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Is there any similar solutions including $\pi$ like $1-\frac{1}{2}+\frac{1}{4}-\frac{1}{5}+\cdots=\frac{\pi}{3\sqrt{3}}$? First equation is very popular - there are only odd numbers. Other words, numbers, which are coprime with $2$.
$$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{7}+\cdots=\frac{\pi}{4}$$
Second is similar, but not very well known. There are numbers coprime with $3$.
$$1-\frac{1}{2}+\frac{1}{4}-\frac{1}{5}+\frac{1}{7}-\frac{1}{8}+\cdots=\frac{\pi}{3\sqrt{3}}$$
With some simple transformation we can say, that
$$\left(1+\frac{1}{2}\right)\left[\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1+\frac{1}{7}\right)\left(1+\frac{1}{11}\right)\left(1-\frac{1}{13}\right)\cdots\right]^{-1}=\frac{\pi}{4}$$
where primes of form $4n+1$ is negative, other is positive. Similar way
$$\left(1+\frac{1}{3}\right)\left[\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\right(1+\frac{1}{5}\left)\left(1-\frac{1}{7}\right)\left(1+\frac{1}{11}\right)\left(1-\frac{1}{13}\right)\cdots \right]^{-1}=\frac{\pi}{3\sqrt{3}}$$
where primes of form $6n+1$ is negative, other is positive. Here we can create a function $s_{m}(r)$, which generates $r$-th coprime with $m$, then
$$f_{k}(m)=\sum\limits_{i=0}^{\infty}\left[\sum\limits_{r=1}^{\varphi(m)/2}\left(\frac{1}{mi+s_{m}(2r-1)}\right)^k-\sum\limits_{r=1}^{\varphi(m)/2}\left(\frac{1}{mi+s_{m}(2r)}\right)^k\right]$$
$$g_{k}(m)=\prod\limits_{p}^{\infty}(1+p^{-k})^{-1}\cdot\prod\limits_{p|m}^{}(1+p^{-k})$$
$$h_{k}(mn+d)=\prod\limits_{p[mn+d]}^{\infty}\frac{p^k+1}{p^k-1}$$
$$j_{k}(m)=\prod\limits_{r=1}^{\varphi(m)/2}h_{k}(mn+s_{m}(2r-1))$$
$$f_{k}(m)=g_{k}(m)\cdot j_{k}(m)$$
Here $\varphi(m)$ - Euler totient function, $p[mn+d]$ - primes of form $mn+d$. Also
$$f_{k}(m)= f_{k}(m^2)= f_{k}(m^3)=\cdots= f_{k}(m^x)$$
so we can work only with $m$, which $|\mu(m)|=1$. If $m=2$, then we need to correct $mi$ in $f_{k}(m)$ to $2mi$ (and if $m=2$ or $m$ odd - $mn$ in $h_{k}(m)$ to $2mn$). Then we have
$$f_{1}(2)=\frac{\pi}{4}, f_{3}(2)=\frac{\pi^3}{32}, f_{5}(2)=\frac{5\pi^5}{1536}$$
Is there any results, including $\pi$, for example
$$f_{k}(5)=\pi^k\frac{a^{\frac{b}{c}}}{d^{\frac{e}{f}}}$$
where $a,b,c,d,e,f$ - integers?
If I made some mistakes, sorry for my English.
|
I was going to write more or less the same answer of Lord Shark, so let us steer in a more elementary direction. The fact that Gregory series equals $\frac{\pi}{4}$ can be seen as a consequence of
$$ \sum_{n\geq 0}\frac{(-1)^n}{2n+1}=\int_{0}^{1}\sum_{n\geq 0}(-1)^n x^{2n}\,dx = \int_{0}^{1}\frac{dx}{1+x^2}=\arctan(1)=\frac{\pi}{4}\tag{A}$$
but since
$$ \int_{0}^{+\infty}\frac{dx}{1+x^k} = \frac{\pi}{k\sin\frac{\pi}{k}}\tag{B}$$
from Euler's Beta function and the reflection formula for the $\Gamma$ function, a lot of similar identities can be simply derived from $(B)$ by reverse engineering. For instance, by taking $k=5$,
$$ \frac{\pi}{5\sin\frac{\pi}{5}}=\int_{0}^{1}\frac{1+x^3}{1+x^5}\,dx =\sum_{n\geq 0}(-1)^n\left(\frac{1}{5n+1}+\frac{1}{5n+4}\right)\tag{C}$$
and by taking $k=12$:
$$ \sum_{n\geq 0}(-1)^n\left(\frac{1}{12n+1}+\frac{1}{12n+11}\right)=\frac{\pi}{3\sqrt{2}(\sqrt{3}-1)}.\tag{D} $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Can either of these two simultaneous equations be solved? I have been struggling with the elimination of the variables.
$\begin{array}{c|c}
\left(1-2^k\right) a+2^k b=-2^k+3^k & \left(-1+2^k\right) a-2^k b=2^k-3^k\\
b+\left(-1+2^k\right) c=-1+3^k & a+2^k c=3^k\\
a+2^k c=3^k & b+\left(-1+2^k\right) c=-1+3^k \\
\end{array}$
Note: each equation has two variables because I haven't figured out how to combine them to make 3 3-variable equations.
Some pointers would be nice.
|
Note that using Cramer’s rule may be much better. But, let us go with a brute-force method.
From the last equation, we have: $c = \frac{3^k-a}{2^k}$. Substituting this in the second equation, gives us: $$b + (2^k-1)[\frac{3^k-a}{2^k}]=3^k -1$$ $$\implies b -a+\frac{a}{2^k}=\frac{3^k}{2^k} -1 \implies (1-2^k)a+2^kb= 3^k -2^k$$ which is the same as the first equation!
This means there are infinite such triples $$(a,b,c) \equiv (a, a - \frac{a}{2^k}+\frac{3^k}{2^k}-1, \frac{3^k-a}{2^k})$$ a fact which can be confirmed using Cramer’s rule as the resultant matrix will have only two non-zero rows, implying the existence of infinite such triples.
|
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|
How to calculate the determinant? I need help with this determinant ($n \times n)$:
$D_n = \begin{vmatrix}
a & x & x & \dots & x & x \\
y & a & x & \dots & x & x \\
y & y & a & \dots & x & x \\
\vdots \\
y & y & y & \dots & a & x \\
y & y & y & \dots & y & a \\
\end{vmatrix}$
I tried to simplify it and derive a reсcurrent formula using Laplace expansion, but it was unsuccessful. Also I had an idea to express the determinant as a sum of determinants.
The answer is: $\displaystyle D_n = \frac{x(a-y)^n - y(a-x)^n}{x-y}$.
|
Subtract the 2nd column from the first, the 3rd from the second, etc.
Here is what you get in the $4\times4$ case:
$$D_4 = \begin{vmatrix}
a-x & 0 & 0 & x \\
y-a & a-x & 0 & x \\
0 & y-a & a-x & x \\
0 & 0 & y-a & a
\end{vmatrix}$$
Laplace expansion on the first column will yield:
$$D_4=(a-x)D_3-(y-a) \begin{vmatrix}
0 & 0 & x \\
y-a & a-x & x \\
0 & y-a & a
\end{vmatrix}$$
Then again Laplace expansion, on the first row:
$$D_4=(a-x)D_3-(y-a)x \begin{vmatrix}
y-a & a-x \\
0 & y-a \end{vmatrix}=(a-x)D_3-x(y-a)^3$$
General case: $D_n=(a-x)D_{n-1}-(-1)^nx(y-a)^{n-1}=(a-x)D_{n-1}+x(a-y)^{n-1}$, with $D_1=a$.
Then
$$\begin{align}
D_n & = &(a-x)^2D_{n-2}+(a-x)x(a-y)^{n-2}+x(a-y)^{n-1}\\
& = &(a-x)^3D_{n-3}+(a-x)^2x(a-y)^{n-3}+(a-x)x(a-y)^{n-2}+x(a-y)^{n-1}\\
& \vdots & \\
& = &(a-x)^{n-1}D_1+(a-x)^{n-2}x(a-y)+\dots+x(a-y)^{n-1}\\
\end{align}$$
Or
$$D_n=a(a-x)^{n-1}+x\sum_{k=0}^{n-2}(a-x)^{k}(a-y)^{n-1-k}$$
(if you don't like the dots, you can prove the previous formula by induction on $n$)
$$D_n=a(a-x)^{n-1}+x\left(\sum_{k=0}^{n-1}(a-x)^{k}(a-y)^{n-1-k}-(a-x)^{n-1}\right)$$
And, if $u\neq v$,
$$\sum_{k=0}^{n-1}u^{k}v^{n-1-k}=\frac{u^n-v^n}{u-v}$$
Hence, if $x\neq y$,
$$\begin{align}
D_n&=a(a-x)^{n-1}-x(a-x)^{n-1}+x\frac{(a-x)^n-(a-y)^n}{(a-x)-(a-y)}\\
&=(a-x)^{n}-x\frac{(a-x)^n-(a-y)^n}{x-y}\\
&=\frac{(x-y)(a-x)^{n}-x(a-x)^n+x(a-y)^n}{x-y}\\
&=\frac{x(a-y)^n-y(a-x)^{n}}{x-y}
\end{align}
$$
And if $x=y$, you take again the formula
$$\begin{align}
D_n&=a(a-x)^{n-1}+x\sum_{k=0}^{n-2}(a-x)^{k}(a-y)^{n-1-k}\\
&=a(a-x)^{n-1}+x(n-1)(a-x)^{n-1}\\
&=(a+(n-1)x)(a-x)^{n-1}
\end{align}$$
|
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|
$\sum_{n=1}^{\infty} \frac{1}{n^2+n+1}$ and other sums of quadratic reciprocals What is the exact value for $$\sum_{n=1}^{\infty} \frac{1}{n^2+n+1}$$ and can other infinite sums of quadratic reciprocals have specific values.
|
According to Maple,
$$ \frac{\pi}{\sqrt{3}} \tanh \left(\frac{\pi \sqrt{3}}{2}\right) - 1 $$
It looks like all the sums $\sum_{n=1}^\infty \frac{1}{n^2+tn+1}$ for integers $t\ne -2 $ have closed forms.
EDIT: Here are a few examples:
$$ \eqalign{\sum_{n=1}^{\infty } \frac{1}{ {n}^{2}+n+1} & =\frac{\pi \tanh(\pi \sqrt{3}/2)}{\sqrt{3}}-1\cr
\sum_{n=1}^\infty \frac{1}{n^2+n+1} &= \frac{\pi^2}{6}-1\cr
\sum_{n=1}^\infty \frac{1}{n^2+3n+1} &= \frac{\pi \tan(\pi \sqrt{5}/2)}{\sqrt{5}}\cr
\sum_{n=1}^\infty \frac{1}{n^2+4n+1} &= -\frac{\pi \cot(\pi \sqrt{3})}{2\sqrt{3}}-\frac{1}{3}\cr
\sum_{n=1}^\infty \frac{1}{n^2+5n+1} &= \frac{\pi \tan(\pi \sqrt{21}/2)}{\sqrt{21}}-\frac{7}{15}\cr
\sum_{n=1}^\infty \frac{1}{n^2+6n+1} &= -\frac{\pi\cot(2 \sqrt{2}\pi)}{4\sqrt{2}}-\frac{61}{112}}$$
|
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|
Finding $\sqrt{a+ib}$ How do to find the square root of a complex number?
$z=a+ib$
$z=\sqrt{3+i4}=a+ib$
remove the square by squaring both side
$a^2+2iab+b^2=3+i4$
$a^2+b^2=3$
$2ab=4$
So I just solve the simultaneous equation. But my friend is telling me this way I am doing is wrong. Can anyone give a hand, thank in advance
|
The method is correct, you identify the real and imaginary parts:
$$\begin{cases}x^2-y^2=a,\\2xy=b.\end{cases}$$
Then multiplying the first equation by $4x^2$,
$$4x^4-4ax^2-4x^2y^2=4x^4-4ax^2-b^2=(2x^2-a)^2-a^2-b^2=0,$$
giving
$$x^2=\frac{\pm\sqrt{a^2+b^2}+a}2.$$
As we need $a$ to be real, only the $+$ sign is useful and
$$x=\pm\sqrt{\frac{\sqrt{a^2+b^2}+a}2}.$$
|
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|
Find the largest value of the expression $P=\frac{3x+2y+1}{x+y+6}$ For real numbers $x, y$ satisfy the condition $x>0, y>0$ and
$$\log _{\sqrt{3}}\left(\frac{x+y}{x^2+y^2+xy+2}\right)=x\left(x-3\right)+y\left(y-3\right)+xy$$
Find the largest value of the expression $$P=\frac{3x+2y+1}{x+y+6}$$
|
I think this is a very nice question.
We have $$P(x,y)=\frac{3x+2y+1}{x+y+6}\tag{1}$$ so by the Quotient Rule $$P'(x,y)=\frac{(3+2y')(x+y+6)-(3x+2y+1)(1+y')}{(x+y+6)^2}=\frac{y+(11-x)y'+17}{(x+y+6)^2}=0$$ for stationary points. Thus $$(11-x)y'+y+17=0\implies y'+\frac1{11-x}y=-\frac{17}{11-x}$$ is a first-order linear ODE, which is easy to solve.
Here $p(x)=\dfrac1{11-x}$ and $q(x)=-\dfrac{17}{11-x}$. The integrating factor is $$h(x)=\exp\left(\int p(x)\, dx\right)=\frac1{11-x}$$ so we solve $$y\cdot p(x)=\int p(x)q(x)\, dx \implies \frac y{11-x}=\int -\frac{17}{(11-x)^2}\, dx=-\frac{17}{11-x}+c$$ where $c$ is a constant. Hence $$y=-17-c(x-11)=-17+11c-cx\tag{2}$$ Plugging this back into $(1)$, we get $$\max P=\frac{3x-34+22c-2cx+1}{x-17+11c-cx+6}=\frac{(3-2c)(x-11)}{(1-c)(x-11)}=\frac{3-2c}{1-c}\tag{3}$$
Now let $x=0$. Then by $(2)$, we have $$y(0)=-17+11c\implies c=\frac1{11}(y(0)+17)\tag{4}$$ We are given the condition that $$\log _{\sqrt{3}}\left(\frac{x+y}{x^2+y^2+xy+2}\right)=x\left(x-3\right)+y\left(y-3\right)+xy$$ so $$\log_{\sqrt3}\left(\frac y{y^2+2}\right)=y(y-3)$$ This has solutions when $y=1, 2$.
Therefore, when $y=1$, $c=\dfrac{18}{11}$ from $(4)$ and so from $(3)$,$$P_{y=1}=\frac{3-2\left(\frac{18}{11}\right)}{1-\frac{18}{11}}=\frac37$$ and when $y=2$, $c=\dfrac{19}{11}$ from $(4)$ and so from $(3)$, $$P_{y=2}=\frac{3-2\left(\frac{19}{11}\right)}{1-\frac{19}{11}}=\frac58$$ Since $\dfrac58>\dfrac37$, $$\boxed{\max P = \frac58.}$$
|
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|
X ways to choose R objects in generating functions I'm currently doing generating functions, and the purpose of those functions is to find how many ways there are to choose an x amount of objects.
I just don't understand how they get to the final conclusion.
Take for example this exercise:
You have to go to the baker and you have to buy r amount of cakes.
The baker has 3 cheese-cakes, 2 apricots-cakes, and 4
strawberry-cakes. How many different ways are there for you to choose?
This gives us the following generating function:
$(1 + x + x^2 + x^3)(1 + x + x^2)(1 + x + x^2 + x^3 + x^4)$
After resolving this function, we get the polynomial:
$1 + 3x + 6x^2 + 9x^3 + 11x^4 + 11x^5 + 9x^6 + 6x^7 + 3x^8 + x^9$
I understand how they get the generating function, and how to resolve it to get the polynomial, but now the solution of the exercise says:
So we can conclude that there are 6 different ways to choose 7 cakes.
I know the 6 and 7 relate to each other because one is the exponent of the other (see the term $6x^7$), but how do they get to the number 6, and 7?
There are 9 different cakes, so why 7?
|
The generating function approach is as follows. We would like to know the number of integer solutions to
$$x_1 + x_2 + x_3 = r$$
subject to $0 \le x_1 \le 3$, $0 \le x_2 \le 2$, and $0 \le x_3 \le4$. To solve the problem with a generating function, let $a_r$ be the number of solutions for $r = 0, 1, 2, 3, \dots$ and define
$$f(x) = \sum_{r=0}^{\infty} a_r x_r$$
(In this case we know that $a_r = 0$ for $r>9$, so $f(x)$ is actually a polynomial.)
You have already found that
$$\begin{align}f(x) &= (1 + x + x^2 + x^3)(1+x+x^2)(1+x+x^2+x^3+x^4) \\
&= 1+3 x+6 x^2+9 x^3+11 x^4+11 x^5+9 x^6+6 x^7+3 x^8+x^9
\end{align}$$
so $a_0 = 1$, $a_1=3$, $a_2=6$, ..., $a_7=6$, ..., $a_9=1$.
The number of ways to choose 7 cakes is just the value of $a_7$, but the generating function gives the number of ways to choose $r$ cakes for all values of $r$.
|
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|
Finding the limit of the sequence $a_n\cdot a_{n+1}=n,\,n=1,2,3,\cdots.$ Let $(a_n)_{n>=1}$ be a sequence of real numbers defined by the below recurrence relation:
$$a_n\cdot a_{n+1}=n,\quad n=1,2,3,\cdots.$$
Prove that $\lim_{n\to \infty}a_n=+\infty.$
Edit: $a_1>0$
|
Since $a_1 > 0$, it follows that $a_n > 0$, for all $n$.
Since $a_n a_{n+1} = n$, at least one of $a_n,a_{n+1}$ is greater or equal to $\sqrt{n}$, hence the sequence $(a_n)$ is unbounded above.
For all positive integers $n > 2$, we have
\begin{align*}
a_n &= \frac{n-1}{a_{n-1}}\\[4pt]
a_{n-2} &= \frac{n-2}{a_{n-1}}\\[4pt]
\end{align*}
hence
$$
\frac
{a_n}
{a_{n-2}}
=
\frac
{n-1}{n-2}
$$
It follows that
$$a_1 < a_3 < a_5 < \cdots$$
$$a_2 < a_4 < a_6 < \cdots$$
Hence, since the sequence $(a_n)$ is unbounded above, at least one of the above subsequences approaches infinity.
But for $m>1$, we have
\begin{align*}
a_{2m+2} &\,=\,
a_2\prod_{k=1}^m \frac{a_{2k+2}}{a_{2k}}
\,=\,
a_2\prod_{k=1}^m \frac{2k+1}{2k}
\\[4pt]
a_{2m+1} &\,=\,
a_1\prod_{k=1}^m \frac{a_{2k+1}}{a_{2k-1}}
\,=\,
a_1\prod_{k=1}^m \frac{2k}{2k-1}\\[4pt]
\end{align*}
so
$$
\frac
{a_{2m+1}}{a_{2m+2}}
= \frac{a_1}{a_2}
\prod_{k=1}^m \frac{4k^2}{4k^2-1} > \frac{a_1}{a_2}
$$
Hence, since at least one of the every-other-term subsequences approaches infinity, it follows that the odd-term subsequence approaches infinity.
But also,
$$
a_{2m+3} \,=\,
a_3\prod_{k=1}^{m} \frac{a_{2k+3}}{a_{2k+1}}
\,=\, a_3\prod_{k=1}^{m} \frac{2k+2}{2k+1}
$$
so
$$
\frac
{a_{2m+2}}{a_{2m+3}}
= \frac{a_2}{a_3}
\prod_{k=1}^m \frac{4k^2+4k+1}{4k^2+4k} > \frac{a_2}{a_3}
$$
Hence, since the odd-term subsequence approaches infinity, the even-term subsequence must also approach infinity.
It follows that the sequence $(a_n)$ approaches infinity.
|
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|
How to prove $2\le\frac{g(x)}{g(y)}+\frac{g(y)}{g(x)}$ using AM-GM? In this question Prove that, if $g(x)$ is concave, for $S = {x : g(x) > 0}$, $f(x) = 1/g(x)$ is convex over $S$. , in the proof of Math536,
how to prove $2\le\frac{g(x)}{g(y)}+\frac{g(y)}{g(x)}?$
This is the proof of Math536:
$$\begin{align}
1 &= (a+(1-a))^2 \\
&= a^2 + 2a(1-a) + (1-a)^2 \\
&\le a^2 + a(1-a)\left(\frac{g(x)}{g(y)} + \frac{g(y)}{g(x)}\right) + (1-a)^2 \\
&= (ag(x)+(1-a)g(y)) \left(\frac{a}{g(x)} + \frac{1-a}{g(y)}\right) \\
&\le g(ax+(1-a)y) \left(\frac{a}{g(x)} + \frac{1-a}{g(y)}\right) \\
&= \frac{af(x) + (1-a)f(y)}{f(ax+(1-a)y)}
\end{align}$$
He said we should use the AM-GM inequality which is $\forall a_i\in\mathbb R^+: \frac{a_1+...+a_n}{n}\ge (a_1\dots a_n)^{1/n}.$
I tried to find a similarity with the LHS of the AM-GM inequality:
$$g(x)f(y)+f(x)g(y)\ge\sqrt{g(x)g(y)f(x)f(y)}$$
I honestly don't know how will I get the $2$ in the inequality?
Can someone help me please?
|
There is no need for the general AM-GM here. Suppose you have two positive real numbers $a, b > 0$ (justified since $g(x), g(y) > 0$). Then,
$$2 \leq \frac{a}{b} + \frac{b}{a} = \frac{a^2+b^2}{ab} \iff$$
$$2ab \leq a^2+b^2 \iff a^2-2ab+b^2 \geq 0 \iff (a-b)^2 \geq 0$$
with equality iff $a=b$.
|
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|
Evaluate $\lim_{n\rightarrow\infty}$ $\left[\frac{\left(n+1\right)\left(n+2\right).....\left(n+n\right)}{n^{n}}\right]^{\frac{1}{n}}$
Question: Evaluate lim$_{n\rightarrow\infty}$$\left[\frac{\left(n+1\right)\left(n+2\right).....\left(n+n\right)}{n^{n}}\right]^{\frac{1}{n}}$
My Approach Let $a_{n}=\frac{\left(n+1\right)\left(n+2\right).....\left(n+n\right)}{n^{n}}$
$$a_{n+1}=\frac{\left(n+2\right)\left(n+3\right).....\left(2n+2\right)}{\left(n+1\right)^{\left(n+1\right)}}$$
$$\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_{n}}=\frac{4}{\exp (1)}$$
Using Cauchy's second theorem On Limits
Lim$_{n\rightarrow\infty}\left[\frac{a_{2}}{a_{1}}\frac{a_{3}}{a_{2}}..........\frac{a_{n+1}}{a_{n}}\right]^{\frac{1}{n}}$=Lim$_{n\rightarrow\infty}$$\left[\frac{a_{n+1}}{a_{1}}\right]^{\frac{1}{n}}=\frac{4}{\exp(1)}$
Now I don't know how to move forward
|
Let the limit be $L$ then:
$$\begin{align}
\ln(L) &= \lim_{n\to \infty}\frac{1}{n} \sum_{i=1}^{n} \ln\left( 1 + \frac{i}{n} \right) \\
&= \int_{0}^{1}\ln( 1+x ) dx \\
&= (x+1)\ln(1+x) -x |_{0}^{1} \\
&= 2\ln(2) - 1
\end{align}$$
So the required limit is $\exp({2\ln (2) -1}) = \color{blue}{\dfrac{4}{e}}$
|
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|
Given $3\cos x - 4 \sin x = 2$, find $3 \sin x + 4 \cos x$ without first solving for $x$ If $$3\cos{x}-4\sin{x}=2$$
find $$3\sin{x} +4\cos{x} $$
I have solved the equation for $x$, then calculated the required value, but I think there is a direct solution without solving the equation.
|
Think of it as a rotation equation. This is equivalent to the matrix formulation $$\begin{bmatrix} \cos x & -\sin x \\ \sin x & \cos x\end{bmatrix} \begin{bmatrix} 3 \\ 4 \end{bmatrix} = \begin{bmatrix} 2 \\ y \end{bmatrix}$$ where y is the unknown. Since the magnitude of a vector is preserved in rotation, $2^2 + y^2 = 3^2 + 4^2$ So, $y = \pm \sqrt {21}$.
|
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|
Partial Fraction Decomposition of $\prod\limits_{j=0}^{n}\frac 1{1+Xa^{j-k}}$
Question: Decompose the fraction$$\frac 1{\left\{1+X\right\}\left\{1+Xa\right\}\left\{1+Xa^2\right\}\cdots\left\{1+Xa^n\right\}}$$where $X=x^2$.
My work: Obviously, the right-hand side has to take the form$$\prod\limits_{j=0}^n\frac 1{1+Xa^j}=\sum\limits_{i=0}^n\frac {C_i}{1+Xa^i}$$Multiplying both sides by $1+Xa^i$ and evaluating at $X=-a^{-i}$ isolates the $C_k$ in the right-hand side. Hence$$\begin{align*}C_k & =\prod\limits_{j=0}^n\frac {1+Xa^i}{1+Xa^j}=\prod\limits_{j=0,\,j\neq k}^n\frac 1{1+Xa^{j-k}}\end{align*}$$Splitting up the product into two separate terms, we have$$C_k=\frac 1{\left(1-\frac X{a^k}\right)\left(1-\frac X{a^{k-1}}\right)\cdots\left(1-\frac Xa\right)}\frac 1{\left(1+aX\right)\left(1+a^2X\right)\cdots\left(1+a^nX\right)}$$However, I'm not sure what to do afterwards. The main goal here is to rewrite each product in terms of the rising/falling factorial and gamma function.
This is because the original problem is to integrate the fraction and I feel more comfortable integrating from $0$ to $\infty$ in terms of gamma functions and factorials, than with the product of terms.
The original problem was$$\int\limits_0^{\infty}dx\,\frac 1{(1+x^2)(1+ax^2)(1+a^2x^2)\ldots}$$And I plan to take the limit as $n\to\infty$.
|
Consider
$$\dfrac{1}{(1+X)(1+Xa)(1+Xa^2)}
=\dfrac{c_0}{1+X} + \dfrac{c_1}{1+aX} + \dfrac{c_2}{1+a^2}$$
Multiplying both sides by $(1+X)(1+Xa)(1+Xa^2)$, we get
$$c_0(1+Xa)(1+Xa^2) + c_0(1+X)(1+Xa^2) + c_0(1+X)(1+Xa) = 1$$
Let $X=-1$, and we get
$$c_0(1-a)(1-a^2) = 1$$
$$c_0 = \dfrac{1}{(1-a)(1-a^2)}$$
Let $X=-\dfrac{1}{a}$, and we get
$$c_1(1-\frac 1a)(1-a) = 1$$
$$c_1 = \dfrac{1}{(1-\frac 1a)(1-a)}$$
Let $X=-\dfrac{1}{a^2}$, and we get
$$c_0(1-\frac{1}{a^2})(1-\frac 1a) = 1$$
$$c_2 = (1-\frac{1}{a^2})(1-\frac 1a)$$
Finding the general answer is just a matter of bookkeeping.
|
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|
Justify that the vectors are linearly dependent Let $V$ be a real vector space and $a,b,c,d,e\in V$.
We have the vectors \begin{align*}&v_1=a+b+c \\ &v_2=2a+2b+2c-d \\ &v_3=a-b-e \\ &v_4=5a+6b-c+d+e \\ &v_5=a-c+3e \\ &v_6=a+b+d+e\end{align*}
I want to justify that these vectors are linearly dependent.
$$$$
These six vectors are linearly dependetn iff we can get the zer0-vector by a linear combination of these vectors, \begin{equation*}\lambda_1v_1+\lambda_2v_2+\lambda_3v_3+\lambda_4v_4+\lambda_5v_5+\lambda_6v_6=0\end{equation*} where at least of the coefficients $\lambda_1, \lambda_2, \lambda_3, \lambda_4, \lambda_4$ or $\lambda_6$ is not equal to zero.
We have the following:
\begin{align*}&\lambda_1v_1+\lambda_2v_2+\lambda_3v_3+\lambda_4v_4+\lambda_5v_5+\lambda_6v_6=0 \\ & \Rightarrow \lambda_1\left (a+b+c\right )+\lambda_2\left (2a+2b+2c-d\right )+\lambda_3\left (a-b-e\right )+\lambda_4\left (5a+6b-c+d+e\right )+\lambda_5\left (a-c+3e\right )+\lambda_6\left (a+b+d+e\right )=0 \\ & \Rightarrow \lambda_1a+\lambda_1b+\lambda_1c+2\lambda_2 a+2\lambda_2 b+2\lambda_2 c-\lambda_2d+\lambda_3 a-\lambda_3 b-\lambda_3 e+5\lambda_4 a+6\lambda_4 b-\lambda_4 c+\lambda_4 d+\lambda_4 e+\lambda_5 a-\lambda_5 c+3\lambda_5 e+\lambda_6a+\lambda_6b+\lambda_6d+\lambda_6e=0 \\ & \Rightarrow a\left (\lambda_1+2\lambda_2+\lambda_3+5\lambda_4+\lambda_5+\lambda_6\right )+b\left (\lambda_1+2\lambda_2-\lambda_3+6\lambda_4+\lambda_6\right )+c\left (\lambda_1+2\lambda_2 -\lambda_4-\lambda_5\right )+d\left (-\lambda_2+\lambda_4+\lambda_6\right ) +e\left (-\lambda_3 +\lambda_4 +3\lambda_5 +\lambda_6\right )=0\end{align*}
Is everything correct? How could we continue?
Do we use the fact that $a,b,c,d,e\in V$ ? But how?
|
If you select the set $\{v_k \}_{k=1}^{\color{red}{5}}$ you can write
$$
\left(\begin{array}{c}v_1\\ v_2 \\ v_3 \\v_4 \\ v_5 \end{array}\right) = \left(\begin{array}{ccccc}
1 & 1 & 1 & 0 & 0 \\
2 & 2 & 2 & -1 & 0 \\
1 & -1 & 0 & 0 & -1 \\
5 & 6 & -1 & 1 & 1 \\
1 & 0 & -1 & 0 & 3
\end{array}\right) \left(\begin{array}{c} a \\ b \\ c \\ d \\ e\end{array}\right)
$$
Which can be solved for $\{a, \cdots, e \}$ since the array as determinant different from zero (-44). This means it is possible to write $a, \cdots, e$ as a linear combination of $v_1, \cdots, v_5$.
This means that $v_6$ can be written as a linear combination of $\{v_k \}_{k=1}^{\color{red}{5}}$
|
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|
Sum of the infinite series The series is $$\frac{5}{1\cdot2}\cdot\frac{1}{3}+\frac{7}{2\cdot3}\cdot\frac{1}{3^2}+\frac{9}{3\cdot4}\cdot\frac{1}{3^3}+\frac{11}{4\cdot5}\cdot\frac{1}{3^4}+\cdots$$
This is my attempt:
$$T_n=\frac{2n+3}{n(n+1)}\cdot\frac{1}{3^n}$$
Assuming $$\frac{2n+3}{n(n+1)}=\frac{A}{n}+\frac{B}{n+1}$$
we find $A=3,B=-1.$ Putting these values in $T_n$ we get,
$$T_n=\frac{1}{n}\cdot\frac{1}{3^{n-1}}-\frac{1}{n+1}\cdot\frac{1}{3^{n}}$$
How do I find the sum of the series from here $?$
|
Just try to write the expression for $T_{n+1}$ and note that the series telescopes.
$$T_n=\frac{1}{n}\cdot\frac{1}{3^{n-1}}-\color{purple}{\frac{1}{n+1}\cdot\frac{1}{3^{n}}} \\
T_{n+1}=\color{purple}{\frac{1}{n+1}\cdot\frac{1}{3^{n}}}-\frac{1}{n+2}\cdot\frac{1}{3^{n+1}}$$
We see the terms are cancelling, so
$$\begin{align}
\sum_{i=1}^{n} T_i &= \frac{1}{1}\frac{1}{3^{0}}
-\color{blue}{\frac{1}{2}\frac{1}{3^{1}}+\frac{1}{2}\frac{1}{3^{1}}}...
-\color{red}{\frac{1}{n-1}\frac{1}{3^{n-2}}+\frac{1}{n-1}\frac{1}{3^{n-2}}}
-\frac{1}{n}\frac{1}{3^{n-1}}\\
&= 1-\frac{1}{n3^{n-1}}
\end{align}$$
Therefore as $n\to \infty$, we see the sum tends to $1$.
|
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|
Find ED and (I am so so sorry for my bad English in advance, it is my math homework problem, I tried to solve it but I've got no idea whether the solution's right or wrong:))) It is given that $∠B=120^{o}$, $AB=6 cm$, $BC=4cm$, $CE=BE$, $AC=CD$. Find the lenght of ED and angle $∠CED$. Look
Solution:
$AC=\sqrt{6^2+4^2-2*6*4*\cos{120^o}}=2\sqrt{19}\;cm$
$\frac{6}{\sin(∠ACB)}=\frac{2\sqrt{19}}{\sin 60^o} \Rightarrow \sin(∠ACB)=\frac{3\sqrt3}{2\sqrt{19}}$
$∠ACB<90^o\Rightarrow ∠DCB>90^o$
$\cos(∠DCB)=-\cos(∠ACB)=-\sqrt{1-(\frac{3\sqrt3}{2\sqrt{19}})^2}=\frac{-7}{2\sqrt{19}}$
$ED=\sqrt{(2\sqrt{19})^2+2^2+2*2\sqrt{19}*2*\frac{7}{2\sqrt{19}}}=6\sqrt{3} \;cm$
$\frac{6\sqrt{3}}{\frac{3\sqrt3}{2\sqrt{19}}}=\frac{2\sqrt{19}}{\sin(∠CED)} \Rightarrow \sin(∠CED)=\frac{1}{2} \Rightarrow ∠CED=30^o$
Is it correct guys? Thank you :)).
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$$ED^2=4+CD^2-4CD\cos(180^{\circ}-\gamma)$$
$$CD^2=AC^2=36+16-48\cos(120^{\circ})$$
$$\frac{\sin(\gamma)}{\sin(120^{\circ})}=\frac{6}{AC}$$
$$AC^2=4+ED^2-4ED\cos(\delta)$$
|
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|
using $x^9 - 1$ as a product of linear and quadratic factors with real coefficients to solve trig So here's where I'm at I know i can find the ninth roots of unity by using the nth roots of unity formula $$ \zeta_n = e^{\frac{i2\pi k}{n}} ,$$ when $k = 1,2,3,...,n-1$
so the roots of unity where $e^{i\frac{2\pi}{9}},e^{i\frac{4\pi}{9}},...,e^{i\frac{8\pi}{9}}$and the other 4 would be the same except the angles would just be negative.
The previous parts of the question was to show that $$(x-e^{i\theta})(x-e^{-i\theta})$$ could be expressed as a quadratic with real coefficients the answer I got was $$(x-e^{i\theta})(x-e^{-i\theta})=x^2-2x\cos(\theta)+1$$so I said well the $x^9 - 1= (x-e^{i\frac{2\pi}{9}})(x-e^{-i\frac{2\pi}{9}})...(x-e^{i\frac{8\pi}{9}})(x-e^{-i\frac{8\pi}{9}})(x-1)$ and then using the formula I proved said that, that was equal to $$(x^2-2x\cos(\frac{2\pi}{9})+1)(x^2-2x\cos(\frac{4\pi}{9})+1)...(x^2+2x\cos(\frac{8\pi}{9})+1)(x-1)$$
Somehow I need to use this to solve
$$\cos(\frac{2\pi}{9})+\cos(\frac{4\pi}{9})+\cos(\frac{6\pi}{9})+\cos(\frac{8\pi}{9})= -\frac{1}{2}$$
Any help ,much appreciated!!
|
Note sure where this problem came from or what kind of solution is expected, but this runs up against the Casus Irreducibilus. Let $\zeta=\zeta_9$ be an irreducible $9$th root of unity, then you want the polynomial
$$(x-\zeta)(x-\zeta^{-1})=x^2-(\zeta+\zeta^{-1})x+1$$
So to find the coefficient we have
$$(\zeta+\zeta^{-1})^3=3(\zeta+\zeta^{-1})+(\zeta^3+\zeta^{-3})$$
So $\zeta+\zeta^{-1}$ is the solution of
$$x^3-3x+1=0.$$
Now the discriminant is $D=3^4$ meaning that the roots are real, but cannot be expressed as cube roots of real numbers. Using Cardano's formula
$$\zeta+\zeta^{-1}=\sqrt[3]{\frac{-1+\sqrt{-3}}{2}}+\sqrt[3]{\frac{-1-\sqrt{-3}}{2}}$$
|
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|
Probability-Tossing of die What's the probability of tossing a die 4 times, and getting a higher value than the previous value in all the 4 outcomes ? How can we I the method for such a case in the case of 3 tosses?
|
There is no method. you have to list all the possibilities.
The first thought is that the first result cannot higher that $3$. So, there are three groups of favorable outcomes:
$$\begin{matrix}
\color{red}1&2&3&4\\
\color{red}1&2&3&5\\
\color{red}1&2&3&6\\
\color{red}1&2&4&5\\
\color{red}1&2&4&6\\
\color{red}1&2&5&6\\
\color{red}1&3&4&5\\
\color{red}1&3&4&6\\
\color{red}1&3&5&6\\
\color{red}1&4&5&6\\
\color{white}1\\
\color{blue}2&3&4&5\\
\color{blue}2&3&4&6\\
\color{blue}2&3&5&6\\
\color{blue}2&4&5&6\\
\color{white}2&\\
\color{green}3&4&5&6
\end{matrix}$$
As a total, we have these $15$ possibilities with the common probability of $\frac{1}{\ 6^4}.$ So, the probability sought for is$$\frac{15}{\ 6^4}.$$
|
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|
lim$_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{3}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{9}}+...+\frac{1}{\sqrt{3n}+\sqrt{3n+3}}\right)$
Question
Edit:
My Approach:
$\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{3}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{9}}+...+\frac{1}{\sqrt{3n}+\sqrt{3n+3}}\right)$
$S_{k}= \sum_{n=1}^{k} \frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{3n}+\sqrt{3n+3}}\right) \frac{\left(\sqrt{3n}-\sqrt{3n+3}\right)}{\left(\sqrt{3n}-\sqrt{3n+3}\right)} = \sum_{n=1}^{k}\frac{1}{\sqrt{n}}\left(\frac{\sqrt{3n}-\sqrt{3n+3}}{-3}\right) = \frac{1}{3\sqrt{k}} \left\{ \sqrt{3k+3}\right\} $
$\lim_{k\rightarrow\infty} S_{k} = \frac{1}{\sqrt{3}}$
|
We have: $$\lim_{n \to \infty} \frac1{\sqrt n} \left( \frac1{\sqrt 6 + \sqrt 3} + \frac1{\sqrt 9 + \sqrt 6} +\ldots \right) = \lim_{n \to \infty} \frac1{\sqrt n} \left( \frac{\sqrt 6- \sqrt 3}3 + \frac{\sqrt 9 - \sqrt 6}3 + \ldots \right) $$
Can you see it is a telescoping sum?
|
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In how many ways can $6$ red and $6$ blue balls be distributed among $10$ persons such that every person gets at least one ball? In how many ways can $6$ red and $6$ blue balls be distributed among $10$ persons such that every person gets at least one ball? (Balls of the same colour are identical).
My Attempt
There appear to be too many cases which we need to consider. Not able to find correct approach.
|
Here is a solution by way of the Principle of Inclusion / Exclusion (PIE).
Suppose we ignore, for now, the restriction that each person must get at least one ball. Then by a bars-and-stars argument, there are $\binom{10+6-1}{6}$ ways to distribute the six red balls among the ten people, and similarly for the six blue balls. Since we can distribute the red and blue balls independently, the number of ways to distribute both the red and blue balls to the people is $N = \binom{10+6-1}{6}^2$.
Now to deal with the restriction that every person must get at least one ball. Say a distribution of the balls has "Property $i$" if the $i$th person receives no balls, for $i=1,2,3,\dots,10$, and let $S_j$ be the number of distributions with $j$ of the properties, for $j=1,2,3,\dots,10$. Then
$$\begin{align}
S_1 &= \binom{10}{1} \binom{9+6-1}{6}^2 \\
S_2 &= \binom{10}{2} \binom{8+6-1}{6}^2 \\
S_3 &= \binom{10}{3} \binom{7+6-1}{6}^2 \\
&\dots \\
S_9 &= \binom{10}{9} \binom{1+6-1}{6}^2
\end{align}$$
By PIE, the number of distributions with none of the properties, i.e. the number of distributions in which every person gets at least one ball, is
$$N_0 = N - S_1 + S_2 - S_3 + \dots - S_9 =26,250$$
|
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|
Behaviour of $\sum\limits_{n=1}^\infty \frac1{n^2}\left(\sqrt{n^2+n} - \sqrt{n^2+1}\right)^n x^n$ for $|x|=2$ I want to determine the radius of convergence for the power series
$$\sum_{n=1}^\infty \frac{(\sqrt{n^2+n} - \sqrt{n^2+1})^n}{n^2} x^n$$
and determine what happens at the boundaries. I determined the ratio of convergence $R$ to be $2$, but I struggle to show that it converges on $x = \pm 2$. Can you give me a hint?
What I have so far: $R = \frac{1}{\limsup_{n \to \infty} (x_n)} = \frac{1}{\frac{1}{1+1}} = \frac{2}{1} = 2$, where
$$\begin{align} x_n := \sqrt[n]{|a_n|} &= \sqrt[n]{| \frac{(\sqrt{n^2+n} - \sqrt{n^2+1})^n}{n^2}|} = \frac{\sqrt{n^2+n} - \sqrt{n^2+1}}{1} \\ &= \frac{(n^2+n) - (n^2+1)}{\sqrt{n^2+n} + \sqrt{n^2+1}} = \frac{(n^2+n) - (n^2+1)}{\sqrt{n^2+n} + \sqrt{n^2+1}} \\ &= \frac{n - 1}{\sqrt{n^2+n} + \sqrt{n^2+1}} = \frac{1 - \frac{1}{n}}{\sqrt{1+\frac{1}{n}} + \sqrt{1+\frac{1}{n^2}}} \end{align}$$
How do I determine if $\sum_{n=1}^\infty \frac{(\sqrt{n^2+n} - \sqrt{n^2+1})^n}{n^2} (\pm 2)^n$ converges? The root test doesn't work and the ratio test doesn't seem to be a smart move. What can I do?
|
$$
\begin{align}
&\sum_{n=1}^\infty \frac{\left(\sqrt{n^2+n} - \sqrt{n^2+1}\right)^n}{n^2}x^n\\
&=\sum_{n=1}^\infty \frac{n^n\left(\sqrt{1+\frac1n} - \sqrt{1+\frac1{n^2}}\right)^n}{n^2}x^n\\
&=\sum_{n=1}^\infty \frac{n^n\left(\left(1+\frac1{2n}-\frac1{8n^2}+O\!\left(\frac1{n^3}\right)\right) - \left(1+\frac1{2n^2}+O\!\left(\frac1{n^4}\right)\right)\right)^n}{n^2}x^n\\
&=\sum_{n=1}^\infty \frac{\left(\frac12-\frac5{8n}+O\!\left(\frac1{n^2}\right)\right)^n}{n^2}x^n\\[3pt]
&=\sum_{n=1}^\infty \frac{e^{-5/4}+O\!\left(\frac1n\right)}{2^nn^2}x^n
\end{align}
$$
Thus, the radius of convergence is $2$, and the series converges absolutely at the boundaries
|
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|
Find $\lim\limits_{n\to+\infty}\sum\limits_{k=1}^{n}\frac{k^{3}+6k^{2}+11k+5}{\left(k+3\right)!}$
Compute
$$\lim_{n\to+\infty}\sum_{k=1}^{n}\frac{k^{3}+6k^{2}+11k+5}{\left(k+3\right)!}.$$
My Approach
Since $k^{3}+6k^{2}+11k+5= \left(k+1\right)\left(k+2\right)\left(k+3\right)-1$
$$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{k^{3}+6k^{2}+11k+5}{\left(k+3\right)!}
= \lim_{n\rightarrow\infty}\sum_{k=1}^{n}\left(\frac{1}{k!}-\frac{1}{\left(k+3\right)!}\right)$$
But now I can't find this limit.
|
Good start!
$$ \begin{align}
\lim_{n \to \infty}\sum_{k=1}^n\left(\frac{1}{k!} - \frac{1}{(k+3)!}\right) &=
\lim_{n \to \infty}\left(\sum_{k=1}^n\frac{1}{k!} - \sum_{k=1}^n\frac{1}{(k+3)!} \right) \\ &= \lim_{n \to \infty}\left(\sum_{k=1}^n\frac{1}{k!} - \sum_{k=4}^{n+3}\frac{1}{k!} \right) \\ &= \lim_{n\to\infty}\left(\frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} - \frac{1}{(n+1)!} - \frac{1}{(n+2)!} - \frac{1}{(n+3)!} \right) \\ &= \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} \\ &= \frac{5}{3}
\end{align} $$
|
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|
Beginner troubleshooting an eigenvector calculation I am having some difficulty identifying the error in my eigenvector calculation. I am trying to calculate the final eigenvector for $\lambda_3 = 1$ and am expecting the result $ X_3 = \left(\begin{smallmatrix}-2\\17\\7\end{smallmatrix}\right)$
To begin with, I set up the following equation (for the purpose of this question I will refer to the leftmost matrix here as A).
$$
\begin{bmatrix}
1 - \lambda & 0 & 0 \\
3 & 3 - \lambda & -4\\
-2 & 1 & -\lambda -2 \\
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2\\
x_3 \\
\end{bmatrix}
=
\begin{bmatrix}
0\\
0\\
0\\
\end{bmatrix}
$$
I) Substitute $\lambda_3 = 1$
$$
\begin{bmatrix}
0 & 0 & 0 \\
3 & 2 & -4\\
-2 & 1 & -3 \\
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2\\
x_3 \\
\end{bmatrix}
=
\begin{bmatrix}
0\\
0\\
0\\
\end{bmatrix}
$$
II) Reduce the matrix with elementary row operations.
$R_2 \leftarrow R_2 - 2R_3$
$$
A =
\begin{bmatrix}
0 & 0 & 0 \\
7 & 0 & 2\\
-2 & 1 & -3 \\
\end{bmatrix}
$$
$R_3 \leftarrow 3R_2 + 2R_3$
$$
A =
\begin{bmatrix}
0 & 0 & 0 \\
7 & 0 & 2\\
17 & 2 & 0 \\
\end{bmatrix}
$$
$R_2 \leftarrow \frac{1}{7} R_2$
$R_3 \leftarrow \frac{1}{17} R_3$$
$$
A =
\begin{bmatrix}
0 & 0 & 0 \\
1 & 0 & 2/7\\
1 & 2/17 & 0 \\
\end{bmatrix}
$$
III) multiply matrices to get a series of equations equal to 0 and rearrange them in terms of a common element.
$x_1 + \frac{2}{7}x_3 = 0 \rightarrow x_1 = -\frac{2}{7}x_3$
$x_1 + \frac{2}{17}x_2 = 0 \rightarrow x_1 = -\frac{2}{17}x_2$
IV) Substitute a value into the vector to get an eigenvector.
Let $\ x_1 = 1 \rightarrow X_3 = \left(\begin{smallmatrix}1\\-2/17\\-2/7\end{smallmatrix}\right)
$
Which at this point we can see is not a multiple of the expected $X_3$. Can anyone highlight my error for me?
Many thanks in advance.
|
You have solved the problem correctly up to $$x_1 + \frac{2}{7}x_3 = 0 \rightarrow x_1 = -\frac{2}{7}x_3$$ and $$x_1 + \frac{2}{17}x_2 = 0 \rightarrow x_1 = -\frac{2}{17}x_2$$ Now when you let $x_1=1$ you get $x_3=-\frac {7}{2}$ and $x_2=-\frac {17}{2}$ which give you the correct eigenvector.
|
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|
How to prove that this triangle is equilateral? Question: If $\cos A +\cos B +\cos C=\frac{3}{2}$, prove that the triangle is equilateral.
My attempt: According to cosine rule, $\cos A=\frac{b^2+c^2-a^2}{2bc}$, $\cos B=\frac{c^2+a^2-b^2}{2ca}$ and $\cos C=\frac{a^2+b^2-c^2}{2ab}$.
$\cos A +\cos B +\cos C=\frac{3}{2}$
$\therefore\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ca}+\frac{a^2+b^2-c^2}{2ab}=\frac{3}{2}$
$\implies\frac{ab^2+ac^2-a^3+bc^2+ba^2-b^3+ca^2+cb^2-c^3}{2abc}=\frac{3}{2}$
I have tried simplifying the given equation using cosine rule but could not get far. Please help.
|
To prove that the triangle is equilateral, i have done the following: I have used the cosine rule to simplify the given equation, algebraic manipulation to bring it into the required form, triangle inequality to deduce that each term in the manipulated equation is non-negative and the fact that when the sum of non-negative terms is equal to zero, each term is equal to zero to set up separate equations to finally deduce that all three sides of the triangle are equal.
According to cosine rule, $\cos A=\frac{b^2+c^2-a^2}{2bc}$, $\cos B=\frac{c^2+a^2-b^2}{2ca}$ and $\cos C=\frac{a^2+b^2-c^2}{2ab}$.
$\cos A +\cos B +\cos C=\frac{3}{2}$
$\implies\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ca}+\frac{a^2+b^2-c^2}{2ab}=\frac{3}{2}$
$\implies\frac{ab^2+ac^2-a^3+bc^2+ba^2-b^3+ca^2+cb^2-c^3}{2abc}=\frac{3}{2}$
$\implies ab^2+ac^2-a^3+bc^2+ba^2-b^3+ca^2+cb^2-c^3=3abc$
$\implies ab^2+ac^2+bc^2+ba^2+ca^2+cb^2=a^3+b^3+c^3+3abc$
$\implies ab^2+ac^2+bc^2+ba^2+ca^2+cb^2-6abc=a^3+b^3+c^3-3abc$
$\implies ca^2-2abc+cb^2+ab^2-2abc+ac^2+bc^2-2abc+ba^2=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$\implies c(a^2-2ab+b^2)+a(b^2-2bc+c^2)+b(c^2-2ca+a^2)=\frac{1}{2}(a+b+c)(2a^2+2b^2+c^2-2ab-2bc-2ca)$
$\implies 2[c(a-b)^2+a(b-c)^2+b(c-a)^2]=(a+b+c)(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2)$
$\implies 2c(a-b)^2+2a(b-c)^2+2b(c-a)^2=(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]$
$\implies 2c(a-b)^2+2a(b-c)^2+2b(c-a)^2=(a+b+c)(a-b)^2+(a+b+c)(b-c)^2+(a+b+c)(c-a)^2$
$\implies 0=(a+b+c)(a-b)^2-2c(a-b)^2+(a+b+c)(b-c)^2-2a(b-c)^2+(a+b+c)(c-a)^2-2b(c-a)^2$
$\implies (a+b+c)(a-b)^2-2c(a-b)^2+(a+b+c)(b-c)^2-2a(b-c)^2+(a+b+c)(c-a)^2-2b(c-a)^2=0$
$\implies (a+b+c-2c)(a-b)^2+(a-2a+b+c)(b-c)^2+(a+b-2b+c)(c-a)^2=0$
$\implies (a+b-c)(a-b)^2+(-a+b+c)(b-c)^2+(a-b+c)(c-a)^2=0$
$\implies (a+b-c)(a-b)^2+(b+c-a)(b-c)^2+(c+a-b)(c-a)^2=0$
According to triangle inequality, $a+b>c, b+c>a$ and $c+a>b$.
$\therefore a+b-c>0, b+c-a>0$ and $c+a-b>0$.
When the sum of non-negative terms is equal to zero, each term$=0$.
$\therefore (a+b-c)(a-b)^2+(b+c-a)(b-c)^2+(c+a-b)(c-a)^2=0$
$\implies (a+b-c)(a-b)^2=0, (b+c-a)(b-c)^2=0$ and $(c+a-b)(c-a)^2=0$
$\implies a=b, b=c$ and $c=a$
$\implies a=b=c$
$\therefore$ The triangle is equilateral.
|
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|
Prove that if three real numbers $a, b, c$ satisfy $a^2+b^2+c^2=ab+bc+ac$, then $a=b=c$. I don't have much to start the problem, as I have trouble approaching it. What method of solving should I use here?
|
$$(a^2+b^2+c^2)-(ab+ac+bc) = \frac{1}{2}\left[(a-b)^2+(a-c)^2+(b-c)^2\right]\geq 0 $$
and the middle term equals zero iff $a=b=c$.
Alternative approach: by the Cauchy-Schwarz inequality
$$\left|ab+ac+bc\right|\leq \sqrt{a^2+b^2+c^2}\sqrt{b^2+c^2+a^2} = a^2+b^2+c^2$$
and the equality holds iff the vectors $(a,b,c)$ and $(b,c,a)$ are linearly dependent, i.e. iff $a=b=c$.
|
{
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|
Find non-zero real numbers $a,b,c,d$ such that $a^2+c^2=b^2+d^2$ and $ab+cd=0$. There are the obvious cases where $a=c=0$ and $b=d$. Are there any cases for this where all four numbers are non-zero?
Note, this problem is equivalent to saying the matrix $A^{-1}JA$ is a rotation invariant linear complex structure where $A=\begin{pmatrix}a & b\\c & d\end{pmatrix}\in GL(2,\mathbb{R})$ and $J=\begin{pmatrix}0 & 1\\-1 & 0\end{pmatrix}$ is the usual linear complex structure.
|
$$d=-\frac{ab}{c}$$ and we obtain:
$$a^2+c^2=b^2+\frac{a^2b^2}{c^2}$$ or
$$(a^2+c^2)(b^2-c^2)=0.$$
If $b=c$ then we get $d=-a$.
if $b=-c$ then we get $d=a$.
Id est, we got the answer:
$$\{(a,b,b,-a),(a,b,-b,a)|a\neq0,b\neq0\}$$
|
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|
Value of $a_2+a_6+a_{10}+\cdots+a_{42}$ If $(1+x+x^2+\cdots+x^9)^4(x+x^2+x^3+\cdots+x^9)=\sum_{r=1}^{45} a_rx^r ,$ then what is the value of $a_2+a_6+a_{10}+\cdots+a_{42}$
|
Calling your expression $f(x)$, we have
$$ a_2 + a_6 + \ldots + a_{42} = \dfrac{f(1) + f(-1) - f(i) - f(-i)}{4}
$$
Now $f(1) = 9 \cdot 10^4$, $f(-1) = 0$, $f(i) = -4 i$ and $f(-i) = 4i$, so
this is $(9 \cdot 10^4)/4 = 22500$.
|
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|
Computing $\int_0^\pi \frac{dx}{1+a^2\cos^2(x)}$
I am trying to compute the following integral
$$\int_0^\pi \frac{dx}{1+a^2\cos^2(x)}$$
But I got stuck on my way.
Indeed, enforcing the change of variables $t =\cos^2x$ leads to
$$\int_0^\pi \frac{d x}{1+a^2\cos^2(x)}= \int_{-1}^1 \frac{d x}{(1+a^2t^2)(1-t^2)^{1/2}}dx =2\int_{0}^1 \frac{d x}{(1+a^2t^2)(1-t^2)^{1/2}}dx$$
then what next? I though this was related the beta and Gamma function but it seems not. Can some help me here?
|
Here is another way. Consider the identity $$\cos^2x=\frac{1+\cos 2x}{2}$$ and the integral is then transformed into $$2\int_{0}^{\pi}\frac{dx}{2+a^2+a^2\cos 2x}$$ and using substitution $$2x=t$$ we get $$\int_{0}^{2\pi}\frac{dt}{2+a^2+a^2\cos t} $$ Since $\cos(2\pi - t) =\cos t$ we can see that the above integral is equal to $$2\int_{0}^{\pi}\frac{dt}{2+a^2+a^2\cos t} $$ and using the standard formula $$\int_{0}^{\pi}\frac{dx}{A+B\cos x}=\frac{\pi} {\sqrt{A^2-B^2}}$$ we can see that the integral in question is equal to $$\frac{2\pi}{\sqrt{(2+a^2)^2-a^4}}=\frac{\pi}{\sqrt{1+a^2}}$$ The standard formula given above can be proved using the substitution $$(A+B\cos x) (A-B\cos y) =A^2-B^2$$ and is valid for $A>|B|$.
|
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|
Prove that $a+b+c\le \frac {a^3}{bc} + \frac {b^3}{ac} + \frac {c^3}{ab}$ One of my friend had just given me an inequality to solve which is stated below.
Consider the three positive reals $a, b, c$ then prove that
$$a+b+c\le \frac {a^3}{bc} + \frac {b^3}{ac} + \frac {c^3}{ab}$$
I have solved this inequality very easily using Muirhead. But my friend has no idea what Muirhead inequality is. So I want to know whether there is any other method to solve this problem except for Muirhead's inequality.
|
This is the same as
$$a^4+b^4+c^4\ge abc(a+b+c).$$
By AM/GM,
$$2a^4+b^4+c^4\ge 4a^2bc.$$
Now add the cyclic permutations of this.
|
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|
Find the value of a third order circulant type determinant To calculate the value of the determinant\begin{pmatrix}
(b+c)^2 &c^2 &b^2 \\
c^2 &(c+a)^2 &a^2 \\
b^2 &a^2&(a+b)^2
\end{pmatrix} .
I multiplied R1,R2,R3 by a²,b²,c² respectively and then used the operations R1=R1-R2,R2=R2-R3,R3=R3-R1 which got me a zero valued determinant eventually. The correct answer given is 2(bc+ca+ab)². What's wrong with the row Operations that I used? Is there any other suitable Operations?
|
It's $$\prod_{cyc}(a+b)^2+2a^2b^2c^2-\sum_{cyc}a^4(b+c)^2=$$
$$=2\sum_{cyc}(a^3b^3+3a^3b^2c+3a^3c^2b+2a^2b^2c^2)=2(ab+ac+bc)^3.$$
I used the following.
$$\prod_{cyc}(a+b)=\sum_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right),$$
$$\left(\sum_{cyc}(a^2b+a^2c)\right)^2=\sum_{cyc}(a^4b^2+a^4c^2+2a^3b^3+2a^4bc+2a^3b^2c+2a^3c^2b+2a^2b^2c^2)$$ and
$$\sum_{cyc}a^4(b+c)^2=\sum_{cyc}(a^4b^2+a^4c^2+2a^4bc).$$
|
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|
Show that if $x+y+z=x^2+y^2+z^2=x^3+y^3+z^3=1$ then $xyz=0$ For any real numbers $x,y,z$ such that $x+y+z = x^2+y^2+z^2 = x^3+y^3+z^3 =1\\$ show that $x \cdot y \cdot z=0$.
I think that because $x \cdot y \cdot z = 0$ at least one of the three number should be equal to zero, but I'm stuck into relating the other things.
|
$$x^3+y^3+z^3-3xyz=(x+y+z)\{(x+y+z)^2-3(xy+yz+zx)\}$$
Now $$2(xy+yz+zx)=(x+y+z)^2-(x^2+y^2+z^2)=?$$
|
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|
Formulas involving pairs of coefficients of expansion of $(1+x+x^2)^n$ If $a_{0},a_{1},a_{2},\cdots\cdots$be coefficient in the expansion of $(1+x+x^2)^n$ in ascending power of $x$.Then prove that
$(1)\;a_{0}\cdot a_{1}-a_{1}\cdot a_{2}+a_{2}\cdot a_{3}-\cdots\cdots -a_{2n-1}\cdot a_{2n}=0$
$(2)\;a_{0}\cdot a_{2}-a_{1}\cdot a_{3}+a_{2}\cdot a_{4}-\cdots\cdots +a_{2n-2}a_{2n}=a_{n+1}$
Try: $(1+x+x^2)^n=a_{0}+a_{1}x+a_{2}x^2+\cdots \cdots +a_{2n}x^{2n}$
Put $\displaystyle x=\frac{1}{x}$,we have
Try: $(1+x+x^2)^n=a_{0}x^{2n}+a_{1x^{2n+1}}+a_{2}x^{2n+2}+\cdots \cdots +a_{2n}x^{4n}$
Could someone help me to solve it? Thanks!
|
Substituting $x$ by $\frac 1x$ into
$$(1+x+x^2)^n=\sum_{j=0}^{2n}a_{j}x^j\tag3$$
we get
$$\left(1+\frac 1x+\frac{1}{x^2}\right)^n=\sum_{j=0}^{2n}a_{j}\left(\frac 1x\right)^j\tag4$$
Multiplying the both sides by $x^{2n}$ gives
$$(x^2+x+1)^n=\sum_{j=0}^{2n}a_{j}x^{2n-j}\tag5$$
Substituing $x$ by $-x$ into $(5)$ gives
$$(x^2-x+1)^n=\sum_{j=0}^{2n}a_{j}(-1)^{2n-j}x^{2n-j}\tag6$$
Multiplying $(6)$ by $(3)$, we get
$$(1+x^2+x^4)^{n}=\left(a_0x^{2n}-a_1x^{2n-1}+a_2x^{2n-2}-\cdots +a_{2n}\right)\left(a_0+a_1x+\cdots +a_{2n}x^{2n}\right)\tag7$$
$(1)$
Let us consider the coefficient of $x^{2n+1}$ in $(7)$.
The coefficient of $x^{2n+1}$ in the LHS of $(7)$ is $0$, so we have
$$0=a_0a_1-a_1a_2+a_2a_3-\cdots -a_{2n-1}a_{2n}$$
$(2)$
Let us consider the coefficient of $x^{2n+2}$ in $(7)$.
The coefficient of $x^{2n+2}$ in the LHS of $(7)$ is the same as the coefficient of $x^{n+1}$ in $(1+x+x^2)^n$, i.e. $a_{n+1}$, so we have
$$a_{n+1}=a_0a_2-a_1a_3+a_2a_4-\cdots +a_{2n-2}a_{2n}$$
|
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|
How to find major & minor axes of the ellipse $10x^2+14xy+10y^2-7=0$?
What are major & minor axes of the ellipse: $10x^2+14xy+10y^2-7=0$ ?
My trial:
from given equation: $10x^2+14xy+10y^2-7=0$
$$10x^2+14xy+10y^2=7$$
$$\frac{x^2}{7/10}+\frac{xy}{7/14}+\frac{y^2}{7/10}=1$$
I know the standard form of ellipse: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
But the term of $xy$ is a bottleneck.
From here I can't proceed.
Can somebody please help me solve this problem?
Thank you.
|
Rewriting the equation in polar coordinates, with $x=r\cos\theta$ and $y=r\sin\theta$, we get
$$r^2(10\cos^2\theta+14\cos\theta\sin\theta+10\sin^2\theta)-7=0$$
Using the trig identities $\sin^2\theta+\cos^2\theta=1$ and $2\sin\theta\cos\theta=\sin2\theta$, we find this simplifies to
$$r^2(10+7\sin2\theta)-7=0$$
or
$$r=\sqrt{7\over10+7\sin2\theta}$$
The largest value of $r$ occurs when $\sin2\theta=-1$ and the smallest when $\sin2\theta=1$. Thus the major axis is $2\sqrt{7/3}$ and the minor axis is $2\sqrt{7/17}$.
|
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|
Calculate the triple integral: $\iiint_W(x^2+y^2)z\ dx\,dy\,dz$ I am trying to solve the following triple integral:
$$
\iiint_W(x^2+y^2)z\,dx\,dy\,dz \\
W=\{(x,y,z) \in \mathbb{R}: x^2+y^2+z^2 \le 9; x^2+y^2\le 1; x \ge 0; y \ge 0; z \ge 0\}
$$
From which I know that there are two surfaces:
$$
x^2+y^2+z^3 = 9 \rightarrow \text{Sphere of radius 3} \\
x^2+y^2=1 \rightarrow \text{Cylinder with a bade radius of 1}
$$
Converting this to Cylindrical Coordinates:
$$
\rho^2+z^2 \le 9 \implies \rho^2 \le 9-z^2 \implies \rho \le \sqrt{9-z^2} \\
\rho^2\le 1
$$
Where do I go from here?
|
Hint:
from $\rho^2\le 1$ you have the limits for $\rho$, that is : $0\le \rho \le 1$
so you can find the limits for $z$ that is: $0\le z \le \sqrt{9-\rho^2}$
Finally, from the condition that $x,y$ are positive, you have the limits $0\le \theta \le \frac{\pi}{2}$
Can you do from this?
|
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|
Is this a correct derivation of completing the square? $x^2 + bx$
$=x^2 + bx + c - c$
$=(x + k)^2 - c$
$=x^2 + 2kx + (k^2 - c) = x^2 + bx + 0$
This implies:
$2k = b$, so $k = b/2$, and:
$k^2 - c = 0$, or $k^2 = c$, or $(b/2)^2 = c$
So to complete the square we are making the transformation:
$x^2 + bx \implies (x + b/2)^2 - (b/2)^2$
|
It's true, but you can get it much more easier:
$$x^2+bx=x^2+bx+\frac{b^2}{4}-\frac{b^2}{4}=\left(x+\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2$$
|
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|
Prove that the intersection of a cylinder and a plane forms an ellipse I've seen the nice proof of this using spheres, but I'm looking for a way to prove it parametrically if possible. Using a cylinder $x^2+y^2=r^2$ and a plane $ax+by+cz+d=0$ I got:
$x=r\cos(\theta), y=r\sin(\theta), z=\dfrac{-ar\cos(\theta)+br\sin(\theta)+d}{c}$
But after this I'm stuck trying different projections and messing with ellipse definitions
|
Without loss of generality let's assume the cylinder has radius $1$, so that it has equation $x^2 + y^2 = 1$. Assuming that the plane is not parallel to the cylinder, we can always rearrange the coordinate system so that the plane goes through the origin, or even better, make the plane go through the $x$-axis after a suitable rotation. Now the plane should have equation $z = y \tan\alpha$ (with $\alpha$ being the slope in the $yz$-plane).
Now your parametrization gives the curve
$$x = \cos\theta, \quad y = \sin\theta, \quad z=\sin\theta\tan\alpha.$$
We think of a rotation (of the whole space) around the $x$-axis, through an angle of $-\alpha$ in the $yz$-plane:
$$\begin{aligned}
\begin{bmatrix} x \\ y \\ z \end{bmatrix}
& \mapsto
\begin{bmatrix}
1 & 0 & 0 \\
0 & \cos(-\alpha) & -\sin(-\alpha) \\
0 & \sin(-\alpha) & \cos(-\alpha)
\end{bmatrix}
\begin{bmatrix} x \\ y \\ z \end{bmatrix} \\
& =
\begin{bmatrix}
x \\
y\cos\alpha + z\sin\alpha \\
-y\sin\alpha + z\cos\alpha
\end{bmatrix}.
\end{aligned}$$
Under this rotation the curve becomes
$$\begin{aligned}
\begin{bmatrix} \cos\theta \\ \sin\theta \\ \sin\theta\tan\alpha \end{bmatrix}
& \mapsto
\begin{bmatrix}
\cos\theta \\
\sin\theta\cos\alpha + \sin\theta\tan\alpha\sin\alpha \\
-\sin\theta\sin\alpha + \sin\theta\tan\alpha\cos\alpha
\end{bmatrix} \\
& =
\begin{bmatrix}
\cos\theta \\
\sin\theta\cos\alpha + \sin\theta\sin^2\alpha/\cos\alpha \\
-\sin\theta\sin\alpha + \sin\theta\sin\alpha
\end{bmatrix} \\
& =
\begin{bmatrix}
\cos\theta \\
\sin\theta/\cos\alpha \\
0
\end{bmatrix}.
\end{aligned}$$
This is
$$x = \cos\theta, \quad y = \frac{1}{\cos\alpha}\sin\theta, \quad z\equiv0,$$
which is an ellipse.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Sum of series $\frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{4 \cdot 5 \cdot 6} + \frac{1}{6 \cdot 7 \cdot 8} + \cdots $ What is the limit of series $\frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{4 \cdot 5 \cdot 6} + \frac{1}{6 \cdot 7 \cdot 8} + \cdots $?
The $n$th summand is $\frac{1}{(2n)(2n + 1)(2n+2)} = \frac{1}{4} \frac{1}{n(2n+1)(n+1)}$.
I have tried expressing this as a telescoping sum, or as the limit of Riemann sums of a partition (the usual methods I normally try when doing this type of question- what are some other strategies?)
|
The $n$-th term is
$$\frac12\left(\frac1{2n}-\frac{2}{2n+1}+\frac1{2n+2}\right).$$
The whole series does not telescope but is
$$\frac12\left(\frac12-\frac23+\frac24-\frac25+\frac26-\frac27+\cdots
\right).$$
This is very similar (not identical) to a well-known series...
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find lengths of tangents drawn from $(3,-5)$ to the Ellipse Find lengths of tangents drawn from $A(3,-5)$ to the Ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$
My try: I assumed the point of tangency of ellipse as $P(5\cos a, 4 \sin a)$
Now Equation of tangent at $P$ is given by
$$\frac{x \cos a}{5}+\frac{ y \sin a}{4}=1$$ whose slope is $$m_1=\frac{-4 \cot a}{5}$$
Also slope of $AP$ is given by
$$m_2=\frac{4 \sin a+5}{5 \cos a-3}$$
So both slopes are equal , with that we get
$$12 \cos a-25 \sin a=20 \tag{1}$$
Now distance $AP$ is given by
$$AP=\sqrt{(3-5 \cos a)^2+(5+4 \sin a)^2}=\sqrt{75-30 \cos a+40 \sin a} \tag{2}$$
Now using $(1)$ we have to find $\cos a$ and $\sin a$ and then substitute in $(2)$ which becomes lengthy.
Any better way?
|
The slope of tangent at a point (x,y) on the ellipse $$\frac{x^2}{25}+\frac{y^2}{16}=1$$ is $$\frac {dy}{dx} = \frac {-16x}{25y}$$
On the other hand slope of the line passing through $(x,y)$ and $(3,-5)$ is $m = \frac {y+5}{x-3} $
$$ \frac {-16x}{25y}= \frac {y+5}{x-3} \implies 48x-125y=400 $$
Solve the system $$ \frac{x^2}{25}+\frac{y^2}{16}=1$$ $$48x-125y=400 $$
to find the two points $(-1.56199, -3.79980)$ and $(4.68293,-1.40176).$
Now we can find the distance from P to these points.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find all solutions in positive integers of the Diophantine equation $x^2+2y^2 =z^2$ I can't solve the following exercise. It's already on this site, but the solution method is not the same as in my solutions manual.
Find all solutions in positive integers of the Diophantine equation $x^2+2y^2 =z^2$.
The solution is given, but I do not understand the italicised parts.
13.1.8 $2y^2=z^2-x^2=(z-x)(z+x)$. $x$ and $y$ have the same parity, so $(z-x)/2$ and $(z+x)/2$ are integers. It suffices to assume $(x,z)=1$. Then either $((z-x)/2,z+x)=1$, and then $y^2=((z-x)/2)(z+x)=m^2n^2$, and solving $(z-x)/2=m^2$ and $z+x=n^2$ for $x$ and $y$ gives $x=(m^2-2n^2)/2,y=mn,z=(m^2+2n^2)/2$. Or $((z+x)/2,z-x)=1$ which gives $x=(2m^2-n^2)/2,y=mn,z=(2m^2+n^2)/2$.
*
*Why have x and y the same parity?
*Why are $(z-x)/2$ and $(z+x)/2$ integers?
*Why do they assume $\gcd((z-x)/2,z+x)=1$ and next $\gcd((z+x)/2,z-x)=1$?
*How to solve $(z-x^2)/2 = m^2$ and $z-x=n^2$ for x and y to become the values for x, y and z?
|
*
*$2y^2=z^2-x^2$, the left-hand side is an even number so $z,x$ must either both be even (even minus even is even) or they must both be odd (odd minus odd is even). They did, however, mistype. They typed $x,y$ have the same parity (false for some solutions) but meant $x,z$ gave the same parity (proven true here).
*$z-x$ must be even (by the argument above), so $\frac{z-x}{2}$ is an integer.
*$z+x$ must also be even by the exact same argument, so $\frac{z+x}{2}$ is also an integer.
*First they assume that $\gcd(x,z)=1$, because if they have a common factor, say $d$, then you have the expression:
$$2y^2=(z'd-x'd)(z'd+x'd)=d^2(z'-x')(z'+x')$$ and you can divide both sides by $d^2$ and get exactly the same expression ($x=x'd$, $y=y'd$, $z=z'd$). At some point this has to stop so you can assume that $x,z$ have no common factor.
*This is perhaps the hardest part:
Why is $$\gcd\left (\frac{(z-x)}{2},z+x\right )=1$$ or $$\gcd\left (z-x,\frac{(z+x)}{2}\right )=1$$ This follows from another result:
If $\gcd(z,x)=1$ then $\gcd(z-x,z+x)$ is $1$ or $2$
The result is proven here. Now, we know that both $z-x,z+x$ are divisible by $2$, and this is the largest factor they can have in common. Atleast one of $z-x,z+x$ is divisible by $2$ and not $4$. This means that we must have either $\gcd\left (\frac{(z-x)}{2},z+x\right )=1$ or $\gcd\left (z-x,\frac{(z+x)}{2}\right )=1$. Now, which one is it? We don't know!
*
*If $\gcd\left (\frac{(z-x)}{2},z+x\right )=1$ then we know that $$y^2=\frac{(z-x)}{2}(z+x)$$
but $y^2$ is a square, and $\frac{(z-x)}{2},z+x$ have no factor in common! This must mean that $\frac{(z-x)}{2}=m^2$ is a square, and $z+x=n^2$ is a square. You can use this information to solve for $x,y,z$.
*If $\gcd\left (z-x,\frac{(z+x)}{2}\right )=1$ then again we know that $$y^2=(z-x)\frac{(z+x)}{2}$$
but $y^2$ is a square, and $z-x,\frac{(z+x)}{2}$ have again no factor in common! This must mean that $z-x=n^2$ is a square, and $\frac{z+x}{2}=m^2$ is a square.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why is $\int_{-\infty}^{\infty}\left\lvert\log\frac{a^2+x^2}{b^2+x^2}\right\rvert dx=2\pi(b-a)$?
If $a,b\in\mathbb R_{>0}$ and $ b>a$, then why is $\int_{-\infty}^{\infty}\left\lvert\log\frac{a^2+x^2}{b^2+x^2}\right\rvert dx=2\pi(b-a)$ ?
Usual change of variables doesnt bring anything, I think. Is there a special function involved here (because of $\pi$) ?
|
\begin{align*}
\int_{-M}^{M}\left|\log\dfrac{a^{2}+x^{2}}{b^{2}+x^{2}}\right|dx&=\int_{-M}^{M}\log\dfrac{b^{2}+x^{2}}{a^{2}+x^{2}}dx\\
&=\int_{-M}^{M}\log(b^{2}+x^{2})dx-\int_{-M}^{M}\log(a^{2}+x^{2})dx,
\end{align*}
where
\begin{align*}
\int_{-M}^{M}\log(b^{2}+x^{2})dx&=2M\log(b^{2}+M^{2})-\int_{-M}^{M}\dfrac{2x}{b^{2}+x^{2}}dx\\
&=2M\log(b^{2}+M^{2})-4M+ 2b^{2}\int_{-M}^{M}\dfrac{1}{x^{2}+b^{2}}dx\\
&=2M\log(b^{2}+M^{2})-4M+4b\tan^{-1}\dfrac{M}{b},
\end{align*}
and so
\begin{align*}
&\int_{-M}^{M}\log(b^{2}+x^{2})dx-\int_{-M}^{M}\log(a^{2}+x^{2})dx\\
&=2M\log\dfrac{b^{2}+M^{2}}{a^{2}+M^{2}}+4b\tan^{-1}\dfrac{M}{b}-4a\tan^{-1}\dfrac{M}{a}\\
&\rightarrow0+4b\cdot\dfrac{\pi}{2}-4a\cdot\dfrac{\pi}{2}\\
&=2\pi(b-a).
\end{align*}
|
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|
how to get the limit of $\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right)$ How to get the limit $\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right) = \frac{1}{2}$ ?
$\begin{align}
\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right) &=
\lim_{n \to \infty}\left( \frac{(\sqrt{n+\sqrt{n}}-\sqrt{n})(\sqrt{n+\sqrt{n}}+\sqrt{n})}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{n+\sqrt{n}}+\sqrt{n}\sqrt{n+\sqrt{n}}-\sqrt{n}\sqrt{n+\sqrt{n}}-n}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{n+\sqrt{n}}-n}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \right) \\
&= \lim_{n \to \infty}\left( \frac{\frac{1}{\sqrt{n}}(\sqrt{n+\sqrt{n}}-n)}{\frac{1}{\sqrt{n}}(\sqrt{n+\sqrt{n}}+\sqrt{n})} \right) \\
&= \lim_{n \to \infty}\left( \frac{(\sqrt{\frac{n}{n}+\frac{\sqrt{n}}{n}}-\frac{n}{\sqrt{n}})}{(\sqrt{\frac{n}{n}+\frac{\sqrt{n}}{n}}+ \frac{\sqrt{n}}{\sqrt{n}})} \right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{1}-\frac{n}{\sqrt{n}}}{2}\right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{1}-\frac{\sqrt{n}\sqrt{n}}{\sqrt{n}}}{2}\right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{1}-\sqrt{n}}{2}\right) \\
&= \lim_{n \to \infty}\left( \frac{1}{2} - \frac{\sqrt{n}}{2} \right) \\
\end{align}$
Which is wrong.
Where could be my mistake?
|
In a simpler way:
Let $n=m^2$.
$$\lim_{m\to\infty}\sqrt{m^2+m}-m=\lim_{m\to\infty}\frac m{\sqrt{m^2+m}+m}=\lim_{m\to\infty}\frac 1{\sqrt{1+\dfrac1m}+1}.$$
|
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|
Compute the 100th power of a given matrix So, I was asked this question in an interview.
Given a matrix
$$M = \begin{bmatrix} 0 & 1& 1 \\ 1& 0& 1 \\ 1&1& 0\end{bmatrix},$$
find $M^{100}$.
How does one approach this question using pen and paper only?
|
Trying to avoid orthogonal diagonalization and square roots. The matrix of all ones has eigenvalues $(3,0,0),$ with eigenvectors (NOT normalized) as the columns of
$$
W =
\left(
\begin{array}{ccc}
1 & -1 & -1 \\
1 & 1 & -1 \\
1 & 0 & 2
\end{array}
\right)
$$
Oh, we need the inverse,
$$
W^{-1} = \frac{1}{6}
\left(
\begin{array}{ccc}
2 & 2 & 2 \\
-3 & 3 & 0 \\
-1 & -1 & 2
\end{array}
\right)
$$
Your matrix $M$ has the same eigenvectors with eigenvalues $(2,-1,-1).$ Therefore $M^{100}$ has the same eigenvectors with eigenvalues $(B,1,1),$ where $B = 2^{100}$ stands for BIG. We have the matrix equation for $M^{100} W,$ namely
$$
M^{100}
\left(
\begin{array}{ccc}
1 & -1 & -1 \\
1 & 1 & -1 \\
1 & 0 & 2
\end{array}
\right) =
\left(
\begin{array}{ccc}
B & -1 & -1 \\
B & 1 & -1 \\
B & 0 & 2
\end{array}
\right)
$$
We multiply both sides on the right by $W^{-1},$ I got it wrong the first time,
$$
M^{100} = \frac{1}{6}
\left(
\begin{array}{ccc}
B & -1 & -1 \\
B & 1 & -1 \\
B & 0 & 2
\end{array}
\right)
\left(
\begin{array}{ccc}
2 & 2 & 2 \\
-3 & 3 & 0 \\
-1 & -1 & 2
\end{array}
\right) =
\frac{1}{6}
\left(
\begin{array}{ccc}
2B+4 & 2B-2 & 2B-2 \\
2B-2 & 2B+4 & 2B-2 \\
2B-2 & 2B-2 & 2B+4
\end{array}
\right) =
\frac{1}{3}
\left(
\begin{array}{ccc}
B+2 & B-1 & B-1 \\
B-1 & B+2 & B-1 \\
B-1 & B-1 & B+2
\end{array}
\right)
$$
This is all integers as $ B = 2^{100} = 4^{50} \equiv 1^{50} \equiv 1 \pmod 3$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove this geometry question on quadrilateral and circles? In quadrilateral $ABCD$, $AC$ and $BD$ crossed at point $E$. Circle $O$ passes through $A, D$, and $E$ and its center is $O$. $P, Q, R$ are the midpoint of $AB$, $BC$ and $CD$, respectively. Circle $O_2$ passes through $P, Q$, and $R$, and crosses $BC$ at $F$.
Prove: $OF$ is perpendicular to $BC$.
|
It suffices to prove that$$
BO^2 - CO^2 = BF^2 - CF^2. \tag{1}
$$
Denote the radius of circle $ADE$ by $r$. The power of point $B$ with respect to circle $ADE$ is$$
BO^2 - r^2 = BE \cdot BD,
$$
and the power of point $C$ with respect to circle $ADE$ is$$
CO^2 - r^2 = CE \cdot CA,
$$
also $BD = 2 QR$, $CA = 2 QP$, therefore$$
BO^2 - CO^2 = BE \cdot BD - CE \cdot CA = 2(BE \cdot QR - CE \cdot QP). \tag{2}
$$
Because $P, F, Q, R$ are concyclic and $BD$ and $QR$ are parallel, then$$
∠FPR = ∠CQR = ∠EBC.
$$
Analogously,$$
∠FRP = ∠BQP = ∠ECB.
$$
Therefore,$$
△BCE \sim △PRF,
$$
which implies$$
\frac{BE}{PF} = \frac{CE}{RF} = \frac{BC}{PR},
$$
or$$
BE = PF \cdot \frac{BC}{PR}, \quad CE = RF \cdot \frac{BC}{PR}. \tag{3}
$$
Now, there are two cases with respect to the relative positions of $F$ and $Q$.
Case 1: If segment $PQ$ and $FR$ intersect, then$$
BF^2 - CF^2 = (BF + CF)(BF - CF) = BC \cdot (-2QF) = -2BC \cdot QF. \tag{4}
$$
Therefore,\begin{align*}
(1) &\stackrel{(2)(4)}{\Longleftrightarrow} BE \cdot QR - CE \cdot QP = -BC \cdot QF\\
&\stackrel{(3)}{\Longleftrightarrow} PF \cdot \frac{BC}{PR} \cdot QR - RF \cdot \frac{BC}{PR} \cdot QP = -BC \cdot QF\\
&\Longleftrightarrow PF \cdot QR - RF \cdot QP = -PR \cdot QF\\
&\Longleftrightarrow PF \cdot QR + PR \cdot QF = RF \cdot QP,
\end{align*}
and the last equality holds due to Ptolemy's theorem.
Hence (1) holds, which implies $OF⊥BC$.
Case 2: If segment $PQ$ and $FR$ do not intersect, then$$
BF^2 - CF^2 = (BF + CF)(BF - CF) = BC \cdot 2QF = 2BC \cdot QF. \tag{4'}
$$
Therefore,\begin{align*}
(1) &\stackrel{(2)(4)}{\Longleftrightarrow} BE \cdot QR - CE \cdot QP = BC \cdot QF\\
&\stackrel{(3)}{\Longleftrightarrow} PF \cdot \frac{BC}{PR} \cdot QR - RF \cdot \frac{BC}{PR} \cdot QP = BC \cdot QF\\
&\Longleftrightarrow PF \cdot QR - RF \cdot QP = PR \cdot QF\\
&\Longleftrightarrow RF \cdot QP + PR \cdot QF = PF \cdot QR,
\end{align*}
and the last equality holds due to Ptolemy's theorem.
Hence (1) holds, which implies $OF⊥BC$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is it always a positive value? The values of $x$ can be from $0$ to $\infty$. So for these values of $x$ is the following function always positive $$f(x)=x^2+x+\exp(\frac{1}{x})(2x+1)Ei(-\frac{1}{x})$$ where $Ei(x)$ is the exponential integral. My checking in the WA shows that it is a positive value for arbitrarily large values of $x$.
|
\begin{align}
\mathrm{Ei}(-1/x) &= - \int_{1/x}^\infty \frac{e^{-t}}{t} \, dt \\
f(x) &> 0 \\
x (x + 1) &> - (2x+1) \exp(1/x) \, \mathrm{Ei}(−1/x) \\
x (x + 1) &> (2x+1) \exp(1/x) \, \int_{1/x}^\infty \frac{e^{-t}}{t} \, dt \\
\end{align}
Since by requirement $x$ is positive, we can bring everything except the integral to the left without changing the direction of the inequality sign
\begin{align}
\frac{x(x + 1)}{(2x + 1)} \exp(-1/x) &> \int_{1/x}^\infty \frac{e^{-t}}{t} \, dt \\
\end{align}
now write everything in terms of $u = 1/x$:
\begin{align}
\frac{(1 + u)}{u(2 + u)} \exp(-u) &> \int_{u}^\infty \frac{e^{-t}}{t} \, dt \\
\end{align}
take the difference of both sides $\Delta(u)$ and take the derivative with respect to $u$
\begin{align}
\Delta'(u) &= \left[ \frac{u(2 + u) - 2(1+u)^2}{u^2(2 + u)^2}- \frac{(1 + u)}{u(2 + u)} \right] \exp(-u) + \frac{e^{-u}}{u} \\
&= \left[ - \frac{(u^2 + 2u + 2) + u(1+u)(2+u)}{u^2(2 + u)^2} \right] \exp(-u) + \frac{e^{-u}}{u} \\
&= \left[ 1 - \frac{2 + 4u + 4u^2 + u^3}{u(2 + u)^2} \right] \frac{e^{-u}}{u} \\
&= \left[ \frac{(2u + 4u^2 + u^3) - (2 + 4u + 4u^2 + u^3)}{u(2 + u)^2} \right] \frac{e^{-u}}{u} \\
&= - \frac{ 2 (1 + u) }{u^2(2 + u)^2} e^{-u} \\
\end{align}
since $x > 0$ we have $u > 0$, so this difference function is strictly monotonically decreasing. The divergence approaches $\Delta'(u) \to 0$ very quickly for $u \to \infty$, so $\Delta(u)$ converges to a minimum at $u \to \infty$ or $x \to 0$.
So all we need to check is whether $f(x)$ is non-negative for $x \to 0$. Clearly we can ignore the polynomial terms, so we get
\begin{align}
f(0) &= \lim_{x\to0} \, e^{1/x} \int_{1/x}^\infty \frac{e^{-t}}{t} \, dt
\end{align}
which is clearly non-negative. Thus $f(x)$ is positive for all $x > 0$.
|
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|
Solving $\cos\big(\tan^{-1}x\big)=\sin\big(\cot^{-1}\frac{3}{4}\big)$ Solve
$$
\cos\big(\tan^{-1}x\big)=\sin\big(\cot^{-1}\frac{3}{4}\big)
$$
My Attempt:
From the domain consideration,
$$
\boxed{0\leq\tan^{-1}\frac{3}{4},\tan^{-1}x\leq\frac{\pi}{2}}
$$
$$
\cos\big(\tan^{-1}x\big)=\cos\big(\frac{\pi}{2}-\cot^{-1}\frac{3}{4}\big)\implies\cos\big(\tan^{-1}x\big)=\cos\big(\tan^{-1}\frac{3}{4}\big)\\\implies\tan^{-1}x=2n\pi\pm\tan^{-1}\frac{3}{4}\\
\implies \tan^{-1}x=\tan^{-1}\frac{3}{4}\quad\text{ as }0\leq\tan^{-1}\frac{3}{4}\leq\frac{\pi}{2}\\
\implies x=\frac{3}{4}
$$
Is it correct or $\frac{-3}{4}$ also is a solutions ?
What about the condition $0\leq\tan^{-1}\frac{3}{4},\tan^{-1}x\leq\frac{\pi}{2}$, does this affect the solutions ?
|
$$\cos(\arctan x)=\cos\left(\arctan\dfrac34\right)$$
$$\implies\arctan x=2m\pi\pm\arctan\dfrac34=2m\pi+\arctan\left(\pm\dfrac34\right)$$ where $m$ is any integer as $\arctan(-a)=-\arctan(a)$
As $-\dfrac\pi2<\arctan y\le\dfrac\pi2,$
$$\arctan x= \arctan\left(\pm\dfrac34\right)$$
Apply tan on both sides
|
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|
Eigenvalues of symmetric $\mathbb{R}^{p\times p}$ matrix I want to prove that
$$A^{(p)}
=
\begin{pmatrix}
a & 1 & 1 & \dots & 1 & 1 & 1 \\
1 & a & 1 & \dots & 1 & 1 & 1 \\
1 & 1 & a & \dots & 1 & 1 & 1 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\
1 & 1 & 1 & \dots & 1 & a & 1 \\
1 & 1 & 1 & \dots & 1 & 1 & a
\end{pmatrix}
\in\mathbb{R}^{p\times p}$$
has eigenvalues $\lambda_+=a+p-1$ and $\lambda_-=a-1$ with degeneracy $p-1$ by using mathematical induction. Evidence that these eigenvalues could be correct were found 'empiricaly' by Mathematica.
Induction Base Case $(p=2)$
$$\det(A^{(2)}-\lambda E_2)
=
\det\begin{pmatrix}a-\lambda & 1\\1 & a-\lambda\end{pmatrix}
=(\lambda-(a-1))(\lambda-(a+1)) $$
Induction Hypothesis $(p\in\mathbb{N})$
$$\det(A^{(p)}-\lambda E_p)
=(\lambda-(a-1))^{p-1}(\lambda-(a+p-1))$$
Induction Step $(p\to p+1)$
$$\det(A^{(p+1)}-\lambda E_{p+1})
=
\det\begin{pmatrix}
a-\lambda & 1 & 1 & \dots & 1 & 1 & 1 \\
1 & a-\lambda & 1 & \dots & 1 & 1 & 1 \\
1 & 1 & a-\lambda & \dots & 1 & 1 & 1 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\
1 & 1 & 1 & \dots & 1 & a-\lambda & 1 \\
1 & 1 & 1 & \dots & 1 & 1 & a-\lambda
\end{pmatrix}$$
We can subtract the last row from the first row
$$\det\begin{pmatrix}
a-\lambda-1 & 0 & 0 & \dots & 0 & 0 & 1-a+\lambda \\
1 & a-\lambda & 1 & \dots & 1 & 1 & 1 \\
1 & 1 & a-\lambda & \dots & 1 & 1 & 1 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\
1 & 1 & 1 & \dots & 1 & a-\lambda & 1 \\
1 & 1 & 1 & \dots & 1 & 1 & a-\lambda
\end{pmatrix}$$
and apply Laplace's formula to the first row. The first entry (upper left corner) gives
$$(a-\lambda-1)
\det\begin{pmatrix}
a-\lambda & 1 & \dots & 1 & 1 & 1 \\
1 & a-\lambda & \dots & 1 & 1 & 1 \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\
1 & 1 & \dots & 1 & a-\lambda & 1 \\
1 & 1 & \dots & 1 & 1 & a-\lambda
\end{pmatrix}$$
which is just the case of our induction hypothesis, thus
$$-(\lambda-(a-1))^p(\lambda-(a+p-1)) $$
is the first contribution to Laplace's formula. The second contribution reads
$$(1-a+\lambda) (-1)^{p+2}
\det\begin{pmatrix}
1 & a-\lambda & 1 & \dots & 1 & 1 \\
1 & 1 & a-\lambda & \dots & 1 & 1 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
1 & 1 & 1 & \dots & 1 & a-\lambda\\
1 & 1 & 1 & \dots & 1 & 1
\end{pmatrix}$$
and we can subtract the first column from all other columns yielding
$$\det\begin{pmatrix}
0 & a-\lambda-1 & 0 & \dots & 0 & 0 \\
0 & 0 & a-\lambda-1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \dots & 0 & a-\lambda-1\\
1 & 1 & 1 & \dots & 1 & 1
\end{pmatrix}$$
on which we can reapply Laplace's formula and use that the determinant of a diagonal matrix equals the product of the diagonal entries, hence
$$(-1)^{p+1}(\lambda-(a-1))^{p+1} $$
is the second contribution to the first application of Laplace's formula. Now combining both I find
$$\det(A^{(p+1)}-\lambda E_{p+1})
=
(-1)(\lambda-(a-1))^p(\lambda-(a+p-1))+(-1)^{p+1}(\lambda-(a-1))^{p+1}
=(-1)(\lambda-(a-1))^p(\lambda-a-p+1+(-1)^p(\lambda-a+1)) $$
which seems faulty: on the one hand side the factor $(-1)^{p}$ should not exist as for odd $p$ the characteristic polynomial should varnish which has been shown to be not the case for i.e. $p=3$ (see Mathematica). On the other hand even if that factor would be wrong the final expression would yield
$$(-2)(\lambda-(a-1))^p(\lambda-(a-p/2+1))$$
which would contradict the induction hypothesis.
Did I made a mistake in evaluating the determinant or is the induction hypothesis wrong?
|
This is an answer not by induction, in response to a comment/answer by OP.
$$
\newcommand{\bu}{ {\mathbf u}}
\newcommand{\bv}{ {\mathbf v}}
\newcommand{\bM}{ {\mathbf M}}
\newcommand{\bI}{ {\mathbf I}}
\newcommand{\bU}{ {\mathbf U}}
\newcommand{\bQ}{ {\mathbf Q}}
$$
Your matrix $\bM$ is
$$
\bM = \bU + (a-1)\bI
$$
where $\bU$ is a matrix of all $1$s. It's clear that
$$
(n-1) + a
$$
is an eigenvalue, for the vector $\bu'$ consisting of all $1$s is a corresponding eigenvector. We can normalize this to a vector $\bu$ consisting of all $\frac{1}{\sqrt{n}}$ entries. Furthermore, we can write $\bU$ as $n \bu \bu^t$, which shows that $U$ has rank $1$, which is no surprise, because all its columns are the same.
Now let $\bv_1, \ldots, \bv_{n-1}$ be an orthonormal basis for the space
$$
H = \{ \bv \mid \bu \cdot \bv = \bu^t \bv = 0 \}
$$
of vectors orthogonal to $\bu$. Let's compute $\bM\bv_i$ for any one of these. It's
\begin{align}
\bM\bv_i
&= ( \bU + (a-1)\bI ) \bv_i \\
&= ( n\bu \bu^t + (a-1)\bI ) \bv_i \\
&= (n\bu \bu^t) \bv_i + (a-1)\bI \bv_i \\
&= n\bu (\bu^t \bv_i) + (a-1)\bv_i \\
&= n\bu (0) + (a-1)\bv_i \\
&= (a-1)\bv_i \\
\end{align}
In other words, each of the vectors $\bv_i$ is an eigenvector of eigenvalue $a-1$.
So the vectors $\{\bu, \bv_1, \bv_2, \ldots, \bv_{n-1}\}$ constitute an orthonormal set of eigenvectors of $\bM$, and if we form a matrix $\bQ$ with those as its columns, we get
$$
\bQ^t \bM \bQ = \pmatrix{(n-1)+ a & & & & & \\
& a-1 & & & & \\
& & a-1 & & & \\
& & & \ldots& & \\
& & & & a-1 & \\
& & & & & a-1}
$$
which is pretty much a complete eigenanalysis of $\bM$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to tell if this integral converges? $$\int_{0}^{\infty} \frac{\sqrt{1-x+x^2}}{1-x^2+x^4} dx$$
What's the method for determining if this integral converges or diverges? The integral seems to converge if I put it into Wolfram Alpha.
But do we assume it's similar to $$\int_{0}^{\infty} \frac{1}{x^4} dx$$
Because if so I can't get that one to converge since it's $-\frac{1}{3x^3}$ where $x=\infty$ and $x=0$ (I don't know how to write the "right bar" notation for integrals) which goes to $-0 + \infty$ which is divergent. So I am not sure what is right.
|
To prove convergence of $\int_0^{\infty}f(x)dx$ for a continuous $f(x)$ it often helps to find a crude but effective upper bound for $|f(x)|$ for large $x$.
When $x>1$ we have $\sqrt {1-x+x^2}\; <\sqrt {1+2x+x^2}=x+1<2x .$
And when $x>2$ we have $x^4-x^2+1>x^4-x^2>x^4/2.$
So when $x>2$ we have $$0<\frac {\sqrt {1-x+x^2}}{1-x^2+x^4}<\frac {2x}{x^4/2}=\frac {4}{x^3}$$ so the integral converges.
We can also observe that when $x>0$ we have $\sqrt {1-x+x^2}=x(1+g(x))$ and $1-x^2+x^4=x^4(1+h(x))$ where $g(x)$ and $h(x)$ converge to $0$ as $x\to \infty.$
So for all sufficiently large $x$ we have $|1+g(x)|<3/2$ and $|1+h(x)|>1/2.$ Thus $$0<\left|\frac {x(1+g(x)}{x^4(1+h(x)}\right|<\frac {x(3/2)}{x^4(1/2)}=\frac {3}{x^3}$$ for all sufficiently large $x.$
|
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|
Reduction formula for $\int\frac{dx}{(x^2+a^2)^n}$ How can I use integration by parts to write $$\int\frac{dx}{(x^2+a^2)^n} = \frac{x}{2a^2(n-1)(x^2+a^2)^{n-1}}+\frac{2n-3}{2a^2(n-1)}\int\frac{dx}{(x^2+a^2)^{n-1}}?$$
I would try $$u=\frac{1}{(x^2+a^2)^n},du=\frac{-2nx}{(x^2+a^2)^{n+1}}dx; dv=dx,v=x.$$ Integration by parts implies $$\int\frac{dx}{(x^2+a^2)^n}=\frac{x}{(x^2+a^2)^n}+2n\int\frac{x^2}{(x^2+a^2)^{n+1}}\\=\frac{x}{(x^2+a^2)^n}+2n\int\frac{(x^2+a^2)-a^2}{(x^2+a^2)^{n+1}}\\=\frac{x}{(x^2+a^2)^n}+2n\int\frac{1}{(x^2+a^2)^{n}}-2na^2\int\frac{dx}{(x^2+a^2)^{n+1}}$$ but I don't think I'm doing this correctly since the power of the denominator $(x^2+a^2)$ is not decreasing.
|
Hint:
Use integration by parts for the computation of $\;\displaystyle\int\frac{\mathrm dx}{(x^2+a^2)^{\color{red}{n-1}}} $, setting
$$ \begin{cases}u=\dfrac{1}{(x^2+a^2)^{n-1}},\\[1ex]\mathrm dv=\mathrm dx,\end{cases}
\;\text{ whence}\quad\begin{cases}\mathrm du=\dfrac{-2(n-1)x\,\mathrm dx}{(x^2+a^2)^{n}},\\[1ex] v=x,\end{cases} $$
so that
\begin{align}
\int\frac{\mathrm dx}{(x^2+a^2)^{n-1}}&=\dfrac{x}{(x^2+a^2)^{n-1}}+2(n-1)\int \dfrac{x^2\,\mathrm dx}{(x^2+a^2)^{n}} \\
&=\dfrac{x}{(x^2+a^2)^{n-1}}+2(n-1)\int \dfrac{\mathrm dx}{(x^2+a^2)^{n-1}}-2(n-1)a^2\int \dfrac{\mathrm dx}{(x^2+a^2)^{n}}\\\text{and finally}\hspace4em\\[1ex]
2(n-1)a^2\int \dfrac{\mathrm dx}{(x^2+a^2)^{n}}&=\dfrac{x}{(x^2+a^2)^{n-1}}+(2n-3)\int \dfrac{\mathrm dx}{(x^2+a^2)^{n-1}}.
\end{align}
|
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|
Find the nth term of the given series The series is $$1^3 + (2^3 +3^3)+(4^3+5^3+6^3)+...$$
i.e., $1^3$, $(2^3 +3^3)$ , $(4^3+5^3+6^3)$ , $(7^3+8^3+9^3+10^3)$ , and so on are the $1^{st}$, $2^{nd}$, $3^{rd}$ and so on terms respectively. ($(4^3+5^3+6^3)$ is the third term). So we need to find the $n^{th}$ term.
So, the first element of every individual term makes a sequence $1, 2, 4, 7...$. $\therefore$ the first element of each term is $(1+\frac{n(n-1)}2)^3$ and this the last element is $(\frac{n(n-1)}2 + 1 + (n-1))^3$. But I don't understand what to do next.
|
Let me see if I understand you correctly for I am a little bit confused about your notation. I guess that by the first term of the sequence you mean $1^3$ and $(2^3 + 3^3)$ is the second term, $(4^3 + 5^3 + 6^3)$ is the third term and so on. If this is so, it seems correct that the firs summand of the $n$-th term is given by
$$(\frac{n(n-1)}{2}+1)^3 \ .$$
Since the $n$-th term contains $n$ summands, the last summand of the $n$-th term is given by
$$(\frac{n(n-1)}{2}+1 + (n-1))^3 = (\frac{n(n+1)}{2})^3 \ , $$
where we added $(n-1)$ because the first summand is to be raised $(n-1)$ times by $1$ to get the last summand of the $n$-th term (your formula is almost the same for it seems that you used the same logic and only forgot to add $+1$). This expression will serve as a checker after we express the general formula for the $n$-th term.
To describe the $n$-th term should be straightforward now for we know the first summand and the number of steps we have to make to get to the last summand for arbitrary $n \in \mathbb{N}$. The $n$-th term should be
$$ \sum_{i=0}^{n-1} (\frac{n(n-1)}{2}+1 +i)^3 \ . $$
The first summand is obviously in accordance with what we know and if we compute the last summand of the above expression we get
$$ (\frac{n(n-1)}{2}+1 + n-1)^3 = (\frac{n(n-1)}{2} + n)^3 = (\frac{n(n+1)}{2} )^3 \ ,$$
which is what we expected.
In case this answer is not what you were looking for, try to specify more carefully what you mean by $n$-th term.
|
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|
If $\sin^{-1} (x) + \sin^{-1} (y)+ \sin^{-1} (z)=\dfrac {\pi}{2}$, prove that $x^2+y^2+z^2+2xyz=1$
If $\sin^{-1} (x) + \sin^{-1} (y)+ \sin^{-1} (z)=\dfrac {\pi}{2}$, prove that:
$$x^2+y^2+z^2+2xyz=1$$
My Attempt:
$$\sin^{-1} (x) + \sin^{-1} (y) + \sin^{-1} (z)=\dfrac {\pi}{2}$$
$$\sin^{-1} (x\sqrt {1-y^2}+y\sqrt {1-x^2})+\sin^{-1} (z)=\dfrac {\pi}{2}$$
$$\sin^{-1} (x\sqrt {1-y^2} + y\sqrt {1-x^2})=\dfrac {\pi}{2} - \sin^{-1} (z)$$
|
Let: $\sin^{-1}x=\alpha$, $\sin^{-1}y=\beta$, $\sin^{-1}z=\gamma$. We know that $\alpha+\beta+\gamma={\pi\over2}$, that is: $\gamma={\pi\over2}-\alpha-\beta$ and $$\sin\gamma=\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.$$
We have then:
$$
\begin{align}
&x^2+y^2+z^2+2xyz=\\
&\sin^2\alpha+\sin^2\beta+\sin^2\gamma+2\sin\alpha\sin\beta\sin\gamma=\\
&\sin^2\alpha+\sin^2\beta+(\cos\alpha\cos\beta-\sin\alpha\sin\beta)^2
+2\sin\alpha\sin\beta(\cos\alpha\cos\beta-\sin\alpha\sin\beta)=\\
&\sin^2\alpha+\sin^2\beta+\cos^2\alpha\cos^2\beta-\sin^2\alpha\sin^2\beta=\\
&\sin^2\alpha(1-\sin^2\beta)+\sin^2\beta+\cos^2\alpha\cos^2\beta=\\
&\sin^2\alpha\cos^2\beta+\sin^2\beta+\cos^2\alpha\cos^2\beta=\\
&(\sin^2\alpha+\cos^2\alpha)\cos^2\beta+\sin^2\beta=\\
&\cos^2\beta+\sin^2\beta=1
\end{align}
$$
|
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|
Limit using Taylor series expansion
Can someone help me with this limit. I know I have to expand in Taylor series
$\sqrt{1+x}=1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3+o(x^3)$
$a^x=e^{x\ln a}=1+x\ln a+\frac{1}{2!}x^2(\ln a)^2+\frac{1}{3!}x^3(\ln a)^3+o(x^3)$.
But how to proceed next?
|
$$\lim_{ x \to 0} \frac{a^{\sqrt{1+x}}-a^{1+x/2-x^2/8}}{x^3}=\lim_{ x \to 0} \;\underbrace{a^{1+x/2-x^2/8}}_{\to a} \cdot\left(\frac{a^{\sqrt{1+x}-1-x/2+x^2/8}-1}{x^3}\right)$$
Using Taylor series $\displaystyle \sqrt{1+x} = 1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3+\mathcal{O}(x^3)$
;
$$\lim_{ x \to 0} \;\underbrace{a^{1+x/2-x^2/8}}_{\to a} \cdot\left(\frac{a^{\sqrt{1+x}-1-x/2+x^2/8}-1}{x^3}\right)=a \lim_{ x \to 0} \left(\frac{a^{x^3/16}-1}{x^3}\right)=\frac{a \ln a}{16}$$
|
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|
Let $f(x)=\big(|x+a|-|x+2|+b|x+5| \big)$ be an odd function then find $a+b$ Let $f(x)=\big(|x+a|-|x+2|+b|x+5| \big)$ be an odd function then find $a+b$.
My try :
$$\begin{align}f(-x)&=|-x+a|-|-x+2|+b|-x+5| \\ &=-f(x) \\&= -\big(|x+a|-|x+2|+b|x+5| \big) \end{align}$$
$$|-x+a|-|-x+2|+b|-x+5|+ \big(|x+a|-|x+2|+b|x+5|) =0$$
Now what do I do?
|
You can equate the slopes for small and large $x$ and
$$-1+1-b=1-1+b$$ so that $b=0$.
Then
$$f(x)=|x+a|-|x+2|$$ odd requires $|a|=2$ to ensure that $f(0)=0$.
By direct check, $f(x)=|x-2|-|x+2|$ or $f(x)=|x+2|-|x+2|=0$ are indeed odd.
|
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|
I need to find the general nth degree polynomial expansion of $x(x+1)(x+2)\dots(x+n)$ in order to solve an integral. Expanding $x(x+1)(x+2)\dots(x+n)$ has been challenging for me.
Say $n=5$, you will notice many patterns like $x^n+(n-1)x^{n-1}+\dotsb$ et cetera, but does there exist a general polynomial expansion for any $n$? This has been driving me nuts. I need a whiteboard.
The integral is $$\frac{1}{x(x+1)(x+2)\dots(x+n)}$$ and will be easy once I expand the polynomial. I am trying to avoid partial fraction decomposition.
|
$f(x) = \frac{1}{(x-a)(x-b)\cdots(x-k)} = \frac {A}{x-a} + \frac {B}{x-b} \cdots \frac {K}{(x-k)}$
$A = \lim_\limits {x\to a} (x-a)f(x) = \frac {1}{(a-b)(a-c)\cdots(a-k)}$
$f(x) = {1}{(x)(x+1)\cdots(x+n)} = \frac 1{n!}\frac {1}{x} - \frac {1}{(n-1)!}\frac {1}{x+1} + \frac {1}{2(n-2)!}\frac {1}{x+2} \cdots\\
f(x) = \sum_{i=0}^n\frac {(-1)^n}{i!(n-i)!} \frac{1}{x+i}$
By the way, supposing you had a nice way to expand the polynomial in the denominator, then what? It still does not integrate without the partial fraction decomposition.
|
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|
Prove that $\sqrt{x^2+yz}+\sqrt{y^2+xz}+\sqrt{z^2+xy} \le 3$ Prove that
$$\sqrt{x^2+yz}+\sqrt{y^2+xz}+\sqrt{z^2+xy} \le 3$$
for $x \ge 0$, $y \ge 0$, $z \ge 0$ and $x+y+z \le 2$.
My work:
\begin{align*}
&\mathrel{\phantom{=}} \sqrt{x^2+yz}+\sqrt{y^2+xz}+\sqrt{z^2+xy}\\
&\le\sqrt3 \sqrt{x^2+yz+y^2+xz+z^2+xy}\\
&\le\sqrt3\sqrt{2x^2+2y^2+2z^2}=\sqrt6\sqrt{x^2+y^2+z^2}.
\end{align*}
|
By C-S we have:
$$\left(\sum_{cyc}\sqrt{x^2+yz}\right)^2\leq\sum_{cyc}\frac{x^2+yz}{2x^2+y^2+z^2+yz}\sum_{cyc}(2x^2+y^2+z^2+yz).$$
Id est, it remains to prove that
$$\sum_{cyc}\frac{x^2+yz}{2x^2+y^2+z^2+yz}\sum_{cyc}(4x^2+yz)\leq\frac{9}{4}(x+y+z)^2,$$ which is
$$\sum_{sym}(x^8+2x^7y+9x^6y^2-23x^5y^3+10x^4y^4)+$$
$$+\sum_{sym}(5.5x^6yz+70x^5y^2z+79x^4y^3z+131.5x^4y^2z^2-97.5x^3y^3z^2)\geq0,$$
which is obvious after using following Schur:
$$2\sum_{cyc}(x^8-x^7y-x^7z+x^6yz)\geq0.$$
|
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|
Solve the equation $ { \sqrt{4+\sqrt{4+\sqrt{4-x}}}}=x $ Prove that one root of the Equation
$${ \sqrt{4+\sqrt{4+\sqrt{4-x}}}}=x $$
is
$$x=2\left(\cos\frac{4\pi}{19}+\cos\frac{6\pi}{19}+\cos\frac{10\pi}{19}\right) $$
My progress:
After simplyfying the given Equation I got
$$ x^8-16x^6+88x^4-192x^2+x+140=0$$
Then I tried to factorise it, got this
$ x^8-16x^6+88x^4-192x^2+x+140=(x^3+x^2-6x-7)(x^5-x^4-9x^3+10x^2+17x-20)$
$ x^8-16x^6+88x^4-192x^2+x+140=(x^3+x^2-6x-7)(x^3-2x^2-3x+5)(x^2+x-4)$
Should I now solve cubic?
|
$\pm 2,\pm 3,\pm 5$ are precisely twice the cubic residues $\!\pmod{19}$. Let $\omega=\exp\left(\frac{2\pi i}{19}\right)$. $\omega$ is al algebraic number over $\mathbb{Q}$ with degree $18$, hence by Galois theory both
$$ \sum_{k\in\{\pm 1,\pm 7,\pm 8\}}\omega^k,\qquad \sum_{k\in\{\pm 2,\pm 3,\pm 5\}}\omega^k $$
(related to the Kummer sums $\!\!\pmod{19}$) are algebraic numbers over $\mathbb{Q}$ with degree $3$. Indeed, the minimal polynomial of the second sum
is $z^3+z^2-6z-7$, and the second sum is exactly $2\left(\cos\frac{4\pi}{19}+\cos\frac{6\pi}{19}+\cos\frac{10\pi}{19}\right)$.
|
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|
Solving logarithmic equation with different bases $\log_2\left(x-5\right)=\log_5\left(2x+7\right)$ I have to solve this but my answer vary from x = 7, x = -9, x = 5/3 and I don't know why this happens.
Here is my approach:
$\frac{\log_5\left(x-5\right)}{\log_5\left(2\right)}=\log_5\left(2x+7\right)$
$\frac{\log_5\left(x-5\right)}{\log_5\left(2x+7\right)}=\log_5\left(2\right)$
$\log_{2x+7}\left(x-5\right)=\log_5\left(2\right)$
Conditions: x > 5
So we are left with:
x - 5 = 2 => x = 7 (verifies the conditions)
2x + 7 = 5 => x = -1 (does not verify the conditions)
Therefore, x = 7.
But when you plug in x = 7 => $\log_2\left(2\right)=\log_5\left(21\right)$ which are not equal.
I guessed a solution that works which is x = 9.
trying to solve that system, i found another way to solve it
x - 5 = 2 | *(-1)
2x + 7 = 5
=> - x + 5 = -2
2x + 7 = 5
We add them up and we have:
x + 12 = 3 => x = -9 which does not verify the condition.
What am I doing wrong?
|
$\log_2(x-5) = \log_5(2x+7)=k$
$2^k = x-5$ so $x = 2^k + 5$. $5^k =2x - 7$ so $x = \frac {5^k -7}2$
Or $2*2^k +10 = 5^k - 7$
$2^{k+1} = 5^k-17$
[cute that $2^{k+1} -8 = 5^k -25$... but, I'll assume I wasn't clever enough to see that.]
At this point we can really only guess but clearly if we increase $k$ by small incrementss the RHS will increase more than the LHS, so there is only one solution and we can hone in on values. If the RHS is larger than the LHS we take smaller values of $k$ and vice versa.
$2^{1+1} > 5^1 - 17$ so $k > 1$
And $2^{3 + 1} < 5^3 - 17$ so $k < 3$
It's simple enough to discover $k = 2$ is a solution..
(Even simpler if we noticed $2^{k+1} - 8 = 5^k - 25$ so $2^3 - 8 = 5^2 - 25 = 0$)
So $x = 2^2 + 5 = \frac{5^2 - 7}2 = 9$.
|
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|
Find all the solutions $\in \Bbb R^+$ that satisfy $x(6-y)=9$, $y(6-z)=9$, $z(6-x)=9$ Find all the solutions $\in \Bbb R^+$ that satisfy $x(6-y)=9$, $y(6-z)=9$, $z(6-x)=9$
My try
I found that $0 \lt x,y,z \lt 6$
Multiplying the equations we got $x(6-y)y(6-z)z(6-x)=9^3$
$x(6-x)y(6-y)z(6-x)=9^3$
And here is the problem, i applied AM-GM inequality for $(x \;, \;6-x)$
$$\Biggl (\frac{(x+(6-x)}{2}\Biggr) \ge \sqrt {x(x-6)}$$
Expanding out we get $$(x-3)^2\ge0$$
Holding the equality when $x=3$
We can do the same with $(y \;, \; 6-y)$ and $(z \;, \; 6-z)$ getting $(y-3)^2\ge0$ and $(z-3)^2\ge0$ holding when $y,z =3$ and getting that one solution for the system is $x=y=z=3$ but i don't know if this is enough for proving that those are the only solutions.
|
Repeated substitution gives
\begin{eqnarray*}
x \left( 6- \frac{9}{6- \frac{9}{6-x}} \right) =9
\end{eqnarray*}
and this simplifies to $(x-3)^2=0$.
|
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|
quadrilateral inside a parallelogram
the diagonals of the parallelogram creates four equal trianges, so △ ABD and △ BOC together becomes 3/4 s. But stuck here, how do I find area of △DEM and △MFC, so as to solve to find out the area of MEOF??
|
Let $DM=xDC$.
Thus, $$S_{\Delta ADM}=xS_{\Delta ADC}=\frac{1}{2}xS$$ and
$$\frac{S_{\Delta ADE}}{\frac{1}{2}xS}=\frac{AE}{AM}=\frac{AE}{AE+EM}=\frac{1}{1+\frac{EM}{AE}}=\frac{1}{1+\frac{DM}{AB}}=\frac{1}{1+x}.$$
Hence, $$S_{\Delta ADE}=\frac{\frac{1}{2}xS}{1+x}.$$
In another hand,
$$S_{\Delta BCM}=(1-x)S_{\Delta BDC}=\frac{1}{2}(1-x)S$$ and
$$\frac{S_{\Delta BCF}}{\frac{1}{2}(1-x)S}=\frac{BF}{BM}=\frac{BF}{BF+FM}=\frac{1}{1+\frac{FM}{BF}}=\frac{1}{1+\frac{CM}{AB}}=\frac{1}{2-x}.$$
Hence, $$S_{\Delta ADE}=\frac{\frac{1}{2}(1-x)S}{2-x}.$$
Id est,
$$\frac{\frac{1}{2}xS}{1+x}+\frac{\frac{1}{2}(1-x)S}{2-x}=\frac{1}{3}S,$$
which gives $$x=\frac{1}{2},$$
$$S_{\Delta DEM}=S_{\Delta MFC}=\frac{1}{12}S$$ and
$$S_{EMFO}=\frac{1}{4}S-2\cdot\frac{1}{12}S=\frac{1}{12}S.$$
|
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|
$(3x- 3)\prod_{n = 1}^∞\frac{\exp(2^{-n})+1}{x^{2^{-n}}+1}=(x^3-1)\prod_{n = 1}^∞\frac{\exp(2^{-n})+1}{x^{3 \cdot 2^{-n}}+1}$ Show that
$$ (3x - 3) \prod_{n = 1}^{\infty} \frac{ \exp(2^{-n}) +1} {x^{2^{-n}} + 1}
= (x^3 - 1) \prod_{n = 1}^{\infty} \frac{ \exp(2^{-n}) +1} {x^{3\cdot2^{-n}} + 1} $$
|
Hint: $\;\;
\left(x^{2^{-n}}-1\right) \cdot \left(x^{2^{-n}}+1\right)\left(x^{2^{-(n-1)}}+1\right)\ldots\left(x^{2^{-1}}+1\right) = x - 1
\,$ by telescoping.
[ EDIT ] It then follows that the $\,n^{th}\,$ partial product on the LHS is:
$$\require{cancel}
3\cancel{(x - 1)} \cdot \frac{\left(x^{2^{-n}} - 1\right)}{\left(e^{2^{-n}} - 1\right)} \cdot \frac{\left(e^{2^{-n}} - 1\right) \prod_{k = 1}^{n} \left(e^{2^{-k}} +1\right)}{\cancel{\left(x^{2^{-n}} - 1\right)\prod_{k = 1}^{n} \left(x^{2^{-k}} +1\right)}} = 3(e-1) \cdot \frac{x^{2^{-n}} - 1}{e^{2^{-n}} - 1}
$$
For $\,n \to \infty\,$ $\,\,2^{-n} \to 0\,$, and the LHS therefore reduces to:
$$
\lim_{a \to 0} \,3(e-1) \cdot \frac{x^a - 1}{e^{a} - 1} = \ldots
$$
|
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|
If $2\tan^2x - 5\sec x = 1$ has exactly $7$ distinct solutions for $x\in[0,\frac{n\pi}{2}]$, $n\in N$, then the greatest value of $n$ is? If $2\tan^2x - 5\sec x = 1$ has exactly $7$ distinct solutions for $x\in[0,\frac{n\pi}{2}]$, $n\in N$, then the greatest value of $n$ is?
My attempt:
Solving the above quadratic equation, we get $\cos x = \frac{1}{3}$
The general solution of the equation is given by $\cos x = 2n\pi \pm \cos^{-1}\frac{1}{3}$
For having $7$ distinct solutions, $n$ can have value = 0,1,2,3
So, from here we can conclude that $n$ is anything but greater than $6$. So, according to the options given in the questions, the greatest value of $n$ should be $13$. But the answer given is $14$. Can anyone justify?
|
I agree with your answer, indeed note that
$$2\tan^2x - 5\sec x = 1\iff2(1-\cos^2x)-5\cos x=\cos^2x\\\iff3\cos^2x+5\cos x-2=0$$
and
$$3t^2+5t-2=0\implies t=\frac{-5\pm\sqrt{25+24}}{6}\implies t=\frac{-5+\sqrt{47}}{6}= \frac13$$
thus we have 2 solution on the interval $[0,2\pi]$ and notably one in the interval $[0,\pi/2]$ and the other in $[3\pi/2,2\pi]$.
Therefore, since the function is periodic with period $2\pi$ we have that $n=13$.
Note that for $x\in\left[0,\frac{13\pi}{2}\right]$ the expression is not defined when $x=\frac{\pi}2+k\pi$ thus this points should be excluded by the solution.
|
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|
Area of a triangle using a matrix Find the area of a triangle whose vertices are (1,0), (2,2), and (4,3)
$$\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=\begin{vmatrix}1&0&1\\2&2&1\\4&3&1\end{vmatrix}$$
$$=1(-1)^2\begin{vmatrix}2&1\\3&1\end{vmatrix}+0(-1)^3\begin{vmatrix}2&1\\4&1\end{vmatrix}+1(-1)^4\begin{vmatrix}2&2\\4&3\end{vmatrix}$$
$$=1(-1)+0+1(-2)$$
$$=-3$$
Then you multiply by $-\frac{1}{2}$ to get a positive area of $\frac{3}{2}$.
My question is where does $-1$ come from in the second line of the equation and why is it being squared first, then cubed, then raised to a power of 4?
|
It is from Laplace expansion for determinant.
As an alternative to calculate the area or to check the result you can also use
$$A=\frac12 |\det(u,v)|$$
with, for example, $u=(2,2)-(1,0)=(1,2)$ and $v=(4,3)-(1,0)=(3,3)$ that is
$$A=\frac12\begin{vmatrix}1&2\\3&3\end{vmatrix}=\frac32$$
|
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|
Using DeMoivre's Theorem to prove some identities regarding trigonometric functions The following question is from a previous post. The reason that I am posting this again is because it had two questions which were not really related and hence, one of the questions was not answered. The question is from a book, "Mathematical Analysis - 2nd Edition" by Apostol. The question has three parts and is as follows:
*
*By equating imaginary parts in DeMoivre's formula, prove that
$$\sin{\left(n\theta\right)} = \sin^n\theta \left\lbrace \binom{n}{1}\cot^{n - 1}\theta - \binom{n}{3} \cot^{n - 3}\theta + \binom{n}{5} \cot^{n - 5}\theta - + \cdots \right\rbrace$$
*If $0 < \theta < \dfrac{\pi}{2}$, prove that
$$\sin{\left(\left( 2m + 1 \right)\theta\right)} = \sin^{2m + 1}\theta P_m(\cot^2\theta)$$
where $P_m$ is the polynomial of degree $m$ given by
$$P_m\left( x \right) = \binom{2m + 1}{1}x^m - \binom{2m + 1}{3}x^{m-1} + \binom{2m + 1}{5}x^{m-2} - + \cdots$$
Use this to show that $P_m$ has zeros at $m$ distinct points $x_k = \cot^2\left( \dfrac{\pi k}{2m + 1} \right)$ for $k = 1, 2, \dots, m$.
*Show that the sum of zeros of $P_m$ is given by
$$\sum\limits_{k=1}^{m} \cot^2\dfrac{\pi k}{2m + 1} = \dfrac{m \left( 2m - 1 \right)}{3}$$
and that the sum of theie squares is given by
$$\sum\limits_{k=1}^{m} \cot^4\dfrac{\pi k}{2m + 1} = \dfrac{m \left( 2m - 1 \right) \left( 4m^2 + 10m - 9 \right)}{45}$$
As far as the solution is concerned, I have been able to prove the first part and upto proving the existence of the polynomial in the second part. For later parts, I do not have any insights in proceeding towards the solution.
Help will be appreciated!
|
Hints
For part 2, since you already showed the existence of the polynomial s.t. $$\sin \left( 2m + 1 \right)\theta = \left(\sin^{2m + 1}\theta \right)P_m(\cot^2\theta)$$
replace $\theta$ with $\frac{k\pi}{2m+1}$ and
$$\left(\sin^{2m + 1}\frac{k\pi}{2m+1} \right)\color{red}{P_m\left(\cot^2\frac{k\pi}{2m+1}\right)}=\sin \left(\left( 2m + 1 \right)\frac{k\pi}{2m+1}\right)=\sin{k\pi}=\color{red}{0}$$
For part 3, have you tried using Vieta's formulas? It is a simple application of
$$\color{red}{\sum\limits_{k=1}^{m}x_k}=\sum\limits_{k=1}^{m}\cot^2\frac{\pi k}{2m+1}=\color{red}{-\frac{a_{m-1}}{a_m}}=-\frac{-\binom{2m+1}{3}}{\binom{2m+1}{1}}=\\
\frac{(2m+1)2m(2m-1)}{1\cdot2\cdot3}\cdot \frac{1}{2m+1}=\frac{m(2m-1)}{3}$$
For the 2nd part of $3$ use the fact that
$$\sum\limits_{k=1}^{m}x_k^2=\left(\sum\limits_{k=1}^{m}x_k\right)^2-2\sum\limits_{k<t}x_kx_t=\left(-\frac{a_{m-1}}{a_m}\right)^2-2\left(\frac{a_{m-2}}{a_m}\right)$$
|
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|
For $x, y \in \mathbb{R}$, prove that $\max(x, y) = \frac{x + y + |x - y|}{2},$ and $\min(x, y) = \frac{x + y - |x - y|}{2}$. Prove that for all real numbers $x$ and $y$,
$$\max(x, y) = \dfrac{x + y + |x - y|}{2},$$
and
$$\min(x, y) = \dfrac{x + y - |x - y|}{2}.$$
For any real number $x$, the absolute value of $x$, denoted $|x|$ is defined as follows:
\begin{equation}
|x| = \begin{cases} x; & \text{ if } x \geq 0 \\
-x; & \text{ if } x< 0
\end{cases}
\end{equation}
What I understand from this is that $|x| = x$ if $x \geq 0$ or $|x| = −x$ if $x<0$. Other than that I don't really know how to start this.
|
If $\displaystyle x>y$ then $x-y>0$:
$|x-y|=x-y$
$\max(x,y) = (x+y+x−y)/2 = x$
$\min(x,y) = (x+y-(x−y))/2 = y$
If $\displaystyle x<y$ then $x-y<0$:
$|x-y|=y-x$
$\max(x,y) = (x+y+y-x)/2 = y$
$\min(x,y) = (x+y-(y-x))/2 = x$
If $x=y$ then:
$|x-y|=0$
$\max(x,y) = (x+y)/2 = x=y$
$\min(x,y) = (x+y)/2 =x=y$
|
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|
Computing $\int_{0}^{+\infty}\arctan\left(\frac{1}{x^n}\right)\,\mathrm{d}x$ Let $n \in (1,+\infty)$, I've shown that the function
$$
I(n)=\int_{0}^{+\infty}\arctan\left(\frac{1}{x^n}\right)\,\mathrm{d}x
$$
is defined on $(1,+\infty)$ and that it can be expressed by
$$
I(n)=\frac{1}{n}\int_{0}^{+\infty}\frac{\arctan\left(x\right)}{x^{(n+1)/n}}\,\mathrm{d}x
$$
I wanted to know if there was a way to compute this kind of integral. I know that
$$
I\left(2\right)=\frac{\pi}{\sqrt{2}}, \quad I\left(3\right)=\frac{\pi}{\sqrt{3}}.
$$
Despite that I did not manage to prove it, it was starting well to find a beautiful formula however it becomes difficult to compute for odd integer value of $n$ and gives strange value for even integer value.
Any help?
|
Here is yet another approach that uses Feynman's trick of differentiating under the integral sign.
We begin by enforcing a substitution of $x \mapsto x^{1/n}$. This gives
$$\int_0^\infty \tan^{-1} \left (\frac{1}{x^n} \right ) \, dx = \frac{1}{n} \int_0^\infty \frac{\tan^{-1} x}{x^{1 + \frac{1}{n}}} \, dx, \quad n > 1.$$
Now let
$$I(a) = \frac{1}{n} \int_0^\infty \frac{\tan^{-1} (ax)}{x^{1 + \frac{1}{n}}} \, dx, \quad a > 0.$$
Note that $I(0) = 0$ and we are required to find the value of $I(1)$.
Differentiating with respect to the parameter $a$ yields
$$I'(a) = \frac{1}{n} \int_0^\infty \frac{x^{-1/n}}{1 + a^2 x^2} \, dx,$$
or
$$I'(a) = \frac{a^{\frac{1}{n} - 1}}{2n} \int_0^\infty \frac{t^{-\frac{1}{2n}-\frac{1}{2}}}{1 + t} \, dt,$$
after a substitution of $x \mapsto \sqrt{t}/a$ has been enforced. The integral appearing above can be expressed in terms of Euler's Beta function $\text{B}(x,y)$. Here
\begin{align*}
I'(a) &= \frac{a^{\frac{1}{n} - 1}}{2n} \int_0^\infty \frac{t^{(\frac{1}{2} - \frac{1}{2n}) - 1}}{(1 + t)^{(\frac{1}{2}-\frac{1}{2n}) + (\frac{1}{2} + \frac{1}{2n})}} \, dt\\
&= \frac{a^{\frac{1}{n}-1}}{2n} \text{B} \left (\frac{1}{2} - \frac{1}{2n}, \frac{1}{2} + \frac{1}{2n} \right )\\
&= \frac{a^{\frac{1}{n} - 1}}{2n} \Gamma \left (\frac{1}{2} - \frac{1}{2n} \right ) \Gamma \left (\frac{1}{2} + \frac{1}{2n} \right )\\
&= \frac{a^{\frac{1}{n} - 1}}{2n} \Gamma \left [1 - \left (\frac{1}{2} + \frac{1}{2n} \right ) \right ] \Gamma \left (\frac{1}{2} + \frac{1}{2n} \right )\\
&= \frac{a^{\frac{1}{n} - 1}}{2n} \frac{\pi}{\sin \pi \left (\frac{1}{2} + \frac{1}{2n} \right )} \tag1\\
&= \frac{\pi}{2n} \sec \left (\frac{\pi}{2n} \right ) a^{\frac{1}{n} - 1},
\end{align*}
where in (1) Euler's reflection formula has been used.
So on integrating up with respect to the parameter $a$ we have
\begin{align*}
I(1) &= \int_0^1 I'(a) \, da = \frac{\pi}{2n} \sec \left (\frac{\pi}{2n} \right ) \int_0^1 a^{\frac{1}{n} - 1} \, da = \frac{\pi}{2} \sec \left (\frac{\pi}{2n} \right ) \Big{[} a^{\frac{1}{n}} \Big{]}_0^1
\end{align*}
giving
$$\int_0^\infty \tan^{-1} \left (\frac{1}{x^n} \right ) \, dx = \frac{\pi}{2} \sec \left (\frac{\pi}{2n} \right ),$$
as expected.
|
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|
Prove that $n^2 - 2n + 7$ is even then $n + 1$ is even I have to use Proof by contradiction to show what if $n^2 - 2n + 7$ is even then $n + 1$ is even.
Assume $n^2 - 2n + 7$ is even then $n + 1$ is odd. By definition of odd integers, we have $n = 2k+1$.
What I have done so far:
\begin{align}
& n + 1 = (2k+1)^2 - 2(2k+1) + 7 \\
\implies & (2k+1) = (4k^2+4k+1) - 4k-2+7-1 \\
\implies & 2k+1 = 4k^2+1-2+7-1 \\
\implies & 2k = 4k^2 + 4 \\
\implies & 2(2k^2-k+2)
\end{align}
Now, this is even but I wanted to prove that this is odd(the contradiction). Can you some help me figure out my mistake?
Thank you.
|
As pointed out bt @Bram28 and others you use contradiction by making the appropriate changes.
However, I suggest that you try to use contrapositive instead of contradiction, whenever possible as it makes fewer assumptions.
The easiest (and cleanest) way to solve this is proving the contrapositive instead i.e. if $n+1$ is odd, then $n^2-2n+7$ is odd.
Since $n+1$ is odd, so $n$ is even and we can write $n=2k$ for some interger $k$.
Note that $n^2-2n+7=4k^2-4k+6+1=2(2k^2-2k+3)+1=2m+1$, where $m=2k^2-2k+3$ is an integer, proving the result.
|
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Variable inequality: Show that $ {(\sum\limits_{i = 1}^{2n + 1} {{a_i}} )^2} \geqslant 4n\sum\limits_{i = 1}^{n + 1} {{a_i}{a_{i + n}}}.$ For any positive integer $n$, and real numbers (not necessarily positive) $a_1\geqslant a_2 \geqslant …\geqslant a_{2n+1}$, show that $$
{(\sum\limits_{i = 1}^{2n + 1} {{a_i}} )^2} \geqslant 4n\sum\limits_{i = 1}^{n + 1} {{a_i}{a_{i + n}}}.$$
What I've tried: I set $x_i :=a_i - a_{i+1}$ for $i\leqslant 2n$ and $x_{2n+1}:=a_{2n+1}$, and calculate the coefficients on both sides, but gradually find it difficult to go further, perhaps it just can't.
Please help.
Something more: if all $a_i=1$ except $a_{2n+1}=0$, the equality holds.
|
I'm close to a solution,
but I can't go all the way,
so I'll show what I've got
in the hope that
someone else
can complete the proof.
Let
$a_i = a-b_i$,
where
$a = a_1$ and
$b_1 = 0$
so $b_i \ge 0$
and
$b_i \le b_{i+1}$.
The inequality becomes
${(\sum\limits_{i = 1}^{2n + 1} {(a-b_i)} )^2}
\ge 4n\sum\limits_{i = 1}^{n + 1} {(a-b_i)(a-b_{i + n})}
$.
The left side is,
if
$B = \sum\limits_{i = 1}^{2n + 1} b_i$,
$\begin{array}\\
(\sum\limits_{i = 1}^{2n + 1} {(a-b_i)} )^2
&=((2n+1)a-\sum\limits_{i = 1}^{2n + 1} b_i )^2\\
&=((2n+1)a-B )^2\\
&=(2n+1)^2a^2-2(2n+1)aB+B^2\\
\end{array}
$
The right side is
$\begin{array}\\
4n\sum\limits_{i = 1}^{n + 1} {(a-b_i)(a-b_{i + n})}
&=4n\sum\limits_{i = 1}^{n + 1} (a^2-a(b_i+b_{i+n})+b_ib_{i + n})\\
&=4n((n+1)a^2-\sum\limits_{i = 1}^{n + 1}a(b_i+b_{i+n})+\sum\limits_{i = 1}^{n + 1}b_ib_{i + n})\\
&=4n(n+1)a^2-4na\sum\limits_{i = 1}^{n + 1}(b_i+b_{i+n})+4n\sum\limits_{i = 1}^{n + 1}b_ib_{i + n}\\
&=4n(n+1)a^2-4na(B+b_{n+1})+4n\sum\limits_{i = 1}^{n + 1}b_ib_{i + n}\\
&=4n(n+1)a^2-4na(B+b_{n+1})+4nS
\qquad\text{where } S=\sum\limits_{i = 1}^{n + 1}b_ib_{i + n}\\
\end{array}
$
The left-right is thus
$((2n+1)^2a^2-2(2n+1)aB+B^2)-
(4n(n+1)a^2-4na(B+b_{n+1})+4nS)\\
\quad=((2n+1)^2-4n(n+1))a^2-(2(2n+1)-4n)aB+B^2+4nab_{n+1}-4nS\\
\quad=a^2-2aB+B^2+4nab_{n+1}-4nS\\
\quad=(a-B)^2+4nab_{n+1}-4nS\\
\quad=(a-B)^2+4n(ab_{n+1}-S)\\
$
So if we can show that
$(a-B)^2+4n(ab_{n+1}-S)
\ge 0$,
or, equivalently,
$a^2-2aB+B^2+4nab_{n+1}-4nS
\ge 0$,
we are done.
At this point,
I'm stuck.
I think that
we somehow need to use
$b_i \le b_{i+1}$
to bound $S$ in relation
to $B$,
but I don't see how.
|
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|
A committee of 4 is to be formed out of 6 superheroes, 7 supervillains, and 4 citizens. How many ways to form a committee? I am, once again, working on a problem which I have a disagreement with the given solution and I was wondering which solution is correct.
A committee of four people is to be formed from a set of 6 superheroes, 7 supervillains,
and 4 ordinary citizens (17 people in total). How many ways are there to form the committee if:
$\textbf{(d)}$ At least one superhero and one supervillain must be on the committee, but Mr. Fantastic (a superhero)
and Dr. Doom (a supervillain) cannot be on the committee together.
$\textbf{Given Solution:}$
$\binom{6}{1}\binom{7}{1}\binom{15}{2}-\binom{15}{2}$
(Here we count the committees with at least one hero and at least one supervillain,
then subtract the committees with Mr Fantastic and Dr Doom on them).
QED
I feel like the solution above is problematic in this way: The term $\binom{6}{1}\binom{7}{1}\binom{15}{2}$ calculates several cases twice. To illustrate this, let us separate the 3 binomial expressions and show that the term overcounts at least one case more than once.
Case 1: in $\binom{6}{1}$ we choose Mr. Fantastic, in $\binom{7}{1}$ we choose Dr. Doom and in $\binom{15}{2}$, we choose Spongebob and Patrick, then in our Comittee we have {Mr. Fantastic, Dr. Doom, Spongebob, Patrick}.
Case 2: Let's say Spongebob is a superhero and Patrick is a supervillain, then in $\binom{6}{1}, \binom{1}{7}, \binom{15}{2}$, we choose Spongebob, Patrick, Mr. Fantastic and Dr. Doom respectively. This forms the committee {Spongebob, Patrick, Mr. Fantastic, Dr. Doom}, which is the same committee as previously but is counted twice.
$\textbf{My Solution:}$
Consider the case where both Dr. Doom and Mr. Fantastic are both in the committee, we have $1*1*\binom{15}{2}$.
Case 1: Consider the case where there are neither superheroes nor supervillains. $\binom{4}{4}$.
Case 2: Consider the case where there are no superheroes, $\binom{11}{4}$.
Case 3: Consider the case where there are no supervillains, $\binom{10}{4}$.
Our final answer is given by $\binom{17}{4} - \binom{15}{2} - \binom{11}{4} - \binom{10}{4} + \binom{4}{4}$
Note that Case 1 is included in both Case 3 and Case 2, so we add $\binom{4}{4}$ to compensate for the exclusion of Case 1 twice (by the exclusion of case 3 and case 2)
QED
|
Your solution looks good.
You can cross-check against the less-elegant method of assembling valid committees and then excluding the $\binom{15}{2}$ of the Fantastic/Doom constraint.
$\binom{6}{1}\binom{7}{1}\binom{4}{2} + \binom{6}{2}\binom{7}{1}\binom{4}{1} + \binom{6}{1}\binom{7}{2}\binom{4}{1} + \binom{6}{3}\binom{7}{1} + \binom{6}{2}\binom{7}{2} + \binom{6}{1}\binom{7}{3} - \binom{15}{2} \\
= 252 + 420 + 504 + 140 + 315 + 210 - 105 = 1736$
|
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|
Binomial Harmonic Numbers Prove this equation for $0 \leq m \leq n$:
$$
\frac{1}{\binom{n}{m}}\sum_{k=1}^m \binom{n-k}{n-m} \frac{1}{k} = H_n - H_{n-m}
$$
where $H_k$ denotes the k-th harmonic number $\left(~H_k := \sum_{n=1}^k \frac{1}{n}~\right)$.
Tried to use Abels partial summation $\big(\sum_{k=1}^m a_k b_k = a_m \sum_{k=1}^m - \sum_{k=1}^{m-1} (a_{k+1}-a_k)\sum_{i=1}^k b_i \big)$, but it leads to nowhere.
|
Prove by Induction:
Base Case: n=0
0 = 0 $\checkmark$
Induction Step:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\begin{eqnarray*}\frac{1}{\binom{n+1}{m}} \sum_{k=1}^m \binom{n+1-k}{n+1-m} \frac{1}{k} &=& \frac{1}{\binom{n+1}{m}} \sum_{k=1}^m \frac{n+1-k}{n+1-m} \binom{n-k}{n-m} \frac{1}{k} \\
&=& \frac{1}{\binom{n+1}{m}} \frac{1}{n+1-m} \sum_{k=1}^m (n+1-k) \binom{n-k}{n-m} \frac{1}{k} \\
&=& \frac{1}{\binom{n+1}{m}} \frac{1}{n+1-m} \Big[ \sum_{k=1}^m (n+1) \binom{n-k}{n-m} \frac{1}{k} -\sum_{k=1}^m \binom{n-k}{n-m} \Big] \\
&=& \frac{1}{\binom{n+1}{m}} \frac{n+1}{n+1-m} \sum_{k=1}^m \binom{n-k}{n-m} \frac{1}{k} - \frac{1}{\binom{n+1}{m}} \frac{1}{n+1-m} \sum_{k=1}^m \binom{n-k}{n-m} \\
&=& H_n - H_{n-m} - \frac{1}{\binom{n+1}{m}} \frac{1}{n+1-m} \frac{n \binom{n-1}{n-m}}{n+1-m} \\
&=& H_n - H_{n-m} - \frac{m}{(n+1)(n+1-m)} \\
&=& H_n - H_{n-m} + \frac{-m+(n+1)-(n+1)}{(n+1)(n+1-m)} \\
&=& H_n - H_{n-m} + \frac{1}{n+1}-\frac{1}{n+1-m} \\
&=& \bbx{H_{n+1} - H_{n+1-m}}
\end{eqnarray*}$
|
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|
Third Moment of Geometric Series? $\sum_{x=1}^{\infty}x(1-p)^{x-1} = \frac{1}{p^2}$
$\sum_{x=1}^{\infty}x^2(1-p)^{x-1} = \frac{2-p}{p^3}$
$\sum_{x=1}^{\infty}x^3(1-p)^{x-1} =$ ... ?
The textbook gives the first 2 moments, but not the third one. I could not find the forumla on Google as well. I think it'll be useful in case I need to find $E(X^3)$ where $X$ is a geometric distribution.
|
Using summation of series.
$$S=\sum_{k=1}^\infty x^3(1-p)^{k-1}\\
-(1-p)S=-\sum_{k=1}^\infty (x-1)^3(1-p)^{k-1}$$
We add these two and get
$$pS=\sum_{k=1}^\infty (x^3-(x-1)^3)(1-p)^{1-k} \\ =\sum_{k=1}^\infty (x^2+x(x-1)+(x-1)^2)(1-p)^{1-k} \\ =\sum_{k=1}^\infty x^2(1-p)^k+\sum_{k=1}^\infty (x-1)^2(1-p)^{1-k}+\sum_{k=1}^\infty x(x-1)(1-p)^{1-k}\\
=E(X^2)+(1-p)E(X^2) + K$$
$$K=\sum_{k=1}^\infty x(x-1)(1-p)^{1-k}=\sum_{k=1}^\infty x(x+1)(1-p)^{k} \\
(1-p)K=\sum_{k=1}^\infty x(x-1)(1-p)^k \\ K-(1-p)K=pK=\sum_{k=1}^\infty (x(x+1)-x(x-1))(1-p)^{k} \\ pK=\sum_{k=1}^\infty 2x(1-p)^{k}=2(1-p)E(X)$$
Therefore $$pS=E(X^2)+(1-p)E(X)^2+\frac{2(1-p)}{p}E(X) \\ = \frac{(2-p)^2}{p^3}+\frac{2(1-p)}{p^3}=\frac{p^2-6p+6}{p^3}$$
$$S=\frac{p^2-6p+6}{p^4}$$
|
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|
Prove that $x^4 + y^4 - 3xy = 2$ is compact The exercise consists of showing that the function $f(x,y)=x^4 + y^4$ has a global minimum and maximum under the constraint $x^4 + y^4 - 2xy = 2$.
In the solution to the exercise, it it follows that the constraint is compact if we can show that $\lim_{x^2 + y^2 \rightarrow \infty} x^4 + y^4 - 3xy - 2 \rightarrow \infty$.Why this is the case?
My intuition tells me that this is because the $x^4$ and $y^4$ terms dominates the other two terms when $x$ and $y$ gets large. This would then imply that $x$ and $y$ cannot get arbitrarily big without violating the constraint. Does this imply that if the limit of the constraint was $0$, that the domain would not be compact? Is my reasoning valid?
Many thanks,
|
If you prove that $\lim_{x^2+y^2\to\infty}(x^4+y^4-2xy)=\infty$, then there exists $K>0$ such that $x^2+y^2-2xy>2$ for $x^2+y^2>K$. Hence the solutions of $x^2+y^2-2xy-2=0$ are contained in the circle $x^2+y^2\le K$ and so they form a bounded set.
For the limit, you can use polar coordinates; given $x=r\cos\varphi$ and $y=r\sin\varphi$, you have, for $r\ge2$,
\begin{align}
x^4+y^4-2xy
&=r^2(r^2\cos^4\varphi+r^2\sin^4\varphi-2\cos\varphi\sin\varphi) \\[6px]
&\ge r^2(4\cos^4\varphi+4\sin^4\varphi-2\cos\varphi\sin\varphi) \\[6px]
&\ge r^2(4-8\cos^2\varphi\sin^2\varphi-2\cos\varphi\sin\varphi) \\[6px]
&=r^2(4-2\sin^22\varphi-\sin\varphi) \\[6px]
&\ge r^2
\end{align}
because $2\sin^22\varphi+\sin\varphi\le 3$.
|
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|
>Finding range of $f(x)=\frac{\sin^2 x+4\sin x+5}{2\sin^2 x+8\sin x+8}$
Finding range of $$f(x)=\frac{\sin^2 x+4\sin x+5}{2\sin^2 x+8\sin x+8}$$
Try: put $\sin x=t$ and $-1\leq t\leq 1$
So $$y=\frac{t^2+4t+5}{2t^2+8t+8}$$
$$2yt^2+8yt+8y=t^2+4t+5$$
$$(2y-1)t^2+4(2y-1)t+(8y-5)=0$$
For real roots $D\geq 0$
So $$16(2y-1)^2-4(2y-1)(8y-5)\geq 0$$
$$4(2y-1)^2-(2y-1)(8y-5)\geq 0$$
$y\geq 0.5$
Could some help me where I have wrong, thanks
|
Let $t=sin(x)$, $$y=\frac{t^2+4t+5}{2t^2+8t+8}=\frac {1}{2}[1+ \frac {1}{(t+2)^2}]$$
Note that $$-1\le t \le 1$$ Thus $$ 1\le y \le5/9$$
|
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|
Find a pattern and prove it by mathematical induction: $1 = 1$
$3 + 5 = 8$
$7 + 9 + 11 = 27$
$13 + 15 + 17 + 19 = 64$
Etc...
I am having trouble seeing a pattern with this, I know it is relatable with Fibonacci Numbers but I am having trouble grasping this topic
|
We can continue the above pattern as follows:
$$ 21 + 23 + 25 + 27 + 29 = 125 = 5^3.$$
$$ 31 + 33 + 35 + 37 + 39 + 41 = 216 = 6^3. $$
We note that the first odd natural number is $1$, which we can write as $$ 1 = 2(1) - 1;$$
the second odd natural number is $3$, which can be written as
$$ 3 = 2(2) - 1; $$
the third odd natural number is $5$, which can be written as
$$ 5 = 2(3) - 1; $$
and so on and so forth.
We also note that, for each $n \in \mathbb{N}$, at step $n$, we add all the odd natural numbers from the ($ 1 + \sum_{k=1}^n k $)th to the ($\sum_{k=1}^{n+1} k$)th to obtain $n^3$.
Now
$$ 1 + \sum_{k=1}^n k = 1 + \frac{n(n+1)}{2} = \frac{n^2 + n + 2}{2}, $$
and
$$ \sum_{k=1}^{n+1} k = \frac{ (n+1)(n+2) }{2} = \frac{ n^2 + 3n + 2 }{2}. $$
So we can write
$$ \sum_{k= \frac{n^2 + n + 2}{2}}^{\frac{ n^2 + 3n + 2 }{2} } (2k-1) = n^3 $$
for every $n \in \mathbb{N}$.
This can be rewritten as
$$ 2 \sum_{k= \frac{n^2 + n + 2}{2}}^{ \frac{ n^2 + 3n + 2 }{2} } k \ - \ \left[ \frac{ n^2 + 3n + 2 }{2} - \frac{n^2 + n + 2}{2} \right] = n^3 $$
for each $n \in \mathbb{N}$.
Or,
$$ 2 \sum_{k= \frac{n^2 + n + 2}{2}}^{ \frac{ n^2 + 3n + 2 }{2} } k \ - \ n = n^3 $$
for each $n \in \mathbb{N}$.
Or,
$$ 2 \sum_{k= \frac{n^2 + n + 2}{2}}^{ \frac{ n^2 + 3n + 2 }{2} } k = n^3 + n $$
for each $n \in \mathbb{N}$.
We can verify the last identity using mathematical induction.
Hope this works out for you now.
|
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|
Equilateral triangle $ABC$ with $O$ is the circumcenter and M is a point on $(O)$, is $MA^x+MB^x+MC^x$ a constant?
Given an inscribed equilateral triangle $ABC$ with circumcenter $O$. M can be a point on any position of the circle $(O)$, except $A,B,C$. Prove or disprove that $MA^x+MB^x+MC^x$ is a constant for all integers $x$.
In a practice test, I got this question and it told me to prove that $MA=MB+MC$ (when chord $MA$ intersects chord $BC$), $MA^2+MB^2+MC^2=6R^2$ and $MA^4+MB^4+MC^4=18R^4$ (put $R=OA$).
I managed to solve them all (by putting an extra point $E$ on $AM$ so that $MB=ME$). However, with the use of computer (to draw the figure above) and calculator, I estimate that $MA^3+MB^3+MC^3=10R^3$; $MA^5+MB^5+MC^5=33R^5$; $MA^6+MB^6+MC^6=62R^6$; $MA^7+MB^7+MC^7=118R^7$.
Of course, that is just estimated, not exact because both the Sketchpad and the calculator don't give out the exact integer number. That's why I want to know if $MA^x+MB^x+MC^x$ is really a constant for all intergers $x$, calculated by $R$.
|
A necessary condition for the equality to hold in general is that the sums when $\,M =A\,$ and when $\,M=A'\,$ be equal, where $\,A'\,$ is the point diametrically opposed to $\,A\,$:
$$
\begin{align}
0^x + \left(R \sqrt{3}\right)^x + \left(R \sqrt{3}\right)^x = (2R)^x + R^x + R^x \;&\iff\; \left(2^x - 2 \left(\sqrt{3}\right)^x+2\right)R^x=0 \\
&\iff\; 2^x - 2 \left(\sqrt{3}\right)^x+2=0
\end{align}
$$
The latter equation has $\,x = 2, 4\,$ as the only two real roots.
Using properties of the centroid, $\,MA^2+MB^2+MC^2$ $= 3 MO^2 + OA^2+OB^2+OC^2$ $= 6 R^2\,$, so the sum is indeed constant for $\,x=2\,$.
For $\,x=4\,$ the sum is also constant, as follows from:
*
*$\,MA=MB+MC\,$ by Ptolemy's theorem for the cyclic quadrilateral $\,ABMC\,$
*by the law of cosines $\begin{cases}MB^2+MC^2+MB\,MC = 3R^2 \\ MA^2+MB^2-MA\,MB = 3R^2 \\ MA^2+MC^2 - MA\,MC = 3R^2 \end{cases}$, then multiplying the equalities by $\,MA^2, MC^2, MB^2\,$, respectively, and adding up:
$$\require{cancel} 2(MA^2MB^2+MB^2MC^2+MC^2MA^2) + \cancel{MA\,MB\,MC(MA-MB-MC)} = 3R^2(MA^2+MB^2+MC^2) = 18 R^4$$
*$MA^4+MB^4+MC^4 = \left(MA^2+MB^2+MC^2\right)^2 -2\left(MA^2MB^2+MB^2MC^2+MC^2MA^2\right) = 36R^4-18R^4=18R^4\,$
|
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How would one solve a linear equation in two integer variables? For example, how would I find integers $a$ and $b$ that satisfy the following equation?
$$5a - 12b = 13$$
I always resorted to trial and error when doing something like this and more often than not I would finally reach my answer. But for this one I just kept going and going to no avail. So it finally brought about the concern that trial and error wasn't always going to work.
So what would be the best way to get two integer solutions to the above equation?
|
The method here is to repeatedly putting new variables:
$5a-12b=13 \Leftrightarrow 5a=12b+13 \Leftrightarrow a=\frac{12b+13}{5}=2b+2+\frac{2b+3}{5}$
Put $\frac{2b+3}{5}=c$ with $c$ is an integer (because $a$ or $2b+2+c$ is an integer)
$\Leftrightarrow 2b+3=5c \Leftrightarrow 2b-5c=-3$ or $b=\frac{5c-3}{2}$.
$b$ is an integer $\Leftrightarrow 5c-3$ is divisible by $2 \Leftrightarrow c$ is an odd number.
Put $c=2k+1$ ($k$ is an integer), we have $b= \frac{5c-3}{2}=\frac{5(2k+1)-3}{2}=5k+1.$
$\Rightarrow a=\frac{12b+13}{5}=\frac{12(5k+1)+13}{5}=12k+5$.
The equation has infinite solutions: $\begin{cases}a=12k+5\\b=5k+1\end{cases}$, $\forall k\in \mathbb{Z}$.
Trial-and-error can still be applied for any numbers multiplied by $a$, $b$, as long as their $GCD$ is $1$.
For example this ugly equation: $73a+89b=3$. You won't have time to do trial-and-error, so here's a way to make it smaller:
*
*Put $a=\frac{3-89b}{73}=\frac{3-16b}{73}-b$, then $\frac{3-16b}{73} \in \mathbb{Z}$ because $a,b\in \mathbb{Z}$.
*Put $c=\frac{3-16b}{73}$, then $3-16b=73c$ or $b=\frac{3-73c}{16}=\frac{3-9c}{16}-4c$, then $\frac{3-9c}{16} \in \mathbb{Z}$ because $b,c\in \mathbb{Z}$.
*Put $d=\frac{3-9c}{16}$, then $3-9c=16d$ or $c=\frac{3-16d}{9}=\frac{3-7d}{9}-d$, then $\frac{3-7d}{9} \in \mathbb{Z}$ because $c,d\in \mathbb{Z}$.
*Put $e=\frac{3-7d}{9}$, then $3-7d=9e$ or $d=\frac{3-9e}{7}=\frac{3-2e}{7}-e$, then $\frac{3-2e}{7} \in \mathbb{Z}$ because $d,e\in \mathbb{Z}$.
*Put $f=\frac{3-2e}{7}$, then $3-2e=7f$ or $e=\frac{3-7f}{2}=\frac{1-f}{2}+1-3f$, then $\frac{1-f}{2} \in \mathbb{Z}$ because $e,f\in \mathbb{Z}$.
$e$ would be an integer if and only if $f=2k+1$ or $f$ odd
$\Rightarrow e=-7k-2$
$\Rightarrow d=9k+3$
$\Rightarrow c=-16k-5$
$\Rightarrow {\begin{cases}a=-89k-28\\b=73k+23\end{cases}}$
|
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|
How to calculate the summation of $w_i$, where $w_i = {n(n-1)\over (n-i)(i+1)}+{i-1\over i+1}w_{i-1}$ I am having trouble to calculate $$\sum_{i=1}^{n-1} w_i$$
where n is constant and $$w_i = {n(n-1)\over (n-i)(i+1)}+{i-1\over i+1}w_{i-1}$$
I am given the hint that
$$ \sum_{i=1}^{n-1} w_i = n(n-1)\sum_{i=1}^{n-1} {1 \over (n-i)(i+1)}(1+ \sum_{j=i+1}^{n-1} \prod_{k=i+1}^{j} {k-1 \over k+1}) $$
I have no idea how to get this equation. If I know it, absolutely I can find the final solution is $$\sum_{i=1}^{n-1} w_i = (n-1)^2$$
|
Let have a look at the first terms.
$w_1=\dfrac{n(n-1)}{2(n-1)}+0=\dfrac{n(n-1)}{2}\left[\dfrac 1{n-1}\right]$
$w_1+w_2=w_1+\dfrac{n(n-1)}{3(n-2)}+\dfrac{1}{3}w_1=\dfrac{n(n-1)}{3(n-2)}+\dfrac 43\times\dfrac{n(n-1)}{2(n-1)}\\\phantom{w_1}=\dfrac{n(n-1)}{3}\left[\dfrac 2{n-1}+\dfrac 1{n-2}\right]$
I let you do the calculation for the next one, you will find
$w_1+w_2+w_3 = \dfrac{n(n-1)}{4}\left[\dfrac 3{n-1}+\dfrac 2{n-2}+\dfrac 1{n-3}\right]$
We can show the relation by induction:
$$\sum\limits_{k=1}^{i}w_k = \dfrac{n(n-1)}{i+1}\sum\limits_{k=1}^{i}\dfrac{k}{n-i-1+k}$$
$\begin{align}w_1&+w_2+\cdots+w_{i+1} \\\\
&= \sum\limits_{k=1}^{i}w_k+\dfrac{n(n-1)}{(n-i-1)(i+2)}+\dfrac{i}{i+2}w_i\\\\
&=\sum\limits_{k=1}^{i}w_k+\dfrac{n(n-1)}{(n-i-1)(i+2)}+\dfrac{i}{i+2}\left(\sum\limits_{k=1}^{i}w_k-\sum\limits_{k=1}^{i-1}w_k\right)\\\\
&=2\,\dfrac{i+1}{i+2}\sum\limits_{k=1}^{i}w_k-\dfrac{i}{i+2}\sum\limits_{k=1}^{i-1}w_k+\dfrac{n(n-1)}{(n-i-1)(i+2)}\\\\
&=2\,\dfrac{i+1}{i+2}\dfrac{n(n-1)}{i+1}\sum\limits_{k=1}^{i}\dfrac{k}{n-i-1+k}-\dfrac{i}{i+2}\dfrac{n(n-1)}{i}\sum\limits_{k=1}^{i-1}\dfrac{k}{n-i+k}+\dfrac{n(n-1)}{(n-i-1)(i+2)}\\\\
&=\dfrac{n(n-1)}{i+2}\left[2 \sum\limits_{k=1}^{i}\dfrac{k}{n-i-1+k}-\sum\limits_{k=1}^{i-1}\dfrac{k}{n-i+k}+\dfrac{1}{(n-i-1)}\right]\\\\
&=\dfrac{n(n-1)}{i+2}\left[2 \sum\limits_{k=1}^{i}\dfrac{k}{n-i-1+k}-\sum\limits_{k=2}^{i}\dfrac{k-1}{n-i+k-1}+\dfrac{1}{(n-i-1)}\right]\\\\
&=\dfrac{n(n-1)}{i+2}\left[\dfrac{2}{n-i}\bigg|_{k=1}+\sum\limits_{k=2}^{i}\dfrac{2k-(k-1)}{n-i-1+k}+\dfrac{1}{(n-i-1)}\right]\\\\
&=\dfrac{n(n-1)}{i+2}\left[\dfrac{2}{n-i}+\sum\limits_{k=2}^{i}\dfrac{k+1}{n-i-2+k+1}+\dfrac{1}{(n-i-1)}\right]\\\\
&=\dfrac{n(n-1)}{i+2}\left[\sum\limits_{k=3}^{i+1}\dfrac{k}{n-i-2+k}+\dfrac{2}{n-i}\bigg|_{k=2}+\dfrac{1}{(n-i-1)}\bigg|_{k=1}\right]\\\\
&=\dfrac{n(n-1)}{i+2}\left[\sum\limits_{k=1}^{i+1}\dfrac{k}{n-i-2+k}\right]
\qquad\checkmark\end{align}$
Finally we can calculate the desired sum
$$\sum\limits_{k=1}^{n-1}w_k = \dfrac{n(n-1)}{(n-1+1)}\sum\limits_{k=1}^{n-1}\underbrace{\dfrac{k}{n-(n-1)-1+k}}_{=\frac kk=1}=(n-1)\sum\limits_{k=1}^{n-1} 1=(n-1)^2$$
|
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|
Counting Principle - Dependent Events There are 7 red and 5 yellow fish in an aquarium. Three fish are randomly caught in a net. Find the probability that the fish were:
a) All red
b) Not all of the same color.
I have solved this question using the method for dependent events as follows:
a) $\frac{7}{12} \cdot \frac{6}{11} \cdot \frac{5}{10} = \frac{210}{1320}$
b) $1 - \left(\frac{7}{12} \cdot \frac{6}{11} \cdot \frac{5}{10} + \frac{5}{12} \cdot \frac{4}{11} \cdot \frac{3}{10}\right) = \frac{1170}{1320}$
Is it possible to use combinations to solve this question? If yes, how?
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a) $$\frac{\binom73\binom50}{\binom{12}3}$$
b) $$1-\frac{\binom73\binom50}{\binom{12}3}-\frac{\binom70\binom53}{\binom{12}3}$$
|
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$\sum_{k=1}^{10} (\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11})=?$ How can we find the value of $$\sum_{k=1}^{10} (\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11})=?$$
My Approach:
By Euler's Theorem: $$\huge \cos \frac{2 \pi k}{11}+i \sin \frac{2 \pi k}{11} =e^{
\frac{i2 \pi k}{11}}$$
$$\therefore \sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11}=-ie^{
\frac{i2 \pi k}{11}}$$
so,$$\sum_{k=1}^{10} (\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11})=-i \sum_{k=1}^{10}e^{
\frac{i2 \pi k}{11}}=-i \times e^{
\frac{i2 \pi }{11}} \times \frac{{\{e^{
\frac{i2 \pi }{11}}\}}^{10} -1}{e^{
\frac{i2 \pi }{11}}-1}$$
$$=-i e^{
\frac{i2 \pi }{11}} \frac{e^{
\frac{i20 \pi }{11}} -1}{e^{
\frac{i2 \pi }{11}}-1}$$
now how can i proceed and simplify the result?
Answer $=i$
|
$$-i e^{
\frac{i2 \pi }{11}} \frac{e^{
\frac{i20 \pi }{11}} -1}{e^{
\frac{i2 \pi }{11}}-1}=
-i \frac{e^{
\frac{i2 \pi }{11}} e^{
\frac{i20 \pi }{11}} -e^{
\frac{i2 \pi }{11}} }{e^{
\frac{i2 \pi }{11}}-1}=-i \frac{1-e^{
\frac{i2 \pi }{11}} }{e^{
\frac{i2 \pi }{11}}-1}=i
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Area of largest inscribed rectangle in an ellipse. Can I take the square of the area to simplify calculations? So say I have an ellipse defined like this:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
I have to find the largest possible area of an inscribed rectangle.
So the area ($A$) of a rectangle is $2x2y=4xy$. Also we can redefine $y$ in terms of $x$:
$$\frac{y^2}{4} = 1 - \frac{x^2}{9}$$
$$y^2 = 4 - \frac{4x^2}{9}$$
$$y = \sqrt{4 - \frac{4x^2}{9}}$$
So the area function is now:
$$A=4x \cdot \sqrt{4 - \frac{4x^2}{9}}$$
$$A' = \frac{4x}{2 \cdot \frac{-8x}{9}} + \sqrt{4 - \frac{4x^2}{9}} \cdot 4$$
So this track seems too difficult, so I'd like to find another approach. Can I square the area first, find the derivative of that to solve for $x$?
So the $\text{Area} = 4x \cdot \sqrt{4 - \frac{4x^2}{9}}$
Is this valid?
$$\text{Area}^2 = 16x^2 \cdot \left(4 - \frac{4x^2}{9}\right)$$
$$= 64x^2 - \frac{64x^4}{9}$$
Derivative:
$$ \frac{d}{dx} \text{Area}^2 = 128x - \frac{256x^3}{9}$$
$$128x\left(1-\frac{2x^2}{9}\right)$$
So critical values: $x = 0, \frac{3}{\sqrt{2}}$
because the derivative equals $0$ when:
$$2x^2 = 9$$
$$x = \frac{3}{\sqrt{2}}$$
Plugging this value of $x$ into $y$ we get that $y = \sqrt{2}$ so the $\text{Area}$ is $3$.
Is this valid? If so why? Does squaring not cause any problems?
|
Yes, it is valid. You want to determine the maximum of a non-negative function $f$. But, since $f$ is non-negative, asserting the $\max f=M$ is equivalent to asserting that $\max f^2=M^2$. Besides, $f(x)=M\iff f^2(x)=M^2$. So, the points at which the functions $f$ and $f^2$ attain their maximal value are the same.
Note that $f$ being non-negative is essential. If $f(x)=x$, with $x\in[-2,1]$, then $f$ has a maximum at $1$, whereas $f^2$ hasn't (its maximum value is attained at $-2$).
|
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|
Inverse of a modular matrix I have the matrix$$
\begin{pmatrix}
1 & 5\\
3 & 4
\end{pmatrix} \pmod{26}
$$
and I need to find its inverse. I do it according to this website.
I find the modular multiplicative inverse (of the matrix determinant, which is $1×4-3×5=-11$) with the extended Euclid algorithm (it is $-7 \equiv 19 \pmod{26}$). Then I have $$\frac{1}{19}×\begin{pmatrix}4 & -5\\-3 & 1\end{pmatrix}.$$ I calculate that$$-5 \equiv 21 \pmod{26},\ -3 \equiv 23 \pmod{26}.$$
No matter what I do I am not able to get the solution they have on the website, which is$$\begin{pmatrix}2 & 17\\5 & 7\end{pmatrix}.$$
Can someone help me with this? What am I doing wrong?
|
The determinant is $-11 \pmod{26}$.
$$(-7)(-11)=77=26\cdot 3-1\equiv -1 \pmod{26}$$
Hence the inverse is suppose to be $7$ rather than $-7$.
$$7 \begin{bmatrix} 4 & -5 \\ -3 & 4\end{bmatrix} \equiv \begin{bmatrix} 2 & 17 \\ 5 & 3\end{bmatrix} \pmod{26}$$
|
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|
Find $\lim_\limits{(x,y)\to (0,0)}\frac{x^\alpha y^4}{x^2+y^4}$ where $\alpha > 0$ How can we find the following limit? $$\lim_{(x,y)\to(0,0)}\frac{x^\alpha y^4}{x^2+y^4}\qquad \alpha>0$$
By using the polar coordinate, we get
$$\lim_{r\to 0}r^{\alpha+2}\frac{\cos^\alpha\theta \sin^4\theta}{\cos^2\theta+r^2\sin^4{\theta}}=0$$ if $\theta\notin\{\frac{\pi}{2}+\pi k:k\in\Bbb Z\}$.
Now, if $\theta = \frac{\pi}{2}+\pi k$ for some $k\in\Bbb Z$, then we get $$\lim_{(0,y)\to (0,0)}\frac{x^\alpha y^4}{x^2+y^4}=0.$$
Can we conclude that $$\lim_{(x,y)\to(0,0)}\frac{x^\alpha y^4}{x^2+y^4}=0?$$
|
Just notice that since $\alpha > 0$ and $x^2 \geq 0$ you have that
$$ \frac{x^\alpha y^4}{x^2+y^4} \leq \frac{x^\alpha y^4}{y^4} = x^\alpha \overset{(x,y) \to 0}{\longrightarrow} 0.$$
Thus it is clear that one has $\lim_{(x,y)\to(0,0)}\frac{x^\alpha y^4}{x^2+y^4} = 0$. You rarely need polar coordinates for these kinds of questions. The most problems I encountered personally are easy to solve by using this trick above. I hope it helps you :)
|
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|
Help with $\int\frac{1+\sin x}{(1+2\cos x)\sin x}\;dx$ $$\int\frac{1+\sin x}{(1+2\cos x)\sin x}\;dx$$
I tried to solve this question of integral many times but I don't understand. How do I solve it?
|
This calls for the hyperbolic Kepler angle! We make the substitution
$$\begin{align}\sinh\theta=\frac{\sqrt{e^2-1}\sin x}{1+e\cos x},&&\cosh\theta=\frac{\cos x+e}{1+e\cos x},&&d\theta=\frac{\sqrt{e^2-1}}{1+e\cos x}dx\end{align}$$
With $e=2$ for us. We will also need the inverse relation:
$$\sin x=\frac{\sqrt{e^2-1}\sinh\theta}{e\cosh\theta-1}$$
Then
$$\begin{align}\int\frac{\csc x+1}{1+e\cos x}dx&=\int\left(\frac{e\cosh\theta-1}{\sqrt{e^2-1}\sinh\theta}+1\right)\frac{d\theta}{\sqrt{e^2-1}}\\
&=\frac{-1}{e^2-1}\int\text{csch}\,\theta\,d\theta+\frac e{e^2-1}\int\coth\theta\,d\theta+\frac1{\sqrt{e^2-1}}d\theta\\
&=\frac1{e^2-1}\ln\left(\text{csch}\,\theta+\coth\theta\right)+\frac e{e^2-1}\ln\sinh\theta+\frac1{\sqrt{e^2-1}}\theta+C_1\\
&=\frac1{e^2-1}\ln\left\{\frac{1+e\cos x}{\sqrt{e^2-1}\sin x}\left(1+\frac{\cos x+e}{1+e\cos x}\right)\right\}\\
&\quad+\frac e{e^2-1}\ln\left(\frac{\sqrt{e^2-1}\sin x}{1+e\cos x}\right)+\frac1{\sqrt{e^2-1}}\sinh^{-1}\left(\frac{\sqrt{e^2-1}\sin x}{1+e\cos x}\right)+C_1\\
&=\frac1{e^2-1}\ln\left(\frac{(1+e)(1+\cos x)}{\sqrt{e^2-1}\sin x}\right)+\frac e{e^2-1}\ln\left(\frac{\sqrt{e^2-1}\sin x}{1+e\cos x}\right)\\
&\quad+\frac1{\sqrt{e^2-1}}\ln\left(\frac{\sqrt{e^2-1}\sin x}{1+e\cos x}+\frac{\sqrt{(e^2-1)\sin^2x+(1+e\cos x)^2}}{(1+e\cos x)^2}\right)+C_1\\
&=\frac1{e^2-1}\ln(1+\cos x)-\frac1{e^2-1}\ln\sin x+\frac e{e^2-1}\ln\sin x\\
&\quad-\frac e{e^2-1}\ln(1+e\cos x)+\frac1{\sqrt{e^2-1}}\ln\left(\sqrt{e^2-1}\sin x+e+\cos x\right)\\
&\quad-\frac1{\sqrt{e^2-1}}\ln(1+e\cos x)+C\\
&=\frac1{e^2-1}\ln(1+\cos x)+\frac1{e+1}\ln\sin x-\frac{e+\sqrt{e^2-1}}{e^2-1}\ln(1+e\cos x)\\
&\quad+\frac1{\sqrt{e^2-1}}\ln\left(\sqrt{e^2-1}\sin x+e+\cos x\right)+C\end{align}$$
Of course I leave it up to the interseted reader to change all those $e$'s back to $2$'s.
|
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|
Number of divisors of the number $2079000$ which are even and divisible by $15$
Find the number of divisors of $2079000$ which are even and divisible by $15$?
My Attempt: Since they are divisible by $15$ and are even, $2$ and $5$ have to included from the numbers prime factors.
$2079000 = 2^3 \cdot 3^3 \cdot 5^3 \cdot 7 \cdot 11$
Therefore, the number of divisors should be
$2 \cdot 2 \cdot (3+1) \cdot (1+1) \cdot (1+1)$
But however this answer is wrong.
Any help would be appreciated.
|
$$2079000 = 2^3*3^3*5^3*7*11$$
You want to have at least one of each numbers $2$,$3$,and $5$
Thus you should have
$$3*3*3*(1+1)*(1+1)=108$$
|
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|
Basis over GF(2^5) with sum of its elements = 1 I need to work over $GF(2^5)$ starting from $GF(2)$ and I would need a basis $\{a_0, a_1, \cdots , a_4\}$ over $GF(2^5)$ such that $a_0+a_1+\cdots +a_4=1$. Do you know how could i proceed?
Thank you!
|
The sums of the roots of the primitive polynomials
$$x^5+x^4+x^3+x^2+1, \quad x^5+x^4+x^3+x+1, \quad x^5+x^4+x^2+x+1\tag{1}$$
in $\mathbb F_2[x]$ all equal $1$ (the coefficient of $x^4$ is $1$ in each case) and the five roots are a normal basis of $\mathbb F_{32}$ over $\mathbb F_2$. If you are using $x^5+x^2+1$ for constructing $\mathbb F_{32}$ and denoting the roots as $\alpha, \alpha^2, \alpha^4, \alpha^8, \alpha^{16}$, then these roots sum to $0$ as you have discovered. The roots of the polynomials in $(1)$ above are respectively
$x^5+x^4+x^3+x^2+1\colon \quad\alpha^3, \alpha^6, \alpha^{12}, \alpha^{24}, \alpha^{17}$
$x^5+x^4+x^3+x+1\colon \quad\alpha^{11}, \alpha^{22}, \alpha^{13}, \alpha^{26}, \alpha^{21}$
$x^5+x^4+x^2+x+1\colon \quad\alpha^5, \alpha^{10}, \alpha^{20}, \alpha^{9}, \alpha^{18}$
and each set is a basis that sums to $1$ as desired. The first set listed above is the one that the OP found for himself, as mentioned in his comment on his own question.
|
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|
Minimum value of $\tan^2\alpha+\tan^2\beta+\tan^2\gamma$
If $\alpha,\beta,\gamma$ be a variable and $k$ be a constant such that $a\tan\alpha+b\tan\beta+c\tan\gamma=k$.Then find minimum value of $\tan^2\alpha+\tan^2\beta+\tan^2\gamma$ is
Try: Using Cauchy Schwarz Inequality:
$(a^2+b^2+c^2)(\tan^2\alpha+\tan^2\beta+\tan^2\gamma)\geq (a\tan\alpha+b\tan\beta+c\tan\gamma)^2$
So we have $$\tan^2\alpha+\tan^2\beta+\tan^2\gamma)\geq \frac{k^2}{a^2+b^2+c^2}$$
Could some help me to solve it without Cauchy Schwarz Inequity .
Please explain, thanks
|
Let
\begin{align}
f(x,y,z)&=\tan^2 x + \tan^2 y+\tan^2 z\\
g(x,y,z)&=a\tan x +b\tan y+c\tan z=k
\end{align}
We want to find the minimum of $f$ constrained to $g$, and we can apply Lagrange Multiplier. Consider the equation $\nabla f=\lambda \nabla g$. Since
\begin{align}
\nabla f(x,y,z)&=(2\tan x\sec^2 x, 2\tan y\sec^2 y, 2\tan z\sec^2 z),\\
\nabla g(x,y,z)&=(a\sec^2 x, b\sec^2 y,c\sec^2 z),
\end{align}
for a root of the equation $(x^*,y^*,z^*)$, this equation holds:
$$\begin{cases}\tag 1
2\tan x^*=\lambda a\\
2\tan y^*=\lambda b\\
2\tan z^*=\lambda c
\end{cases}$$
Substitite (1) into $g(x,y,z)=k$, then $\lambda=\dfrac{2k}{a^2+b^2+c^2}$ and
$$
f(x^*,y^*,z^*)=\frac{\lambda^2 (a^2+b^2+c^2)}{4}=\frac{k^2}{a^2+b^2+c^2}.
$$
This is what we want.
|
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|
Trigonometric identity. How can I prove that $\sin A = 2\sin\frac{A}2\cos\frac{A}2$? How can I prove that $\sin A = 2\sin\frac{A}2\cos\frac{A}2$ ?
My failed take on this matter is:
$$
\sin A =
\sin\frac{A}2 + \sin\frac{A}2 = 2\sin\Big(\frac{\frac{A}2+\frac{A}2}2\Big)\cos\Big(\frac{\frac{A}2-\frac{A}2}2\Big)=
2\sin\Big(\frac{\frac{2A}2}2\Big)\cos\frac02 =
2\sin\frac{A}2
$$
where $\cos\frac02 = \cos0 = 1$
|
In your work, you used the sum-to-product formula, $$\sin\frac{A}2 + \sin\frac{A}2 = 2\sin\Big(\frac{\frac{A}2+\frac{A}2}2\Big)\cos\Big(\frac{\frac{A}2-\frac{A}2}2\Big),$$
which is correct. The mistake is that $$\sin A\ne\sin\frac{A}2 + \sin\frac{A}2$$
You can still use the sum-to-product formula.
$$\sin A + \sin 0 = 2\sin\Big(\frac{A+0}2\Big)\cos\Big(\frac{A-0}2\Big),$$
|
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|
Evaluate $\int_0^1 \frac 1 {\lfloor 1-\log_2(x) \rfloor}\,\mathrm{d}x$ Evaluate $$\int_{0}^{1} \frac{1}{\lfloor 1-\log_2(x) \rfloor}\,\mathrm{d}x$$ where $\lfloor \cdot \rfloor$ is the Greatest Integer Function (GIF)
I learnt that the solution is probably $\displaystyle\sum_{r=1}^\infty \frac 1 {r \cdot 2^r}.$
But how does that happen?
|
Let $$f(x)=\frac {1}{\lfloor 1-\log_2 x\rfloor}$$
For $x\in \left(1/2,1\right)$, $f(x)=1$
For $x\in \left(1/4,1/2\right)$, $f(x)=1/2$
For $x\in \left(1/8,1/4\right)$, $f(x)=1/3$
So the integral turns out to be $$\frac 11\cdot\frac 12+\frac 12\cdot\frac 14+\frac 13\cdot\frac 18\cdots= \frac {1}{1\cdot 2^1}+\frac {1}{2\cdot 2^2}+\frac {1}{3\cdot 2^3}\cdots
= \sum_{r=1}^{\infty} \frac {1}{r\cdot 2^r}$$
On further computation you might notice that the answer is simply $\ln 2$
|
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|
Formula for the large derivative Is there any formula for the large number derivative?
I need to find $y^{(100)}$ at $x=0$, if $y=(x+1)2^{x+1}$
I tried to find a pattern, but 2nd and 3rd derivatives are already too hairy. I see no pattern, how it evolves.
1st derivative $2^{x+1}+\ln \left(2\right)\cdot \:2^{x+1}\left(x+1\right)$
or $2^{x+1}(1+ln(2)(x+1))$
2nd $\ln ^2\left(2\right)\cdot \:2^{x+1}x+\ln ^2\left(2\right)\cdot \:2^{x+1}+\ln \left(2\right)\cdot \:2^{x+2}$
3rd $\ln ^2\left(2\right)\left(\ln \left(2\right)\cdot \:2^{x+1}x+2^{x+1}\right)+\ln ^3\left(2\right)\cdot \:2^{x+1}+\ln ^2\left(2\right)\cdot \:2^{x+2}$
|
Use Taylor series:
$$y=2(x+1)\left(1+x\ln 2+\frac{x^2}{2!}\ln^2 2+\cdots \right)=\\
[2x+\cdots +\frac{2x^{100}}{99!}\ln^{99} 2+O(x^{101})]+\\
[2+\cdots+\frac{2x^{100}}{100!}\ln^{100} 2+O(x^{101})].$$
$$y^{(100)}(0)=200\ln^{99} 2+2\ln^{100} 2.$$
|
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|
Double integral of maximum function. Let $D:= \lbrace (x,y) \in [0,\infty)^2: 1 \le x^2+y^2\le9 \rbrace$.
Determine the integral :
$$\int\int_D \max(3x^2,y^2)\;dx\,dy.$$
I have a little problem, because I'm not sure where the maximum is $3x^2$ or $y^2$.
|
Let $x=r\cos\theta$ and $y=r\sin\theta$, then
$$
\iint_D \max\{3x^2,y^2)\ \mathsf dx\ \mathsf dy = \iint_D \det J\cdot\max\{3r^2\cos^2\theta, r^2\sin^2\theta \}\ \mathsf d\theta\ \mathsf d r,
$$
where
$$
J = \begin{bmatrix}\frac{\partial x}{\partial r}&\frac{\partial x}{\partial \theta}\\ \frac{\partial y}{\partial r}&\frac{\partial y}{\partial \theta}
\end{bmatrix}
= \begin{bmatrix} \cos\theta&-r\sin\theta\\ \sin\theta& r\cos\theta
\end{bmatrix}.
$$
The determinant of the Jacobian matrix is
$$
\det J = \cos\theta\cdot r\cos\theta - (-r\sin\theta\cdot\sin\theta) = r(\sin^2\theta + \cos^2\theta) = r,
$$
and so the integral in question is
$$
\int_1^3\int_0^{2\pi} r\cdot\max\{3r^2\cos^2\theta,r^2\sin^2\theta\}\ \mathsf d\theta\ \mathsf dr = \int_1^3 r^3\int_0^{2\pi} \max\{3\cos^2\theta,\sin^2\theta\}\ \mathsf d\theta\ \mathsf dr.
$$
To compute $\max\{3\cos^2\theta,\sin^2\theta\}$, first note that
\begin{align}
3\cos^2\theta = \sin^2\theta &\iff \sqrt 3|\cos\theta| = |\sin\theta|\\
&\iff \frac{|\sin\theta|}{|\cos\theta|} = \sqrt 3\\
&\iff \theta \in \left\{\frac\pi3, \frac{2\pi}3,\frac{4\pi}3,\frac{5\pi}3 \right\}.
\end{align}
Hence
\begin{align}
&\quad\int_1^3r^3\int_0^{2\pi} \max\{3\cos^2\theta,\sin^2\theta\}\ \mathsf d\theta\ \mathsf dr\\
&= \int_1^3 r^3 \bigg(\int_0^{\frac\pi3}3\cos^2\theta\ \mathsf d\theta + \int_{\frac\pi3}^{\frac{2\pi}3} \sin^2\theta\ \mathsf d\theta + \int_{\frac{2\pi}3}^{\frac{4\pi}3}3\cos^2\theta\ \mathsf d\theta \\
&\quad\quad\quad+\int_{\frac{4\pi}3}^{\frac{5\pi}3}\sin^2\theta\ \mathsf d\theta + \int_{\frac{5\pi}3}^{2\pi} 3\cos^2\theta \bigg) \mathsf d\theta\\
&=40\sqrt3 + \frac{140\pi}3.
\end{align}
|
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|
What is the limit of $\frac{\left ( 2n \right )!}{(n!)^{2}4^{n}}$? What is the limit of the following sequence?.
$$\frac{\left ( 2n \right )!}{(n!)^{2}4^{n}}$$
I guess the limit is zero but I don't know exactly how to prove it.
Any help please.
|
Because I think there are people interested in an elementary solution:
The ratio between terms $a_n=\binom{2n}{n}\frac{1}{4^n}$ is given by
$$\begin{array}{ll} \displaystyle \frac{a_{n+1}}{a_n} & =\frac{\displaystyle\frac{(2n+2)!}{(n+1)!(n+1)!}\frac{1}{4^{n+1}}}{\displaystyle\frac{(2n)!}{n!n!}\frac{1}{4^n}} \\[6pt] & \displaystyle = \frac{(2n+2)(2n+1)}{(n+1)(n+1)4} \\[3pt] & \displaystyle =\frac{2n+1}{2n+2} \\[2pt] & \displaystyle =1-\frac{1}{2n+2}. \end{array} $$
Therefore, the term $a_n$ is
$$\begin{array}{ll} a_n & \displaystyle =a_0\cdot\frac{a_1}{a_0}\cdot\frac{a_2}{a_1}\cdots\frac{a_n}{a_{n-1}} \\ & = \left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right)\cdots\left(1-\frac{1}{2n}\right) \\ & \le \left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\cdots\left(1-\frac{1}{2n+1}\right) \end{array} $$
and therefore
$$ \begin{array}{ll} a_n^2 & \le \left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\cdots\left(1-\frac{1}{2n}\right)\left(1-\frac{1}{2n+1}\right) \\ & = \left(\frac{1}{2}\right)\left(\frac{2}{3}\right)\cdots\left(\frac{2n-1}{2n}\right)\left(\frac{2n}{2n+1}\right)=\frac{1}{2n+1}. \end{array} $$
From $a_n\le\frac{1}{\sqrt{2n+1}}$ we conclude the limit is $0$.
|
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|
How to approximate a function involving arc-tangent for a sufficiently large argument? I have a function, $f(x) = {\tan ^{ - 1}}(\sqrt {1 + {x^2}} - x)$. Let's assume the argument $x$ is sufficiently high. If so, how can I approximate $f(x)$ in terms of $x$? I have computed $f(x)$ with MATLAB. It looks like that $f(x)$ can be approximated to ${1 \over {2x}}$. That is, ${\tan ^{ - 1}}(\sqrt {1 + {x^2}} - x) \approx {1 \over {2x}}$ for a large $x$. It this right?
|
As Martin Argerami already answered, you are totally write.
If you want more accuracy, you could use Taylor series
$$\sqrt{1+x^2}=x+\frac{1}{2 x}-\frac{1}{8 x^3}+\frac{1}{16
x^5}+O\left(\frac{1}{x^7}\right)$$
$$\sqrt{1+x^2}-x=\frac{1}{2 x}-\frac{1}{8 x^3}+\frac{1}{16
x^5}+O\left(\frac{1}{x^7}\right)$$ Now, using
$$\tan^1(\epsilon)=\epsilon -\frac{\epsilon ^3}{3}+\frac{\epsilon ^5}{5}+O\left(\epsilon ^7\right)$$ you should arrive to
$$f(x) = {\tan ^{ - 1}}(\sqrt {1 + {x^2}} - x)=\frac{1}{2 x}-\frac{1}{6 x^3}+\frac{1}{10
x^5}+O\left(\frac{1}{x^7}\right)$$ or more generally
$$f(x) = {\tan ^{ - 1}}(\sqrt {1 + {x^2}} - x)=\sum_{i=1}^\infty \frac{(-1)^{i-1} }{(4 i-2)x^{2 i-1}}$$
Using $x=5$ (which is quite small), the truncated series given above leads to $\frac{9253}{93750}\approx 0.0986987$ while the "exact" value would be $\approx 0.0986978$.
|
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If $\frac{\cos \alpha}{\cos \beta}+\frac{\sin \alpha}{\sin \beta}=-1$, find $\frac{\cos^3\beta}{\cos \alpha}+\frac{\sin^3\beta}{\sin \alpha}$.
If $\displaystyle \frac{\cos \alpha}{\cos \beta}+\frac{\sin \alpha}{\sin \beta}=-1$, find $\displaystyle \frac{\cos^3\beta}{\cos \alpha}+\frac{\sin^3\beta}{\sin \alpha}$.
I tried$$
\sin(\alpha+\beta)=-\sin \beta \cos \beta,\\
2\sin(\alpha+\beta)=-\sin (2\beta),$$
and $$\frac{\cos^3\beta}{\cos \alpha}+\frac{\sin^3\beta}{\sin \alpha} =\frac{\sin\alpha\cos^3\beta+\sin^3\beta \cos \alpha}{\sin \alpha \sin \alpha},$$
but unable to find that ratio.
Any help, please.
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\begin{cases}
\dfrac{\cos\alpha}{\cos\beta}+\dfrac{\sin\alpha}{\sin\beta}+1=0\\[4pt]
\sin2\alpha + \sin2\beta = 2\sin(\alpha+\beta)\cos(\alpha - \beta),
\end{cases}
\begin{cases}
2\sin(\alpha+\beta) = - \sin2\beta\\
2\sin(\alpha+\beta)(1+\cos(\alpha - \beta)) = \sin2\alpha.
\end{cases}
\begin{align}
\dfrac{\cos^3\beta}{\cos\alpha}+\dfrac{\sin^3\beta}{\sin\alpha} = \dfrac{3cos\beta + \cos3\beta}{4\cos\alpha}+\dfrac{3\sin\beta-\sin3\beta}{4\sin\alpha} = \dfrac{3\sin(\alpha+\beta)+\sin(\alpha-3\beta)}{2\sin2\alpha} = \dfrac{4\sin(\alpha+\beta) - (\sin(\alpha+\beta) - \sin(\alpha-3\beta))}{2\sin2\alpha} = \dfrac{4\sin(\alpha+\beta) - 2\sin2\beta\cos(\alpha -\beta))}{2\sin2\alpha} = \dfrac{2\sin(\alpha+\beta)(1+\cos(\alpha-\beta))}{\sin2\alpha} = \dfrac{\sin2\alpha}{\sin2\alpha} \color{\red}{ = 1}.
\end{align}
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Evaluating $\lim_{x\to \infty } (x +\sqrt[3]{1-{x^3}} ) $ $$\lim_{x\to \infty } (x +\sqrt[3]{1-{x^3}} ) $$
What method should I use to evaluate it. I can't use the ${a^3}$-${b^3}$ formula because it is positive. I also tried to separate limits and tried multiplying with $\frac {\sqrt[3]{(1-x^3)^2}}{\sqrt[3]{(1-x^3)^2}}$ , but still didn't get an answer. I got -$\infty$, and everytime I am getting $\infty -\infty$ .
|
How about without calculus? Does $x+\sqrt[3]{1-x^3}$ ever cross the x axis?
$$\begin{align}
x+\sqrt[3]{1-x^3}&=0 \\
x&=-\sqrt[3]{1-x^3} \\
x&=\sqrt[3]{x^3-1} \\
x^3&=x^3-1 \\
0&=-1
\end{align}$$
It doesn't. It's also defined for all $x$. And one point on the curve $f(x)=x+\sqrt[3]{1-x^3}$ is $f(1)=1\ge0$. Therefore $x+\sqrt[3]{1-x^3} \geq 0 \quad \forall x$. The line $y=0$ serves as the greatest lower bound for $f(x)$
The goal now being to bound the function from the top, let there be some arbitrary value, $a>0$, that we wish to find all $x$ such that $f(x)$ is less than $a$. Firstly we need to know when $f(x)=a$
$$\begin{align}
x+\sqrt[3]{1-x^3}&=a \implies 1-x^3=(a-x)^3=a^3-3a^2x+3ax^2-x^3 \implies \\
0&=3ax^2-3a^2x+a^3-1 \implies a \neq 0 \quad \land \\
x&=\frac{3a^2 \pm D}{6a} =\frac{a}{2} \pm \frac{D}{6a}\\
&\quad \text{discriminant} \ D=9a^4-4(3a)(a^3-1)=(3a)(4-a^3) \\
&\quad D=0 \implies [a=0] \lor \left[a=\sqrt[3]{4}=2^\frac{2}{3}\right] \\
&\quad \quad [a \neq 0] \implies \\
&\quad \quad [a=\sqrt[3]{4}] \implies x=\frac{\sqrt[3]{4}}{2} \pm 0=2^\frac{-1}{3}\\
&\quad \quad f(2^\frac{-1}{3})=2 \cdot 2^\frac{-1}{3} =2^\frac{2}{3} \ \text{which is possibly} \max{f(x)} \\
&\quad D>0 \implies 0<a<2^\frac{2}{3} \\
&\quad D<0 \implies a \in \lbrace (-\infty, 0) \lor (2^\frac{2}{3}, \infty) \rbrace \\
\end{align}$$
3 points
*
*Seeing that the total range of intersection is the $y$ interval $(0, 2^\frac{2}{3}]$, and how the only place where the line $y=a$ within this range intersects with $f(x)$ only $1$ time is when $x=2^\frac{-1}{3}$ i.e. when $a=2^\frac{2}{3}$, the maximum point of $f(x)$ is $(2^\frac{-1}{3},2^\frac{2}{3})$.
*On the $x$ interval $(2^\frac{-1}{3}, \infty)$, $f(x)$ can either stay the same value or decrease, since there will be no increasing. Our point $(1,1)$ from earlier shows that $f(x)$ decreases on this interval
*with no least upper bound, yet a greatest lower bound of $0$, the end behavior of $f(x)$ tends to $0$
Probably easier with calculus
|
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|
Minimum of $\frac{x}{\sqrt{1-x^2}}+\frac{y}{\sqrt{1-y^2}}+\frac{z}{\sqrt{1-z^2}}$ Let $0\leq x,y,z<1$ and $x^2+y^2+z^2=1$. What is the minimum value of
$$\frac{x}{\sqrt{1-x^2}}+\frac{y}{\sqrt{1-y^2}}+\frac{z}{\sqrt{1-z^2}}?$$
From the condition, the point $(x,y,z)$ lies on a sphere with radius $1$. At the equality point, all three variables are equal to $1/\sqrt{3}$, and the value taken is $3/\sqrt{2}$. If one variable is $0$ and the other two are equal, the two variables are equal to $1/\sqrt{2}$, and the value taken is $2$.
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By AM-GM $$\sum_{cyc}\frac{x}{\sqrt{1-x^2}}=\sum_{cyc}\frac{2x^2}{2\sqrt{x^2(1-x^2)}}\geq\sum_{cyc}\frac{2x^2}{x^2+1-x^2}=2.$$
The equality occurs for $z=0$ and $x=y=\frac{1}{\sqrt2},$ which says that $2$ is a minimal value.
|
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Prove $\left ( a+ b \right )^{2}\left ( b+ c \right )^{2}\left ( c+ a \right )^{2}\geq 64abc$ Give $a, b, c$ be positive real numbers such that $a+ b+ c= 3$. Prove that:
$$\left ( a+ b \right )^{2}\left ( b+ c \right )^{2}\left ( c+ a \right )^{2}\geq 64abc$$
My try
We have:
$$\left ( a+ b \right )^{2}\geq 4ab$$
$$\left ( b+ c \right )^{2}\geq 4bc$$
$$\left ( c+ a \right )^{2}\geq 4ca$$
So I want to prove $abc\geq 1$. I need to the help. Thanks!
|
Set $$f=27(a+b)^2(b+c)^2(c+a)^2-64abc(a+b+c)^3$$
Assume $a=\max \{a,b,c\}$ thus $$f=27\,(a-b)^2(b-c)^2(c-a)^2+4abc\left[11(a^3+b^3+c^3-3abc)+6\sum_{cyc} c(a-b)^2\right]+108\left\{a^2\Big[ab+c(a-b)\Big](b-c)^2+b^2c^2(a-c)(a-b)\right\} \ge 0$$
|
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|
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, prove that $\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, prove that $\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$
My Attempt
$$
\frac{-2x}{2\sqrt{1-x^2}}-\frac{2y}{2\sqrt{1-y^2}}.\frac{dy}{dx}=a-a\frac{dy}{dx}\\
\implies \frac{dy}{dx}\bigg[a-\frac{y}{\sqrt{1-y^2}}\bigg]=a+\frac{x}{\sqrt{1-x^2}}\\
\frac{dy}{dx}=\frac{a\sqrt{1-x^2}+x}{\sqrt{1-x^2}}.\frac{\sqrt{1-y^2}}{a\sqrt{1-y^2}+x}=\sqrt{\frac{1-y^2}{1-x^2}}.\frac{a\sqrt{1-x^2}+x}{a\sqrt{1-y^2}-y}
$$
How do I poceed further and find the derivative ?
|
Both $x,y\in[-1,1]$. So, $x=\sin\theta$ and $y=\sin\phi$ for some $\displaystyle \theta,\phi\in\left[\frac{-\pi}{2},\frac{\pi}{2}\right]$. Note that $\cos\theta, \cos\phi\ge0$.
We have $$\cos\theta+\cos\phi=a(\sin\theta-\sin\phi)$$
So, $$2\cos\frac{\theta+\phi}{2}\cos\frac{\theta-\phi}{2}=2a\cos\frac{\theta+\phi}{2}\sin\frac{\theta-\phi}{2}$$
$\displaystyle \tan\frac{\theta-\phi}{2}=\frac{1}{a}$ is a constant. So, $\displaystyle \frac{d\phi}{d\theta}=1$.
$$\frac{dy}{dx}=\frac{\cos\phi}{\cos\theta}\frac{d\phi}{d\theta}=\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$$
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|
Interesting 3 Variable Inequality for Real Numbers I recently saw an olympiad style inequality that seemed very difficult. I tried to use elementary inequalities such as AM-GM or Cauchy-Schwarz, but neither have helped in making significant progress. Could anyone provide a rigorous proof, preferably using more elementary inequalities?
Problem: "Prove the inequality $\frac{1}{y(x+y)}+\frac{1}{z(y+z)}+\frac{1}{x(z+x)}\geq\frac{27}{2(x+y+z)^2}$ if $x,y,z$ are positive reals."
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This follows almost immediately by a direct application of Hölder's inequality (generalised Cauchy-Schwarz):
$$\begin{aligned} \left ( \sum \limits_{\text{cyc}} \frac{1}{y(x+y)} \right ) (y + z + x) ((x+y) + (y+z) + (z+x)) &\geq (1^{\frac{1}{3}} + 1^{\frac{1}{3}} + 1^{\frac{1}{3}})^{3}\\
\iff \sum \limits_{\text{cyc}} \frac{1}{y(x+y)} &\geq \frac{27}{(x+y+z)(2x+2y+2z)}\\
&= \frac{27}{2(x+y+z)^2}\\
\end{aligned}$$
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If $8R^2=a^2+b^2+c^2$ then prove that the triangle is right angled. If $8R^2=a^2+b^2+c^2$ then prove that the triangle is right angled. Where $a,b,c$ are the sides of triangle and $R$ is the circum radius
My Attempt
From sine law,
$$\dfrac {a}{\sin A}=\dfrac {b}{\sin B}=\dfrac {c}{\sin C}=2R$$
So,
$$a=2R \sin A$$
$$b=2R \sin B$$
$$c=2R \sin C$$
Then,
$$8R^2=a^2+b^2+c^2$$
$$8R^2=4R^2 \sin^2 (A)+ 4R^2 \sin^2 (B) + 4R^2 \sin^2 C$$
$$8R^2=4R^2(\sin^2 (A)+\sin^2 (B) +\sin^2 (C)$$
$$2=\sin^2 (A)+\sin^2 (B)+\sin^2 (C)$$
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HINT:
Use the equality valid for any $A$, $B$, $C$ with sum $\pi$
$$1-(\cos^2 A + \cos^2 B + \cos^2 C)- 2 \cos A \cos B \cos C = 0$$
ADDED: The equality follows from the following formula valid for any angles $\alpha$, $\beta$, $\gamma$ with sum $2s$
$$1-(\cos^2\alpha + \cos^2 \beta + \cos^2 \gamma)- 2 \cos \alpha \cos \beta \cos \gamma = - 4 \cos s\cos (s-\alpha) \cos (s-\beta) \cos (s - \gamma)$$
|
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Where’s the error in my factorial proof? On one of our tests, the extra credit was to find which number you would take out from the set $\{1!,2!,3!,...(N-1)!,N!\}$ such that the product of the set is a perfect square, for even $N$ My answer was as follows:
Assume $N$ is even. First note that $(n!)=(n-1)!\cdot n$.
Apply this to the odd numbers to get the product:
$$(2!)(2!)\cdot 3 \cdot(4!)(4!)\cdot 5\cdots ((N-2)!)((N-2)!)\cdot (N-1)\cdot N!$$
Let $ 2!4!6!...(N-2)!=E$. Then our equation is equal to:
$$E^23\cdot 5\cdot 7\cdots (N-1)\cdot (N!)$$
Expand $N!$:
$$E^2\cdot 3\cdot 5\cdot 7\cdots (N-1)\cdot 2\cdot 3\cdot 4\cdots N$$
Group the odd terms together:
$$E^2\cdot 3\cdot 3\cdot 5\cdot 5\cdot 7 \cdot 7\cdots (N-1)\cdot 2\cdot 4\cdot 6\cdots N$$
Let $O=1\cdot3\cdot5...\cdot (N-1)$:
$$E^2O^2 2\cdot 4\cdot 6\cdots N = E^2O^2\cdot (2\cdot 2)\cdot (2\cdot 3)\cdot (2\cdot 4)\cdots (2\cdot (N/2))$$
Group together the $2$'s:
$$E^2O^22^{(N/2)}1\cdot2\cdot3...(N/2)=E^2O^22^{(N/2)}\cdot(N/2)!$$
So, if $N/2$ is even, it can be expressed as $2m$ for some $m$. So we have:
$$E^2O^22^{2m}(N/2)!=(EO2^m)^2(N/2)!$$
Therefore, if $N$ is even, the number missing is $(N/2)!$ if $N/2$ is even. For example, for $N=4$, $2!$ is missing, $N=6$ is impossible ($3$ is odd), and for $N=100$, $50!$ is missing.
However, this turned out not to be correct - indeed, for $N=8$ we have solutions of $3!$ and $4!$. So where did I go wrong in my proof, or what did I leave out?
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You are only showing the case $N/2$ is even, but missing when it is odd.
Alternative to your method, the shortcut is:
$$1!\cdot 2!\cdot 3!\cdot 4!\cdots (N-1)!\cdot N!=(1!)^2\cdot 2\cdot (3!)^2\cdot 4\cdots ((N-1)!)^2\cdot N=(1!3!\cdots(N-1)!)^2\cdot 2^{N/2}\cdot \left(\frac{N}{2}\right)!$$
When $\frac{N}{2}$ is even, removing $\left(\frac{N}{2}\right)!$ is sufficient, though there can be other solutions when $N$ is a multiple of $4$, because when $N=4M$, then $(N/2)!=(2M)!$ is even, then it is an iteration of the problem.
When $\frac{N}{2}$ is odd, there is no solution, because neither $2^{N/2}$ nor $(N/2)!$ nor both are square. For example, $6!$, $10!$ do not have solutions.
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Two disjoint random events You roll twice with four-sided die in which the numbers one and two occur with probability $\frac{1}{3}$, and the numbers three and four each with probability $\frac{1}{6}$. Let X be the number of singles and Y the number of fours that occurred after two throws.
How do I create a table of probability function $p_{x,y}(x,y)=P\left \{X=x \wedge Y=y\right \}$?
This symbols at the end of this quations are little bit confusing to me.
$P(X=1)=\frac{1}{3}$, $P(X=2)=\frac{1}{3}$, $P(X=3)=\frac{1}{6}$ and $P(X=4)=\frac{1}{6}$.
$P(X,Y)=P(x)P(Y)=\frac{1}{3}\frac{1}{6}=\frac{1}{18}$
So do I just write Table:
$x_i$ $P(X=x_i)$
1 $2*1/3$
4 $2*1/6$
Because there are two throws or?
How do I calculate $P\left \{X+Y>0\right \}$?
Do I just add them?
$P\left \{X+Y>0\right \}=\frac{1}{3}+\frac{1}{6}$
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You cannot say $P(X,Y)=P(X)\cdot P(Y)$ as these events are not independent.
Further, you need to account for different orderings. For example,
$$P(X=1, Y=1)=2\cdot\left(\frac{1}{3}\cdot\frac{1}{6}\right)=\frac{1}{9}$$
since we can get a one and then a four or a four and then a one.
Similarly
$$P(X=0, Y=1)=2\cdot\left(\frac{1}{2}\cdot\frac{1}{6}\right)=\frac{1}{6}$$
since we must get a two or three and then a four, or a four and then a two or a three.
You should find that the joint probability mass function is
$$\begin{array}{|c|c|c|c|}
\hline
X/Y& 0 & 1 & 2 \\ \hline
0& \frac{1}{4}& \frac{1}{6}&\frac{1}{36}\\ \hline
1& \frac{1}{3}& \frac{1}{9}&0\\ \hline
2& \frac{1}{9}& 0&0\\ \hline
\end{array}$$
As a check, the probabilities in this table do indeed sum to one, as required.
|
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Complete sequence of sequences Who can continue (complete) the following sequence:
$$1,n-1,\frac{(n-2)(n-1)}{2},\frac{(n-4)(n-3)(n+1)}{6},\frac{(n-7)(n-4)(n-2)(n+3)}{24},\dots$$
This was emerging in the course of this question as crucial coefficients in the transformation of Fibonacci polynomials.
I am pretty sure I have seen this before in another context, but I cannot remember it exactly, I have the vague memory that it contains some more complicated structure than factorials, maybe something like rising factorials/Pochhammer-symbols.
Edit
The next sequence of this row is numerically $$a_6 = \{-4,12,-21,-24,-9,42,154,360,702,1232\dots\}$$
This is no 5$^{th}$ degree polynomial anymore and I am unable to detect the law.. If fact it is one, this can be seen from the fifth difference sequence which gets constant (when calculated correctly).
By interpolation and factorisation I finally got the law (see below).
Edit-2
Meanwhile I found an alternative way to generate these series, but however still got stuck at the 6$^{th}$ one.
They can be expressed as the coefficients of the series expansion of polynomial fractions $f_i(x)$ at $x=0$, the functions are in detail:
\begin{array}{c|c|c|c|c|c|c}
i& 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline
a_i&1 & n-1 & \frac{(n-2)(n-1)}{2}& \frac{(n-4)(n-3)(n+1)}{6}&\frac{(n-7)(n-4)(n-2)(n+3)}{24}&v.i.&v.i.\\ \hline
f_i(x)&\frac{1}{x-1}& \frac{1}{(x-1)^2}&\frac{1}{(x-1)^3}&\frac{(x-2)x}{(x-1)^4}&\frac{(x-2)(1+x(x-3))}{(x-1)^5}&?&\\
\end{array}
Edit-3
So, the expression for the next sequences are:
$$ a_6 = \frac{(n-5)(n-4)(n-1)(n^2-5n-54)}{120} $$
$$ a_7 = \frac{(n-6)(n-3)(n^4-12n^3-71n^2+642n+160)}{720} $$
I am still unable to guess a general law for $f_i(x)$ or $a_i$.
Edit-4
I will try to get some recursion relation now from the consideration of the difference series.
|
As you reference in another question, $$F_n(x) = \sum_{k=0}^n \binom{\frac{n+k-1}{2}}{k} x^k$$
If we consider $F_n(x+1)$ directly we get $$\begin{eqnarray}
F_n(x+1) & = & \sum_{k=0}^n \binom{\frac{n+k-1}{2}}{k} (x+1)^k \\
& = & \sum_{k=0}^n \binom{\frac{n+k-1}{2}}{k} \sum_{i=0}^k \binom{k}{i} x^i \\
& = & \sum_{i=0}^n x^i \sum_{k=i}^n \binom{\frac{n+k-1}{2}}{k} \binom{k}{i} \\
\end{eqnarray}$$
If we consider $F_n(x+1)$ as a sum of weighted $F_m(x)$ we get $$\begin{eqnarray}
F_n(x+1) & = & \sum_{m=0}^n \lambda_m \sum_{k=0}^m \binom{\frac{m+k-1}{2}}{k} x^k \\
& = & \sum_{k=0}^n x^k \sum_{m=k}^n \binom{\frac{m+k-1}{2}}{k} \lambda_m \\
\end{eqnarray}$$
By equating coefficients in $x^i$ we get
$$\sum_{m=i}^n \binom{\frac{m+i-1}{2}}{i} \lambda_m = \sum_{k=i}^n \binom{\frac{n+k-1}{2}}{k} \binom{k}{i}$$
On the left-hand side, discard the term $m=i$ (since $2i-1$ is odd) and extract the term $m=i+1$ from the sum, and the sum can be passed to the right:
$$\lambda_{i+1} = \left(\sum_{k=i}^n \binom{\frac{n+k-1}{2}}{k} \binom{k}{i}\right) - \sum_{m=i+2}^n \binom{\frac{m+i-1}{2}}{i} \lambda_m$$
And voilà: one recurrence.
NB your $a_j$ is $\lambda_{n-j+1}$, so
$$a_j = \left(\sum_{k=n-j}^n \binom{\frac{n+k-1}{2}}{k} \binom{k}{n-j}\right) - \sum_{m=n-j+2}^n \binom{\frac{n+m-j-1}{2}}{n-j} a_{n-m+1}$$
and it makes sense to reindex at least the second sum as
$$a_j = \left(\sum_{k=n-j}^n \binom{\frac{n+k-1}{2}}{k} \binom{k}{n-j}\right) - \sum_{m=1}^{j-1} \binom{n-\frac{m+j}{2}}{n-j} a_{m}$$
|
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Solve the Goursat problem $xy^3u_{xx} - x^3yu_{yy} -y^3u_x + x^3u_y=0$
Solve the following Goursat problem
$xy^3u_{xx} - x^3yu_{yy} -y^3u_x + x^3u_y=0,$
$u(x,y)=f(x) \; \; on \; \; y^2+x^2=16 \; \; for \; \; 0 \leq x \leq 4
> $-- eq 1
$u(x,y)=g(y) \; \; on \; \; x=0 \; \; for \; \; 0 \leq y \leq4$ -- eq
2
$f(0) = g(4)$
My attempt :
Using $\alpha = y^2 - x^2 \; \; and \; \; \beta = y^2 +x^2 $ I've reduced the given problem to
$u_{\alpha \beta} = 0 \Rightarrow u= \phi(\alpha) + \gamma(\beta)$
$ \Rightarrow u(x,y) = \phi(y^2-x^2) + \gamma(y^2+x^2)$ - eq 3
Using 1 $\Rightarrow f(x) = \phi(16 - 2x^2) + \gamma(16) \Rightarrow \phi(x)= f(\sqrt{\frac{16-x}{2}}) - \gamma(16)$-- eq 4
Using 2 $\Rightarrow g(y) = \phi(y^2) + \gamma(y^2) \Rightarrow \gamma(y) = g(\sqrt{y}) - \phi(y) = g(\sqrt(y)) - f(\sqrt{\frac{16-x}{2}}) + \gamma(16)$ -- eq 5
$f(0)= \phi(16) + \gamma(16) = g(4)$ -- eq 6
using 4 and 5
$u(x,y)= \phi(y^2-x^2) + \gamma(y^2-x^2) = f(\sqrt{\frac{16-y^2+x^2}{2}}) - \gamma(16) + g(\sqrt{y^2+x^2}) - f(\sqrt{\frac{16-x^2-y^2}{2}}) + \gamma(16)$
$\Rightarrow u(x,y)= f(\sqrt{\frac{16-y^2+x^2}{2}}) - f(\sqrt{\frac{16-x^2-y^2}{2}}) + g(\sqrt{y^2+x^2})$
Is this correct ?
I didn't get to use the given condition f(0)=g(4) anywhere.
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$$xy^3u_{xx} - x^3yu_{yy} -y^3u_x + x^3u_y=0$$
HINT :
We observe that the change of $x$ into $-x$ doesn't change the equation, as well as the change of $y$ into $-y$. This draw us to change of variables :
$\begin{cases}X=x^2\\Y=y^2\end{cases} \quad;\quad u_x=2xu_X\quad;\quad u_y=2yu_Y$
$u_{xx}=2u_X+4x^2u_{XX}\quad;\quad u_{yy}=2u_Y+4y^2u_{YY}$
$$xy^3(2u_X+4x^2u_{XX}) - x^3y(2u_Y+4y^2u_{YY}) -y^3(2xu_X) + x^3(2yu_Y)=0$$
$$Y(2u_X+4Xu_{XX}) - X(2u_Y+4Yu_{YY}) -2Yu_X + 2Xu_Y=0$$
$$u_{XX}-u_{YY}=0$$
The well-known general solution is :
$$u(X,Y)=F(X+Y)+G(X-Y)$$
with arbitrary functions $F$ and $G$.
$$u(x,y)=F(x^2+y^2)+G(x^2-y^2)$$
Then, determine the functions $F$ and $G$ according to the boundary conditions.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2732869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
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