^{i\arg(z^{-1})} = e^{i(\arg(z)+\arg(z^{-1}))} \Rightarrow \arg(z)+\arg(z^{-1}) = 0$$
| {
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$(5 + \sqrt{2})(2-\sqrt{2})=(11-7\sqrt{2})(2+\sqrt{2})$ - Unique factorization? Personal question : We know that $5 + \sqrt{2}$, $2-\sqrt{2}$, $11-7\sqrt{2}$ and $2+\sqrt{2}$ are irreductible in $\mathbb{Z}[\sqrt{2}]$ and that $$(5 + \sqrt{2})(2-\sqrt{2})=(11-7\sqrt{2})(2+\sqrt{2}).$$ Why this fact doesn't contradict the unique factorization in $\mathbb{Z}[\sqrt{2}]$?
Is it because $(5 + \sqrt{2})(2-\sqrt{2})=(5 + \sqrt{2})(2+\sqrt{2})(3-2\sqrt{2})=(11 -7\sqrt{2})(2+\sqrt{2})$?
Is anyone could give me a full explication in ''Answer the question''?
| This is because the factorisation is unique up to units, and $\Bbb Z[\sqrt 2]$ has units, for example $(\sqrt 2+1)$.
In $\Bbb Z$, you had $15 = 3 \times 5 = (-3) \times (-5)$. This doesn't contradict the unique factorisation theorem either because each factor of one factorisation differs from a factor in the other by a unit (here, $-1$)
In your case you can tell by looking at the norms, that $(5+\sqrt 2)$ and $(11-7\sqrt 2)$ may be associates, and we can easily check this :
$(5+\sqrt 2)/(11-7\sqrt 2) = (5+\sqrt 2)(11+7\sqrt 2)/23 = (69+46\sqrt 2)/23 = 3+2\sqrt 2$.
Meanwhile, $(2-\sqrt 2)/(2+\sqrt 2) = (2-\sqrt 2)^2/2 = (6-4\sqrt 2)/2 = 3-2\sqrt 2$.
And we have $(3+2\sqrt 2)(3-2\sqrt 2) = 1$ so really, you go from one factorisation to the other by taking a $(3\pm 2\sqrt 2)$ factor from one irreducible and giving it to the other :
$(5+2\sqrt 2)(2-\sqrt 2) = (5+2\sqrt 2)(3-2\sqrt 2)(2+\sqrt 2) = (11-7\sqrt 2)(2+\sqrt 2)$
In $\Bbb Z$ there are only $2$ units, so you can just decide to only use positive representatives of primes and put a $(-1)$ to the side, and then you can immediately tell if two factorisations are different.
In $\Bbb Z[\sqrt 2]$ there isn't really a nice choice of representatives to be made.
| {
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Integral $\int \frac{\sqrt{x}}{(x+1)^2}dx$ First, i used substitution $x=t^2$ then $dx=2tdt$ so this integral becomes $I=\int \frac{2t^2}{(1+t^2)^2}dt$ then i used partial fraction decomposition the following way:
$$\frac{2t^2}{(1+t^2)^2}= \frac{At+B}{(1+t^2)} + \frac{Ct + D}{(1+t^2)^2} \Rightarrow 2t^2=(At+B)(1+t^2)+Ct+D=At+At^3+B+Bt^2 +Ct+D$$
for this I have that $A=0, B=2, C=0, D=-2$
so now I have
$I=\int \frac{2t^2}{(1+t^2)^2}dt= \int\frac{2}{1+t^2}dt - \int\frac{2}{(1+t^2)^2}dt$
Now,
$$ \int\frac{2}{1+t^2}dt = 2\arctan t$$
and
$$\int\frac{2}{(1+t^2)^2}dt$$
using partial integration we have:
$$u=\frac{1}{(1+t^2)^2} \Rightarrow du= \frac{-4t}{1+t^2}$$
and $$dt=dv \Rightarrow t=v$$
so now we have:
$$\int\frac{2}{(1+t^2)^2}dt =\frac{t}{(1+t^2)^2} + 4\int\frac{t^2}{1+t^2}dt = \frac{t}{(1+t^2)^2} + 4\int\frac{t^2 + 1 -1}{1+t^2}dt = \frac{t}{(1+t^2)^2} + 4t -4\arctan t$$
so, the final solution should be:
$$I=2\arctan t - \frac{t}{(1+t^2)^2} - 4t +4\arctan t$$
since the original variable was $x$ we have
$$I= 6\arctan \sqrt{x} - \frac{\sqrt{x}}{(1+x)^2} - 4\sqrt{x} $$
But, the problem is that the solution to this in my workbook is different, it says that solution to this integral is $$I=\arctan \sqrt{x} - \frac{\sqrt{x}}{x+1}$$
I checked my work and I couldn't find any mistakes, so i am wondering which solution is correct?
| One may observe that
$$
\int \frac{2t}{\left(1+t^2\right)^2} \, dt=\int \frac{\left(1+t^2\right)'}{\left(1+t^2\right)^2} \, dt=-\frac{1}{ 1+t^2}
$$ then, using an integration by parts, one has
$$
\int \frac{2t^2}{\left(1+t^2\right)^2} \, dt=t \times \left(-\frac{1}{ 1+t^2} \right)+\int \frac1{\left(1+t^2\right)} \, dt=-\frac{t}{ 1+t^2}+\arctan t+C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1674188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
row exchange results in wrong answer? This is probably a very simple, but why is it that when I do row exchanges I end up with the wrong answer. I have the matrix equation $Ax = B$, as:
$$
\underbrace{
\begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & -2 \\ -1 & -1 & 1 \end{bmatrix}
}_A
\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}
=
\underbrace{
\begin{bmatrix} 3 \\ 1 \\ -6 \end{bmatrix}
}_B$$
I know the answer is $x = \begin{bmatrix} 7 \\ -3 \\ -2 \end{bmatrix}$, but when I do a row exchange (whether needed or not) I end up with the wrong answer. Why? NOTE: I exchanged rows 2 and 3.
| $$\left[ \begin{array} {rrr|r}
1 & 0 & 2 & 3 \\
0 & 1 & -2 & 1 \\
-1 & -1 & 1 & -6 \\
\end{array}\right]$$
Exchange row 2 and 3 :
$$\left[ \begin{array} {rrr|r}
1 & 0 & 2 & 3 \\
-1 & -1 & 1 & -6 \\
0 & 1 & -2 & 1 \\
\end{array}\right]$$
Add row 1 to row 2:
$$\left[ \begin{array} {rrr|r}
1 & 0 & 2 & 3 \\
0 & -1 & 3 & -3 \\
0 & 1 & -2 & 1 \\
\end{array}\right]$$
Add row 2 to row 3:
$$\left[ \begin{array} {rrr|r}
1 & 0 & 2 & 3 \\
0 & -1 & 3 & -3 \\
0 & 0 & 1 & -2 \\
\end{array}\right]$$
Multiply row 2 by -1:
$$\left[ \begin{array} {rrr|r}
1 & 0 & 2 & 3 \\
0 & 1 & -3 & 3 \\
0 & 0 & 1 & -2 \\
\end{array}\right]$$
Add 3 times row 3 to row 2:
$$\left[ \begin{array} {rrr|r}
1 & 0 & 2 & 3 \\
0 & 1 & 0 & -3 \\
0 & 0 & 1 & -2 \\
\end{array}\right]$$
Add -2 times row 3 to row 1:
$$\left[ \begin{array} {rrr|r}
1 & 0 & 0 & 7 \\
0 & 1 & 0 & -3 \\
0 & 0 & 1 & -2 \\
\end{array}\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Why does this pattern occur? I first saw this pattern when I was trying to factor quadratics.
let $a + b = k$ where $k$ is any constant,
Now let $(a - b)/2 = x$
It appears to be that $(k^2/4) - x^2 = a*b$
For example, let's say $k = 8$
then,
$4*4 = 16$
$3*5 = 15$
$2*6 = 12$
$1*7 = 7$
$0*8 = 0$
$-1*9 = -9$
If we subtract 16 from all of those, we get $0, -1, -4,-9, -16, and -25$.
My question is, why does this pattern occur? Maybe I'm missing something obvious, but there are so many variables that I don't know where to start
| In the original post, $x=a-b$ and was changed to $x=\frac{a-b}{2}$. This answer addresses both formulae.
Consider
$$
\frac{k^2}{4}-x^2=\frac{(a+b)^2}{4}-(a-b)^2=\frac{a^2}{4}+\frac{ab}{2}+\frac{b^2}{4}-a^2+2ab-b^2=-\frac{3}{4}a^2+\frac{5}{2}ab-\frac{3}{4}b^2=-\frac{3}{4}(a-b)^2+ab
$$
This is not $ab$, except in fairly special situations.
Now, if $x=\frac{a+b}{2}$ (as suggested in the comments and what appears to be the OP's original intent from the line starting with if we subtract 16...)
$$
\frac{k^2}{4}-x^2=\frac{(a+b)^2}{4}-\frac{(a-b)^2}{4}=\frac{a^2}{4}+\frac{ab}{2}+\frac{b^2}{4}-\frac{a^2}{4}+\frac{ab}{2}-\frac{b^2}{4}=ab,
$$
as desired.
This happens because $(a+b)^2=a^2+2ab+b^2$ and $(a-b)^2=a^2-2ab+b^2$. Therefore, $(a+b)^2-(a-b)^2=4ab$, which can be scaled to $ab$. Tricks like this often show up on calculus exams, especially in arc-length situations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1676170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sum of the series $\sum_{k=1}^\infty ((\frac{x+1}{2})^2)^k$ Find the sum of the series for those values of $$\sum_{k=1}^\infty ((\frac{x+1}{2})^2)^k$$
I have found that $$a=\frac{x^2+2x+1}{4}$$ and $$r=\frac{x^2+2x+1}{4}$$
The sum of the series would than equal $$=\frac{a}{1-r}=\frac{(\frac{x^2+2x+1}{4})}{1-(\frac{x^2+2x+1}{4})}=\frac{x^2+2x+1}{-x^2-2x+3}$$
I am now trying to figure out where this would be correct, but I am having trouble. I know that it has to be $|r|<1$ to be correct. I did
$$\frac{x^2+2x+1}{4}<1, which = x < 1$$, however, when I do the other side, I run into a problem
$$\frac{x^2+2x+1}{4}>-1$$ $$x^2+2x+1>-4$$ $$x^2+2x+5>0$$
This does not simplify anymore. Does that mean that it is correct for all x < 1 or is there something else I have to do?
| Note that $r=\left(\frac{x+1}{2}\right)^2$. We have
$$\left(\frac{x+1}{2}\right)^2\lt 1$$
if and only if
$$-1\lt \frac{x+1}{2}\lt 1.$$
The right-hand inequality holds if and only if $x\lt 1$.
The left-hand inequality holds if and only if $-3\lt x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1676567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving this $\gcd(2^n-1,3^n+2)=1$ for all postive integers $n$ I have found
$$\gcd(2-1,3+2)=\gcd(1,5)=1$$
$$\gcd(2^2-1,3^2+2)=\gcd(3,11)=1$$
$$\gcd(2^3-1,3^3+2)=\gcd(7,29)=1$$
$$\gcd(2^4-1,3^4+2)=\gcd(15,83)=1$$
$$\gcd(2^5-1,3^5+2)=\gcd(31,245)=1$$
$$\cdots\cdots$$
I conjecture $\gcd(2^n-1,3^n+2)=1$,I can't prove this?
| I have checked with Sage that, for $n=176$, we have
\begin{align*}
2^n-1 &= 95780971304118053647396689196894323976171195136475135 \\
3^n+2 &= 940461086986004843694934910131056317906479029659199959555574885740211572136210345923,
\end{align*}
and $gcd(2^n-1,3^n+2) = 257$.
$n=176$ is the smallest. The next ones are: 432, 688, 944.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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The indefinite integral $\int x^{13}\sqrt{x^7+1}dx$ I used integration by parts technique to solve this problem
so $u = \sqrt{x^7+1}$ ,$du = \frac{1}{2}\frac{7x^6}{\sqrt{x^7+1}}dx$
$v =\frac{x^{14}}{14}$ $dv=x^{13}dx$
then it becomes
$\frac{x^{14}}{14}\sqrt{x^7+1}-\int\frac{x^{14}}{28}\frac{7x^6}{\sqrt{x^7+1}}dx$
and this is where i got stuck at. I tried to substitute $u =x^7$
but then the integral become $\int \frac{u^2}{28\sqrt{u+1}}du$
the final answer that I found using wolffram calculator is $\frac{2}{105}(x^7+1)^{3/2}(3x^7-2)$
| \begin{align}
\int x^{13}\sqrt{x^7+1}\,dx&=\int x^6(x^7+1)\sqrt{x^7+1}\,dx-\int x^6\sqrt{x^7+1}\,dx\\
&=\int x^6(x^7+1)^{3/2}\,dx-\int x^6(x^7+1)^{1/2}\,dx
\end{align}
Then, use $u=x^7+1$ in order to evaluate these integrals.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1679142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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2,3,7 is the only triple $\geq 2$ that satisfies this property The property being that "taking the product of two of the numbers and adding one yields a number that is divisible by the third".
Clearly, this holds for 2,3,7 since
*
*$2\cdot 3 + 1 = 1\cdot 7$
*$2\cdot7 + 1 = 5\cdot 3$
*$3\cdot 7 + 1 = 11\cdot 2$
but why is this the only such triple?
This is from a (German) textbook on algebraic number theory, reviewing elementary number theory in chapter 1. Up to this point, only divisibility, the gcd and the euclidean algorithm have been introduced (no prime numbers yet). I took an elementary number theory course last semester, but I'm completely puzzled by this problem and don't know where to start. Trying to set up some equations and to manipulate them doesn't lead me anywhere.
My solution attempt
We can, without loss of generality, assume $2 \leq a \leq b \leq c$. From MXYMXY's answer below we have $abc|ab+bc+ca+1$ (which I'll denote (*)) and also that $a < 4$.
Let's then consider the two possibilities for $a$ in turn.
*
*$a =3$: We can show that $b$ and $c$ can't both be $\geq 4$, for if they were we would have $$\frac{3}{c}+1+\frac{3}{b}+\frac{1}{bc} \leq \frac{3}{4}+1+\frac{3}{4}+\frac{1}{16} < 3 \Leftrightarrow 3b + bc + 3c + 1 < 3bc,$$
but from (*) we have $3bc|3b+bc+3c+1$, which is a contradiction.
It follows that $b = 3$ and then we have $ab+1=10$, so $c|10$. Hence $c = 5$ or $c = 10$, but $a = 3$ divides neither $3\cdot5 + 1 = 16$ nor $3\cdot10 + 1 = 31$. Thus, this case is impossible.
*$a = 2$. We will show that $b$ and $c$ can't be both $\geq 5$. Otherwise we'd have $$\frac{2}{c}+1+\frac{2}{b}+\frac{1}{bc} \leq \frac{2}{5}+1+\frac{2}{5}+\frac{1}{25} < 2 \Leftrightarrow 2b + bc + 2c + 1 < 2bc,$$ but from (*) we have that $2bc|2b+bc+2c+1$, which is a contradiction. This implies that $2 \leq b < 5$. Let's consider the different cases for $b$:
*
*$b=4$: Then we have $ab+1=9$, so $c|9$, i.e. $c = 9$. But $a = 2$ does not divide $4\cdot 9 + 1 = 37$, so this is a contradiction.
*$b=2$: Then $ab+1=5$, so $c|5$, i.e. $c = 5$. But $a = 2$ does not divide $2\cdot 5 + 1 = 11$, so we also have a contradiction.
The only case left is where $a = 2, b = 3$. Then, we have $ab+1=7$, so $c|7$. Hence, $c=7$.
Hence, either $2,3,7$ is the only solution or there is none. That it is indeed a solution was verified above.
| HINT
Since $a|bc+1$, $b|ca+1$, $c|ab+1$, multiplying these together gives us that $$abc|ab+bc+ca+1$$
Note that $$a,b,c \ge 4 \Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{abc}<1 \Leftrightarrow abc>ab+bc+ca+1$$
This gives us that one of $a,b,c$ must be smaller than $4$.
I think you can continue to divide cases from here. It requires slightly complicatd calculations, but not very.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1681635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Establish the identity $\frac{\cot\theta + \sec\theta}{\cos\theta + \tan\theta} = \sec\theta \cot\theta\$ Establish the identity:
$$\dfrac{\cot\theta + \sec\theta}{\cos\theta + \tan\theta} = \sec\theta \cot\theta$$
The first step I got was:
$$\sec\theta \cot\theta = \dfrac{\sec\theta \cot\theta\,\big(\cos\theta + \tan\theta\big)}{\cos\theta + \tan\theta}$$
Then it tells me to rewrite the factor $$\cos\theta + \tan\theta$$
in the numerator using reciprocal identities.
How would I do that?
Here is what the assignment looked like:
| Hint:
$$\cot\theta + \sec\theta = \frac{1}{\tan\theta} + \frac{1}{\cos\theta} = \frac{\cos\theta + \tan\theta}{\tan\theta\cos\theta}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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If $A=\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}\;,$ Then $\lfloor A \rfloor\;\;,$
If $$A=\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}\;,$$ Then $\lfloor A \rfloor\;\;,$ Where $\lfloor x \rfloor $ Represent floor fiunction of $x$
My Try:: Using $\bf{A.M\geq H.M\;,}$ We get
$$\frac{1980+1891+1982+....+2012}{33}>\frac{33}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}$$
So $$\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}<\frac{1980+1981+....+2012}{(33)^2}=\frac{1996}{33}\approx 60.5<61$$
Now how can i prove that the above expression $A$ is $>60$
Help me, Thanks
| Let $$A=\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}=\frac 1K$$
$$K=\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}<\frac{1}{1980}+\frac{1}{1980}+\frac{1}{1980}+........+\frac{1}{1980}=\frac{33}{1980}=\frac{1}{60}$$
$$A=\frac 1K >60$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Find all integer solutions for $x*y = 5x+5y$ For this equation $x*y = 5x + 5y$ find all possible pairs.
The way I did it was: $x=5y/(y-5)$
And for this I wrote a program to brute force a couple of solutions. If it helps, some possibilities are: [4,-20], [6, 30], [10, 10]
So my question is: What is a mathematically correct way to solve this problem?
| $$5x+5y-xy=0$$
$$5x+5y-xy-25=-25$$
$$x(5-y)-5(5-y)=-25$$
$$(x-5)(y-5)=25$$
1) $x-5=1; y-5=25; \Rightarrow x=6; y=30$;
2) $x-5=-1; y-5=-25; \Rightarrow x=4; y=-20$;
3) $x-5=5; y-5=5; \Rightarrow x=10; y=10$
4) $x-5=-5; y-5=-5; \Rightarrow x=0; y=0$
5) $x-5=25; y-5=1; \Rightarrow x=30; y=6$
6) $x-5=-25; y-5=-1; \Rightarrow x=-20; y=4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Integrate square root of quotient polynomials I need to estimate this integral:
$$\displaystyle\int_a^\lambda \sqrt{\frac{(z-a)(z-b)}{z(z-1)}}dz$$
where $a$ and $b$ are closed to 1/2, and $z$ is complex.
Mathematica does not know how to do that. Any idea or reference ?
| Not quite what you asked but using Mathematica 10.2 with the input:
FullSimplify[Integrate[Sqrt[(z - a) (z - b)/(z (z - 1))], z]]
I got the following output:
$$\left(\sqrt{\frac{(-a+z)(-b+z)}{(-1+z)z}}\left((a-b)(b-z)\sqrt{\frac{(-1+b)(a-z)}{(a-b)(-1+z)}}z\sqrt{\frac{b-z}{b-bz}}+(a-b)b\sqrt{\frac{(-1+a)(b-z)}{(-a+b)(-1+z)}}(-1+z)\sqrt{-\frac{(-1+b)(b-z)z}{b^2(-1+z)^2}}\text{EllipticE}\left[\text{ArcSin}\left[\sqrt{\frac{(-1+a)(b-z)}{(-a+b)(-1+z)}}\right],\frac{a-b}{(-1+a)b}\right]-(-1+b)\left((a-b)\sqrt{\frac{(-1+a)(b-z)}{(-a+b)(-1+z)}}(-1+z)\sqrt{-\frac{(-1+b)(b-z)z}{b^2(-1+z)^2}}\text{EllipticF}\left[\text{ArcSin}\left[\sqrt{\frac{(-1+a)(b-z)}{(-a+b)(-1 + z)}}\right],\frac{a-b}{(-1+a)b}\right]+2(-1+a)(b-z)\sqrt{\frac{(-1+b)z}{b(-1+z)}}\text{EllipticF}\left[\text{ArcSin}\left[\sqrt{\frac{b-z}{b-bz}}\right],\frac{(-1+a)b}{a-b}\right]+b(-1+a+b)(-1+z)\sqrt{-\frac{(-1+b)(b-z)z}{b^2(-1+z)^2}}\sqrt{\frac{b-z}{b-bz}}\text{EllipticPi}\left[b,\text{ArcSin}\left[\sqrt{\frac{b-z}{b-bz}}\right],\frac{(-1+a)b}{a-b}\right]\right)\right)\right)\bigg/\left((a-b)(b-z)\sqrt{\frac{(-1+b)(a-z)}{(a-b)(-1+z)}}\sqrt{\frac{b-z}{b-bz}}\right)$$
You could hopefully then manually apply your limits of integration.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Cauchy integral formula- Evaluate the integrals The contour $\Gamma$ is parameterised by $r:[-\pi,\pi] \rightarrow \mathbb{C}$ is given by $r(\theta)=3e^{i\theta}+1$.
(a) $\int_{\Gamma} \frac{\sin(z)}{z-1}\ dz = 2 \pi i \sin(1)$?
(b) $\int_{\Gamma} \frac{\sin(\pi z^2)+\cos(\pi z^2)}{(z-1)(z-2)}\ dz = \frac{\frac{\sin(\pi z^2)+\cos(\pi z^2)}{(z-2)}}{(z-1)}\ dz = 2 \pi i \frac{\sin (\pi)+\cos(\pi)}{-1} = 2 \pi i$?
(c) $\int_{\Gamma} \frac{e^z}{z^2+4}\ dz = \int_{\Gamma} \frac{\frac{e^z}{z+2i}}{z-2i}\ dz = 2\pi i \frac{e^{2i}}{4i} = \frac{\pi}{2}e^{2i}$?
(d) $\int_{\Gamma} \frac{e^z}{z(z-1)(z-2)}\ dz $
Do I need partial fractions or can do this: $\int_{\Gamma} \frac{\frac{e^z}{z(z-1)}}{(z-2)}\ dz =2 \pi i \frac{e^2}{2} $
I need some clarification on my answers.
Thanks.
| The first one is correct, but you forgot to take into account all of the isolated singular points. Let's work through the problems one by one.
Throughout this post I will use the theorem that if we can write $f(z)=\frac{p(z)}{q(z)}$ where $p(z_0)\neq 0$, $q(z_0)=0$, and $q'(z_0)\neq0$ for some isolated singularity $z_0$, then
$$\underset{z=z_0}{\mbox{Res}} \, \frac{p(z)}{q(z)} = \frac{p(z_0)}{q'(z_0)}$$
A) Using the Cauchy residue theorem and identifying the isolated singular point to be $z_0=1$, we can solve the integral
$$\int_{\Gamma} \frac{\sin(z)}{z-1} dz = 2 \pi i \bigg( \underset{z=1}{\mbox{Res}} \, \frac{\sin(z)}{z-1} \bigg)=2 \pi i \big(\frac{\sin(1)}{1} \big)=2 \pi i \sin(1)$$
You got this question right because you used the appropriate equation and there was only one pole.
B) Using the Cauchy residue theorem and identifying the isolated singular points to be $z_0=1$ and $z_1=2$, we can solve the integral
$$\begin{aligned} \int_{\Gamma} \frac{\sin(z^2 \pi)+\cos(z^2 \pi)}{(z-1)(z-2)} dz =& 2 \pi i \bigg( \underset{z=1}{\mbox{Res}} \, \frac{\sin(z^2 \pi)+\cos(z^2 \pi)}{(z-1)(z-2)} + \underset{z=2}{\mbox{Res}} \, \frac{\sin(z^2 \pi)+\cos(z^2 \pi)}{(z-1)(z-2)} \bigg) \\ =& 2 \pi i \bigg(\frac{\frac{\sin(1 \pi)+\cos(1\pi)}{1-2}}{1} + \frac{\frac{\sin(4 \pi)+\cos(4\pi)}{2-1}}{1} \bigg) \\ =& 2 \pi i (1+1) \\ =& 4 \pi i\end{aligned}$$
C) Using the Cauchy residue theorem and identifying the isolated singular points to be $z_0=2i$ and $z_1=-2i$, we can solve the integral
$$\begin{aligned} \int_{\Gamma} \frac{e^z}{z^2+4} dz =& 2 \pi i \bigg( \underset{z=2i}{\mbox{Res}} \, \frac{e^z}{(z-2i)(z+2i)} + \underset{z=-2i}{\mbox{Res}} \, \frac{e^z}{(z-2i)(z+2i)} \bigg) \\ =& 2 \pi i \bigg(\frac{\frac{e^{2i}}{(2i+2i)}}{1} + \frac{\frac{e^{-2i}}{(-2i-2i)}}{1} \bigg) \\ =& 2 \pi i \big(\frac{e^{2i}}{4i} + \frac{e^{-2i}}{-4i} \big) \\ =& \pi i \sin(2)\end{aligned}$$
Note at the end we used the identity $\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$.
D) Using the Cauchy residue theorem and identifying the isolated singular points to be $z_0=0$, $z_1=1$ and $z_2=2$, we can solve the integral
$$\begin{aligned} \int_{\Gamma} \frac{e^z}{z(z-1)(z-2)} dz =& 2 \pi i \bigg( \underset{z=0}{\mbox{Res}} \, \frac{e^z}{z(z-1)(z-2)} + \underset{z=1}{\mbox{Res}} \, \frac{e^z}{z(z-1)(z-2)} + \underset{z=2}{\mbox{Res}} \, \frac{e^z}{z(z-1)(z-2)} \bigg) \\ =& 2 \pi i \bigg(\frac{\frac{e^{0}}{(0-1)(0-2)}}{1} + \frac{\frac{e^{1}}{(1)(1-2)}}{1} + \frac{\frac{e^{2}}{(2)(2-1)}}{1} \bigg) \\ =& 2 \pi i \big(\frac{1}{2} + \frac{e}{-1} + \frac{e^2}{2} \big) \\ =& \pi i (e-1)^2 \end{aligned}$$
It seems from these questions that when using the Cauchy residue theorem
$$\int_{C} f(z) \, dz = 2 \pi i \sum_{k=0}^{n} \underset{z=z_k}{\mbox{Res}} \, f(z)$$
You forgot that you had to calculate the residue for $\textit{every}$ isolated singular point inside of our contour. Since the contour $\Gamma$ is a circle of radius $3$ centered at $z=1$, it contains all of the isolated singularities for each function in parts A) through D).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does $a_{n} = \frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + ... + \frac{1}{\sqrt{n^2+2n-1}}$ converge? $a_{n} = \frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + ... + \frac{1}{\sqrt{n^2+2n-1}}$
and I need to check whether this sequence converges to a limit without finding the limit itself. I think about using the squeeze theorem that converges to something (I suspect '$1$').
But I wrote $a_{n+1}$ and $a_{n-1}$ and it doesn't get me anywhere...
| $$\frac{1}{\sqrt{n^2+2n-1}}+\frac{1}{\sqrt{n^2+2n-1}}\cdots+\frac{1}{\sqrt{n^2+2n-1}}\le\frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + \cdots + \frac{1}{\sqrt{n^2+2n-1}}\le\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+n}}\cdots+\frac{1}{\sqrt{n^2+n}}$$
$$\frac{n}{\sqrt{n^2+2n-1}}\le\frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + \cdots + \frac{1}{\sqrt{n^2+2n-1}}\le\frac{n}{\sqrt{n^2+n}}$$
$$\lim_{n\rightarrow\infty}\frac{n}{\sqrt{n^2+2n-1}}\le\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + \cdots + \frac{1}{\sqrt{n^2+2n-1}}\le\lim_{n\rightarrow\infty}\frac{n}{\sqrt{n^2+n}}$$
$$\lim_{n\rightarrow\infty}\frac{n}{n\sqrt{1+\frac2n-\frac1{n^2}}}\le\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + \cdots + \frac{1}{\sqrt{n^2+2n-1}}\le\lim_{n\rightarrow\infty}\frac{n}{n\sqrt{1+\frac1n}}$$
$$1\le\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + \cdots + \frac{1}{\sqrt{n^2+2n-1}}\le1$$
$$\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + \cdots + \frac{1}{\sqrt{n^2+2n-1}}=1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving for $z$ given 3 constraints: $\DeclareMathOperator{\Re}{Re}\Re[z^4]=1/2$ , $z\bar{z}+2|z|-3=0$, $\arg z \leq \frac{\pi}{4}.$
Let $z$ be a complex number satisfying
$$\DeclareMathOperator{\Re}{Re}\Re[z^4]=\frac{1}{2}$$
$$z\bar{z}+2|z|-3=0$$
$$\arg z \leq \frac{\pi}{4}.$$
Find $z$
As a side note this is the same as this question here: Finding complex number defined by 3 equations , but I'm looking for a way to finish it using my approach.
I'm going to let $z = x+iy$ and attempt to solve for $x$ and $y$
$$\DeclareMathOperator{\Re}{Re}\Re[z^4]=\frac{1}{2}$$
$$\DeclareMathOperator{\Re}{Re}\Re[(x+iy)^4]=\frac{1}{2}$$
$$\DeclareMathOperator{\Re}{Re}\Re[x^4+y^4-6x^2y^2 +i(4x^3y-4xy^3)]=\frac{1}{2}$$
Considering real only
$$\DeclareMathOperator{\Re}{Re}\Re[x^4+y^4-6x^2y^2]=\frac{1}{2}$$
$$ x^4+y^4-6x^2y^2 = \frac{1}{2} $$
Now with the second equation
$$z\bar{z}+2|z|-3=0$$
$$ (x+iy)(x-iy) + 2\sqrt{x^2+y^2} - 3 = 0 $$
$$ x^2 + y^2 + 2\sqrt{x^2+y^2} - 3 = 0 $$
I will worry about the argument after I solve for $x$ and $y$.
I now have $2$ equations:
$$ x^4+y^4-6x^2y^2 = \frac{1}{2} $$
$$ x^2 + y^2 + 2\sqrt{x^2+y^2} - 3 = 0 $$
Working on the second one
$$ x^2 + y^2 + 2\sqrt{x^2+y^2} - 3 = 0 $$
Let $x^2+y^2$ $=$ $p$
$$ p + 2\sqrt{p} - 3 = 0 $$
$$ (\sqrt{p}-1)(\sqrt{p}+3) = 0 $$
$$ \sqrt{p} = 1 , \sqrt{p} \neq -3 $$
$$ x^2+y^2 =1 $$
Squaring both sides
$$ x^4 + y^4 + 2x^2y^2 = 1 $$
$$ x^4 + y^4 = 1-2x^2y^2 $$
Subbing this into the first equation
$$ x^4+y^4-6x^2y^2 = \frac{1}{2} $$
$$ 1-2x^2y^2-6x^2y^2 = \frac{1}{2} $$
$$ \frac{1}{2} = 8x^2y^2 $$
$$ \frac{1}{16} = x^2y^2 $$
$$ y^2 = \frac{1}{16x^2} $$
Subbing this into
$$ x^2+\frac{1}{16x^2} =1 $$
$$ x^4 - x^2 + \frac{1}{16} = 0 $$
$$ x^2 = \frac{2+\sqrt{3}}{4} , x^2 = \frac{2-\sqrt{3}}{4}$$
$$ x = ± \frac{\sqrt{6}+\sqrt{2}}{4} , x= ± \frac{\sqrt{6}-\sqrt{2}}{4} $$
$$ y = ± \frac{\sqrt{6}-\sqrt{2}}{4} , y= ± \frac{\sqrt{6}+\sqrt{2}}{4}$$
Now I am stuck which set is correct to represent $z=x+iy$?
| The third condition gives you - Both $x$ and $y$ are positive and $x>y$
So, from your set of answers(assuming they are right), the answer would be
$x=\frac{\sqrt 6+\sqrt 2}{4}$
$y=\frac{\sqrt 6-\sqrt 2}{4}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $4k^3 + 17k^2 - 228k -1116 = 0$ The equation given to me is $$4x^4 + 16x^3 - 17x^2 - 102x -45 = 0$$
I'm asked to find it's resolvent cubic which is not so difficult to find. But the problem is that the question further asks to find the solution of resolvent cubic.
I have found resolvent cubic using Ferrari's method. The resolvent cubic came out to be $$4k^3 + 17k^2 - 228k -1116 = 0$$
The trouble begins now when I'm trying to solve this cubic it's getting very exhaustive and long calculations. But since the question asks to find the solution of the cubic.
Now can someone help me solve this cubic easily or there's no way out to this problem. I want to escape the tedious calculations while solving this cubic.
Kindly help me if you can.
| Ferrari's method: If $a$, $b$, $c$, $d$ are the roots of the quartic, form the equation with roots
$$a b + c d, a c + b d, a d + b c$$
The coefficients of this equation are symmetric in $a$, $b$, $c$, $d$ ( since we have $a b + c d$ and all its symmetric transforms).
After we solve the cubic, the quadric is solved using the equalities
$$(a+b+c+d)^2 - 4 ( a b + c d) - 4 ( a d + b c) = (a- b + c - d)^2$$
Now, with $u = a b + c d$, $v = a c + b d$, $w= a d + b c$ we have
$(u-v)(u-w)(v-w) = (a-b)(a-c)(a-d)(b-c)(b-d)(c-d)$$
so the discriminant of the cubic resolvent equals the discriminant of the quadric.
If the quartic is
$$x^ 4 + s x^3 + p x^2 + q x + r$$
then the Ferrari cubic resolvent is
$$x^3- p x^2 + (q s - 4 r)x + (4 p r - r s^2 + 4 p r)$$
So, if the particular quartic is
$$4(x^4 + 4x^3 - \frac{17}{4}x^2 - \frac{51}{2}x -\frac{45}{4})$$
then the Ferrari resolvent is
$$x^3 + \frac{17}{4}x^2 - 57 x - 279$$
or
$$4 x^3 + 17 x^2 - 228 x - 1116$$
Calculating the discriminant of $4 x^3 + 17 x^2 - 228 x - 1116$ we get $0$. That means that the cubic has multiple roots. To find a multiple root, calculate the GCD of the cubic and its derivative
$$\textrm{gcd}(x^3 + \frac{17}{4}x^2 - 57 x - 279, 3 x^2 + \frac{17}{2}x - 57) = x+6$$
Now, divide the cubic $x^3 + \frac{17}{4}x^2 - 57 x - 279$ by $(x+6)^2$ and find $x-\frac{31}{4}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Presentation of the unit Let $N -$ number of fives $(a_1, a_2, a_3, a_4, a_5)$ positive integers,
satisfying the condition
$$\frac {1}{a_1}+\frac {1}{a_2}+\frac {1}{a_3}+\frac {1}{a_4}+\frac {1}{a_5}=1.$$
Find out even or odd number is $N$.
My work so far:
$$\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=1 -$$ first five $\left({5};{5};{5};{5};{5} \right)$
$$\frac{1}{2}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=1 -$$ the following five fives $\left(2;8;{8};{8};{8} \right)$; $\left({8};{2};{8};{8};{8} \right)$;...; $\left({8};{8};{8};{8};{2} \right)$
$$\frac{1}{3}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=1 -$$ the following five fives $\left({3};{6};{6};{6};{6} \right)$; $\left({6};{3};{6};{6};{6} \right)$;...; $\left({6};{6};{6};{6};{3} \right)$.
$$N=1+5+5+k?$$
$$k=?$$
| HINT
You just have to determine if $N$ is even or odd.
If $a_1, a_2, a_3, a_4, a_5$ are all distinct, there are $120$ ways to arrange them.
Similarly, if two of them were the same there are $60$ ways to arrange them.
In both cases, there are an even number of ways to arrange them.
However, these cases, which are part of $N$ will make no difference in whether of not $N$ is even or odd.
Note the only cases where there are an odd number of ways to arrange them
is when at least $4$ numbers are the same.
I think you can continue from here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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indefinite integration $ \int \frac { x^2 dx} {x^4 + x^2 -2}$ problem : $ \int \frac { x^2 dx} {x^4 + x^2 -2}$
solution : divide numerator and denominator by $x^2$
$ \int \frac { dx} {x^2 + 1 -\frac{1}{x^2}}$
Now whats the next step $?$
Am I doing right $?$
| You are better off using partial fractions, which you should know if you have been given this integral to evaluate. You have the following factors for the denominator:
$x^4+x^2-2
= (x^2-1)(x^2+2)
= (x+1)(x-1)(x^2+2)$
And then
$x^2/(x^4+x^2-2) = a/(x+1)+b/(x-1)+(cx+d)/(x^2+2)$
You are to find the condtants a, b, c, d that fit this equation. Then you integrate the resulting partial fractions using logatlrithms and inverse tangents, and add those partial integrals up.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Simultaneous Diagonalization of two bilinear forms I need to diagonalize this two bilinear forms in the same basis (such that $f=I$ and $g$=diagonal matrix):
$f(x,y,z)=x^2+y^2+z^2+xy-yz $
$g(x,y,z)=y^2-4xy+8xz+4yz$
I know that it is possible because f is positive-definite, but I don't know how can I do it
| As I commented, a more difficult version is at Congruence and diagonalizations
This one is easier, the matrix $C$ has three distinct eigenvalues that are integers.
$$
A =
\left(
\begin{array}{rrr}
1 & \frac{1}{2} & 0 \\
\frac{1}{2} & 1 & -\frac{1}{2} \\
0 & -\frac{1}{2} & 1
\end{array}
\right)
$$
$$
B =
\left(
\begin{array}{rrr}
0 & -2 & 4 \\
-2 & 1 & 2 \\
4 & 2 & 0
\end{array}
\right)
$$
$$
A^{-1} =
\left(
\begin{array}{rrr}
\frac{3}{2} & -1 & -\frac{1}{2} \\
-1 & 2 & 1 \\
-\frac{1}{2} & 1 & \frac{3}{2}
\end{array}
\right)
$$
$$
C = A^{-1}B =
\left(
\begin{array}{rrr}
0 & -5 & 4 \\
0 & 6 & 0 \\
4 & 5 & 0
\end{array}
\right)
$$
The theorem from Horn and Johnson (first edition hardcover was 1985, paperback 1990) is that we can continue if and only if $C$ is diagonalizable in that $R^{-1} C R =D$ is diagonal.
They were extremely careful: the eigenvalues of $C$ are $6,4,-4$ and we can make the matrix $R$ with columns as eigenvectors with
$$
R =
\left(
\begin{array}{rrr}
1 & 1 & -1 \\
-2 & 0 & 0 \\
-1 & 1 & 1
\end{array}
\right).
$$
Confirm
$$
CR =
\left(
\begin{array}{rrr}
6 & 4 & 4 \\
-12 & 0 & 0 \\
-6 & 4 & -4
\end{array}
\right).
$$
It follows that $R^{-1}CR$ is the diagonal matrix with entries $6,4,-4.$
$$
D =
\left(
\begin{array}{rrr}
6 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & -4
\end{array}
\right).
$$
Finally,
$$
R^TBR =
\left(
\begin{array}{rrr}
12 & 0 & 0 \\
0 & 8 & 0 \\
0 & 0 & -8
\end{array}
\right),
$$
and
$$
R^TAR =
\left(
\begin{array}{rrr}
2 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2
\end{array}
\right).
$$
The reason this works is that we have arranged $R^T AR D = R^T BR.$
Let's see, this is in the first edition of Horn and Johnson, table 4.5.15T on page 229, then detail for case II on pages 231-232. The technique gives the full problem when $C$ has all eigenvalues distinct, as in this problem. With repeat eigenvalues, one needs to continue working, that is half of page 232. Indeed, when this happens, we are guaranteed to have square diagonal blocks in both the revised $A$ and $B$ matrices, such that each $B$ block is just a scalar multiple of each corresponding $A$ block. The extra work is then to diagonalize the $A$ block, the same thing will work on the $B$ block. That is exactly what is shown at Congruence and diagonalizations
Oh, you wanted the identity matrix. Do another step with $R_2 = I/ \sqrt 2$ on right and left, the result is just to halve each diagonal matrix. The final overall matrix is my $R/\sqrt 2.$
Tuesday: to continue, take $Q = R/\sqrt 2.$ Then $Q^{-1} = R^{-1} \sqrt 2$
$$
Q^{-1} = R^{-1} \sqrt 2 = \; \;
\left(
\begin{array}{rrr}
0 & \frac{-1}{\sqrt 2} & 0 \\
\frac{1}{\sqrt 2} & 0 & \frac{1}{\sqrt 2} \\
\frac{-1}{\sqrt 2} & \frac{-1}{\sqrt 2} & \frac{1}{\sqrt 2}
\end{array}
\right).
$$
The rows of $Q^{-1}$ give us
$$ \left(\frac{-1}{\sqrt 2}y \right)^2 + \left(\frac{1}{\sqrt 2} x + \frac{1}{\sqrt 2} z \right)^2 + \left(\frac{-1}{\sqrt 2} x +\frac{-1}{\sqrt 2}y + \frac{1}{\sqrt 2} z \right)^2 = x^2 + y^2 + z^2 -yz +xy $$
$$ 6\left(\frac{-1}{\sqrt 2}y \right)^2 + 4 \left(\frac{1}{\sqrt 2} x + \frac{1}{\sqrt 2} z \right)^2 -4 \left(\frac{-1}{\sqrt 2} x +\frac{-1}{\sqrt 2}y + \frac{1}{\sqrt 2} z \right)^2 = y^2 + 4yz + 8zx - 4 xy $$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to expand $f(z)$ as a Laurent Series Given
$$\frac{z}{(z-1)(z+2i)}$$
expand $f(z)$ in the following regions: $|z|<1$, $1<|z|<2$, $|z|>2$
I'm preparing for an exam and Laurent Series are a weakness of mine. I would love advice regarding interpretation, understanding, and solution of the problem.
| Here is a somewhat detailed description of the Laurent expansion of $f(z)$ around $z=0$.
The function
\begin{align*}
f(z)&=\frac{z}{(z-1)(z+2i)}
=\frac{1-2i}{5}\cdot\frac{1}{z-1}+\frac{4+2i}{5}\cdot\frac{1}{z+2i}\\
\end{align*}
has two simple poles at $1$ and $-2i$.
Since we want to find a Laurent expansion with center $0$, we look at the poles $1$ and $-2i$ and see they determine three regions.
\begin{align*}
|z|<1,\qquad\quad
1<|z|<2,\qquad\quad
2<|z|
\end{align*}
*
*The first region $ |z|<1$ is a disc with center $0$, radius $1$ and the pole $-2i$ at the boundary of the disc. In the interior of this disc all two fractions with poles $0$ and $-2i$ admit a representation as power series at $z=0$.
*The second region $1<|z|<2$ is the annulus with center $0$, inner radius $1$ and outer radius $2$. Here we have a representation of the fraction with pole $1$ as principal part of a Laurent series at $z=0$, while the fraction with pole at $-2i$ admits a representation as power series.
*The third region $|z|>2$ containing all points outside the disc with center $0$ and radius $2$ admits for all fractions a representation as principal part of a Laurent series at $z=0$.
A power series expansion of $\frac{1}{z+a}$ at $z=0$ is using geometric series expansion
\begin{align*}
\frac{1}{z+a}&=\frac{1}{a}\cdot\frac{1}{1+\frac{z}{a}}
=\frac{1}{a}\sum_{n=0}^{\infty}\left(-\frac{z}{a}\right)^n\\
&=\sum_{n=0}^{\infty}\frac{(-1)^n}{a^{n+1}}z^n
\end{align*}
The principal part of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{z}\cdot\frac{1}{1+\frac{a}{z}}
=\frac{1}{z}\sum_{n=0}^{\infty}\left(-\frac{a}{z}\right)^n
=\sum_{n=0}^{\infty}(-a)^n\frac{1}{z^{n+1}}\\
&=\sum_{n=1}^{\infty}\left(-a\right)^{n-1}\frac{1}{z^n}
\end{align*}
We can now obtain the Laurent expansion of $f(z)$ at $z=0$ for all three regions. Here we consider the second region
*
*Region 2: $1<|z|<2$
\begin{align*}
f(z)&=\frac{1-2i}{5}\frac{1}{z-1}+\frac{4+2i}{5}\frac{1}{z+2i}\\
&=\frac{1-2i}{5}\sum_{n=1}^{\infty}\frac{1}{z^n}
+\frac{4+2i}{5}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2i)^{n+1}}z^{n}\\
&=\frac{1-2i}{5}\sum_{n=1}^{\infty}\frac{1}{z^n}
+\frac{4+2i}{5}\cdot\frac{1}{2i}\sum_{n=0}^{\infty}\left(-\frac{1}{2i}\right)^nz^n\\
&=\frac{1-2i}{5}\left(\sum_{n=1}^{\infty}\frac{1}{z^n}
+\sum_{n=0}^{\infty}\left(\frac{i}{2}\right)^nz^n\right)\\
\end{align*}
The other regions can be calculated similarly.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $
Question: Solve $$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 0 \le x \le 360^{\circ} $$
My attempt:
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 6(\frac{1}{2} - \frac{\cos(2x)}{2}) + \sin(x)\cos(x) -(\frac{1}{2} + \frac{\cos(2x)}{2}) = 5 $$
$$ 3 - 3\cos(2x)+ \sin(x)\cos(x) - \frac{1}{2} - \frac{\cos(2x)}{2} = 5$$
$$ \frac{7\cos(2x)}{2} - \sin(x)\cos(x) + \frac{5}{2} = 0 $$
$$ 7\cos(2x) - 2\sin(x)\cos(x) + 5 = 0 $$
$$ 7\cos(2x) - \sin(2x) + 5 = 0 $$
So at this point I am stuck what to do, I have attempted a Weierstrass sub of $\tan(\frac{x}{2}) = y$ and $\cos(x) = \frac{1-y^2}{1+y^2}$ and $\sin(x)=\frac{2y}{1+y^2} $ but I got a quartic and I was not able to solve it.
| $$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 \cdot 1 $$
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 \cdot (\sin^2(x)+\cos^2(x)) $$
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5\sin^2(x)+5\cos^2(x) $$
$$ \sin^2(x)+\sin(x)\cos(x)-6\cos^2(x)=0$$
$$\tan^2(x)+\tan(x)-6=0$$
$\tan(x)=2$ or $\tan (x)=-3$
$x=\arctan2+\pi n, n \in \mathbb Z$
$x=-\arctan3+\pi k, k \in \mathbb Z$
| {
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Which two digit number when you find the product of the digits yields a number that is half the original? Which two digit number when you find the product of the digits yields a number that is half the original?
Let x=$ab$ be the $2$-digit number. So $x=10a+b$.
Then $ab=\frac{x}{2} \implies ab=\frac{10a+b}{2} \implies 2ab=10a+b \implies b=\frac{10a}{2a-1}$. I guess $a=3$ and get $b=6$. So the answer is $36$.
But how can this be done without guessing.
| From what you figured out so far, $ab=5a+\frac{b}{2}$. If there is an integer solution, $b$ is even and greater than $5$, so $b$ is either $6$ or $8$. Also (see the picture), $\frac{b}{2}$ is divisible by $b-5$, which rules out $b=8$. If $b=6$, then the rightmost rectangle below is $1$ unit across and has area $\frac{6}{2}=3$, so $a=3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $x^2+y^4=1994$ Let $x$ and $y$ positive integers with $y>3$, and $$x^2+y^4=2(x-6)^2+2(y+1)^2$$
Prove that $x^2+y^4=1994$.
I've tried finding an upper bound on the value of $x$ or $y$, but without sucess. Can anyone help me prove this problem? Note that $x^2+y^4=1994$ is the result we are trying to prove, not an assumption.
| Rearrange the equation given into
$$
(x-12)^2 = 71 + (y^2-1)^2 - 4y.
$$
$(y^2-1)^2$ is a square. The square before $(y^2-1)^2$ is $(y^2-2)^2$, which is $2y^2-1$ less. For $y > 2$, $2y^2-1 > 4y > 4y-71$. The square after $(y^2-1)^2$ is $y^4$, which is $2y^2+1$ greater. For $y\geq 6$, $2y^2+1 > 71 > 71-4y$.
You therefore only need to consider $y=4$ and $y=5$. It happens that $y=5$ gives you $x=37$.
| {
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Unable to find solution for $a^2+b^2-ab$, given $a^2+b^2-ab$ is a prime number of form $3x+1$ I have a list of prime numbers which can be expressed in the form of $3x+1$. One such prime of form $3x+1$ satisfies the expression: $a^2+b^2-ab$.
Now I am having list of prime numbers of form $3x+1$ (i.e., $7,19 \ldots$). But I am unable to find the $a$ and $b$ which satisfy the above expression.
Thanks for your help in advance.
| Use the identity $a^3+b^3=(a+b)(a^2-ab+b^2)$, and note that, by Little Fermat, $x^3\equiv x\mod 3$, so $a^3+b^3\equiv a+b \mod3$. Hence, if $a+b\not\equiv 0\mod 3$, necessarily $a^2-ab+b^2\equiv 1\mod3$.
Now suppose you've found $a$ and $b$ such that $a^3+b^3$ is the product of two primes. Then one of them will be congruent to $1\bmod 3$, and have the required form.
Computing some values yields $1^3+4^3=(1+4)(1^2-1\cdot 4+4^2)=5\cdot 13$. Thus $13$ is a solution.
| {
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If the range of the function $f(x)=\frac{x^2+ax+b}{x^2+2x+3}$ is $[-5,4],a,b\in N$,then find the value of $a^2+b^2.$ If the range of the function $f(x)=\frac{x^2+ax+b}{x^2+2x+3}$ is $[-5,4],a,b\in N$,then find the value of $a^2+b^2.$
Let $y=\frac{x^2+ax+b}{x^2+2x+3}$
$$x^2y+2xy+3y=x^2+ax+b$$
$$x^2(y-1)+x(2y-a)+3y-b=0$$
As $x$ is real,so the discriminant of the above quadratic equation has to be greater than or equal to zero.
$$(2y-a)^2-4(y-1)(3y-b)\geq0$$
$$-8y^2+y(-4a+4b+12)+a^2-4b\geq0$$
As above quadratic inequality is greater than or equal to zero,so its discriminant is less than or equal to zero.
$$(-4a+4b+12)^2+32(a^2-4b)\leq0$$
I am stuck here,i cant find $a^2+b^2$.
| Begin with
$$g(y)=-8y^2+y(-4a+4b+12)+a^2-4b\geq 0,\quad\forall y\in[-5,4].$$
Then we wish to solve $g(-5)=g(4)=0$, that is,
\begin{align}
-200-5(-4a+4b+12)+a^2-4b=0,\\
-128+4(-4a+4b+12)+a^2-4b=0,
\end{align}
or
\begin{align}
a^2+20a-24b&=260,\\
a^2-16a+12b&=80.
\end{align}
Thus
\begin{align}
a-b=5\quad\mbox{and}\quad
0=a^2-4a-140=(a-14)(a+10)
\end{align}
and we obtain two sets of solutions
$$a=14,\,b=9\quad\mbox{and}\quad
a=-10,\,b=-15.$$
where the second solution does not fit because $a,$ $b$ are assumed to be positive integers. Hence $a^2+b^2=277.$
| {
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solve $3^x +4^x = 5^x$ in $\Bbb R$ solve the following equation :
$$3^x +4^x = 5^x \text{ in } \Bbb R$$
the trivial answer is $x=2$ .
| $$\left( \frac{3}{5} \right)^x+\left( \frac{4}{5} \right)^x=1$$
Then $x=2$
$f(x)=\left( \frac{3}{5} \right)^x+\left( \frac{4}{5} \right)^x$decreasing function
| {
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Find the largest integer $n$ such that $n^2$ is the difference of two consecutive cubes and $2n +79$ is a perfect square.
Find the largest integer $n$ such that $n^2$ is the difference of two consecutive cubes and $2n +79$ is a perfect square.
This is an AIME problem. I have been trying and have been going round in circles.
First of all, let the smaller cube be $k$.
Difference in cubes $= 3k^2 + 3k +1 = 3k(k+1) + 1=n^2$. It's clear that this $n$ is always odd.
So,
$$2\sqrt{3k(k+1)+1}+79=m^2$$
Solving this, we get:
$$3k(k+1) = \frac{(m^2-81)(m^2-77)}{4}$$
or
$$3k(k+1)+1 = \frac{(m^2-81)^2 + 4(m^2-81)+4}{4}$$
I feel that I'm getting nowhere near the solution.
Can you please give me some hints on how to proceed. Thanks.
| Let $$n^2 = (m + 1)^3 - m^3 = 3m^2 + 3m + 1$$
Note that $$(2n + 1)(2n - 1) = 4n^2 - 1 = 12m^2 + 12m + 3 = 3(2m + 1)^2$$
Since $$\gcd(2n-1,2n+1)=1$$
Since their product is three times a square, one of them must be a square and the other three times a square. If we have $2n - 1=3a^2$, then $2n + 1=b^2=3a^2+2 \equiv 2 \pmod 3$ which is impossible since $2$ is a quadratic non-residue of $3$.
Thus $2n - 1$ is a square, let this be $b^2$. But since $2n + 79$ is also a square, say $a^2$. Then $(a + b)(a - b) = a^2 - b^2 = 80$.
Since $a + b$ and $a - b$ have the same parity and their product is even, they are both even.
To find the largest value of $n$, it suffices to maximize $2b$, which is $(a + b) - (a - b)$.
However , it is not difficult to see this occurs when $a + b = 40$ and $a - b = 2$, that is, when $a = 21$ and $b = 19$. This yields $n = 181$ and $m = 104$, so the answer is $181$.
| {
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What type of functional equation is this? I'm trying to solve the following functional equation
$f\left(x\right)=A\mbox{ exp}\left\{ \int\frac{1}{f\left(x\right)x^{2}+Bx}dx\right\}$
where $f\left(x\right):\mathbb{R}_{+}\rightarrow\mathbb{R}_{\geq0}$, A and B are constants in $\mathbb{R}$.
Does anybody recognize this type of functional equation so I can look up for the solution?
Alternatively, does anybody know how to solve it? Or a suggestion on how to start tackling this problem?
Thank you!
After the initial comments I realized the problem is equivalent to solving the following first-order nonlinear ODE: $f^{\prime}f=\frac{1}{x^{2}}\left(\left(Ax+Bx^{2}\right)f^{\prime}+Bf\right)$
This equation seems similar to an Abel differential equation of the second kind, i.e.
$ff^{\prime}=g\left(x\right)f+h\left(x\right)$
although it's not quite the same. If anybody has an idea how to deal with it I would appreciate it!
| First, there is no concept about "indefinite integral equation" , so you should modify the question as $f(x)=Ae^{\int_k^x\frac{1}{f(x)x^2+Bx}dx}$
$\ln\dfrac{f(x)}{A}=\int_k^x\dfrac{1}{f(x)x^2+Bx}dx$
$\dfrac{1}{f(x)}\dfrac{df(x)}{dx}=\dfrac{1}{f(x)x^2+Bx}$ with $f(k)=A$
$f\dfrac{dx}{df}=fx^2+Bx$ with $x(A)=k$
Case $1$: $B=0$
Then $f\dfrac{dx}{df}=fx^2$ with $x(A)=k$
$df=\dfrac{dx}{x^2}$ with $f(k)=A$
$\int_A^fdf=\int_k^x\dfrac{dx}{x^2}$
$[f]_A^f=\left[-\dfrac{1}{x}\right]_k^x$
$f-A=\dfrac{1}{k}-\dfrac{1}{x}$
$f(x)=A+\dfrac{1}{k}-\dfrac{1}{x}$
Case $1$: $B=0$
Then $f\dfrac{dx}{df}=fx^2+Bx$ with $x(A)=k$
$\dfrac{dx}{df}-\dfrac{Bx}{f}=x^2$ with $x(A)=k$
Let $x=\dfrac{1}{y}$ ,
Then $\dfrac{dx}{df}=-\dfrac{1}{y^2}\dfrac{dy}{df}$
$\therefore-\dfrac{1}{y^2}\dfrac{dy}{df}-\dfrac{B}{fy}=\dfrac{1}{y^2}$ with $y(A)=\dfrac{1}{k}$
$\dfrac{dy}{df}+\dfrac{By}{f}=-1$ with $y(A)=\dfrac{1}{k}$
I.F. $=e^{\int\frac{B}{f}df}=e^{B\ln f}=f^B$
$\therefore\dfrac{d(f^By)}{df}=-f^B$
$f^By=-\int f^B~df$
$f^By=\begin{cases}-\dfrac{f^{B+1}}{B+1}+C&\text{when}~B\neq0,-1\\-\ln f+C&\text{when}~B=-1\end{cases}$
$y(A)=\dfrac{1}{k}$ :
$\dfrac{A^B}{k}=\begin{cases}-\dfrac{A^{B+1}}{B+1}+C&\text{when}~B\neq0,-1\\-\ln A+C&\text{when}~B=-1\end{cases}$
$C=\begin{cases}\dfrac{A^B}{k}+\dfrac{A^{B+1}}{B+1}&\text{when}~B\neq0,-1\\\dfrac{1}{kA}+\ln A&\text{when}~B=-1\end{cases}$
$\therefore f^By=\begin{cases}\dfrac{A^B}{k}+\dfrac{A^{B+1}}{B+1}-\dfrac{f^{B+1}}{B+1}&\text{when}~B\neq0,-1\\\dfrac{1}{kA}+\ln A-\ln f&\text{when}~B=-1\end{cases}$
$\dfrac{f^B}{x}=\begin{cases}\dfrac{A^B}{k}+\dfrac{A^{B+1}}{B+1}-\dfrac{f^{B+1}}{B+1}&\text{when}~B\neq0,-1\\\dfrac{1}{kA}+\ln A-\ln f&\text{when}~B=-1\end{cases}$
$x=\begin{cases}\dfrac{1}{\dfrac{A^B}{kf^B}+\dfrac{A^{B+1}}{(B+1)f^B}-\dfrac{f}{B+1}}&\text{when}~B\neq0,-1\\\dfrac{1}{\dfrac{f}{kA}+f\ln A-f\ln f}&\text{when}~B=-1\end{cases}$
| {
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Equation of sphere through variable points and origin Question : Find the equation of the sphere through four points $(0,0,0) , (-a,b,c) , (a,-b,c)$ and $(a,b,-c)$. Also find the centre and the radius of the sphere.
Now I know that the sphere passes through origin. Therefore, the constant term in the equation of the sphere will be $0$.
According to standard equation the equation of sphere will look like :
$$x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$$
Here the centre is $(u,v,w)$ is the centre and $d = 0$ because the sphere passes through origin.
Now while trying to get the equation by using points I'm getting nowhere. Is there any way to solve this easily?
I don't know what to do in this. Kindly help.
| The four points form two triangles on the sphere. The planes on which these triangles lie, intersect the sphere in two circles, which are the circumscribed circles of the two triangles. The two lines perpendicular to the planes, through the center of the circles, intersect each other in the center of the sphere. The center of each circle is the intersection point of the perpendicular lines on the three sides of the triangle.
First look at the triangle composed by $(0,0,0), (-a,b,c)$ and $(a,-b,c)$: the perpendicular vector is the vector product of two of the vectors composing the triangle sides, thus
$$\vec{n_1} = \left(\begin{matrix}-a\\ b \\ c\end{matrix}\right) \times \left(\begin{matrix}a\\ -b \\ c\end{matrix}\right) = \left(\begin{matrix}2bc\\ 2ac \\ 0\end{matrix}\right) \equiv \left(\begin{matrix}b\\ a \\ 0\end{matrix}\right)$$
The intersection of the perpendicular lines can be derived from the fact that the sides itself are the perpendiculars to the planes that intersect into that same point. These perpendicular vectors and points on these three planes are:
$$
\begin{cases}
\left(\begin{matrix}-a\\ b \\ c\end{matrix}\right) passing through \frac{1}{2}(-a,b,c)\\
\left(\begin{matrix}a\\ -b \\ c\end{matrix}\right) passing through \frac{1}{2}(a,-b,c)\\
\left(\begin{matrix}a\\ -b \\ 0\end{matrix}\right) passing through (0,0,c)
\end{cases}
$$
Thus that point is a solution to the following set of equations (as the three planes perpendicular to the triangle sides intersect into a line, the equation for the triangle plane itself, which passes through (0,0,0), has to be included):
$$
\begin{cases}
-ax+by+cz &= \frac{1}{2}(a^2+b^2+c^2)\\
ax-by+cz &= \frac{1}{2}(a^2+b^2+c^2)\\
ax-by &= 0\\
bx+ay &= 0
\end{cases}
$$
Which leads to the following point that is the center of the circumscribed circle: $(0,0,\frac{a^2+b^2+c^2}{2c})$
The first line through the sphere's center thus has the following vector specification:
$$\left(\begin{matrix}x\\ y \\ z\end{matrix}\right)=\left(\begin{matrix}0\\ 0 \\ \frac{a^2+b^2+c^2}{2c}\end{matrix}\right) + \lambda\left(\begin{matrix}b\\ a \\ 0\end{matrix}\right)$$
Similar for the other intersection point of the three perpendicular lines of the second triangle (composed by $(a,b,-c), (-a,b,c)$ and $(a,-b,c)$): perpendicular vector is
$$\vec{n_2} = \left(\begin{matrix}0\\ -b \\ c\end{matrix}\right) \times \left(\begin{matrix}a\\ -b \\ 0\end{matrix}\right) = \left(\begin{matrix}bc\\ ac \\ ab\end{matrix}\right)$$
$$
\begin{cases}
\left(\begin{matrix}a\\ -b \\ 0\end{matrix}\right) passing through (0,0,c)\\
\left(\begin{matrix}0\\ -b \\ c\end{matrix}\right) passing through (a,0,0)\\
\left(\begin{matrix}a\\ 0 \\ -c\end{matrix}\right) passing through (0,b,0)
\end{cases}
$$
Thus that point is a solution to all of the following set of equations (including the equation for the triangle plane itself, which passes through (a,0,0)):
$$
\begin{cases}
-by+cz &= 0\\
ax-cz &= 0\\
ax-by &= 0\\
bcx+acy+abz &= abc
\end{cases}
$$
Which you may wish to solve into the point that is the center of the second circumscribed circle, but in this case we can see that the radius line also passes through (0,0,0), and thus the second line through the sphere's center has the following vector specification:
$$\left(\begin{matrix}x\\ y \\ z\end{matrix}\right)= \mu\left(\begin{matrix}bc\\ ac \\ ab\end{matrix}\right)$$
Intersecting the two lines gives $\mu=\frac{a^2+b^2+c^2}{2abc}$, and hence the center of the sphere is
$$\vec{M}= \frac{a^2+b^2+c^2}{2abc}\left(\begin{matrix}bc\\ ac \\ ab\end{matrix}\right)$$ and the length of that vector is the radius (as the sphere passes through the origin), hence
$$r=\frac{a^2+b^2+c^2}{2abc} \cdot \sqrt{b^2c^2+a^2c^2+a^2b^2}$$
EDIT
The points of intersection of the perpendicular lines for each triangle actually do not have to be derived, not even for the first triangle above. It is simpler to state that the planes perpendicular to the sides intersect into the sphere radius. One triangle side is shared among the two triangles, so instead of the above, we can solve the following set of equations:
$$
\begin{cases}
-ax+by+cz &= \frac{1}{2}(a^2+b^2+c^2)\\
ax-by+cz &= \frac{1}{2}(a^2+b^2+c^2)\\
ax-by &= 0\\
-by+cz &= 0\\
ax-cz &= 0\\
\end{cases}
$$
from which it follows that $ax=by=cz$ and $ax=\frac{a^2+b^2+c^2}{2}$ thus $x=\frac{a^2+b^2+c^2}{2a}$, etcetera.
| {
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Why does $F(\sqrt{a+b+2\sqrt{ab}}) = F(\sqrt{a},\sqrt{b})$? Let $F$ be a field of characteristic $\neq 2$. Let $a \neq b \in F$, and $F(\sqrt{a},\sqrt{b})$ is of degree 4 over $F$. I've shown that $F(\sqrt{a}+\sqrt{b}) = F(\sqrt{a},\sqrt{b})$.
Observe that $x=\sqrt{a} + \sqrt{b}$ and $x = \sqrt{a+b+2\sqrt{ab}}$ both satisfy $p(x) = x^4 - 2(a+b)x^2 + (a-b)^2 = 0$. The problem is to conclude that $F(\sqrt{a+b+2\sqrt{ab}}) = F(\sqrt{a},\sqrt{b})$.
I can see that LHS $\cong$ RHS, but cannot see why they are literally the same field. For one thing, it's not clear to me that $\sqrt{a}, \sqrt{b} \in$ LHS.
| As you and @D_S showed, $F(\sqrt{a+b+2\sqrt ab})=F(\sqrt a+\sqrt b)$ since they have the same minimal polynomial, so it suffices to show that $F(\sqrt a+\sqrt b)=F(\sqrt a, \sqrt b)$, which can be shown by showing that $\sqrt a \in F(\sqrt a+\sqrt b)$ (showing that $\sqrt b \in F(\sqrt a+\sqrt b)$ is very similar and showing that $\sqrt a+\sqrt b \in F(\sqrt a, \sqrt b)$ is trivial).
We start with $\sqrt a+\sqrt b$. The reciprocal of this element is $\frac{\sqrt a-\sqrt b}{a^2-b^2}$. We can multiply this by the scalar $a^2-b^2$ to get $\sqrt a-\sqrt b$ and then add this with our original $\sqrt a+\sqrt b$ to get $2\sqrt a$. Finally, divide by the scalar of $2$ to get $\sqrt a$.
| {
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$a$,$b$ and $c$ are roots of the equation $x^3-x^2-x-1=0$ The roots of the equation $x^3-x^2-x-1=0$ are $a$,$b$ and $c$.
if $n \gt 21 $ and $n \in \mathbb{N}$ The find the possible values of $$E=\frac{a^n-b^n}{a-b}+\frac{b^n-c^n}{b-c}+\frac{c^n-a^n}{c-a}$$ in $[0 \: 2]$ are?
Since $x^3=x^2+x+1$ from the graphs of $x^3$ and $x^2+x+1$ its clear that they meet at one point. So number of real roots of $x^3-x^2-x-1=0$ is one which is positive and remaining two are complex conjugates. So let the roots be $a$ and $b=re^{i\theta}$ and $c=re^{-i\theta}$ But product of the roots is $$ar^2=1$$ hence $$r=\frac{1}{\sqrt{a}}$$
Now assuming $b=z$ and $c=\bar{z}$ then
$$E=\frac{a^n-z^n}{a-z}+\frac{z^n-\bar{z}^n}{z-\bar{z}}+\frac{\bar{z}^n-a^n}{\bar{z}-a}$$ so
$$E=2 \Re\left(\frac{a^n-z^n}{a-z}\right)+r^{n-1}\frac{\sin(n\theta)}{\sin\theta}$$
I am not able to proceed from here
| HINT: If
$E_n=\frac{a^n-b^n}{a-b}+\frac{b^n-c^n}{b-c}+\frac{c^n-a^n}{c-a}$
then, because of $x^3=x^2+x+1$ you have $x^n=x^{n-1}+x^{n-2}+x^{n-3}$ and:
$E_n=E_{n-1} + E_{n-2} + E_{n-3}$.
It is easy to find out the first values of this recurrence.
| {
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How to prove $1+\cos \left(\frac{2\pi}{7}\right)-4\cos^2 \left(\frac{2\pi}{7}\right)-8\cos^3 \left(\frac{2\pi}{7}\right) \neq 0$ The task is to prove the following non-equality by hand:
$$1+\cos \left(\frac{2\pi}{7}\right)-4\cos^2 \left(\frac{2\pi}{7}\right)-8\cos^3 \left(\frac{2\pi}{7}\right) \neq 0$$
Wolframalpha shows this, but I can't prove it.
http://www.wolframalpha.com/input/?i=1%2Bcos(2pi%2F7)-4cos%5E2(2pi%2F7)-8cos%5E3(2pi%2F7)%3D0
| We consider $$f(x) = x - 4x^2 - 8x^3 $$
Notice that for $x \ge \frac{1}{2}, \>\> f(x) < -1$ siince $f(\frac{1}{2}) = -\frac{3}{2} $ and $f'(\frac{1}{2}) < 0$ and $f''(x) < 0$ for all $x > -1$
Now it's only important to show that $\cos \frac{2 \pi}{7} \ge \frac{1}{2}$
It should be relatively easy to show $\cos \frac{2 \pi}{7} \ge \cos \frac{\pi}{3}$
| {
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Riemann Sum to show convergence help? I have the function $f(x)=\sin\left(\dfrac{\pi x}{2}\right)$ on the partition of $[0,1]$ given by $$P_{n}: 0 < \frac{1}{n} < \frac{2}{n} < ... < \frac{n-1}{n} < 1$$
I have shown that $$L(f,P_{n})= \sum^{n-1}_{j=0} \frac{1}{n}\sin\left(\frac{j \pi}{2n}\right)$$
However i don't know how to use this to show:$$\frac{1}{n}\left(\sin\left(\frac{\pi}{2n}\right)+\sin\left(\frac{2 \pi}{2n}\right)+...+\sin\left(\frac{(n-1)\pi}{2n}\right) \rightarrow \sin\left(\frac{2}{\pi}\right)\right)$$
Any help would be appreciated.
| It appears you need to compute the limit of the lower Riemann sum using some analytical tools -- as opposed to declaring simply that it converges to the integral.
The following identity is useful
$$\sum_{j=1}^n \sin (jx) = \frac{\sin\left( \frac{nx}{2}\right)\sin\left( \frac{(n+1)x}{2}\right)}{\sin\left( \frac{x}{2}\right)},$$
and is proved easily by taking the imaginary part of the geometric sum $\sum_{j=1}^n (e^{ix})^j.$
Consequently,
$$\begin{align}\frac{1}{n} \sum_{j=0}^{n-1}\sin\left(\frac{j \pi}{2n}\right) &= \frac{1}{n} \sum_{j=1}^{n-1}\sin\left(\frac{j \pi}{2n}\right) \\ &= \frac{\sin\left( \frac{(n-1)\pi}{4n}\right)\sin\left( \frac{n\pi}{4n}\right)}{n\sin\left( \frac{\pi}{4n}\right)} \\ &= \frac{4}{\pi} \left(\frac{\sin\left( \frac{\pi}{4n}\right) }{\frac{\pi}{4n}}\right)^{-1}\sin\left(\frac{\pi}{4}(1-1/n)\right)\sin\left(\frac{\pi}{4}\right)\end{align},$$
and
$$\lim_{n \to \infty}\frac{1}{n} \sum_{j=0}^{n-1}\sin\left(\frac{j \pi}{2n}\right) = \frac{4}{\pi}\lim_{n \to \infty}\left(\frac{\sin\left( \frac{\pi}{4n}\right) }{\frac{\pi}{4n}}\right)^{-1} \lim_{n \to \infty}\sin\left(\frac{\pi}{4}(1-1/n)\right) \sin\left(\frac{\pi}{4}\right) \\ = \frac{4}{\pi} \cdot 1 \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \\ = \frac{2}{\pi}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1720706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving the following trigonometric equation: $\sin x + \cos x = \frac{1}{3} $ I have to solve the following equation:
$$\sin x + \cos x = \dfrac{1}{3} $$
I use the following substitution:
$$\sin^2 x + \cos^2 x = 1 \longrightarrow \sin x = \sqrt{1-\cos^2 x}$$
And by operating, I obtain:
$$ \sqrt{(1-\cos^2 x)} = \dfrac{1}{3}-\cos x$$
$$ 1 - \cos^2 x = \dfrac{1}{9} + \cos^2 x - \dfrac{2}{3}\cos x$$
$$ -2\cos^2 x + 2/3\cos x +\dfrac{8}{9}=0$$
$$ \boxed{\cos^2 x -\dfrac{1}{3}\cos x -\dfrac{4}{9} = 0}$$
Can I just substitute $\cos x$ by $z$ and solve as if it was a simple second degree equation and then obtain $x$ by taking the inverse cosine? I have tried to do this but I cannot get the right result. If I do this, I obtain the following results:
$$ z_1 = -0.520517 \longrightarrow x_1 = 121.4º\\
z_2= 0.8538509 \longrightarrow x_2 = 31.37º$$
I obtain $x$ from $z$ by taking the inverse cosine.
The correct result should be around 329º which corresponds to 4.165 rad. My question is if what I am doing is wrong because I have tried multiple times and I obtain the same result (or in the worst case, I have done the same mistake multiple times).
| $$(\sin(x) + \cos(x))^2 = 1/9$$
$$\sin^2(x) + \cos^2(x) + 2\sin(x)\cos(x) = 1/9$$
$$1 + \sin(2x) = 1/9$$
$$\sin(2x) = -8/9$$
$$x = \arcsin(-8/9) / 2$$
$$ x = -0.547457... $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 5
} |
Show that the group is abelian Let $M$ be a field and $G$ the multiplicative group of matrices of the form $\begin{pmatrix}
1 & x & y \\
0 & 1 & z \\
0 & 0 & 1
\end{pmatrix}$ with $x,y,z\in M$.
I have shown that all the elements of the center $Z(G)$ are the matrices of the form $\begin{pmatrix}
1 & 0 & \tilde{y} \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}$.
How could I show that $G/Z(G)$ is abelian?
We have that $$G/Z(G)=\{gZ(G)\mid g\in G\}$$
Do we have to take $A=g_1Z(G)$ and $B=g_2Z(G)$ and show that $AB=BA$ ?
Or do we have to take the same $g$ just an other element of the center?
| The multiplication is
$$
\begin{pmatrix}
1 & a & b \\
0 & 1 & c \\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
1 & x & y \\
0 & 1 & z \\
0 & 0 & 1
\end{pmatrix}
=
\begin{pmatrix}
1 & x+a & y+az+b \\
0 & 1 & z+c \\
0 & 0 & 1
\end{pmatrix}
$$
Similarly
$$
\begin{pmatrix}
1 & x & y \\
0 & 1 & z \\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
1 & a & b \\
0 & 1 & c \\
0 & 0 & 1
\end{pmatrix}
=
\begin{pmatrix}
1 & a+x & b+xc+y \\
0 & 1 & c+z \\
0 & 0 & 1
\end{pmatrix}
$$
Denote by $\mu(a,b,c)$ the matrix
$$
\begin{pmatrix}
1 & a & b \\
0 & 1 & c \\
0 & 0 & 1
\end{pmatrix}
$$
Saying that $\mu(a,b,c)$ is in the center is the same as saying that, for every $x,y,z$,
$$
y+az+b=b+xc+y
$$
that is, $az=xc$. Taking $z=1$ and $x=0$ we have $a=0$; with $z=0$ and $x=1$ we get $c=0$. It follows easily that the center consists of the matrices of the form $\mu(0,b,0)$.
Now consider the map $f\colon G\to M^2$ (considering $M^2$ a group with respect to componentwise addition) defined by
$$
f(\mu(x,y,z))=(x,z)
$$
The computation above shows that
$$
f(\mu(a,b,c)\mu(x,y,z))=(a+x,c+z)=f(\mu(a,b,c))+f(\mu(x,y,z))
$$
so $f$ is a homomorphism, obviously surjective. It's also clear that $\ker f=Z(G)$. By the isomorphism theorem,
$$
G/Z(G)=G/\ker f\cong M^2
$$
is abelian.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find last two digits of $33^{100} $ Find last two digits of $33^{100}$.
My try:
So I have to compute $33^{100}\mod 100$
Now by Euler's Function $a^{\phi(n)}\equiv 1\pmod{n}$
So we have $33^{40}\equiv 1 \pmod{100}$
Again by Carmichael Function : $33^{20}\equiv 1 \pmod{100}$
Since $100=2\cdot40+20$ so we have $33^{100}=1\pmod{100}$
So last two digits are $01$
Is it right?
| $33^{100} \equiv (33^2)^{50} \equiv 1^{50} \equiv 1 \bmod 4$, because $\varphi(4)=2$.
And $33^{100} \equiv (33^{20})^5 \equiv 1^5\equiv 1 \bmod 25$, because $\varphi(25)=20$.
So $33^{100} \equiv 1 \bmod 100$.
Similarly, $a^{100} \equiv 1 \bmod 100$ for all $a$ coprime to $100$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Closed formula for finite product series I need to solve the recurrence:
$$
\begin{align*}
T(n) &= kT\left(\frac{n}{2}\right) + (k - 2)n^3 \\
&\textit{where}\; k \in \mathbb{Z}: k \geq 2 \\
&= k(kT\left(\frac{n}{4} + \frac{(k - 2)n^3}{2}\right) + (k - 2)n^3 \\
&= k^2T\left(\frac{n}{4}\right) + \frac{(k + 2)(k - 2)n^3}{2} \\
&= k^2\left(kT\left(\frac{n}{8}\right) + \frac{(k + 2)(k - 2)n^3}{8}\right) + \frac{(k + 2)(k - 2)n^3}{2} \\
&= k^3T\left(\frac{n}{8}\right) + \frac{(k^2 + 4)(k + 2)(k - 2)n^3}{8} \\
&\dots \\
&= k^iT\left(\frac{n}{2^i}\right) + \frac{(k - 2)n^3}{2^i}\prod_{j=1}^i(k^j+2^j)
\end{align*}
$$
But I don't know how can I develop this further. I know that the recursion will end at $i = \lg n$, but I don't know what to do with the product term.
| Here is a quick way to bound many kinds of functions defined by a recurrence if one is interested in the function's behaviour for large $n$.
Let $\lg x=\log_2 x$.
Suppose we are given a recurrence of the form $f(n)=cf(n/2)+p(n), f(1)=1$, where $p(n)$ is a function depending on $n$ and $c$ is a constant.
If $p(n)\ge 0$ for all $n\ge 1$, then
$f(n)\ge cf(n/2)$,
so $f(n)\ge c^{\lg n}= n^{\lg c}$.
Further, if $p(n)\ge p(n/2)$ for every $n\ge 2$, then
$f(n)\le c^{\lg n} + p(n)(c^{\lg n}-1)/(c-1)\le n^{\lg c}(1+p(n)/(c-1))$.
So $f(n)=\Omega(n^{\lg c})$ and $f(n)=O(n^{\lg c}p(n))$,
and it is easy to see that these expressions also hold for other constant values of $f(1)>0$.
The specific recurrence has $p(n)=(k-2)n^3$ and $c=k\ge 2$, which satisfy the conditions assumed for the inequalities to hold.
In particular, $f(n)=O(n^{3+\lg k})$ and $f(n)=\Omega(n^{\lg k})$ which is a polynomial in $n$ for fixed $k$.
(As stated the sequence of equations in the question has errors, and the final expression is $\Omega(n^{\lg n})$ which is not polynomial in $n$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Evaluate the integral using the theory of residues: $\int_0^{2\pi} \frac{(\cos \theta)^2 d \theta}{3-\sin \theta}$ $$\int_0^{2\pi} \frac{(\cos \theta)^2 d \theta}{3-\sin \theta}$$
I''m having trouble simpliyfing this into a form that will allow me to use Residue Theorem. I got it to the point where the integrand looks like this:
$$=\frac{i(e^{2i\theta}+1)^2}{-2e^{i\theta}(e^{2i\theta}-6ie^{i\theta}-1)}$$
Then I found that the roots of the denominator are
$$\theta=-i\ln(3i+2i\sqrt{2})+i\arg(3i+2i\sqrt{2})+2i\pi k$$
$$\theta=-i\ln(3i-2i\sqrt{2})+i\arg(3i-2i\sqrt{2})+2i\pi k$$
But this doesn't seem likely to be the correct way of solving this. Any tips?
Update:
Ok so I tried something else:
Let $z=e^{i\theta}$. thus,
$$\oint_C (\frac{z^2+2+z^{-2}}{4})(\frac{1}{3-(\frac{z-z^{-1}}{2i})})\frac{dz}{iz}$$
$$=\oint_C (\frac{z^4+2z^2+1}{4z^2})(\frac{2}{6i-z^2+1})dz$$
$$=\oint_C (\frac{z^4+2z^2+1}{4z^2})(\frac{2}{(z+\sqrt{1+6i})(z-\sqrt{1+6i})})$$
Is this on the right track?? I guess the next step would just be to apply the residue theory...
| On the unit circle, Using the substitution $z=e^{i\theta}$, we get
$$
\begin{align}
\sin(\theta)&=\frac1{2i}\left(z-\frac1z\right)\\
\cos(\theta)&=\frac12\left(z+\frac1z\right)\\
\mathrm{d}\theta&=\frac{\mathrm{d}z}{iz}
\end{align}
$$
Therefore, on a counter-clockwise contour along the unit circle,
$$
\begin{align}
\int_0^{2\pi}\frac{\cos^2(\theta)\,\mathrm{d}\theta}{3-\sin(\theta)}
&=\frac12\oint\frac{(z^2+1)^2}{6iz-(z^2-1)}\frac{\mathrm{d}z}{z^2}\\
&=\frac12\oint\frac{1+2z^2+z^4}{1+6iz-z^2}\frac{\mathrm{d}z}{z^2}\\
&=\frac12\oint\left(-1+\frac1{z^2}\color{#C00000}{-\frac{6i}z}+4\sqrt2\,i\left(\color{#C00000}{\frac1{z-\left(3-2\sqrt2\right)i}}-\frac1{z-\left(3+2\sqrt2\right)i}\right)\right)\mathrm{d}z\\[3pt]
&=2\pi i\left(-3i+2\sqrt2\,i\right)\\[9pt]
&=\left(6-4\sqrt2\right)\pi\\[3pt]
&=\frac{2\pi}{3+2\sqrt2}
\end{align}
$$
Since the red terms have non-zero residue inside the unit circle.
| {
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"url": "https://math.stackexchange.com/questions/1725134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Evaluate: $\int x\sqrt{\frac{a-x}{a+x}}\hspace{1mm}dx$
Evaluate: $$\int x\sqrt{\dfrac{a-x}{a+x}}\hspace{1mm}\mathrm{d}x$$
I don't know where to start. Hints/suggestions will be appreciated
| $$
\frac{a-x}{a+x} = u^2 \rightarrow x = a\frac{1-u^2}{1+u^2} \rightarrow dx = \frac{-4au}{(1+u^2)^2}du
$$
$$
\int x\sqrt{\dfrac{a-x}{a+x}}\hspace{1mm}\mathrm{d}x = \int a\frac{1-u^2}{1+u^2} u\hspace{1mm}\frac{-4au}{(1+u^2)^2}du = -4a^2 \int \frac{u^2(1-u^2)}{(1+u^2)^3}du =
$$
$$
-4a^2 \int \frac{u^2(1-u^2)}{(1+u^2)^3}du = -4a^2[\int \frac{du}{1+u^2} + \int \frac{3du}{(1+u^2)^2} + \int \frac{-2du}{(1+u^2)^3}]
$$
For each integral apply $u=\tan\theta$
$$
\int \frac{du}{1+u^2} = \arctan(u)
$$
$$
\int \frac{3du}{(1+u^2)^2} = 1/2 (u/(u^2+1)+\arctan(u))
$$
$$
\int \frac{-2du}{(1+u^2)^3} = 1/8 ((u (3 u^2+5))/(u^2+1)^2+3 \arctan(u))
$$
Now substitute $u = \sqrt{\frac{a-x}{a+x}} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the number of words each consisting of $3$ consonants and $3$ vowels that can be formed from the letters of the word CIRCUMFERENCE. Find the number of words each consisting of $3$ consonants and $3$ vowels that can be formed from the letters of the word CIRCUMFERENCE.In how many of these $c$'s will be together?
The consonants are $c,r,c,m,f,r,n,c$ and the vowels are $i,u,e,e,e$.
Number of ways of arranging 3 vowels$=\frac{3!}{2!}+1+3!=10$
I am stuck in counting the number of ways of arranging 3 consonants because $c$ is coming 3 times and $r$ is coming 2 times.
| Arrangements of the vowels:
You correctly calculated that the number of distinguishable ways to arrange three E's is $1$ and that the number of ways to arrange three different vowels is $3!$. However, you calculated the number of arrangements with two E's incorrectly.
There are $\binom{3}{2}$ ways of selecting the positions of the two $E$'s and two ways of selecting the other vowel. Hence, the number of arrangements with two E's is
$$\binom{3}{2}\binom{2}{1}$$
Hence, the number of ways of arranging the three vowels in the three positions selected for the vowels is
$$3! + \binom{3}{2}\binom{2}{1} + 1$$
Arrangements of the consonants:
Case 1: Three different consonants.
We can choose three of the five consonants, then arrange them in order in
$$\binom{5}{3} \cdot 3!$$
ways.
Case 2: Two different consonants.
We choose which of the two letters C or R is repeated, choose two of the three positions in which to place the repeated letter, then choose one of the other four letters to fill the open slot, which yields
$$\binom{2}{1}\binom{3}{2}\binom{4}{1}$$
possible arrangements.
Case 3: One consonant is used.
There is only one way to fill the three positions with a C.
Hence, the number of ways of arranging the consonants within their three selected positions is
$$\binom{5}{3}\cdot 3! + \binom{2}{1}\binom{3}{2}\binom{4}{1} + 1$$
Number of words consisting of three consonants and three vowels from the letters of the word CIRCUMFERENCE
We choose three of the six positions for the vowels, arrange the vowels in the selected locations, then arrange the consonants in the three remaining positions, which yields
$$\binom{6}{3}\left[3! + \binom{3}{2}\binom{2}{1} + 1\right]\left[\binom{5}{3}\cdot 3! + \binom{2}{1}\binom{3}{2}\binom{4}{1} + 1\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1729142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
An integral calculus I know that $\displaystyle \frac{1}{2i\pi}\int_0^{2\pi}\frac{ib \cos t - a \sin t}{a\cos t+ib\sin t} \, dt =1.$
I'm trying to use this to find the value of $\displaystyle \int_0^{2\pi}\frac{dt}{a^2\cos^2t+b^2\sin^2t}.$
Is it possible? Any hint would be much appreciated.
| Note that we can write
$$\begin{align}
\frac{ib\cos(t)-a\sin(t)}{a\cos(t)+ib\sin(t)}&=\frac{iab }{a^2\cos^2(t)+b^2\sin^2(t)}+\frac{\frac12(b^2-a^2)\sin(2t) }{a^2\cos^2(t)+b^2\sin^2(t)}\end{align} \tag 1$$
Therefore, integrating $(1)$ reveals
$$\frac{1}{2\pi i}\int_0^{2\pi}\frac{ib\cos(t)-a\sin(t)}{a\cos(t)+ib\sin(t)}\,dt=\frac{ab}{2\pi}\int_0^{2\pi}\frac{1 }{a^2\cos^2(t)+b^2\sin^2(t)}\,dt \tag 2$$
since
$$\begin{align}
\int_0^{2\pi}\frac{\frac12(b^2-a^2)\sin(2t) }{a^2\cos^2(t)+b^2\sin^2(t)}\,dt&=\int_{-\pi}^{\pi}\frac{\frac12(b^2-a^2)\sin(2t) }{a^2\cos^2(t)+b^2\sin^2(t)}\,dt \tag 3\\\\
&=0
\end{align}$$
due to the odd symmetry and $2\pi$-periodicity of the integrand in $(3)$!
Therefore, rearranging $(2)$ and using $\frac{1}{2\pi i}\int_0^{2\pi}\frac{ib\cos(t)-a\sin(t)}{a\cos(t)+ib\sin(t)}\,dt=1$ yields
$$\bbox[5px,border:2px solid #C0A000]{\int_0^{2\pi}\frac{1 }{a^2\cos^2(t)+b^2\sin^2(t)}\,dt=\frac{2\pi}{ab}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1729259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
definite integrals with trigonometric substitution How do you integrate $$\int_{-7}^{-5} \frac{2}{x^4\sqrt{x^2-25}}dx,$$ I seem to be getting my answer off by -1 all the time.
| Note that since the integrand is an even function, we can write
$$\int_{-7}^{-5}\frac{2}{x^4\sqrt{x^2-25}}\,dx=\int_5^7\frac{2}{x^4\sqrt{x^2-25}}\,dx$$
Now, enforcing the substitution $x\to 5\sec(x)$ reveals
$$\begin{align}
\int_5^7\frac{2}{x^4\sqrt{x^2-25}}\,dx&=\int_{0}^{\arccos(5/7)}\frac{2}{3125\sec^4(x)\tan(x)}\,5\sec(x)\tan(x)\,dx\\\\
&=\frac2{625}\int_0^{\arccos(5/7)}\cos^3(x)\,dx\\\\
&=\frac2{625}\int_0^{\arccos(5/7)}\left(\frac34 \cos(x)+\frac14\cos(3x)\right)\,dx\\\\
&=\frac3{1250}\sin(\arccos(5/7))+\frac{1}{3750}\sin(3\arccos(5/7))\\\\
&=\frac3{1250}\frac{2\sqrt{6}}{7}+\frac{1}{3750}\frac{102\sqrt{6}}{343}\\\\
&=\frac{164\sqrt{6}}{214375}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1731218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove with some AM-GM inequality? I have proved the following inequality:
Let $a,b,c>0$
$$\dfrac{(a+\sqrt{ab}+\sqrt[3]{abc})}{3}\le \sqrt[3]{a\cdot\dfrac{a+b}{2}\cdot\dfrac{a+b+c}{3}}$$
My solution is:$$a\dfrac{a+b}{2}\cdot\dfrac{a+b+c}{3} =\dfrac{1}{3^3}(a+a+a)(a+\dfrac{a+b}{2}+b)(a+b+c)\ge\dfrac{1}{3^3}(a+a+a)(a+\sqrt{ab}+b)(a+b+c)\ge \dfrac{1}{3}(a+\sqrt{ab}+\sqrt[3]{abc})^3$$
But I met the following, harder inequality:
$$3(a+\sqrt{ab}+\sqrt[3]{abc})\le \left(7+\dfrac{4\sqrt{ab}}{a+b}\right)\sqrt[3]{a\cdot\dfrac{a+b}{2}\cdot\dfrac{a+b+c}{3}}$$
How to prove this inequality? Thanks
| This is a proof by computer.
Substitute by $a=x^3, b=y^3, c=z^3$, the original inequality is equivalent to
show a polynomial in $x,y,z$ of degree $30$ is non-negative. A Cylindrical Decomposition shows this is true.
The code is the following
ieq1 = 3 (a + Sqrt[a b] + (a b c)^(1/3)) <= (7 + (4 Sqrt[a b])/(
a + b)) (a (a + b)/2 (a + b + c)/3)^(1/3);
assume = a >= 0 && b >= 0 && c >= 0;
f1[a_, b_, c_] =
ieq1[[2]]^3 - ieq1[[1]]^3 // Together // Numerator //
Simplify[#, assume] &;
assume1 = x >= 0 && y >= 0 && z >= 0;
f2[x_, y_, z_] = f1[x^6, y^6, z^6] // Simplify[#, assume1] &;
ieq2 = ForAll[{x, y, z}, assume1, f2[x, y, z] >= 0];
CylindricalDecomposition[ieq2, {}]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1733055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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Is there an elegant way to solve $\int \frac{(\sin^2(x)\cdot \cos(x))}{\sin(x)+\cos(x)}dx$? The integral is:
$$\int \frac{(\sin^2(x)\cdot \cos(x))}{\sin(x)+\cos(x)}dx$$
I used weierstraß substitution
$$t:=\tan(\frac{x}{2})$$
$$\sin(x)=\frac{2t}{1+t^2}$$
$$\cos(x)=\frac{1-t^2}{1+t^2}$$
$$dx=\frac{2}{1+t^2}dt$$
Got this:
$$\int \frac{8t^4-8t^2}{t^8−2t^7+2t^6−6t^5−6t^3−2t^2−2t−1}dt$$
and with partial fraction expansion the final answer is:
$$\frac{1}{4}[\ln(\sin(x)+\cos(x))-\cos(x)*(\sin(x)+\cos(x))]+C$$
It is a long way and I am very convinced, there is a shorter way, maybe you know one? Thanks
| \begin{align}
\int\frac{\sin^2 x\cos x}{\sin x+\cos x}\,\mathrm dx
&=
\int\frac{\sin^2 x\sin\left(\frac\pi2-x\right)}{\sin x+\sin\left(\frac\pi2-x\right)}\,\mathrm dx
\\
&=
\int\frac{\sin^2\left(\frac\pi4+u\right)\sin\left(\frac\pi4-u\right)}{\sin \left(\frac\pi4+u\right)+\sin\left(\frac\pi4-u\right)}\,\mathrm d\left(\frac\pi4+u\right)
\\
&=
\int\frac{\left(\sin\frac\pi4\cos u+\cos\frac\pi4\sin u\right)^2\left(\sin\frac\pi4\cos u-\cos\frac\pi4\sin u\right)}{\left(\sin\frac\pi4\cos u+\cos\frac\pi4\sin u\right)+\left(\sin\frac\pi4\cos u-\cos\frac\pi4\sin u\right)}\,\mathrm d\left(\frac\pi4+u\right)
\\
&=
\frac14\int\frac{\left(\cos u+\sin u\right)^2\left(\cos u-\sin u\right)}{\cos u}\,\mathrm d\left(\frac\pi4+u\right)
\\
&=
\frac14\int\frac{\left(\cos u+\sin u\right)\left(\cos^2 u-\sin^2 u\right)}{\cos u}\,\mathrm d\left(\frac\pi4+u\right)
\\
&=
\frac14\int\left(\cos2u+\sin2u-\tan u\right)\,\mathrm d\left(\frac\pi4+u\right)
\\
&=
\frac14\left(\frac12\sin\left(2x-\frac\pi2\right)-\frac12\cos\left(2x-\frac\pi2\right)+\log\cos\left(x-\frac\pi4\right)\right)+\textsf{const.}
\\
&=
\frac14\left(\log\cos\left(x-\frac\pi4\right)-\frac12\left(\sin2x+\cos2x\right)\right)+\textsf{const.}
\\
&=
\frac14\left(\log\cos\left(x-\frac\pi4\right)-\frac1{\sqrt2}\cos\left(2x-\frac\pi4\right)\right)+\textsf{const.}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1733789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Function that maps the "pureness" of a rational number? By pureness I mean a number that shows how much the numerator and denominator are small.
E.g. $\frac{1}{1}$ is purest, $\frac{1}{2}$ is less pure (but the same as $\frac{2}{1}$), $\frac{2}{3}$ is less pure than the previous examples, $\frac{53}{41}$ is worse, .... $\pi$ isn't pure at all (as well as e...).
| You can take,
$$\frac{a}{b}\mapsto \frac{a+b}{\gcd(a,b)},$$
Note that this is independent of the choice of representative since $\gcd(na,nb)=n\gcd(a,b)$ for non-negative integers $n$.
For your examples,
$$\frac{1}{1}\mapsto 2,\quad \frac{1}{2}\mapsto 3,\quad\frac{2}{3}\mapsto 5,\quad\frac{53}{41}\mapsto 94.$$
Another possibility is
$$\frac{a}{b}\mapsto \frac{ab}{(\gcd(a,b))^2},$$
which yields
$$\frac{1}{1}\mapsto 1,\quad\frac{1}{2}\mapsto 2,\quad\frac{2}{3}\mapsto 6,\quad\frac{53}{41}\mapsto 2173.$$
Both these "pureness" functions have the property that $a/b$ is as pure as $b/a$, as we would expect.
The following table shows how the first choice partitions the positive rationals into "pureness classes". Each row corresponds to rationals of the same pureness.
$$
\begin{align}
& \frac{1}{1} \\
& \frac{1}{2}\quad\frac{2}{1} \\
& \frac{1}{3}\quad\frac{3}{1} \\
& \frac{1}{4}\quad\frac{2}{3}\quad\frac{3}{2}\quad\frac{4}{1} \\
& \frac{1}{5}\quad\frac{5}{1} \\
& \frac{1}{6}\quad\frac{2}{5}\quad\frac{3}{4}\quad\frac{4}{3}\quad\frac{2}{5}\quad\frac{6}{1} \\
& \frac{1}{7}\quad\frac{3}{5}\quad\frac{5}{3}\quad\frac{7}{1} \\
& \frac{1}{8}\quad\frac{2}{7}\quad\frac{4}{5}\quad\frac{5}{4}\quad\frac{7}{2}\quad\frac{8}{1} \\
& \frac{1}{9}\quad\frac{3}{7}\quad\frac{7}{3}\quad\frac{9}{1} \\
& \frac{1}{10}\quad\frac{2}{9}\quad\frac{3}{8}\quad\frac{4}{7}\quad\frac{5}{6}\quad\frac{6}{5}\quad\frac{7}{4}\quad\frac{8}{3}\quad\frac{9}{2}\quad\frac{10}{1}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1734697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 9,
"answer_id": 4
} |
Area between two trigonometric curves I need to find the area between two curves:
$$\begin{cases} x=\sqrt { 2 } \cos { t } \\ y=4\sqrt { 2 } \sin { t } \end{cases}\\ y=4\quad (y\ge 4)$$
I came up with:
$$\frac { 1 }{ 4 } \left( 8\pi n +\pi \right) \le t\le \frac { 1 }{ 4 } \left( 8\pi n + 3\pi \right) $$
$$\int _{ \frac { 1 }{ 4 } (8\pi n+\pi ) }^{ \frac { 1 }{ 4 } (8\pi n+3\pi ) }{ 4\sqrt { 2 } \sin { (t) } } -\sqrt { 2 } \cos { (t) } \quad dx =2\left( \sin { (2\pi n) } +4\cos { (2\pi n) } \right)$$
So the area is a function of $n$, though I was supposed to get a finite solution. What am I doing wrong?
| As I said in my comment, you need to restrict your angles, both in general, to have only one revolution around the origin, and when you want to meet your additional restriction on $y$. For the first case, it's obvious that $t \in [0, 2\pi)$. For the second one, solve an easy equation
$$
y = 4 \implies 4\sqrt 2 \sin t = 4 \implies \sin t = \frac 1{\sqrt 2} \implies t_1 = \frac \pi 4,\ t_2 = \frac {3\pi}4
$$
If you visualize the analysis above, you get
You need to find the area of the shape with red dome and black straight base,
Finding the area of the red curve and $y = 0$ is as easy as
$$
A_f = \int_{\frac {3\pi} 4}^{\frac \pi 4} y(t)\ x'(t)\ dt = 8 \int_{\frac \pi 4}^{\frac {3\pi}4} \sin^2 t\ dt = 4 \left . \left( t - \frac {\cos 2t}2\right) \right |_{\frac \pi 4}^{\frac {3\pi}4} = 2(2 + \pi)
$$
since you know your angles.
And the area you need is the difference between the area above and rectangle with dashed sides, solid black top and bottom side on $x$ axis, which is
$$
A_r = 2 \cdot 4 = 8
$$
and finally, $A = A_f - A_r = 2\pi - 4$
PS
In integration, I used the positive direction of $x$ to put upper and lower bounds for $t$, but then because of the negative sign that comes from $x'(t)$ I switched them again to get increasing order of angles.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1742810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate the value of the following integral: $ \int^{\frac{\pi}{2}}_0 \frac{\sin^n(x)}{\sin^n(x) + \cos^n(x)} $
The question is show that $$ \int_0^{a} f(x) dx = \int_0^{a} f(a-x) dx $$
Hence or otherwise, calculate the value of the following integral
$$ \int^{\frac{\pi}{2}}_0 \frac{\sin^n(x)}{\sin^n(x) + \cos^n(x)} $$
What I have done for the first part
$$ \int_0^{a} f(x) dx = \int_0^{a} f(a-x) dx $$
$$ \left[F(x) \right]^a_0 dx = \left[-F(a-x) \right]^a_0 $$
$$ F(a) - F(0) = \left[-F(a-a) - -F(a-0) \right] $$
$$ F(a) - F(0) =F(a) - F(0) $$
$$ LHS = RHS $$
Now I am stuck applying this to the integral. I have attempted this:
$$ \int^{\frac{\pi}{2}}_0 \frac{\sin^n(x)}{\sin^n(x) + \cos^n(x)} $$
Because $ \int_0^{a} f(x) dx = \int_0^{a} f(a-x) dx $ The integral is transformed to
$$ \int^{\frac{\pi}{2}}_0 \frac{\sin^n(\frac{\pi}{2}-x)}{\sin^n(\frac{\pi}{2}-x) + \cos^n(\frac{\pi}{2}-x)} $$
Which then becomes
$$ \int^{\frac{\pi}{2}}_0 \frac{\cos^n(x)}{\cos^n(x) + \sin^n(x)} $$
Now I am stuck...
| So you have shown that:
$\int^{\frac{\pi}{2}}_0 \frac{\sin^n(x)}{\sin^n(x) + \cos^n(x)} dx= \int^{\frac{\pi}{2}}_0 \frac{\cos^n(x)}{\cos^n(x) + \sin^n(x)}dx$
$2\int^{\frac{\pi}{2}}_0 \frac{\sin^n(x)}{\sin^n(x) + \cos^n(x)}dx =$$\int^{\frac{\pi}{2}}_0 \frac{\sin^n(x)}{\sin^n(x) + \cos^n(x)}dx + \int^{\frac{\pi}{2}}_0 \frac{\cos^n(x)}{\cos^n(x) + \sin^n(x)}dx\\
\int^{\frac{\pi}{2}}_0 \frac{\sin^n(x)+cos^n(x)}{\sin^n(x) + \cos^n(x)}dx\\\pi/2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Characteristic polynomials of matrices related How to show that the characteristic polynomials of matrices A and B are $\lambda^{n-1}(\lambda ^2-\lambda -n)=0$ and $\lambda^{n-1}(\lambda^2+\lambda-n)=0$ respectively by applying elementary row or column operations.
$A=\begin{bmatrix}
1 & 1 & 1 & \cdots & 1 \\
1 & 0 & 0 & \cdots & 0 \\
1 & 0 & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 0 & 0 & \cdots & 0 \\
\end{bmatrix}$
$B=\begin{bmatrix}
-1 & 1 & 1 & \cdots & 1 \\
1 & 0 & 0 & \cdots & 0 \\
1 & 0 & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 0 & 0 & \cdots & 0 \\
\end{bmatrix}$
Where $A$ and $B$ are symmetric matrices of order $n+1$.
| About first matrix.
Lets take the bottom line. As we know :
$$\det{A} = \sum{(-1)^{i+j}\cdot a_{i,j}\cdot M_{i,j}},$$ so we got:
$$S_{n+1} = a_{n+1,n+1}\cdot (-1)^{2n+2}\cdot S_{n} + (-1)^{n+2}a_{n+1,1}S'_{n},$$ where $S'_{n}= (-1)^{1 + n} \lambda^{n-1}(-1)^{n-1}$, because of :
$S'_{n} = \begin{bmatrix}
1 & 1 & \cdots & 1 & 1\\
-\lambda & 0 & \cdots & 0 & 0 \\
0 & -\lambda & \cdots & 0 & 0\\
\vdots & \vdots & \cdots &\ddots & \vdots \\
0 & 0 & \cdots & -\lambda & 0 \\
\end{bmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1744698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Simplify this vector expression I have a question which I know the answer to but am having difficulty showing it. It's about simplifying the following vector equation (I'm aware I've grouped the terms in an arguably strange way):
$ (p^2 + q^2 |\mathbf{P}|^2 + (\mathbf{P} \bullet \mathbf{Q})^2)\mathbf{X} + (\mathbf{Q} \bullet \mathbf{X})(|\mathbf{P}|^2 \mathbf{Q} - (\mathbf{P} \bullet \mathbf{Q})\mathbf{P}) + (\mathbf{P} \bullet \mathbf{X})(|\mathbf{Q}|^2 \mathbf{P} - (\mathbf{P} \bullet \mathbf{Q})\mathbf{Q}) + (\mathbf{X} \bullet (\mathbf{P} \times \mathbf{Q}))(\mathbf{P} \times \mathbf{Q}), $
where $p$ and $q$ are scalars and $\mathbf{P} , \mathbf{Q} , \mathbf{X} $ are three dimensional vectors equipped with the standard dot, $\bullet$, and cross, $\times$, products. Also, $|\mathbf{P}|^2 \equiv (\mathbf{P} \bullet \mathbf{P}) $.
Furthermore, we have the condition that $p^2 + |\mathbf{P}|^2 = q^2 + |\mathbf{Q}|^2 = 1 $.
I'm quite sure that this expression should just simplify to $\mathbf{X}$, particularly having run different values of each variable and vector through some code. Maybe the last three terms always linearly combine to be parallel to $\mathbf{X}$?
Any help would be much appreciated!
| Many thanks for the reply. After a few hours straight of trying to tackle this, I finally solved the problem. I'm going to use these two vector identities (which I've pulled from Wikipedia and trust to be correct).
$ (\mathbf{A} \times \mathbf{B}) \bullet (\mathbf{C} \times \mathbf{D}) = (\mathbf{A} \bullet \mathbf{C})(\mathbf{B} \bullet \mathbf{D}) - (\mathbf{B} \bullet \mathbf{C})(\mathbf{A} \bullet \mathbf{D}) \qquad (1) $
$ \mathbf{D} = \frac{\mathbf{D} \bullet (\mathbf{B} \times \mathbf{C})}{[\mathbf{A},\mathbf{B},\mathbf{C}]}\mathbf{A} + \frac{\mathbf{D} \bullet (\mathbf{C} \times \mathbf{A})}{[\mathbf{A},\mathbf{B},\mathbf{C}]}\mathbf{B} + \frac{\mathbf{D} \bullet (\mathbf{A} \times \mathbf{B})}{[\mathbf{A},\mathbf{B},\mathbf{C}]}\mathbf{C} \qquad (2) $
Where $[\mathbf{A},\mathbf{B},\mathbf{C}] = \mathbf{A} \bullet (\mathbf{B} \times \mathbf{C}) $ and $(2)$ describes how to describe a vector $\mathbf{D}$ in terms of three arbitrary vectors $\mathbf{A},\mathbf{B},\mathbf{C}$.
So let's begin!
So first, I rearrange the expression into a more useful form:
$ (p^2 + q^2 |\mathbf{P}|^2 + (\mathbf{P}\bullet\mathbf{Q})^2)\mathbf{X} + ((\mathbf{Q}\bullet\mathbf{X})|\mathbf{P}|^2 - (\mathbf{P}\bullet\mathbf{X})(\mathbf{P}\bullet\mathbf{Q}))\mathbf{Q} + ((\mathbf{P}\bullet\mathbf{X})|\mathbf{Q}|^2 - (\mathbf{Q}\bullet\mathbf{X})(\mathbf{P}\bullet\mathbf{Q}))\mathbf{P} +(\mathbf{X}\bullet(\mathbf{P}\times\mathbf{Q}))\mathbf{P}\times\mathbf{Q} $
Using relation $(1)$, we can rewrite the coefficients of $\mathbf{P}$ and $\mathbf{Q}$ such that we have
$ (p^2 + q^2 |\mathbf{P}|^2 + (\mathbf{P}\bullet\mathbf{Q})^2)\mathbf{X} - ((\mathbf{P}\times\mathbf{Q})\bullet(\mathbf{X}\times\mathbf{P}))\mathbf{Q} + ((\mathbf{P}\times\mathbf{Q})\bullet(\mathbf{X}\times\mathbf{Q}))\mathbf{P} +(\mathbf{X}\bullet(\mathbf{P}\times\mathbf{Q}))\mathbf{P}\times\mathbf{Q} $
Now, we use relation $(2)$ to rewrite $\mathbf{P}\times\mathbf{Q}$ in terms of $\mathbf{X} , \mathbf{P}$ and $\mathbf{Q}$ such that
$ \mathbf{P}\times\mathbf{Q} = \frac{(\mathbf{P}\times\mathbf{Q}) \bullet (\mathbf{P} \times \mathbf{Q})}{\mathbf{X}\bullet(\mathbf{P}\times\mathbf{Q})}\mathbf{X} + \frac{(\mathbf{P}\times\mathbf{Q}) \bullet (\mathbf{Q} \times \mathbf{X})}{\mathbf{X}\bullet(\mathbf{P}\times\mathbf{Q})}\mathbf{P} + \frac{(\mathbf{P}\times\mathbf{Q}) \bullet (\mathbf{X} \times \mathbf{P})}{\mathbf{X}\bullet(\mathbf{P}\times\mathbf{Q})}\mathbf{Q} = \frac{(\mathbf{P}\times\mathbf{Q}) \bullet (\mathbf{P} \times \mathbf{Q})}{\mathbf{X}\bullet(\mathbf{P}\times\mathbf{Q})}\mathbf{X} - \frac{(\mathbf{P}\times\mathbf{Q}) \bullet (\mathbf{X} \times \mathbf{Q})}{\mathbf{X}\bullet(\mathbf{P}\times\mathbf{Q})}\mathbf{P} + \frac{(\mathbf{P}\times\mathbf{Q}) \bullet (\mathbf{X} \times \mathbf{P})}{\mathbf{X}\bullet(\mathbf{P}\times\mathbf{Q})}\mathbf{Q}$
Substituting this back, we find that terms cancel and we're left with
$ (p^2 + q^2 |\mathbf{P}|^2 + (\mathbf{P}\bullet\mathbf{Q})^2 + |\mathbf{P}\times\mathbf{Q}|^2)\mathbf{X} = (p^2 + q^2 |\mathbf{P}|^2 +|\mathbf{P}|^2 |\mathbf{Q}|^2)\mathbf{X} = \mathbf{X} $
Many thanks for the help, really appreciated it; hope you guys don't feel like I've wasted your time... do let me know if you agree with what I've written!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1745224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Testing whether the circumcenter of a cyclic quadrilateral lies inside it For a triangle with sides $a, b, c$ (where $c$ is the biggest side) there is a simple check to see whether it's circumcenter lies inside of it:
$$a^2 + b^2 < c^2$$
Is there such an inequality for a cyclic quadrilateral, given its side lengths $a, b, c, d$ (with longest side $d$)?
Can this be generalized to a cyclic convex $n$-gon?
| The circumcenter of a triangle is inside the triangle if and only if
$$
(a^2+b^2-c^2)(c^2+a^2-b^2)(b^2+c^2-a^2)\gt0\tag{1}
$$
The diagonal with sides $a$ and $b$ on one side and $c$ and $d$ on the other is
$$
e^2=\frac{\frac{a^2+b^2}{ab}+\frac{c^2+d^2}{cd}}{\frac1{ab}+\frac1{cd}}\tag{2}
$$
Then the circumcenter is inside the quadrilateral if and only if
$$
\hspace{-10pt}\small(a^2+b^2-e^2)(e^2+a^2-b^2)(b^2+e^2-a^2)(c^2+d^2-e^2)(e^2+c^2-d^2)(d^2+e^2-c^2)\lt0\tag{3}
$$
or
$$
a^2+b^2=c^2+d^2\tag{4}
$$
Note that $(4)$ implies equality in $(3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1745840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to compute taylor series for $f(x)=\frac{1}{1-x}$ about $a=3$ and $f(x)=\sin{x}$ about $a=\frac{\pi}{4}$? How to compute taylor series for $f(x)=\frac{1}{1-x}$ about $a=3$ and $f(x)=\sin{x}$ about $a=\frac{\pi}{4}$ ?
For the first one, using substitution, let $t=x-3$, then $x=t+3$. Then $f(x)=\frac{1}{1-(t+3)}=\sum^{\infty}_{n=0}(t+3)^n=\sum^{\infty}_{n=0}(x)^n$. Apparently my solution wrong as it returns to the original series.
For the second one, let $t=x-\frac{\pi}{4}$. Then $\sin({t+\frac{\pi}{4}})=\sum^{\infty}_{n=0}\frac{(-1)^n}{(2n+1)!}(t+\frac{\pi}{4})^{2n+1}$. Then I fail to countinue.
Could someone suggest a solution?
| Hint. One may write
$$
f(x):=\frac{1}{1-x}=-\frac12\frac{1}{1+\frac{(x-3)}2}=-\frac12\sum_{n=0}^\infty(-1)^n\frac{(x-3)^n}{2^n}, \quad |x-3|<2,
$$ and one may write, for any real number $x$,
$$
\begin{align}
f(x)&:=\sin x
\\\\&=\frac{\sqrt{2}}2 \left(\sin\left(x-\frac{\pi}4\right)+ \cos\left(x-\frac{\pi}4\right)\right)
\\\\&=\frac{\sqrt{2}}2\left(\sum^{\infty}_{n=0}\frac{(-1)^n}{(2n+1)!}\left(x-\frac{\pi}{4}\right)^{2n+1}+\sum^{\infty}_{n=0}\frac{(-1)^n}{(2n)!}\left(x-\frac{\pi}{4}\right)^{2n}\right).
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to compute $\sum^{\infty}_{n=2}\frac{(4n^2+8n+3)2^n}{n!}$? How to compute $\sum^{\infty}_{n=2}\frac{(4n^2+8n+3)2^n}{n!}$?
I am trying to connect the series to $e^x$
My try: $\sum^{\infty}_{n=2}\frac{(4n^2+8n+3)2^n}{n!}=\sum^{\infty}_{n=2}\frac{(2n+1)(2n+3)2^n}{n!}$
Let $x=\sqrt{2}$, then the series becomes $\sum^{\infty}_{n=2}\frac{(2n+1)(2n+3)x^{2n}}{n!}$.
$\sum^{\infty}_{n=2}\frac{(2n+1)(2n+3)x^{2n}}{n!}=\frac{d}{dx}\sum^{\infty}_{n=2}\frac{(2n+3)x^{2n+1}}{n!}=\frac{d}{dx}\frac{1}{x}\sum^{\infty}_{n=2}\frac{(2n+3)x^{2n+2}}{n!}=\frac{d}{dx}\frac{1}{x}\frac{d}{dx}\sum^{\infty}_{n=2}\frac{x^{2n+3}}{n!}=\frac{d}{dx}\frac{1}{x}\frac{d}{dx}x^3\sum^{\infty}_{n=2}\frac{(x^{2})^n}{n!}=\frac{d}{dx}\frac{1}{x}\frac{d}{dx}x^3e^{x^{2}}$
In the end we subtract $\sum^{1}_{n=0}\frac{(4n^2+8n+3)2^n}{n!}$ since above we assumed $\sum^{\infty}_{n=0}$
| HINT:
Let $f(x)=e^x-(1+x)$. Then, note that
$$f(x)=e^x-(1+x)=\sum_{n=2}^\infty \frac{x^n }{n!}$$
$$xf'(x)=x(e^x-1)=\sum_{n=2}^\infty \frac{n\,x^{n} }{n!}$$
$$x(xf'(x))'=x(x+1)e^x-x=\sum_{n=2}^\infty \frac{n^2\,x^{n} }{n!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How can you prove this by strong induction? The sequence $b_1,b_2,...$ is defined recursively as:\begin{align} b_1&=0;\\ b_2&=1;\\ b_n&=2b_{n-1}-2b_{n-2}-1 \ \text{for} \ n\geq3. \end{align} Prove that this means: $$\forall n\geq1: b_n=(\sqrt{2})^n \sin{\left(\frac{1}{4}\pi n \right)}-1$$
Edit:
I have tried to prove this by strong induction and have verified than $P(1)$ and $P(2)$ is true, where $P(n)$ is the statement $b_n=(\sqrt{2})^n \sin{\left(\frac{1}{4}\pi n \right)}-1$.
edit: I have assumed that $P(k-2)$ and $P(k-1)$ is true for $k\in\mathbb{N}$. Then I have managed to simplify to: $$b_n=(\sqrt{2})^n \left[\sqrt{2}\sin{\left(\frac{1}{4}\pi (n-1) \right)}- \sin{\left(\frac{1}{4}\pi (n-2) \right)}\right]-5$$
I can't simplify any further.
| Comment on the Approach in the Question
The inductive step needs to show
$$
\begin{align}
2b_{n-1}-2b_{n-2}-1
&=2\left(2^{(n-1)/2}\sin\left(\tfrac{(n-1)\pi}4\right)-1\right)-2\left(2^{(n-2)/2}\sin\left(\tfrac{(n-2)\pi}4\right)-1\right)-1\\
&=2^{(n+1)/2}\sin\left(\tfrac{(n-1)\pi}4\right)-2^{n/2}\sin\left(\tfrac{(n-2)\pi}4\right)-1\\
&=2^{n/2}\left[\sqrt2\sin\left(\frac{n\pi}4\right)\cos\left(\frac\pi4\right)-\sqrt2\cos\left(\frac{n\pi}4\right)\sin\left(\frac\pi4\right)\right]\\
&-2^{n/2}\left[\sin\left(\frac{n\pi}4\right)\cos\left(\frac\pi2\right)-\cos\left(\frac{n\pi}4\right)\sin\left(\frac\pi2\right)\right]-1\\
&=2^{n/2}\sin\left(\frac{n\pi}4\right)-1\\[3pt]
&=b_n
\end{align}
$$
A Different Approach
To solve
$$
b_n=2b_{n-1}-2b_{n-2}-1\tag{1}
$$
let $a_n=b_n+1$. Then
$$
a_n=2a_{n-1}-2a_{n-2}\tag{2}
$$
which has the characteristic polynomial $x^2-2x+2$, whose roots are $1\pm i$. Therefore, the solutions to $(2)$ are
$$
a_n=\alpha(1+i)^n+\beta(1-i)^n\tag{3}
$$
and the solutions to $(1)$ are
$$
b_n=\alpha(1+i)^n+\beta(1-i)^n-1\tag{4}
$$
To satisfy $b_1=0$ and $b_2=1$, we compute $\alpha=-\frac i2$ and $\beta=\frac i2$.
Thus, we get
$$
b_n=-\frac i2(1+i)^n+\frac i2(1-i)^n-1\tag{5}
$$
Changing the Form
Starting from $(5)$, we get
$$
\begin{align}
b_n
&=-\frac i2(1+i)^n+\frac i2(1-i)^n-1\\
&=\frac{e^{-i\pi/2}}22^{n/2}e^{in\pi/4}+\frac{e^{i\pi/2}}22^{n/2}e^{-in\pi/4}-1\\
&=2^{n/2}\cos\left(\frac{(n-2)\pi}4\right)-1\\
&=2^{n/2}\sin\left(\frac{n\pi}4\right)-1\tag{6}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How to square both the sides of an equation? Question: $x^2 \sqrt{(x + 3)} = (x + 3)^{3/2}$
My solution: $x^4 (x + 3) = (x + 3)^3$
$=> (x + 3)^2 = x^4$
$=> (x + 3) = x^2$
$=> x^2 -x - 3 = 0$
$=> x = (1 \pm \sqrt{1 + 12})/2$
I understand that you can't really square on both the sides like I did in the first step, however, if this is not the way to do it, then how can you really solve an equation like this one (in which there's a square root on the LHS) without substitution?
| Why can't you square both sides like you did? You absolutely can. If you do this you just need to make sure you didn't introduce any extraneous solutions. You can make sure of this by checking each solution you get. Admittedly that may be a little difficult with answers like $x = (1 \pm \sqrt{13})/2$, but it is what it is.
What I'm not sure of is how your $x^4$ became an $x^2$ when you divided both sides by $x+3$. Actually now that I look closer that $x^2$ just looks like a typo. Anyway.. I would generally not divide like that because then you lose solutions. In this case you lost the solution $x = -3$. It's better to factor rather than divide.
Recall that $y^{3/2} = y\sqrt{y}$, so in particular we have $(x+3)^{3/2} = (x+3)\sqrt{x+3}$. If you want to do this without squaring both sides, I'd proceed like this:
\begin{align}
x^2\sqrt{x+3} &= (x+3)^{3/2}\\[0.3cm]
x^2\sqrt{x+3} - (x+3)^{3/2} &= 0\\[0.3cm]
x^2\sqrt{x+3} - (x+3)\sqrt{x+3} &= 0\\[0.3cm]
\sqrt{x+3}\left(x^2 - (x+3)\right) &= 0\\[0.3cm]
\sqrt{x+3}(x^2-x-3) &= 0
\end{align}
So either $\sqrt{x+3} = 0$ or $x^2 - x - 3 = 0$. The first gives $x = -3$ and the second gives $x = (1 \pm \sqrt{13})/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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} |
Any hint to solve given integral $\int_0^{2{\pi}}{{d\theta}\over{a^2\cos^2\theta+b^2\sin^2\theta}}$? Show that for $ab>0$ $$\int_0^{2{\pi}}{{d\theta}\over{a^2\cos^2\theta+b^2\sin^2\theta}}={{2\pi}\over ab}$$
I'm not sure how to go about this. Any solutions or hints are greatly appreciated.
| We use the fact that the 1-form
$$\eta = \frac{x dy - y dx}{x^2 + y^2}$$
has integral of $2 \pi$ over $\gamma_r(t) = (r \cos t, r \sin t)$.
Furthermore, if $\Gamma(0) = \Gamma(2\pi)$ and if the intervals $[\gamma(t), \Gamma(t)]$ do not contain $\mathbf{0}$ for any $t \in [0, 2 \pi]$, then the integral over $\Gamma$ is also zero.
Now take $\Gamma(t) = (a \cos t, b \sin t)$. We have
\begin{align}
2\pi = \int_{\Gamma}\eta &= \int_{0}^{2\pi} \frac{a \cos t}{a^2 \cos^2 t + b^2 \sin^2 t} b \cos t + \frac{-b \sin t}{a^2 \cos^2 t + b^2 \sin^2 t} (-a \sin t) \\
&=\int_0^{2\pi} \frac{ab}{a^2 \cos^2 t + b^2 \sin^2 t}.
\end{align}
Proof of the statements stated above:
\begin{align} \int_{\gamma} \eta &= \int_0^{2\pi} \sum_{i=1}^2 a_i(\gamma(t)) \frac{\partial\gamma_i}{\partial t} \, dt\\ &= \int_0^{2\pi} -\frac{\sin t}{r} (-r \sin t) + \frac{\cos t}{r} r \cos t \, dt \\
&= \int_0^{2\pi} \sin^2 t + \cos^2 t \, dt = 2\pi.\end{align}
with $a_1 = \dfrac{-y}{x^2 + y^2}, a_2 = \dfrac{x}{x^2 + y^2}$.
Now, \begin{align*}d \eta &= (da_1) \wedge dx_1 + (da_2) \wedge dx_2\\
&= D_2 a_1 \, dx_2 \wedge dx_1 + D_1 a_2 \, dx_1 \wedge dx_2\\ &= ((D_1 a_2)(x, y) - (D_2a_1)(x, y)) \, \wedge dx_1 \wedge dx_2\\
&= \frac{y^2 - x^2}{(x^2 + y^2)^2} - \frac{y^2 - x^2}{(x^2 + y^2)^2} \, dx_1 \wedge dx_2 = 0.
\end{align*}
Next, let $\Gamma$ be as described. Take $$\Phi(t, u) = (1-u)\Gamma(t) + u\gamma(t).$$
We get $\partial \Phi = \Gamma -\gamma$
Hence $$0 = \int_{\Phi} d\eta = \int_{d \Phi} \eta = \int_{\Gamma - \gamma} \eta$$
by Stokes' theorem
and so \begin{equation} \int_\Gamma \eta = \int_\gamma \eta.\end{equation}
| {
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"url": "https://math.stackexchange.com/questions/1752721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrating $\int \frac{\sqrt{x^2-x+1}}{x^2}dx$ Evaluate $$I=\int\frac{\sqrt{x^2-x+1}}{x^2}dx$$ I first Rationalized the numerator and got as
$$I=\int\frac{(x^2-x+1)dx}{x^2\sqrt{x^2-x+1}}$$ and splitting we get
$$I=\int\frac{dx}{\sqrt{x^2-x+1}}+\int\frac{\frac{1}{x^2}-\frac{1}{x}}{\sqrt{x^2-x+1}}dx$$ i.e.,
$$I=\int\frac{dx}{\sqrt{x^2-x+1}}+\int\frac{\frac{1}{x^3}-\frac{1}{x^2}}{\sqrt{1-\frac{1}{x}+\frac{1}{x^2}}}dx$$
First Integral can be evaluated using standard integral. But second one i am not able to do since numerator is not differential of expression inside square root in denominator.
| So the last step should be changing variable $t=\frac{2}{\sqrt3}(x-\frac{1}{2})$ and use the identity
$$\int \frac{1}{\sqrt{x^2+1}}dx=arcsinh(x)$$
so we get
$$I=-\frac{\sqrt{x^2-x+1}}{x}+\frac{3}{2}\int\frac{1}{\sqrt{x^2-x+1}}dx=
-\frac{\sqrt{x^2-x+1}}{x}+\frac{3}{2}arcsinh[\frac{2}{\sqrt3}(x-\frac{1}{2})]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find a limit for Doublet Stream function In fluid Mechanics,
The superimposed stream function of point source and sink is:
$\psi=-\frac{Qcos\theta_1}{4\pi}+\frac{Qcos\theta_2}{4\pi}$
Graphical image of the function
and for a sink - source doublet, we need to show that as: $l\rightarrow 0$
we get:
$\psi=\frac{m}{r}sin^2\theta$
where: $m=\lim_{l\rightarrow0}\frac{Ql}{4\pi}$
I know it's a simple geometric limit problem , but somehow I can't derive the desired result.
Thanks!
| The positions of the sink and source are $(-l/2,0)$ and $(l/2,0),$ respectively.
Hence,
$$\psi = \frac{Q}{4 \pi}(\cos \theta_2 - \cos \theta_1) \\= \frac{Q}{4 \pi}\left( \frac{x-l/2}{\sqrt{(x-l/2)^2 + y^2}}- \frac{x+l/2}{\sqrt{(x+l/2)^2 + y^2}}\right) \\ = \frac{Q}{4 \pi}\left( \frac{x-l/2}{\sqrt{x^2 + y^2 -lx +l^2/4}}- \frac{x+l/2}{\sqrt{x^2 + y^2 +lx + l^2/4}}\right) $$
In terms of polar coordinates, $r = \sqrt{x^2 +y^2}$ and
$$\psi = \frac{Q}{4 \pi}\left( \frac{x-l/2}{r}\left(1 - \frac{lx -l^2/4}{r^2}\right)^{-1/2}- \frac{x+l/2}{r}\left(1 + \frac{lx +l^2/4}{r^2}\right)^{-1/2}\right).$$
Using the Taylor expansion of the square root terms for small $l$ we obtain
$$\psi = \frac{Q}{4 \pi}\left( \frac{x-l/2}{r}\left(1 + \frac{1}{2}\frac{lx}{r^2}+ O(l^2)\right)- \frac{x+l/2}{r}\left(1 - \frac{1}{2}\frac{lx}{r^2}+ O(l^2)\right)\right) \\ =\frac{Q}{4 \pi}\left(-\frac{l}{r}+ \frac{lx^2}{r^3} +O(l^2) \right) \\= \frac{Ql}{4 \pi r}\left(-1+ \frac{x^2}{r^2} \right)+O(l^2),$$
and with $x = r\cos \theta,$
$$\psi = \frac{Ql}{4 \pi r}\left(-1+ \cos^2 \theta \right)+O(l^2) \\ = -\frac{Ql}{4 \pi r}\sin^2 \theta+O(l^2).$$
Taking the limit as $l \to 0$ we get
$$\lim_{l \to 0} \psi = -\frac{m}{r}\sin^2 \theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1760931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using the $\epsilon-\delta$ definition show that $f(x) = \frac 1 {x^2}$ is a continuous function at any $x_0 = a, a > 0$ Using the $\epsilon-\delta$ definition show that $f(x) = \frac 1 {x^2}$
is a continuous function at any
$x_0 = a, a > 0$
I have expressed in the form:
$$lim_{x\to a}\frac1{x^2}=\frac1{a^2}$$
and thus, $$|f(x)-f(a)|=|\frac{1}{x^2}-\frac{1}{a^2}|=|\frac{a^2-x^2}{x^2a^2}|=|\frac{(x-a)(x+a)}{x^2a^2}| $$
let $\delta=1$, thus $$|\frac{(x-a)(x+a)}{x^2a^2}|< \frac{(x+a)}{x^2a^2}$$
I am stuck as to what I should let my delta be to finish off this proof
| There were some (really) stupid errors in my other answer.
We want
$$|\frac{(x-a)(x+a)}{x^2a^2}|<\epsilon$$ We need to bound $x$ about $a$, that is of course what makes this problem harder.
If we set at least $\delta <a/2$ then
$$-\frac{a}{2}<x-a<\frac{a}{2}$$ and so $$\frac{a}{2}<x<\frac{3a}{2}$$
this gives
$$|\frac{(x-a)(x+a)}{x^2a^2}|<|x-a|\frac{10}{a^3}$$
Now to achive
$$|\frac{(x-a)(x+a)}{x^2a^2}|<|x-a|\frac{10}{a^3} <\epsilon$$
we need only $$|x-a|<\frac{a^3}{10} \epsilon$$
So we set $\delta=\min(a/2, \frac{a^3}{10} \epsilon)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1762674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
inequality involving heights and bisectors Let $a,b,c,a \le b \le c$ be the sides of the triangle $ABC$, $l_a,l_b,l_c$ the lengths of its bisectors and $h_a,h_b,h_c$ the lengths of its heights. Prove that:
$$\frac {h_a+h_c} {h_b} \ge \frac {l_a+l_c} {l_b}$$
| Here's a proof.
The formulae for the lenghts of the angular bisectors and the heights are known. (see e.g. https://en.wikipedia.org/wiki/Triangle ). Let $T$ be the area of the triangle. We have
$$ l_a = \sqrt{bc (1 - \frac{a^2}{(b+c)^2})}$$
and
$$h_a = \frac{2 T}{a}$$
and cyclic shifts of those.
With these formulae, the required inequality gets
$$
\frac{1/a + 1/c}{1/b} \geq \frac{\sqrt{bc (1 - \frac{a^2}{(b+c)^2})}
+ \sqrt{ab (1 - \frac{c^2}{(a+b)^2})}}{\sqrt{ac (1 - \frac{b^2}{(a+c)^2})}}
$$
Multiplying the RHS by
$$1 = \frac{1/\sqrt{abc}}{1/\sqrt{abc}}
$$
gives
$$
\frac{1/a + 1/c}{1/b} \geq \frac{\frac{1}{a}\sqrt{a (1 - \frac{a^2}{(b+c)^2})}
+ \frac{1}{c}\sqrt{c(1 - \frac{c^2}{(a+b)^2})}}{\frac{1}{b}\sqrt{b (1 - \frac{b^2}{(a+c)^2})}}
$$
Due to homogeneity, we can set $b=1$. This gives the condition $a\leq 1\leq c$ and the inequality gets
$$
\frac{1}{a} + \frac{1}{c} \geq \frac{1}{a}\sqrt{
a \frac{
1 - \frac{a^2}{(1+c)^2}
}{
1 - \frac{1}{(a+c)^2}
}
}
+ \frac{1}{c}\sqrt{
c\frac{
1 - \frac{c^2}{(a+1)^2}
}{
1 - \frac{1}{(a+c)^2}
}
}
$$
We now first prove that the first root is less or equal than 1, i.e. we need
$$
{a (1 - \frac{a^2}{(1+c)^2})} \leq
1 - \frac{1}{(a+c)^2}
$$
which can be transformed into
$$
(a+c)^2 (1-a) \geq 1 - a^3 (\frac{a+c}{1+c})^2
$$
The last bracket can be expanded and it suffices to prove
$$
(a+c)^2 (1-a) \geq 1 - a^3 (1 - 2 \frac{1-a}{1+c})
$$
which shortens to
$$(1+c) a (1-a) \geq 2 a^3 (1-a)$$ or $$(1+c) \geq 2 a^2$$ which is certainly true under the given condition $a\leq 1\leq c$.
Likewise, we prove that the second root is less or equal than 1, i.e. we need
$$
{c (1 - \frac{c^2}{(1+a)^2})} \leq
1 - \frac{1}{(a+c)^2}
$$
which can be transformed into
$$
(a+c)^2 (1-c) \geq 1 - c^3 (\frac{a+c}{1+a})^2
$$
We can expand the last bracket and it suffices to prove
$$
(a+c)^2 (1-c) \geq 1 - c^3 (1 + 2 \frac{c-1}{1+a})
$$
or in positive terms
$$
(a+c)^2 (c-1) \leq c^3 -1 + 2 c^3 \frac{c-1}{1+a}$$
Since $a\leq1$, it suffices to prove
$$
(1+c)^2 (c-1) - c^3 +1 \leq 2 c^3 \frac{c-1}{1+a}$$
or
$$
(1+a)(c^2 - c) \leq 2 c^2 (c^2-c)
$$
which shortens to
$$1+a \leq 2 c^2$$ which is certainly true under the given condition $a\leq 1\leq c$.
Hence both roots are less or equal to 1, which proves the inequality.
| {
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"url": "https://math.stackexchange.com/questions/1764412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Bronstein Integral 21.42 Good morning.
I came across the following integral in some field theory calculation:
$\int_0^\pi dx\,\log\left(a^2+b^2-2ab\cos x\right)=2\pi\log\left(\max\lbrace a,b\rbrace\right)$
for $0<a,b\in\mathbb{R}$. The notation has been adapated to make contact with Bronstein's integral 21.42.
Despite 6 pp.of calculations, I was not able to figure out a way to prove the identity (integration by parts, substitution, derivative trick, contour integration...). Does anyone have a proof?
Best regards,
David
| Suppose WLOG $b > a$. Then with $c = b/a$ we have
$$\begin{align} I &= \int_0^\pi \log(a^2 + b^2 - 2ab \cos x) \, dx \\ &=\int_0^\pi \log a^2 \, dx + \int_0^\pi \log(1 + (b/a)^2 - 2(b/a) \cos x) \, dx \\ &= 2\pi \log a + \int_0^\pi \log(1 + c^2 - 2c \cos x) \, dx \end{align}$$
Now we can evaluate the second integral on the RHS as the limit of a Riemann sum
$$\begin{align}J &= \int_0^\pi \log(1 + c^2 - 2c \cos x) \, dx \\ &= \lim_{n \to \infty} \frac{\pi}{n}\sum_{j=1}^n\log(1 + c^2 - 2c\cos(\pi(j-1)/n)) \\ &= \lim_{n \to \infty} \frac{\pi}{n}\log\prod_{j=1}^n(1 + c^2 - 2c\cos(\pi(j-1)/n)) \\ &=\lim_{n \to \infty} \frac{\pi}{n}\log(1-c)^2+\lim_{n \to \infty} \frac{\pi}{n}\log\prod_{j=2}^{n}(1 + c^2 - 2c\cos(\pi(j-1)/n)) \\ &= \lim_{n \to \infty} \frac{\pi}{n}\log\prod_{j=1}^{n-1}(1 + c^2 - 2c\cos(\pi j/n))\end{align} .$$
Upon factoring, we obtain
$$1 + c^2 - 2c\cos(\pi j/n) = [c - \exp(i \pi j /n)][c - \exp(-i \pi j /n)],$$
and
$$c^{2n} -1 = (c-1)(c+1)\prod_{j=1}^{n-1}[c - \exp(i \pi j /n)][c - \exp(-i \pi j /n)].$$
Hence, recalling that $c = b/a > 1$, it follows that
$$\begin{align}J &= \lim_{n \to \infty} \frac{\pi}{n} \log\left(\frac{c^{2n}-1}{c^2-1} \right) \\ &= \pi \lim_{n \to \infty} \log\left(\frac{c^{2n}-1}{c^2-1} \right)^{1/n} \\ &= \pi \lim_{n \to \infty} \log\left[c^2 \left(\frac{1-c^{-2n}}{c^2-1} \right)^{1/n}\right] \\ &= \pi \log c^2 + \log \left[\lim_{n \to \infty} \left(\frac{1-c^{-2n}}{c^2-1} \right)^{1/n}\right]\\ &= \pi \log c^2\end{align}$$
Thus,
$$I = 2\pi \log a + \pi \log c^2 = 2\pi \log a + \pi \log (b/a)^2 \\ = 2\pi \log b $$
| {
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"url": "https://math.stackexchange.com/questions/1764999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Find $\sin\frac{\pi}{3}+\frac{1}{2}\sin\frac{2\pi}{3}+\frac{1}{3}\sin\frac{3\pi}{3}+\cdots$ Find $$\sin\frac{\pi}{3}+\frac{1}{2}\sin\frac{2\pi}{3}+\frac{1}{3}\sin\frac{3\pi}{3}+\cdots$$
The general term is $\frac{1}{r}\sin\frac{r\pi}{3}$
Let $z=e^{i\frac{\pi}{3}}$
Then, $$\frac{1}{r}z^r=\frac{1}{r}e^{i\frac{r\pi}{3}}$$
I have to find the imaginary part of $$P=\sum_{r=1}^\infty \frac{1}{r}z^r$$
Let $$S=1+z+z^2+\cdots$$
Hence, $$P=\int_0^z Sdz=z+\frac{z^2}{2}+\frac{z^3}{3}+\cdots$$
which is the required sum.
$$S=\frac{1}{1-z}$$
$$P=\int_0^z \frac{1}{1-z}dz$$
$$P=\ln\left(\frac{1}{1-z}\right)=\ln\left(\frac{1}{1-e^{i\frac{\pi}{3}}}\right)=i\frac{\pi}{3}$$
Hence, the imaginary part of $P$ is $\frac{\pi}{3}$
Is this method correct? Is there any method that does not require complex numbers?
| In your line
$$P=-\ln(1-e^{i\pi/3})$$
Now $$\ln(1-e^{i\pi/3})=\ln(-1)+\ln(e^{i\pi/6})+\ln(2i\sin\dfrac\pi6)\equiv i\pi+\dfrac{i\pi}6+\ln(i)\pmod{2\pi i}$$
$$\equiv i\left(\pi+\dfrac\pi6+\dfrac\pi2\right)\equiv-\dfrac{i\pi}3$$
So, the principal value of the imaginary part of $P$ is $$\dfrac\pi3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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System of equations that can be solved by inequalities: $(x^3+y^3)(y^3+z^3)(z^3+x^3)=8$ and $\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=\frac32$
S367. Solve in positive real numbers the system of equations:
\begin{gather*}
(x^3+y^3)(y^3+z^3)(z^3+x^3)=8,\\
\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=\frac32.
\end{gather*}
Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam
From https://www.awesomemath.org/wp-pdf-files/math-reflections/mr-2016-02/mr_2_2016_problems.pdf
I think I am smelling inequalities here. In the first equation I used Holder's inequality to show, $xyz \le 1$ , But in the second equation I used Titu's Lemma to get $x+y+z \le 3$ .But I think there would an equality case in one of the two equations. Can anyone help? The original source is Facebook
| The inequality $$\frac{a^2+b^2}{a+b}\geq\sqrt[3]{\frac{a^3+b^3}{2}}$$ we can prove also by AM-GM:
$$2(a^2+b^2)^3=\frac{2}{27}(3a^2+3b^2)^3=$$
$$=\frac{2}{27}\left(2(a^2-ab+b^2)+2\cdot\frac{(a+b)^2}{2}\right)^3\geq$$
$$\geq\frac{2}{27}\left(3\sqrt[3]{2(a^2-ab+b^2)\cdot\left(\frac{(a+b)^2}{2}\right)^2}\right)^3=$$
$$=(a^2-ab+b^2)(a+b)^4=(a^3+b^3)(a+b)^3$$
and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum_{k=1}^n \frac{1}{n+k} = \sum_{k=1}^{2n} \frac{1}{k}(-1)^{k-1}$ using induction I'm trying to prove (using induction) that:
$$\sum_{k=1}^n \frac{1}{n+k} = \sum_{k=1}^{2n} \frac{1}{k}(-1)^{k-1}.$$
I have found problems when I tried to establish an induction hypothesis and solving this because I've learned to do things like:
$$ \sum_{k=1}^{n+1} \frac{1}{k}= \sum_{k=1}^{n} \frac{1}{k} + \frac{1}{n+1}.$$
But, in this case, $n$ appears in both parts of summation and I have no idea how make a relation with
$$\sum_{k=1}^n \frac{1}{n+k} $$ and $$\sum_{k=1}^{n+1} \frac{1}{n+1+k}. $$
Because I've seen tha, the case with "$n+1$" should be like:
$$\sum_{k=1}^{n+1} \frac{1}{n+1+k} = \sum_{k=1}^{2n+2} \frac{1}{k}(-1)^{k-1}$$
and I cant find a connection between
$$\sum_{k=1}^{n+1} \frac{1}{n+1+k} $$ and $$\sum_{k=1}^{n} \frac{1}{n+k}.$$
Could anyone help me with this?
| We can write
$$\begin{align}\sum_{k=1}^{n+1}\frac{1}{n+1+k}&=\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n+2}\\&=\left(-\frac{1}{n+1}+\frac{1}{n+1}\right)+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n+2}\\&=\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots +\frac{1}{2n}\right)-\frac{1}{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2}\\&=\left(\sum_{k=1}^{n}\frac{1}{n+k}\right)-\frac{1}{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2}\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that : $\frac{a+b+c+d}{a+b+c+d+f+g}+\frac{c+d+e+f}{c+d+e+f+b+g}>\frac{e+f+a+b}{e+f+a+b+d+g}$
Prove inequality for positive numbers:
$$\frac{a+b+c+d}{a+b+c+d+f+g}+\frac{c+d+e+f}{c+d+e+f+b+g}>\frac{e+f+a+b}{e+f+a+b+d+g}$$
My work so far:
Lemma: If $x>y>0, t>z>0$, then $$\frac{x}{x+z}>\frac y{y+t}.$$
Proof. Really, $\frac zx< \frac ty\Rightarrow \frac{x+z}x<\frac{t+y}y.$
| First, let's write the left side of the inequality as:
$L(c)=\dfrac{c+A}{c+B}+\dfrac{c+D}{c+E}$,
the derivative of $L(c)$ is:
$L'(c)=\dfrac{B-A}{(c+B)^2}+\dfrac{E-D}{(c+E)^2}$,
which is always positive for $c\geq 0$, because $B>A$ and $E>D$. Hence, L(c) is cresent for $c\geq 0$ and so, $L(c)>L(0)$ for $c\geq 0$, i.e.
$\dfrac{a+b+c+d}{a+b+c+d+f+g}+\dfrac{c+d+e+f}{c+d+e+f+b+g}>\dfrac{a+b+d}{a+b+d+f+g}+\dfrac{d+e+f}{d+e+f+b+g}$ $\cdots (1)$.
Now, working with the right side of the inequality,
$\dfrac{e+f+a+b}{e+f+a+b+d+g}=\dfrac{a+b}{e+f+a+b+d+g}+\dfrac{e+f}{e+f+a+b+d+g}$ $\cdots (2)$,
and by the positive condition in the numbers, it is not difficult to prove that the first term in $(1)$ is greater than the first term in $(2)$, and that the same happens with the second term of $(1)$ and $(2)$. Therefore, it is demostrated.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1769787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove that exist a $a_n<0$ by the recursion $a_{n+2}-Pa_{n+1}+Qa_n = 0$ with the condition $\Delta:= P^2-4Q < 0$ Suppose $P, Q > 0$ such that $\Delta:= P^2-4Q < 0$. The sequence $\{a_n\}$ is defined by the recursion $a_{n+2}-Pa_{n+1}+Qa_n = 0$ where both $a_1$ and $a_2$ are real and at least one is non-zero.
Then, which is needed to be proved, there exists n > 0 such that $a_n < 0$.
| As user1952009
said,
if
$a_{n+2}-Pa_{n+1}+Qa_n = 0
$,
by assuming that
$a_n
=r^n
$,
we get,
after dividing by
$r^n$,
$0
=r^2-Pr+Q
$
or
$r
=\frac12(P \pm \sqrt{D})
$
where
$D
= P^2-4Q
< 0
$.
By considering the two possible
values of $r$,
which I will call
$u$ and $v$,
we can find
$A$ and $B$ such that
$A+B = a_0$
and
$Au+Bv = a_1$.
By induction,
this will allow us to show that
$Au^n+Bv^n
= a_n
$
for all $n$.
For an explicit formula for $A$ and $B$,
we have
$B = a_0-A$,
so
$a_1
=Au+(a_0-A)v
=A(u-v)+a_0v
$
or
$A
=\dfrac{a_1-a_0v}{u-v}
$
and
$B
=a_0-A
=a_0-\dfrac{a_1-a_0v}{u-v}
=\dfrac{a_0(u-v)-a_1+a_0v}{u-v}
=\dfrac{a_0u-a_1}{u-v}
$.
Note that
$u-v
=i\sqrt{-D}
$.
Now consider
$u
=\frac12(P + \sqrt{D})
=\frac12(P + i\sqrt{-D})
$.
$|u|
=\frac14\sqrt{P^2+(-D)}
=\frac14\sqrt{P^2-P^2+4Q}
=\frac12\sqrt{Q}
$.
If
$\dfrac{u}{|u|}
=c
$,
then
$|c| = 1$
and $c$ is in the first quadrant,
so
$c = e^{it}
$
where
$0 < t < \pi/2$.
Therefore
$u^n
=|u|^n c^n
=|u|^n e^{int}
$
will take values
in all four quadrants.
Similarly,
$v
=\frac12(P - \sqrt{D})
=\frac12(P - i\sqrt{-D})
=\bar{u}
$,
so that
$v^n
=|u|^n e^{-int}
$
and,
since
$A+B = a_0$
and
$\begin{array}\\
A-B
&=\dfrac{a_1-a_0v}{u-v}-\dfrac{a_0u-a_1}{u-v}\\
&=\dfrac{a_1-a_0v-a_0u+a_1}{u-v}\\
&=\dfrac{2a_1-a_0(u+v)}{u-v}\\
&=\dfrac{2a_1-a_0P}{i\sqrt{-D}}\\
\end{array}
$
$\begin{array}\\
Au^n+Bv^n
&=A|u|^ne^{int}+B|u|^n e^{-int}\\
&=|u|^n(Ae^{int}+B e^{-int})\\
&=|u|^n(A(\cos(nt)+i\sin(nt))+B(\cos(nt)-i\sin(nt)))\\
&=|u|^n(\cos(nt)(A+B)+i\sin(nt)(A-B))\\
&=|u|^n(\cos(nt)a_0+i\sin(nt)\dfrac{2a_1-a_0P}{i\sqrt{-D}})\\
&=|u|^n(\cos(nt)a_0+\sin(nt)\dfrac{2a_1-a_0P}{\sqrt{-D}})\\
&=|u|^nR\cos(nt+\theta)\\
\end{array}
$
with $R$ and $\theta$
being gotten by
the usual method of expressing
a linear combination
of
$\sin(z)$
and
$\cos(z)$
as a constant times
$\cos(z+\theta)$
for some $\theta$.
Since
$0 < t < \pi/2$,
$\cos(nt+\theta)$
will take both positive
and negative values
for large enough $n$.
And we are done.
(Whew!)
(added later)
Explicitly,
$\begin{array}\\
R^2
&=a_0^2+\left(\dfrac{2a_1-a_0P}{\sqrt{-D}}\right)^2\\
&=a_0^2+\dfrac{(2a_1-a_0P)^2}{-P^2+4Q}\\
&=\dfrac{a_0^2(-P^2+4Q)+4a_1^2-4a_0a_1P+a_0^2P^2}{-P^2+4Q}\\
&=\dfrac{4a_0^2Q+4a_1^2-4a_0a_1P}{-P^2+4Q}\\
&=4\dfrac{a_0^2Q+a_1^2-4a_0a_1P}{-P^2+4Q}\\
\end{array}
$
and a related expression for
$\tan(\theta)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Random points on a sphere — expected angular distance Suppose we randomly select $n>1$ points on a sphere (all independent and uniformly distributed).
*
*What is the expected angular distance from a point to its closest neighbor?
*What is the expected angular distance from a point to its $m^{\text{th}}$ closest neighbor (where $m<n$)?
| Let
*
*$a = m -1$, $b = n - m - 1$, $d = a + b = n - 2$.
*$\theta_{ij}$, $1 \le i \ne j \le n$ be the angular separation between point $x_i$ and $x_j$.
*$\ell_m$ be the expected angular separation among a pair of $m^{th}$ nearest neighbors.
*$\mathcal{E}_m$ be the event that $x_2$ is the $m^{th}$ nearest neighbor of
$x_1$.
*$\mathbf{1}_m$ be the indicator function for event $\mathcal{E}_m$.
Since all points are equal, we can use any pair of points, say $x_1$ and $x_2$, as probe and $\ell_m$ will be equal to the expected value of $\theta_{12}$
subject to the constraint that event $\mathcal{E}_m$ happens. i.e.
$$\ell_m = \mathbf{E}( \theta_{12} | \mathcal{E}_m )
= \frac{\mathbf{E} ( \theta_{12}\mathbf{1}_m )}{\mathbf{P}( \mathcal{E}_m )}
= \frac{\mathbf{E} ( \theta_{12}\mathbf{1}_m )}{\mathbf{E}( \mathbf{1}_m )}
$$
For any $\theta \in [0,\pi]$, let $$p = \frac{1-\cos\theta}{2} \iff \theta = 2\sin^{-1}(p^{1/2})$$
When $\theta_{12} = \theta$, we have
$$\mathbf{P}( \theta_{1k} < \theta | \theta_{12} = \theta ) = \frac{1-\cos\theta}{2} = p
\quad\text{ for any } 2 \le k \le n
$$
Since these $n-2$ angular separations $\theta_{1k}$ are independent and there are
$\displaystyle\;\binom{d}{a}$ ways of picking $a$ out of $d$ points. we have
$$\mathbf{E}( \mathbf{1}_m | \theta_{12} = \theta ) = \binom{d}{a}p^a(1-p)^b$$
Since $x_2$ are distributed uniformly over the sphere, the probability for $\theta \le \theta_{12} \le \theta + d\theta$ is proportional to $\sin\theta_{12} d\theta_{12} \propto dp$. We have
$$
\mathbf{E}( \theta_{12}\mathbf{1}_m ) = \displaystyle\;\binom{d}{a}\int_0^1 p^a (1-p)^b \theta_{12} dp
\quad\text{ and }\quad
\mathbf{E}( \mathbf{1}_m ) = \displaystyle\;\binom{d}{a}\int_0^1 p^a (1-p)^b dp
$$
As a result,
$$\begin{align}
\ell_m
&= m\binom{n-1}{m}\int_0^1 p^a (1-p)^b 2\sin^{-1}(p^{1/2})\,dp\\
&= m\binom{n-1}{m}\sum_{k=0}^a(-1)^k \binom{a}{k} \int_0^1 (1-p)^{b+k} 2\sin^{-1}(p^{1/2})\, dp\\
&= m\binom{n-1}{m}\sum_{k=0}^a(-1)^{a-k} \binom{a}{k} \int_0^1 (1-p)^{d-k} 2\sin^{-1}(p^{1/2})\, dp
\end{align}
$$
For any $c \in \mathbb{N}$, we have
$$\begin{align}
\int_0^1 (1-p)^c 2\sin^{-1}(p^{1/2}) dp
&= \frac{-1}{c+1}\int_0^1 2\sin^{-1}(p^{1/2})\, d(1-p)^{c+1}\\
&= \frac{-1}{c+1}\left\{\bigg[ 2\sin^{-1}(p^{1/2}) (1-p)^{c+1} \bigg]_0^1 - \int_0^1 \frac{(1-p)^{c+1}}{\sqrt{p(1-p)}} dp \right\}\\
&= \frac{1}{c+1}\frac{\Gamma(\frac12)\Gamma(c+\frac32)}{\Gamma(c+2)}
= \frac{\pi}{(c+1)2^{2c+1}}\binom{2c+1}{c}
\end{align}
$$
Form this, we get
$$
\ell_m
= \pi m\binom{n-1}{m}\sum_{k=0}^{m-1}
\frac{(-1)^{m-1-k}}{(n-k-1)2^{2n-2k-3}}\binom{m-1}{k}
\binom{2n-2k-3}{n-k-2}
$$
In particular, the expected nearest neighbor angular separation $\displaystyle\;\ell_1 = \frac{\pi}{2^{2n-3}}\binom{2n-3}{n-2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1771835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
numerical values of points in cantor set Let $C$ be the standard middle thirds Cantor set in the interval $[0,1]$. The "endpoints" of $C$ have very simple numerical values that can be listed off: $$0,1,1/3,2/3,1/9,2/9,6/9,7/9,1/27,2/27,7/27,8/27,... $$
What I am looking for is some numerical values of the "nonendpoints" of $C$. Specifically, is there a dense subset of the nonendpoints which has some nice pattern to its values? Can you list off some of these for me?
| For example,
$$\frac 14 = \frac{0}{3} + \frac{2}{3^2} + \frac{0}{3^3} + \frac{2}{3^4} + \frac{0}{3^5} + \frac{2}{3^6} + \cdots$$
is in the Cantor Set.
Edit: Actually, elements of the Cantor Set are of the form
$$\sum_{n=1}^\infty \frac{a_n}{3^n}$$
where $a_n$ is any sequence consisting of only $0$ and $2$'s.
So, when $a_n$ is not almost constant (an almost constant sequence is a sequence which is constant from some point on) then the corresponding number is a non-endpoint.
Edit2:
$$\frac{2}{25} = \frac{0}{3} + \frac{0}{3^2} + \frac{2}{3^3} + \frac{0}{3^4} + \frac{0}{3^5} + \frac{2}{3^6} + \cdots$$
$$\frac{11}{12} = \frac{2}{3} + \frac{2}{3^2} + \frac{0}{3^3} + \frac{2}{3^4} + \frac{0}{3^5} + \frac{2}{3^6} + \cdots$$
For further examples, you can take:
$a_n = 0,0,0,2,0,2,0,2,0,2,0,2,0, \dots$
$a_n = 2,0,2,0,2,2,0,2,0,2,0,2,0, \dots$
$a_n = 0,2,2,0,0,2,2,0,0,2,2,0,0, \dots$
$a_n = 0,0,0,0,2,2,2,2,0,0,0,0,2,2,2,2, \dots$
$a_n = 0,0,0,2,2,0,0,0,2,2,0,0,0,2,2, \dots$
etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Maximum value of the sum of absolute values of cubic polynomial coefficients $a,b,c,d$
If $p(x) = ax^3+bx^2+cx+d$ and $|p(x)|\leq 1\forall |x|\leq 1$, what is the $\max$ value of $|a|+|b|+|c|+|d|$?
My try:
*
*Put $x=0$, we get $p(0)=d$,
*Similarly put $x=1$, we get $p(1)=a+b+c+d$,
*similarly put $x=-1$, we get $p(-1)=-a+b-c+d$,
*similarly put $\displaystyle x=\frac{1}{2}$, we get $\displaystyle p\left(\frac{1}{2}\right)=\frac{a}{8}+\frac{b}{4}+\frac{c}{2}+d$
So, given that $|p(x)|\leq 1\forall |x|\leq 1$, we get $|d|\leq 1$.
Similarly $$\displaystyle |b|=\left|\frac{p(1)+p(-1)}{2}-p(0)\right|\leq \left|\frac{p(1)}{2}\right|+\left|\frac{p(1)}{2}\right|+|p(0)|\leq 2$$
Now I do not understand how can I calculate the $\max$ of $|a|$ and $|c|$.
| I'll assume that $x$ is real.
Consider system of four equations with four variables.
$$
a + b + c + d = p(1),
$$
$$
-a+b-c+d = p(-1),
$$
$$
\frac{a}{8}+\frac{b}{4} + \frac{c}{2} + d = p(1/2),
$$
$$
-\frac{a}{8}+\frac{b}{4} - \frac{c}{2} + d = p(-1/2).
$$
You can just solve it and find values of $a,b,c,d$.
From this equations we get
$$
a+c = \frac{p(1)-p(-1)}{2},
$$
$$
\frac{a}{4} + c = p(1/2) - p(-1/2)
$$
Thus
$$
\frac{3a}{4} = \frac{p(1)-p(-1)}{2} - (p(1/2) - p(-1/2)),
$$
and $|a|\le4$. In the same way we get
$$
3c = 4(p(1/2) - p(-1/2)) - \frac{p(1)-p(-1)}{2},
$$
and $|c|\le 3$.
The example $T(x) = 4x^3 - 3x$ shows existence of polynomial with $a=4$ and $c=3$.
So, $|a|+|b|+|c|+|d|\le 1 + 2 +3 +4 = 10$, though I don't know if such a polynomial (with $|a|+|b|+|c|+|d|\le 10$ and $p(x)\le 1$) exists.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Use polar coordinates to evaluate function
To find:
$$\iint_R\frac{1}{1+x^2+y^2}\,dA$$
where R is first quadrant bounded by $y=0$, $y=x$, $x^2+y^2=4$
$$\iint_Rf(r\cos \theta,r\sin\theta)r\,dA$$
which is as far as I got -
$$\iint\frac{1}{1+r^2\cos^2\theta +r^2\sin^2\theta}\,dA$$
$$I=\iint \frac{1}{1+r^2}\,dr\,d\theta$$
$$I=\int\tan{^{-1}}r \,d\theta$$
Where to from here?
| So
$$\iint_R\frac{1}{1+x^2+y^2}\,dA$$
Since $r^2 = x^2 + y^2$ and $dA = r \,dr \,d\theta$
$$\iint_R\frac{1}{1+r^2}r \,dr \,d\theta$$
Now, $R$ is the region bounded by $r = [0, 2]$ and $\theta = [0, \pi/4]$. So
$$\int_{0}^{\pi/4} \int_{0}^{2} \frac{r}{1+r^2} \,dr \,d\theta$$
$$\int_{0}^{\pi/4} \,d\theta \times \int_{0}^{2} \frac{r}{1+r^2} \,dr$$
$$\left( \frac{\pi}{4} \right) \left. \frac{1}{2} \ln(r^2+1)\right|_{r=0}^{2}$$
$$\left( \frac{\pi}{8} \right) \left[ \ln(5) - \ln(1)\right]$$
$$\left( \frac{\pi}{8} \right) \ln(5)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding the Exponential of a Matrix that is not Diagonalizable Consider the $3 \times 3$ matrix
$$A =
\begin{pmatrix}
1 & 1 & 2 \\
0 & 1 & -4 \\
0 & 0 & 1
\end{pmatrix}.$$
I am trying to find $e^{At}$.
The only tool I have to find the exponential of a matrix is to diagonalize it. $A$'s eigenvalue is 1. Therefore, $A$ is not diagonalizable.
How does one find the exponential of a non-diagonalizable matrix?
My attempt:
Write
$\begin{pmatrix}
1 & 1 & 2 \\
0 & 1 & -4 \\
0 & 0 & 1
\end{pmatrix} = M + N$,
with $M = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}$ and $N = \begin{pmatrix}
0 & 1 & 2 \\
0 & 0 & -4 \\
0 & 0 & 0
\end{pmatrix}$.
We have $N^3 = 0$, and therefore $\forall x > 3$, $N^x = 0$. Thus:
$$\begin{aligned}
e^{At}
&= e^{(M+N)t} = e^{Mt} e^{Nt} \\
&= \begin{pmatrix}
e^t & 0 & 0 \\
0 & e^t & 0 \\
0 & 0 & e^t
\end{pmatrix} \left(I + \begin{pmatrix}
0 & t & 2t \\
0 & 0 & -4t \\
0 & 0 & 0
\end{pmatrix}+\begin{pmatrix}
0 & 0 & -2t^2 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}\right) \\
&= e^t \begin{pmatrix}
1 & t & 2t \\
0 & 1 & -4t \\
0 & 0 & 1
\end{pmatrix} \\
&= \begin{pmatrix}
e^t & te^t & 2t(1-t)e^t \\
0 & e^t & -4te^t \\
0 & 0 & e^t
\end{pmatrix}.
\end{aligned}$$
Is that the right answer?
| If you know about the Jordan Canonical Form you can use that.
Another method, probably more elementary, was mentioned in a comment. The comment was deleted; I don't know why. Note that $A=I+N$, where $N^3=0$. It follows that $$A^k=I+kN+\frac{k(k-1)}{2}N^2.$$You can use that to calculate $e^{At}=\sum t^kA^k/k!$.
Edit: Oh, that comment was converted to an answer. I'll leave this here anyway, being more detailed (at least regarding one approach).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 1
} |
If $3^x +3^y +3^z=9^{13}$.Find value of $x+y+z$ Problem: If $3^x +3^y +3^z=9^{13}$.Find value of $x+y+z$.
Solution: $3^x +3^y +3^z=9^{13}$
$3^x +3^y +3^z=3^{26}$
I am unable to continue from here.
Any assistance is appreciated.
Edited
$9^{13} =3^{26}$
$=3^{25} (3)$
$=3^{25} (1+1+1)$
$=3^{25} + 3^{25} + 3^{25}$
So $x+y+z =75$
| Let $x\ge y\ge z$. Then
$$3^x +3^y +3^z=3^{26}$$
$$3^z(3^{x-z}+3^{y-z}+1)=3^{z}\cdot3^{26-z}$$
$$3^{x-z}+3^{y-z}+1=3^{26-z} \Leftrightarrow x=z=y=25$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1779912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
I have a problem when I go to calculate $\lim_{x\to\infty}\left( \frac {2x+a}{2x+a-1}\right)^{x}.$ The limit:
$\lim_{x\rightarrow\infty}
\left( {\frac {2\,x+a}{2\,x+a-1}} \right) ^{x}$
I make this:
$\left( {\frac {2\,x+a}{2\,x+a-1}} \right) ^{x}$=${{\rm e}^{{\it x\ln} \left( {\frac {2\,x+a}{2\,x+a-1}} \right) }}$
Then:
${{\rm e}^{{\it \lim_{x\rightarrow\infty}
x\ln} \left( {\frac {2\,x+a}{2\,x+a-1}} \right) }}$ = $0\cdot\infty$
Note: I cannot use L'Hopital
| $(\frac{2x+a}{2x+a-1})^x=(\frac{2x+a+1-1}{2x+a-1})^x=(1+\frac{1}{\alpha})^{\frac{\alpha}{2}+\frac{1-a}{2}}\longrightarrow e^{\frac{1}{2}}$. Where $\alpha=2x+a-1$ and $\alpha\rightarrow \infty,$ when $x\rightarrow\infty$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
What is the probability of selecting five of the winning balls and one of the supplementary balls? So I'm just doing a bit of probability questions and wanted to make sure I got it right.
I have $50$ balls numbered $1-50$, and we pick $6$ winning balls and $2$ supplementary without replacement.
So the chance to get the $6$ winning balls would simply be: $$\frac{6}{50} \cdot \frac{5}{49} \cdot \frac{4}{48} \cdot \frac{3}{47} \cdot \frac{2}{46} \cdot \frac{1}{45} = \frac{1}{15890700}$$
and for $5$ winning balls it would be, same process as above: $$\frac{3}{1059380}$$
Now the part that confuses me is $5$ winning, $1$ supplementary. Would this be given by: $$\frac{6}{50} \cdot \frac{5}{49} \cdot \frac{4}{48} \cdot \frac{3}{47} \cdot \frac{2}{46} \cdot \frac{44}{45} \cdot \frac{2}{44} = \frac{1}{7945350}?$$
Can someone please check if I did this right, I have a sense that I did not, but not sure where I went wrong.
| I will do it another way. The Tax on the Poor Corporation chooses $6$ "main" numbers and $2$ supplementary numbers. There are $\binom{50}{6}$ ways to choose the main numbers. Presumably they are all equally likely.
We find the probability that you get all $6$ main numbers. There is only $1$ hand that will do the job, so the probability is $\frac{1}{\binom{50}{6}}$.
Now we find the probability that your hand has exactly $5$ main numbers, and none of the supplementary numbers. There are $2$ supplementary numbers, so you must have $5$ of the $6$ main numbers, plus one of the $42$ useless numbers. Thus there are $42\binom{6}{5}$ hands that have $5$ main numbers and no supplementary. That yields probability $\frac{42\binom{6}{5}}{\binom{50}{6}}$.
Finally, we count the hands that have $5$ main numbers and $1$ supplementary. There are $\binom{6}{5}\cdot 2$ such hands, and we compute the probability as before.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find the limit of $\lim_{n\to \infty}\frac{2^n n^2}{n!}$ I am trying to find the limit of the following, $$\lim_{n\to \infty}\frac{2^n n^2}{n!}$$
L'Hospital is not going to work. Hoping for a squeeze, by the observation that $2^n<n!$ for $n\geq 4$ does not help either as one side of the limit goes to $\infty$. How can I solve this? Any hints?
| Note that:
$$n!\ge1\cdot 2\cdot3\cdot3\cdots\cdot3\cdot(n-2)\cdot(n-2)\cdot(n-2)=2\cdot3^{n-5}(n-2)^3$$
So for $n>5$:
$$\frac{2^nn^2}{n!}\le\frac{2^nn^2}{2\cdot3^{n-5}(n-2)^3}=\frac{3^5}{2}\left(\frac{2}{3}\right)^n\left(\frac{n}{n-2}\right)^3\frac{1}{n}<\frac{3^5}{2}\left(\frac{2}{3}\right)^5\left(\frac{5}{3}\right)^3\frac{1}{n}\to0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1782490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Find the range of $y = \sqrt{x} + \sqrt{3 -x}$ I have the function $y = \sqrt{x} + \sqrt{3 -x}$. The range in wolfram is $y \in\mathbb R: \sqrt{3} \leq y \leq \sqrt{6}$
(solution after correction of @mathlove)
$\sqrt{x} + \sqrt{3 -x} = y$
$$
\begin{cases}
x \geq 0\\
x \leq 3
\end{cases}
$$
then
$(\sqrt{x} + \sqrt{3 -x})^2 = y^2$
$x+3-x+2\sqrt{x(3-x)}=y^2$
$2\sqrt{x(3-x)}=y^2-3$
irrational equation, therefore:
$$
\begin{cases}
y^2-3 \geq 0\\
4(3x-x^2)=(y^2-3)^2
\end{cases}
$$
The $y^2 \geq 3$ is verified when $y \leq -\sqrt{3}$ or $y \geq \sqrt{3}$
regarding the second element of the system $4(3x-x^2)=(y^2-3)^2$
$4(3x-x^2)=(y^2-3)^2$
$12x-4x^2=y^4+9-6y^2$
$12x-4x^2-y^4-9+6y^2 = 0$
$12x-4x^2-y^4-9+6y^2 = 0$
$4x^2-12x+y^4-6y^2+9 = 0$
the quadratic equation is verified when the discriminant is $\geq 0$, then
$b^2 - 4ac = (-12)^2-16(y^4-6y^2+9) \geq 0$
$144-16y^4+96y^2-144 \geq 0$
$16y^4-96y^2 \leq 0$
change $t=y^2$ and $t^2=y^4$
$16t^2-96t \leq 0$
the inequality is verified when $t_1 \leq t \leq t_2$ because the discriminant in t is $\geq 0$
$t(16t^2-96) = 0$
then
$t_1 = 0$ and $t_2 = 6$ but $t=y^2$ and
$\sqrt(0) \leq y \leq \sqrt{6}$
$-\sqrt{6} \leq y \leq \sqrt{6}$
Finally, the solutions in $y$ in the system are:
$y\leq -\sqrt{3}$ or $y\geq \sqrt{3}$
and
$-\sqrt{6} \leq y \leq \sqrt{6}$
the solutions in $y$ (=range) of the system is:
$\sqrt{3} \leq y \leq \sqrt{6}$ because the function in the domain is satisfy only for $0 \leq x \leq 3$
| $f(x) = \sqrt{x} + \sqrt{3-x}$. From this you have correctly deduced that the domain is $[0,3]$.
\begin{align*}
f'(x) &= \frac{1}{2\sqrt x} - \frac{1}{2\sqrt{3-x}}\\[0.3cm]
&= \frac{\sqrt{3-x}}{2\sqrt x \sqrt{3-x}} - \frac{\sqrt{x}}{2\sqrt x\sqrt{3-x}}
\end{align*}
$f'(x) = 0$ when $\sqrt{3-x} = \sqrt{x}$, which is when $x = 3/2$. Also, $f'(x)$ is undefined when $x = 0$ or when $x=3$. Luckily these coincide with the endpoints of the domain (which we'd also need to check separately if they didn't coincide).
So we have $f(0) = f(3) = \sqrt{3}$, and
$$ f(3/2) = \sqrt\frac{3}{2} + \sqrt\frac{3}{2} = \frac{2\sqrt{3}}{\sqrt{2}} = \sqrt{6}.$$
So the absolute min is $\sqrt{3}$ and the absolute max is $\sqrt{6}$. Also, since the function is continuous on $[0,3]$ then by the Intermediate Value Theorem, the function attains every value between $\sqrt{3}$ and $\sqrt{6}$ for input values coming from $[0,3]$. Therefore the range is $[\sqrt{3}, \sqrt{6}]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1783925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to find the integral of a quotient of rational functions? How do I compute the following integral: $$\int \dfrac{x^4+1}{x^3+x^2}\,dx$$
My attempt:
We can write $$\dfrac{x^4+1}{x^3+x^2} = \dfrac{A}{x^2} + \dfrac{B}{x} + \dfrac{C}{x+1}$$
It is easy to find that
$A=1$,
$B=2$, and
$C=-1$.
Therefore
$$\frac{x^4+1}{x^3+x^2} = \frac{1}{x^2} + \frac{2}{x} - \frac{1}{x+1}$$
Therefore:
$$\int \frac{x^4+1}{x^3+x^2}\,dx = \int \frac{dx}{x^2} + \int \dfrac{2\,dx}{x} - \int \frac{dx}{x+1} = -\frac{1}{x} +2\log \vert x\vert - \log \vert x+1 \vert + C$$
The problem is I was supposed to find: $$\int \frac{x^4+1}{x^3+x^2}\,dx = \frac{x^2}{2} - x - \frac{1}{2} - \log \vert x \vert + 2 \log \vert x+1 \vert + C$$
Where is my mistake?
| Your A, B and C are wrong. Assuming your A,B and C are right, the integraiotn is right
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
Express last equation of system as sum of multiples of first two equations The question says to 'Express the last equation of each system as a sum of multiples of the first two equations."
System in question being:
$ x_1+x_2+x_3=1 $
$ 2x_1-x_2+3x_3=3 $
$ x_1-2x_2+2x_3=2 $
The question gives a hint saying "Label the equations, use the gaussian algorithm" and the answer is 'Eqn 3 = Eqn 2 - Eqn 1' but short of eye-balling it, I'm not sure how they deduce that after row-reducing to REF.
| You can do row reduction on the transpose:
\begin{align}
\begin{bmatrix}
1 & 2 & 1\\
1 & -1 & -2\\
1 & 3 & 2\\
1 & 3 & 2\\
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 2 & 1\\
0 & -3 & -3\\
0 & 1 & 1\\
0 & 1 & 1\\
\end{bmatrix}
&&\begin{aligned}
R_2&\gets R_2-R_1\\
R_3&\gets R_3-R_1\\
R_4&\gets R_4-R_1
\end{aligned}
\\[6px]
&\to
\begin{bmatrix}
1 & 2 & 1\\
0 & 1 & 1\\
0 & 1 & 1\\
0 & 1 & 1\\
\end{bmatrix}
&&R_2\gets-R_2/3
\\[6px]
&\to
\begin{bmatrix}
1 & 2 & 1\\
0 & 1 & 1\\
0 & 0 & 0\\
0 & 0 & 0\\
\end{bmatrix}
&&\begin{aligned}
R_3&\gets R_3-R_2\\
R_4&\gets R_4-R_2
\end{aligned}
\\[6px]
&\to
\begin{bmatrix}
1 & 0 & -1\\
0 & 1 & 1\\
0 & 0 & 0\\
0 & 0 & 0\\
\end{bmatrix}
&&R_1\gets R_1-2R_2
\end{align}
which makes clear that $C_3=-C_1+C_2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1785444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
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