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Find length BC of triangle with incircle and circumcircle
Some thoughts... some chord theorem to get the angle between PQ and BC... AB and AC are tangent to the circle, there has to be another theorem about that, perhaps ADE is isosceles and that helps looking at the angled PDE and AEQ, and with all this one should be able to figure the angle at A, and once I have that probably I can use yet another theorem on chords to determine the length BC from the length PQ...idk
|
Let $r$ and $R$ be the radii of the incircle and circumcircle, respectively. Then
$$\frac r{4R} =\sin \frac A2 \sin \frac B2 \sin \frac C2\tag1$$
Note that $ab= PD\cdot DQ =6$ and $ac= PE\cdot EQ =4$
$$r\cos \frac A2 =a \sin\frac A2 =\frac{DE}2=1$$
$$2R\sin A = BC = b+ c=\frac{6+4}{a}=10\sin\frac A2\tag2
$$
which leads to the ratio $\frac rR = \frac25$. Also
$$\tan \frac B2 = \frac r{b} = \frac{ra}6 = \frac1{3\sin A}, \>\>\>\>\>\>\>\tan \frac C2 = \frac1{2\sin A}
$$
Substitute above into (1)
$$\frac1{10} =\frac{\sin\frac A2}{\sqrt{(9\sin^2A+1)(4\sin^2A+1)}}
=\frac{\sqrt{\frac{1-\cos A}2}}{\sqrt{(10-9\cos^2 A)(5-4\cos^2A)}}\tag3
$$
which has the solution $\cos A=0$, or $\sin\frac A2 = \frac1{\sqrt2}$. Thus, per (2)
$$BC = 10\sin \frac A2 = 10\cdot \frac1{\sqrt2} = 5\sqrt2$$
Edit: as pointed out by @Intelligenti pauca, $\cos A= \frac56$ is also a valid solution to (3), which leads to
$BC = \frac5{\sqrt3}$.
|
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|
Given $x^2 +px + q$ has roots -1 & 4, find the values for p & q.
Given $x^2 +px + q$ has roots $-1$ & $4$, find the values for $p$ & $q$.
Attempt:
$$
x = \frac{-p\pm\sqrt{p^2-4q}}{2}\\(p + 2x)^2 = p^2 - 4q\\
p^2 + 4px + 4x^2 = p^2 -4q
\\
q = -x^2 - px
\\
q = -(-1)^2 -p(-1)
\\
q = -1 - p
\\
q = -(4)^2-p(4)
\\
q = -16 - 4p
\\
-1 - p = - 16 - 4p
\\
3p = -15
\\
p = -5
\\
q = -1 - (-5)
\\
q = 4
$$
According to the book the answer should be $-3$ & $-4$
|
Instead of substituting the roots later, you could have substituted them right at the start:
$(-1)^2+p·(-1)+q = 0$.
$(4)^2+p·(4)+q = 0$.
And it is easy from there.
|
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|
How to evaluate $ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$? I was given the series:
$$ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$$
Making some observations I realized that the $ a_{n} $ term would be the following:
$$ a_{n} = \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right)$$
What I wanted to do is to find the result of the series, so the answer would be:
$$\sum_{n=1}^{ \infty } \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right) = \frac{3}{2} \sum_{n=1}^{ \infty } \left[ \frac{1}{3^{n} } \right]+ \sum_{n=1}^{ \infty } \left[\frac{1}{3^{n} } \frac{(-1)^{n}}{2}\right]$$
I can tell that the first term is convergent because it is a geometric series, in fact, the result is $\frac{3}{4}$. However, I have no clue in how to solve the second term series. I should say that the series given in the beginning is convert and its result is 5/8. How to arrive to it is a mystery to me.
|
It is acceptable to split one convergent series in two as long as both are convergent.
$$ 0 < \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ... < \frac{2}{3} + \frac{2}{3^{2} } + \frac{2}{3^{3}} + \frac{2}{3^{4}} + \frac{2}{3^{5}} + ...$$
and since
$$\frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$$ is monotonically increasing as you are adding terms, and it is bounded by another convergent series (above), it is convergent itself.
Now:
$$ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...=\sum_{n=1}^{\infty} \frac{1}{3^{2n-1}}+\sum_{n=1}^{ \infty }\frac{2}{3^{2n} }=3\sum_{n=1}^{\infty} \frac{1}{3^{2n}}+2\sum_{n=1}^{ \infty }\frac{1}{3^{2n} }=5\sum_{n=1}^{ \infty }\frac{1}{9^{n} }=\frac{5}{8}$$
|
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|
Computing $\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$ I'm trying to compute
$$I=\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$$
The following is my effort,
$$I(a)=\int_0^\infty\frac{\ln x}{x^2+a^2}dx$$
Let $x=a^2/y$ so that $dx=-(a^2/y^2)dy$ which leads to
$$I(a)=\int_0^\infty \frac{\ln(a/y)}{a^2+y^2}dy=\int_0^\infty\frac{\ln a}{y^2+a^2}dy-I(a)$$
$$I(a)=\frac{1}{2}\int_0^\infty\frac{\ln a}{y^2+a^2}dy=\frac{1}{2}\frac{\ln a}{a}\arctan\left( \frac{y}{a}\right)_0^\infty=\frac{\ln a}{a}\frac{\pi }{4}$$
Differentiating with respect to $a$ then
$$\frac{dI(a)}{a}=-2aI'(a)=\frac{\pi}{4}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
where $$I'(a)=\int_0^\infty \frac{\ln y}{(y^2+a^2)^2}dx$$
$$I'(a)=\frac{\pi}{-8a}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
$$I'(a=1)=-\frac{\pi}{8}$$
But the correct answer is $-\pi/4$.
Can you help me figure where I mistake? Please give some method if there is which is much better than what I have done?
|
I would like to find a more general integral using trigonometric substitution $x=a\tan \theta.$
$\displaystyle \begin{aligned} \int_0^{\infty} \frac{\ln x}{\left(a^2+x^2\right)^2} d x &=\frac{1}{a^3} \int_0^{\frac{\pi}{2}} \cos ^2 \theta \ln (a \tan \theta) d \theta\\ &=\frac{\ln a}{a^3}\int _0^\frac{\pi}{2} \cos ^2 \theta d \theta+\frac{1}{a^3} \int^{\frac{\pi}{2}} \cos ^2 \theta \ln (\tan \theta) d \theta\\ &=\frac{\ln a}{2 a^3} \int_0^{\frac{\pi}{2}}\left(1+\cos 2 \theta\right) d \theta+\frac{1}{2a^3} \int_0^{\frac{\pi}{2}}(1+\cos 2 \theta) \ln (\tan \theta) d \theta \\ &=\frac{\pi\ln a}{4 a^3}+\frac{1}{4a^3} \int_0^{\frac{\pi}{2}} \ln (\tan \theta) d(\sin 2 \theta)\quad \left(\textrm{ By }\int_0^{\frac{\pi}{2}} \ln (\tan \theta) d \theta=0 \right)\\&=\frac{\pi \ln a}{4a^3}+\frac{1}{4a^3}[\sin 2 \theta \ln (\tan \theta)]_0^{\frac{\pi}{2}}-\frac{1}{4a^3} \int_0^{\frac{\pi}{2}} \frac{\sin 2 \theta \sec ^2 \theta}{\tan \theta} d \theta \\&=\frac{\pi}{4 a^3}(\ln a-1)\end{aligned} \tag*{} $
|
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|
Why can the Taylor formula be used to evaluate the limit,Why does its remainder converge to zero？ A simple example $\lim\limits_{x\to0}\frac{\cos(x)-1}{x^2}=\frac{1-\frac{1}{2}x^2+o(x^2)-1}{x^2}=-\frac{1}{2}$
If taken out alone, its limit is zero.How can I judge that the sum of infinite limits converges to zero.
Such as:$\lim\limits_{x\to0}\frac{o(x^2)}{x^2}=\lim\limits_{n\to\infty}[\sum\limits_{i=3}^{n} (\lim\limits_{x\to0}\frac{x^i}{x^2})]=?$
Clarification of the problem
For example $\lim\limits_{n\to\infty}(\frac{1}{n}+\frac{1}{n}+.....+\frac{1}{n})=\lim\limits_{n\to\infty} n*\frac{1}{n}=1$
In the case of an infinite number of items,I cannot simply think of adding an infinite number of zero
|
Apply the rules of asymptotic analysis:
$$\frac{\cos x-1}{x^2}=\frac{1-\frac{1}{2}x^2+o(x^2)-1}{x^2}=-\frac 12+o(1).$$
Added: an elementary way to determine the limit:
$$\frac{\cos x-1}{x^2}=\frac{\cos^2 x-1}{x^2(\cos x +1)}=-\underbrace{\frac{\sin^2x}{x^2}}_{\substack{\downarrow\\1^2=1}}\,\underbrace{\frac1{\cos x+1}}_{\substack{\downarrow\\1+1=2}}$$
|
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|
Evaluate the following limit using Taylor Evaluate the following limit:
\begin{equation*}
\lim_{x\to 0} \frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}.
\end{equation*}
I know the Taylor series of $e^x$ at $a=0$ is $\sum_{k=0}^{\infty} \frac{x^k}{k!}$. And if we substitute $x$ with $x^2$ we get $e^{x^2}=\sum_{k=0}^{\infty} \frac{x^{2k}}{k!}$. Also $\cos x=\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!}x^{2k}$ I am struggeling with finding the Taylor series for $\sin^2x^2$. So far I have this:
\begin{align*}
\sin^2 x=\frac{1}{2}-\frac{1}{2}\cos 2x.
\end{align*}
And by substituting $x$ with $2x$ in the Taylor series of $\cos x$ we get $\cos 2x=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}(2x)^{2k}$.
This gives us
\begin{align*}
\sin^2 x&=\frac{1}{2}-\frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}(2x)^{2k}\\
&=\frac{1}{2}-\frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}2^{2k}x^{2k}.
\end{align*}
And then, by substituting $x$ with $x^2$ we get
\begin{align*}
\sin^2 x^2 =\frac{1}{2}-\frac{1}{2}\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!}2^{2k}x^{4k}.
\end{align*}
Is this correct?
This would give:
\begin{align*}
&\lim_{x\to 0}\frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}\\
&=\lim_{x\to 0}\frac{1+x^2+O(|x|^4)+2-\frac{2x^2}{2!}+O(|x|^4)-3}{\frac{1}{2}-\frac{1}{2}(1-\frac{4x^4}{3!}+O(|x|^8))}\\
&=\lim_{x\to 0}\frac{O(|x|^4)}{\frac{2x^4}{3!}-O(|x|^8)}\\
&=\lim_{x\to 0}\frac{1}{\frac{2}{6}-O(|x|^4)}\\
&=\frac{1}{\frac{1}{3}}\\
&=3
\end{align*}
Could someone tell me if this is correct? Thanks!
|
Neither the numerator nor the denominator has been correctly analysed.
The numerator is $O(x^4)$, but is asymptotic not to $x^4$ as your calculation incorrectly assumes, but $\tfrac12x^4+\tfrac{1}{12}x^4=\tfrac{7}{12}x^4$.
The denominator $\sin^2x^2=\tfrac12(1-\cos2x^2)$ has leading order term $\tfrac14(2x^2)^2=x^4$. You seem to have used the Taylor series for $\sin y$, rather than $\cos y$, with $y=2x^2$. An easier approach is $\sin^2x^2\sim(x^2)^2=x^4$.
So the correct limit is $\tfrac{7}{12}$, as @Atmos said.
|
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|
Find the value of $T=\mathop {\lim }\limits_{n \to \infty } {\left( {1+ \frac{{1+\frac{1}{2}+ \frac{1}{3}+ . +\frac{1}{n}}}{{{n^2}}}} \right)^n}$ I am trying to evaluate
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n}.$$
My solution is as follow
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n} \Rightarrow T = {e^{\mathop {\lim }\limits_{n \to \infty } n\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}} - 1} \right)}} = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{n}} \right)}}$$
$$T = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{2n}} + \frac{1}{{3n}} + \cdots + \frac{1}{{{n^2}}}} \right)}} = {e^{\left( {0 + 0 + \cdots + 0} \right)}} = {e^0} = 1$$
The solution is correct but I presume my approach $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdot + \frac{1}{n}}}{n}} \right) \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{2n}} + \frac{1}{{3n}} + \cdot + \frac{1}{{{n^2}}}} \right) = 0$ is wrong.
Is there any generalized method
|
Your doubt is right - it is not possible to split limit in variable amount of summands, as shows example $1=\lim 1=\lim\left(\frac{1}{n}+\cdots + \frac{1}{n}\right)=\lim \frac{1}{n} +\cdots + \lim\frac{1}{n} = 0$.
On another way you can use representation
$$\sum_{i=1}^{n}\frac{1}{i} = \ln n + \gamma + \varepsilon_{n}= \ln n +O(1), n\rightarrow \infty$$
where $ \varepsilon_{n} \sim \frac{1}{2n}, n\rightarrow \infty$ based on Euler–Mascheroni constant $\gamma$. This also gives you $\frac{{1 + \frac{1}{2} + \frac{1}{3} + . + \frac{1}{n}}}{n} \to 0, n\rightarrow \infty$.
|
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|
Why do we multiply by $-1$ here? WolframAlpha solves $$\sin\left(x-\fracπ4\right)=\frac{1+\sqrt3}{2\sqrt2}$$ by multiplying by $-1$ as such:
$$\sin\left(-x+\fracπ4\right)=-\frac{1+\sqrt3}{2\sqrt2}$$ then arcsin both sides, etc. but why multiplying by $-1$ and not just directly find the answer? Mainly say this because I've tried without multiplying by $-1$ and haven't managed to got $\frac56 π$ which is the answer (in $0<x≤2π$)
|
Apply trig addition identity,we can do this step by step as the following (there is no requirement to multiply -1 for both sides):
\begin{align}\sin\left(x-\fracπ4\right)=\frac{1+\sqrt3}{2\sqrt2} \end{align}
\begin{align} \sin x\cos\frac{\pi}{4}-\cos x\sin \frac{π}{4}=\frac{1+\sqrt3}{2\sqrt2} \end{align}\begin{align}(\sin x-\cos x)\frac{\sqrt 2}{2}=\frac{1+\sqrt3}{2\sqrt2}\end{align} \begin{align}\sin x-\cos x=\frac{1}{2}+\frac{\sqrt3}{2} \end{align}
\begin{align}x = \frac{2\pi}{3},\frac{5\pi}{6} ( x\in (0,2\pi]) \end{align}
|
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|
Prove for every $1 \leq n$, that $47\mid3^n4^{2n}-1$ I tried to solve with induction, base $n=1$ is correct such that, $\frac{3*16-1}{47}=1$.
Then assuming that for $n$ the statement is correct such that, $3^n4^{2n} \equiv 0 \pmod {47}$.
We need prove that for $n+1$ is true such that, $\frac{(3*16*48^n)-1}{47}$ .
Then I tried to simplify it to $\frac{48}{47}*48^n-\frac{1}{47}= 1\frac{1}{47}*48^n-\frac{1}{47}=\frac{(3^n4^{2n})+47(3^n4^{2n})}{47}-\frac{1}{47}=\frac{(3^n4^{2n})}{47}-\frac{1}{47}+3^n4^{2n}$
We get that from $\frac{(3^n4^{2n})}{47}-\frac{1}{47}+3^n4^{2n}$, the expression $\frac{(3^n4^{2n})} {47}-\frac{1}{47}$ is from the assumption so we are left with $3^n4^{2n}$ which is a natural number.
Is that proof enough to show that $47\mid3^n4^{2n}-1$?
|
Transform ($3^n4^{2n}-1) \to$ ($3^n16^n-1) \to (48^n-1)$
For sum of geometric series
$1+x+x^2+x^3+...+x^{n-1}=\frac{x^n-1}{x-1}$
or
$x^n-1=(x-1)(1+x+x^2+x^3+...+x^{n-1})$
Set x=48,
$48^n-1=(48-1)(1+48+48^2+48^3+...+48^{n-1})$
=
$47 (1+48+48^2+48^3+...+48^{n-1}) \to 47\mid3^n4^{2n}-1$
|
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|
Finding $p$ and $t$ such that the system $px+y+z=1$, $x+2y+4z=t$, $x+4y+10z=t^2$ has one, infinitely-many, or no solutions
Consider the following equations:
$$\begin{align}
px+\phantom{2}y+\phantom{10}z &=1 \\
x+2y+\phantom{1}4z &=t \\
x+4y+10z &=t^2
\end{align}$$
Now find the values of $p$ and $t$ for which
(a) there is a unique solution
(b) there are infinitely many solutions
(c) there is no solution
My attempt: For (a) we require the determinant $\begin{vmatrix}p&1&1\\1&2&4\\1&4&10\end{vmatrix}\neq 0$. Hence $p\neq 1$.
For (b) we need two of the equations to represent the same line - I tried subbing in the difference between the second and third equation into the first but got nowhere. Similarly for (c).
|
System of equations give the below matrix:
$\left(\begin{array}{rrr|r}
p & 1 & 1 & 1\\
1 & 2 & 4 & t\\
1 & 4 & 10 & t^2
\end{array}\right)$
We set third row as $(R_3 - 2 \cdot R_2)$ and second row as $(R_2 - 2 \cdot R_1)$.
$\left(\begin{array}{rrr|r}
p & 1 & 1 & 1\\
1-2p & 0 & 2 & t-2\\
-1 & 0 & 2 & t^2-2t
\end{array}\right)$
If $p = 1$, $t$ must be either $1$ or $2$ for row $2$ and $3$ to be consistent. Also note that it leads to only two equations in $3$ variables.
So we have no solutions, if $p = 1, t \notin (1,2)$
If $p = 1, t \in (1,2)$, we have infinite solutions
If $p \ne 1$, we have unique solution.
|
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|
In checking whether $\tan^{-1}(2\sin^2x-1)+C$ agrees with $\tan^{-1}(\tan^2x)+C$, why am I getting $\sin^{2} 2x = 2$? So this began with the integral-
$$\int \frac{\sin 2x}{\sin^4x + \cos^4x}dx$$
The integral is simple, I solved it as following-
$$\int \frac{\sin 2x}{\sin^4x + \cos^4x}dx=\int \frac{2\sin x\cos x}{\sin^4x + (1-\sin^2x)^2}dx$$
Put $\sin^2 x=t$
$2\sin x\cos x\ dx = dt$
$$\int \frac{dt}{t^2+(1-t)^2}=\frac{1}{2}\int \frac{dt}{(t-\frac{1}{2})^2 + (\frac{1}{2})^2}$$
Now put
$t-1/2 = m$
$dt=dm$
$$\frac{1}{2}\int \frac{dm}{m^2 + (\frac{1}{2})^2}$$
Which by standard formula is equal to
$$\frac{1}{2}\left( \frac{1}{\frac{1}{2}}tan^{-1}\frac{m}{\frac{1}{2}}\right)+C=tan^{-1}(2t-1)+C=tan^{-1}(2\sin^2 x-1)+C$$
Now the answer according to my worksheet and this page the answer is $tan^{-1}(\tan ^2 x)+C$
Now my answer clearly doesn't match with the given answer so I check whether these two are equal or not-
$2\sin^2 x-1=tan^2 x$
$2\sin^2 x=1+tan^2 x$
$2\sin^ 2 x=\sec^2 x$
$2\sin^ 2 x=\frac{1}{\cos^2 x}$
$2\sin^ 2 x \cos^2 x=1$
$4\sin^ 2 x \cos^2 x=2$
$\sin^2 2x=2$
This is impossible and I know I have done something wrong but I cross checked it every time but couldn't find the error. Kindly help me out.
EDIT-The standard formula I used above-
$$\int \frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+C$$
|
Using
$$\tan(A-B) = \frac{\tan A - \tan B}{1+\tan A \tan B}$$
we have that
$$\tan\left(\tan^{-1}(\tan^2 x) - \tan^{-1}\left(2\sin^2 x - 1\right)\right) = 1$$
So the two solutions are off by a $\frac{\pi}{4}$
|
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|
Given a triangle $\triangle ABC$, which fulfills the equations $6\cos{A}+2\cos{B}=7$ and $6\sin{A}-2\sin{B}=\sqrt{3}$, find the value of $\angle C$ Given a triangle $\triangle ABC$, which fulfills the equations $6\cos{A}+2\cos{B}=7$ and $6\sin{A}-2\sin{B}=\sqrt{3}$, find the value of $\angle C$.
I though of solving it as follows:
$36\cos^2{A}+24\cos{B}\cos{A}+4\cos^2{B}=49$
$36\sin^2{A}-24\sin{B}\sin{A}+4\sin^2{B}=3$
Hence $12=24(\cos{A}\cos{B}-\sin{B}\sin{A})$
And I got stuck here. I then tried putting in different values for $\angle A, \angle B$ to find a pair that works, but I couldn't find one. Could you please explain to me how to solve the question?
|
$6\cos{A}+2\cos{B}=7$ and $6\sin{A}-2\sin{B}=\sqrt{3}$
Squaring and adding these two equations, we get
$$36(\sin^2 A+\cos^2 A)+4(\cos^2 B+\sin^2 b)+24(\cos A \cos B-\sin A \sin B)=52$$
$$\implies 36+4+24\sin (A+B)=52 \implies \cos(A+B)=1/2=-\cos C \implies C= 2\pi/3$$
|
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|
Where does the plus-minus come from in the quadratic formula? In the formula
$$x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
for solving quadratic equations, where does the $\pm$ come from?
The square root already results in both a positive and negative term, is the $\pm$ not therefore extraneous?
|
What about completing squares?
\begin{align*}
ax^{2} + bx + c & = a\left(x^{2} + \frac{bx}{a}\right) + c\\\\
& = a\left(x^{2} + \frac{bx}{a} + \frac{b^{2}}{4a^{2}}\right) + c - \frac{b^{2}}{4a}\\\\
& = a\left(x + \frac{b}{2a}\right)^{2} + \frac{4ac - b^{2}}{4a}
\end{align*}
Hence we deduce that
\begin{align*}
ax^{2} + bx + c = 0 & \Longleftrightarrow a\left(x + \frac{b}{2a}\right)^{2} + \frac{4ac - b^{2}}{4a} = 0\\\\
& \Longleftrightarrow \left(x + \frac{b}{2a}\right)^{2} = \frac{b^{2} - 4ac}{4a^{2}}\\\\\
& \Longleftrightarrow x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}
\end{align*}
and we are done.
Hopefully this helps!
|
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|
How can I prove this formula $\int_0^\infty f\left( a^2x^2+\frac{b^2}{x^2}\right)\mathrm{d}x=\frac{1}{a}\int_0^\infty f(x^2+2ab)\mathrm{d}x$? While solving this trigonometric integral :
$$\int_0^\infty \sin\left(a^2x^2+\frac{b^2}{x^2}\right)\mathrm{d}x$$
I came across this formula :
$$\int_0^\infty f\left( a^2x^2+\frac{b^2}{x^2}\right)\mathrm{d}x=\frac{1}{a}\int_0^\infty f(x^2+2ab)\mathrm{d}x$$
And I'm wondering if I can prove it, but I hadn't any idea, because it seems like it depends on the properties of the function $f$.
|
With $a,\>b>0$
\begin{align}
\int_0^\infty f\left( a^2x^2+\frac{b^2}{x^2}\right){d}x
\overset{x\to \frac b{ax}}= &
\int_0^\infty f\left( a^2x^2+\frac{b^2}{x^2}\right) \frac b{ax^2} dx\\
= & \frac12 \int_0^\infty f\left( a^2x^2+\frac{b^2}{x^2}\right) \left(1+\frac b{ax^2}\right) dx\\
= & \frac1{2a}\int_0^\infty f\left( (ax-\frac bx)^2+2ab\right) d\left(ax-\frac b{x}\right)\\
=& \frac1{2a}\int_{-\infty}^\infty f(x^2+2ab)dx\\
=&\frac1{a}\int_{0}^\infty f(x^2+2ab)dx
\end{align}
|
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|
How many five-digit numbers can be formed using digits $1,2,3,4,5,6$ which are divisible by $3$ and $5$? How many five-digit numbers can be formed using digits $1,2,3,4,5,6$ which are divisible by $3$ and $5$, without any of the digits repeating?
A number is divisible by $5$ if and only if last digit is $0$ or $5$ and; divisible by $3$ if and only if the sum of the digits add to a multiple of $3$ ($3,6,9,12,15,18,21$).
Divisible by $5$ means last digit is $5$. That's the only option.
Divisible by $3$ means that the sum of the digits add up to a multiple of $3$.
What can we do next? It's difficult for me to find the 5-tuples just by looking at the given digits? What can I do to systematize them and not miss some?
|
hint
The mostright digit should be $ 5$.
now, we have to choose $ 4 $ digits from $5 $. there are five possibilities
$$\{1,2,3,4\}\;;\;\{1,3,4,6\}\;;\;\{1,2,4,6\}\;;\;\{2,3,4,6\}\;;\;\{1,2,3,6\}$$
take those for which the (sum +5) is divisible by $ 3$.
we find
$$\{1,2,4,6,5\}\;;\;\{1,2,3,4,5\}$$
The final answer is $2\times 4!=48$.
|
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|
Geometric Proof of $a^2-b^2=(a+b)(a-b )$ and its applications I am teaching a student on the subject of factoring. One commonly used formula is $a^2-b^2=(a+b)(a-b) $. It is easy to prove it from RHS to LHS algebraically, but how to prove it geometrically? I would also like to find some of its applications, such as this:$(\sin\theta)^2+(\cos\theta)^2=( \cos \theta +i\sin \theta)(\cos\theta -i\sin \theta )=e^{i\theta}\cdot e^{-i\theta } =e^0=1$
Are there any other good examples?
EDIT: some comments for applications
(Albus) $\frac{1}{\sqrt a+b}=\frac{\sqrt a-b}{a-b^2}$
(Paul) $a\times b=(\frac{a+b}{2})^2-(\frac{a-b}{2})^2$
(other ) $x=x’+y’, y=x’-y’, xy=x’^2-y’^2=1$ Hyperbola equation
|
Hint:
$$a^2 - b^2 = (a-b)b + (a-b)b + (a-b)^2 $$
$$ = (a-b)[b + b + (a-b)] $$
or
$$a^2 - b^2 = (a-b)a + (a-b)b$$
$$ = (a-b)(a+ b) $$
Draw two squares inside each other ($a>b$) and look at the three or two leftover rectangular areas.
|
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|
Find the value of $A=\frac{3\sqrt{8+2\sqrt7}}{\sqrt{8-2\sqrt{7}}}-\frac{\sqrt{2\left(3+\sqrt7\right)}}{\sqrt{3-\sqrt7}}$ Find the value of $$A=\dfrac{3\sqrt{8+2\sqrt7}}{\sqrt{8-2\sqrt{7}}}-\dfrac{\sqrt{2\left(3+\sqrt7\right)}}{\sqrt{3-\sqrt7}}.$$
Answer: $A=1$
Since $$8+2\sqrt7=8+2\cdot1\cdot\sqrt{7}=1^2+2\cdot1\cdot\sqrt7+\left(\sqrt{7}\right)^2=(1+\sqrt{7})^2,\\8-2\sqrt{7}=(1-\sqrt{7})^2,$$
then $$A=\dfrac{3(1+\sqrt{7})}{\sqrt{7}-1}-\dfrac{\sqrt{2\left(3+\sqrt7\right)}}{\sqrt{3-\sqrt7}}.$$
What can we do with the second fraction?
|
$\begin{align}&A = \frac{3\sqrt{8+2\sqrt7}}{\sqrt{8-2\sqrt7}}\frac{\sqrt{8+2\sqrt7}}{\sqrt{8+2\sqrt7}}- \frac{\sqrt2\sqrt{3+\sqrt7}}{\sqrt{3-\sqrt7}}\frac{\sqrt{3+\sqrt7}}{\sqrt{3+\sqrt7}}\\\Rightarrow&A=\frac{3(8+2\sqrt7)}{6}-\sqrt2\left(\frac{3 +\sqrt7}{\sqrt2}\right) = 4+\sqrt7-3-\sqrt7=1\end{align}$
|
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|
If x, y, w, z >0 and $x^4$+$y^4$+$w^4$+$z^4$ <=4 prove 1/$x^4$+1/$y^4$+1/$w^4$+1/$z^4$>=4 I would appreciate suggestions to solve:
If x, y, w, z > 0 and $x^4$ + $y^4$ + $w^4$ + $z^4$ <=4 prove the following:
1/$x^4$ + 1/$y^4$ + 1/$w^4$ + 1/$z^4$ >= 4
From plugging in numbers into Excel, it looks like x, y, w, z must be numbers near 1.
I tried to do a simple case:
If a, b, k > 0 and a + b < k prove the following: 1/a + 1/b > k
a + b < k so 1/(a+b) > 1/k
But 1/a + 1/b > 1/(a+b) so 1/a + 1/b > 1/k which is not what I want to prove.
It seems that I would have to impose a condition for the possible values of k.
|
Remember the Cauchy-Schwarz inequality:
$$(x_1y_1+ \ldots +x_ny_n)^2\leq (x_1^2+ \ldots + x_n^2)(y_1^2+\ldots +y_n^2)$$
In this case we have $n=4$. Set $x_1=x^2, x_2=y^2, x_3=z^2, x_4=w^2$ and $y_1=1/x^2, y_2=1/y^2, y_3=1/z^2, y_4=1/w^2$. Now, applying the inequality we have
$$4^2 \leq (x^4+y^4+z^4+w^4)\left( \frac{1}{x^4}+\frac{1}{y^4} + \frac{1}{z^4}+\frac{1}{w^4} \right).$$
Hence,
$$\frac{16}{x^4+y^4+z^4+w^4} \leq \left( \frac{1}{x^4}+\frac{1}{y^4} + \frac{1}{z^4}+\frac{1}{w^4} \right)$$
Now, by hypotesis, we have $x^4+y^4+z^4+w^4\leq 4$, hence $$\frac{1}{x^4+y^4+z^4+w^4}\geq \frac14.$$
Can you conclude?
|
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|
Calculate $\sum_{i=1}^{n-1} i\alpha^{2i}$ I try to calculate $\sum_{i=1}^{n-1} i\alpha^{2i}$
I think one week but still have no ideas to this sigma,thank everyone
|
Call $S = \sum_{i=1}^{n-1} i \alpha^{2i}=\sum_{i=1}^{n-1} i (\alpha^2)^{i}$ and $\alpha^2 = \beta$. If $\beta \neq 1$, then \begin{align} \beta S = \sum_{i=1}^{n-1} i \beta^{i+1} &= \sum_{i=1}^{n-1}\left((i+1)-1\right) \beta^{i+1} \\ &=\sum_{i=1}^{n-1} (i+1)\beta^{i+1} - \sum_{i=1}^{n-1} \beta^{i+1} \\ &= \sum_{i=2}^{n}i \beta^{i} - \sum_{i=2}^{n}\beta^{i} \\ &= \left( n\beta^n-\beta \right) +\sum_{i=1}^{n-1}i\beta^{i} - \sum_{i=2}^{n}\beta^{i} \\ &= \left( n\beta^n-\beta \right) + S - \frac{\beta^{n+1} - \beta^2 }{\beta -1 }
\end{align}
Thus $$ (\beta - 1)S = n\beta^{n}-\beta - \frac{\beta^{n+1} - \beta^2}{\beta -1 } $$
or \begin{align} S &= \frac{n\beta^n -\beta}{\beta -1} - \frac{\beta^{n+1} - \beta^2 }{(\beta - 1)^2
} \\ &=\frac{(n-1)\beta^{n+1} - n\beta^n + \beta}{ (\beta - 1)^2} \\& = \frac{(n-1)\alpha^{2n+2} - n\alpha^{2n} + \alpha^2}{ (\alpha^2 - 1)^2}
\end{align}
If $\beta = \alpha^2 = 1$, then $S=\sum_{i=1}^{n-1} i (\alpha^2)^{i} = \sum_{i=1}^{n-1}i=\frac{(n-1)n}{2}$.
|
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|
How can I solve $x\left(y^2+z\right)z_x-y\left(x^2+z\right)z_y=\left(x^2-y^2\right)z$? I have this right now:
$$x\left(y^2+z\right)z_x-y\left(x^2+z\right)z_y=\left(x^2-y^2\right)z$$
$$\frac{dx}{x\left(y^2+z\right)}=\frac{dy}{-y\left(x^2+z\right)}=\frac{dz}{\left(x^2-y^2\right)z}$$
I get the first first integral like this:
$$\frac{xdx}{x^2\left(y^2+z\right)}=\frac{ydy}{-y^2\left(x^2+z\right)}=\frac{-dz}{-\left(x^2-y^2\right)z}$$
$$\frac{d\left(\frac{1}{2}x^2\right)}{x^2y^2+x^2z}=\frac{d\left(\frac{1}{2}y^2\right)}{-y^2x^2-y^2z}=\frac{d\left(-z\right)}{-x^2z+y^2z}$$
$$\frac{d\left(\frac{1}{2}x^2+\frac{1}{2}y^2-z\right)}{0}=ds$$
$$d\left(\frac{1}{2}x^2+\frac{1}{2}y^2-z\right)=0$$
$$\frac{1}{2}x^2+\frac{1}{2}y^2-z=C$$
$$x^2+y^2-2z=C_1$$
$$\Psi _1\left(x,y,z\right)=x^2+y^2-2z$$
But I am not sure how to get the second first integral
I tried using $z=\frac{1}{2}x^2+\frac{1}{2}y^2-C$ when doing:
$$\frac{dx-dy}{x\left(y^2+z\right)+y\left(x^2+z\right)}=\frac{dz}{\left(x^2-y^2\right)z}$$
$$\frac{d\left(x-y\right)}{\frac{1}{2}\left(x+y\right)^3-C\left(x+y\right)}=\frac{dz}{\left(x+y\right)\left(x-y\right)z}$$
$$\frac{2d\left(x-y\right)}{\left(x+y\right)^2-2C}=\frac{dz}{\left(x-y\right)z}$$
$$\frac{2\left(x-y\right)d\left(x-y\right)}{\left(x+y\right)^2-2C}=\frac{dz}{z}$$
Let $w=\left(x-y\right)$ and $C=\frac{1}{2}x^2+\frac{1}{2}y^2-z$, then:
$$\frac{2w\:dw}{4z-w^2}=\frac{dz}{z}$$
Then we do for $v=w^2$:
$$\frac{\:dv}{4z-v}=\frac{dz}{z}$$
Which we then solve:
$$\frac{\:dv}{dz}=4-\frac{v}{z}$$
Using $p=\frac{v}{z}$ we get:
$$p+p'z=4-p$$
$$-\frac{1}{2}\frac{dp}{p-2}=\frac{dz}{z}$$
Integrating:
$$-\frac{1}{2}ln\left|p-2\right|+C_2=ln\left|z\right|$$
$$ln\left|\frac{1}{p+2}\right|+C_2=ln\left|z^2\right|$$
$$\frac{C_2}{p+2}=z^2$$
$$C_2=z^2\left(\left(x-y\right)^2+2\right)$$
Is there an easier way to calculate the second first integral? I don't see any mistake in my calculations but still it's very very long
|
In the same way you found the first identity you can also sum up
$$
ds=\frac{dx/x }{y^2+z}=-\frac{dy/y}{x^2+z}=\frac{dz/z}{x^2-y^2}
$$
to get
$$
ds=\frac{dx/x+dy/y+dz/z}{0}\implies xyz=C_2
$$
|
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|
Minimize $3\sqrt{5-2x}+\sqrt{13-6y}$ subject to $x^2+y^2=4$
If $x, y \in \mathbb{R}$ such that $x^2+y^2=4$, find the minimum value of $3\sqrt{5-2x}+\sqrt{13-6y}$.
I could observe that we can write
$$3\sqrt{5-2x}+\sqrt{13-6y}=3\sqrt{x^2+y^2+1-2x}+\sqrt{x^2+y^2+9-6y}$$
$\implies$
$$3\sqrt{5-2x}+\sqrt{13-6y}=3\sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y-3)^2}=3PA+PB$$
Where $P$ is a generic point on $x^2+y^2=4$ and $A(1,0),B(0,3)$.
So the problem essentially means which point on the circle $x^2+y^2=4$ minimizes $3PA+PB$.
I am really struggling to find the geometrical notion of this and hence unable to solve.
|
Comment: you may use this modification:
We rewrite final relation as:
$$A=\sqrt{[3(x-1)=a]^2+(3y=b)^2}+\sqrt{(x=a')^2+(y-3=b')^2}$$
and use this inequality:
$\sqrt{a^2+b^2}+\sqrt{a'^2+b'^2}\geq\sqrt{(a+a')^2+(b+b'^2)}$
we get:
$A\geq\sqrt{16(x^2+y^2)-24(x+y)+18}$
$x^2+y^2=4$
$\Rightarrow$
$A\geq \sqrt{82-24(x+y)}$
If $x=y=\sqrt 2$ then $A\geq\sqrt{84-67.7}\approx 4$
Update:Wolfram says minimum is $2\sqrt {10}=6.32$ at $(x,y)=(\frac25+\frac{3\sqrt6}5=1.87, \frac 65-\frac{\sqrt6}5)=0.71$. If we put this in $\sqrt{82-24(x+y)}$ we get 4.47. mind you x and y must suffice $x^2+y^2=4$, and what Wolfram gives does; $1.87^2+0.71^2=4$.Hence 40 can not be minimum .
|
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|
Integrate $\int\frac{x}{(x^4-1)\cdot\sqrt{x^2+1}} \, dx$ Integrate $\int\frac{x}{(x^4-1)\cdot\sqrt{x^2+1}} \, dx$
I tried substituting $x=\tan(t)$ in order to get away with square root. ($\:dx=\frac{1}{\cos^2(t)}dt\:$)
$\sqrt{x^2+1}=\frac{1}{\cos(t)}\:\:$ and $\:\:x^4-1=\frac{\sin^4(t)-\cos^4(t)}{\cos^4(t)}$
Now after putting both into main Integral and by simplifying I have :
$$
\int\frac{\sin(t)\cdot\cos^2(t)}{\sin^4(t)-\cos^4(t)} \, dt=\text{?}
$$
Now need a bit help if possible.
Thank you in advance :)
|
Note
\begin{align}
\int\frac{\sin t\cos^2t}{\sin^4t-\cos^4t} \, dt
= &\int\frac{\sin t\cos^2t}{(1-\cos^2t)^2-\cos^4t} \, dt
=\int\frac{\sin t\cos^2t}{1-2\cos^2t} \, dt\\
= & \frac12 \int\left( 1- \frac{1}{1-2\cos^2t} \right) \, d(\cos t)\\
=&\frac12\cos t -\frac1{4\sqrt2}\ln \frac{1+\sqrt2\cos t}{1-\sqrt2\cos t}
\end{align}
|
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|
Integrate $\int\frac{3x^2-1}{\sqrt{x^2+x-1}}dx$ Integrate $\int\frac{3x^2-1}{\sqrt{x^2+x-1}}dx$
I solved this integral by euler substitution by replacing
$\sqrt{x^2+x-1}=x+t$
but it's not allowed by the problem.
p.s Is there any other method to solve with?
Thank you in advance :)
|
Hint:
We observe that
$$p':=\left(\sqrt{x^2+x-1}\right)'=\frac{2x+1}{2\sqrt{x^2+x-1}},$$
$$q':=\left(x{\sqrt{x^2+x-1}}\right)'=\frac{4x^2+3x-2}{2\sqrt{x^2+x-1}}.$$
Also, by completing the square,
$$x^2+x-1=\left(x+\frac12\right)^2-\frac54=\frac54\left(\left(\frac{2x+1}{\sqrt5}\right)^2-1\right)$$
and
$$r':=\left(\text{arcosh}\frac{2x+1}{\sqrt5}\right)'=\frac1{\sqrt{x^2+x-1}}$$
Hence we form a linear combination to match the numerator of the integrand and we find
$$\frac{3q'}2-\frac{9p'}4+\frac{13r'}8=\frac{3x^2-1}{\sqrt{x^2+x-1}}.$$
|
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|
Evaluate $\int \frac{x^3+4x^2+x-1}{x^3+x^2}dx$ Evaluate $\int \frac{x^3+4x^2+x-1}{x^3+x^2}dx$.
Where do I start with this integral? I can easily see that it is possible to fator $x^{2}$ out on the denominator and use partial fractions. The numerator is also factorable but it does not have any integer roots. Can someone show me how to integrate this (either using what I said about partial fractions), or another method? Thanks
|
\begin{align}
\frac{x^3+4x^2+x-1}{x^3+x^2}
&= \frac{\color{red}{x^3 + x^2} + 3x^2+x-1}{x^3+x^2} \\
&= \color{red}{1} + \frac{3x^2+x-1}{x^3+x^2} \\
\end{align}
Now notice that the derivative of the denominator is almost the numerator (but not quite). But let's start with:
\begin{align}
\int \frac{3x^2+x-1}{x^3+x^2}dx
&= \int\frac{3x^2+2x}{x^3+x^2} - \frac{x+1}{x^3+x^2}dx \\
&= \ln\left( x^3 + x^2 \right) - \int \frac{\cancel{x+1}}{x^2\cancel{(x+1)}} dx
\end{align}
That's a scattered answer, but it has all the ingredients.
|
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|
How do I solve for z for this inverse matrix? How do I go about solving this for z?
$$
\begin{pmatrix}
a & b & c \\
1 & 2 & 3 \\
d & e & f \\
\end{pmatrix}
^{-1}= \begin{pmatrix}
1 & 2 & 3 \\
x & y & z \\
4 & 5 & 6 \\
\end{pmatrix}
$$
|
As :
$$\underbrace{\begin{pmatrix}
a & b & c \\
\color{red}{1} & \color{red}{2} & \color{red}{3} \\
d & e & f \\
\end{pmatrix}}_A
\underbrace{\begin{pmatrix}
1 & 2 & \color{red}{3} \\
x & y & \color{red}{z} \\
4 & 5 & \color{red}{6} \\
\end{pmatrix}}_B=
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & \color{red}{0} \\
0 & 0 & 1 \\
\end{pmatrix}$$
multiplication of the second row of $A$ by the third column of $B$ will give $\color{red}{0}$, i.e.,
$$21+2z=0 \implies z=-21/2$$
Besides, it isn't evident a priori that a global solution exists with these constraints. I have checked using a Computer Algebra System that there is indeed a unique solution with this data:
$$a=-3/2, b=-1, c=-1, d=1/6, e=-1, f=-5/3, x=-13/2, y=-8, z=-21/2$$
|
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|
Two form surface integral over sphere I'm trying to compute $\int_M \omega$ with
\begin{align*}
\omega &= x^4 dy \wedge dz + y^4 dz \wedge dx + z^4 dx \wedge dy, \\
M &: x^2 + y^2 + z^2 = R^2.
\end{align*}
I have done this in two ways: Stokes' theorem and direct computation of the wedge product via a spherical coordinate transformation. Using $\phi$ as azimuthal and $\theta$ as polar angle and applying Stokes I obtain
\begin{align*}
4 \int_0^{2 \pi}\int_0^{\pi} (x^3 + y^3 + z^3) R^2 \sin \theta d\theta d\phi = 0,
\end{align*}
where I used the standard
\begin{align*}
x &= R \sin \theta \cos \phi \\
y &= R \sin \theta \sin \phi \\
z &= R \cos \theta.
\end{align*}
By directly computing the wedge products and plugging them in I obtain
\begin{align*}
dy \wedge dz &= R^2 \sin^2 \theta \cos \phi d\theta \wedge d\phi \\
dz \wedge dx &= R^2 \sin^2 \theta \sin \phi d\theta \wedge d\phi \\
dx \wedge dy &= R^2 \cos \theta \sin \theta d\theta \wedge d\phi \\
\int_M x^4 dy &\wedge dz + y^4 dz \wedge dx + z^4 dx \wedge dy = 0.
\end{align*}
The solution given by my professor was $\frac{12}{5} \pi^2 R^5$, which neither method agrees with. Then I repeat this procedure but instead of taking a sphere, I take a hemisphere. Changing the limits in the $\theta$ integrals to $[0, \frac{\pi}{2}]$, the Stokes' method yields $2 \pi R^5$ and the direct wedge computation yields $\frac{\pi R^6}{3}$. The solution given then is just $\frac{6}{5} \pi^2 R^5$, which also does not agree with my solutions. Any help with my mistakes would be greatly appreciated.
|
In case of sphere, the surface integral is indeed zero.
Please note in below working of mine, I use $\theta$ as azimuthal angle and $\phi$ as polar. I think you have the other way round. So want to call out to avoid confusion.
In case of hemisphere above $z = 0$,
Parametrization of sphere is $r(\phi, \theta) = (R \cos \theta \sin \phi, R \sin\theta \sin\phi, R \cos\phi)$
For outward normal vector, $\displaystyle r'_{\phi} \times r'_\theta = R^2 \sin\phi(\cos\theta \sin\phi, \sin\theta \sin\phi, \cos\phi)$
The integral becomes,
$\displaystyle \small \int_0^{2\pi}\int_0^{\pi/2} R^2 \sin\phi \ (R^4 \cos^4\theta\sin^4\phi, R^4 \sin^4\theta\sin^4\phi, R^4 \cos^4\phi) \cdot (\cos\theta \sin\phi, \sin\theta \sin\phi, \cos\phi) \ d\phi \ d\theta$
$= \displaystyle \frac{\pi R^6}{3} \ $, which is same as your working.
In case of applying divergence theorem, we close the surface with a disk at $z = 0$.
$\small \nabla \cdot (x^4, y^4, z^4) = 4(x^3+y^3+z^3) = 4r^3(\cos^3\theta \sin^3\phi+ \sin^3\theta \sin^3\phi + \cos^3\phi)$
where $0 \leq r \leq R$.
So volume integral is,
$\displaystyle \small \int_0^{2\pi} \int_0^{\pi/2} \int_0^R 4r^5 \sin\phi (\cos^3\theta \sin^3\phi+ \sin^3\theta \sin^3\phi + \cos^3\phi) \ dr \ d\phi \ d\theta = \frac{\pi R^6}{3}$.
Now note the flux through the bottom disk is zero so flux through spherical surface is $\frac{\pi R^6}{3}$, the same answer we received through direct surface integral.
|
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"url": "https://math.stackexchange.com/questions/4111353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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|
Solve the equation $\frac{x-13}{x-14}-\frac{x-15}{x-16}=-\frac{1}{12}$ Solve the equation $$\dfrac{x-13}{x-14}-\dfrac{x-15}{x-16}=-\dfrac{1}{12}.$$
For $x\ne14$ and $x\ne 16$ by multiplying the whole equation by $$12(x-14)(x-16)$$ we get: $$12(x-16)(x-13)-12(x-14)(x-15)=-(x-14)(x-16).$$ This doesn't look very nice. Can we do something else at the beginning? $$x-14=(x-13)-1\\x-15=(x-14)-1\\..?$$
|
$$\frac{x-13}{x-14} - \frac{x-15}{x-16} = -\frac{1}{12}$$
Let $x = t+ 15$
\begin{align}
\frac{t+2}{t+1} - \frac{t}{t-1} &= -\frac{1}{12} \\
\dfrac{(t^2+t-2) - (t^2+t)}{t^2-1} &= -\frac{1}{12} \\
\dfrac{-2}{t^2-1} &= -\frac{1}{12} \\
t^2-1 &= 24 \\
t^2 &= 25 \\
t &= \pm 5 \\
x &= 20, 10
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4117579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
How the right term of the derivative is gained? This deduction is one of the typical ones I think.
What I want to deduce is the right term from the left term of the below equation.
$$\frac{d}{dx}\left(\log\left(\frac{a+\sqrt{a^{2}+x^{2}}}{x}\right)\right)=\frac{-a}{x\sqrt{a^{2}+x^{2}}}$$
$\frac{d}{dx}\left(\log\left(\frac{a+\sqrt{a^{2}+x^{2}}}{x}\right)\right)$
$=\left(\frac{x}{a+\sqrt{a^{2}+x^{2}}}\right)\frac{d}{dx}\left(\frac{a+\sqrt{a^{2}+x^{2}}}{x}\right)$
$=\left(\frac{x}{a+\sqrt{a^{2}+x^{2}}}\right)\frac{d}{dx}\left(x^{-1}\left(a+\sqrt{a^{2}+x^{2}}\right)\right)$
$=\left(\frac{x}{a+\sqrt{a^{2}+x^{2}}}\right)\left((x^{-1})'\left(a+\sqrt{a^{2}+x^{2}}\right)\right)\left(x^{-1}\frac{d}{dx}\left(a+\sqrt{a^{2}+x^{2}}\right)\right)$
$=\left(\frac{x}{a+\sqrt{a^{2}+x^{2}}}\right)\left((-1\cdot x^{-2})\left(a+\sqrt{a^{2}+x^{2}}\right)\right)\left(x^{-1}\frac{d}{dx}\left(\sqrt{a^{2}+x^{2}}\right)\right)$
$=\left(\frac{x}{a+\sqrt{a^{2}+x^{2}}}\right)\left(-x^{-2}\left(a+\sqrt{a^{2}+x^{2}}\right)\right)\left(x^{-1}\frac{d}{dx}\left(\left(a^{2}+x^{2}\right)^{\frac{1}{2}}\right)\right)$
Anyone deduced it in someday?
ps. I have to go to work. Back after about 7hours.
|
$$f(x)=\frac{d}{dx}\left(\log\left(\frac{a+\sqrt{a^{2}+x^{2}}}{x}\right)\right)=\frac{d}{dx}\left(\log\left(a+\sqrt{a^{2}+x^{2}}\right)\right)-\frac 1x$$
$$\frac{d}{dx}\left(\log\left(a+\sqrt{a^{2}+x^{2}}\right)\right)=\frac{\Big[a+\sqrt{a^{2}+x^{2}}\Big]' }{ a+\sqrt{a^{2}+x^{2}}}$$
$$\Big[a+\sqrt{a^{2}+x^{2}}\Big]'=\frac{x}{\sqrt{a^2+x^2}}$$
$$f(x)=\frac{x}{\sqrt{a^2+x^2} \left(\sqrt{a^2+x^2}+a\right)}-\frac{1}{x}$$
$$f(x)=\frac{x\left(\sqrt{a^2+x^2}-a\right)}{\sqrt{a^2+x^2} \left(\sqrt{a^2+x^2}+a\right)\left(\sqrt{a^2+x^2}-a\right)}-\frac{1}{x}$$
Just finish.
|
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"url": "https://math.stackexchange.com/questions/4120409",
"timestamp": "2023-03-29T00:00:00",
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|
When ${u_n}$ is converge, find $u_0$ and solve $\lim(nu_n)$ with $ {u_n} : 2u_{n+1}-2u_n+u_n^2=0$ With ${u_n}: 2u_{n+1}-2u_n+u_n^2=0$. its show $${u_{n + 1}} - {u_n} = \dfrac{{ - u_n^2 + 2{u_n}}}{2} - {u_n} = \dfrac{- u_n^2}{2} \leqslant 0$$
${u_n}$ is decrease.
When $n \to \infty $ we have $\lim(u_n)=0$ (it's happen when ${u_n}$ is converge).
Also have
$${u_{n + 1}} = \frac{{ - u_n^2 + 2{u_n} - 1}}{2} + \frac{1}{2} = - \frac{{{{\left( {{u_n} - 1} \right)}^2}}}{2} + \frac{1}{2} \leqslant \frac{1}{2}$$
It's true with ${u_{n+1}}$ so $u_n\leqslant\dfrac{1}{2}$. Therefore I guest $u_0=\dfrac{1}{2}$ but i have no idea no prove it and how to find $\lim(nu_n)$.
|
If the $\left\{ u_n \right\} $ is converge,let $\displaystyle\lim_{n\rightarrow \infty} u_n=A$,we can get that$$A=-\frac{A^2}{2}-A\Rightarrow A=0 \text{or} -4$$
i.e. $\displaystyle\lim_{n\rightarrow \infty} u_n=0$ or $\displaystyle\lim_{n\rightarrow \infty} u_n=-4$.
Due to $\displaystyle u_{n+1}-u_n=-\frac{u_{n}^{2}}{2}\leqslant 0$
If we want to the $\left\{ u_n \right\} $ is converge,we should let the $\left\{ u_n \right\} $ has the lower bound.
$$u_{n+1}=-\frac{u_{n}^{2}}{2}+u_n=-\frac{\left( u_n-1 \right) ^2-1}{2}=-\frac{\left( u_n-1 \right) ^2}{2}+\frac{1}{2}$$
If $\displaystyle 0<u_0<2$,then $\displaystyle 0<u_1<\frac{1}{2}$,so on and so forth,we can see that $\displaystyle 0<u_n<\frac{1}{2}$.
In this case,the $\left\{ u_n \right\} $ is converge($\displaystyle \lim_{n\rightarrow \infty} u_n=0$),and we come to calculate the $\displaystyle\lim_{n\rightarrow \infty} nu_n$($\text{Stolz}$).
$$
\begin{align*}
\lim_{n\rightarrow \infty} nu_n&=\lim_{n\rightarrow \infty} \frac{n}{\frac{1}{u_n}}
\\
&=\lim_{n\rightarrow \infty} \frac{1}{\frac{1}{u_{n+1}}-\frac{1}{u_n}}
\\
&=\lim_{n\rightarrow \infty} \frac{1}{\frac{2}{2u_n-u_{n}^{2}}-\frac{1}{u_n}}
\\
&=\lim_{n\rightarrow \infty} \frac{1}{\frac{2}{u_n\left( 2-u_n \right)}-\frac{1}{u_n}}
\\
&=\lim_{n\rightarrow \infty} \frac{u_n\left( 2-u_n \right)}{2-2+u_n}
\\
&=\lim_{n\rightarrow \infty} \left( 2-u_n \right)
\\
&=2
\end{align*}
$$
|
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"url": "https://math.stackexchange.com/questions/4122846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find the sum of $\frac{1}{\sqrt1+\sqrt3} + \frac{1}{\sqrt3+\sqrt5} + \frac{1}{\sqrt5+\sqrt7} + ... \frac{1}{\sqrt79+\sqrt81}$ Find the sum of $\frac{1}{\sqrt1+\sqrt3} + \frac{1}{\sqrt3+\sqrt5} + \frac{1}{\sqrt5+\sqrt7} + ... \frac{1}{\sqrt{79}+\sqrt{81}}$
I've thought about multiplying every fraction by 1, but like this $\frac{1} {\sqrt1+\sqrt3} *\frac{\sqrt1+\sqrt3}{\sqrt1+\sqrt3}$, $\frac{1} {\sqrt3+\sqrt5} *\frac{\sqrt3+\sqrt5}{\sqrt3+\sqrt5}...$
After multiplying I end up with $\frac{\sqrt1+\sqrt3} {(\sqrt1+\sqrt3)^2} + \frac{\sqrt3+\sqrt5} {(\sqrt3+\sqrt5)^2} ...$ but I don't have any more ideas than this, am I on the right path?
I apologize for the question's simplicity, but I'm a younger math enthusiast so questions like this are harder for me.
|
Hint: $\frac 1{\sqrt{n} + \sqrt{n+2}} \cdot \frac {\sqrt{n+2} - \sqrt{n}}{\sqrt{n+2} - \sqrt{n}} = \frac{\sqrt{n+2} - \sqrt{n}}{2}$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4126580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
$\sum\left(\frac{ n^2 + 1}{n^2 +n + 1}\right)^{n^2} $ converges or diverges? The original question is to show that $\;\sum\left(\dfrac{ n^2 + 1}{n^2 +n + 1}\right)^{n^2} $ either converges or diverges.
I know it diverges but I'm having difficulty arriving at something useful for $ a_n $.
Here's what I did:
$ a_n = \left(\dfrac{ n^2 + 1}{n^2 +n + 1}\right)^{n^2} \geq\left( \dfrac{ n^2 + 1}{n^2 +2n + 1}\right)^{n^2} \geq\left( \dfrac{ n^2 - 1}{n^2 +2n + 1}\right)^{n^2} = \left(\dfrac{(n-1)(n+1)}{(n+1)^2}\right)^{n^2} = \left(\dfrac{n-1}{n+1}\right)^{n^2} = \left( 1 - \dfrac{2}{n+1}\right)^{n^2} $
Do you have idea how to continue from here? ( my idea was to arrive to some expression that goes to infinity and deduce by the comparison test that $ a_n $ diverges ). I feel like something is right infront of my eyes but I can't see it.
Edit: The series converges. However, how do you propose I should continue from where I've left?
|
$$a_n=\left(\frac{ n^2 + 1}{n^2 +n + 1}\right)^{n^2}\implies \log(a_n)=n^2\log\left(\frac{ n^2 + 1}{n^2 +n + 1}\right)$$
For large $n$
$$\log(a_n)=-n+\frac{1}{2}+\frac{2}{3 n}-\frac{3}{4 n^2}-\frac{1}{5
n^3}+O\left(\frac{1}{n^4}\right)$$ Repeat it for $a_{n+1}$ to obtain
$$\log(a_{n+1})-\log(a_n)=-1-\frac{2}{3 n^2}+\frac{13}{6 n^3}+O\left(\frac{1}{n^4}\right)$$
$$\frac{a_{n+1}}{a_{n}}=e^{\log(a_{n+1})-\log(a_n)}=\frac 1e \left(1-\frac{2}{3 n^2}+\frac{13}{6 n^3} \right)+O\left(\frac{1}{n^4}\right)$$ So, it converges.
Now, if you make
$$\sum_{n=0}^\infty a_n=\sum_{n=0}^p a_n+\sum_{n=p+1}^\infty a_n$$
$$\sum_{n=p+1}^\infty a_n\sim \sum_{n=p+1}^\infty e^{\frac{1}{2}-n}=\frac{e^{\frac{1}{2}-p}}{e-1}$$
If you make $p=4$, the approximation gives
$$\frac{31179700687749571657625354575093}{15170464778704564632455243642733}+\frac{1}{(e-
1) e^{7/2}}\sim 2.072864$$ while the "excat" result is $2.074636$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Distribution of a function of a random variable and the steps/range that is needed to take. The Question that I need to solve is:
Let $X\sim~\text{Unif}[-1,2]$. Find the probability density function of the random variable $Y = X^2$.
Now I have solved it as follows, but the steps I have to take after that are unclear to me. I need to find some kind of range and take 'certain?' steps, and I just do not know how big of steps I have to take or what is most logical.
$F_Y(y) = P(Y \leq y) = P(X^2\leq y) = P(\sqrt y \leq x \leq \sqrt -y) = F_X(\sqrt y) - F_X(\sqrt-y) $
Now you take derivative and you get
$f_y(y) = \frac{1}{2 \sqrt y} f_X(\sqrt y) + \frac{1}{2 \sqrt y} f_X(\sqrt -y) $.
So, I made a drawing of the uniform distribution and because of $Y=X^2$ it goes from $[0,2]$. I know that for $y < 0, f_Y(y) = 0.$ But what do I need to do next? How do I know how much steps I need to take (in this case two more, but why?) and how big must those steps be? Which ranges are important? How do I find those?
|
We have that $X \in [-1, 2]$.
Consider the graph $Y = X^2$. This is not a one-to-one function. (Observe that for $X \in [-1, 1]$ that there are two $X$ values that yield the same $Y$ value.)
Now, by definition, for each $y \in \mathbb{R}$,
$$F_{Y}(y) = \mathbb{P}(Y \leq y) = \mathbb{P}(X^2 \leq y)\text{.}$$
Partition the values of $Y$ into the following:
*
*If $y < 0$, obviously $F_{Y}(y) = 0$.
*Suppose $0 \leq y \leq 1$ (since this corresponds to where $X \in [-1, 1]$; i.e., $Y$ is not one-to-one). Choose any $y$ between $0$ and $1$ inclusive and plot the line $Y = y$; it follows that $X^2 \leq y$ if and only if $-\sqrt{y} \leq X \leq \sqrt{y}$. Hence, $$
\begin{align}
F_{Y}(y) &= \mathbb{P}(X^2 \leq y) \\
&= \mathbb{P}(-\sqrt{y} \leq X \leq \sqrt{y}) \\
&= F_{X}(\sqrt{y})-F_{X}(-\sqrt{y}) \\
&= \dfrac{\sqrt{y}-(-1)}{2-(-1)} - \dfrac{-\sqrt{y}-(-1)}{2-(-1)} \\
&= \dfrac{\sqrt{y}+1}{3} + \dfrac{\sqrt{y}-1}{3} \\
&= \dfrac{2\sqrt{y}}{3}\text{.}
\end{align}$$
*Now suppose $1 < y \leq 4$ (where $X \in (1, 2]$). Then
$$\{X^2 \leq y\} = \{X^2 \leq 1\} \cup \{1 < X^2 \leq y\}\text{.}$$
It follows that
$$\begin{align}
F_{Y}(y) &= \mathbb{P}(X^2 \leq 1) + \mathbb{P}(1 < X^2 \leq y) \\
&= F_{Y}(1) + \mathbb{P}(1 < X < \sqrt{y}) \\
&= \dfrac{2\sqrt{1}}{3} + F_{X}(\sqrt{y}) - F_{X}(1) \\
&= \dfrac{2}{3}+\dfrac{\sqrt{y} - (-1)}{2-(-1)} - \dfrac{1-(-1)}{2-(-1)} \\
&= \dfrac{2}{3}+\dfrac{\sqrt{y}+1}{3} - \dfrac{2}{3} \\
&= \dfrac{\sqrt{y}+1}{3}\text{.}
\end{align}$$
*If $y > 4$, obviously $F_{Y}(y) = 1$.
Hence, the density is given by the derivative of $F_Y$, or
$$f_{Y}(y) = \begin{cases}
\dfrac{1}{3\sqrt{y}}, & 0 \leq y \leq 1 \\
\dfrac{1}{6\sqrt{y}}, & 1 < y \leq 4 \\
0, & \text{otherwise.}
\end{cases}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Convergence of $ \sum\limits_{n=1}^{\infty}\frac{1}{n(n+m)} $ Let $ m \in \mathbb{N} $. Find to what does the series $ \sum\limits_{n=1}^{\infty}\frac{1}{n(n+m)} $ converge to
My attempt: ( I did as discussed in Infinite Series $\sum 1/(n(n+1))$ )
Let $ N>m $.
$
\begin{align}
S_N & = \sum\limits_{n=1}^{N}\frac{1}{n(n+m)} = \frac{1}{m}\sum_{n=1}^N \left(\dfrac1n - \dfrac1{n+m}\right) = \frac{1}{m}( \sum_{n=1}^N \dfrac1 n - \sum_{n=1}^N \dfrac1{n+m}) = \frac{1}{m}(\sum_{n=1}^N \dfrac1n - \sum_{n=m+1}^{N+m} \dfrac1{n} )\\
& = \frac{1}{m}(\sum_{n=1}^m \dfrac1n + \sum_{n=m+1}^N \dfrac1n - \sum_{n=m+1}^N\dfrac1n - \sum_{n=N+1}^{N+m}\dfrac1n ) = \frac{1}{m}(\sum_{n=1}^m \dfrac1n - \sum_{n=N+1}^{N+m}\dfrac1n )
~~\text{ [ from here I got stuck] } \end{align}
$
|
You are almost done. We have $$\begin{align}
S_N &= \frac{1}{m} \sum_{n=1}^m \frac{1}{n} - \frac{1}{m} \sum_{n = N+1}^{N+m} \frac{1}{n} \\
&= \frac{1}{m} \sum_{n=1}^m \frac{1}{n} - \frac{1}{m} \sum_{n = 1}^{m} \frac{1}{n+N} \\
&= \frac{1}{m} H_m - \frac{1}{m} \sum_{n = 1}^{m} \frac{1}{n+N},
\end{align}$$
where I have used $H_m = \sum_{n=1}^m \frac{1}{n}$ to represent the $m^{\rm th}$ harmonic number. hence for a fixed $m$, as $N \to \infty$,
$$\lim_{N \to \infty} S_N = \frac{H_m}{m} - \frac{1}{m} \lim_{N \to \infty} \sum_{n=1}^m \frac{1}{n + N}.$$ But since the remaining sum contains only finitely many terms, we can interchange the order of summation and limit to obtain
$$\lim_{N \to \infty} S_N = \frac{H_m}{m} - \frac{1}{m} \sum_{n=1}^m \lim_{N \to \infty} \frac{1}{n + N} = \frac{H_m}{m} - \frac{1}{m} \sum_{n=1}^m 0 = \frac{H_m}{m}.$$
|
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|
Finding distance from Local maximum of $f(x)=x+\sqrt{4x-x^2}$ to bisector of first quadrant
What is the distance of the local maximum of the function
$f(x)=x+\sqrt{4x-x^2}$ to bisector of first quadrant?
$1)1\qquad\qquad2)\sqrt2\qquad\qquad3)2\qquad\qquad3)2\sqrt2$
It is a question from timed exam so the fastest answers are the best.
My approach:
First I realized the function defined for $x\in[0,4]$. to find local maximum I put derivative equal to zero :
$$f'(x)=1+\frac{4-2x}{2\sqrt{4x-x^2}}=0$$
$$x-2=\sqrt{4x-x^2}$$
$$2x^2-8x+4=0\rightarrow x^2-4x+2=0\rightarrow x=2\pm\sqrt2$$
From here I noticed that the quadratic $-x^2+4x$ is symmetric along $x=2$ line so after plugging $x=2\pm\sqrt2$ in the $\sqrt{4x-x^2}$, hence $f(x)$ is maximum for the value $x=2+\sqrt2$ and $f(2+\sqrt2)=2+\sqrt2+\sqrt{2}=2+2\sqrt2$
Now we should find the distance from the point $(2+\sqrt2,2+2\sqrt2)$ from the line $y=x$. Now by recognizing that $(2+\sqrt2,2+\sqrt2)$ is a point on the line $y=x$, I guess the distance might be $1$ and because it is the least value in the choices then I conclude it is the answer.
Is it possible to solve this problem quickly with other approaches?
|
Let $u$ be the maximum value of $f(x)$. Then:
$$\sqrt{4x-x^2} = u-x \Rightarrow 4x-x^2 = u^2 - 2ux + x^2$$
$$\Rightarrow 2x^2 - x(2u + 4) + u^2 = 0 \tag{*}$$
$$\Delta = 0 \Rightarrow (2u+4)^2-4(2)(u^2) = 0$$
$$\Rightarrow -4u^2+16u+16=0 \Rightarrow u^2 - 4u - 4 = 0$$
$$\Rightarrow u = 2 ± 2\sqrt{2}$$
hence the maximum value ($y$) is $2 + 2 \sqrt{2}$.
Since $\Delta = 0$, quadratic $(*)$ is a perfect square. Let this be $(ax-b)^2$: thus $b = 2 + 2 \sqrt{2}$ being the only constant term, and $a^2 = 2 \Rightarrow a = \sqrt{2}$ (concave up). Thus when $(ax-b)^2 = 0$, $\sqrt{2}x - (2 + 2\sqrt{2})=0 \Rightarrow x = \frac{\sqrt2 + 2 \sqrt{2}}{\sqrt{2}} = 2 + \sqrt{2}$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How to use polar coordinate in ODE? I don't understand how to use polar coordinate.
\begin{cases} \frac{dx(t)}{dt}=2x-y \\ \frac{dy(t)}{dt}=5x-2y \\ \end{cases}
$$ \frac{d}{dt} \left( \begin{array}{c} x \\ y \end{array} \right) = \begin{pmatrix} 2 & -1\\ 5 & -2 \end{pmatrix} \left( \begin{array}{c} x \\ y \end{array} \right)$$
let $A = \begin{bmatrix} 2 & -1\\ 5 & -2 \end{bmatrix}$, we have $\det(A) = 1, \operatorname{tr}(A) = 0, P(\lambda) = \lambda^2+1 = 0 \Rightarrow \lambda_{1,2}=\pm i $
\begin{cases} 2x-y-ix = 0 \\ 5x-2y-iy = 0 \\ \end{cases}
x = 1, y = 0
$\vec v_1 = (x,y) = (x, x(2-i)) = x(1,(2-i)) = \binom{1}{2-i} $
$E_{\lambda1} = \binom{1}{0}+i \binom{0}{-1} = \begin{pmatrix} 1 & 0\\ 2 & -1 \end{pmatrix} = P $ which give us $P^{-1}= \begin{pmatrix} -1 & 0\\ -2 & 1 \end{pmatrix}$ and $J = \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}$
$X(t)= e^{0t} \begin{pmatrix} \cos(t) & \sin(t)\\ -\sin(t) & \cos(t) \end{pmatrix}\begin{pmatrix} x_1\\ x_2 \end{pmatrix}$
$Y(t) = \begin{pmatrix} 1 & 0\\ 2 & -1 \end{pmatrix}\begin{pmatrix} \cos(t) & \sin(t)\\ -\sin(t) & \cos(t) \end{pmatrix}\begin{pmatrix} x_1\\ x_2 \end{pmatrix}$
It's a center stable
So the solution is pretty simple but i want to try to use the polar coordinate and i can't do it
$r^2= x^2 + y^2$ and $x = r\cos(\theta)$, $y = r \sin(\theta)$
$2rr'= 2xx'+ 2yy' = 2x(2x-y) + 2y(5x-2y) = 4x^2 +8xy -4y^2 $
$\theta ' = \frac{y(2x-y)+x(5x-2y)}{x^2+y^2} = \frac{-y^2+5x^2}{x^2+y^2}$
maybe if the exercise doesnt ask polar coordinate, we don't do it ?
i've check this : Polar coordinates differential equation but i still can't do it
|
You can tackle it, but it's not necessarily pretty, and as you've noticed - you can solve it perfectly fine in Cartesian coordinates so that's probably how you're expected to approach it.
However, if you're keen, then you can start like this (I'm using dots for time derivatives by convention and because $\theta'$ looks weird to me):
First, use $\dot{x} = \dot{r} \frac{\partial x}{\partial r} + \dot{\theta} \frac{\partial x}{\partial \theta} = \dot{r} \cos \theta - r \dot{\theta} \sin \theta$ and similarly $\dot{y} = \dot{r} \sin \theta + r \dot{\theta} \cos \theta$. Substituting those into the DE gives:
$$\begin{eqnarray}
\dot{r} \cos \theta - r \dot{\theta} \sin \theta & = & 2x - y & = & 2r \cos \theta - r \sin \theta \\
\dot{r} \sin \theta + r \dot{\theta} \cos \theta & = & 5x - 2y & = & 5r \cos \theta - 2r \sin \theta
\end{eqnarray}$$
Then you solve these as simultaneous equations for $\dot{r}$ and $\dot{\theta}$. I'll start with $\dot{\theta}$, dividing the first equation by $\cos \theta$ and the second by $\sin \theta$ and subtracting the first from the second (and being happy that I can cancel an $r$ term all around):
$$\begin{eqnarray}
\dot{r} - r \dot{\theta}\frac{\sin \theta}{\cos \theta} & = & 2r - r \frac{\sin \theta}{\cos \theta} \\
\dot{r} + r \dot{\theta}\frac{\cos \theta}{\sin \theta} & = & 5r \frac{\cos \theta}{\sin \theta} - 2r \\
r \dot{\theta} \left(\cot \theta - \tan \theta\right) & = & 5r \cot \theta - r \tan \theta - 4r \\
\dot{\theta}\frac{\cot \theta - \tan \theta}{5 \cot \theta - \tan \theta - 4} & = & 1 \\
\int \frac{\cot \theta - \tan \theta}{5 \cot \theta - \tan \theta - 4} \ d \theta & = & \int dt \\
t & = & \int \frac{\cot \theta - \tan \theta}{5 \cot \theta - \tan \theta - 4} \ d \theta
\end{eqnarray}$$
and then you just have to calculate that integral, invert it to get $theta$ in terms of $t$, and then substitute that back into the DE somewhere to solve for $r$. And then, if you really want, you can then try to transform those back into $x$ and $y$.
Is it going to be nice? Uh, probably not. And there are also some singularities that you'll want to be careful about. Is it a good exercise? It probably depends a lot on how much you enjoy this kind of thing.
|
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|
Solve the inequality $\frac{3-x}{x^2-2x-3}\le\frac{3-x}{x^2+2x-3}$ Solve the inequality $$\dfrac{3-x}{x^2-2x-3}\le\dfrac{3-x}{x^2+2x-3}$$
We have $D: \begin{cases}x^2-2x-3\ne0\Rightarrow x\ne -1;3 \\ x^2+2x-3\ne0\Rightarrow x\ne-3;1\end{cases}$
Is the given equality equivalent (in $D$) to $$x^2-2x-3\ge x^2+2x-3\\\iff x\le0$$ So the solutions are $x\le0\cap D$? It seems like the inequalities I wrote aren't equivalent, because I don't get the answer.
|
hint
For $ x\notin\{-1,1,-3,3\}$, the inequation is equivalent to
$$(3-x)\Bigl(\frac{1}{(x-1)(x+3)}-\frac{1}{(x+1)(x-3)}\Bigr)\ge 0$$
$$\iff x(x-1)(x+1)(x+3)\ge 0$$
You will find that the solution is
$$S=(-\infty,-3)\cup(-1,0]\cup(1,3)\cup(3,+\infty)$$
|
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|
Could someone check my work for this problem for vectors? Find the angle between the two planes. (Round your answer to two decimal places. If the planes do not intersect, enter DNE.)
I found this to be 65.28°
(b) Find parametric equations of their line of intersection. (Let $z = t$, then solve for $x(t)$ and $y(t)$ in terms of $t$. If the planes do not intersect, enter DNE.)
$x − 3y + z = −5$
$3x + 2z + 5 = 0$
My try: $3x + 2z + 5 = 0 ⇒ x = \frac{1}{3}(-2z-5)$
Then $3y = x + z + 5$
$3y= \frac{1}{3}(2z-5)+z+5 = \frac{z+10}{3} \implies y = \frac{z+10}{9}$
Let $z = t$. Parametric equations: $x=-\frac{2t}{3}-\frac{5}{3},\ y=\frac{t}{9}+\frac{10}{9},\ z=t$
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$$\cos^{-1} \left( \frac{3+2}{\sqrt{(1^2+3^2+1)(3^2+2^2)}}\right)=\cos^{-1}\left( \frac{5}{\sqrt{143}}\right)\approx 65.28^\circ$$
To verify your solution, we just have to substitute them in:
$$x-3y+z=-\frac{2t}{3}-\frac53-\frac{t}3-\frac{10}3+t=-5$$
$$3x+2z+5=-2t-5+2t+5=0$$
It is fine.
|
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|
Flaw in my reasoning for the maximum of $ab$ if $a,b\ge0$ and $a+2b=3$? Problem statement: What is the maximum value of the product $ab$ if $a$,$b$ are non-negative numbers such that $a+2b=3$?
What is the flaw in my solution?
We know that $\sqrt{ab} ≤ \frac{ (a+b)}{2}$ and that $a=3-2b$. The product $ab$ will be maximum when it is equal to the square of the RHS of the inequality above. Plugging in for $a$, and squaring both sides we get the equation: $(3-2b)(b)=\frac{(((3-2b)+b)^2}{4}$. Which gives $b=1$. And plugging $b=1$ into the the equation in the problem statement gives $a=1$. So, the max product is $1*1=1$.
What am I doing wrong? What concepts could I be I misunderstanding? Can you please explain? The actual answer is $9/8$. Thank you.
|
Building upon OP's attempt to use AM-GM, the following will work, instead, because it is arranged such that the right-hand side of the inequality matches the known constant sum.
$$
\sqrt{a \cdot 2b} \le \frac{a + 2b}{2}=\frac{3}{2} \;\implies\; 2ab \le \left(\frac{3}{2}\right)^2 = \frac{9}{4} \;\implies\; ab \le \frac{9}{8}
$$
The maximum value of $\frac{9}{8}$ is attained when $a = 2b$, which is $a=\frac{3}{2}, b=\frac{3}{4}$.
|
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|
Find $x$ for $3^x = 5/3$ Why my answer is wrong? $3^x = 5/3$
$e^{\ln 3^x} = 5/3$
$e^{x \ln 3}=5/3$
$e^x \times e^{\ln 3} = 5/3$
$e^x \times 3 = 5/3$
$e^x = 5/9$
$x = \ln(5/9)$
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$3^x=\frac{5}{3}\Leftrightarrow e^{x ln 3}=\frac{5}{3}$; as you see you have a mistake right after the first equal sign.
Try instead $$3^x=\frac{5}{3}|\log_3 \Leftrightarrow \log_3(3^x)=\log_3\left( \frac{5}{3}\right )$$
|
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|
Find the equations of circles passing through $(1, -1)$ touching the lines $4x+3y+5=0$ and $3x-4y-10=0$
Find the equations of circles passing through $(1,-1)$ touching the lines $4x+3y+5=0$ and $3x-4y-10=0$
The point of intersection of the lines is $(\frac25,-\frac{11}5)$
If we want this point of intersection to be $(0,0)$ (because the given lines are perpendicular) then define the new $X=x-\frac25,Y=y+\frac{11}5$, where $x,y$ are the original coordinates.
So, the point $(1,-1)$ in the new system becomes $(\frac35,\frac65)$. Since it's in the first quadrant, so, the center of the circle will be $(r,r)$ where $r$ is the radius of the circle.
So, $(\frac35-r)^2+(\frac65-r)^2=r^2\implies r=\frac35,3$
So, the equations of circles are $$(X-\frac35)^2+(Y-\frac35)^2=\frac9{25}\implies(x-1)^2+(y+\frac85)^2=\frac9{25}$$
and $$(X-3)^2+(Y-3)^2=9\implies(x-\frac{17}5)^2+(y-\frac45)^2=9$$
Is this correct?
|
As pointed out in comments, the original lines are not parallel to coordinate axes. So if you want them to be coordinate axes, you will need rotation, in addition to shifting of origin. Instead you can find the equation of lines in new coordinate system after shifting the origin and complete the work using your approach.
In my answer, I am sharing another approach for this specific problem -
First, given that both lines are tangent to the circle and intersect at a point, we know that the center of the possible circles is on angle bisector of both lines.
Angle bisector of both lines is given by,
$4x+3y+5 = \pm(3x-4y-10)$ and we get two equations,
$y = - \dfrac{x}{7} - \dfrac{15}{7} \ $ and $ \ y = 7x-5$
Now we know x-intercept of $4x+3y+5=0 \ $ is $ - \dfrac{5}{4}$ and of $3x-4y-10=0$ is $\dfrac{10}{3}$. Given x-coordinate of point $(1, -1)$ being between the two x-intercepts, we know that the slope of angle bisector that we choose has to be between the slopes of tangent lines. That leads to the center of circles being on $y = 7x-5$.
So if radius of circle is $r$ and center is $(h, k)$, equating distance between center and $(1, -1)$,
$(h-1)^2 + (7h-4)^2 = r^2 \ \ $ ...$(i)$
$k = 7h-5 \ \ $ ...$(ii)$
Second we find distance from point $(1, -1)$ to both tangent lines. We have, $|d_1| = \dfrac{6}{5}, |d_2| = \dfrac{3}{5}$
As both tangent lines are perpendicular to each other, we know that (refer diagram for clarification)
$\left(\dfrac{6}{5}- r\right)^2 + \left(\dfrac{3}{5}-r\right)^2 = r^2 \implies r = 3, \dfrac{3}{5}$
Plugging both values of $r$ into $(i)$, we can find values of $h$ and then $k$ from $(ii)$.
Now for each value of $r$, $(i)$ gives us a quadratic in $h$ so that will return two values. We can test which one works. To do that, we can check distance to tangent lines from $(h, k)$ should be equal to radius. But as we know the intersection point of tangent lines is $\left(\dfrac{2}{5}, -\dfrac{11}{5}\right)$, we can simply test
$ \left(h - \dfrac{2}{5}\right)^2 + \left(k + \dfrac{11}{5}\right)^2 = 2 r^2$ should be true (refer diagram for clarification).
Finally we get equation of two circles as,
$(x-1)^2+(y-2)^2 = 9 \ $ and $ \ \left(x-\dfrac{13}{25}\right)^2 + \left(y + \dfrac{34}{25}\right)^2 = \dfrac{9}{25}$
|
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Can there be a triangle ABC if $\frac{\cos A}{1}=\frac{\cos B}{2}=\frac{\cos C}{3}$? Can there be a triangle ABC if $$\frac{\cos A}{1}=\frac{\cos B}{2}=\frac{\cos C}{3}\;?$$ Equating the ratios to $k$ we get $\cos A=k$, $\cos B=2k$, $\cos C=3k$.
Then the identity $$\cos^2A+\cos^2B+\cos^2C+2\cos A \cos B \cos C=1 \implies 12k^3+14k^2-1=0$$
$f(k)=12k^3+14k^2-1$ being monotonic for $k>0$ can have at most one real positive root. Further, $f(0)=-1, f(1/3)=1>0$, so there will be one real root in $(0,1/3)$. Hence all three cosines will be positive and less that 1, for a unque triangle to be possible.
What can be other ways to solve this question?
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Well, not as sophisticated.
But for any angle $\frac \pi 2 > A > \frac \pi 3$ we can have $\cos A$ be any value from $0 < \cos A < \frac 1 2$ we can have $B_A= \cos^{-1} (2\cos A)$ so $\cos B_A = 2\cos A$ and $\frac \pi 2 > B_A > 0$ and $0 < \cos B < 1$.
These can be two angles of a triangle with the third angle being $C_A= \pi - A - B_A=\pi -A - \cos^{-1}(2\cos A)$.
We need $\cos C_A = 3\cos A$ or $\cos C_A - 3\cos A = 0$. Is that possible?
Well, If $A = \frac \pi 2$ then $\cos A =0$ and $B_A = \cos^{-1} 0 = \frac \pi 2$ and $C_A = 0$ and $\cos C_A = 1$ and $\cos C_A - 3\cos A = 1$.
And If $A=\frac \pi 3$ then $\cos A = \frac 12$ and $B_ = \cos^{-1} 1 = 0$ and $C_A = \frac {2\pi}3$ and $\cos C_A = -\frac 12 $ and $\cos C_1 - 3\cos A = -2$.
But $\cos C_A -3\cos A$ is continuous, so there must be a value of $A$ between $\frac \pi 2$ and $\frac \pi 3$ where $\cos C_A - 3\cos A = 0$.
And for that $A$ we will have $B_A = \cos^{-1}(2\cos A)$ and $\cos B_A = 2\cos A$ and $\cos C_A = 3\cos A$.
Ta-da....
(You never said you wanted to find it; just that you wanted to know one exists. Well, one must exist by continuity and intermediate value theorem but this argument gives utterly no method of finding it.)
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Solve the equation: $ 3^{2x+1}+4 \cdot 3^x = 15 $ Solve the equation: $ 3^{2x+1}+4 \cdot 3^x = 15 $ where $x$ is a real number.
Background: Doing Olympiad question and got one from the book.
Attempt:
Let $3^x$ be $u$.
\begin{align*}
3^{2x+1} + 4 \cdot 3x - 15 &= 0 \\
3^{2x} \cdot 3^1 + 4 \cdot 3^x - 15 &= 0 \\
3^{2x} \cdot 3 + 4 \cdot 3^x - 15 &= 0 \\
3 \cdot 3^{2x} + 4 \cdot 3^x - 15 &= 0 \\
3u^2 + 4u - 15 &= 0
\end{align*}
Factoring the equation we get
$$ u=\frac{5}{3} $$
(We eliminate $u=-3$ as $x$ is real.)
$$ 3^x=\frac{5}{3} $$
Taking $\log$ both sides
$$ x \log 3 = \log\left(\frac{5}{3}\right) $$
Now I want to know how to further solve this. Also is there any easier way to solve this?
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HINT:
You want to find value of $x$
It can be proceeded as
$x\log3=\log(\frac{5}{3})$
$x=\frac{\log(\frac{5}{3})}{\log3}$
$x=\log_3(\frac{5}{3})$ by converse of base change theorem
$x=\log_3 5-1$
To make your solution short by two or three steps take $\log_3()$ on both sides
$3^x=5/3$
$\log_3 3^x=\log_3 (5/3)$
$x=\log_3(5/3)$
$x=\frac{\log 5 - \log 3}{\log3}$
$x=\frac{0.698 - 0.477}{0.477}$
$x=\frac{0.221}{0.477}$ which can nearly be approximated to $0.5$
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Given that $\int \frac{1}{x\sqrt{x^2-1}}dx=\arccos(\frac{1}{x})+C$, what is $\int\frac{1}{x\sqrt{x^2-a^2}}dx$? The following is clear:
$x\sqrt{x^2-a^2}=x\sqrt{a^2}\sqrt{\frac{x^2}{a^2}-1}=ax\sqrt{\frac{x^2}{a^2}-1}= a^2\frac{x}{a}\sqrt{\frac{x^2}{a^2}-1}$.
So I get that $$\int\frac{1}{x\sqrt{x^2-a^2}}dx=\frac{1}{a^2}\int\frac{1}{\frac{x}{a}\sqrt{\frac{x^2}{a^2}-1}}dx=\frac{1}{a^2}\arccos(\frac{a}{x})+C$$
Is this correct?
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$$ \frac{1}{a^2}\int\frac{1}{\frac{x}{a}\sqrt{\frac{x^2}{a^2}-1}}dx $$
Substitute $u=\frac{x}{a} \Longrightarrow dx = a*du $
$$ \frac{1}{a}\int\frac{1}{u\sqrt{u^{2}-1}}du = \frac{1}{a} \arccos(\frac{1}{u})= \frac{1}{a} \arccos(\frac{a}{x})$$
|
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|
Spot the mistake when computing the density function of $Y = 1/X$ when the density function of $X$ is known. The random variable $X$ has density function $f_{X}$ given by
\begin{equation}
f_{X}(x) =
\begin{cases}
\frac{1}{2} \ \ \ \ \ x \in (0,1] \\
\frac{1}{2x^2} \ \ x \in (1,\infty) \\
0 \ \ \ \ \ \ x \in (-\infty,0]
\end{cases}
\end{equation}
Let $F_{X}$ be the distribution function of $X$. The random variable $Y$ is defined as follows $Y = 1/X$. I want to compute the density function of $Y$.
Here's my attempt:
Let $y \in \mathbb{R}$ and let $F_{Y}(y)$ be the distribution function of $Y$.
\begin{equation}
F_{Y}(y) = P(Y\leq y) = P(1/X \leq y) = P(1/y \leq X) = P(X \geq 1/y)
\end{equation}
case 1: if $y \in (0,1) \implies 1/y \in (1,\infty) $. Then
\begin{equation}
F_{Y}(y) = P(X \geq 1/y) = 1 - P(X < 1/y) = 1 - F_{X}(1/y) = 1- \int_{1}^{1/y} \frac{1}{2x^2} dx = \frac{1+y}{2}
\end{equation}
case 2: if $y \in [1,\infty) \implies 1/y \in (0,1) $. Then
\begin{equation}
F_{Y}(y) = P(X \geq 1/y) = 1 - P(X < 1/y) = 1 - F_{X}(1/y) = 1- \int_{0}^{1/y} \frac{1}{2} dx =1-\frac{1}{2y}
\end{equation}
case 3: if $y \in (-\infty,0) \implies 1/y \in (-\infty,0) $. Then
\begin{equation}
F_{Y}(y) = P(X \geq 1/y) = 1 - P(X < 1/y) = 1 - F_{X}(1/y) = 1- \int_{\infty}^{1/y} 0 dx =1
\end{equation}
\begin{equation}
F_{Y}(y) =
\begin{cases}
1, \ \ \ \ \ \ \ \ \ \ y \in (-\infty,0) \\
\frac{1+y}{2}, \ \ \ \ \ \ \ y \in (0,1) \\
1-\frac{1}{2y}, \ \ \ \ y \in [1,\infty)
\end{cases}
\end{equation}
But the function $F_{Y}(y)$ is not non-decreasing, hence is not a distribution. If $F_{Y}$ were a distribution then it would be easy to compute $f_{Y}$
\begin{equation}
f_{Y}(y) =
\begin{cases}
0, \ \ \ \ \ \ \ \ \ y \in (-\infty,0) \\
\frac{1}{2}, \ \ \ \ \ \ \ y \in (0,1) \\
\frac{1}{2y^{2}}, \ \ \ \ y \in [1,\infty)
\end{cases}
\end{equation}
questions:
*
*Do you know where I made a mistake?
*Do you know another strategy to compute $f_{Y}$?
Thanks in advance.
|
A simpler approach is possible via the formula $$f_Y(y) = f_X(g^{-1}(y)) \left|\frac{dg^{-1}}{dy}\right|,$$ where $Y = g(X) = 1/X$, since $g$ is an injective function. Then since the density of $X$ may be more compactly written as
$$f_X(x) = \frac{1}{2 \max(1, x^2)} \mathbb 1 (x > 0)$$
then for $g^{-1}(y) = 1/y$ we immediately have $$f_Y(y) = \frac{1}{2 \max(1, 1/y^2)} \mathbb 1 (1/y > 0) \cdot \frac{1}{y^2} = \frac{\min(1, y^2)}{2 y^2} \mathbb 1 (0 < y < \infty) = \frac{\min(1/y^2, 1)}{2} \mathbb 1 (0 < y < \infty)$$ which in piecewise form is $$f_Y(y) = \begin{cases} 0, & y < 0 \\ \frac{1}{2}, & 0 \le y \le 1 \\ \frac{1}{2y^2}, & 1 < y < \infty .\end{cases}$$
This shows that $1/X$ has the same distribution as $X$ itself, which makes sense since the conditional distribution $X \mid 0 < X < 1$, and $1/X \mid X > 1$, are both the same uniform distribution.
|
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$P(p^2,p^3)$ and $Q(q^2,q^3)$ are two points on the curve $y^2=x^3$. Find the values of p and q. I am working through a pure maths book as a hobby. I am tackling this problem.
$P(p^2,p^3)$ and $Q(q^2,q^3)$ are two points on the curve $y^2=x^3$. Find the equation of the chord PQ and deduce the equation of the tangent at P. Given that the tangent at P passes through Q and is normal to the curve at Q, find the values of p and q.
I have said gradient of chord =:
$\frac{q^3-p^3}{q^2-p^2} = \frac{q^2+pq+p^2}{p+q}$
$\rightarrow \frac{y-p^3}{x-p^2} = \frac{q^2+pq+p^2}{p+q}$
$\rightarrow y(p+q) = x(x^2+pq+p^2)-p^2q^2$
For equation of tangent let $q \rightarrow p$
Then $2py=x3p^2-p^4 \rightarrow 2y=3px-p^3$
But the last part has me stumped. I imagine the scenario as illustrated below but can't think of a way to find p and q.
|
Given curve is $y^2 = x^3$. Slope of its tangent is given by,
$\dfrac{dy}{dx} = \dfrac{3x^2}{2y}$
Now as you found, the gradient of line $PQ$ is $\dfrac{p^2+pq+q^2}{p+q}$.
But we also know that $PQ$ is tangent to the curve at point $P$ and is normal to the curve at point $Q$.
Slope of tangent at point $P$ is $\dfrac{3 p^4}{2p^3} = \dfrac{3p}{2}$ (for $p \ne 0)$
Slope of normal line to the curve at $Q$ is then $-\dfrac{2}{3q}$ (for $q \ne 0$)
Equating $\dfrac{p^2+pq+q^2}{p+q} = \dfrac{3p}{2} = -\dfrac{2}{3q} \ $ and solving,
$p = \pm \dfrac{2\sqrt2}{3}, q = \mp \dfrac{\sqrt2}{3}$
This gives you two such pairs of points $P, Q$ as shown in your diagram.
Edit: to your later question on steps to solve the equations -
$\dfrac{p^2+pq+q^2}{p+q} = \dfrac{3p}{2} = -\dfrac{2}{3q}$
From $\dfrac{3p}{2} = -\dfrac{2}{3q}, \ pq = - \dfrac{4}{9} \ $...$(i)$
From $\dfrac{p^2+pq+q^2}{p+q} = \dfrac{3p}{2}, \ p^2 - 2q^2 + pq = 0 \ $ ...$(ii)$
Substituting $q$ from $(i)$ in $(ii)$,
$p^2 - 2 \left(-\dfrac{4}{9p}\right)^2 - \dfrac{4}{9} = 0$
Simplifying, $81p^4 - 36p^2 - 32 = 0$
Completing the square, $(9p^2 - 2)^2 = 36$
$9p^2 = 2 \pm 6$ and given $p$ is real, we get $p = \pm \dfrac{2\sqrt2}{3}$. Now you can easily find $q$ from equation $(i)$.
|
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|
Solvability of a particular Negative Pell's Equation How to show that the equation $x^2-(k^2-4)y^2=-1$ is solvable only when k=3?
I can show that 3 must divide d otherwise 3 divides $k^2-4$ and the equation will not be solvable.Again k=3(2m+1) because other wise $x^2\equiv -1(mod8)$ which is not possible.Now how should I show that m=0?
|
Another approach:
Consider equation $x^2-Dy=-1$
Let $D=m^2+1$. we have:
$x^2-m^2y^2-y^2=-1$
$\Rightarrow (x-y)(x+y)=(my-1)(my+1)$
Suppose:
$\begin{cases} x-y=my-1\\x+y=my+1\end{cases}$
which gives :
($x=my$) and ($y=1$) and we have:
$D=m^2-1=k^2-4\Rightarrow (m-k)(m+k)=-5$
This is possible only if ($m=2$) and ($k=3$).
|
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|
Prove that $\frac{a}{a+\sqrt{2013a+bc}}+\frac{b}{b+\sqrt{2013b+ca}}+\frac{c}{c+\sqrt{2013c+ab}}\leq 1$ For positive real numbers satisfying $a+b+c=2013$. Prove that
$$\frac{a}{a+\sqrt{2013a+bc}}+\frac{b}{b+\sqrt{2013b+ca}}+\frac{c}{c+\sqrt{2013c+ab}}\leq 1$$
This is my attempt.
We have
$$\frac{a}{a+\sqrt{2013a+bc}}+\frac{b}{b+\sqrt{2013b+ca}}+\frac{c}{c+\sqrt{2013c+ab}}=\frac{a}{a+\sqrt{(a+b+c)a+bc}}+\frac{b}{b+\sqrt{(a+b+c)b+ca}}+\frac{c}{c+\sqrt{(a+b+c)c+ab}}=\frac{a}{a+\sqrt{(a+b)(a+c)}}+\frac{b}{b+\sqrt{(b+c)(b+a)}}+\frac{c}{c+\sqrt{(c+a)(c+b)}}=1-\frac{\sqrt{(a+b)(a+c)}}{a+\sqrt{(a+b)(a+c)}}+1-\frac{\sqrt{(b+c)(b+a)}}{b+\sqrt{(b+c)(b+a)}}+1-\frac{\sqrt{(c+a)(c+b)}}{c+\sqrt{(c+a)(c+b)}}$$
|
$$\sum_\text{cyclic} \frac{a}{a+\sqrt{2013a+bc}}=\sum_\text{cyclic} \frac{a}{a+\sqrt{(a+b+c)a+bc}}=\sum_\text{cyclic}\frac{a}{a+\sqrt{(a+b)(a+c)}} $$
Using AM-GM inequlaity,
$$a^2+bc\ge 2a{\sqrt{bc}}\;\;\Longleftrightarrow \;\;\sqrt{(a+b)(a+c)}\ge \sqrt{ab}+\sqrt{ac} $$
Therefore,
$$\sum_\text{cyclic}\frac{a}{a+\sqrt{(a+b)(a+c)}}\le\sum_\text{cyclic}\frac{a}{a+\sqrt{ab}+\sqrt{ac}}=\sum_\text{cyclic}\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1 $$
|
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|
Show that $(a^2-b^2)(a^2-c^2)(b^2-c^2)$ is divisible by $12$
Let $a,b,c\in\Bbb N$ such that $a>b>c$. Then $K:=(a^2-b^2)(a^2-c^2)(b^2-c^2)$ is divisible by $12$.
My attempt : Since each $a,b,c$ are either even or odd, WLOG we may assume $a,b$ are both even or odd. For both cases, $a+b$ and $a-b$ are divisible by $2$ so $K$ is divisible by $4$. Note that any $n\in\Bbb N$ is one of $\overline{0},\overline{1},\overline{2}$ in $\operatorname{mod}3$. Well from this, I can argue anyway but I want to show $K$ is divisible by $3$ more easier or nicer way. Could you help?
|
Note that $x^2\equiv 0 \quad\textrm{or}\quad 1 \mod 3$ for $x\in\mathbb{N}$.
By Pigeonhole Principle, at least two among $a^2$, $b^2$ and $c^2$ have the same remainder when divided by $3$.
|
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|
Evaluate $\int\frac{1}{x\sqrt{1-x^2}}dx$ without using trigonometric substitution. The integral is
$$\int\frac{1}{x\sqrt{1-x^2}}dx\tag{1}$$
I tried solving it by parts, but that didn't work out. I couldn't integrate the result of substituting $t=1-x^2$ either.
The answer is
$$\ln\left|\dfrac{1-\sqrt{1-x^2}}{x}\right|$$
|
With the substitution $t=\sqrt{1-x^2}$, the integral can be integrated as follows
$$\int\frac{1}{x\sqrt{1-x^2}}dx=-\int \frac1{1-t^2}dt= \frac12 \ln\frac{1-t}{1+t}= \frac12\ln \frac{1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}}+C
$$
which is the same as $\ln\frac{1-\sqrt{1-x^2}}{x}$ after rationalizing the denominator.
|
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|
high order integer equation
Find all tuple $(x,y)$ such that $x,y$ are integers and $(x^2-y^2)^2=20y+1$.
First i see that $x^2-y^2$ is odd and from the fact that a difference between square of two odd is multiple of $8$ and thus $y$ is a multiple of $2$.
Moreover, we have $(x^2-y^2+1)(x^2-y^2-1)=20y$.
Somebody can give some hint! Whether the first information is useful?
|
Since $x^2-y^2$ is odd then $x\not=y$ and
$$20y+1=(x^2-y^2)^2\geq ((y-1)^2-y^2)^2=(-2y+1)^2=1-4y+4y^2.$$
Hence
$$0\geq 4y^2-24y\Leftrightarrow y(y-6)\leq 0 $$
which implies that $0\leq y\leq 6$.
Moreover $20y+1$ is a perfect square, and therefore $y\in\{0,4,6\}$:
*
*if $y=0$ then $(x^2-0)^2=1^2$ and $x=\pm 1$;
*if $y=4$ then $(x^2-16)^2=9^2$ and $x=\pm 5$;
*if $y=6$ then $(x^2-36)^2=11^2$ and $x=\pm 5$.
Therefore the integer solutions are $(\pm1,0),(\pm5,4),(\pm 5,6).$
|
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|
In this problem of a moving point in a graph, why is $aθ+θ/3=π/2$ being used to find the angle in which the points $O$,$P$, and $Q$ meet? In this problem the circles $C_1:x^2+y^2=1$ and $C_2:x^2+y^2=4$ are given along with the points $P(\cos(a\theta), \sin(a\theta))$ and $Q(2\cos(\frac{\pi}2-\frac{\theta}3), 2\sin(\frac{\pi}2-\frac{\theta}3)) $, which are on the circunference of the circles respectively.
The problem asks you to find the first value of $\theta$ in which the origin and the points $P$ and $Q$ will meet.
In the explanation, this image is given:
Representation of $C_1$ and $C_2$
and it is stated that when the blue and the red line meet $a\theta+\frac{\theta}3=\frac{\pi}2$ and therefore $\theta=\frac{3}{6a+2}\pi$.
My question is why is $a\theta+\frac{\theta}3=\frac{\pi}2$ being used to find the angle in which the points O,P, and Q meet?(What I mean by this is why is $\frac{\theta}3$ being used instead of $-\frac{\theta}3$ and why is the equation equal to $\frac{\pi}2$ and not $0$ since it is also near the two lines).
I would also like to know if the constant of the trigonometric function is the starting point and the variable is the direction (For example in $Q(2\cos(\frac{\pi}2-\frac{\theta}3), 2\sin(\frac{\pi}2-\frac{\theta}3)) $) the line starts at $\frac{\pi}2$ and goes right because the variable $-\frac{\theta}3$ is negative right?).
Thank you.
|
Since the points P, Q, O are collinear, so the area of the triangle formed by them will be zero.
$\mathrm{P\equiv\left(cos(a\theta),sin(a\theta)\right)\,\,\,\,\&\,\,\,\,Q\equiv\left(2\,cos\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}\right),2\,sin\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}\right)\right)}$
So,
$\left|\begin{array}{ccc}\mathrm{cos(a\theta)}&\mathrm{sin(a\theta)}&\mathrm{1}\\\mathrm{2\,cos\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}\right)}&\mathrm{2\,sin\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}\right)}&\mathrm{1}\\\mathrm{0}&\mathrm{0}&\mathrm{1}\end{array}\right|=0$
Expanding only 3rd row,
$\implies\,1\cdot\left|\begin{array}{cc}\mathrm{cos(a\theta)}&\mathrm{sin(a\theta)}\\\mathrm{2\,cos\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}\right)}&\mathrm{2\,sin\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}\right)}\end{array}\right|=0$
$\mathrm{\implies\,2\,sin\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}\right)\,cos(a\theta)-2\,sin(a\theta)\,cos\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}\right)=0}$
$\mathrm{\implies\,2\left\{sin\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}\right)\,cos(a\theta)-cos\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}\right)\,sin(a\theta)\right\}=0}$
$\mathrm{\implies\,2\,sin\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}-a\theta\right)=0}$
$\mathrm{\implies\,sin\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}-a\theta\right)=0}$
Since the question asked to find the first value of $\,\,\theta\,\,$, so,
$\mathrm{\implies\,\dfrac{\pi}{2}-\dfrac{\theta}{3}-a\theta=0}$
$\mathrm{\implies\,\dfrac{\theta}{3}+a\theta=\dfrac{\pi}{2}}$
$\mathrm{\implies\,\left(\dfrac{1+3a}{3}\right)\theta=\dfrac{\pi}{2}}$
$\mathrm{\implies\,\theta=\dfrac{3}{1+3a}\cdot\dfrac{\pi}{2}}$
$\mathrm{\implies\,\theta=\dfrac{3\pi}{6a+2}}$
|
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|
Showing$ \int_{0}^{\infty} \frac{1-3\cos 2x+2\cos 3x}{x^2} dx=0$ Showing $$\int_{0}^{\infty} \frac{1-3\cos 2x+2\cos 3x}{x^2} dx=0$$
We can show this by re-writing $I$ as
$$
\implies I=6\int_{0}^{\infty}\frac{\frac{1-\cos(2x)}{2x}-\frac{1-\cos(3x)}{3x}}{x}\,\mathrm dx,
$$
which is Frullani Integral.
$$J=\int_{0}^{\infty} \frac{f(ax)-f(bx)}{x} dx=[f(\infty)-f(0)]\ln(a/b).$$ Here, $f(x)=\frac{1-\cos(x)}{x},$ hence $I=0.$
So the question is how to show (1), otherwise?
|
Here is an approach that uses the following identity for Laplace transforms
$$\int_0^\infty f(x) g(x) \, dx = \int_0^\infty \mathcal{L} \{f(x)\} (s)\mathcal{L}^{-1} \{g(x)\} (s) \, ds.$$
Recently this identity has been referred as the Maz identity for Laplace transforms.
Setting $f(x) = 1 - 3\cos 2x + 3 \cos 3x$ and $g(x) = \frac{1}{x^2}$, it is easy to see that
\begin{align*}
\mathcal{L}\{f(x)\} &= \mathcal{L}\{1 - 3\cos 2x + 2 \cos 3x\} = \mathcal{L}\{1\} - 3 \mathcal{L}\{\cos 2x \} + 2 \mathcal{L}\{\cos 3x \}\\
&= \frac{1}{s} - \frac{3s}{s^2 + 4} + \frac{2s}{s^2 + 9},
\end{align*}
and
$$\mathcal{L}^{-1} \{g(x)\} = \mathcal{L}^{-1} \left \{\frac{1}{x^2} \right \} = s.$$
So from the Maz identity one has
\begin{align*}
\int_0^\infty \frac{1 - 3\cos 2x + 2 \cos 3x}{x^2} \, dx &= \int_0^\infty \left [\frac{1}{s} - \frac{3s}{s^2 + 4} + \frac{2s}{s^2 + 9} \right ] s \, ds\\
&= \int_0^\infty \left [1 - 3 \frac{s^2}{s^2 + 4} + 2 \frac{s^2}{s^2 + 9} \right ] \, ds\\
&= \int_0^\infty \left [1 - 3 \frac{(s^2 + 4) - 4}{s^2 + 4} + 2 \frac{(s^2 + 9) - 9}{s^2 + 9} \right ] \, ds\\
&= \int_0^\infty \left [1 - 3 + \frac{12}{s^2 + 4} + 2 - \frac{18}{s^2 + 9} \right ] \, ds\\
&= 12 \int_0^\infty \frac{ds}{s^2 + 4} - 18 \int_0^\infty \frac{ds}{s^2 + 9}\\
&= 6 \arctan \left (\frac{s}{2} \right ) \Big{|}_0^\infty - 6 \arctan \left (\frac{s}{3} \right ) \Big{|}_0^\infty\\
&= 6 \cdot \frac{\pi}{2} - 6 \cdot \frac{\pi}{2} = 0,
\end{align*}
as required to show.
|
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|
A proof in $\varepsilon$-language for $\lim \sqrt[n]{1^2+2^2+...+n^2} = 1$ I found a proof that $\lim \sqrt[n]{1^2+2^2+...+n^2}=1$ by $\varepsilon$-language, but I think it's quite complicated and not sure that it's correct.
My question is:
1- Is my proof correct?
2- Is there another simpler proof in the sense of $\varepsilon$-language?
Please help me. Thanks.
Solution:
Let $a_n=\sqrt[n]{1^2+2^2+...+n^2}-1$. Then $a_n>0$ and:
$$ (a_n+1)^n= 1^2+2^2+...+n^2 = \frac{n(n+1)(2n+1)}{6}$$
By the binomial theorem we have:
$ (a_n+1)^n = 1 + na_n+\frac{n(n-1)}{2}a_n^2+\frac{n(n-1)(n-2)}{6}a_n^3+\frac{n(n-1)(n-2)(n-3)}{24}a_n^4+\cdots+a_n^n $
Since $a_n>0$, then $(a_n+1)^n>\dfrac{n(n-1)(n-2)(n-3)}{24}a_n^4$ and therefore:
$$\dfrac{n(n+1)(2n+1)}{6}>\dfrac{n(n-1)(n-2)(n-3)}{24}a_n^4$$
This is equivalent to
$$(n+1)(2n+1)>\dfrac{(n-1)(n-2)(n-3)}{4}a_n^4 \Leftrightarrow a_n^4<\dfrac{4(n+1)(2n+1)}{(n-1)(n-2)(n-3)}$$
$$\Rightarrow a_n^4< \dfrac{8(n+1)^2}{(n-2)^2(n-3)}=\dfrac{8}{n-3}\Big(1+\dfrac{3}{n-2}\Big)^2<\dfrac{128}{n-3}.$$
Thus: $a_n<\sqrt[4]{\dfrac{128}{n-3}}$. For every $\varepsilon >0$, take $N>3+\dfrac{128}{\varepsilon^4}$, then for all $n\ge N$,
$$ n-3 \ge N -3 > \dfrac{128}{\varepsilon^4} $$
Thus: $ \varepsilon^4 > \dfrac{128}{n-3} \Rightarrow \varepsilon > \sqrt[4]{\dfrac{128}{n-3}} >a_n $.
Hence, $\lim a_n=0$ or equivalently, $\lim \sqrt[n]{1^2+2^2+...+n^2} =1$.
|
because
\begin{gather*}
1\leq \sqrt[n]{1^2+2^2+\cdots+n^2}\leq \sqrt[n]{n\cdot n^2}
=(\sqrt[n]{n})^3,\end{gather*}
and
\begin{gather*}
\lim_{n\to\infty}1=1=\lim_{n\to\infty}(\sqrt[n]{n})^3=\left(\lim_{n\to\infty}\sqrt[n]{n}\right)^3=1,
\end{gather*}
we have, by the squeeze test,
$\lim_{n\to\infty}\sqrt[n]{1^2+2^2+\cdots+n^2}=1.$
|
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|
How many points are common to the graphs of the two equations $(x-y+2)(3x+y-4)=0$ and $(x+y-2)(2x-5y+7)=0$? How many points are common to the graphs of the two equations $(x-y+2)(3x+y-4)=0$ and $(x+y-2)(2x-5y+7)=0$?
\begin{align*}
(x-y+2)(3x+y-4) &= 0\tag{1}\\
(x+y-2)(2x-5y+7) &= 0\tag{2}
\end{align*}
In equation $1$, it is not possible for both $x-y+2$ and $3x+y-4$ to be nonzero. Like-wise in equation $2$, it is not possible for both $x+y-2$ and $2x-5y+7$ to be nonzero.
Therefore the LHS of either equation must involve a product of $0$ and some number.
This means there are infinitely many solutions to both equations.
However this doesn't shead any light on which points are common between the two equations. I'm confused. It seems I've made an error in my judgement. What should I reconsider?
|
If $x-y+2=0,y=?$
Replace the value of$y$ in terms of $x,$ in the second equation to find
$$0=(x+x+2-2)(3x+x+2-4)=2x(4x-2)$$
$$x=?,?$$
Had there been $n$ factors, we would have $n$ degree equation in $x$
Similarly for $3x+y-4=0\implies y=4-3x$
Can you take it from here?
|
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|
Question about the properties of radicals all. I was solving a simple problem when I realized I had something seemingly contradictory come up. I completed the square on a polynomial as such:
$$
\begin{eqnarray}
y^2-3y&=&-1\\
y^2-3y+\frac{9}{4}&=&-1+\frac{9}{4}\\
(y-\frac{3}{2})^2&=&\frac{5}{4}\\
y-\frac{3}{2}&=& \pm \sqrt{\frac{5}{4}}\\
y&=& \frac{3}{2} \pm \sqrt{\frac{5}{4}}\\
y&=& \frac{3}{2} \pm \frac{\sqrt{5}}{2}\\
y&=& \frac{3 \pm \sqrt{5}}{2}
\end{eqnarray}
$$
However, here it seems that I got confused. Knowing the Square Root Property, the solution set of $x^2 = k \hspace{1mm}is \hspace{1mm}\{\pm \sqrt{k}\}.$
But in other instances, the radical sign is used to designate the principal square root, which only is used to designate a nonnegative square root.
Which leads to my main question. Ignoring the positive root of 4, I decided to take the negative square root of 4 as such:
$$
\begin{eqnarray}
y-\frac{3}{2}&=& \pm \sqrt{\frac{5}{4}}\\
y&\stackrel{?}{=}& \frac{3}{2} \pm \frac{\sqrt{5}}{-2} \hspace{2 cm}(\sqrt{4}=-2)\\
y&\stackrel{?}{=}& \frac{3 \pm \sqrt{5}}{-2}
\end{eqnarray}
$$
Plugging $y=\frac{3 \pm \sqrt{5}}{-2}$ back into $y^2-3y$, yields $8+3\sqrt{5}, 8-3\sqrt{5}$, respectively, so,
$$
\begin{eqnarray}
y&\neq& \frac{3 \pm \sqrt{5}}{-2}
\end{eqnarray}
$$
I'm wondering where my reasoning fell short and how I can fix it when I come across something similar in the future. Why are we assuming the square root of 4 in the denominator to be a positive square root, i.e., the principal square root, and why does the equation become false when we take the negative square root? Thank you, I greatly appreciate it.
|
The way you are adding fractions is not correct. Note that$$\frac32+\frac{\sqrt5}{-2}=\frac32+\frac{-\sqrt5}2=\frac{3-\sqrt5}2$$and that$$\frac32-\frac{\sqrt5}{-2}=\frac32+\frac{\sqrt5}2=\frac{3+\sqrt5}2.$$So, you get$$\frac32\pm\frac{\sqrt5}{-2}=\frac{3\mp\sqrt5}2,$$which are the same numbers that you had got before.
|
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|
How to solve $5^{x + 1} = 3^{x + 2}$
Solve $5^{x + 1} = 3^{x + 2}$.
I got this far, but I'm not sure how to continue:
\begin{align}
5^{x + 1} &= 3^{x + 2} \\
(5^x)(5^1) &= (3^x)(3^2) \\
5(5^x) &= 9(3^x)
\end{align}
Where do I go from here?
|
Your equation is equivalent to
$$\left(\frac{5}{3}\right)^x = \frac{9}{5}$$
Therefore, $x = \text{log}_{\frac{5}{3}} \frac{9}{5}$.
|
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|
Determine the conditional extreme for $u = x^2 + y^2 + z^2$ for $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$. Determine the conditional extreme for $u = x^2 + y^2 + z^2$ for $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$.
Using Lagrange's multipliers we get:
$$<2x, 2y, 2z> = \lambda<\frac{2x}{a^2}, \frac{2y}{b^2}, \frac{2z}{c^2}>$$
concluding $\lambda = \frac{1}{a^2} = \frac{1}{b^2} = \frac{1}{c^2} \rightarrow a^2=b^2=c^2$. If so, then this provides the condition that :
$$<x,y,z> = \frac{\lambda}{a^2}<{x}, {y}, z>$$
Then either $<x,y,z> = <0,0,0>$ or $\lambda = a^2$ (or both). I can't seem to conclude an extrema condition from this. What am I missing?
|
$a,b,c$ are all constants so you cannot conclude that they are all equal. What you can conclude from your four equations is that either
*
*$\lambda = a^2, y = z = 0$, $x = \pm a$
*$\lambda = b^2, x = z = 0$, $y = \pm b$
*$\lambda = c^2, x = y = 0$, $z = \pm c$
Then, you can take the maximum of $a,b,c$ and get your extrema conditions.
|
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How does $\sin(\frac{\pi}{2}+h)$ become $\cos h$, and $\cos h-1$ become $-2 \sin^2 (\frac{h}{2})$? I have questions regarding trigonometry used in the solution of this problem:
Discuss the differentiability of $f(x)$ at the point $x=\frac{\pi}{2}$.
$$f(x) = \begin{cases}
1, & x<0 \\[4pt]
1+\sin x, &0\leq x<\frac\pi2 \\[2pt]
2+\left(x-\frac{\pi}{2}\right)^2, &x\geq\frac\pi2
\end{cases}$$
$$\begin{align}
Lf'\left(\frac{\pi}{2}\right)&=\lim_{h\to0^-} \frac{f(\frac{\pi}{2}+h)-f(\frac{\pi}{2})}{h} \\[4pt]
&=\lim_{h\to0^-} \frac{1+\sin(\frac{\pi}{2}+h)-2}{h} \\[4pt]
&=\lim_{h\to0^-} \frac{\cos h-1}{h} \\[4pt]
&=\lim_{h\to0^-} \frac{-2 \sin^2 (\frac{h}{2})}{h} \\[4pt]
&=-\lim_{h\to0^-} \left(\frac{\sin\frac{h}{2}}{\frac{h}{2}}\right)^2\cdot\frac{h}{2} \\[4pt]
&=0
\end{align}$$
The last two line was little bit weird (It was looking just like magic) to me.
*
*How they converted $$\sin(\frac{\pi}{2}+h)=\cos h$$?
*How they wrote $$\cos h-1 = -2 \sin^2 (\frac{h}{2})$$
I think I am missing something on trigonometry.
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Both of your questions are answered by formulas which can be derived from the angle sum formulas for sine and cosine given by
\begin{align}
\sin(\alpha \pm \beta) &= \sin \alpha \cos \beta \pm \cos \alpha \sin \beta \\
\cos(\alpha \pm \beta) &= \cos \alpha \cos \beta \mp \sin \alpha \sin \beta
\end{align}
For the first part, using the angle sum formula for sine we get
$$
\sin\left(\frac{\pi}{2}+h\right) = \underbrace{\sin\left(\frac{\pi}{2}\right)}_{\color{blue}{1}}\cos(h) + \sin(h) \underbrace{\cos\left(\frac{\pi}{2}\right)}_{\color{blue}{0}} = \cos(h)
$$
For the second part, you can use the double angle formula for cosine (which is just the angle sum formula when $\alpha = \beta$) to obtain said relationship:
\begin{align*}
&\cos(2u) = \underbrace{\cos^2(u)}_{\color{blue}{1 - \sin^2(u)}} - \sin^2(u) = 1-2\sin^2(u) \\
\implies& -2\sin^2(u) =\cos(2u) - 1\overset{\color{purple}{u = h/2}}{\implies}-2\sin^2\left(\frac{h}{2}\right) =\cos(h) - 1
\end{align*}
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|
Orthogonally diagonalizable matrix If $A$ a $3\times3$ matrix, where
$A \begin{bmatrix} 1 \\ 2\\4\\ \end{bmatrix}=3\begin{bmatrix} 1 \\ 2\\4\\ \end{bmatrix}$, $A \begin{bmatrix} 2 \\ -5\\ 2\\ \end{bmatrix}=3\begin{bmatrix} 2 \\ -5\\ 2\\ \end{bmatrix}$, $A \begin{bmatrix} -6 \\ 1\\ 1\\ \end{bmatrix}=2\begin{bmatrix} -6 \\ 1\\ 1\\ \end{bmatrix}$.
I need to check if $A$ is orthogonally diagonalizable. How would I do that by these relations?
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Recall that a real square matrix is orthogonally diagonalizable if and only if $A$ is a symmetric matrix. (The forwards implication is easy, the reverse implication is the spectral theorem).
Note that $$A\begin{bmatrix} 1&2&-6 \\ 2&-5&1\\ 4&2&1\\ \end{bmatrix}=\begin{bmatrix} 1&2&-6 \\ 2&-5&1\\ 4&2&1\\ \end{bmatrix}\begin{bmatrix} 3&0&0\\0&3&0\\0&0&2 \end{bmatrix}.$$
From this you can calculate $A$ explicitly by multiplying with the inverse (on the right) of the matrix consisting of eigenvectors.
Alternatively, you can try to verify directly whether or not the eigenvectors are orthogonal to each other. If so, the jobs done (and this is clearly the case here).
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|
Evaluate $\int_{0}^{\frac{\pi}{2}} \ln(1+\sin^3 x)\text{d}x$ Here's the integral that I would like to solve. Purely for recreational purposes:
$$I=\int_{0}^{\frac{\pi}{2}} \ln(1+\sin^3x)\text{d}x$$
Here's my shot at it. I would like to stick to this method if possible. Let $I(\alpha)$ be defined as follows:
$$I(\alpha) = \int_{0}^{\frac{\pi}{2}} \ln(1+\alpha\sin^3 x)\text{d}x$$
$$\implies I'(\alpha) = \int_{0}^{\frac{\pi}{2}} \frac{\sin^3x}{1+\alpha \sin^3x}\text{d}x$$
Let $t = \tan(\frac{x}{2})$, and we have
$$I'(\alpha) = \int_{0}^{1}\frac{\frac{8t^3}{(1+t^2)^3}}{1 + \alpha\frac{8t^3}{(1+t^2)^3}}\cdot\frac{2}{1+t^2}\text{d}t$$
I will now get rid of the fraction
$$I'(\alpha) = \int_{0}^{1}\frac{\frac{8t^3}{(1+t^2)^3}}{1 + \alpha\frac{8t^3}{(1+t^2)^3}}\cdot\frac{2}{1+t^2}\cdot\frac{(1+t^2)^3}{(1+t^2)^3}\text{d}t = \int_{0}^{1} \frac{16t^3}{(1+t^2)(t^6 + 3t^4 + 8\alpha t^3 + 3t^2 + 1)}\text{d}t$$
Perform partial fractions (which was so time consuming):
$$I'(\alpha) = \frac{-2}{\alpha}\int_{0}^{1} \left(\frac{(t^2+1)^2}{t^6+3t^4+8\alpha t^3+3t^2+1}-\frac{1}{1+t^2}\right)\text{d}t$$
The second integral is inverse tangent. How would I go about doing the first integral? It's a $4^{th}$-degree polynomial over a $6^{th}$ degree polynomial, but parts of the polynomials look kind of simple.
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Not a finished answer, just the beginning of a variation of the $I(\alpha)$ approach from the OP.
Using $$I(\alpha)=\int_{0}^{\pi/2} \log(1+\alpha^3\sin^3x)\,dx$$
Then use $$1+z^3=(1+z)\left(1+\omega z\right) \left(1+\omega^2z\right)$$
where $\omega=e^{2\pi i/3}.$
Then, for $k=0,1,2,$ let$$I_k(\alpha)=\int\log(1+\omega^k\alpha\sin x)\,dx.$$
Then $$\begin{align}I_k’(\alpha)&=\omega^k\int_{0}^{\pi/2}\frac{\sin x}{1+\omega^k\alpha \sin x}\,dx\\
&=\omega^k\int_0^2 \dfrac{\frac{2t}{1+t^2}}{1+\omega^k\alpha\frac{2t}{1+t^2}}\cdot\dfrac2{1+t^2}\,dt\\
&=-\frac{2}{\alpha}\int_0^1\left(\frac{1}{t^2+2\omega^k\alpha t+1}-\frac{1}{1+t^2}\right)\,dt
\end{align}$$
Now:
$$\int_0^1\frac{1}{1+t^2}\,dt =\arctan(t)\Bigg\vert_0^1=\frac{\pi}4.$$
So now you need to compute:
$$I_k’(\alpha)-\frac\pi{2\alpha}=-\frac2{\alpha}\int_{0}^1\frac{1}
{t^2+2\omega^k\alpha t+1}\,dt$$
The roots of the denominator are $$r_k^{\pm}=-\omega^k\alpha \pm\sqrt{\omega^{2k}\alpha^2-1}$$
And we get the partial fractions for this integrand:
$$2\sqrt{\omega^{2k}\alpha^2-1}\left(\frac1{t-r_k^+}-\frac1{t-r_k^-}\right)$$
So
$$\int_0^1\frac{dt}{1+2\omega^k\alpha t+t^2}\\=2 \sqrt{\omega^{2k}\alpha^2-1} \left(\log(t-r_k^+)-\log(t-r_k^-)\right)\Bigg\vert_0^1\\ =2 \sqrt{\omega^{2k}\alpha^2-1} \left(\log(1-r_k^- )-\log(1-r_k^+)\right)
$$
A cursory look indicates it might be tough to take the antiderivative of the resulting expression. Remember, the $r_k^\pm$ are functions of $\alpha.$
When $k=0$ and $\alpha=\cos\theta$ then $r_k^\pm=e^{\pm i\theta}$ and $$-\frac{2}{\alpha}\int_0^1\frac{dt}{1+2\alpha t+t^2} =4i\theta\tan(\theta)$$
But there isn’t any nice way that I can see to get $k=1,2,$ and not seeing much nice cancellation when summing, either.
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|
We have $(x^2+y^2)^2-3(x^2+y^2)+1=0$. What is the value of $\frac{d^2y}{dx^2}$?
We have $(x^2+y^2)^2-3(x^2+y^2)+1=0$. What is the value of
$\frac{d^2y}{dx^2}$?
$1)-\frac{x^2+y^2}{y^2}\qquad\qquad2)-\frac{x^2+y^2}{y^3}\qquad\qquad3)\frac{x+y}{x^2+y^2}\qquad\qquad4)\frac{xy}{x^2+y^2}$
Here is my approach:
We have a quadratic equation in $x^2+y^2$. So $x^2+y^2=\frac{3\pm\sqrt5}{2}$ (RHS is a constant). Taking differentiate with respect to $x$,
$$2x+2yy'=0\Rightarrow\quad y'=\frac{-x}{y}$$
$$2+2y'^2+2yy''=0\Rightarrow \quad y''=\frac{-y'^2}{y}\Rightarrow y''=\frac{-x^2}{y^3}$$
Edit:
The second option can be written as $-(\frac{x^2}{y^3}+\frac{1}y)$. It seems it is the correct answer. But why I missed $-\frac1y$ in this approach?
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You missed the $-\dfrac{1}{y} $ because you forgot the $2$ on the left-side of the equation.
$$ y'' = \dfrac{-2(y')^2 \color{red}{-2}}{2y} \ne \dfrac{-2(y')^2 }{2y} $$
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Evaluate $\lim\limits_{n\to\infty}\frac{1\cdot3+2\cdot4+\dots+n(n+2)}{n^3}.$
Evaluate: $$\lim\limits_{n\to\infty}\frac{1\cdot3+2\cdot4+\dots+n(n+2)}{n^3}.$$
I'm learning limits for the first time and this is an exercise problem from my book. Here is my solution to the problem:
Let $S=1\cdot3+2\cdot4+\dots+n(n+2)\\ =(1^2+2)+(2^2+4)+\dots+(n^2+2n)\\ =(1^2+2^2+\dots+n^2)+2(1+2+\dots+n)\\ =\frac{n(n+1)(2n+1)}{6}+2\cdot\frac{n(n+1)}2\\ =\frac13n^3+\frac32n^2+\frac76n$
Hence, $\lim\limits_{n\to\infty}\frac{1\cdot3+2\cdot4+\dots+n(n+2)}{n^3}\\ =\lim\limits_{n\to\infty}\frac13+\frac3{2n}+\frac7{6n^2}\\ =\frac13.$
I'm quite sure about the solution. But my book says the answer is $\frac16$. So, is the answer in the book wrong, or am I missing something? And can the problem be solved with L'Hôpital's rule? (I've just started learning the rule and I don't know how to solve this using this). Some other methods to solve the problem are also welcome.
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Yet another solution, using Stolz Cesaro theorem, which states that if $(b_n)_{n \in \mathbb N_+}$ is an increasing sequence diverging to $+\infty$ and if limit $\lim_{n \to \infty} \frac{a_n-a_{n-1}}{b_n - b_{n-1}}$ exists, then $\lim_n \frac{a_n}{b_n} = \lim_n \frac{a_n - a_{n-1}}{b_n-b_{n-1}}$.
Using this with $b_n = n^3$ and $a_n = 1\cdot 3 + ... + n \cdot (n+2)$ you arrive at $$ \frac{a_n - a_{n-1}}{b_n - b_{n-1}} = \frac{n(n+2)}{n^3 - (n-1)^3} = \frac{n(n+2)}{n^3-n^3+3n^2 - 3n + 1} = \frac{n^2+2n}{3n^2-3n+1} \xrightarrow[n \to \infty]{}\frac{1}{3}$$
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Proving on the equation $(x^2+mx+n)(x^2+px+q)=0$ Find all real numbers k such that if $a,b,c,d \in \mathbb R$ and $a>b>c>d \geq k$ then there exist permutations $( m ,n ,p,q)$ of $(a,b,c,d)$ so that the following equation has 4 distinct real solutions :
$(x^2+mx+n)(x^2+px+q)=0$
Here all i did :
$(x^2+mx+n)(x^2+px+q)=0$
$\Leftrightarrow x^2+mx+n=0$ or $x^2+px+q=0$
so I think the four real solutions, if any, of the equation can only be :
*
*$ x= \frac{\sqrt{m^2-4n} -m}{2} $
*$ x= \frac{-\sqrt{m^2-4n} -m}{2} $
*$ x= \frac{\sqrt{p^2-4q} -p}{2} $
*$ x= \frac{-\sqrt{p^2-4q} -p}{2} $
So I think to solve the problem we just need to find all real numbers k such that if $a,b,c,d \in \mathbb R$ and $a>b>c>d \geq k$ then there exist permutations $( m ,n ,p,q)$ of $(a,b,c,d)$ so that
$p^2 \geq 4q$ and $ m^2 \geq 4n$
We can prove that for $k \geq 4 $ it is absolutely true. So are there any other satisfying $k $ values ? I'm not entirely sure. Hope to get help from everyone. Thanks very much !
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The question needs to be answered in three steps: First, we need to place constraints on $m,n,p,q$ so that all four roots are real. Second, we need to add constraints so that all four roots are distinct. Lastly, we need to find $k$ such that for all $a>b>c>d>k$, at least one permutation exists where all the constraints hold true. Note that by symmetry, permuting $m$ with $p$ and $n$ with $q$ will not change the constraints.
As stated in the post, in order for all four roots to be real, we require $m^2\geq 4n$ and $p^2\geq 4q$. If equality holds in either constraint, we won't have distinct roots, so our constraints are $m^2>4n$ and $p^2>4q$. Suppose $k=4-\delta$ for some $\delta>0$. Then, If all four of $m,n,p,q\in(4-\delta,4)$ these constraints cannot hold. As a result, we require at the very least $k\geq 4$, along with possibly further constraints.
The above constraint guarantees that the two roots defined by $m,n$ are different from each other, and that the two roots defined by $p,q$ are different from each other. We still need to exclude the remaining pairs of roots. Let $y,z\in\{-1,1\}$ so that the remaining pair of roots equations can be set up as $\frac{-y\sqrt{m^2-4n}-m}{2}=\frac{-z\sqrt{p^2-4q}-p}{2}$. We will then develop this:
$$-y\sqrt{m^2-4n}-m=-z\sqrt{p^2-4q}-p$$
$$y\sqrt{m^2-4n}+m=z\sqrt{p^2-4q}+p$$
$$y\sqrt{m^2-4n}-z\sqrt{p^2-4q}=p-m$$
$$m^2-4n+p^2-4q-2yz\sqrt{(m^2-4n)(p^2-4q)}=m^2-mp+p^2$$
$$-2yz\sqrt{(m^2-4n)(p^2-4q)}=4n+4q-mp$$
$$4m^2p^2-16m^2q-16np^2+64nq=16n^2+16q^2+m^2p^2+32nq-8mnp-8mpq$$
$$3m^2p^2-16m^2q-16np^2+32nq-16n^2-16q^2+8mnp+8mpq=0$$
$$-16q^2+(-16m^2+32n+8mp)q+(3m^2p^2-16np^2-16n^2+8mnp)=0$$
Let $\beta=\sqrt{m^2-4n}$. Our quadratic is now (after some algebra): $2q^2+(2\beta^2+4n-mp)q+(-\frac{3}{8}\beta^2p^2+\frac{1}{2}np^2+2n^2-mnp)=0$. The discriminant is $4\beta^4+16n^2+m^2p^2+16\beta^2n-4\beta^2mp-8mnp+3\beta^2p^2-4np^2-16n^2+8mnp=4\beta^4+16\beta^2n-4\beta^2mp+3\beta^2p^2+m^2p^2-4np^2=4\beta^2(4n-mp)$. This has no real roots if $mp>4n$, which means that the constraints $mp>4n, mp>4q, m^2>4n, p^2>4q$ are sufficient to guarantee that the roots are real and distinct. Now, for $k\geq4$, note that the permutations of constraints $4\leq q,n<m,p$ are sufficient to guarantee that each of the four constraints holds true. As such, 4 is the minimum possible value of k.
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Base $11$ representation of a number must equal to another representation in base $8$. Credit: 2020 AIME I (Problem #$3$ on the test)
Question: A positive integer $N$ has base-eleven representation $\underline{a}\kern 0.1em\underline{b}\kern 0.1em\underline{c}$ and base-eight representation $\underline1\kern 0.1em\underline{b}\kern 0.1em\underline{c}\kern 0.1em\underline{a},$ where $a,b,$ and $c$ represent (not necessarily distinct) digits. Find the least such $N$ expressed in base ten.
My solution: The base-$11$ representation of $abc$ has a value of $11^2(a) + 11b + c = 121a + 11b + c$. (In base $10$)
The base-$8$ representation of the number of $1bca$ has a value of $8^3 + 8^2(b) + 8c + a = 512 + 64b + 8c + a$. (In base $10$)
Therefore, $121a + 11b + c = 64b + 8c + a + 512 \Longrightarrow 120a - 53b - 7c = 512$. So, $120a > 512$, because $a, b$, and $c$ are all positive. $a, b,$ and $c$ are all less than or equal to $7$ because they are in base $8$, and $120a > 512$, the minimum possible value of $a$ is $5$, and $a = 5$ to create the minimum possible $N$. ($6$ and $7$ would be too high).
So, $53b + 7c = 88$. The only pair of positive integers $b$ and $c$ that works is $b = 1, c = 5$.
Therefore, Substituting into either of the $2$ equations; $N = 121a + 11b + c = 512 + 64b + 8c + a = \boxed{621}.$
I was wondering, is this solution valid and/or is there a better way to solve it, if it is.
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Yes your solution is correct. Another solution also suggested by AoPS is
The conditions of the problem imply that $121a + 11b + c = 512 + 64b + 8 c + a$, so $120 a = 512+ 53b+7c$. The maximum digit in base eight is $7,$ and because $120a \ge 512$, it must be that $a$ is $5, 6,$ or $7.$ When $a = 5$, it follows that $600=512 + 53b+7c$, which implies that $88 = 53b+7c$. Then $b$ must be $0$ or $1.$ If $b = 0$, then $c$ is not an integer, and if $b = 1$, then $7c = 35$, so $c = 5$. Thus $N = 515_{11}$, and $N=5\cdot 121 + 1\cdot 11 + 5 = 621$. The number $637_{11} =1376_{8} = 766$ also satisfies the conditions of the problem, but $621$ is the least such number.
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Prove that $2^n + 5^n + 56$ is divisible by $9$, where $n$ is an odd integer
Prove that $9 \mid2^n + 5^n + 56$ where n is odd
I have proved this by using division into cases based on the value of $n\bmod3$ but it seems a little bit clumsy to me and I wonder if there are other ways to prove it, probably by using modular arithmetic or induction? Below is my proof:
$\text{Case 1, }n\bmod3=0,\text{then $n=3k$ for some odd integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k}+5^{3k}+56
\\ & = 8^k+125^k+56
\\ & \equiv (-1)^k+(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv 0\quad&\left(\bmod9\right)
\end{align}$$
$\text{Case 2, }n\bmod3=1,\text{then $n=3k+1$ for some even integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k+1}+5^{3k+1}+56
\\ & = 2\cdot8^k+5\cdot125^k+56
\\ & \equiv 2\cdot(-1)^k+5\cdot(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv 9\equiv0\quad&\left(\bmod9\right)
\end{align}$$
$\text{Case 3, }n\bmod3=2,\text{then $n=3k+2$ for some odd integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k+2}+5^{3k+2}+56
\\ & = 4\cdot8^k+25\cdot125^k+56
\\ & \equiv 4\cdot(-1)^k+25\cdot(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv -27\equiv0\quad&\left(\bmod9\right)
\end{align}$$
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Mathematical induction approach
We know that $n$ is an odd integer, i.e. we can use $n= 2k - 1, k \in \mathbb{N}$. Hence, we can reformulate a starting statement:
$$9 \mid 2^{2k-1} + 5^{2k-1} + 56, \quad k \in \mathbb{N}.$$
$\text{1.}$ Basis of Induction:
For $\ k=1 \ $ we have:
$$\ 2^1 + 5^1 + 56 = 63 = 9 \cdot 7, \ $$
i.e. the statement is true for $\ k=1$.
$\text{2.}$ Induction Hypothesis:
Suppose the statement holds for some $\ k>1, \ $i.e. that it holds:
$$\ 2^{2k-1} + 5^{2k-1} + 56 = 9 \cdot m, \ m \in \mathbb{N}.$$
$\text{3.}$ The proof:
Let's check if the statement holds for $\ k+1. \ $ We have:
$2^{2(k+1)-1} + 5^{2(k+1)-1}+56 = 2^{2k+2-1} + 5^{2k+2-1}+56=4\cdot 2^{2k-1} + 25 \cdot 5^{2k-1} + 56 =
\\ 4\cdot (2^{2k-1}+5^{2k-1}+56) + 21\cdot 5^{2k-1} - 168 = 4\cdot 9 \cdot m + 21 \cdot (5^{2k-1} - 8).$
The first summand on the right side (i.e. $\ 4\cdot 9 \cdot m$) is obviously divisible by $\ 9, \ $ and because we know $\ 21 = 3 \cdot 7, \ $ it remains to prove that it holds: $\ 3 \mid 5^{2k-1} - 8, \ k \in \mathbb{N}.$ For that purpose we will use a mathematical induction too!
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Any trick for evaluating $\left(\frac{\sqrt{3}}{2}\cos(\theta) + \frac{i}{2}\sin(\theta)\right)^7$? Expressions of the form $(a\cos(\theta) + bi\sin(\theta))^n$ come up from time to time in applications of complex analysis, but to my knowledge the De Moivre's formula can only be applied with $a = b$. Is there some trick to deal with the case of $a \neq b$, for example when the expression is $\left(\frac{\sqrt{3}}{2}\cos(\theta) + \frac{i}{2}\sin(\theta)\right)^7$?
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Long Comment:
For your particular problem there is a small tiny tiny algebraic simplification as
$$T=\sin \left(\frac{\pi }{3}\right) \cos (\theta )-i \cos \left(\frac{\pi }{3}\right) \sin (\theta )=\frac{1}{2} \sqrt{3} \cos (\theta )-\frac{1}{2} i \sin (\theta )\tag1$$
or
$$T=\cos \left(\frac{\pi }{6}\right) \cos (\theta )-i \sin \left(\frac{\pi }{6}\right) \sin (\theta )=\frac{1}{2} \sqrt{3} \cos (\theta )-\frac{1}{2} i \sin (\theta )\tag2$$
Rewriting (1) as
$$S=\left( \sin \left(\omega\right) \cos (\theta )-i \cos \left(\omega\right) \sin (\theta )\right)^n \tag3$$
we can get to
$$S= \left(\frac{1}{2}+\frac{i}{2}\right)^n (\sin (\theta +\omega )+i \sin (\theta -\omega ))^n \tag4$$
and likewise for (2)
$$S=\left(\cos (\theta ) \cos \left(\frac{\omega }{2}\right)+i \sin (\theta ) \sin \left(\frac{\omega }{2}\right) \right)^n$$
we get
$$S= \left(\frac{1}{2}-\frac{i}{2}\right)^n \left(\cos \left(\theta +\frac{\omega }{2}\right)+i \cos \left(\theta -\frac{\omega }{2}\right)\right)^n$$
Using @Buraian 's methods we have the interesting general result for (3)
$$S=\left( \sin \left(\omega\right) \cos (\theta )-i \cos \left(\omega\right) \sin (\theta )\right)^n=\left(\frac{1}{2}+\frac{i}{2}\right)^n\left(1-\cos (2 \theta ) \cos (2 \omega )\right)^{n/2} e^{ \left(i \,n \tan ^{-1}(\sin (\theta +\omega ),\sin (\theta -\omega ))\right)}\tag5$$
where $\tan ^{-1}(x,y)=\tan ^{-1}(\frac{y}{x})$ gives the angle of the point $[x,y]$ and where $\sin ^2(\theta -\omega )+\sin ^2(\theta +\omega )$ simplifies to $1-\cos (2 \theta ) \cos (2 \omega )$.
|
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|
Count the 4-digit integers that are multiple of 9 I want to count the total number of 4-digit integers that are multiple of 9, without any zero digit.
I was wondering if the best strategy is to treat it as a count problem or just apply some properties from number theory.
Naive way is to solve $a+b+c+d = 9, 18, 27, 36$ with $1 \leq a, b, c, d \leq 9$ but I feel this takes much time with too many case works to consider.
Any way we could do faster?
|
Okay, so you have determined that $abcd = 1000a + 100b + 10c + d$ is a multiple of $9$ if and only if $a+b+c+d$ is a multiple of $9$.
But now the trick is to realize that $a+b+c+d$ is a multiple of $9$ if and only if
$a+b+c+d \equiv 0 \pmod 9$ if and only if
$d\equiv -(a+b+c) \pmod 9$.
Now for any possible value $a+b+c$ may take, $-(a+b+c)\equiv$ to a specific digit $\pmod 9$ and there is exactly 1 digit between $1$ and $9$ that $d$ can take to be equivalent to $-(a+b+c)$.
(For example, if I just randomly pick $a= 5; b=7; c=2$ then $a+b+c\equiv 5\pmod 9$ and in order to have $572d$ divisible by $9$ we must have $a+b+c + d\equiv 5+d\equiv 0 \pmod 9$ or in other words $d \equiv -5 \equiv 4\pmod 9$. We must have $d=4$. It is the only choice and whatever choices we have for $a,b,c$ there will always bu exactly one choice for $d$).
So the number of four digit numbers equals
$(\text{number of choices for }a)\cdot(\text{number of choices for }b)\cdot(\text{number of choices for }c)\cdot(\text{number of choices for }d)=$
$9\cdot 9\cdot 9\cdot 1 = 9^3=729$
|
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|
What are the values of $x$, $y$ and $r$ if $\cot\theta=\frac{5}{6}$ and $\csc\theta <0$? I solved this problem, where $r$ is a radius of a circle. Is my solution right?
What are the values of $x$, $y$ and $r$ if $\cot\theta=\frac{5}{6}$ and $\csc\theta <0$?
Since $\cot\theta$ is positive and $\csc\theta$ is negative, $\theta$ must be on the third quadrant. Therefore, the values of $x$ and $y$ must be both negative. Since $\cot\theta=\frac{\cos\theta}{\sin\theta}$, $\sin\theta=y$ and $\cos\theta=x$, then $x=-5$ and $y=-6$. I can use the Pythagorean Theorem to solve for $r$.
$r^2=x^2+y^2$
$r^2=25+36$
$r^2=61$
$r=\sqrt{61}$
|
You can't conclude that $x=-5$ and $y=-6$.
You are right that it is in the third quadrant.
$$\cot \theta = \frac{r\cos \theta}{r \sin \theta}=\frac{x}y=\frac{5}{6}.$$
Hence, we have the half-line:
$$6x=5y, x < 0.$$
Any point on the half line satisfies the condition.
|
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|
Where am I going wrong with this proof for partial fraction decomposition? We have to prove that $ \frac{F}{G} $ can be written as $\frac{F_1}{G_1}+ \frac{F_2}{G_2}$ if $G=G_1G_2$ and $G_1,G_2$ are co-prime polynomials. This was the proof given in my textbook:
$$There \space exists \space polynomials \space C,D \space such \space that \space CG_1+DG_2=1$$
$$Let \space DF=G_1Q+F_1$$
$$Then \space CF=\frac{C}{D}{\left (\frac{Q(1-DG_2)}{C}+F_1 \right)} $$
$$CF= \frac{Q}{D}-QG_2+\frac{C}{D}F_1$$
$$CF+QG_2= \frac{Q}{D}+\frac{C}{D}F_1$$
The next line states that $\frac{Q}{D}+\frac{CF_1}{D}$ can be written as $F_2$. How do we prove that $Q+CF_1$ is divisible by $D$, otherwise we can't make this assumption
|
That is a very strange proof. It seems easier this way: From $CG_1 + DG_2 = 1$ we have
$$
FC\cdot G_1 + FD \cdot G_2 = F
$$
Now let $F_1 = FD$ and $F_2 = FC$. Then
\begin{align*}
\frac{F_1}{G_1} + \frac{F_2}{G_2} &= \frac{F_1 G_2}{G_1 G_2} + \frac{F_2 G_1}{G_2 G_1} \\&= \frac{F_1 G_2 + F_2 G_1}{G_1 G_2} = \frac{F}{G}
\end{align*}
After this answer was accepted I began to think more about the problem, and I decided there must be other conditions not stated. The question title mentions partial fraction decomposition, after all.
I think the full statement is this:
Suppose that $G_1$ and $G_2$ are relatively prime polynomials and $G=G_1G_2$. Suppose further that $F$ is a polynomial with $\deg F < \deg G$. Then there exist polynomials $F_1$ and $F_2$ with $\deg F_1 < \deg G_1$, $\deg F_2 < \deg G_2$, and
$$ \frac{F_1}{G_1} + \frac{F_2}{G_2} = \frac{F}{G}$$
I don't like the quoted proof; the use of quotients makes it hard to read. Here is how I would organize it:
Since $G_1$ and $G_2$ are relatively prime, there exist polynomials $C$ and $D$ such that $C G_1 + D G_2 = 1$. By the division algorithm for polynomials, there exist polynomials $Q$ and $F_1$ such that $DF = QG_1 + F_1$, and $\deg F_1 < \deg G_1$. Notice
\begin{align*}
Q + CF_1 &= (CG_1 + D G_2)Q + CF_1
\\&= CG_1 Q + C F_1 + D Q G_2
\\&= C(QG_1 + F_1) + D Q G_2
\\&= CDF + D Q G_2 = D(CF + QG_2)
\end{align*}
Set $F_2 = CF + QG_2$. Notice
\begin{align*}
F_1 G_2 + F_2 G_1
&= (DF - QG_1)G_2 + (CF + QG_2)G_1
\\&= DG_2 F - QG_1 G_2 + CG_1 F + Q G_1 G_2
\\&= (CG_1 + DG_2)F = F
\end{align*}
Therefore the equation $ \frac{F_1}{G_1} + \frac{F_2}{G_2} = \frac{F}{G}$ is satisfied.
It remains to show that $\deg F_2 < \deg G_2$. Since $F_2 G_1 = F - F_1 G_2$, we know that
$$
\deg(F_2 G_1) \leq \max \{\deg F, \deg(F_1 G_2)\}
$$
Now
\begin{align*}
\deg(F_1 G_2) &= \deg F_1 + \deg G_2 < \deg G_1 + \deg G_2 \\
\deg F &< \deg G = \deg G_1 + \deg G_2
\end{align*}
So in either case,
$$
\deg F_2 + \deg G_1 < \deg G_1 + \deg G_2
$$
and we can cancel $\deg G_1$ from both sides.
Here is an example which compares the approaches. You can see that the second is more “interesting” than the first.
Let $G_1 = x+1$, $G_2 = x^2 + 1$, and $F = x$. So we're looking for a decomposition of $\frac{x}{(x+1)(x^2+1)}$. The Bézout coefficients $C = \frac{1}{2}(1-x)$ and $D = \frac{1}{2}$ satisfy $CG_1 + DG_2 = 1$.
In the first answer, we let $F_1 = FD = \frac{1}{2} x$ and $F_2 = FC = \frac{1}{2}x(1-x)$. The resulting decomposition is
$$
\frac{x}{(x+1)(x^2+1)}
= \frac{\frac{1}{2} x}{x+1}+ \frac{\frac{1}{2}x(1-x)}{x^2+1}
$$
Both of the fractions on the right are “improper” in the sense that the degree of the numerator equals that of the denominator.
In the second answer, we divide:
$$
DF = \frac{1}{2} x = \frac{1}{2}(x+1) - \frac{1}{2}
$$
So $Q=\frac{1}{2}$ and $F_1 = - \frac{1}{2}$. Then
$$
F_2 = CF + QG_2 = \frac{1}{2}(1-x)x + \frac{1}{2}(x^2+1)
= \frac{1}{2} x + \frac{1}{2}
$$
Check:
\begin{align*}
F_1 G_2 + F_2 G_1
= -\frac{1}{2}(x^2+1) + \frac{1}{2}(x+1)(x+1) = x
\end{align*}
This time the decomposition is
$$
\frac{x}{(x+1)(x^2+1)}
= \frac{-\frac{1}{2}}{x+1}+ \frac{\frac{1}{2}(x+1)}{x^2+1}
$$
and both fractions on the right are “proper.”
|
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|
A JEE Exam problem on determinants and matrices I am first stating the question:
Let $A=\{a_{ij}\}$ be a $3\times 3$ matrix, where
$$a_{ij}=\begin{cases}
(-1)^{j-i}&\text{if $i<j$,}\\
2&\text{if $i=j$,}\\
(-1)^{i-j}&\text{if $i>j$,}
\end{cases}$$
then $\det(3\,\text{adj}(2A^{-1}))$ is equal to __________
I solved this in the following manner:
$$
A=\left[\begin{array}{lcc}
2 & (-1)^{2-1} & (-1)^{3-1} \\
(-1)^{2+1} & 2 & (-1)^{3-2} \\
(-1)^{3+1} & (-1)^{3 + 2} & 2
\end{array}\right]=\left[\begin{array}{ccc}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right]
$$
$$\begin{aligned}|A| &=2(4-1)+(-2+1)+(1-2) \\ &=6-1-1=4 \end{aligned}$$
$$
\begin{aligned}
& \operatorname{det}\left(3 \operatorname{adj}\left(2 A^{-1}\right)\right) \\
=& 3^{3}\left|\operatorname{adj}\left(2 A^{-1}\right)\right| \\
=& 3^{3}\left|2^{3} \operatorname{adj}\left(A^{-1}\right)\right| \\
=&(3 \times 2)^{3} \times\left(\left|A^{-1}\right|\right)^{2}\\=&6^3\times\Big(\frac14\Big)\\=&13.5
\end{aligned}
$$
Original image
Is my solution correct?
Note: The problem came in the JEE Main Exam of India, on the 20th of July. The answer given for this question in the Answer Key is 108.
|
Alternatively, use the properties:
$$\begin{align}\text{adj}(cA)&=c^{n-1}\text{adj}(A) \quad (1)\\
\det(cA)&=c^n\det(A) \quad (2)\\
\det(\text{adj}(A))&=(\det(A))^{n-1} \quad (3)\\
\det(A^{-1})&=(\det(A))^{-1} \quad (4)
\end{align}$$
to get:
$$\det(3\,\text{adj}(2A^{-1}))\stackrel{(1)}{=}\\
\det(3\cdot 2^2\,\text{adj}(A^{-1}))\stackrel{(2)}{=}\\
12^3\det(\text{adj}(A^{-1}))\stackrel{(3)}{=}\\
12^3(\det(A^{-1}))^2\stackrel{(4)}{=}\\
12^3((\det(A))^{-1})^2=\\
12^3\cdot 4^{-2}=\\
108.$$
|
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|
What is the sum of the coefficients of the terms containing $x^2t^3$ in the given expansion.
It is given that $(x^2 +y +2t +3k)^{10}$. What is the sum of the coefficients of the terms containing $x^2t^3$ in the given expansion. For example the terms are $x^2y^4t^3k^2$ etc.
I said that let $(x^2 +2t)$ be $"a"$ and $(y +3k)$ be $"b"$. Then, $(x^2 +y +2t +3k)^{10}=(a+b)^{10}$.
If we are looking for $x^2t^3$, then the exponential of $a$ must be $4$. Then, the exponential of $b$ is $6$.
Then, the coefficient of $x^2t^3$ is $4 \times 2^3=32$. Now, we have $(y +3k)^6$ as $b$. If we find the sum of the all coefficients in $(y +3k)^6$, we can handle the question.
The sum of the all coefficients in $(y +3k)^6$ is $4^6$.
Then, answer is $32 \times 4^6$
Is my solution correct ?
|
Following OPs approach somewhat more formally we obtain
\begin{align*}
\color{blue}{[x^2t^3]}&\color{blue}{\left(\left(x^2+2t\right)+(y+3k)\right)^{10}}\\
&=[x^2t^3]\sum_{q=0}^{10}\binom{10}{q}\left(x^2+2t\right)^q\left(y+3k\right)^{10-q}\\
&=\sum_{q=0}^{10}\binom{10}{q}\left([x^2t^3]\sum_{r=0}^q\binom{q}{r}x^{2r}(2t)^{q-r}\right)(y+3k)^{10-q}\tag{1}\\
&=\binom{10}{4}\binom{4}{1}2^3(y+3k)^6\\
&\,\,\color{blue}{=6\,720\cdot(y+3k)^6}\tag{2}
\end{align*}
We see in (1) in order to select the coefficient of $x^2$ and $t^3$ we have to choose $r=1$ and $q=4$. Evaluating (2) at $y=1$ and $k=1$ we obtain
\begin{align*}
\color{blue}{6\,720\cdot4^6}
\end{align*}
|
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|
$a^a\cdot{b^b}\ge \bigl(\frac{a+b}{2}\bigl)^{a+b}\ge{a^b}\cdot{b^a}$ If $a$ and $b$ are positive rational numbers, prove that
$$a^a\cdot b^b\ge \left(\frac{a+b}{2}\right)^{a+b} \ge a^b \cdot{b^a}$$
My try:
consider $\frac{a}{b}$ and $\frac{b}{a}$ be two positive numbers with associated weights $b$ and $a$.
Then $\displaystyle\frac{b\cdot\frac{a}{b}+a\cdot\frac{b}{a}}{a+b}\ge \biggl[\left(\frac{a}{b}\right)^b\cdot \left(\frac{b}{a} \right)^a\biggl]^\frac{1}{a+b}$ implies $a^a\cdot{b^b}\ge a^b\cdot{b^a}$
Please help me to solve this problem. Thanks
|
Part 1: by GM-HM:
$\left(\underbrace{(a.a.a\dots a)}_{\text{a times}}\underbrace{(b.b.b\dots b)}_{\text{b times}}\right)^{\frac{1}{a+b}} \ge \frac{a + b}{\left(\underbrace{\frac{1}{a}+ .... \frac{1}{a}}_{\text{a times}}\right)+\left(\underbrace{\frac{1}{b}+ .... \frac{1}{b}}_{\text{b times}}\right)}$
$\implies a^ab^b \ge \left(\frac{a+b}{2}\right)^{a+b}$
Part 2: by AM-GM:
$\frac{\underbrace{(a+a+a+\dots +a)}_{\text{b times}}+\underbrace{(b+b+b\dots +b)}_{\text{a times}}}{a+b} \ge \left(a^bb^a\right)^{\frac{1}{a+b}}$
$\implies \frac{2ab}{a+b} \ge \left(a^bb^a\right)^{\frac{1}{a+b}}$
$\implies \frac{a+b}{2} \ge \frac{2ab}{a+b} \ge \left(a^bb^a\right)^{\frac{1}{a+b}}$ (using AM-HM for the first inequality)
$\implies \left(\frac{a+b}{2}\right)^{a+b} \ge a^bb^a $
|
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|
Show that $ \frac{x^2}{\sqrt{x^2+y^2}} + \frac{y^2}{\sqrt{y^2+z^2}} + \frac{z^2}{\sqrt{z^2+x^2}} \geq \frac{1}{\sqrt{2}}(x+y+z) $ Show that for positive reals $x,y,z$ the following inequality holds and that the constant cannot be improved
$$
\frac{x^2}{\sqrt{x^2+y^2}} + \frac{y^2}{\sqrt{y^2+z^2}} + \frac{z^2}{\sqrt{z^2+x^2}} \geq \frac{1}{\sqrt{2}}(x+y+z)
$$
Background: I was digging through some old correspondence and found a letter from a very young me to Professor Love at the University of Melbourne. I had apparently ask via a letter (yes it was back when we wrote letters) how one could prove the above inequality (my version had $1/\sqrt{3}$ in it). He kindly wrote back but without a full proof. I just found the correspondence today and thought that this was a good question for this site.
Based on his letter and my old writings you can transform the above inequality as follows. First note that
$$
\text{g.l.b.}f(x,y,z) = \text{g.l.b.}f(x,z,y) = k \quad (say)
$$
where g.l.b is the greatest lower bound and $f(x,y,z)$ is the function
$$
\left(\frac{x^2}{\sqrt{x^2+y^2}} + \frac{y^2}{\sqrt{y^2+z^2}} + \frac{z^2}{\sqrt{z^2+x^2}}\right)\bigg/(x+y+z).
$$
and so
$$
\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2\phantom{y}} \geq 2k (x+y+z)
$$
Thus we need to prove that for positive reals $x,y$ and $x$ the following is true and tight:
$$
\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2\phantom{y}} \geq \sqrt{2} (x+y+z)
$$
One approach to prove this (used by Prof. Love) was to apply Hölder's inequality but this unfortunately only gives:
$$
\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2\phantom{y}} \geq \frac{2}{\sqrt{3}} (x+y+z)
$$
Geometric View: From a geometric view point this inequality can be viewed as stating that the perimeter of $\Delta PQR$ is not less than $\sqrt{2}$ times the sum of the the three edge-lengths of the box of sides $x,y,$ and $z$ and the points $P,Q$ and $R$ are three corners of the box that are not adjacent to each other.
I suspect that this is a "well known" inequality in the right circles but it is still not known to me. Thought that is was a nice problem for lovers of inequalities.
|
For $x=y=z$ the inequality $\sum\limits_{cyc}\frac{x^2}{\sqrt{x^2+y^2}}\geq k(x+y+z)$ gives $k\leq\frac{1}{\sqrt2}$.
The Peter Scholze's solution for $k=\frac{1}{\sqrt2}.$:
By Rearrangement $$\sum_{cyc}\frac{x^2}{\sqrt{x^2+y^2}}=\sqrt{\sum_{cyc}\left(\frac{x^4}{x^2+y^2}+\frac{2x^2y^2}{\sqrt{x^2+y^2}}\cdot\frac{1}{\sqrt{y^2+z^2}}\right)}\geq$$
$$\geq\sqrt{\sum_{cyc}\left(\frac{x^4}{x^2+y^2}+\frac{2x^2y^2}{\sqrt{x^2+y^2}}\cdot\frac{1}{\sqrt{x^2+y^2}}\right)}=$$
$$=\sqrt{\sum_{cyc}\left(\frac{x^4}{x^2+y^2}+\frac{2x^2y^2}{x^2+y^2}\right)}\geq\frac{x+y+z}{\sqrt2},$$ where the last inequality it's just $$\sum_{cyc}\frac{(x-y)^4}{x^2+y^2}\geq0.$$
In the making of Rearrangement we used the following reasoning.
The triples $\left(\frac{x^2y^2}{\sqrt{x^2+y^2}},\frac{x^2z^2}{\sqrt{x^2+z^2}},\frac{y^2z^2}{\sqrt{y^2+z^2}}\right)$ and $\left(\frac{1}{\sqrt{x^2+y^2}},\frac{1}{\sqrt{x^2+z^2}},\frac{1}{\sqrt{y^2+z^2}}\right)$ have the opposite ordering, which gives a possibility to use Rearrangement.
|
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|
Proof-verification: of numbers such that $(a^2+b^2)(c^2+d^2)=u^2+w^2$ I would like to ask if someone can verify that my solution to the following question is correct:
Question:
Prove that for any $a,b,c,d\in\mathbb{Z}$ there exists integers $u,v$ such that
$$(a^2+b^2)(c^2+d^2)=u^2+v^2$$
My solution:
Let $z_1 = a+ib$ and $z_2 = c+id$ then one has that
$$(a^2+b^2)(c^2+d^2)=|z_1|^2|z_2|^2=|z_1z_2|^2\\
=|(ac-bd)+(ad+bc)i|^2\\
=(ac-bd)^2+(ad+bc)^2$$
Let $u=ac-bd$ and $v=ad+bc$. And since $a,b,c,d\in\mathbb{Z}$, $u,v\in\mathbb{Z}$. QED
My Question: Is this correct? And if not, where is my mistake?
|
There are an infinite number of solutions where the right side is a hypotenuse and the left side is the product of hypotenuses of smaller Pythagorean triples. One example is
$$(3,4,5)\rightarrow 3^2+4^2=5^2\quad (5,12,13)\rightarrow 5^2+12^2=13^2\\
5\times13=65\quad \implies\quad a=2, b=1, c=3, d=2\\
\implies u=7, v=4\quad\lor\quad u=8, v=1$$
There will be $2^{n-1} $ solutions where $n$ is the number of unique prime factors of $u^2+v^2$ provided that these [prime] factors are of the form $4n+1.$ In this example the $2^{2-1}=2$ prime factors were $5$ and $13$.
Other solutions are
$$a=2, b=1, c=4, d=1\qquad u=7, v=6\quad\lor\quad u=9, v=2$$
$$a=2, b=1, c=5, d=2\qquad u=9, v=8\quad\lor\quad u=12, v=1$$
|
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|
Determinant of a Pascal Matrix, sort of Let $A_{n}$ be the $(n+1) \times(n+1)$ matrix with coefficients
$$
a_{i j}={i+j \choose i}
$$
(binomial coefficients), where the rows and columns are indexed by the numbers from
0 to $n$ are indexed.
Now I want to determine the Determinant and with the first 5 matrices i found out that it is $n+1$ if i did not make a mistake.
The Matrix looks like this:
$$
\left(\begin{matrix}
{1+1 \choose 1} & {1+2 \choose 1} & {1+3 \choose 1} & \dots & {1+n+1 \choose 1} \\
{2+1 \choose 2} & {2+2 \choose 2} & {2+3 \choose 2} & \dots & {2+n+1 \choose 2} \\
{3+1 \choose 3} &{3+2 \choose 3} & {3+3 \choose 3} & \dots & {3+n+1 \choose 3} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
{n+1+1 \choose n+1} & {n+1+2 \choose n+1} & {n+1+3 \choose n+1} & \dots & {n+1+n+1 \choose n+1}
\end{matrix}\right)
$$
The Problem is to bring this Matrix into an upper or lower triangle matrix. If anyone has hints or ideas that can help, please help, thanks in advance. Maybe the approach is not even good. If I make progress at all i will update this question.
|
Let's consider the version where the columns are indexed from $0$ to $n$. Consider for example
$$ A_3 = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 \\ 1 & 3 & 6 & 10 \\ 1 & 4 & 10 & 20 \end{bmatrix}. $$
Note that each element in the matrix (except the elements in the $0$th row and column) is the sum of the element above it and the element to the left of it. This is the "Pascal Triangle" property of the matrix.
How can we reduce $A_4$ to an upper triangular matrix? By the Pascal Triangle property, if we replace row $R_i$ by $R_i - R_{i-1}$, the resulting row is just $R_i$, shifted to the right (with the first element padded by zero and the last element dropped). So for example,
$$ \begin{bmatrix} 1 & 1 & 1 & 1 \\ \color{pink}{1} & \color{\pink}{2} & \color{\pink}{3} & 4 \\ \color{red}{1} & \color{red}{3} & \color{red}{6} & 10 \\ \color{blue}{1} & \color{blue}{4} & \color{blue}{10} & 20 \end{bmatrix} \xrightarrow{R_4 = R_4 - R_3}
\begin{bmatrix} 1 & 1 & 1 & 1 \\ \color{pink}{1} & \color{pink}{2} & \color{pink}{3} & 4 \\ \color{red}{1} & \color{red}{3} & \color{red}{6} & 10 \\ 0 & \color{blue}{1} & \color{blue}{4} & \color{blue}{10} \end{bmatrix} \xrightarrow{R_3 = R_3 - R_2}
\begin{bmatrix} 1 & 1 & 1 & 1 \\ \color{pink}{1} & \color{pink}{2} & \color{pink}{3} & 4 \\ 0 & \color{red}{1} & \color{red}{3} & \color{red}{6} \\ 0 & \color{blue}{1} & \color{blue}{4} & \color{blue}{10} \end{bmatrix} \xrightarrow{R_2 = R_2 - R_1}
\begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & \color{pink}{1} & \color{pink}{2} & \color{pink}{3} \\ 0 & \color{red}{1} & \color{red}{3} & \color{red}{6} \\ 0 & \color{blue}{1} & \color{blue}{4} & \color{blue}{10} \end{bmatrix}. $$
Now the corner $3 \times 3$ matrix again satisfies the "Pascal Triangle" property so you can repeat this process and get
$$ \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & \color{pink}{1} & \color{pink}{2} & \color{pink}{3} \\ 0 & \color{red}{1} & \color{red}{3} & \color{red}{6} \\ 0 & \color{blue}{1} & \color{blue}{4} & \color{blue}{10} \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 1 & 1 & 1 \\
0 & \color{pink}{1} & \color{pink}{2} & \color{pink}{3} \\
0 & 0 & \color{red}{1} & \color{red}{3} \\
0 & 0 & \color{blue}{1} & \color{blue}{4} \end{bmatrix}. $$
Repeating this process once again for the corner $2 \times 2$ matrix, we get
$$ \begin{bmatrix} 1 & 1 & 1 & 1 \\
0 & \color{pink}{1} & \color{pink}{2} & \color{pink}{3} \\
0 & 0 & \color{red}{1} & \color{red}{3} \\
0 & 0 & \color{blue}{1} & \color{blue}{4} \end{bmatrix} \rightarrow
\begin{bmatrix} 1 & 1 & 1 & 1 \\
0 & \color{pink}{1} & \color{pink}{2} & \color{pink}{3} \\
0 & 0 & \color{red}{1} & \color{red}{3} \\
0 & 0 & 0 & \color{blue}{1} \end{bmatrix} $$
which is an upper triangular matrix with $1$'s on the diagonal so the determinant of the matrix is one.
I'll leave you the details of generalizing this argument to the $(n+1)\times(n+1)$ case and how one can use it to calculate the shifted determinant.
|
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|
Can the formula for the cubes $a^3 + b^3 + c^3 - 3abc$ be generalized for powers other than 3? I recently learnt out about this formula:
$$a^3 + b^3 + c^3 - 3abc = (a + b + c) (a^2 + b^2 + c^2 - ab - ac - bc)$$
Is there a way of generalizing for powers other than $3$, i.e.
$$a^n + b^n + c^n + \mathop{???} = \mathop{???} $$
where the RHS is in terms of $(a^{n-1} + b^{n-1} + c^{n-1})$, $(a^{n-2} + b^{n-2} + c^{n-2})$, ... , $(a^{1} + b^{1} + c^{1})$, just like the formula for $n=3$.
|
For cases of $n=5$ and $n=7$, let's have a look at the following identities
$$(x+y+z)^3 - (x^3+y^3+z^3) = 3(x+y)(x+z)(y+z)$$
$$(x+y+z)^5 - (x^5+y^5+z^5) = 5(x+y)(x+z)(y+z)(x^2+y^2+z^2+xy+xz+yz)$$
$$(x+y+z)^7 - (x^7+y^7+z^7) = 7(x+y)(x+z)(y+z)((x^2+y^2+z^2+xy+xz+yz)^2+xyz(x+y+z))$$
The three identities listed above are called Lame-type identities (see here).
Algebraically manipulating the second and third identities give;
Let
$$A = x+y+z,$$
$$B = x^2+y^2+z^2,$$
$$C = x^3+y^3+z^3,$$
$$E = x^5+y^5+z^5,$$
$$G = x^7+y^7+z^7,$$
then
\begin{equation}
6E=A^5-5BA^3+5CA^2+5BC \tag{1}
\end{equation}
\begin{equation}
36G=A^7+7CA^4-21B^2A^3+28C^2A+21B^2C \tag{2}
\end{equation}
By the way, I think they have nice applications. For example, if the values of $A, $ $B, $ and $E$ are given, $x $, $y $ and $z$ can be found with ease since $C$ can be generated from (1) and that, solving $A, $ $B, $ and $C$ simultaneously won't be difficult. Same thing goes for $A, $ $B, $ and $G$.
|
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|
To prove: $\cot^{-1}7+\cot^{-1}8+\cot^{-1}18 = \cot^{-1}3$ To prove: $\cot^{-1}7+\cot^{-1}8+\cot^{-1}18 = \cot^{-1}3$
My Attempt:
First Method: we know that $\cot^{-1}x = \tan^{-1}\frac{1}{x}$ for $x>0$ and $\tan^{-1}x+\tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy}, xy<1$
Now $\cot^{-1}7+\cot^{-1}8+\cot^{-1}18$ = $\tan^{-1}\frac{1}{7}+\tan^{-1}\frac{1}{8}+\tan^{-1}\frac{1}{18}$ = $(\tan^{-1}\frac{1}{7}+\tan^{-1}\frac{1}{8})+\tan^{-1}\frac{1}{18}$ = $\tan^{-1}\frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7}\frac{1}{8}}+\tan^{-1}\frac{1}{18}$ = $\tan^{-1}\frac{\frac{15}{56}}{\frac{55}{56}}+\tan^{-1}\frac{1}{18}$ = $\tan^{-1}\frac{3}{11} +\tan^{-1}\frac{1}{18}$ = $\tan^{-1}\frac{\frac{3}{11}+\frac{1}{18}}{1-\frac{3}{11}\frac{1}{18}}$ = $\tan^{-1}\frac{\frac{65}{198}}{\frac{195}{198}}$ = $\tan^{-1}\frac{1}{3}$ = $\cot^{-1}3$
Second Method: we know that $\tan^{-1}x+\cot^{-1}x = \frac{π}{2}$ for $x \in \Bbb R$. So $\cot^{-1}x = \frac{π}{2} - \tan^{-1}x$.
Now $$\cot^{-1}7+\cot^{-1}8+\cot^{-1}18 = \frac{π}{2} - \tan^{-1}7+ \frac{π}{2} - \tan^{-1}8+ \frac{π}{2} - \tan^{-1}18 \implies \\ \frac{3π}{2} - (\tan^{-1}7+\tan^{-1}8+\tan^{-1}18) = \frac{3π}{2} - \tan^{-1}\frac{7+8+18-7×8×18}{1-7×8-8×18-7×18} \\ = \frac{3π}{2} - \tan^{-1}\frac{-975}{-325} = \frac{3π}{2} - \tan^{-1}3 = π + (\frac{π}{2} - \tan^{-1}3) = π + \cot^{-1}3 .$$ Also we know that $\tan x$ and $\cot x$ are periodic function with period $π$. Please help me in Second Method. Is $π + \cot^{-1}3 = \cot^{-1}3$ ?. If yes, then elaborate it.
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Obviously $\pi+\cot^{-1} 3 \neq \cot^{-1} 3$, since $\pi \neq 0$. Your second method is incorrect because $$\tan^{-1} a+\tan^{-1}b+\tan^{-1}c=\tan^{-1} \left(\frac {a+b+c-abc}{1-ab-bc-ac}\right)$$ is only true for $a,b,c>0$ if $ab+bc+ac<1$.
It is better for you to combine the $\arctan$s two at a time. We have:
$$\tan^{-1}7+\tan^{-1}8=\pi-\tan^{-1} \frac {3}{11} {\tag 1}$$
So $$\tan^{-1}7+\tan^{-1}8+\tan^{-1}18=\left(\pi-\tan^{-1} \frac {3}{11}\right)+\tan^{-1}18=\pi+\left(\tan^{-1}18-\tan^{-1} \frac {3}{11}\right)=\pi+\left(\tan^{-1}\frac {195}{65}\right)=\pi+\tan^{-1} 3$$
Thus we get $$\frac {3\pi}{2}-(\tan^{-1}7+\tan^{-1}8+\tan^{-1} 18)=\frac {3\pi}{2}-(\pi+\tan^{-1}3)=\frac {\pi}{2}-\tan^{-1}3=\cot^{-1}3$$ as expected.
Note:
$(1)$: I used the formula $\tan^{-1}a+\tan^{-1}b=\pi+\tan^{-1} \left( \frac {a+b}{1-ab}\right)$ which is true iff $ab>1, a>0, b>0$.
●For calculating $\tan^{-1}18-\tan^{-1}\frac {3}{11}$, I used $\tan^{-1}a-\tan^{-1}b=\tan^{-1}\left(\frac {a-b}{1+ab}\right)$ which is true iff $ab>-1$.
|
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|
Find the second real root for cubic $x^3+1-x/b=0$ A cubic of the form $$x^3+1-x/b=0$$ has has three real roots
Using the Lagrange inversion theorem one of the roots is given by
$$x = \sum_{k=0}^\infty \binom{3k}{k} \frac{b^{3k+1} }{(2)k+1} $$
How do you find the second one? I cannot find any info online. I am looking for a simialr series solution. I suspect is is of the form
$$x = \sum_{k=0}^\infty \binom{3k+2}{k} \frac{b^{3k+2} }{(3)k+2} $$
|
The related OEIS sequences A206300 and
A224884 help to answer your question. Define
$$ a_n := \frac{2^{2n-1}}{n!}\frac{\Gamma(3n/2-1/2)}{\Gamma(n/2+1/2)}. \tag{1} $$
Now define
$$ y_2 := \sum_{n=0}^\infty -a_n z^{3n-1},\quad
y_3 := \sum_{n=0}^\infty\, (-1)^na_n z^{3n-1},\quad
y_1 := -y_2-y_3. \tag{2} $$
These are the three roots of the cubic $$ y^3 - y/z^2 + 4 = 0. \tag{3}$$
The power series expansions are
$$ y_1 = 4z^2 + 64z^8 + 3072z^{14} + 196608z^{20} +\dots, \tag{4} $$
$$ y_2 = z^{-1} - 2z^2 - 6z^5 - 32z^8 - 210z^{11} - \dots, \tag{5}$$
$$ y_3 = -z^{-1} - 2z^2 + 6z^5 - 32z^8 +210z^{11} - \dots. \tag{6}$$
The roots of the cubic
$$ x^3 + 1 - x/b = 0 \tag{7} $$ are the roots of the previous cubic
divided by $\,2^{2/3}\,$ and where $\,z = b^{1/2}/2^{2/3}.$
The power series expansions are
$$ x_1 = b + b^4 + 3b^7 + 12b^{10} + 55b^{13} + \dots, \tag{8} $$
$$x_2 = b^{-1/2}-\frac12 b - \frac38 b^{5/2} - \frac12 b^4+\dots, \tag{9}$$
$$x_3 =-b^{-1/2}-\frac12 b+\frac38 b^{5/2}-\frac12 b^4+\dots. \tag{10}$$
Note that the coefficient of $\,x^{3n+1}\,$ in $\,x_1\,$ is
OEIS sequence A001764.
Note that $\,x_1 = b(1+x_1^3).\,$ Use it iteratively to generate
power series truncations of $\,x_1.$
Note that
$$ (x-x_1)(x-x_2)(x-x_3) = x^3-x/b+1, \tag{11}$$
which implies
$$ x_1+x_2+x_3 = 0,\quad x_1 x_2 x_3 = -1. \tag{12}$$
Thus, given a value for $\,x_1\,$ the other two roots $\,x_2,x_3\,$
satisfy a quadratic.
|
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|
Derivative of $\tan^{-1}\left(\sqrt{\frac{a-b}{a+b}}\tan \frac x2\right)$.
Find the derivative of $\tan^{-1}\left(\sqrt{\frac{a-b}{a+b}}\tan \frac x2\right)$.
I'm learning differentiation and this is an exercise problem from my book. I used chain rule and got the following:
$\begin{align}
\dfrac d{dx}\left[\tan^{-1}\left(\sqrt{\frac{a-b}{a+b}}\tan\frac x2\right)\right] &= \dfrac{1}{1+\frac{a-b}{a+b}\tan^2\frac x 2}\cdot\dfrac{d}{dx}\left(\sqrt{\frac{a-b}{a+b}}\tan \frac x2\right)\\ &= \dfrac{1}{1+\frac{a-b}{a+b}\tan^2\frac x 2}\cdot\frac 1 2\sqrt{\frac{a-b}{a+b}}\sec^2\frac x 2
\end{align}$
But this doesn't match the answer in the book. The given answer is $\frac{\sqrt{a^2-b^2}}{2(a+b\cos x)}$. So, where did I go wrong and what is the correct solution?
|
Power of $t$-formula
Let $t=\tan \dfrac{x}{2}$, then
\begin{align}
\frac{dt}{dx} &= \frac{1}{2} \sec^2 \frac{x}{2} \\
&= \frac{1+t^2}{2} \\
\tan y &= t\sqrt{\frac{a-b}{a+b}} \\
\sec^2 y \times \frac{dy}{dx} &=
\frac{dt}{dx} \times \sqrt{\frac{a-b}{a+b}} \\
\left( 1+\frac{a-b}{a+b} t^2 \right) \frac{dy}{dx}
&= \frac{1+t^2}{2} \times \sqrt{\frac{a-b}{a+b}} \\
\frac{dy}{dx} &=
\sqrt{\frac{a-b}{a+b}} \times \frac{(a+b)(1+t^2)}{2[a+b+(a-b)t^2]} \\
&= \frac{\sqrt{a^2-b^2}}{2} \times \frac{1+t^2}{a(1+t^2)+b(1-t^2)} \\
&= \frac{\frac{1}{2} \sqrt{a^2-b^2}}
{a+b\left( \frac{1-t^2}{1+t^2} \right)} \\
&= \frac{\sqrt{a^2-b^2}}{2(a+b\cos x)}
\end{align}
|
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Approximate Trigonometric Integral Let the function $I(x)$ be defined as
$$I(x) = \int_{-\pi/2}^{\pi/2}{\rm d} \theta \ \frac{\cos^2{\theta}}{\sqrt{\cos^2\theta+x^2\sin^2\theta}} \ , $$
Then we have for instance $I(0)=2$. Is there a nice answer for $I(x)$ for $x$ small? The most obvious thing to do would be to binomially expand
$$I(x) \approx \int_{-\pi/2}^{\pi/2}{\rm d} \theta \ \left( \cos \theta - \frac{x^2}{2}\frac{\sin^2\theta}{\cos\theta}\right) \ , $$
but this doesn't seem promising since the $\cos \theta$ in the denominator of the second term renders it no longer small near $\pm \pi/2$. For what it's worth, Mathematica gives an answer in terms of elliptic functions for $I(x)$ so I'm expecting $I(x)$ to be smooth enough that some approximation should exist.
|
There is something interesting : starting from the result given by Wofram Alpha
$$I(x)=\frac{2 \left(x^2 K\left(1-x^2\right)-E\left(1-x^2\right)\right)}{x^2-1}$$ Expanding as a series around $x=0$
$$I(x)=2 +\sum_{n=1}^\infty \Big[a_n-b_n \log(2)+c_n \log(x^2)\Big]\, x^{2n}$$ and the coefficients are
$$\left(
\begin{array}{cccc}
n & a_n & b_n & c_n \\
1 & \frac{3}{2} & 2 & \frac{1}{2} \\
2 & \frac{51}{32} & \frac{9}{4} & \frac{9}{16} \\
3 & \frac{105}{64} & \frac{75}{32} & \frac{75}{128} \\
4 & \frac{40985}{24576} & \frac{1225}{512} & \frac{1225}{2048} \\
5 & \frac{110439}{65536} & \frac{19845}{8192} & \frac{19845}{32768} \\
6 & \frac{2224761}{1310720} & \frac{160083}{65536} & \frac{160083}{262144}
\end{array}
\right)$$ where interesting patterns seem to appear.
Using the terms in the table, there is a good fit for $0 \leq x \leq \frac 34$ (the maximum error being $0.014$ at the right bound; it is only $6.92 \times 10^{-5}$ for $x=\frac 12$.
|
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|
How to solve for when this trigonometric function intersects the line $y=1$? How can I solve for $\alpha$ in
$$4\sin\left(\frac{\alpha}{2}\right)\cos^3\left(\frac{\alpha}{2}\right)\left(t-r\right)+\sin\left(\frac{\alpha}{2}\right)=1$$
on the domain $0\leq\alpha\leq\pi$? Clearly, one solution is when $\alpha=\pi$, but through plotting, it seems to only hold true when $t-r$ is less than a value around $0.3$. When $t-r$ is greater than this value, it seems to have different solutions.
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Excluding the trivial $a=\pi$, reset the problem as
$$(t-r)=\frac{1-\sin\left(\frac{\alpha}{2}\right)}{4\sin\left(\frac{\alpha}{2}\right)\cos^3\left(\frac{\alpha}{2}\right) }=\frac{1}{4} \left(1-\sin \left(\frac{a}{2}\right)\right) \csc \left(\frac{a}{2}\right)
\sec ^3\left(\frac{a}{2}\right)\tag 1$$
Let $a=2 \csc ^{-1}(x)$ and $(1)$ becomes
$$(t-r)=\frac{x-1}{4 \left(1-\frac{1}{x^2}\right)^{3/2}}$$
The derivative of the rhs is
$$\frac{x^2 \left(x^2+x-3\right)}{4 (x-1) (x+1)^2 \sqrt{x^2-1}}$$
So, in the given range, the zero of the first derivative corresponds to
$$x^2+x-3=0 \implies x=\frac{\sqrt{13}-1}{2}\implies a=2 \csc ^{-1}\left(\frac{\sqrt{13}-1}{2} \right)$$
At this point, the rhs of $(1)$
$$\frac{\left(\sqrt{13}-3\right) \left(\sqrt{13}-1\right)^3}{8 \left(10-2
\sqrt{13}\right)^{3/2}} =0.287482$$ which is a minimum value.
So, if $(t-r)$ is greater than this value, there are two roots.
|
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The sequence $(x_n)$ : $x_{n+2}=\frac{x_{n+1}\sqrt{x_n^2+1}+x_{n}\sqrt{x_{n+1}^2+1}-x_n-x_{n+1}}{x_{n+1}x_n-(\sqrt{x_{n+1}^2+1}-1)(\sqrt{x_n^2+1}-1)}$ The sequence $(x_n)$ is defined by the formula:
$$\left\{\begin{array}{cc}
x_1=1, x_2=\sqrt{3}\\
x_{n+2}=\frac{x_{n+1}\sqrt{x_n^2+1}+x_{n}\sqrt{x_{n+1}^2+1}-x_n-x_{n+1}}{x_{n+1}x_n-\big(\sqrt{x_{n+1}^2+1}-1\big)\big(\sqrt{x_{n}^2+1}-1\big)}, \quad n=1,2,3\dots
\end{array} \right.$$
Find $\lim\limits_{n\to \infty}x_n$.
I see:
$$\begin{array}{rl}
x_{n+2}&=\frac{x_{n+1}\sqrt{x_n^2+1}+x_{n}\sqrt{x_{n+1}^2+1}-x_n-x_{n+1}}{x_{n+1}x_n-\big(\sqrt{x_{n+1}^2+1}-1\big)\big(\sqrt{x_{n}^2+1}-1\big)}\\
&=\frac{x_{n}\big(\sqrt{x_{n+1}^2+1}-1 \big)+x_{n+1}\big(\sqrt{x_{n}^2+1}-1 \big)}{x_{n+1}x_n-\frac{x_n^2x_{n+1}^2}{\big(\sqrt{x_{n+1}^2+1}+1\big)\big(\sqrt{x_{n}^2+1}+1\big)}}\\
&=\frac{x_{n+1}\big(\sqrt{x_{n}^2+1}+1\big)+x_{n}\big(\sqrt{x_{n+1}^2+1}+1\big)}{\big(\sqrt{x_{n+1}^2+1}+1\big)\big(\sqrt{x_{n}^2+1}+1\big)-x_nx_{n+1}}
\end{array}$$
That's all I can do. So, I hope hints from you. Thank you.
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By rationalising the denominator, $x_{n+2}$ simplifies to $$\frac{\sqrt{x_n^2+1}\sqrt{x_{n+1}^2+1}+x_nx_{n+1}-1}{x_n+x_{n+1}}.$$
Letting $$x_n=\frac12\left(p_n-\frac1{p_n}\right),$$ that is (say) $$p_n=x_n+\sqrt{x_n^2+1}$$ and simplifying further, you get $$x_{n+2}=\frac{p_np_{n+1}-1}{p_n+p_{n+1}}$$
So, if $p_n=\cot\theta_n$, then $$x_{n+2}=\cot(\theta_n+\theta_{n+1}).$$
Furthermore, $$x_n=\frac12\left(\cot\theta_n-\frac1{\cot\theta_n}\right)=\cot{2\theta_n}.$$
It follows that $2\theta_{n+2}=\theta_n+\theta_{n+1},$ so that you can solve for $\theta_n$ explicitly from the initial conditions. The solution is
$$\frac{2\theta_2+\theta_1}{3}+\frac{4\theta_2-4\theta_1}{3}\left(-\frac12\right)^n,$$ which has limit $$\frac{2\theta_2+\theta_1}{3}$$.
You know that $\cot2\theta_1=1$ and $\cot2\theta_2=\sqrt{3}$, so that $\theta_1=\frac\pi8$ and $\theta_2=\frac\pi{12}$. Therefore, the limit of $\theta_n$ is $\frac{7\pi}{72}$, and so the limit of $x_n$ is $\cot\frac{7\pi}{36}$.
|
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|
If $T_1=7^7,T_2=7^{7^{7}},T_3=7^{7^{7^{7}}}$ and so on, what will be the tens digit of $T_{1000}$? $7^4$ ends with $2301$, so $7^{4k+r}$ ends with $7^r$ digits
$7^2\equiv1 \mod(4) $, and $7^7\equiv 3 \mod(4) $
Can we use the modulo function in exponent form? I think we will use these two properties, how can I proceed further?
|
If i correctly understand the question, we have to compute the value of $T_{1000}$ modulo $100$.
Long solution:
This is done in a few steps applying at each one the Euler indicator function $\varphi$, noting that basis $7$ is relatively prime to each of the numbers $\varphi(100)=40$, $\varphi(40)=16$, $\varphi(16)=8$ and so on, if we really need to so deeper...
We have for some odd power $N$ the relation
$$7^N=(8-1)^N\equiv(-1)^N\equiv -1\pmod8\ $$
Computations have been done in the units group of the ring $\Bbb Z/8$.
The above implies
$$
7^{7^N}
\equiv 7^{7^N\pmod {\varphi(16)}}
\equiv 7^{7^N\pmod 8}
\equiv 7^{-1}
\equiv 7\pmod{16}\ .
$$
Computations have been done in the units group of the ring $\Bbb Z/16$.
The above implies
$$
7^{7^{7^N}}
\equiv 7^{7^{7^N}\pmod {\varphi(40)}}
\equiv 7^{7^{7^N}\pmod {16}}
\equiv 7^7
\equiv 7\cdot 49^3
\equiv 7\cdot 9^3
\equiv 7\cdot 9\cdot 81
\equiv 7\cdot 9\cdot 1
\equiv 63
\equiv 23\pmod{40}\ .
$$
Computations have been done in the units group of the ring $\Bbb Z/40$.
The above implies finally
$$
7^{7^{7^{7^N}}}
\equiv 7^{7^{7^{7^N}}\pmod {\varphi(100)}}
\equiv 7^{7^{7^{7^N}}\pmod{40}}
\equiv 7^{23}
\equiv 7^3\cdot (7^4)^5
\equiv 7^3\cdot 2401^5
\equiv 7^3\cdot 1^5
\equiv 7^3
\equiv 343\equiv 63
\equiv 43\pmod{100}\ .
$$
Computations have been done in the units group of the ring $\Bbb Z/100$.
$\square$
Short solution: The above procedure is designed to work in general, this is written so for didactic purposes. In our case however, we finally need the power of an element of multiplicative order $4$ in $\Bbb Z/100$. So we need in fact in $T_{1000}=7^{7^{\text{et caetera}}}$ only the value of
$7^{\text{et caetera}}$ modulo four. This is $-1$ (as computed even modulo $8$). So we need $7^{-1}$ in $\Bbb Z/100$. This is $43$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Remainder Theorem Technique
Determine the remainder when $(x^4-1)(x^2-1)$ is divided by $1 + x + x^2$ (HMMT 2000, Guts Round)
A. Write the division in the form:
$$(x^4-1)(x^2-1)= (1 + x + x^2)Q(x) + R(x)$$
B. Multiply both sides by $x-1$:
$$(x-1)(x^4-1)(x^2-1)= (x^3-1)Q(x) + R(x)(x-1)$$
C. Substitute $x^3=1,x\neq1$, and reduce the resulting equation:
$$(x-1)(x-1)(x^2-1)= R(x)(x-1)$$
D. Divide both sides by $x-1$:
$$R(x)=(x-1)(x^2-1)=x^3 -x - x^2 + 1=-(x^2+x+1)+3=3$$
For someone who knows the method, is it valid to skip Steps B and D, directly substitute $x^3=1,x\neq1$ and use the fact that $x$ is a cube root of unity to get $x^2+x+1=0$.
|
Yes, you can skip those steps.
The verbose equivalent is that $$\begin{align}(x^4-1)(x^2-1)&=(x^4-x)(x^2-1)+(x-1)(x^2-1)\\
&=(x^2+x+1)x(x-1)(x^2-1)+(x-1)(x^2-1)
\end{align} $$
So $(x-1)(x^2-1)$ has the same remainder as $(x^4-1)(x^2-1)$ when dividing by $x^2+x+1.$
Answering the question in comments, yes, you can replace $x^2+x+1$ with zero.
Here, you can write:
$$(x-1)(x^2-1)=(x-1)(x^2+x+1)-(x-1)(x+2)$$
So you get the same remainder for $-(x-1)(x+2)$ as $(x-1)(x^2-1).$
|
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"url": "https://math.stackexchange.com/questions/4237066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Coefficient of $x^{10}$ in $f(f(x))$ Let $f\left( x \right) = x + {x^2} + {x^4} + {x^8} + {x^{16}} + {x^{32}}+ ..$, then the coefficient of $x^{10}$ in $f(f(x))$ is _____.
My approach is as follow
$f\left( {f\left( x \right)} \right) = f\left( x \right) + {\left( {f\left( x \right)} \right)^2} + {\left( {f\left( x \right)} \right)^4} + {\left( {f\left( x \right)} \right)^8} + ..$
Let $T = f\left( x \right);U = {\left( {f\left( x \right)} \right)^2};V = {\left( {f\left( x \right)} \right)^4};W = {\left( {f\left( x \right)} \right)^8}$
Let the coefficient of $x^{10}$ in $T $ is zero
Taking U
${\left( {f\left( x \right)} \right)^2} = {\left( {x + {x^2} + {x^4} + {x^8} + ..} \right)^2}$
Hence the coefficient of $x^{10}$ in $U$ is $2x^{10}$ hence $2$
For $V$ and $W$ it is getting complicated hence any short cut or easy method to solve it
|
Let's replace $10$ with a lower number, $5$, and illustrate the role of partitions in the result.
Only $f+f^2+f^4$ contributes to the coefficient of $x^5$. The $f$ term has none, of course.
The $f^2=f\cdot f$ term contributes a coefficient of $2$ from $x^1\cdot x^4$ and $x^4\cdot x^1$. Note that we get two terms because one contribution has $x^4$ from the first factor and the other has the $x^4$ factor coming from the second $f$ factor. Ordering of the partitions is important.
The $f^4$ term contributes a coefficient of $4$ which comes from $x^2\cdot x^3$ and $x^3\cdot x^2$ in the square of $f^2$. We get a contribution of $4$ here because the $x^3$ factor from $f^2$ had a coefficient of $2$ in $f^2$, which in turn came from $x^1\cdot x^2$ and $x^2\cdot x^1$ in the square of $f$.
So, the total of six terms in $x^5$, therefore a coefficient of $6$, came from the following partitions:
$5 = 1+4$ (from $f^2$)
$5 = 4+1$ (from $f^2$)
$5 = (1+1)+(1+2)$ (from $f^4$, with the parentheses showing the separate inputs to $f^4$ from $f^2$)
$5 = (1+1)+(2+1)$ (from $f^4$)
$5 = (1+2)+(1+1)$ (from $f^4$)
$5 = (2+1)+(1+1)$ (from $f^4$)
In other words:
*
*We picked partitions into powers of $2$, corresponding to the powers of $x$ in $f$.
*We picked those partitions where the number of terms is also a power of $2$, from the powers of $f$ you use in the outer function evaluation.
*For each partitioning satisfying the criteria above, we count all the distinguishable orderings.
Now try this partition scheme using $10$ as the input. There could be $1,2,4,$ or $8$ powers of $2$ in the partition, and for each partition that has different powers of $2$ there will be multiple orderings to count as we saw with the $x^5$ coefficient above.
We have $8+2$ with $2$ orderings, $4+4+1+1$ with six orderings, $4+2+2+2$ with four orderings, and $2+2+1+1+1+1+1+1$ with $28$ orderings. Thus the coeficient of $x^{10}$ will be $40$.
The entire series begins as
$x+2x^2+2x^3+3x^4+6x^5+8x^6+8x^7+16x^8+22x^9+\text{[spoiler]}x^{10}+...$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to calculate the residue of this function quickly (or by mathematica)? While looking at this post (The second answer), I find that it is to tedious to calculate the Residue of
$$ f(z)=-\left(\frac{z-1}{z+1}\right)^2\frac{2n/z}{z^{2n}-1} $$
at $-1$.
I do know that we can do this:
$$\operatorname*{Res}_{z=-1}f(z)=\frac{1}{2!}\frac{d^2}{dz^2}(f(z)\cdot(z+1)^3).$$
But I am not satisfied with this method. I have the following questions:
1. How many methods do we have to calculate this, can you provide me with an ingenious one?
2. I tried to calculate this via Mathematica like this:[]
If I let $n$ take concrete integers, say, $n=5$, we do get the right answer $-34$. However, it seems that we cann't get the right answer if we let $n$ be a variable. What's wrong here, how can we use mathematica to get a general answer (instead of concrete examples by letting n be some integers). I faced with this sort of problems in similar situations. Can you tell me what shoud I do, or just let me know that mathematica cannot do this! Thank you !
Addition As for Sangchul Lee's answer for my second question, I have another quesion. Why does the following code does not work, what's the difference between "Element[n, Integers]" and "Assumptions -> n \in Integers":
|
1. Note that we have
\begin{align*}
-\left(\frac{z-1}{z+1}\right)^2\frac{1}{z(z^2-1)}
&= -\frac{z-1}{(z+1)^3 z} \\
&= -\frac{2}{(z+1)^3} - \frac{1}{(z+1)^2} - \frac{1}{z+1} + \frac{1}{z}.
\end{align*}
Also, if we write $g(z) = \frac{z^{2n}-1}{z^2-1} = 1 + z^2 + z^4 + \cdots + z^{2(n-1)}$, then the Taylor series for $\frac{1}{g(z)}$ about $z = -1$ begins with
\begin{align*}
\frac{1}{g(z)}
&= \frac{1}{g(-1)} - \frac{g'(-1)}{g(-1)^2}(z+1) + \left(\frac{2g'(-1)^2}{g(-1)^3} - \frac{g''(-1)}{g(-1)^2} \right)\frac{(z+1)^2}{2} + \cdots \\
&= \frac{1}{n} + \frac{n-1}{n} (z+1) + \frac{(n-1)(2n-1)}{6n}(z+1)^2 + \cdots.
\end{align*}
Altogether, the residue of $f(z)$ at $z=-1$ can be computed by reading out the coefficient of $(z+1)^{-1}$ in the Laurent expansion of
$$ f(z) = -\left(\frac{z-1}{z+1}\right)^2\frac{1}{z(z^2-1)} \cdot \frac{2n}{g(z)}, $$
about $z=-1$, which is
\begin{align*}
\mathop{\underset{z=-1}{\mathrm{Res}}}f(z)
&= (2n) \biggl[ (-2)\left(\frac{(n-1)(2n-1)}{6n}\right) + (-1)\left(\frac{n-1}{n}\right) + (-1)\left(\frac{1}{n}\right) \biggr] \\
&= -\frac{2}{3}(2n^2+1).
\end{align*}
2. To compute this residue using Mathematica, you may do as follow:
|
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How to integrate $ \int e^{\cos( x)} \ dx$ I mainly did this for fun of it and am posting it here to have it reviewed and corrected if I made a mistake. This is a considerably simpler version of this solution I posted a couple months back.
Please comment with any corrections, questions, comments or concerns and hopefully you'll find it as interesting as I have! Also, I'm still fairly new to the community and I'm not certain I'm using the answer your own question format appropriately, so feel free to advise if that is the case.
Lastly, here's a link to the graphs on Desmos and a Mathematica function intended to be compared to NIntegrate!
eci[x_] := N[
Table[Sum[
Sum[Sin[x*(n - 2 k)]/((n - k)!k!(n - 2 k)*2^(n - 1)), {k, 0,
Floor[(n - 1)/2]}], {n, 1, L}] + BesselI[0, 1]*x, {L, 100, 100}]];
eciN[x_] := NIntegrate[E^Cos[z], {z, 0, x}];
|
$\displaystyle \begin{array}{{>{\displaystyle}l}}
Evaluating\ \int e^{cos( x)} \ dx:\\
Starting\ with\ the\ taylor\ series\ definition\ we\ have\ e^{cos( x)} =\ \sum _{n=0}^{\infty }\frac{cos^{n}( x)}{n!}\\
Now,\ using\ the\ complex\ exponential\ definition\ of\ cosine,\ we\ can\ obtain\ a\ generalized\ summation\\
for\ expanding\ integer\ powers\ of\ cos( x)\\
\\
cos( x) =\frac{e^{i\ x} +e^{-ix} \ }{2} \ \therefore \ cos^{n}( x) =\left(\frac{e^{i\ x} +e^{-i\ x} \ }{2}\right)^{n} =\frac{\left( e^{i\ x} +e^{-i\ x}\right)^{n}}{2^{n}}\\
\\
By\ evaluating\ the\ expansions\ we\ can\ obtain:\\
\\
\frac{\left( e^{i\ x} +e^{-i\ x}\right)^{n}}{2^{n}} =\sum _{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }\binom{n}{k}\frac{cos( x\cdot ( n-2k))}{2^{n-1}} \ +\frac{cos^{2}\left(\frac{\pi n}{2}\right) \cdot \left( 2\left( n-\left\lfloor \frac{n}{2}\right\rfloor \right)\right) !}{2^{n} \cdot \left(\left( n-\left\lfloor \frac{n}{2}\right\rfloor \right) !\right)^{2}}\\
\\
Now\ substituting\ the\ summation\ back\ into\ the\ taylor\ series,\ the\ function\ f( x) =\ e^{cos( x)} \ can\ now\\
be\ expressed\ by\ the\ following\ infinite\ summations:\\
\\
f( x) =\ e^{cos( x)} =1+\sum _{n=1}^{\infty }\sum _{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }\frac{cos( x( n-2k))}{2^{n-1} \cdot k!\cdot ( n-k) !} +\sum _{n=1}^{\infty }\frac{cos^{2}\left(\frac{\pi n}{2}\right) \cdot \left( 2\left( n-\left\lfloor \frac{n}{2}\right\rfloor \right)\right) !}{2^{n} \cdot \left(\left( n-\left\lfloor \frac{n}{2}\right\rfloor \right) !\right)^{2}}\\
\\
Now\ separating\ the\ constant\ terms\ not\ dependent\ on\ x\ we\ have:\ 1+\sum _{n=1}^{\infty }\frac{cos^{2}\left(\frac{\pi n}{2}\right) \cdot \left( 2\left( n-\left\lfloor \frac{n}{2}\right\rfloor \right)\right) !}{2^{n} \cdot \left(\left( n-\left\lfloor \frac{n}{2}\right\rfloor \right) !\right)^{2}}\\
Becuase\ the\ partial\ sums\ of\ \sum _{n=1}^{\infty }\frac{cos^{2}\left(\frac{\pi n}{2}\right) \cdot \left( 2\left( n-\left\lfloor \frac{n}{2}\right\rfloor \right)\right) !}{2^{n} \cdot \left(\left( n-\left\lfloor \frac{n}{2}\right\rfloor \right) !\right)^{2}} \ are\ only\ non\ zero\ for\ even\ n,\\
we\ can\ rewrite\ this\ as\ 1+\ \ \sum _{n=1}^{\infty }\frac{1}{2^{2n} \cdot n!^{2}} =\sum _{n=0}^{\infty }\frac{1}{2^{2n} \cdot n!^{2}} =I_{0}( 1) ,\\
where\ I_{n}( z) \ is\ the\ modified\ Bessel\ function\ of\ the\ first\ kind.\\
\\
We\ can\ now\ express\ the\ function\ f( x) =e^{cos( x)} as:\ f( x) =\ I_{0}( 1) +\sum _{n=1}^{\infty }\sum _{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }\frac{cos( x( n-2k))}{2^{n-1} \cdot k!\cdot ( n-k) !}\\
\\
This\ now\ becomes\ the\ fairly\ simple\ integration:\\
F( x) =\int f( x) \ dx=\int I_{0}( 1) \ dx+\ \sum _{n=1}^{\infty }\sum _{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }\int \frac{cos( x\cdot ( n-2k))}{( n-k) !\cdot k!\cdot 2^{n-1}} dx\\
\\
Finally,\ F( x) =\int e^{cos( x)} \ dx=x\cdot I_{0}( 1) \ +\ \sum _{n=1}^{\infty }\sum _{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }\frac{sin( x\cdot ( n-2k))}{( n-k) !\cdot k!\cdot ( n-2k) \cdot 2^{n-1}} +C\\
\\
Additionally,\ the\ sinusoid\ produced\ by\ the\ function\ g( x) =e^{cos( x)} -I_{0}( 1) \ can\ be\ expressed\ by:\\
g( x) =\sum _{n=1}^{\infty }\sum _{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }\frac{cos( x( n-2k))}{2^{n-1} \cdot k!\cdot ( n-k) !} \ and\ likewise;\\
G( x) =\int g( x) \ dx=\ \sum _{n=1}^{\infty }\sum _{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }\frac{sin( x\cdot ( n-2k))}{( n-k) !\cdot k!\cdot ( n-2k) \cdot 2^{n-1}} +C,\ shown\ in\ the\ figure\ below:\\
\end{array}$
$\displaystyle \begin{array}{{>{\displaystyle}l}}
Finally,\ some\ other\ interesting\ take\ aways:\ \\
\\
-\ The\ maximum\ and\ minimum\ points\ of\ the\ sinusoid\ occur\ at:\ x=2\pi n\pm \left( \ 1+\frac{2}{\pi } -\frac{3}{\pi ^{2}}\right)\\
-\ F\left(\frac{\pi }{2}\right) =\ \int _{0}^{\frac{\pi }{2}} e^{cos( x)} \ dx\ =\frac{\pi ( I_{0}( 1) +L_{0}( 1))}{2} \ and\ G\left(\frac{\pi }{2}\right) =\int _{0}^{\frac{\pi }{2}}\left( e^{cos( x)} -I_{0}( 1)\right) \ dx=\ \frac{\pi \cdot L_{0}( 1)}{2}\\
\ \ \ \ ( \ Here,\ L_{n}( z) \ is\ the\ modified\ Struve\ function.)\\
\\
-\ Similarly\ to\ \int e^{cos( x)} \ dx,\ \int e^{sin( x)} \ dx\ can\ be\ evaulated\ by\ making\ the\ substitution\ x\rightarrow \left( x-\frac{\pi }{2}\right)\\
\ \ \ and\ adding\ \ \int _{0}^{\frac{\pi }{2}}\left( e^{cos( x)} -I_{0}( 1)\right) \ dx\ to\ obtain:\ \\
\ \ \ \int e^{sin( x)} \ dx=\ \frac{\pi \cdot L_{0}( 1)}{2} +x\cdot I_{0}( 1) +\ \sum _{n=1}^{\infty }\sum _{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }\frac{sin\left(\left( x-\frac{\pi }{2}\right) \cdot ( n-2k)\right)}{( n-k) !\cdot k!\cdot ( n-2k) \cdot 2^{n-1}} +C\\
\\
-\ A\ somewhat\ interesting\ take\ away\ may\ be\ that\ \\
\ \ \ L_{0}( 1) =\frac{2}{\pi }\sum _{n=1}^{\infty }\sum _{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }\frac{cos\left(\frac{\pi }{2} \cdot ( n-2k)\right)}{2^{n-1} \cdot k!\cdot ( n-k) !} =\sum _{n=0}^{\infty }\frac{\left(\frac{1}{2}\right)^{2n+1}}{\left(\left( n+\frac{1}{2}\right) !\right)^{2}}
\end{array}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove by induction $1^2 + 2^2 + ... + n^2 = \frac{n(n+1)(2n+1)}{6}$ I am practicing mathematical induction and I got this question.
$$1^2 + 2^2 + ... + n^2 = \frac{n(n+1)(2n+1)}{6}$$
I am wondering, if this is correct way to do it? Would my answer get accepted if it was an exam?
I put n+1 in the places of n. So Im trying to get this final result:
$$\frac{(n+1)(n+2)(2n+3)}{6}$$
Getting the left side done:
$$\frac{n(n+1)(2n+1)}{6} + \frac{6(n+1)^2}{6} = \frac{n(n+1)(2n+1) +6(n+1)^2}{6} = \frac{(n+1)(n+2)(2n+3)}{6}$$
Am I doing this completely wrong, can someone guide me?
|
It is quite well done, but you should check first that the statement is true when $n=1$, and then to say that if$$1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}6,\tag1$$then$$1^2+2^2+\cdots+n^2+(n+1)^2=\frac{n(n+1)(2n+1)}6+(n+1)^2$$(due to $(1)$), followed by the computations that you have shown us at the end of your question.
|
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"timestamp": "2023-03-29T00:00:00",
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Solving an analytic geometry problem with euclidean geometry
$\mathrm{OABC}$ is a tetrahedron with $\overline{\mathrm{OA}}=1$.
There is a point $P$ on $\triangle \mathrm{ABC}$ such that $\cos^2 \alpha+\cos^2 \beta + \cos^2 \gamma = \frac{11}{6}$ where $\alpha=\angle\mathrm{AOP}$, $\beta=\angle\mathrm{BOP}$ and $\gamma=\angle\mathrm{COP}$.
Describe the trace of such $P$.
We can solve this problem by using the following lemma:
Let's imagine the tetrahedron with four vertices $O(0,0,0)$, $A(a,b,c)$, $B(b,c,a)$ and $C(c,a,b)$.
Since $\overline{\mathrm{OA}}=\overline{\mathrm{OB}}=\overline{\mathrm{OC}}=\overline{\mathrm{AB}}=\overline{\mathrm{BC}}=\overline{\mathrm{CA}}$, we have
$a^2+b^2+c^2=(a-b)^2+(b-c)^2+(c-a)^2\Rightarrow a^2+b^2+c^2=2(ab+bc+ca)$.
And since $\vec{\mathrm{OA}}=(a,b,c)$, $\vec{\mathrm{OP}}=(x,y,z)$, we have $\cos^2 \alpha=\left(\frac{\vec{\mathrm{OA}} \cdot \vec{\mathrm{OP}}}{\vert \vec{\mathrm{OA}} \vert \vert\vec{\mathrm{OP}}\vert}\right)^2=\frac{(ax+by+cz)^2}{(a^2+b^2+c^2)(x^2+y^2+z^2)}$.
WLOG $\cos^2 \beta=\frac{(bx+cy+az)^2}{(a^2+b^2+c^2)(x^2+y^2+z^2)}$, $\cos^2 \gamma=\frac{(cx+ay+bz)^2}{(a^2+b^2+c^2)(x^2+y^2+z^2)}$.
Then by the given equation, $\frac{(ax+by+cz)^2+(bx+cy+az)^2+(cx+ay+bz)^2}{(a^2+b^2+c^2)(x^2+y^2+z^2)}=1+\frac{2(ab+bc+ca)(xy+yz+zx)}{(a^2+b^2+c^2)(x^2+y^2+z^2)}=\frac{11}{6}$.
By substituting $a^2+b^2+c^2=2(ab+bc+ca)$, we obtain $5(x^2+y^2+z^2)=6(xy+yz+zx)$.
Now, observe that $P$ is on the plane $x+y+z=a+b+c$.
So $2(xy+yz+zx)=(x+y+z)^2-(x^2+y^2+z^2)=(a+b+c)^2-(x^2+y^2+z^2)$ and $8(x^2+y^2+z^2)=3(a+b+c)^2$.
As a conclusion, the trace of $P$ is a circle which is an intersection between the sphere $x^2+y^2+z^2=\frac{3}{8}(a+b+c)^2$ and the plane $x+y+z=a+b+c$.
I think it is a quite simple and nice solution, but I do not want to use such analytic methods.
I tried many times to solve this only with euclidean geometry, but I could not find a better strategy.
Would you help me?
|
Here's a much shorter answer.
Assuming $OABC$ is a regular tetrahedron of edge length 1.
Place point O at the origin $(0,0,0)$, and let face $ABC$ be parallel to the $xy$ plane, then points $A, B, C$ can be taken as
$A = (\sin \theta , 0, \cos \theta )$
$B = (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta) $
$C = (\sin \theta \cos(- \phi) , \sin \theta \sin(-\phi) , \cos \theta ) $
where $\cos \theta = \sqrt{\dfrac{2}{3}} $ and $\sin \theta = \dfrac{1}{\sqrt{3}} $
and $\phi = \dfrac{2 \pi}{3} $
Hence,
$A = (\dfrac{1}{\sqrt{3}} , 0, \sqrt{\dfrac{2}{3}} )$
$B = (-\dfrac{1}{2 \sqrt{3}}, \dfrac{1}{2}, \sqrt{\dfrac{2}{3}} ) $
$C = (-\dfrac{1}{2 \sqrt{3}} , -\dfrac{1}{2} , \sqrt{\dfrac{2}{3}}) $
Now let $Q$ be a unit vector (lying on the unit sphere that has point $A, B,C$ on it), then $Q$ can be written as
$Q = (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)$
Since vectors $A, B, C, Q$ all have unit length, it follows that
$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = (A.Q)^2 + (B.Q)^2 + (C.Q)^2 $
In matrix-vector notation, we have,
$ (A.Q)^2 + (B.Q)^2 + (C.Q)^2 = Q^T ( A A^T + B B ^ T + C C^T ) Q $
The $3 \times 3 $ matrix $(A A^T + B B ^ T + C C^T )$ can be evaluated directly, and it comes to:
$(A A^T + B B ^ T + C C^T ) = \begin{bmatrix} \frac{1}{2} && 0 && 0 \\ 0 && \frac{1}{2} && 0 \\ 0 && 0 && 2 \end{bmatrix} $
Hence the condition becomes
$ \dfrac{1}{2} \sin^2 \theta + 2 \cos^2 \theta = \dfrac{11}{6} $
So that,
$ 3 \sin^2 \theta + 12 \cos^2 \theta = 11 $
which becomes,
$ \dfrac{1}{2} ( 15 + 9 \cos 2 \theta ) = 11 $
whose solution is $ \theta = \frac{1}{2} \cos^{-1} \dfrac{7}{9} $
Note that $\cos 2 \theta = \dfrac{7}{9}$ and that $\sin 2 \theta = \dfrac{4 \sqrt{2}}{9} $
The above implies that $Q$ lies on a cone whose semi-vertical angle is $\theta$ as found above. Therefore, point $P$ lies on a circle whose radius is
$R = \sqrt{\dfrac{2}{3}} \tan \theta = \sqrt{\dfrac{2}{3}} \dfrac{\sin 2 \theta}{1 + \cos 2 \theta} = \sqrt{\dfrac{2}{3}} \dfrac{ 4 \sqrt{2} }{ 16 } = \dfrac{1}{\sqrt{12}}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Applying binomial expansion and using algebra to find the value of $a+b$ Q) In the expansion of
$f(x)=(1 + ax)^4 (1 + bx)^5$
where $a$ and $b$ are positive integers,
the coefficient of $x^2$ is 66.
Evaluate $a+b$.
My working:
After expanding the expression I simplified it and got
$5b^2+10ab+3a^2=33$
After further simplification, I managed to get
$5(a+b)^2-2a^2=33$
but I'm not sure what to do next.
|
Your result is correct.
If $a,b\in\mathbb Z^{+}$ then we can write,
$$\begin{align}&5b^2+3a^2≥2\sqrt {15}ab\\
\implies &5b^2+3a^2+10ab≥ab \left(10+2\sqrt {15}\right)\\
\implies &33≥\left(10+2\sqrt {15}\right)ab\\
\implies &ab≤\frac{33}{10+2\sqrt {15}}<2\\
\implies &a=b=1.\end{align}$$
This means, the solution doesn't exist.
|
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|
All Unique Three Digit Combinations using four 8s, three 9s, seven 1's and three 5s So basically I have find the unique possible combinations of 3 digit number using four 8s, three 9s, seven 1's and three 5s, e.g.
"888"
"819"
"891"
"855", ..., etc
Because of having same number present multiple times I am not able to determine the answer. Some of the approaches tried were $\frac{17!}{4!3!7!3!}$. All answers are appreciated.
Thanks
|
These types of questions can be easily solved by exponential generating functions. It will give you power.
It is said that there are four $8's$ , three $9's$ , seven $1's$ , and three $5's$ .In other words , we have $8,8,8,8,9,9,9,1,1,1,1,1,1,1,5,5,5.$
We want to find the number of all possible $3$ digits numbers using these digits without any restriction. So , we should firstly write exponential generating function forms of each digits.
Exponential generating function of $8$ : $$\bigg(1+ \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}\bigg)$$
Exponential generating function of $9$ : $$\bigg(1+ \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} \bigg)$$
Exponential generating function of $1$ : $$\bigg(1+ \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} +\frac{x^6}{6!} +\frac{x^7}{7!}\bigg)$$
Exponential generating function of $5$ : $$\bigg(1+ \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} \bigg)$$
Now , to find the number of all possible $3$ digits , we should find the expansion of these exponential generating functions such that $$\bigg(1+ \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}\bigg) \times \bigg(1+ \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} \bigg)^2 \times \bigg(1+ \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} +\frac{x^6}{6!} +\frac{x^7}{7!}\bigg) $$
After that , we should find the coefficient of $\frac{x^3}{3!}$ or find $x^3$ and multiply it by $3!$. It will give you the result.
Moreover , you can find the number of all possible arrangements of any lenght using this method . For example , if you wnder about lenght $5$ , then we should find the coefficient of $\frac{x^5}{5!}$ or find $x^5$ and multiply it by $5!$.
|
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|
Intuition behind getting two straight lines as result Question:
Find the equation of the straight line that passes through $(6,7)$ and makes an angle $45^{\circ}$ with the straight line $3x+4y=11$.
My solution (if you want, you can skip to the bottom):
Manipulating the given equation to get it to the slope-intercept form,
$$3x+4y=11...(i)$$
$$\implies 4y=-3x+11$$
$$\implies y=\frac{-3}{4}x+\frac{11}{4}$$
Let, the slope of (i) is $m_1=\frac{-3}{4}$, and the slope of our desired equation is $m_2$. Now, according to the question,
$$\tan(45^{\circ})=\pm\frac{-\frac{3}{4}-m_2}{1-\frac{3}{4}m_2}...(1)$$
$$\implies 1=\pm\frac{-\frac{3}{4}-m_2}{1-\frac{3}{4}m_2}$$
$$\implies \pm \frac{3}{4}+m_2=1-\frac{3}{4}m_2...(ii)$$
Picking positive value from (ii),
$$\frac{3}{4}+m_2=1-\frac{3}{4}m_2$$
$$\implies m_2(1+\frac{3}{4})=1-\frac{3}{4}$$
$$\implies m_2=\frac{1-\frac{3}{4}}{1+\frac{3}{4}}$$
$$\implies m_2=\frac{1}{7}$$
Picking negative value from (ii),
$$-\frac{3}{4}-m_2=1-\frac{3}{4}m_2$$
$$\implies -\frac{3}{4}-m_2=1-\frac{3}{4}m_2$$
$$\implies -m_2(1-\frac{3}{4})=1+\frac{3}{4}$$
$$\implies m_2=-\frac{1+\frac{3}{4}}{1-\frac{3}{4}}$$
$$\implies m_2=-7$$
Picking $m_2=\frac{1}{7}$, the equation of the straight line that passes through $(6,7)$,
$$\frac{y-7}{x-6}=\frac{1}{7}$$
$$\implies 7y-49=x-6$$
$$\implies -x+7y-43=0$$
$$\implies x-7y+43=0...(iii)$$
Picking $m_2=-7$, the equation of the straight line that passes through $(6,7)$,
$$\frac{y-7}{x-6}=-7$$
$$\implies -7x+42=y-7$$
$$\implies 7x+y-49=0...(iv)$$
The general form of equation (1) is,
$$\tan\theta=\pm \frac{m_1-m_2}{1+m_1m_2}$$
Here, $\pm$ has been included to include both the acute and the obtuse angles that are formed when two lines with slopes $m_1$ and $m_2$ intersect each other.
Now, I used this equation to find the straight line that makes $45^{\circ}$ with (i). Why am I getting 2 values of $m_2$ when there is only one value of $m_2$ in the general form of the equation? How can I reconcile between my getting of 2 values of $m_2$ with the $\pm$ sign arising due to the acute and obtuse angles?
|
We have to be mindful about sign of slope
To the given inclination $ \tan^{-1} (-3/4) = - 36. 87 ^{\circ}$ add, resulting in $ 45 ^{\circ}- 36.87 ^{\circ} = 8.13^{\circ}$ positive ( counter clockwise) to x-axis and you have only one straight line solution AB. Other solution is spurious, needs to be discarded.
|
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|
Prove that $\frac{1}{2(n+2)}<\int_0^1\frac{x^{n+1}}{x+1}dx$ $\displaystyle\int\limits_0^1\frac{x^{n+1}}{x+1}dx=\left[\frac{x^{n+2}}{(n+2)(x+1)}\right]_0^1+\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$
$\displaystyle\int\limits_0^1\frac{x^{n+1}}{x+1}dx=\frac{1}{2(n+2)}+\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$
If we can prove that $\displaystyle\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$ is always greater than $0$ we can find the minima of the function. How can we prove that the term is always positive?
And how can we prove that $\displaystyle \int\limits_0^1\frac{x^{n+1}}{x+1}dx<\frac{1}{2(n+1)}$
|
Hint
You know that $x \in \left[0;1\right]$ so
$$
\frac{1}{x+1}>\frac{1}{2}
$$
What does this imply for $\displaystyle \int_{0}^{1}\frac{x^n}{1+x}$ ?
|
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|
Independent chances of $3$ events ${a\over{a+x}}$, ${b\over{b+x}}$, ${c\over{c+x}}$ Here's a problem from my probability textbook:
Of three independent events the chance that the first only should happen is a; the chance of the second only is $b$; the chance of the third only is $c$. Show that the independent chances of the three events are respectively$${a\over{a+x}}, \quad {b\over{b+x}}, \quad {c\over{c+x}},$$where $x$ is a root of the equation$$(a+x)(b+x)(c+x) = x^2.$$
Here's what I did. We have$$a = p_1(1 - p_2)(1 - p_3), \quad b = (1-p_1)p_2(1-p_3), \quad c = (1-p_1)(1-p_2)p_3.$$Without loss of generality let's consider $a$. We have$$p_1 = {a\over{(1 - p_2)(1 - p_3)}}.$$If we assume the result we want to show, then this equals$$p_1 = {a\over{\left(1 - {b\over{b+x}}\right)\left(1 - {c\over{c+x}}\right)}} = {a\over{{{x^2}\over{(b+x)(c+x)}}}} = {a\over{a+x}}.$$However, we assumed in part what we wanted to show, which possibly makes this circular.
Another observation I noticed is that$$p_1p_2p_3 + p_1p_2(1-p_3) + p_1(1-p_2)p_3 + (1-p_1)p_2p_3 + (1-p_1)(1-p_2)(1-p_3) + a + b + c = 1.$$However, despite what I've tried, I'm stuck and do not know how to proceed further. Could anybody help me? Is there a way to turn my circular approach into a noncircular one?
|
Let $\alpha$ be a root of $(a+x)(b+x)(c+x)=x^2$
$\implies (a+\alpha)(b+\alpha)(c+\alpha)=\alpha^2$
Now, $(p_1, p_2, p_3)$ satisfies the system of equations
$\begin{align}
(1) \quad p_1(1 - p_2)(1 - p_3) = a \\
(2) \quad (1-p_1)p_2(1-p_3) = b \\
(3) \quad (1-p_1)(1-p_2)p_3 = c
\end{align}$
It's sufficient to show that $\left(\frac{a}{a+\alpha},\frac{b}{b+\alpha}, \frac{c}{c+\alpha}\right)$ satisfies the above system of equations.
$(1) \quad \frac{a}{a+\alpha}\left(1 - \frac{b}{b+\alpha}\right)\left(1 - \frac{c}{c+\alpha}\right)=\frac{a}{a+\alpha}.\frac{\alpha}{b+\alpha}.\frac{\alpha}{c+\alpha}= \frac{a.\alpha^2}{(a+\alpha)(b+\alpha)(c+\alpha)} = \frac{a.\alpha^2}{\alpha^2}=a, \quad$ assuming $\alpha \neq 0$
$(2) \quad \left(1 - \frac{a}{a+\alpha}\right)\frac{b}{b+\alpha}\left(1 - \frac{c}{c+\alpha}\right)=\frac{\alpha}{a+\alpha}.\frac{b}{b+\alpha}.\frac{\alpha}{c+\alpha}= \frac{b.\alpha^2}{(a+\alpha)(b+\alpha)(c+\alpha)} = \frac{b.\alpha^2}{\alpha^2}=b, \quad$ assuming $\alpha \neq 0$
$(3) \quad \left(1 - \frac{a}{a+\alpha}\right)\left(1 - \frac{b}{b+\alpha}\right)\frac{c}{c+\alpha}=\frac{\alpha}{a+\alpha}.\frac{\alpha}{b+\alpha}.\frac{c}{c+\alpha}= \frac{c.\alpha^2}{(a+\alpha)(b+\alpha)(c+\alpha)} = \frac{c.\alpha^2}{\alpha^2}=c, \quad$ assuming $\alpha \neq 0$
|
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|
Definite Integral $\int_{1-\sqrt{a}}^{1+\sqrt{a}} \ln(1-z t^2)\frac{\sqrt{4a-(t^2-1-a)^2}}{t}\, dt$ Suppose $z<\frac{1}{(1+\sqrt{a})^2}$, $0<a<1$. I need to compute the following integral as function of z:
$$
I(z)= \int_{1-\sqrt{a}}^{1+\sqrt{a}} \ln(1-z t^2)\frac{\sqrt{4a-(t^2-1-a)^2}}{t}\, dt
$$
So far, following the trick in this question I defined
$$
J'(p)=\int_{1-\sqrt{a}}^{1+\sqrt{a}} \frac{t}{p t^2 -1}\sqrt{4a-(t^2-1-a)^2}\, dt
$$
With the change of variable $t^2-1-a = x$,
$$
J'(p)=\frac{1}{2} \int_{-2\sqrt{a}}^{2\sqrt{a}} \frac{\sqrt{4a-x^2}}{px + p(a+1)-1}\, dx
$$
Doing the Euler substitution $x = 2\sqrt{a}\frac{t^2-1}{t^2+1}$, we have (using Mathematica)
$$
J'(p) = 16 a \int_0^{\infty}\frac{t^2}{(t^2+1)^2 \bigg( \Big[p\big(\sqrt{a}+1\big)^2-1\Big]t^2+p\big(\sqrt{a}-1)^2-1\bigg)} \, dt
$$
\begin{equation}
\begin{split}
J'(p) &= 16 a \int_0^{\infty}\frac{t^2}{(t^2+1)^2 \bigg( \Big[p\big(\sqrt{a}+1\big)^2-1\Big]t^2+p\big(\sqrt{a}-1)^2-1\bigg)} \, dt \\
&= \frac{\pi}{2} \frac{-1+p(a+1)+\Big(1-\big(\sqrt{a}-1\big)^2p\Big)\sqrt{\frac{-1+\big(\sqrt{a}+1\big)^2p}{-1+\big(\sqrt{a}-1\big)^2p}}}{p^2}
\end{split}
\end{equation}
Now, to find $I(z) = \int_0^z J'(p) \, dp$, there is a problem that in the $J'(p)$ there is a logarithm of $p$ which makes it undefined at 0.
|
With the substitution $x=\frac{ t^2-1-a}{2\sqrt a}$, along with the shorthands $p= \frac{2z\sqrt a}{1-z-az}$, $q= \frac{1+a}{2\sqrt a} $, the integral simplifies to
\begin{align}
I(z)=& \int_{1-\sqrt{a}}^{1+\sqrt{a}} \ln(1-z t^2)\frac{\sqrt{4a-(t^2-1-a)^2}}{t}\, dt\\
=& \sqrt a\int_{-1}^1 [\ln(1-z-az)+\ln (1-p x)]\frac{\sqrt{1-x^2}}{q+x}dx \\
= & \>\pi a\ln(1-z-az)+\sqrt a \>J(p)
\end{align}
where $\int_{-1}^1 \frac{ \sqrt{1-x^2}}{q+x}dx = \pi \sqrt a$ is used and
\begin{align}
J(s) =&\int_{-1}^1 \frac{\ln (1-s x) \sqrt{1-x^2}}{q+x}dx \\
J’(s)=& -\int_{-1}^1
\frac{x \sqrt{1-x^2}}{(1-s x)(q+x)}dx \\
=& \frac1{1+sq}\int_{-1}^1 \left(\frac{q\sqrt{1-x^2}}{x+q}- \frac{\sqrt{1-x^2}}{1-s x}\right)dx\\
= & \frac\pi{1+sq}\left(\frac{1+a}2
-\frac{1-\sqrt{1-s^2}}{s^2} \right)
\end{align}
Then
\begin{align}
I(z)=& \>a\pi \ln(1-z-az)+\sqrt a \int_0^p J’(s)ds
= - \pi\sqrt{a} \int_0^p \frac{1-\sqrt{1-s^2}}{s^2(1+sq)} ds\\
= &- \pi\sqrt{a}\bigg[ \sqrt{q^2-1} \> \bigg(\tanh^{-1}\frac{p+q}{\sqrt{(q^2-1)(1-p^2)}}
- \tanh^{-1} \frac{q}{\sqrt{q^2-1}}\bigg)
\\& \hspace{9mm}-q\,{\rm \tanh^{-1}}\sqrt{1-p^2}
+ q \ln\frac{2(1+pq)}{p} +\frac{\sqrt{1-p^2}-1}p \bigg]
\end{align}
|
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|
How can I find the limit of the following? What is the limit of this
$$\lim_{x\to+\infty}\left(1+\frac{4}{2x+3}\right)^x$$
I know that $$\lim_{x\to+\infty}\left(1+\frac{4}{2x}\right)^x$$ will give me $$e^2$$ but the I dont know what to do with the 3.
I have tried bringing them to a common denominator so I got
$$\lim_{x\to+\infty}\left(\frac{2x+7}{2x+3}\right)^x=\lim_{x\to+\infty}\left(e^{x\ln{(\frac{2x+7}{2x+3})}}\right)$$
And then Im stuck again
|
Let $f(x) = 1 + \frac{4}{ 2x+3}$
And $g(x) = x $
Your limit is of the form $(1)^{\infty}$ , whose value is equal to $$e^{ \lim_{ x \to {\infty}} {(f(x) - 1 )}{g(x)}} $$
Therefore , $$L = e^{ \lim_{ x \to {\infty}} {(\frac{4x}{2x+3} )}} $$
Now $$\frac{4x}{2x+3} = {\frac{4x+6}{2x+3} } - \frac{6}{2x+3} = 2 - \frac{6}{2x+3}$$
Therefore , L becomes
$$ L = e^{ \lim_{x \to \infty} { 2 - \frac{6}{2x+3}} } = e^2$$
|
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|
Prove that $ \int_0^1\frac{x\ln(x)}{1+x^2+x^4}dx=\frac{1}{36}\Big(\psi^{(1)}(2/3)-\psi^{(1)}(1/3)\Big) $ I am having trouble with the following integral
Prove that $$ \int_0^1\frac{x\ln(x)}{1+x^2+x^4}dx=\frac{1}{36}\Big(\psi^{(1)}(2/3)-\psi^{(1)}(1/3)\Big)$$
$$I=\int_0^1\frac{x\ln(x)}{1+x^2+x^4}dx=\int_0^1\frac{\ln(u^2)}{2(1+u+u^2)}du=\int_0^1\frac{\ln(u)}{(1+u+u^2)}du$$
let $x^2=u\rightarrow \frac{du}{dx}=2x$
How does one proceed from here? Is my approach correct? Thank you for your time
|
Alternatively
\begin{align} \int_0^1\frac{x\ln x}{1+x^2+x^4}dx
= &\int_0^1\frac{x\ln x}{(e^{i\frac\pi3}+x^2)(e^{-i\frac\pi3}+x^2)}dx \\
=& \>\frac1{\sin\frac\pi3 }\>\Im \int_0^1 \frac{e^{i\frac\pi3}x\ln x}{1+ e^{i\frac\pi3}x^2 }dx
\overset{x^2\to x} =\frac{1}{2\sqrt3}\Im \text{Li}_2(- e^{i\frac\pi3})
\end{align}
|
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|
Show that $n(x-y)(xy)^{(n-1)/2} \leq x^n - y^n$ for nonnegative $x,y$ and integers $n$ Show for $x,y\geq 0$ and $n \in \mathbb N$ that:
$$n(x-y)(xy)^{(n-1)/2} \leq x^n - y^n$$
What I've tried
Proving by induction seems natural since the case for $n=2$ and $n=3$ are straightforward to show.
Assuming the claim is true for $n = k-1$, we want to show it for $n = k$:
$$(k-1)(x-y)(xy)^{(k-1-1)/2} \leq x^{(k-1)} - y^{(k-1)}$$
Now, multiplying both sides by $x+y$:
$$x^k - y^k - y^{k-1}x + x^{k-1}y \geq (k-1)(x+y)(x-y)(xy)^{(k-1-1)/2}$$
Recall that $(x+y)(x-y) = x^2 - y^2 \geq 2(x-y)\sqrt{xy}$
So then the whole LHS in the previous display is:
$$\geq 2k(x-y)(xy)^{(k-1)/2} - x^{k/2+1}y^{(k-1-1)/2} +x^{(k-1-1)/2}y^{n/2 + 1}$$
So if I could show that:
$$y^{k-1}x - x^{k-1}y - x^{k/2+1}y^{(k-1-1)/2} +x^{(k-1-1)/2}y^{k/2 + 1} \geq k(x-y)(xy)^{(k-1)/2}$$
I would be done. I think it should be possible to factor this expression or perhaps apply AM-GM in a clever way to show this, but I'm having trouble seeing it.
EDIT: as pointed out in the comments, we need the condition $x \geq y$. Notice on the LHS we have:
$$- y^{k-1}x + x^{k-1}y $$
but applying the condition on $x$ and $y$ we know that expression is at least $0$ since $-y^{k-1}x \geq -y^k$ and $x^{k-1}y \geq y^k$.
Now, applying AM-GM to $x+y$ we have that:
$$(k-1)(x+y)(x-y)(xy)^{(k-1-1)/2} \geq 2(k-1)(x-y)(xy)^{(k-1)/2}$$
$$= 2k(x-y)(xy)^{(k-1)/2} - 2(x-y)(xy)^{(k-1)/2}$$
But $2(x-y)(xy)^{(k-1)/2} \leq k(x-y)(xy)^{(k-1)/2}$ for all $k \geq 2$.
|
Yes, AM-GM. Let $x\ge y\ge0$, $n\in\mathbb N$. Then
\begin{align*}
x^n-y^n&=(x-y)\sum_{k=0}^{n-1}x^{n-1-k}y^k\\
&=\frac12(x-y)\sum_{k=0}^{n-1}\left(x^ky^{n-1-k}+x^{n-1-k}y^k\right)\\
&\ge\frac12(x-y)\sum_{k=0}^{n-1}2\sqrt{x^ky^{n-1-k}\cdot x^{n-1-k}y^k}\\
&=(x-y)\sum_{k=0}^{n-1}(xy)^{\frac{n-1}2}\\
&=n(x-y)(xy)^{\frac{n-1}2}.
\end{align*}
|
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|
Find the sum of series $ \sum_{n=1}^{\infty}\frac{1}{n^3(n+1)^3}$ Let it be known that $$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}
{6}.$$ Given such—find $$\sum_{n=1}^{\infty}\frac{1}{n^3(n+1)^3}$$
Attempt:
I have tried using the fact that $\displaystyle \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ and then expanding or using known sum types as $\displaystyle \sum_{k=1}^{n} k=\frac{n(n+1)}{2}$ or $\displaystyle \sum_{k=1}^{n} k^3=\frac{n^2(n+1)^2}{4}$
but nothing seems to lead to anything!
|
Hint:
$$
\frac{1}{n^3} - \frac{1}{{(n+1)}^3} \;=\;
\frac{3n(n+1) + 1}{n^3{(n+1)}^3}
$$
|
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|
If $ax+by = a^n + b^n$ then $\left[\frac{x}{b}\right]+\left[\frac{y}{a}\right]=\left[\frac{a^{n-1}}{b}\right]+\left[\frac{b^{n-1}}{a}\right]$
Let a,b,n be positive integers such that $(a,b) = 1$.
Prove that if $(x,y)$ is a solution of the equation $ax+by = a^n + b^n$ then
$$\left[\frac{x}{b}\right]+\left[\frac{y}{a}\right]=\left[\frac{a^{n-1}}{b}\right]+\left[\frac{b^{n-1}}{a}\right]$$
So I was trying this question which is a Romania TST problem.
$$ax+by=a^n+b^n\implies a^n-ax=b^n-by\implies \frac{a^n-ax}{ab}=\frac{b^n-by}{ba}$$
$$\frac{a^{n-1}-x}{b}=\frac{b^{n-1}-y}{a}\implies \frac{a^{n-1}}{b}-\frac{x}{b}=\frac{b^{n-1}}{a}-\frac{y}{a}. $$
I don't think so we can proceed more.
We also have $a(a^n-ax)=b(b^n-y)\implies a^{n-1}\equiv x\mod b, b^{n-1}\equiv y\mod a.$
Any hints on how to proceed?
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As I stated in my comment, if $(x, y)$ is allowed to be any real solution, then \eqref{eq6A} will not always hold, so I'll assume that $x$ and $y$ must be integers. Also, I believe the square brackets are meant to represent the floor function, as I've seen being used elsewhere on this site.
Given the above, we can proceed from where you left off, i.e.,
$$a^{n-1} \equiv x \pmod{b} \tag{1}\label{eq1A}$$
$$b^{n-1} \equiv y \pmod{a} \tag{2}\label{eq2A}$$
Note \eqref{eq1A} means that $a^{n-1}$ and $x$ have the same remainder when divided by $b$, i.e., there's integers $i_1$, $i_2$ and $q$ where
$$a^{n-1} = i_{1}b + q, \; \; x = i_{2}b + q, \; \; 0 \le q \le b - 1 \tag{3}\label{eq3A}$$
Similarly with \eqref{eq2A}, there's integers $j_1$, $j_2$ and $r$ where
$$b^{n-1} = j_{1}a + r, \; \; y = j_{2}a + r, \; \; 0 \le r \le a - 1 \tag{4}\label{eq4A}$$
Using \eqref{eq3A} and \eqref{eq4A} in the given equation that $x$ and $y$ are a solution of gives
$$\begin{equation}\begin{aligned}
ax + by & = a^n + b^n \\
a(i_{2}b + q) + b(j_{2}a + r) & = a(i_{1}b + q) + b(j_{1}a + r) \\
ab(i_{2}) + aq + ab(j_{2}) + br & = ab(i_{1}) + aq + ab(j_{1}) + br \\
ab(i_{2}) + ab(j_{2}) & = ab(i_{1}) + ab(j_{1}) \\
i_{2} + j_{2} & = i_{1} + j_{1}
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
The equation you're asked to prove is
$$\left\lfloor \frac{x}{b} \right\rfloor + \left\lfloor \frac{y}{a} \right\rfloor = \left\lfloor \frac{a^{n-1}}{b} \right\rfloor + \left\lfloor \frac{b^{n-1}}{a} \right\rfloor \tag{6}\label{eq6A}$$
Using the second parts of \eqref{eq3A} and \eqref{eq4A} in the left side of \eqref{eq6A} gives
$$\begin{equation}\begin{aligned}
\left\lfloor \frac{x}{b} \right\rfloor + \left\lfloor \frac{y}{a} \right\rfloor & = \left\lfloor \frac{i_{2}b + q}{b} \right\rfloor + \left\lfloor \frac{j_{2}a + r}{a} \right\rfloor \\
& = \left\lfloor i_2 + \frac{q}{b} \right\rfloor + \left\lfloor j_2 + \frac{r}{a} \right\rfloor \\
& = i_2 + j_2
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$
while using the first parts of \eqref{eq3A} and \eqref{eq4A} in the right side of \eqref{eq6A} gives
$$\begin{equation}\begin{aligned}
\left\lfloor \frac{a^{n-1}}{b} \right\rfloor + \left\lfloor \frac{b^{n-1}}{a} \right\rfloor & = \left\lfloor \frac{i_{1}b + q}{b} \right\rfloor + \left\lfloor \frac{j_{1}a + r}{a} \right\rfloor \\
& = \left\lfloor i_1 + \frac{q}{b} \right\rfloor + \left\lfloor j_1 + \frac{r}{a} \right\rfloor \\
& = i_1 + j_1
\end{aligned}\end{equation}\tag{8}\label{eq8A}$$
The above $2$ equations show, using \eqref{eq5A}, that the $2$ sides of \eqref{eq6A} are always equal to each other.
Update: The problem can be generalized to, for any fixed integers $c$ and $d$, then for all integer solutions $(x, y)$ of
$$ax + by = ac + bd \tag{9}\label{eq9A}$$
by following the same procedure as in your question and in my answer above, and replacing $a^{n-1}$ with $c$ and $b^{n-1}$ with $d$, we can show that
$$\left\lfloor \frac{x}{b} \right\rfloor + \left\lfloor \frac{y}{a} \right\rfloor = \left\lfloor \frac{c}{b} \right\rfloor + \left\lfloor \frac{d}{a} \right\rfloor \tag{10}\label{eq10A}$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/4255983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
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