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The largest integer dividing these $(n+19)(n+18)(n+17)(n+16)$
The largest integer that divides $(n+19)(n+18)(n+17)(n+16) ~ \forall ~ n \in \mathbb{N}? $ is?
It struck me that we should write this as:
$\dbinom{n+19}{4}\times 4!$
which is always divisible by $4!$. Now I am confused as to how do I take it from here? Because the question asks for largest positive integer.
Edit:
Is there no alternative to brute force method i.e. checking for few values then assuming the event would never occur?
|
Let $f(n) = (n + 16)(n + 17)(n + 18)(n + 19)$.
Then we have that
$$
\gcd(f(1), f(4), f(5)) = \gcd(17 \cdot 18 \cdot 19 \cdot 20, 20 \cdot 21 \cdot 22 \cdot 23, 21 \cdot 22 \cdot 23 \cdot 24) = \gcd(2^3 \cdot 3^2 \cdot 5 \cdot 17 \cdot 19, 2^3 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 23, 2^4 \cdot 3^2
\cdot 7 \cdot 11 \cdot 23) = 2^3 \cdot 3 = 24.
$$
Thus any number that divides $f(n)$ for all $n$ can be at most $24$. You've already shown that $24$ works, so $24$ is the largest such number.
Edit: Why did I consider $n = 1$, $n = 4$, and $n = 5$ in particular?
I took $n = 1$ because it is the smallest value that $n$ is allowed to be, and I'd prefer to work with fairly small numbers.
We now continue with the conjecture that the answer should be $24$. The number $f(1)$ has $17$ and $19$ as factors, so to rule out that these factors will divide $f(n)$ for all $n$, we need to find a $n$ such that $f(n)$ is not divisible by $17$ or $19$. If $n$ is less than $4$, then we can see that one of the numebrs that we are multiplying by to get $f(n)$ is going to be $19$, (since the smallest of the four numbers is $n + 16$, we need $n + 16 > 19$), we see that any $n$ such that $f(n)$ is not divisible by $19$ must be at least $5$. In accordance with our preference for using small numbers, we consider the value of $f(5)$. The other reason that we would like to use a number that is at least $5$ is that otherwise $20$ will divide $f(n)$, and we want to rule out the possibility of $5$ dividing $f(n)$ for all $n$.
The problem we now have is that both $f(1)$ and $f(5)$ are divisible by $9$, so we need to find an $n$ such that $f(n)$ is divisible by $3$, but not by $9$. Now if $(n + 16)$ is divisible by $3$, then so is $(n + 19)$, and so to avoid having a factor of $9$, we need to consider an $n$ such that $(n + 16)$ is not divisible by $3$. This rules out $n = 2$. It turns out that $n = 3$ works, but I overlooked this for some reason and went with $n = 4$ instead.
|
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|
Solve the system of linear equations for all values of a When solving the following matrix for $a$:
$$ \left[
\begin{array}{ccc|c}
x&2y&az&3\\
(2a-4)x&(2-a)y&(-2a+4)z&0\\
4x&4y&2az&8
\end{array}
\right]
$$
When do I put restrictions on which values $a$ can take? Say I divide the second row by $(2-a)$, does this mean that $a$ cannot be 2, or can I only say something like that when the entire matrix is in row echelon form, and I would be left with something like:
$$ \left[
\begin{array}{ccc|c}
1&0&0&\frac{2a}{a-2}\\
0&1&0&\frac{4-5a}{a-2}\\
0&0&1&\frac{a^2}{a-2}
\end{array}
\right]
$$
|
I didn't check the detail of calculation but, if it is correct, the solution you have found is valid for $a\neq 2$ (i.e. the system has exactly a solution $\forall a\neq2$).
Now you need also to verify the solution/s for the case $a=2$ in the original system, that is
$$\left[
\begin{array}{ccc|c}
1&2&2&3\\
4&4&4&8
\end{array}
\right]\implies \left[
\begin{array}{ccc|c}
1&2&2&3\\
1&1&1&2
\end{array}
\right]\implies \left[
\begin{array}{ccc|c}
1&2&2&3\\
0&1&1&1
\end{array}
\right]\implies\left[
\begin{array}{ccc|c}
1&0&0&1\\
0&1&1&1
\end{array}
\right]$$
thus for a=2 the system has infinitely many solutions
*
*$x=1$
*$y+z=1$
|
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|
Show that $\frac{1}{3}<\int_0^1\frac{1}{1+x+x^2} \, dx <\frac{\pi}{4}$ Show that $$\frac{1}{3}<\int_0^1\frac{1}{1+x+x^2}\,dx <\frac{\pi}{4}$$
I want to use if $f<g<h$ then $\int f<\int g<\int h$ formula for Riemann integration.
$1+x^2<1+x+x^2$ and it will give RHS as $$\frac{1}{1+x+x^2}<\frac{1}{1+x^2}$$
How to choose function $f$ and $h.$
|
Note that for $x\ge0$, $1+x\le x^2+x+1\le(x+1)^2$.
Hence, we have
$$\frac13<\frac12=\int_0^1 \frac{1}{(1+x)^2}\,dx\le \int_0^1 \frac{1}{x^2+x+1}\,dx\le \int_0^1 \frac{1}{1+x}\,dx=\log(2)<\frac\pi4$$
which provide tighter bounds then those that were requested.
|
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|
What kind of matrix is this and why does this happen? So I was studying Markov chains and I came across this matrix \begin{align*}P=\left( \begin{array}{ccccc}
0 & \frac{1}{4} & \frac{3}{4} & 0 & 0\\
\frac{1}{4} & 0 & 0 & \frac{1}{4} & \frac{1}{2}\\
\frac{1}{2} & 0 & 0 & \frac{1}{4}& \frac{1}{4}\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
\end{array} \right).\end{align*}
I noticed (by brute force) that \begin{align*}P^2=\left( \begin{array}{ccccc}
\frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
\frac{3}{8} & 0 & 0 & \frac{1}{4}& \frac{1}{2}\\
\frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
\end{array} \right),\end{align*}
and \begin{align*}P^3=\left( \begin{array}{ccccc}
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
\frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
\frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
\end{array} \right).\end{align*}
In fact; using a computer I found that every even power takes the form of the $P^2$ matrix and every odd power takes the form of the $P^3$ matrix.
I just wanted to know why that oscillation occurs? Is there a special name for the kind of matrix that $P$ is for it to exhibit that kind of behaviour?
|
The entries are all between 0 and 1 and in each row the entries add up to 1. They are called stochastic matrices. They have 1 as an eigen value with $(1,1,\ldots,1)^T$ as corresponding eigenvector. This property is inherited by all the powers of that matrix.
|
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|
Given the function $f(x)=\frac {1}{\sqrt[3] {1-x^3}}$ find $\underbrace {f(f(\cdots f(19)\cdots))}_{95}$ Given the function $$f(x)=\frac {1}{\sqrt[3] {1-x^3}}$$ find $$\underbrace {f(f(\cdots f(19)\cdots))}_{95}$$
My try:
Define $$f^n(x)=\underbrace {f(f(\cdots f(x)\cdots))}_{n}$$
We see that $$f(x)=\frac {1}{\sqrt[3] {1-x^3}}$$
$$f^2(x)=\frac {\sqrt[3] {1-x^3}}{-x}$$
$$f^3(x)=x$$
And again $$f^4(x)=\frac {1}{\sqrt[3] {1-x^3}}$$
Hence $$f^{3k+2}(x)=\frac {\sqrt[3] {1-x^3}}{-x}$$
Substituting $k=31$ and $x=19$ I can find $$\underbrace {f(f(\cdots f(19)\cdots))}_{95}$$
So am I going on the right path.
|
As you have already said there are only 3 interactions, then it goes back to the original function:
\begin{align}
f_0 = \underbrace{f(x)}_1&=\frac{1}{\sqrt[3] {1-x^3}} = \left(\frac{1}{1-x^3}\right)^{\frac{1}{3}}\\
f_1 = \underbrace{f(f(x))}_2&= \left(\frac{1}{1-\left(\frac{1}{1-x^3}\right)}\right)^{\frac{1}{3}} = \left(1 - \frac1{x^3}\right)^{\frac{1}{3}}\\
f_2 =\underbrace{f(f(f(x)))}_3 &= \left(\frac{1}{1-\left(1 - \frac1{x^3}\right)}\right)^{\frac{1}{3}} = x
\end{align}
If you count from $0$, so it's the $95$-th interaction equals $f_{94}$. It follows $94 \pmod 3 = 1$,which implies $f_{94} = f_1$:
$$\underbrace{f(f(\cdots f(19)\cdots))}_{95} = f_{94} = f_1 =\left(1 - \frac1{19^3}\right)^{\frac{1}{3}} $$
|
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|
Algebraic Proof on equivalence $$\frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-c)(b-a)}+\frac{c^3}{(c-a)(c-b)}\equiv{a+b+c}$$
Demonstrate the identities above considering a, b & c real numbers and distinct from each other.
I'm stucked on this problem. Take a look on what I did:
$x=(a-b)\therefore -x=(b-a)$; $y=(a-c);-y=(c-a)$; $z=(b-c);-z=(c-b)$.
___________________________________________________________________$$\frac{a^3}{xy}+\frac{b^3}{(-x)z}+\frac{c^3}{(-y)(-z)}$$
$$\frac{a^3z-b^3y+c^3x}{xyz}$$
$$\frac{a^3(b-c)+b^3(a-c)+c^3(a-b)}{xyz}$$
$$\frac{a^3(b-c)+b^3(\color{red}{-b}+a-c\color{red}{+b})+c^3(a-b)}{xyz}$$
$$\frac{a^3b+ab^3-a^3c+ac^3-b^3c-bc^3}{xyz}$$
$$\frac{ab(a^2+b^2)-ac(a^2-c^2)-bc(b^2+c^2)}{xyz}$$
after this step I was thinking about doing the same technique on the second term, $-ac(\color{red}{b^2}+a^2-c^2\color{red}{-b^2})$, but after doing some more algebraic factorization I got stucked, could you help me to demonstrate?
|
Assuming $b\neq c$ the function
$$ f(z) = \frac{z^3}{(z-b)(z-c)}+\frac{b^3}{(b-c)(b-z)}+\frac{c^3}{(c-z)(c-b)} $$
is holomorphic over $\mathbb{C}\setminus\{b,c\}$. In the worst scenario $b$ and/or $c$ are simple poles, but $\operatorname*{Res}_{z=c}f(z)=\operatorname*{Res}_{z=b}f(z)=0$ imply that $f$ is entire (once its removable discontinuities are removed). Given that, by inspecting the definition above it is clear that $f(z)$ is a polynomial with degree $\leq 1$, and actually a linear polynomial of the form $f(z)=z+q$. Since
$$ f(0) = \frac{b^2}{b-c}+\frac{c^2}{c-b} = \frac{b^2-c^2}{b-c} = b+c $$
we have $f(z)=z+b+c$ for any $z\in\mathbb{C}\setminus\{b,c\}$, in particular for $z=a$.
|
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|
On $3+\sqrt{11+\sqrt{11+\sqrt{11+\sqrt{11+\dots}}}}=\phi^4$ and friends Let $\phi$ be the golden ratio. We know it has a beautiful infinite nested radical,
$$\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}=\phi$$
However, it is also the case that,
$$3+\sqrt{11+\sqrt{11+\sqrt{11+\sqrt{11+\dots}}}}=\phi^4$$
$$5+\sqrt{31+\sqrt{31+\sqrt{31+\sqrt{31+\dots}}}}=\phi^5$$
$$\tfrac{17}2+\sqrt{\tfrac{319}4+\sqrt{\tfrac{319}4+\sqrt{\tfrac{319}4+\sqrt{\tfrac{319}4+\dots}}}}=\phi^6$$
Q: How do we show that, in general
$$a_n+\sqrt{b_n+\sqrt{b_n+\sqrt{b_n+\sqrt{b_n+\dots}}}}=\phi^n$$
where,
$$a_n = \frac{L_n-1}2,\quad b_n = \frac{5F_n^2-1}4$$
with Lucas numbers $L_n$ and Fibonacci numbers $F_n$?
|
If we plug in the given formulas we get the famous formula. We note:
$$\sqrt{b_n+x}=x \Rightarrow x=\frac{1+\sqrt{1+4b_n}}{2}=\frac{1+F_n\sqrt{5}}{2};$$
Hence:
$$\frac{L_n-1}{2}+\frac{1+F_n\sqrt{5}}{2}=\frac{L_n+F_n\sqrt{5}}{2}=\phi^n.$$
This is a famous formula that relates the two sequences. See: https://en.m.wikipedia.org/wiki/Lucas_number
|
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|
Complex numbers $\left(\frac{1+i}{1-i}\right)^k = 1$ what is $k$? The smallest possible integer $k$ for which $\left(\frac{1+i}{1-i}\right)^k = 1$ is?
I tried solving this, but my answer doesn't match the given answer. Correct me if I'm wrong at some place
My solution:
\begin{align}
\left(\frac{1+i}{1-i}.\frac{1+i}{1+i}\right)^k&= 1\\
\left(\frac{1+2i+i^2}{1-i^2}\right)^k&= 1\\
\left(\frac{1+2i-1}{1-(-1)}\right)^k&= 1\\
\left(\frac{2i}{2}\right)^k&= 1\\
i^k&= 1\\
i^4&= 1\\
\end{align}
EDIT: The question is part of the multiple choice section and the answer is 2.
Other options include: 4, 8, 16
|
The result $k=4$ is correct. The given multiple choice answer $k=2$ is wrong, since
\begin{align*}
\left(\frac{1+i}{1-i}\right)^2=i^2=-1\ne 1
\end{align*}
|
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|
If $\gcd(m,15)=\gcd(n,15)=1,$ show that $15\mid (m^4+n^4)$ or $15\mid (m^4-n^4)$. If $\gcd(m,15)=\gcd(n,15)=1$ show $15\mid (m^4+n^4)$ or $15\mid (m^4-n^4)$.
This is what I have so far but I'm stuck on essentially the last step.
Proof: Assume $\gcd(m,15)=\gcd(n,15)=1$. By the Euler's Phi Function: If $\gcd(a,m)=1$ then $a^{\phi (m)}\equiv 1 \pmod m$.
$\phi(15)=\phi(5)\phi(3)=(5-1)(3-1)=8$
Hence, $m^8\equiv 1 \pmod{15}$ and $n^8\equiv 1 \pmod{15}$.
Then, $m^8\equiv 1 \pmod{15}\iff 15\mid m^8-1$. Similarly $15\mid n^8-1$.
Divisibility property sates if $a\mid b$ and $a\mid c$ then $a\mid b\pm c$.
So $15\mid (m^8-1)-(n^8-1) \Rightarrow 15\mid m^8-n^8$
So I believe I'm right up to this point, please correct if I'm not. I know that $m^8-n^8$ factors to $(m^4+n^4)(m^4-n^4)$ however since $15$ is not a prime number I can't assume $15\mid m^4+n^4$ or $15\mid m^4-n^4$.
Please help on these last few steps. Thanks.
|
Since none of the existing answers is built-up on OP's work, I posted another one.
OP has proved $15\mid m^8-n^8 = (m^4-n^4)(m^4+n^4)$.
Since $15$ is a composite number, to apply Euclid's Lemma, we have to do so in two steps (on prime number $3$ and $5$).
Note that (without Fermat's Little Theorem)
*
*$a^2 \equiv 0, 1 \pmod 3$, so $a^4 \equiv 0,1 \pmod 3$.
*
*If $3 \mid m^4 + n^4$, it's easy to see that the only possibility is that $3 \mid m,n$, which contradicts $\gcd(m,15) = \gcd(n,15) = 1$
*Use Euclid's Lemma to show that $3 \mid m^4 - n^4$
*$a^2 \equiv 0, 1, 4 \pmod 5$, so $a^4 \equiv 0,1 \pmod 5$, and $m^4 + n^4 \equiv 0,1,2 \pmod 5$.
*
*If $5 \mid m^4 + n^4$, its again easy to see that the only possibility is $5 \mid m,n$, which contradicts $\gcd(m,15) = \gcd(n,15) = 1$
*Use Euclid's Lemma to show that $5 \mid m^4 - n^4$
Since $\gcd(3,5) = 1$, $15 \mid m^4 - n^4$.
|
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|
solving $x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$ I have to solve $x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$
Dividing by $dx$ we have
$x + xy^2 + yy' + yy'x^2=0$
From where,
$$\frac{yy'}{1+y^2}+\frac{x}{1+x^2}=\frac{y\,dy}{1+y^2}+\frac{x\,dx}{1+x^2}=\\ =\frac{d(y^2+1)}{1+y^2}+\frac{d(x^2+1)}{1+x^2}= \frac{1}{2}d\ln(1+y^2)+\frac{1}{2} d\ln(1+x^2)=\frac{1}{2}d\ln(1+y^2)(1+x^2)=0$$
Let $c=(1+y^2)(1+x^2)$, so our equation becomes:
$$
d\ln c=0
$$
So what should I do here, should I integrate, or should I divide by $dx$?
If I divide by dx I get the expression $2x+2yy'+2xy^2+2x^2yy'=0$ which has $x$, $y$ and $y'$ and doesn't help me get anywhere.
Thanks in advance.
|
$$x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$$
$$(1+x^2)y\,dy =-(1+y^2)x\,dx $$
If $x^2,y^2\ne-1$, above equation can be written as (after multiplying both side by $2$):
$$\frac{2y}{1+y^2}\,dy =-\frac{2x}{1+x^2}\,dx $$
$$\int \frac{2y}{1+y^2}\,dy =-\int \frac{2x}{1+x^2}\,dx $$
$$\ln (1+y^2)=-\ln (1+x^2)+\ln c=\ln \frac {c}{1+x^2}$$
where $c$ is a constant. Hence,
$$1+y^2= \frac {c}{1+x^2}$$
$$y^2= \frac {c}{1+x^2}-1$$ Or:
$$y^2+x^2+y^2x^2= C$$ where $C$ is also a constant.
|
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|
Use Ramanujan’s method to denest $\sqrt[3]{7\sqrt[3]{20}-1}$ and $\sqrt[3]{7\sqrt[3]{20}-19}$ A possible way to denest $(2^{1/3}-1)^{1/3}$ is by first setting $x=\sqrt[3]{2}$ so$$x^3-1=1\implies x-1=\frac 1{1+x+x^2}=\frac 3{1+3x+3x^2+x^3}$$Multiply both sides by $9$ so$$9(x-1)=\left(\frac 3{1+x}\right)^3\implies\sqrt[3]{9(\sqrt[3]2-1)}=1-\sqrt[3]2+\sqrt[3]4$$However, I'm having trouble adapting this method to denest
both
$$\begin{align*}\sqrt[3]{7\sqrt[3]{20}-1} & =\sqrt[3]{\frac {16}9}-\sqrt[3]{\frac 59}+\sqrt[3]{\frac {100}9}\tag1\\\sqrt[3]{7\sqrt[3]{20}-19} & =\sqrt[3]{\frac 49}-\sqrt[3]{\frac {80}9}+\sqrt[3]{\frac {25}9}\tag2\end{align*}$$
I've only started on (1) so far. Here's my work.
My work: Let $x^3=6860$ so $x^3-1=19^3$. Hence$$x-1=\frac {19^3}{1+x+x^2}=\frac {19^3\cdot3}{3+3x+3x^2}$$However, I'm not quite sure what to do after that. I don't see an easy relationship between three and $x$.
Perhaps you guys can help?
|
$x^3 +a x^2 +bx +c $ is called a Ramanujan cubic polynomial if
$$b +a c^{1/3} +3c^{2/3}=0\tag1 $$
Its three roots then satisfy
$$\sqrt[3]{x_1}+ \sqrt[3]{x_2 }+ \sqrt[3]{x_3 }=\sqrt[3]{ 3\sqrt[3]{9c-ab }-a-6\sqrt[3]c} \tag2$$
Note that (2) contains nested cubic radicals of the form $\sqrt[3]{\sqrt[3] A -B}$ and it can be used to denest such radicals. Take $\sqrt[3]{7\sqrt[3]{20}-1}$ for example, we set
$$a+6c^{1/3}=1,\>\>\>\>\>27(9c-ab)=7^3\cdot 20$$
Along with (1), we solve for $a=-\frac{37}3$, $b=\frac{340}{27}$, $c=\frac{8000}{729}$ and the corresponding polynomial
$$x^3 -\frac{37}3x^2+\frac{340}{27}x + \frac{8000}{729}
= (x+\frac59)(x-\frac{16}9)(x-\frac{100}9)$$
from which the Ramanujan’s property (2) yields the denesting
$$
\sqrt[3]{7\sqrt[3]{20}-1} =\sqrt[3]{\frac {16}9}-\sqrt[3]{\frac 59}+\sqrt[3]{\frac {100}9}
$$
Above procedure can be used to denest $\sqrt[3]{7\sqrt[3]{20}-19}$ as well. Note that the method is effective without prior knowledge of results.
|
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Help with a trigonometry problem. I have encountered some problems to solve the left side to the right.
$$ \cos^ 2x \sin x = \frac{\sin 3x + \sin x}{4}$$
I trying to solve a differential equation on the form $$ y'' + y = \cos^ 2x * \sin x$$ and need to rewrite it to $$ y'' + y = \frac{\sin 3x + \sin x}{4}$$
Have tried with different combinations but not really got it to $$\frac{\sin 3x + \sin x}{4}$$
I have also tried to start backwards, but it seems not to be the method I would have chosen if I started from the left.
\begin{align}
\frac{\sin 3x+\sin x}{4}
& =\frac{\sin x+\sin 3x}{4} \\
& =\frac{\sin x+\sin \left( 2x+x \right)}{4} \\
& =\frac{\sin x+\left( \sin 2x\cos x+\sin x\cos 2x \right)}{4} \\
& =\frac{\sin x+\left( \left( 2\sin x\cos x \right)\cos x+\sin x\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right) \right)}{4} \\
& =\frac{\sin x+\left( 2\sin x{{\cos }^{2}}x+\sin x\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right) \right)}{4} \\
& =\frac{\sin x+\left( 2\sin x\left( 1-{{\sin }^{2}}x \right)+\sin x\left( 1-2{{\sin }^{2}}x \right) \right)}{4} \\
& =\frac{\sin x+\left( (2\sin x-2{{\sin }^{3}}x)+(\sin x-2{{\sin }^{3}}x) \right)}{4} \\
& =\frac{\sin x+\left( 3\sin x-4{{\sin }^{3}}x \right)}{4} \\
& =\frac{4\sin x-4{{\sin }^{3}}x}{4} \\
& =\frac{4\sin x(1-{{\sin }^{2}}x)}{4} \\
& =\sin x{{\cos }^{2}}x \\
\end{align}
Would be grateful if you could give out some kind of guidance.
|
Consider
\begin{align}
\cos^2x\sin x&=(1-\sin^2x)\sin x\\
&=\sin x-\sin^3 x\\
&=\sin x-\frac14(3\sin x-\sin 3x) &(*)\\
&=\frac{\sin3x-\sin x}{4}
\end{align}
For $(*)$,
\begin{align}
\sin 3x&=\sin(x+2x)\\
&=\sin x\cos 2x+\sin 2x\cos x \\
&= \sin x (1-2\sin^2x)+2\sin x\cos^2x\\
&= \sin x-2\sin^3x+2\sin x(1-\sin^2x)\\
&=3\sin x-4\sin^3x
\end{align}
Hence,
$$\sin^3x=\frac14(3\sin x-\sin3x)$$
Basically, you will have to make some sort of detour around $\sin 3x$ and $\sin^3x$ no matter how you approach this problem...
|
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|
$ \lim_{x \to \infty} [\frac{x^2+1}{x+1}-ax-b]=0 \ $ then show that $ \ a=1, \ b=-1 \ $ $ \lim_{x \to \infty} [\frac{x^2+1}{x+1}-ax-b]=0 \ $ then show that $ \ a=1, \ b=-1 \ $
Answer:
$ \lim_{x \to \infty} [\frac{x^2+1}{x+1}-ax-b]=0 \\ \Rightarrow \lim_{x \to \infty} [\frac{x^2+1-ax^2-ax-bx-b}{x+1}]=0 \\ \Rightarrow \lim_{x \to \infty} \frac{2x-2ax-a-b}{1}=0 \\ \Rightarrow 2x-2ax-a-b=0 \ \ (?) $
Comparing both sides , we get
$ 2-2a=0 \\ a+b=0 \ $
Solving , we get
$ a=1 , \ b=-1 \ $
But I am not sure about the above line where question mark is there.
Can you help me?
|
Note that
$$\frac{x^2+1}{x+1}-ax-b=\frac{x^2+1-ax^2-bx-ax-b}{x+1}=\frac{x^2(1-a)-x(a+b)-b+1}{x+1}$$
and in order to have limi zero we need
*
*$(1-a)=0 \implies a=1$
*$(a+b)=0\implies b=-1$
indeed
$$\frac{x^2(1-1)-x(1-1)-(-1)+1}{x+1}=\frac{2}{x+1}\to 0$$
|
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|
Use Triangle Inequality to Solve Inequality $\left|x+\frac{1}{2}\right| > \frac{\sqrt{5}}{2}$ So this is a really simple question, but I can't seem to work it out. I started with the equation
$$
x^2 + x + 1 > 2
$$
and, by completing the square and taking the square root, was able to simplify it to
$$
\left|x+\frac{1}{2}\right| > \frac{\sqrt{5}}{2} \tag{1}
$$
Next, I want to use the triangle inequality to solve for $x$. I noted that
$$
|a + b| \le |a| + |b| \iff |a| + |b| \ge |a + b|
$$
Now, letting $a = x$ and $b = \frac{1}{2}$, I got
$$
|x| + \frac{1}{2} \ge \left|x + \frac{1}{2}\right|
$$
Thus, combining this with (1), I reached
$$
|x| + \frac{1}{2} > \frac{\sqrt{5}}{2}, \tag{2}
$$
which is easily solvable:
$$
x < \frac{1-\sqrt{5}}{2} \text{ or } x> \frac{\sqrt{5} - 1}{2}
$$
By plugging each step into Wolfram Alpha, I've determined that while (1) is correct, (2) produces a different solution, which means the error must be somewhere in my use of the triangle inequality. The solution that both Wolfram Alpha and my textbook give is
$$
x < \frac{-1-\sqrt{5}}{2} \text{ or } x> \frac{\sqrt{5} - 1}{2},
$$
the difference being the negative in front of the $1$ in the first term.
|
You shouldn't use the triangle inequality here. If $|x+\frac{1}{2}|>\frac{\sqrt{5}}{2}$, then you are really considering two inequalities:
$$
x+\frac{1}{2}>\frac{\sqrt{5}}{2} \text{ when } x+1/2\geq 0
$$
$$
-\left(x+\frac{1}{2}\right)>\frac{\sqrt{5}}{2} \text{ when } x+1/2< 0
$$
These two inequalities then read $x>\frac{\sqrt{5}-1}{2}$ and $x<\frac{-\sqrt{5}-1}{2}$, as the other two inequalities become redundant.
|
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|
Find the value of the constant c such that $\sum_{n=2}^\infty(1+c)^{-n} = 2$
Find the value of the constant c such that $$\sum_{n=2}^\infty(1+c)^{-n} = 2 $$
For this question,I'm not sure if I'm doing it right. If I am doing it right, I'm not sure how to get further. Here is what I have so far. Can anyone please help me out?
Comparing the series to the conventional form $a + a r + a r^2 + \cdots = a/(1-r)$, we have
$$r = 1+c \qquad\text{and}\qquad a = (1+c)^{-2} \qquad\text{and}\qquad 2 = \frac{a}{1-r}$$
So ...
$$\begin{align}
2 &= \frac{(1+c)^{-2}}{1-(1+c)} \\
2 &= \frac{(1+c)^{-2}}{-c} \\
-2c &= (1+c)^{-2} \\
-2c &= \frac{1}{(1+c)^2} \\
-2c(1+c)^2 &= 1 \\
-2c(1+2c+c^2) &= 1
\end{align}$$
|
HINT: you must solve the equation $$\frac{1}{c(1+c)}=2$$ for $c$
this comes from $$\sum_{i=2}^n(1+c)^{-i}=\frac{(c+1)^{-n-1} \left((c+1)^n-c-1\right)}{c}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to quickly find the $x^{24}$ term in this expansion? Is there a swift way to find the $x^{24}$ coefficient in the expansion of
$$ \left(1-x^6\right)^{-2} \left(1-x^3\right)^{-1} \left(1-x\right)^{-1} $$
The general term of each bracket is $(r+1)x^{6r}$, $x^{3r}$ and $x^{r}$ respectively.
|
The given rational function can be written as
$$(1+x^6+x^{12}+\ldots)(1+x^6+x^{12}+\ldots)(1+x^3+x^6+\ldots)(1+x^2+x^3+\ldots)\ .$$
The coefficient $c_{24}$ we are after therefore is the number of nonnegative integer solutions to
$$6k_1+6k_2+3k_3+k_4=24\ ,$$
or $6k_1+6k_2+3k_3\leq24$, which is the same as
$$2(k_1+k_2)+k_3\leq8\ .$$
If $k_1+k_2=:j\geq0$ each given value of $j$ ca be realized by $k_1$ and $k_2$ in $j+1$ ways, and then $k_3$ may take $8-2j+1$ different values. It follows that
$$c_{24}=\sum_{j=0}^4(j+1)\cdot(9-2j)=1\cdot9+2\cdot7+3\cdot5+4\cdot 3+5\cdot1=55\ .$$
|
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|
Show $Z_{ab}\simeq Z_a\oplus Z_b$ by elementary row and column operations Let $a,b\in\mathbb N$ be coprime. By manipulating with the matrix $\operatorname{diag}(a,b)$, prove that the cyclic group $Z_{ab}$ is isomorphic to the direct sum $Z_a\oplus Z_b$.
I guess I should obtain the matrix $\operatorname{diag}(1,ab)$ from the matrix $\operatorname{diag}(a,b)$ by elementary (integer) row and column operations (which can be done by a theorem on Smith's normal form), but I don't know how to do that since $a,b$ needn't be invertible, so I cannot multiply rows/columns by them.
|
You can certainly multiply rows and columns by $a$ and $b$, if all you do is add that multiple to another row or column. Less cryptically put:
$$
\det\begin{pmatrix}
1 & a \\ 0 & 1
\end{pmatrix} = 1
$$
so it is safe to multiply by it. First however, let $s,t\in\Bbb Z$ be so that $sa+tb=1$, which you can do because they are coprime. So, consider
$$
\begin{pmatrix} s & -t \\ b & a \end{pmatrix}
\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}
\begin{pmatrix} 1 & 0 \\ -1 & 1\end{pmatrix}
\begin{pmatrix} 1 & tb \\0 & 1 \end{pmatrix}
=\begin{pmatrix} sa & -tb \\ ba & ab \end{pmatrix}
\begin{pmatrix} 1 & 0 \\ -1 & 1\end{pmatrix}
\begin{pmatrix} 1 & tb \\0 & 1 \end{pmatrix}
=\begin{pmatrix} 1 & -tb \\ 0 & ab \end{pmatrix}
\begin{pmatrix} 1 & tb \\0 & 1 \end{pmatrix}
=\begin{pmatrix} 1 & 0 \\ 0 & ab \end{pmatrix}
$$
where we have multiplied from the left and from the right only by invertible matrices.
|
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|
Arrangement of definite integral in increasing order
If $\displaystyle I=\int^{\frac{\pi}{2}}_{0}\cos(\cos x)dx$ and
If $\displaystyle J=\int^{\frac{\pi}{2}}_{0}\sin(\cos x)dx$ and
If $\displaystyle K=\int^{\frac{\pi}{2}}_{0}\cos xdx$. Then Arrangement of $I, J,K$ in increasing order is
Try: Using $\sin x<x$ for all $x\in\bigg(0,\frac{\pi}{2}\bigg)$
So $\sin(\cos x)<\cos x$ . Means
If $$I=\int^{\frac{\pi}{2}}_{0}\sin(\cos x)dx<\int^{\frac{\pi}{2}}_{0}\cos xdx$$
Could sime help me to solve it , Thanks
|
Let $\displaystyle x=\frac{\pi}{2}-u$. Then
$$\int_\frac{\pi}{4}^\frac{\pi}{2}\cos(\cos(x))dx=\int_{\frac{\pi}{4}}^0\cos(\sin u)(-1)du=\int_0^\frac{\pi}{4}\cos(\sin(x))dx$$
$$\int_\frac{\pi}{4}^\frac{\pi}{2}\cos(x)dx=\int_{\frac{\pi}{4}}^0\sin u(-1)du=\int_0^\frac{\pi}{4}\sin(x)dx$$
For $\displaystyle x\in\left(0,\frac{\pi}{4}\right)$, $\sin(x)<x$ and hence $\cos(\sin(x))>\cos(x)$.
Let $\displaystyle f(x)=\frac{\pi}{2}-x-\cos(x)$. Then $f'(x)=-1+\sin(x)<0$ for $\displaystyle x\in\left(0,\frac{\pi}{4}\right)$. $f$ is strictly decreasing for $x\in\displaystyle x\in\left[0,\frac{\pi}{4}\right]$. So for $\displaystyle x\in\left(0,\frac{\pi}{4}\right)$, $\displaystyle f(x)>f\left(\frac{\pi}{4}\right)=\frac{\pi}{4}-\frac{1}{\sqrt{2}}>0$.
So, for $\displaystyle x\in\left(0,\frac{\pi}{4}\right)$, $\displaystyle \frac{\pi}{2}-x>\cos(x)$ and hence $\displaystyle \sin(x)=\cos\left(\frac{\pi}{2}-x\right)<\cos(\cos(x))$.
\begin{align*}
&\;\int_0^\frac{\pi}{2}\cos(\cos(x))dx-\int_0^\frac{\pi}{2}\cos(x)dx\\
=&\;\int_0^\frac{\pi}{4}[\cos(\cos(x))+\cos(\sin(x))-\cos(x)-\sin(x) ]dx\\
>&\;0
\end{align*}
For $\displaystyle x\in\left(0,\frac{\pi}{2}\right)$, $\displaystyle 0<\cos(x)<1<\frac{\pi}{2}$. So, $\sin(\cos(x))<\cos (x)$.
$$\int_0^\frac{\pi}{2}\sin(\cos(x))dx<\int_0^\frac{\pi}{2}\cos (x)dx$$
Therefore, $I>K>J$.
|
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|
Integration Problem: $\frac{1}{\sqrt{2\pi}}\int^{\infty}_0\sqrt{w}e^{-w-\frac{wx^2}{2}}dw$ I know that:$$\frac{1}{\sqrt{2\pi}}\int^{\infty}_0\sqrt{w}e^{-w-\frac{wx^2}{2}}dw=\frac{1}{{(x^2+2)}^{3/2}}$$
I've attempted to solve this with integration by parts $\int{fdg}=fg-\int{dfg}$
I let
$$\begin{array} ff=\sqrt{w} & dg=e^{-w-\frac{wx^2}{2}}\\df=\frac{1}{2\sqrt{w}}& g=\frac{-2e^{-w-\frac{wx^2}{2}}}{2+x^2}{}\end{array}$$
Which is:$$\frac{1}{2\pi}\bigr[\frac{-2\sqrt{w}e^{-w-\frac{wx^2}{2}}}{2+x^2}\Bigr|^{\infty}_{0}+\frac{1}{2+x^2}\int^{\infty}_0\frac{e^{-w-\frac{wx^2}{2}}}{\sqrt{w}}dw\bigr]=\frac{1}{2\pi}\bigr[\frac{1}{2+x^2}\int^{\infty}_0\frac{e^{-w-\frac{wx^2}{2}}}{\sqrt{w}}dw\bigr]$$
I have to integrate by parts again, this time I have:
$$\begin{array} ff=e^{-w-\frac{wx^2}{2}} & dg=\frac{1}{\sqrt{w}}\\df=\frac{(x^2+2)^2e^{-w-\frac{wx^2}{2}}}{2}& g=2\sqrt{w}\end{array}$$
Which is:
$$\frac{1}{\sqrt{2\pi}}\bigr[\frac{1}{x^2+2}\bigr(2\sqrt{w}e^{-w-\frac{wx^2}{2}}\Bigr|^{\infty}_{0}+(x^2+2)\int^{\infty}_{0}\sqrt{w}e^{-w-\frac{wx^2}{2}}dw\bigr)\bigr]=\frac{1}{\sqrt{2\pi}}\int^{\infty}_0\sqrt{w}e^{-w-\frac{wx^2}{2}}dw$$
So, I'm back to where I started. Can somebody spot my mistake?
|
Hint
you have nearly with $w=u^2 \implies dw=2udu$ this:
$$I=\int \sqrt w e^{-w}dw=2\int u^2e^{-u^2}du$$
$$I=2\int u\times ue^{-u^2}du=\int u\times (-e^{-u^2})'du$$
Integrate by part once...
$$\int_0^{\infty} u\times (-e^{-u^2})'du=(u\times (-e^{-u^2}))|_0^{\infty}-\int_0^{\infty} (-e^{-u^2})du$$
And use gaussian integral to get the value of the last integral
$$\int_0^{\infty} e^{-u^2}du=\frac {\sqrt {\pi}}2$$
|
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|
Evaluating trigonometric limit $\csc^2(2x) - \frac{1}{4x^2}$
$$\lim_{x\rightarrow 0} \left[ \csc^2(2x) - \frac{1}{4x^2} \right]$$
I've tried to use l'Hôpital's rules but still can't find the answer. Here's my approach:
$$
\begin{aligned}
&\lim_{x\rightarrow 0} \left[ \csc^2(2x) - \frac{1}{4x^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{1}{\sin^2(2x)} - \frac{1}{4x^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{4x^2-\sin^2(2x)}{4x^2(\sin^2(2x))} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{8x-4\sin(2x)\cos(2x)}{8x\sin^2(2x)+16x^2\sin(2x)\cos(2x)} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{8x-2\sin(4x)}{8x\sin^2(2x)+8x^2\sin(4x)} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{4x-\sin(4x)}{4x\sin^2(2x)+4x^2\sin(4x)} \right] \\
\end{aligned}
$$
Am I using the correct way? How to solve it correctly?
P. S. I tried to use calculator and it outputs one third $1/3$.
|
Based on DeepSea's hint, I've managed to solve by myself.
$$
\begin{aligned}
&\lim_{x\rightarrow 0} \left[ \csc^2(2x) - \frac{1}{4x^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{1}{\sin^2(2x)} - \frac{1}{4x^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{4x^2-\sin^2(2x)}{4x^2(\sin^2(2x))} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{4x^2-\sin^2(2x)}{4x^2\times 4x^2\left(\frac{\sin(2x)}{2x}\right)^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{4x^2-\sin^2(2x)}{4x^2\times 4x^2} \right] \times \lim_{x\rightarrow 0} \left[ \frac{1}{\left(\frac{\sin(2x)}{2x}\right)^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{4x^2-\sin^2(2x)}{16x^4} \right] \times 1 \\
=& \lim_{x\rightarrow 0} \left[ \frac{8x-2\sin(4x)}{64x^3} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{8-8\cos(4x)}{192x^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{1-\cos(4x)}{24x^2} \right] \\
=& \frac{1}{24} \times \lim_{x\rightarrow 0} \left[ \frac{1-\cos(4x)}{x^2} \right] \\
=& \frac{1}{24} \times \frac{1}{2} \times 4^2 \\
=& \frac{1}{3} \\
\end{aligned}
$$
|
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|
Integral part of sum of huge powers
Question: What is the integral part of the following expression?$$(a+\sqrt{b})^{2n}+(a-\sqrt{b})^{2n}$$
The question has specific values of $a=2,b=5$ and $2n=2016$.
I was able to simplify (or complexify) it to:$$\sum_{i=0}^n\binom{2n}{2i}a^{2i}b^{n-i}.$$
I think I need to use $$\binom{2n}{2i}=\binom{2n-1}{2i-1}+\binom{2n-1}{2i},$$
but because the powers of $a$ and $\sqrt{b}$ dont change I cant make a closed form.
Please help.
|
Define $c_0 = 2$, $c_1 = 4$, and $c_{n + 2} = 4c_{n + 1} + c_n$ for $n \geqslant 0$, then$$
c_n = (2 + \sqrt{5})^n + (2 - \sqrt{5})^n. \quad \forall n \in \mathbb{N}
$$
Note that$$
\begin{pmatrix}c_{n + 2}\\c_{n + 1}\end{pmatrix} = \begin{pmatrix}4 & 1\\ 1 & 0\end{pmatrix} \begin{pmatrix}c_{n + 1}\\c_n\end{pmatrix}.
$$
Denote $\displaystyle A = \begin{pmatrix}4 & 1 \\ 1 & 0\end{pmatrix}$, then$$
\begin{pmatrix}c_n\\c_{n - 1}\end{pmatrix} = A^{n - 1} \begin{pmatrix}c_1\\c_0\end{pmatrix}.
$$
(Tedious computation starts.) Because$$
2016 = 2^{10} + 2^9 + 2^8 + 2^7 + 2^6 + 2^5,
$$
and$$
A^2 = \begin{pmatrix}17 & 4 \\ 4 & 1\end{pmatrix}, A^{2^2} = \begin{pmatrix}305 & 72 \\ 72 & 17\end{pmatrix}, A^{2^3} = \begin{pmatrix}98209 & 23184 \\ 23184 & 5473\end{pmatrix}, \cdots
$$
then the explicit answer can be computed, but it is too hard to continue computing manually.
|
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|
Is it possible to use partitions of an odd integer to generate primes in a given interval? We start with the partition of $N=5$.
$$5$$
$$4+1$$
$$3+2$$
$$3+1+1$$
$$2+2+1$$
$$2+1+1+1$$
$$1+1+1+1+1$$
Then we form the sum of squares (no limit on the number of elements) to get:
$$4^2+1^2=17$$
$$3^2+2^2=13$$
$$3^2+1^2+1^2=11$$
$$2^2+2^2+1^2=9$$
$$2^2+1^2+1^2+1^2=7$$
$$1^2+1^2+1^2+1^2+1^2=5$$
In this simple case we see that all primes from $N=5$ to $N=17$ have been generated. This is not true in general. If we consider the partition of $N=7$ and calculate the sum of squares of individual partitions, we will see that the primes $23,31$ weren't generated. If we consider the partition of $N=9$, we will see that the primes $37,43,59,61$ are missing (but not the previous $23,31$).
Is it enough to consider only the partitions of two odd integers $N$ and $N+2$ to find all the primes between $N$ and $(N-1)^2 + 1$, assuming N is a prime? If $N$ is not a prime, the question will apply to the nearest prime above $N$ and $(N-1)^2+1$.
|
One counterexample is when $p=31$ and $q=853.$ No partition of $31$ or $33$ yields the prime $q=853.$
For $p\geq 7,$ the only values greater than $(p-2)^2$ you can get from partitions of $p$ are:
$$(p-2)^2+2,(p-2)^2+4,(p-1)^2+1$$
This means that $31$ cannot be gotten from a partition of $p=7$, for example.
(This is actually true for $p=5,$ too, which is the underlying reason why $15=(5-2)^2+6$ cannot be reached in your original example.)
If $p\equiv 1,13\pmod{15}$ then $(p-2)^2+2$ is divisible by $3$ and $(p-2)^2+4$ is divisible by $5$, in whch case, any prime between $(p-2)^2$ and $(p-1)^2$ would be a counterexample.
Thus, for $p=13,$ there is no way to partition $13$ to get any of the primes $127, 131, 137,139.$
Allowing also partitions of $p+2$ gives a bunch more values between $(p-2)^2$ and $(p-1)^2, but still a finite list of values.
If we allow partitions of $p$ and $p+2$, for $p\geq 15$, the only values we get strictly between $(p-2)^2$ and $(p-1)^2$ are:
$$(p-2)^2+2,\\(p-2)^2+4,\\(p-2)^2+6,\\(p-2)^2+8,\\(p-2)^2+10,\\(p-2)^2+16.$$
When $p=31,$ $(p-2)^2+12=853$ is not in this set.
If $p\equiv 1\pmod{3}$ and $(p-2)^2\equiv 1\pmod{\cdot 5\cdot 7\cdot 11\cdot 17}$ then all of the above are composite, and thus, if your conjecture were true, we'd have infinitely many $n=p-2$ with no primes between $n^2$ and $(n+1)^2.$ (At the moment is unknown if there exists any such $n$.)
More generally, for fixed $k,$ if you allow partitions of $p,p+2,\dots,p+2k,$ then you can still get, for $p\geq 2k^2+6k+7,$ finitely many values in the range $(p-2)^2$ and $(p-1)^2.$ If these covered all primes, you could find infinitely many $p$ such that there is no prime in that range. (That doesn't mean that it is impossible to find such $k,$ only that finding such $k$ would solve an unsolved problem with a big set of counterexamples.)
In particular, if $p\equiv 1,3\pmod{q}$ for all primes $q<4(k+1)^2,$ then any prime between $(p-2)^2$ and $(p-1)^2$ would be a counterexample.
|
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How to demonstrate this inequality? I have this statement:
If $x > 0 , y > 0$, prove that $\frac{1}{x}+\frac{1}{y} > \frac{2}{x+y}$
I will get the hypothesis? (That's what my teacher calls him, I do not know if it's correct)
$\frac{1}{x}+\frac{1}{y} - \frac{2}{x+y}> 0$
$\frac{x^2 + y^2 + xy + xy - 2xy}{(x+y)xy} > 0$
The hypothesis: $\frac{x^2 + y^2}{(x+y)xy} > 0$
Now I am going to prove, i mean, I need to go back to the origin, which is:
$ \frac {1} {x} + \frac {1} {y}> \frac {2} {x + y} $
Here is my problem, and it is hard for me to go back to the origin, step by step, could you help me? I have 5 sheets worn.
PD: I tried this:
|
Use a better hypothesis, and get a better inequality.
\begin{align}
\frac{(x-y)^2}{(x+y)xy} &\ge 0 \\
\frac{x^2 - 2xy + y^2}{(x+y)xy} &\ge 0 \\
\frac{x^2 + 2xy + y^2}{(x+y)xy} &\ge \frac{4xy}{(x+y)xy} \\
\frac{(x+y)^2}{(x+y)xy} &\ge \frac{4}{x+y} \\
\frac{x+y}{xy} &\ge \frac{4}{x+y} \\
\frac{1}{x} + \frac{1}{y} &\ge \frac{4}{x+y} > \frac{2}{x+y} \\
\end{align}
|
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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 ?
|
Move $a(x-y)$ to the LHS to make $F(x,y)=0$. Then:
$$\frac{dy}{dx}=-\frac{F'_x}{F'_y}=-\frac{-\frac{x}{\sqrt{1-x^2}}-a}{-\frac{y}{\sqrt{1-y^2}}+a}=\frac{a\sqrt{1-x^2}+x}{a\sqrt{1-y^2}-y}\cdot \frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}=\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}},$$
because:
$$\begin{align}&\frac{a\sqrt{1-x^2}+x}{a\sqrt{1-y^2}-y}= \\
&1 \iff a(\sqrt{1-x^2}-\sqrt{1-y^2})=-(x+y) \iff a(y^2-x^2)= \\
&-(x+y)(\sqrt{1-x^2}+\sqrt{1-y^2}) \iff a(x-y)=\\
&\sqrt{1-x^2}+\sqrt{1-y^2}.\end{align}$$
|
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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,
$$8R^2=a^2+b^2+c^2$$
$$8R^2=4R^2 \sin^2 (A)+ 4R^2 \sin^2 (B) + 4R^2 \sin^2 C$$
$$8R^2=4R^2(\sin^2 (A)+\sin^2 (B) +\sin^2 (C)$$
$$2=\sin^2 (A)+\sin^2 (B)+\sin^2 (C)$$
|
$$F=\sin^2A+\sin^2B+\sin^2C=1-\cos^2A+1-(\cos^2B-\sin^2C)$$
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$
and $\cos(B+C)=\cos(\pi-A)=?$
$$2-F=\cos^2A+\cos(B+C)\cos(B-C)$$
$$=\cos^2A-\cos A\cos(B-C)$$
$$=\cos A\{\cos A-\cos(B-C)\}$$
$$=-\cos A\{\cos(B+C)+\cos(B-C)\}=?$$
|
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|
Proving any integer cube is equal to a difference of two squares I know this problem has been asked before by someone. However, my problem is a bit different. I found the following problem in Elementary number theory by Burton:
Prove that the cube of any integer can be written as the difference of two squares. Notice that $n^3 = (1^3+2^3+3^3+.......+n^3) - (1^3+2^3+3^3+.......+(n-1)^3).$
My question: can this be done using the induction method? I know that:
$n^3 = (\frac{n(n+1)}{2})^2 - (\frac{(n-1)(n)}{2})^2$
I am stuck in the induction step. Can someone help?
|
Well, what happens when you try?
$(n+1)^3 = n^3 + 3n^2 + 3n + 1 =$
$(\frac {n(n+1)}2)^2 - (\frac {n(n-1)}2)^2 + 3n^2 + 3n + 1$
Meanwhile $(\frac{(n+1)(n+2)}2)^2 - (\frac {(n+1)n}2)^2 =$
$(\frac{(n+1)n}2 + \frac {2(n+1)}{2})^2 - (\frac {n(n-1)}2 + \frac {2n}{2})^2=$
$(\frac {(n+1)n}2)^2 + (n+1)^2n +(n+1)^2 - (\frac {n(n-1)}2)^2 -n^2(n-1) - n^2=$
$(\frac {(n+1)n}2)^2- (\frac {n(n-1)}2)^2 + (n+1)^2(n+1) - n^2(n)$
$(\frac {(n+1)n}2)^2- (\frac {n(n-1)}2)^2 + (n+1)^3 - n^3$.
$(\frac {n(n+1)}2)^2 - (\frac {n(n-1)}2)^2 + n^3 +3n^2 + 3n + 1- n^3=$
$(\frac {n(n+1)}2)^2 - (\frac {n(n-1)}2)^2 + +3n^2 + 3n + 1$.
So that's that.
|
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|
Divisibility property for sequence $a_{n+2}=-2(n-1)(n+3)a_n-(2n+3)a_{n+1}$ Let $(a_n)$ be the sequence uniquely defined by $a_1=0,a_2=1$ and
$$
a_{n+2}=-2(n-1)(n+3)a_n-(2n+3)a_{n+1}
$$
Can anybody show (or provide a counterexample) that $p|a_{p-2}$ and
$p|a_{p-1}$ for any prime $p\geq 5$ ? I have checked this fact for
$p\leq 200$.
|
The sequence starts
$$0, 1, -5, 25, -105, 105, 5355, \dots$$
We can observe that the statement is true not only for primes, but for odd numbers in general. Even though recurrences might be better for solving this kind of problem, here we can go for closed form formula (luckily there is one!). I have used approach inspired by http://mathforum.org/library/drmath/view/67314.html.
Currently the coefficients contain quadratic polynomials in $n$, but we can make them linear, which makes it a bit easier to work with. With substitution $b_n=\frac{a_n}{(n+2)!}$ and some algebra we get
$$(n+4)b_{n+2}+(2n+3)b_{n+1}+(2n-2)b_n=0$$
Now that the coefficients are linear, let's try to find its generating function $y(x)=\sum_{n \geq 1}b_n x^n$. By summing the whole equation and again some technical steps, we arrive at
$$
y'(x)(2x^3+2x^2+x)=\frac{1}{6}x^2+(2x^2-x-2)y(x)
$$
which is a linear differential equation with variable coefficients, and can be solved to
$$
y(x)=-\frac{1}{3x}-\frac{1}{9x^2}-\frac{5x}{18}-\frac{1}{2}+\frac{(2x^2+2x+1)^{3/2}}{9x^2}
$$
(I have used CAS system for solving this differential equation, but it is just a technicality to show, also it is trivial to verify by differentiation).
Now to get information about the coefficients, let's expand it into the series. The power on the right is by Binomial series
$$
(2x^2+2x+1)^{3/2}=\sum_{k=0}^{\infty}\binom{3/2}{k}2^k(x^2+x)^k
$$
Using Binomial theorem for the inner sum and playing with the indicies, we obtain
$$
[x^n](2x^2+2x+1)^{3/2} = \sum_{n/2 \leq k \leq n}\binom{3/2}{k}\binom{k}{2k-n}2^k
$$
Dividing by $9x^2$ and noticing first few terms in result series are equal to $\frac{1}{9x^2}+\frac{1}{3x}+\frac{1}{2}+\frac{5x}{18}$, we can cancel those out in the $y(x)$, and so
$$
b_n=\sum_{(n+2)/2 \leq k \leq n+2}\binom{3/2}{k}\binom{k}{2k-n-2}2^k
$$
and in turn
$$
a_n=(n+2)!\sum_{(n+2)/2 \leq k \leq n+2}\binom{3/2}{k}\binom{k}{2k-n-2}2^k
$$
Looking at first few individual terms in a sum for couple values of $n$, we can observe those are all integers. This suggests we can further simplify the expression. Writing out the definitions and all the factorials, one eventually finds for $n>1$:
$$
\boxed{a_n = \sum_{(n+2)/2 \leq k \leq n+2}\binom{k}{n+2-k}\frac{(n+2)!}{k!}\frac{(2k-5)!!}{3}(-1)^k}
$$
We can see that for $k<n+2$ the terms above are divisible by $n+2$. So we have
\begin{align}
a_n &\equiv \binom{n+2}{0}\frac{(n+2)!}{(n+2)!}\frac{(2(n+2)-5)!!}{3}(-1)^{n+2}\\
&= \frac{(2n-1)!!}{3}(-1)^{n+2}\\
&= (2n-1)(2n-3)\cdots 5 \cdot (-1)^{n+2} \pmod{n+2}
\end{align}
If $n\geq 3$ is odd, the $n+2$ is odd as well, so the product above will contain it and thus will be divisible by it. In other words $a_n \equiv 0 \pmod {n+2}$, which is what we wanted to prove.
The claim that $a_{n+1} \equiv 0 \pmod {n+2}$ now follows for example by writing the original equation as
$$
a_{n+1}=-2(n-2)(n+2)a_{n-1}-(2n+1)a_n
$$
which modulo $n+2$ yields
$$
a_{n+1} \equiv 3a_n \equiv 0 \pmod {n+2}
$$
I might have omitted few details here and there, but main idea is hopefully clear.
|
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|
Let $a,b,c $ be positive real numbers such that $\sqrt{a}+9\sqrt{b}+44\sqrt{c}=\sqrt{2018(a+b+c)}$ Let $a,b,c $ be positive real numbers such that $\sqrt{a}+9\sqrt{b}+44\sqrt{c}=\sqrt{2018(a+b+c)}$ then find $\frac{b+c}{a}$
So given $\sqrt{a}+9\sqrt{b}+44\sqrt{c}=\sqrt{2018(a+b+c)}$
So squaring both sides $(\sqrt{a}+9\sqrt{b}+44\sqrt{c})^2=(\sqrt{2018(a+b+c)})^2$
then $a+81b+1936c+18\sqrt{ab}+88\sqrt{ac}+792\sqrt{bc}=2018(a+b+c)$
but it looks much difficult is any other method
|
By Cauchy Inequality,
$\sqrt{a}+9\sqrt{b}+44\sqrt{c}\le\sqrt{1^2+9^2+44^2}\sqrt{(\sqrt{a})^2+(\sqrt{b})^2+(\sqrt{c})^2}=\sqrt{2018(a+b+c)}$
with equality holds if and only if $\displaystyle \frac{\sqrt{a}}{1}=\frac{\sqrt{b}}{9}=\frac{\sqrt{c}}{44}$, i.e. $$a:b:c=1:81:1936$$
|
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|
Lagrange Multiplier: Distance to the Origin Find the points on the curve $x^2+xy+y^2=2$ that are closest to the origin.
Is there a way to use Lagrange multipliers to answer this question?
|
Using Lagrange multipliers, consider the function
$$f=x^2+y^2+\lambda \left(x^2+x y+y^2-2\right)$$ Compute the partial derivatives and set them equal to $0$ to get the equations
$$\frac{\partial f}{\partial x}=2x+\lambda (2 x+y)=0$$
$$\frac{\partial f}{\partial y}=2y+\lambda ( x+2y)=0$$
$$\frac{\partial f}{\partial \lambda }=x^2+x y+y^2-2=0$$ and get, as possible solutions
$$\lambda =-\frac 23 \qquad x=-\sqrt{\frac{2}{3}}\qquad y=-\sqrt{\frac{2}{3}}\implies x^2+y^2=\frac 43$$
$$\lambda =-\frac 23 \qquad x=\sqrt{\frac{2}{3}}\qquad y=\sqrt{\frac{2}{3}}\implies x^2+y^2=\frac 43$$
$$\lambda =-2 \qquad x=-\sqrt{2} \qquad y=\sqrt{{2}} \implies x^2+y^2= 4$$
$$\lambda =-2 \qquad x=\sqrt{2} \qquad y=-\sqrt{{2}} \implies x^2+y^2= 4$$
|
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|
Degree of splitting field of $x^6+tx^3+t$ in $\mathbb{F}_3(t)[x]$ I'm trying to find the degree splitting field of $x^6+tx^3+t$ in $\mathbb{F}_3(t)[x]$. After substituting $z= x^3$, and using the qudratic formula, and thensubstituting $x$ back in, I get the roots
$$\frac{\sqrt[3]{-t\pm\sqrt{t^2-4 t}}}{\sqrt[3]{2}},$$
but how do I find the minimal polynomial for this? No matter what power I take, I can't seem to get rid of the radicals. Is there another way to think about square/cube roots when you are in a finite field? Not sure how to proceed.
|
As @jyre says, the polynomial $X^6 + tX^3 + t$ is irreducible, so you can make a first field extension of degree $6$. $\Bbb F_3(t) \to k = \Bbb F_3(t)[x]/(x^6+tx^3+t)$.
Over this field $k$, $x$ is a root of $X^6 + tX^3 + t$ so the polynomial factors a bit. It turns out $x$ is a multiple root :
$X^6 + tX^3 + t = X^6 + tX^3 +t - (x^6 + tx^3 + t)
\\= X^6-x^6 + t(X^3-x^3) = (X^3-x^3)(X^3+x^3+t)
\\= (X-x)^3(X^3+x^3+t)$
Now you need to check if $X^3+x^3+t$ is further reducible over $k$.
But putting $t = -x^6/(x^3+1)$, $k$ is isomorphic to $\Bbb F_3(x)$, and now the remaining cubic factor is $X^3 + x^3-x^6/(x^3+1) = X^3+x^3/(x^3+1) = (X+x/(x+1))^3$
Therefore, your polynomial splits in $k$ as $(X-x)^3(X+x/(x+1))^3$
And so the splitting field is $k$, and the degree is $6$.
|
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|
Area of $\triangle ABC$, given $AB=7$, $AC=15$, and median $AM=10$ I've been working on this interesting problem for a while already, and here it is:
In $\triangle ABC$, $AB = 7$, $AC = 15$, and median $AM = 10$. Find the area of $\triangle ABC$.
I have figured out that $BM$ and $CM$ are both $4\sqrt2$ using Stewart's Theorem. Now, I tried to use Heron's Formula to calculate the area, which was a mess.
Any help is appreciated. Thanks.
|
Are you sure? $p = \dfrac{7+15+8\sqrt{2}}{2} = 11+4\sqrt{2}\implies S^2 = p(p-15)(p-7)(p-8\sqrt{2}) = (11+4\sqrt{2})(11-4\sqrt{2})(-4+4\sqrt{2})(4+4\sqrt{2}) = (121-32)(32-16) = 89\cdot 16.$
So $S = 4\sqrt{89}.$
|
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|
Using the $N - \varepsilon$ definition to find the limit of a sequence Yesterday I had a post about this, and it cleared a lot up; however, even though I feel like I understand how to go about solving problems like this, I don't seem to get the right answers. For example:
$u_n = \frac{2n+3}{2n+1}$. We know that this sequence converges to 1, but I want to prove it using the definition.
So I want to show that , $\forall \varepsilon > 0$, $\exists N$ where,$ n > N \implies u_n - 1 < \varepsilon$.
We can manipulate the expression such that $u_n = \frac{2n+3}{2n+1} =\frac{n}n \frac{2+\frac{3}n}{2+\frac{1}n} = \frac{2+\frac{3}n}{2+\frac{1}n}$
Thus, we can show that $\forall \varepsilon > 0$, $\exists N$ where,$ n > N \implies \frac{2+\frac{3}n}{2+\frac{1}n} - 1 < \varepsilon$
Now, we can say that 0 < $\frac{2+\frac{3}n}{2+\frac{1}n}$ < $\frac{2+\frac{3}n}{2} = 1 + \frac{3}{2n}$.
We can use this logic to say that
$\frac{2+\frac{3}n}{2+\frac{1}n} - 1 < 1 + \frac{3}{2n} - 1 < \varepsilon$ (since $1 + \frac{3}{2n}$ converges to 1).
So, finding an $N$ for $\frac{3}{2n} - 1$ would be the same as finding an $N$ for $u_n$, so we do:
$1 + \frac{3}{2n} - 1 < \varepsilon$ is the same as $\frac{3}{2n} < \varepsilon$.
Thus, if $ N > \frac{3}{2\varepsilon}$, then the initial implication follows. However, I'm not sure how to complete things from here; am I close, or did I fall off the rails at some point? Thanks.
|
You have already shown that your $N$ works, I think it's all the notation that is giving you a headache. I would write like this (these are your words, but rewritten):
Given $\varepsilon$, take any $N > \frac{3}{2\epsilon}$. Now, for any $n >N$, we have $$ \frac{3}{2n} < \varepsilon$$
and so,
$$\left|\frac{2+\frac{3}n}{2+\frac{1}n} - 1\right| < \left|1 + \frac{3}{2n} - 1\right| = \frac{3}{2n} < \varepsilon $$
|
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|
Find the extrema of $\sin(x)\cos(y)$ with the Hessian I've got the following function: $f(x,y) = \sin(x)\cos(y)$.
I performed the Hessian matrix correctly, with the second derivatives:
\begin{bmatrix}
-\sin(x)\cos(y) & -\cos(x)\sin(y)\\
-\cos(x)\sin(y) & -\sin(x)\cos(y)
\end{bmatrix}
But the trouble comes when I want to find the extrema (critical points and saddle points). By solving $\cos(x)\cos(y) = 0$ and $-\sin(x)\sin(y) = 0$ (the first partial derivatives) I obtain:
$\cos(x) = 0$ or $\cos(y) = 0$
$\sin(x) = 0$ or $\sin(y) = 0$.
If $\cos(x) = 0$, then $\sin(x) = \pm 1$ and if $\sin(x) = 0$ then $\cos(x) = \pm1$. Well, here I think that we have got just two possibilities:
*
*$\cos(x) = \sin(y) = 0$;
*$\sin(x) = \cos(y) = 0$.
Now, how do I find the extrema?
I mean, with $\cos(x) = \sin(y) = 0$, which is the point $(x, y)$ and how do I substitute it in the Hessian matrix? (The solution says it must a saddle point but I don't get it, why?)
|
Find the extrema of $f(x,y) = \sin(x)\cos(y)$.
$$\begin{cases} f_x=\cos x\cos y=0 \\ f_y=-\sin x\sin y=0\end{cases} \Rightarrow 1)\begin{cases}\cos x=0 \\ \sin y=0\end{cases} \ \ \text{or} \ \ 2)\begin{cases}\cos y=0 \\ \sin x=0\end{cases}.$$
The Hessian is:
$$H=\begin{bmatrix}
-\sin x \cos y & -\cos x \sin y \\
-\cos x \sin y & -\sin x \cos y
\end{bmatrix}\\
H_1=-\sin x\cos y; \ \ \ H_2=\sin^2 x\cos^2y-\cos^2x\sin^2y.$$
For $1)$:
$$a) \begin{cases}\cos x=0 \\ \sin y=0 \\ \sin x=1 \\ \cos y=1\end{cases} \Rightarrow H_1=-1<0, H_2=1>0 \Rightarrow max; \\
b) \begin{cases}\cos x=0 \\ \sin y=0 \\ \sin x=-1 \\ \cos y=1\end{cases} \Rightarrow H_1=1>0, H_2=1>0 \Rightarrow min; \\
c) \begin{cases}\cos x=0 \\ \sin y=0 \\ \sin x=1 \\ \cos y=-1\end{cases} \Rightarrow H_1=1>0, H_2=1>0 \Rightarrow min; \\
d) \begin{cases}\cos x=0 \\ \sin y=0 \\ \sin x=-1 \\ \cos y=-1\end{cases} \Rightarrow H_1=-1<0, H_2=1>0 \Rightarrow max. \\$$
Can you handle $2)$?
|
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|
Factoring and expanding $1-x^8$ It's been a while since I've studied factoring, and I need it for a question. How did this go from $1-x^8$ to $(1-x^4)(1+x^4)$ and then to $(1-x)(1+x+...+x^7)$? I remember studying this a few years go but unfortunately I don't remember.
$$1-x^8=(1-x^4)(1+x^4)=(1-x)(1+x+\dots + x^7)$$
|
You may remember $$(a+b)(a-b) = a^2 - b^2.$$ To see this, just multiply out. You get $a^2 + ba - ab - b^2$, and the middle terms cancel, leaving you with $a^2-b^2$.
Then if you take $a=1, b=x^4$, you get $$(1+x^4)(1-x^4) = 1-x^8.$$ That's your first identity.
You may also remember $$(1+x+x^2+\ldots+x^n)(x-1) = x^{n+1}-1.$$ It's possible that you remember the formula for the sum of a geometric series: $$1+x+x^2+\ldots+x^n = \frac{x^{n+1}-1}{x-1}$$
but if not, just do the multiplication: distribute the right-hand factor $x-1$ over the left-hand factor:
$$\begin{align}
(1+x+x^2+\ldots+x^n)(x-1)
& = (x-1) + x(x-1) + x^2(x-1) + \ldots + x^n(x-1) \\
& = (\color{darkgreen}{x} - 1) + (\color{maroon}{x^2}-\color{darkgreen}{x}) + (\color{darkblue}{x^3}-\color{maroon}{x^2}) + \ldots + (x^{n+1}-x^n) \tag{$\star$}\\
& = x^{n+1} -1
\end{align}$$
Notice that in line $(\star)$ the two $\color{darkgreen}{x}$ terms cancel, the two $\color{maroon}{x^2}$ terms cancel, and so on, leaving only $-1$ and $x^{n+1}$. Taking $n=7$ we get $$(1+x+x^2+\ldots+x^7)(x-1) = x^8-1.$$
Does this help?
|
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"question_score": "1",
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|
Kummer solution to second order ODE I need to solve a second order linear ODE with non-constant coefficients of the form
$$
\frac{d^2Z}{dt^2}+(a+be^{-ct})\frac{dZ}{dt}+dZ=0
$$
where $a,b,c,f$ are positive real constants.
Mathematica provides me with this solution
$$
i^{(a - \sqrt{a^2 - 4 d})/c}\, b^{(a - \sqrt{a^2 - 4 d})/2 c}\,
c^{-(a - \sqrt{a^2 - 4 d})/2 c} \,{e^{-c t}}^{(
a - \sqrt{a^2 - 4 d})/2 c}\, C[1]\, \text{Hypergeometric1F1}\left[\frac{a}{2c} - \frac{\sqrt{a^2 - 4 d}}{2c}, 1 - \frac{\sqrt{a^2 - 4 d}}{c}, \frac{be^{-ct}}{c}\right] +\\
i^{(a + \sqrt{a^2 - 4 d})/c}\, b^{(a + \sqrt{a^2 - 4 d})/2 c}\,
c^{-(a + \sqrt{a^2 - 4 d})/2 c} \,{e^{-c t}}^{(
a + \sqrt{a^2 - 4 d})/2 c}\, C[2]\, \text{Hypergeometric1F1}\left[\frac{a}{2c} + \frac{\sqrt{a^2 - 4 d}}{2c}, 1 + \frac{\sqrt{a^2 - 4 d}}{c}, \frac{be^{-ct}}{c}\right]
$$
but I don't quite get how to derive such an expression. Any hint?
|
For
$$y'' + (a + b e^{-ct}) \, y' + d \, y = 0$$
let $c x = b \, e^{-c t}$ to obtain
\begin{align}
x &= \frac{b}{c} \, e^{-c t} \\
\frac{dx}{dt} &= - c x \\
\frac{dy}{dt} &= - c x \, \frac{dy}{dx} = - c x y' \\
\frac{d^{2} y}{d x^{2}} &= c^{2} \, x \, \frac{d}{dx} \left( x \, \frac{dy}{dx} \right) = c^{2} \, (x^2 y'' + x y' )
\end{align}
and
$$x^2 \, y'' + x \left(1 - \frac{a}{c} - x \right) \, y' + \frac{d}{c^{2}} \, y = 0.$$
Let $y = x^{\alpha} \, f$ to obtain
\begin{align}
y &= x^{\alpha} \, f \\
y' &= x^{\alpha} \, \left(f' + \frac{\alpha}{x} \, f \right) \\
y'' &= x^{\alpha} \, \left( f'' + \frac{2 \alpha}{x} \, f' + \frac{\alpha(\alpha -1)}{x^2} \, y \right)
\end{align}
and
$$x f'' + \left( 1 - \frac{a}{c} + 2 \alpha - x \right) \, f' + \left(\frac{c^{2} \alpha^{2} - a c \alpha + d}{c^{2} \, x} - \alpha \right) y = 0.$$
Now let $c^{2} \alpha^{2} - a c \alpha + d = 0$, $p = 1 - (a/c) + 2 \alpha$, such that the last equation becomes
$$x f'' + (p - x) f' - \alpha f = 0$$
and has solution
$$f(x) = A_{0} \, {}_{1}F_{1}(\alpha; p; x).$$
Backward sunstitutions lead to
$$y(x) = A_{0} \, \left( \frac{b}{c} \, e^{- c t} \right)^{\alpha} \, {}_{1}F_{1}\left(\alpha; 1 - \frac{a}{c} + 2 \alpha; \frac{b}{c} \, e^{-c t} \right).$$
Note that $c^{2} \alpha^{2} - a c \alpha + d = 0$ has solutions
$$\alpha = \frac{a \pm \sqrt{a^2 - 4d}}{2 c}$$
which assist in the two possible solutions. With initial conditions, or asymptotic forms, the solution can further be reduced.
|
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"url": "https://math.stackexchange.com/questions/2745254",
"timestamp": "2023-03-29T00:00:00",
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|
Generating function of $\log(\log(n))$ What do we know about the following function?
$$f(x)=\sum_{n=2}^\infty \log{(\log{(n)})}x^n$$
Do we know any closed form for it?
I have tried to work with it through different means, but, as it is difficult both to calculate sums over $\log{(\log{(n)})}$ and evaluate integrals involving that function, I have not reached any conclusion.
Thank you.
|
We can write
$$
\eqalign{
& F(x) = \sum\limits_{2\, \le \,n} {\ln \left( {\ln \left( n \right)} \right)x^{\,n} } = x^{\,2} \sum\limits_{0\, \le \,n} {\ln \left( {\ln \left( {n + 2} \right)} \right)x^{\,n} } = \cr
& = x^{\,2} \ln \left( {\ln \left( 2 \right)} \right) + x^{\,3} \sum\limits_{0\, \le \,n} {\ln \left( {\ln \left( {n + 3} \right)} \right)x^{\,n} } \cr}
$$
Then we write
$$
\eqalign{
& \ln \left( {\ln \left( {n + 3} \right)} \right) = \cr
& = \ln \left( {\ln \left( {0 + 3} \right)} \right) - \ln \left( {\ln \left( {0 + 3} \right)} \right)
+ \ln \left( {\ln \left( {1 + 3} \right)} \right) + \cdots - \ln \left( {\ln \left( {n - 1 + 3} \right)} \right) + \ln \left( {\ln \left( {n + 3} \right)} \right) = \cr
& = \ln \left( {\ln \left( 3 \right)} \right) + \sum\limits_{0\, \le \,k\, \le \,n - 1}
{\ln \left( {{{\ln \left( {k + 4} \right)} \over {\ln \left( {k + 3} \right)}}} \right)} \cr}
$$
so that
$$
\eqalign{
& \sum\limits_{0\, \le \,n} {\ln \left( {\ln \left( {n + 3} \right)} \right)x^{\,n} } \quad \left| {\;\left| x \right| < 1} \right.\quad = \cr
& = \ln \left( {\ln \left( 3 \right)} \right)\sum\limits_{0\, \le \,n} {x^{\,n} }
+ \sum\limits_{0\, \le \,n} {\sum\limits_{0\, \le \,k\, \le \,n - 1} {\ln \left( {{{\ln \left( {k + 4} \right)} \over {\ln \left( {k + 3} \right)}}} \right)x^{\,n} } } = \cr
& = \ln \left( {\ln \left( 3 \right)} \right)\sum\limits_{0\, \le \,n} {x^{\,n} }
+ \sum\limits_{1\, \le \,n} {\sum\limits_{0\, \le \,k\, \le \,n - 1} {\ln \left( {{{\ln \left( {k + 4} \right)} \over {\ln \left( {k + 3} \right)}}} \right)x^{\,n} } } = \cr
& = {{\ln \left( {\ln \left( 3 \right)} \right)} \over {1 - x}} + x\sum\limits_{0\, \le \,n}
{\left( {\sum\limits_{0\, \le \,k\, \le \,n} {\ln \left( {{{\ln \left( {k + 4} \right)} \over {\ln \left( {k + 3} \right)}}} \right)} } \right)x^{\,n} } \cr}
$$
It is known that
$$
A(z) = \sum\limits_{n\, \ge \,0} {a_n } \;z^n \quad \Leftrightarrow \quad {{A(z)} \over {\left( {1 - z} \right)}} = \sum\limits_{0\, \le \,n} {\left( {\sum\limits_{0\, \le \,k\, \le \,n} {a_k } } \right)z^n }
$$
therefore
$$
\eqalign{
& \sum\limits_{0\, \le \,n} {\ln \left( {\ln \left( {n + 3} \right)} \right)x^{\,n} } \quad \left| {\;\left| x \right| < 1} \right.\quad = \cr
& = {{\ln \left( {\ln \left( 3 \right)} \right)} \over {1 - x}} + x\sum\limits_{0\, \le \,n}
{\left( {\sum\limits_{0\, \le \,k\, \le \,n} {\ln \left( {{{\ln \left( {k + 4} \right)} \over {\ln \left( {k + 3} \right)}}} \right)} } \right)x^{\,n} } = \cr
& = {{\ln \left( {\ln \left( 3 \right)} \right)} \over {1 - x}} + {x \over {1 - x}}\sum\limits_{0\, \le \,n}
{\ln \left( {{{\ln \left( {n + 4} \right)} \over {\ln \left( {n + 3} \right)}}} \right)x^{\,n} } \cr}
$$
which finally gives:
$$ \bbox[lightyellow] {
\eqalign{
& F(x) = \sum\limits_{2\, \le \,n} {\ln \left( {\ln \left( n \right)} \right)x^{\,n} } \quad \left| {\;\left| x \right| < 1} \right.\quad = \cr
& = x^{\,2} \ln \left( {\ln \left( 2 \right)} \right) + {{\ln \left( {\ln \left( 3 \right)} \right)x^{\,3} } \over {1 - x}}
+ {{x^{\,4} } \over {1 - x}}\sum\limits_{0\, \le \,n} {\ln \left( {{{\ln \left( {n + 4} \right)} \over {\ln \left( {n + 3} \right)}}} \right)x^{\,n} } \cr}
}$$
which is convergent (for $|x| < 1$) and checks to be correct.
Eventually, it remains to fix some appropriate bounds on
the coefficient of the last sum, which are positive, less than
$\ln (\ln (4) / \ln (3)) \approx 0.2326$
and decreasing to zero, and with those to sandwich $F(z)$ between two known functions ...
|
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|
Why is the sum of outer products equal to the matrix product of a matrix and its transpose , so $A^TA = \sum_{i=1}^n a_i a_i^T$? Why is the sum of outer products equal to the matrix product of a matrix and its transpose? So $A^TA = \sum_{i=1}^n a_i a_i^T$, where $A = [ a_0, a_1 , ... , a_n ] $, $a_i \in \mathbb{R}^k$. An answer or a link to the answer would be greatly appriciated. It seems so simple but I just cannot figure it out.
It is essentially this question of which I do not understand the answer:
Matrix $A^T A$ as sum of outer products
|
I think you got the order wrong. $ A A^T = \sum_{i=1}^n a_i a_i^T $. What you should do is to is to look at a sequence of examples:
$$ \left( \begin{array}{c c c}
1 \\
2 \\
3
\end{array} \right)
\left( \begin{array}{c c c}
1 \\
2 \\
3
\end{array} \right) ^T =
$$
$$ \left( \begin{array}{c c c}
1 & -1\\
2 & -2\\
3 & -3
\end{array} \right)
\left( \begin{array}{c c c}
1 & -1\\
2 & -2\\
3 & -3
\end{array} \right) ^T =
$$
Maybe you will see the pattern.
Then, look at
$$ \left( \begin{array}{c c c}
a_{1,1} \\
a_{2,1} \\
a_{3,1}
\end{array} \right)
\left( \begin{array}{c c c}
a_{1,1} \\
a_{2,1} \\
a_{3,1}
\end{array} \right) ^T =
$$
$$ \left( \begin{array}{c c c}
a_{1,1} & a_{1,2} \\
a_{2,1} & a_{2,2}\\
a_{3,1}& a_{3,2}
\end{array} \right)
\left( \begin{array}{c c c}
a_{1,1} & a_{1,2} \\
a_{2,1} & a_{2,2}\\
a_{3,1}& a_{3,2}
\end{array} \right) ^T =
$$
and see if you get the pattern from that.
|
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|
Find all positive integers $x$ such that $[\frac{x}{5}]-[\frac{x}{7}]=1$, where, for any real number $t$, $[t]$ denotes the greatest integer $\le t$ I have tried the following steps:
$[\dfrac{x}{5}]-[\dfrac{x}{7}]=1$
@Berci suggested this:
$[\dfrac{x}{5}]\le\dfrac{x}{5}$
$-[\dfrac{x}{7}]<-\dfrac{x}{7}+1$
$\implies [\dfrac{x}{5}]-[\dfrac{x}{7}]<\dfrac{x}{5}-\dfrac{x}{7}+1 $
$\implies x>0$
$\implies x> 0$
$\implies x \in [1,.... \infty]$
Next is a manual observation, we will see that numbers lets say $n\equiv1(mod 5)\implies n\equiv 5k+1 $ where $k\in \{1,2,3,4\}$
why $k\in \{1,2,3,4\}$? why not $k\in \mathbb{N}$?
Lets see:
Hence if we try numbers like $6,11,16,21$ we will see that
$[\dfrac{x}{5}]-[\dfrac{x}{7}]=1$
But the moment one tries with $x=26,36$ we will see $[\dfrac{x}{5}]-[\dfrac{x}{7}]=2$
Again with $x=41,46,51,56$ we see $[\dfrac{x}{5}]-[\dfrac{x}{7}]=2$
And so on as will proceed and this $[\dfrac{x}{5}]-[\dfrac{x}{7}]>1$ and increases more and more.
Again if we try $x=12,13$,we get $[\dfrac{x}{5}]-[\dfrac{x}{7}]=1$
Also $x=5$ is a solution.
So is it the answer that $x \in {5,6,11,12,13,16,21}$?
|
To obtain a meaningful bound, you need the inequalities in the other way: $\left[\dfrac{x}{5}\right]\ge \dfrac{x}{5}-1$ and $\left[\dfrac{x}{7}\right]\le \dfrac{x}{7}$. Hence, $$\left[\dfrac{x}{5}\right]-\left[\dfrac{x}{7}\right]\ge \frac{x}{5}-1-\frac{x}7=\frac{2x}{35}-1$$ Now, for any $x\ge 36$, you have that $$x\ge 36\implies \frac{2x}{35}-1\ge \frac{2\cdot36}{35}-1>1$$ which implies that $$\left[\dfrac{x}{5}\right]-\left[\dfrac{x}{7}\right]>1$$ for any $x\ge 36$. So, your candidate solutions are the integers $1,\dots, 35$. Trial and error will work. A direct substitution shows that $35$ does not work. So, to do it more efficiently, observe that for $x\in \{1,\dots,34\}$, $\left[\dfrac{x}{5}\right]$ takes only $7$ values and $\left[\dfrac{x}{7}\right]$ only $5$.
A systematic way to do it is the following:
*
*For $x=1,\dots,6$: $\left[\dfrac{x}{7}\right]=0$. Hence, $x=5,6$ will do, since for these values $\left[\dfrac{x}{5}\right]=1$.
*For $x=7,\dots,13$: $\left[\dfrac{x}{7}\right]=1$. Hence, $x=10,\dots,13$ will do, since for these values $\left[\dfrac{x}{5}\right]=2$.
*For $x=14,\dots,20$: $\left[\dfrac{x}{7}\right]=2$. Hence, $x=15,\dots,19$ will do, since for these values $\left[\dfrac{x}{5}\right]=3$.
*For $x=21,\dots,27$: $\left[\dfrac{x}{7}\right]=3$. Hence, $x=21,\dots,24$ will do, since for these values $\left[\dfrac{x}{5}\right]=4$.
*For $x=28,\dots,34$: $\left[\dfrac{x}{7}\right]=4$. Hence, $x=28,29$ will do, since for these values $\left[\dfrac{x}{5}\right]=5$.
This gives you the complete set of solutions, which is $$\{5,6,10,11,12,13,15,16,17,18,19,21,22,23,24,28,29\}$$
|
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|
Find $f_{10}$ of the sequence defined by $f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$ Find $f_{10}$ of the sequence defined by $$f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$$ given $f_0=0$
My Approach:
Letting $f_n=2^n b_n$ we get
$$b_{n+1}=\frac{4b_n}{5}+\frac{3 \sqrt{1-b_n^2}}{5}$$
Now letting $b_n=\cos(x_n)$ we get
$$\cos(x_{n+1})=\cos(x_n-\theta)$$ where $\cos(\theta)=\frac{4}{5}$
Now Since $f_0=0$ we have $b_0=0$ and $x_0=\frac{\pi}{2}$
Now we have
$$x_{n+1}=x_n-\theta$$
Putting $n=0,1,2,3 \cdots 10$ and adding all we get
$$x_{10}=\frac{\pi}{2}-10\theta$$
Hence
$$b_{10}=\cos\left(\frac{\pi}{2}-10\theta\right)=\sin(10\theta)=\sin\left(10\arcsin\left(\frac{3}{5}\right)\right)$$
How to proceed further?
|
Note
$$ \sin(2\theta)=2\sin\theta\cos\theta,\cos(2\theta)=2\cos^2\theta-1.$$
Let $\theta=2\arcsin(\frac35)$. It is easy to get
$$ \sin\theta=\frac{24}{25},\cos\theta=\frac{7}{25} $$
and hence
$$\sin(2\theta)=2\cdot\frac{24}{25}\cdot\frac{7}{25}=\frac{336}{625},\cos(2\theta)=2(\frac7{25})^2-1=\frac{527}{625} $$
and
$$\sin(4\theta)=2\sin(2\theta)\cos(2\theta)=\frac{354144}{390625},\cos(4\theta)=2\cos^2(2\theta)-1=\frac{164833}{390625}. $$
Thus
\begin{eqnarray}
&&\sin(10\arcsin(\frac35))\\
&=&\sin(5\theta)\\
&=&\sin(4\theta)\cos(\theta)+\cos(4\theta)\sin(\theta)\\
&=&\frac{354144}{390625}\cdot\frac7{25}+\frac{164833}{390625}\cdot\frac{24}{25}\\
&=&\frac{10296}{15625}.
\end{eqnarray}
|
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|
How to find the area of triangle? $A =(3, 1)$ is reflected over line $y = 2x$, become $A'$.
$O = (0,0)$
Find area of triangle OAA'
I find that $A'$ is $(-1, 3)$
I find it by mutiply $A$ to matrix of $y = 2x$
There is rule Area = $$\sqrt{s(s-a)(s-b)(s-c)}$$
$S = \frac{a+b+c}{2}$
$a$ = distance between $A$ and $O=\sqrt {10}$
$b$ = between $A'$ and $O = \sqrt {10}$
$c$ = between $A$ and $A' = \sqrt {20}$.
Am i on the right way? Because it looks so complicated to solve. I am afraid i am on the wrong way.
|
The long way:
The formula for the area of a triangle with sides $a$, $b$ and $c$ is given by
$$A=\sqrt{s(s-a)(s-b)(s-c)}$$
where
$$s=\frac{a+b+c}{2}$$
With $a=\sqrt{10}$, $b=\sqrt{10}$ and $c=\sqrt{20}=2\sqrt{5}$,
$$s=\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}=\frac{2\sqrt{10}+2\sqrt{5}}{2}=\sqrt{10}+\sqrt{5}$$
The area is then
$$A=\sqrt{(\sqrt{10}+\sqrt{5})((\sqrt{10}+\sqrt{5})-\sqrt{10})((\sqrt{10}+\sqrt{5})-\sqrt{10})((\sqrt{10}+\sqrt{5})-2\sqrt{5})}$$
$$=\sqrt{(\sqrt{10}+\sqrt{5})(\sqrt{10}+\sqrt{5}-\sqrt{10})(\sqrt{10}+\sqrt{5}-\sqrt{10})(\sqrt{10}+\sqrt{5}-2\sqrt{5})}$$
$$=\sqrt{(\sqrt{10}+\sqrt{5})(\sqrt{5})(\sqrt{5})(\sqrt{10}-\sqrt{5})}=\sqrt{5(\sqrt{10}+\sqrt{5})(\sqrt{10}-\sqrt{5})}$$
$$=\sqrt{5(10-\sqrt{50}+\sqrt{50}-5)}=\sqrt{5(5)}=5$$
|
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|
Weird infinite sum Evaluate
$\sum_{n=0}^{\infty}{\frac{(-1)^n}{3n+1}=1-\frac{1}{4}+\frac{1}{7}-\frac{1}{10}+...}
$
This looks a lot like the series expansion for $\ln(1+x)$ when $x=1$, but I cannot find the relationship.
|
We will use imprecise argument to guess the value of the sum. We know that
$$\sum_{n=0}^\infty x^{3n} = \frac{1}{1-x^3}.$$
By integrating it both sides we have $$\sum_{n=0}^\infty \frac{x^{3n+1}}{3n+1} = \int_0^x \frac{1}{(1-u)(1+u+u^2)}du = \frac{1}{6}\left(\log\left(\frac{x^2+x+1}{(1-x)^2}\right)+ 2\sqrt{3}\arctan\left(\frac{2x+1}{\sqrt{3}}\right)\right).$$
Hence we can guess
$$\sum_{n=0}^\infty \frac{(-1)^{3n+1}}{3n+1} = \frac{1}{6}\left(\log\left(\frac{1}{4}\right)- \frac{2\sqrt{3}\pi}{6}\right).$$
Our argument is not strict as we need to justify we can exchange the integral and the infinite sum. Moreover, we don't know our sum is coincide with the result of the integration.
(Note. The sum in left-hand-side is not equal to your sum, but it is in fact same as the negation of your sum.)
We can justisfy our result as follows: note that
$$\frac{1}{1-x^3} = \sum_{n=0}^N x^{3n} + \frac{x^{3N+3}}{1-x^3}$$
We can integrate from 0 to -1 it both sides then we have
$$\frac{1}{6}\left(\log\left(\frac{1}{4}\right)- \frac{2\sqrt{3}\pi}{6}\right) = \sum_{n=0}^N \frac{-1}{3n+1} + \int_0^{-1}\frac{x^{3N+3}}{1-x^3}dx$$
Here the last term can be regarded as an errer term. It satisfies
$$
\begin{align}
\left|\int_0^{-1}\frac{x^{3N+3}}{1-x^3}dx\right|&
\le \int_{-1}^0\left|\frac{x^{3N+3}}{1-x^3}\right|dx \\
&\le \int_{-1}^0 |x|^{3N+3}dx\\
&= \frac{1}{3N+4} \xrightarrow[N\to\infty]{} 0,
\end{align}$$
since $|1/(1-x^3)|\le 1$ if $-1\le x\le 0$. Now take $N\to\infty$.
|
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|
Legendre symbol of $(-2/p)$. Question. Show $(-2/p)$ equals $1$ when $p\equiv 1,3\bmod 8$ and $-1$ when $p\equiv 5,7\bmod 8$.
So using the multiplicativity of the symbol; we have $$\Big(\frac{-1}{p}\Big)\Big(\frac{2}{p}\Big),$$ and I know the rules for $(2/p)$ and $(-1/p)$. But then what? Usually I use Chinese remainder theorem to try and obtain a bigger modulus but I can't here since $8$ and $4$ aren't co-prime.
Edit. So to combine the congruences we have $$\Big(\frac{-2}{p}\Big)=1,$$ when $p\equiv 1\bmod 8$ and $p\equiv 3\bmod 8$ since we can't have $p$ being congruent to both $7$ and $5$ modulo $8$ at the same time.
Similarly, we have $$\Big(\frac{-2}{p}\Big)=-1,$$ when $p\equiv 7\bmod 8$ or $5\bmod 8$ because this is where the overlaps are when comparing $-1$ and $1$ from the results gained from $(2/p)$ and $(-1/p)$.
|
The proof that
*
*$\Big(\frac{-1}{p}\Big) = \begin{cases} 1 & \text{if } p \equiv 1 \pmod 4 \\ -1 & \text{if } p \equiv 3 \pmod 4 \end{cases}$ and
*$\Big(\frac{2}{p}\Big) = \begin{cases} 1 & \text{if } p \equiv \pm1 \pmod 8 \\ -1 & \text{if } p \equiv \pm3 \pmod 8 \end{cases}$
are classic. The first bullet point becomes $\Big(\frac{-1}{p}\Big) = \begin{cases} 1 & \text{if } p \equiv 1,5 \pmod 8 \\ -1 & \text{if } p \equiv 3,7 \pmod 8 \end{cases}$.
Multiplying the two Legendre symbols gives
$$\Big(\frac{-2}{p}\Big) = \begin{cases} 1 & \text{if } p \equiv 1,3 \pmod 8 \\ -1 & \text{if } p \equiv 5,7 \pmod 8 \end{cases}.$$
|
{
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"url": "https://math.stackexchange.com/questions/2754958",
"timestamp": "2023-03-29T00:00:00",
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|
Why did not I find all the solutions to this equation? $\sin{3x}+\cos{2x}-\sin{x}=1$ $x$ needs to be in $[0,\pi]$
What I tried
$\sin{3x}+\cos{2x}-\sin{x}=1$
$\sin{3x}-\sin{x}=1-\cos{2x}$
$2\sin{x}\cos{2x}=1-(\cos^2{x}-\sin^2{x})$
$2\sin{x}\cos{2x}=\sin^2{x}+\cos^2{x}-\cos^2{x}+\sin^2{x}$
$2\sin{x}\cos{2x}=2\sin^2{x}$
$\cos{2x}=\sin{x}$
$\cos{2x}=\cos{\dfrac{\pi}{2}-x}~~\Longrightarrow~~2x=\dfrac{\pi}{2}-x+2k\pi~~\lor~~2x=x-\dfrac{\pi}{2}+2k\pi$
$x=\dfrac{\pi}{6}+\dfrac{2k\pi}{3}~~~~(k\in\mathbb{Z})~~(1)~\lor~x=-\dfrac{\pi}{2}+2k\pi~~~~(k\in\mathbb{Z})~~(2)$
$(1):~~k=0~\Longrightarrow~x=\dfrac{\pi}{6},~~~~~~k=1~\Longrightarrow~x=\dfrac{5\pi}{6}$
But $0$ and $\pi$ are solutions too. Why did not I find $0$ and $\pi$?
|
Your idea is correct except when you simplify $\sin x$ in
$$2\sin{x}\cos{2x}=2\sin^2{x}$$
A good way is:
$$2\sin{x}\cos{2x}-2\sin^2{x}=0$$
$$(\sin x)(2\cos{2x}-2\sin{x})=0$$
what give you
$$\sin x=0$$
or
$$2\cos{2x}-2\sin{x}=0$$
then, solve both equations and get the complete solution.
|
{
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"url": "https://math.stackexchange.com/questions/2755341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Eigenvalues for $4\times 4$ matrix Show that $0,2,4$ are the eigenvalues for the matrix $A$:
$$A=\pmatrix{
2 & -1 & -1 & 0 \\
-1 & 3 & -1 & -1 \\
-1 & -1 & 3 & -1 \\
0 & -1 & -1 & 2 \\
}$$
and conclude that $0,2,4$ are the only eigenvalues for $A$.
I know that you can find the eigenvalues by finding the $\det(A-\lambda \cdot I)$, but it seems to me that the computation will be rather difficult to compute as it is a $4 \times 4$ matrix.
My question: is there an easier method to calculate the eigenvalues of $A$?
And if I have to conclude that these are the only eigenvalues, is there a theorem that argues how many eigenvalues a matrix can have?
|
You can easily guess some eigenvectors:
$$
\begin{pmatrix}
2 & -1 & -1 & 0 \\
-1 & 3 & -1 & -1 \\
-1 & -1 & 3 & -1 \\
0 & -1 & -1 & 2
\end{pmatrix}
\begin{pmatrix} 1\\1\\1\\1 \end{pmatrix} =
\begin{pmatrix} 0\\0\\0\\0 \end{pmatrix}
\\
\begin{pmatrix}
2 & -1 & -1 & 0 \\
-1 & 3 & -1 & -1 \\
-1 & -1 & 3 & -1 \\
0 & -1 & -1 & 2
\end{pmatrix}
\begin{pmatrix} 1\\0\\0\\-1 \end{pmatrix} =
\begin{pmatrix} 2\\0\\0\\-2 \end{pmatrix}
\\
\begin{pmatrix}
2 & -1 & -1 & 0 \\
-1 & 3 & -1 & -1 \\
-1 & -1 & 3 & -1 \\
0 & -1 & -1 & 2
\end{pmatrix}
\begin{pmatrix} 1\\-1\\-1\\1 \end{pmatrix} =
\begin{pmatrix} 4\\-4\\-4\\4 \end{pmatrix}
\\
\begin{pmatrix}
2 & -1 & -1 & 0 \\
-1 & 3 & -1 & -1 \\
-1 & -1 & 3 & -1 \\
0 & -1 & -1 & 2
\end{pmatrix}
\begin{pmatrix} 0\\1\\-1\\0 \end{pmatrix} =
\begin{pmatrix} 0\\4\\-4\\0 \end{pmatrix}
$$
The corresponding eigenvalues are $0,2,4,4$. It is not hard to check that the latter two vectors are linearly independent, and so these are all eigenvalues of the matrix.
|
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|
$\sqrt[5]{2^4\sqrt[3]{16}} = ? $ $$\sqrt[5]{2^4\sqrt[3]{16}} = ? $$
Rewriting this and we have
$$\sqrt[5\cdot 3]{2^{4\cdot 3}4}$$
$$\sqrt[15]{2^{12}2^2}$$
Finally we get
$$\sqrt[15]{2^{12}2^2} = \sqrt[15]{2^{14}} = 2^{\frac{15}{14}}$$
Am I right?
|
You wrote $$\sqrt[5]{2^4\sqrt[3]{16}}=\sqrt[5\cdot 3]{2^{4\cdot 3}4}$$
which is not true because
$$\sqrt[3]{16}=16^\frac{1}{3}$$
That there is the third root of $16$, not $2^{4\cdot 3}$. Also you can not write $$\sqrt[5]{something\cdot \sqrt[3]{something}}=\sqrt[5\cdot 3]{something\cdot something}$$
The way to do it is:
$$\sqrt[5]{2^4\sqrt[3]{16}}=\sqrt[5]{2^4\cdot16^\frac{1}{3}}=\sqrt[5]{2^4\cdot2^\frac{4}{3}}=\sqrt[5]{2^\frac{16}{3}}=2^\frac{\frac{16}{3}}{5}=2^\frac{16}{15}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Legendre symbol of $(-3/p)$. Question. Show $$\Big(\frac{-3}{p}\Big)=\begin{cases}+1 & p\equiv 1\bmod 3,\\
-1 & p\equiv 2\bmod 3. \end{cases}$$
Attempt. So, using the established results of $$\Big(\frac{3}{p} \Big) = \begin{cases}+1 & p\equiv\pm1\bmod 12,\\
-1 & p\equiv\pm5\bmod 12,\end{cases}$$ $$\Big(\frac{-1}{p}\Big)=\begin{cases}+1 & p\equiv1\bmod 4,\\
-1 & p\equiv3\bmod 4,\end{cases}$$ and observing that $$\Big(\frac{-1}{p} \Big) = \begin{cases}+1 & p\equiv1,5\bmod 12,\\
-1 & p\equiv7,11\bmod 12,\end{cases}$$ we get $$\Big(\frac{-3}{p}\Big)=\begin{cases}+1 & p\equiv 1,7\bmod 12,\\
-1 & p\equiv 5,11\bmod 12. \end{cases}$$ Providing I've carried this out right; how can I reduce this further so it is in the form of modulo $3$?
|
You have that $p\equiv 1,7\pmod{12}$ if and only if $p\equiv 1\pmod{3}$ and $p$ odd.
Similarly for $p\equiv 5,11\pmod{12}$ and $p$ odd with $p\equiv 2\pmod{3}.$\
There is an easier approach if you know quadratic reciprocity:
$$\begin{align}\left(\frac{-3}{p}\right)&=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)\\
&=\left(-1\right)^{\frac{p-1}{2}}\left(-1\right)^{\frac{p-1}{2}\frac{3-1}{2}}\left(\frac{p}{3}\right)\\
&=(-1)^{p-1}\left(\frac p3\right)\\&=\left(\frac{p}{3}\right)
\end{align}$$
More generally, when $q\equiv 3\pmod{4}$ is a prime you get that:
$$\left(\frac{-q}{p}\right)=\left(\frac{p}{q}\right)$$
|
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|
A Problem Dealing with putting balls in bin and Expected Value - possible wrong answer Please consider the problem below. Is my answer correct. If is not correct then where did I go wrong?
Problem:
You keep tossing balls into $n$ bins until one of the bins has two balls. For each toss there is a $\frac{1}{n}$ probability that the ball you toss lands in any one of the bins. What is the expected number of tosses?
Answer:
Let $p_i$ be the probability that after $i$ tosses we have at least one bin
with two balls.
\begin{eqnarray*}
p_1 &=& 0 \\
p_2 &=& 1 - ( 1 - \frac{1}{n} ) = \frac{1}{n} \\
p_3 &=& 1 - (1 - \frac{1}{n})(1 - \frac{1}{n-1}) \\
p_3 &=& 1 - (\frac{n-1}{n})(\frac{n-1-1}{n-1}) \\
p_3 &=& 1 - (\frac{n-2}{n}) = \frac{2}{n} \\
p_4 &=& 1 - (1 - \frac{1}{n})(1 - \frac{1}{n-1})(1 - \frac{1}{n-2}) \\
p_4 &=& 1 - ( \frac{n-1}{n} )( \frac{n-2}{n-1} )( \frac{n - 2 -1}{n - 2} ) \\
p_4 &=& 1 - \frac{n-3}{n} = \frac{3}{n} \\
\end{eqnarray*}
Now for $1 <= i <= n$ we have: $p_i = \frac{i-1}{n}$.
\begin{eqnarray*}
E &=& 2p_2 + 3p_3 + 4p_4 + ... (n+1)p_{n+1} \\
E &=& \sum_{i = 1}^{n} \frac{i(i+1)}{n} = \frac{1}{2n} \sum_{i=1}^{n} i^2 + i \\
E &=& \frac{1}{2n}(\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} ) \\
E &=& \frac{n+1}{4n} ( \frac{2n+1}{3} + 1 ) \\
\end{eqnarray*}
Here is an update to my answer:
Let $p_i$ be the probability that after $i$ tosses we have at least one bin
with two balls.
\newline
\begin{eqnarray*}
p_1 &=& 0 \\
p_2 &=& 1 - ( 1 - \frac{1}{n} ) = \frac{1}{n} \\
p_3 &=& 1 - (\frac{n-1}{n})( \frac{n-2}{n}) \\
p_3 &=& 1 - \frac{(n-1)(n-2)}{n^2} = \frac{n^2 - (n^2 - 3n + 2)}{n^2} \\
p_3 &=& \frac{3n-2}{n^2} \\
p_4 &=& 1 - (\frac{n-1}{n})(\frac{n-2}{n})(\frac{n-3}{n}) \\
p_4 &=& 1 - \frac{(n^2-3n+2)(n-3)}{n^3}\\
p_4 &=& 1 - \frac{n^3-3n^2+2n - 3n^2 +9n - 6}{n^3}\\
p_4 &=& \frac{3n^2-2n + 3n^2 - 9n + 6}{n^3}\\
p_4 &=& \frac{3n^2 + 3n^2 - 11n + 6}{n^3}\\
\end{eqnarray*}
\begin{eqnarray*}
E &=& 2p_2 + 3p_3 + 4p_4 + ... (n+1)p_{n+1} \\
\end{eqnarray*}
Now, am on the right track? That is, is what I have so far correct?
Thanks,
Bob
|
Various answers have been already provided, but let me show a different approach to get it.
At the first toss you have $p_1=0$ and you end up for sure with 1 one-ball bin.
At the second, either you hit the one-ball (prob. $p_2=1/n=(1-p_1)1/n$) or you end up with 2 one-ball bins (prob. $q_2=1-p_2=(n-1)/n$).
At this point, pay attention to the fact that
$$ \bbox[lightyellow] {
\pi _{\,k} = {{k - 1} \over n}\quad \quad \mu _{\,k} = 1 - \pi _{\,k} = {{n - \left( {k - 1} \right)} \over n}
}$$
are transitional probabilities, i.e. conditional probabilities, and precisely:
- $\pi _{\,k}$ is the prob. of having a "hit" at the $k$-th toss, given that you did not have any hit before;
- $\mu _{\,k}$ is the prob. of not having a "hit" at the $k$-th toss, given that you did not have any hit before.
while $p_k$ and $q_k$ are the overall probabilities that, starting with all bins empty,
- we have a hit at (exactly) the $k$-th toss ($p_k$);
- we have (exactly) $k$ one-ball bins at the $k$-th toss, i.e. no hit up to that toss ($q_k$);
Thus it shall be made clear that
- $\pi_k \ne p_k$ and $\mu_k \ne q_k$ notwithstanding that in the first two tosses they coincide;
- $q_k \ne 1-p_k$, because $1-p_k$ is the probability to have a hit at a toss different from $k$, while $1-q_k$ is the probability to have a hit after the $k$-th toss;
So
$$
\left\{ \matrix{
q_{\,3} = \mu _{\,1} \mu _{\,2} \mu _{\,3} = \left( {1 - \pi _{\,1} } \right)\left( {1 - \pi _{\,2} } \right)\left( {1 - \pi _{\,3} } \right) = q_{\,2} \left( {1 - \pi _{\,3} } \right) = \hfill \cr
= {{n\left( {n - 1} \right)\left( {n - 2} \right)} \over {n^{\,3} }} = {{n^{\,\underline {\,3\,} } } \over {n^{\,3} }} \hfill \cr
p_{\,3} = \left( {1 - \pi _{\,2} } \right)\pi _{\,3} = \left( {1 - \pi _{\,1} } \right)\left( {1 - \pi _{\,2} } \right)\pi _{\,3} = q_{\,2} \pi _{\,3} = \hfill \cr
= q_{\,2} - q_{\,3} = {{n - 1} \over n}{2 \over n} \hfill \cr
p_{\,3} \quad \ne \quad 1 - q_{\,3} \hfill \cr} \right.
$$
and in general
$$ \bbox[lightyellow] {
\left\{ \matrix{
q_{\,k} = {{n^{\,\underline {\,k\,} } } \over {n^{\,k} }}\quad \left| {\;0 \le k} \right. \hfill \cr
p_{\,k} = q_{\,k - 1} - q_{\,k} = q_{\,k - 1} {{k - 1} \over n} = {{\left( {k - 1} \right)\,n^{\,\underline {\,k - 1\,} } } \over {n^{\,k} }}\quad \left| {\;1 \le k} \right. \hfill \cr} \right.
}$$
where $x^{\,\underline {\,k\,} }$ represents the Falling Factorial
From that we obtain
$$
\left\{ \matrix{
q_{\,0} = q_{\,1} = 1 \hfill \cr
q_{\,m} = 0\quad \left| {\;n < m} \right. \hfill \cr
\sum\limits_{1\, \le \;k} {p_{\,k} } = \sum\limits_{1\, \le \;k\; \le \,n + 1} {p_{\,k} } = \sum\limits_{1\, \le \;k\; \le \,n + 1} {\left( {q_{\,k - 1} - q_{\,k} } \right)} = q_{\,0} - q_{\,n + 1} = 1 \hfill \cr} \right.
$$
since it is clear that at toss $n+1$ we have a hit for sure, if we did not have before ($\pi_{n+1}=1$).
Coming to the computation of the expected value, we have
$$ \bbox[lightyellow] {
\eqalign{
& E(k;\,n) = \sum\limits_{1\, \le \;k\; \le \,n + 1} {k\,p(k,n)} = \sum\limits_{1\, \le \;k\; \le \,n + 1} {\,{{k\left( {k - 1} \right)\,n^{\,\underline {\,k - 1\,} } } \over {n^{\,k} }}} = \cr
& = \sum\limits_{1\, \le \;k\; \le \,n + 1} {\left( {k\,q(k - 1,n) - k\,q(k,n)} \right)} = \cr
& = \sum\limits_{1\, \le \;k\; \le \,n + 1} {\left( {\left( {k - 1} \right)\,q(k - 1,n) - k\,q(k,n) + q(k - 1,n)} \right)} = \cr
& = - \left( {n + 1} \right)q(n + 1,n) + \sum\limits_{1\, \le \;k\; \le \,n + 1} {q(k - 1,n)} = \sum\limits_{0\, \le \;k\; \le \,n} {q(k,n)} \cr}
}$$
where the change of notation is obvious.
It coincides with answer already given by C. Blatter.
Now, $q(k,n)$ can be written as
$$
q(k,n) = {{\,n^{\,\underline {\,k\,} } } \over {n^{\,k} }} = k!\left( \matrix{
n \cr
k \cr} \right)\left( {{1 \over n}} \right)^{\,k} = \prod\limits_{0\, \le \;j\; \le \,k - 1} {\left( {1 - {j \over n}} \right)}
= {{\,\Gamma (n + 1)} \over {n^{\,k} \,\Gamma (n - k + 1)}}
$$
but unfortunately I could not find (and doubt there is) a closed formula for the sum.
An asymptotic approximation for large $n$ could however be constructed, following various
approaches, depending on the goals that you may have.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\frac{1}{n + 1}{2n\choose n}$ is a positive integer for $n \ge 0$. I attempted to use Pascal's triangle identity to help out, but I do not know how to deal with $\frac{1}{n+1}$.
|
Approach 1:
Note that
$$
\frac{2n+1}{n+1}\overbrace{\ \ \binom{2n}{n}\ \ }^{\large\frac{(2n)!}{n!\,n!}}=\overbrace{\binom{2n+1}{n+1}}^{\large\frac{(2n+1)!}{(n+1)!\,n!}}
$$
Then, because $\frac1{n+1}=2-\frac{2n+1}{n+1}$, we have
$$
\begin{align}
\frac1{n+1}\binom{2n}{n}
&=2\binom{2n}{n}-\frac{2n+1}{n+1}\binom{2n}{n}\\
&=2\binom{2n}{n}-\binom{2n+1}{n+1}
\end{align}
$$
Approach 2:
Note that
$$
\frac{n}{n+1}\overbrace{\ \ \binom{2n}{n}\ \ }^{\large\frac{(2n)!}{n!\,n!}}=\overbrace{\binom{2n}{n+1}}^{\large\frac{(2n)!}{(n+1)!\,(n-1)!}}
$$
Then, because $\frac1{n+1}=1-\frac{n}{n+1}$, we have
$$
\begin{align}
\frac1{n+1}\binom{2n}{n}
&=\binom{2n}{n}-\frac{n}{n+1}\binom{2n}{n}\\
&=\binom{2n}{n}-\binom{2n}{n+1}
\end{align}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Probability of drawing from a urn with and without replacement An urn contains $4$ balls: $1$ red, $1$ black, $1$ white, $1$ yellow.
Two balls are drawn one at a time, compute the probability that:
*
*first is red OR second is white (with replacement)
*first is red OR second is white (without replacement)
(1) $P(1R \cup 2W) = P(1R) + P(2W) - P(1R \cap 2W) = \frac{1}{4}+\frac{1}{4}-\frac{1}{4}\frac{1}{4}=\frac{4+4-1}{16} = \frac{7}{16}$
(2) $P(1R \cup 2W) = P(1R) + P(2W) - P(1R \cap 2W) = \frac{1}{4}+\frac{1}{3}-\frac{1}{4}\frac{1}{3} = \frac{3+4-1}{12} = \frac{6}{12}$
$1R$ means first ball drawn is red, $2W$ means second ball drawn is white
The book says the solution for the second question is $\frac{5}{12}$ but I don't understand why, it seems like it computes as $\frac{1}{4}+\frac{1}{4}-\frac{1}{4}\frac{1}{3}$.
|
Using the law of total probability, we have $P(2W)=P(2W|1W)P(1W) + P(2W|\overline{1W}) P(\overline{1W}) = 0 + \frac{1}{3}\frac{3}{4}$, where $1W$ denotes the event "first taken is white" and $\overline{1W}$ its negation. Plugging that into the equation, you get $\frac{5}{12}$.
|
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|
Surface Integrals - Parametric Representation
Find the parametric representation for the parts of the plane $$2x+3y+z=4$$ where $$1\leq x+y+z\leq 7$$ and $$2\leq x-y\leq4$$.
My attempt:
I thought to let $u=x+y+z$ and $v=x-y$ such that $1\leq u \leq 7$ and $2\leq v\leq4$. But I'm unable to find a suitable parametric representation.
|
All the points that lie on the relevant parts of the plane satisfy the following equations,
\begin{align*}
2x + 3y + z &= 4 \\
x+y+z &= u \\
x-y &= v
\end{align*}
where $1 \leq u \leq 7$ and $2 \leq v \leq 4$.
If we can find expressions for $x$, $y$ and $z$ in terms of $u$ and $v$, we have found a parametrisation. Doing so involves solving the system of three linear equations above for $x$, $y$ and $z$. To do this, we can solve the equation $$\begin{bmatrix}
2 & 3 & 1 \\
1 & 1 & 1 \\
1 & -1 & 0
\end{bmatrix} \begin{bmatrix}
x \\
y\\
z
\end{bmatrix}=
\begin{bmatrix}
4 \\
u\\
v
\end{bmatrix}$$
The solution ends up being $$\begin{bmatrix}
x \\
y\\
z
\end{bmatrix} = \begin{bmatrix}
\frac{-u+2v+4}{3} \\
\frac{-u-v+4}{3}\\
\frac{5u-v-8}{3}
\end{bmatrix}$$
Hence, we can say that your parametrisation is defined by the function $\textbf{x}:\mathbb{R}^2 \rightarrow \mathbb{R}^3$ given by $$\textbf{x}(u,v) = \begin{bmatrix}
\frac{-u+2v+4}{3} \\
\frac{-u-v+4}{3}\\
\frac{5u-v-8}{3}
\end{bmatrix}$$ Where $1 \leq u \leq 7$ and $2 \leq v \leq 4$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Two numbers $x$ and $y$ are chosen at random from the numbers $1,2,3,4,\ldots,2004$. The probability that $x^3+y^3$ is divisible by $3$ is?
Two numbers $x$ and $y$ are chosen at random from the numbers $1,2,3,4,\ldots,2004$. The probability that $x^3+y^3$ is divisible by $3$ is?
The correct answer is $\dfrac13$ while mine is $\dfrac{445}{2003}$
My attempt:
For $x^3+y^3$ to be divisible by $3$, EITHER both $x$ and $y$ should be a multiple of $3$ OR one of them should leave remainder $1$ when divided by $3$ and the other should leave remainder $2$.
Therefore, $$\text{no. of ways} = \frac{668 \times 667 + 668\times 668}{2004\times 2003} = \frac{445}{2003}$$
|
It is immediate (by using arithmetic progression) to get that there are equally $668$ numbers of residues $0,1,2$ modulo $3$. We have
$$(3n)^3+(3n)^3\equiv 0\pmod3\\(3n)^3+(3n+1)^3\equiv 1\pmod3\\(3n)^3+(3n+2)^3\equiv 2\pmod3\\(3n+1)^3+(3n+1)^3\equiv 2\pmod3\\(3n+1)^3+(3n+2)^3\equiv 0\pmod3$$
It follows$$\binom{668}{2}+668^2=669002\space\space\text{favorable cases.}\\\binom{2004}{2}=2007006\space\space\text{possibilities}$$
Thus the probability is $$\frac{669002}{2007006}=\frac13$$
|
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|
Rotate the standard basis and determine the bases that arise from it
Given is $\mathbb{R}^2$ with standard basis $B^2_0$
A "new" basis $B=\left\{\vec{b_1}; \vec{b_2}\right\}$ arises from
$B^2_0$ by rotation of $30°$ (clockwise). Determine the change of
basis $T^{B}_{B^2_0}$ and the basis vectors $\vec{b_1}$ and
$\vec{b_2}$.
So we are in $\mathbb{R}^2$ and we have a standard basis $B^2_0$. Standard matrix is another word for unit matrix I think, so we have
$B^2_0 =\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}$.
Rotate this by $30°$, we have $B=
\begin{pmatrix}
\cos(30°) & -\sin(30°)\\
\sin(30°) & \cos(30°)
\end{pmatrix} =
\begin{pmatrix}
\frac{\sqrt{3}}{2} & -\frac{1}{2}\\
\frac{1}{2} & \frac{\sqrt{3}}{2}
\end{pmatrix}$
So $\vec{b_1} = \begin{pmatrix}
\frac{\sqrt{3}}{2}\\
\frac{1}{2}
\end{pmatrix}$ and $\vec{b_2} =
\begin{pmatrix}
-\frac{1}{2}\\
\frac{\sqrt{3}}{2}
\end{pmatrix}$
And $T^{B}_{B^2_0}$ we can solve with Gaussian elimination, right? But I'm more interested to know if I calculated the rotation matrix and the resulting bases correctly?
|
To calculate the rotation matrix, with reference to the standard basis, it suffices to consider what are the transformed vectors for $\vec e_1$ and $\vec e_2$ after the rotation, that is
$$\vec{e_1} = \begin{pmatrix}
1\\
0
\end{pmatrix}\to \vec{b_1} = \begin{pmatrix}
\frac{\sqrt{3}}{2}\\
\frac{1}{2}
\end{pmatrix}
$$
$$\vec{e_2} = \begin{pmatrix}
0\\
1
\end{pmatrix}\to \vec{b_2} =
\begin{pmatrix}
-\frac{1}{2}\\
\frac{\sqrt{3}}{2}
\end{pmatrix}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Let $f$ be a real polynomial function find relation between the coefficients such that its roots are in an progression
Let $f(x)=ax^3+bx^2+cx+d$, be a polynomial function, find relation between $a,b,c,d$ such that it's roots are in an arithmetic/geometric progression. (separate relations)
So for the arithmetic progression I took let $\alpha = x_2$ and $r$ be the ratio of the arithmetic progression.
We have:
$$x_1=\alpha-2r, \quad x_2=\alpha, \quad x_3=\alpha +2r$$
Therefore:
$$x_1+x_2+x_3=-\frac ba=3\alpha$$
$$x_1^2+x_2^2+x_3^2 = 9\alpha^2-2\frac ca \to 4r^2=\frac {b^2-3ac}{3a^2}$$
$$x_1x_2x_3=\alpha(\alpha^2-4r^2)=-\frac da$$
and we get the final result $2b^3+27a^2d-9abc=0$.
How should I take the ratio at the geometric progression for roots?
I tried something like
$$x_1=\frac {\alpha}q, \quad x_2=\alpha, \quad x_3=\alpha q$$
To get $x_1x_2x_3=\alpha^3$ but it doesn't really work out..
Note:
I have to choose from this set of answers:
$$\text{(a)} \ a^2b=c^2d \quad\text{(b)}\ a^2b^2=c^2d \quad\text{(c)}\ ab^3=c^3d$$
$$\text{(d)}\ ac^3=b^3d \quad\text{(e)}\ ac=bd \quad\text{(f)}\ a^3c=b^3d$$
|
Using your notations
$$ax^3+bx^2+c x+d=a(x-\frac \alpha q)(x-\alpha)(x-\alpha q)$$
Expand the rhs to get after simplifications
$$a x^3 -\frac{a \alpha \left(q^2+q+1\right)}{q}x^2+\frac{a \alpha ^2 \left(q^2+q+1\right)}{q}x-a \alpha ^3$$ Compare the coefficients to get
$$b=-\frac{a \alpha \left(q^2+q+1\right)}{q}$$
$$c=\frac{a \alpha ^2 \left(q^2+q+1\right)}{q}$$
$$d=-a \alpha ^3$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find a corresponding eigenvector for each eigenvalue Let A =$\begin{pmatrix} 6 & 1 \\ 2 & 7 \end{pmatrix}$
(a) Find the eigenvalues of A.
(b) Find a corresponding eigenvector for each eigenvalue in part (a).
My attempt
a) Eigenvalues:
$$\begin{vmatrix} 6-\lambda& 1 \\ 2 & 7-\lambda \end{vmatrix}=0 \Rightarrow \lambda^2-13\lambda +40=0 \Rightarrow \lambda_1=5, \lambda_2=8.$$
b) If $\lambda=5$, then
$\begin{pmatrix} 6-5 & 1 \\ 2 & 7-5 \end{pmatrix}$ = $\begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix}$ $\Rightarrow $ Assuming this as B.
Then $ B\bar { x } =\bar { 0 } $,
$\begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix}$ $\begin{pmatrix}X_1\\X_2\end{pmatrix}=0$
$\begin{pmatrix} 1 & 1 & 0 \\ 2 & 2 & 0 \end{pmatrix}$ By doing row reduction $=>$ $R_2->R_2-2R_1$
$\begin{pmatrix} 1 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$ By doing row reduction $
$X_1+X_2=0$
$X_1=-X_2$
Let $X_2=1$, then $X_1=-1$
$\begin{pmatrix}X_1\\X_2\end{pmatrix}=$ $\begin{pmatrix}-1\\1\end{pmatrix}$
But, the answer is $\begin{pmatrix}X_1\\X_2\end{pmatrix}=$ $\begin{pmatrix}1\\-1\end{pmatrix}$..
I verified many times but I ended up getting the answer as $\begin{pmatrix}X_1\\X_2\end{pmatrix}=$ $\begin{pmatrix}-1\\1\end{pmatrix}$
Can anyone please verify which is the correct answer.
|
Both answers are correct, that is,$$A.\begin{pmatrix}1\\-1\end{pmatrix}=5\begin{pmatrix}1\\-1\end{pmatrix}\iff A.\begin{pmatrix}-1\\1\end{pmatrix}=5\begin{pmatrix}-1\\1\end{pmatrix}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
A limit involving $\cos x$ and $x^2$ The question is finding the value of $L = \lim_{x \to 0}\frac{1 - \cos x \cos 2x \cos 3x}{x^2}$ if it exists . I've found the answer using taylor's formula but I'm looking for other solutions maybe using trigonometric identities .
|
Note that by product to sum formula
$$\cos x \cos 3x = \frac12 (\cos 2x+\cos 4x)\\\implies \cos x \cos 2x \cos 3x = \frac12 (\cos^2 2x+\cos 2x \cos 4x)=\frac12(2\cos^3 2x+\cos^2 2x - \cos 2x )$$
then
$$\frac{1 - \cos x \cos 2x \cos 3x}{x^2}=
\frac{1 - \cos^3 2x-\frac12\cos^2 2x + \frac12\cos 2x }{x^2}=\\\frac{\frac12(1-\cos 2x)(2\cos^2 2x+3\cos 2x+2)}{x^2}=\\2\frac{1-\cos 2x}{4x^2}(2\cos^2 2x+3\cos 2x+2)\to 1 \cdot 7=7$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $\cos(x) \le e^{-x^2/2}$ for $0 \le x \lt \pi/2$. Show that $\cos(x) \le e^{-x^2/2}$ for $0 \le x \lt \pi/2$.
This inequality came up
in my solution to
Show that the sequence $\sum\limits_{k=1}^n\cos\left(\frac kn\right)^{2n^2/k}$ converges.
This is in the category of
"There should be
a number of ways
to prove this."
Here is one way
I came up with.
I am interested in seeing
how many others
there are.
$(\ln(\cos(x))'
=-\tan(x)
\le -x$
for $0 \le x \lt \pi/2$.
Integrating from $0$ to $x$,
since the two sides are $0$
at $x=0$,
$\ln(\cos(x))
\le -x^2/2$
so
$\cos(x)
\le e^{-x^2/2}
$.
|
By Taylor's expansion
$$\cos(x) \le 1-\frac{x^2}2+\frac{x^4}{24} \le 1-\frac{x^2}2+\frac{x^4}{8}-\frac{x^6}{48} \le e^{-x^2/2}$$
indeed
$$1-\frac{x^2}2+\frac{x^4}{24} \le 1-\frac{x^2}2+\frac{x^4}{8}-\frac{x^6}{48}\iff \frac{x^4}{12}-\frac{x^6}{48}\ge 0 \iff x^4(4-x^2)\ge 0$$
|
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|
inclusion-exclusion: class distribution so that professor teaches the same two courses both semesters
How many ways are there to assign each of five professors in a math department to two courses in the fall semester (i.e., $10$ different math courses in all) and then assign each professor two courses in the spring semester such that no professor teaches the same two courses both semesters?
My answer is:
$ P(C(10,2),5) - C(5,1) \cdot 10 \cdot P(C(9,2),4) + C(5,2) \cdot (10 \cdot 9) \cdot P(C(8,2),3) - ... $
And I am wrong. Why am I wrong? You have 10 courses, and 5 professors are supposed to pick one combination of the courses such that (fall, spring) - so it's $P(C(10,2),5)$. By excluding each case of one professor having the same courses for two semesters in a row, it becomes the following sections of the equation $ - C(5,1) \cdot 10 \cdot P(C(9,2),4) + C(5,2) \cdot (10 \cdot 9) \cdot P(C(8,2),3)$ and so on.
What's wrong with this logic?
|
The number of ways five professors can each be assigned two courses in the fall semester is
$$\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
To see this, list the professors in some order, say by seniority. Assign the first professor on the list two of the ten courses, which leaves eight courses available. Assign the next professor on the list two of the remaining eight courses and so forth.
We need to multiply this by the number of ways of assigning each professor two courses in the spring semester without assigning any of those professors the same two courses that the professor taught in the fall semester.
A professor teaches the same two courses both semesters: There are five ways to choose which professor is assigned the same two courses that he or she taught in the fall semester. The remaining eight courses are distributed among the remaining four professors in
$$\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
ways. Thus, there are
$$\binom{5}{1}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
such spring semester assignments.
There are
$$\binom{5}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
ways to assign two professors to teach the same two courses they taught in the fall semester.
There are
$$\binom{5}{3}\binom{4}{2}\binom{2}{2}$$
ways to assign three professors to teach the same two courses they taught in the fall semester.
There are
$$\binom{5}{4}\binom{2}{2}$$
ways to assign four professors to teach the same two courses they taught in the fall semester.
There are
$$\binom{5}{5}$$
ways to assign all five professors to teach the same two courses they taught in the fall semester.
By the Inclusion-Exclusion Principle, the number of admissible spring semester teaching assignments is
$$\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} - \binom{5}{1}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} + \binom{5}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} - \binom{5}{3}\binom{4}{2}\binom{2}{2} + \binom{5}{4}\binom{2}{2} - \binom{5}{5}$$
Therefore, the number of ways five professors can each be assigned two courses in the fall semester and two courses in the spring professor so that no professor teaches the same two courses both semesters is
$$\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}\left[\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} - \binom{5}{1}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} + \binom{5}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} - \binom{5}{3}\binom{4}{2}\binom{2}{2} + \binom{5}{4}\binom{2}{2} - \binom{5}{5}\right]$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Limit of a Function involving tangent function and limits at infinity Determine $$\lim_{x \to \infty}\left(\tan{\frac{\pi x}{2x+1}}\right)^\frac{1}{x}$$.
Attempt
Let $$y=\left(\tan{\frac{\pi x}{2x+1}}\right)^\frac{1}{x}$$
Put $\frac{1}{x}=p$.
$$\lim_{p \to 0}\left(\tan{\frac{\pi}{2+p}}\right)^p$$.
We have
$$\lim_{x \to \infty} y=\lim_{p \to 0}\left(\tan{\frac{\pi}{2+p}}\right)^p$$.
Now consider the function $y$ in variable $p$
Taking $ln$ both sides
$$ln\left(y\right)=p.ln\left(\tan{\frac{\pi}{2+p}}\right)$$.
$$ln\left(y\right)=p.\frac{ln\left(\tan{\frac{\pi}{2+p}}\right)}{\tan{\frac{\pi}{2+p}}}.\tan{\frac{\pi}{2+p}}$$.
Putting $\tan{\frac{\pi}{2+p}}=m$
We have
$$ln\left(y\right)=p.\frac{ln\left(m\right)}{m}.\tan{\frac{\pi}{2+p}}$$.
As $x \to \infty$ we have $p \to 0$ and hence $m \to \infty$
Hence the limit of $\frac{ln\left(m\right)}{m}$ is $0$.
But I am unable to show the limit of other to part of the product.
Please help me out.
|
Note that
$$\left(\tan{\frac{\pi x}{2x+1}}\right)^\frac{1}{x}=\left(\tan{\frac{\pi x+\frac{\pi}2-\frac{\pi}2}{2x+1}}\right)^\frac{1}{x}=\left(\tan{\left(\frac{\pi}2-\frac{\frac{\pi}2}{2x+1}\right)}\right)^\frac{1}{x}=\left(\tan{\frac{\pi}{4x+2}}\right)^{-\frac{1}{x}}=e^{-\frac{\log{\left(\tan{\frac{\pi}{4x+2}}\right)}}{x}}\to 1$$
Indeed
$$-\frac{\log{\left(\tan{\frac{\pi}{4x+2}}\right)}}{x}=-\left(\tan{\frac{\pi}{4x+2}}\right)\log{\left(\tan{\frac{\pi}{4x+2}}\right)}\frac{\frac{\pi}{4x+2}}{\tan{\frac{\pi}{4x+2}}}\frac{\frac{4x+2}{\pi}}{x}\to0\cdot1\cdot \frac4 \pi=0$$
As an alternative, according to the change of variable $p=\frac 1 x \to 0$, from here
$$\left(\tan{\frac{\pi}{2+p}}\right)^p=e^{p\log \left(\tan{\frac{\pi}{2+p}}\right)}\to 1$$
indeed
$$p\log \left(\tan{\frac{\pi}{2+p}}\right)=p\tan{\frac{\pi}{2+p}}\frac{\log \left(\tan{\frac{\pi}{2+p}}\right)}{\tan{\frac{\pi}{2+p}}}\to 0$$
indeed
$$\frac{\log \left(\tan{\frac{\pi}{2+p}}\right)}{\tan{\frac{\pi}{2+p}}}\to 0$$
$$p\tan{\frac{\pi}{2+p}}=\frac{p}{\tan\left({\frac{\pi}2-\frac{\pi}{2+p}}\right)}=\frac{p}{\tan\left({\frac{\pi p}{2p+4}}\right)}=\frac{\frac{\pi p}{2p+4}}{\tan\left({\frac{\pi p}{2p+4}}\right)}\frac{2p+4}{\pi}\to \frac 4 \pi$$
|
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|
What order of operation should I use calculating $2^3 \times 4 - 6 \times 3 ÷ (4 + 3) - 16 ÷ 4$? I am struggling if I should use PEMDAS or BODMAS in this equation. What is the right method to get around these type of equations?
$$2^3 \times 4 - 6 \times 3 ÷ (4 + 3) - 16 ÷ 4 = \ ?$$
|
You silly humans with your mnemonics that just wind up confusing you even more! Mwahahaha! Mwahahahahahahahahaha!
Now, let's see: PEMDAS is Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. And BODMAS is Brackets, Orders, Division, Multiplication, Addition, Subtraction.
Parentheses are pretty much the same thing as brackets. Exponents and orders might be different things, in that all exponents are orders but not all orders are exponents. Or am I trying to confuse you further? Behold: $$\sqrt x = x^{\frac{1}{2}}, \root 3 \of x = x^{\frac{1}{3}}, \root 4 \of x = x^{\frac{1}{4}}, \root 5 \of x = x^{\frac{1}{5}}, \ldots$$
After that, the two mnemonics are the same except that they switch the order of multiplication and division. This was on purpose, to create further confusion, and obfuscate the fact that multiplication and division are basically the same thing, just notated differently, e.g., $$\frac{x}{2} = x \times \frac{1}{2}, \frac{x}{3} = x \times \frac{1}{3}, \frac{x}{4} = x \times \frac{1}{4}, \ldots$$
Then $$2^3 \times 4 - 6 \times 3 ÷ (4 + 3) - 16 ÷ 4$$ $$= 2^3 \times 4 - 6 \times 3 ÷ 7 - 16 ÷ 4$$ $$= 8 \times 4 - 6 \times 3 ÷ 7 - 16 ÷ 4$$ $$= 8 \times 4 - 6 \times \frac{3}{7} - 4 \textrm{ or } 32 - 18 ÷ 7 - 16 ÷ 4$$ $$= 32 - \frac{18}{7} - 4 \textrm{ or } 32 - \frac{18}{7} - 4$$
You see, they're the same.
|
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|
When does $\frac{a+b}{2}$ and $\sqrt{ab}$ have inversed tens digits and ones digits? Let $a$ and $b$ be natural numbers, and
$$A = \frac{a+b}{2}$$
$$B = \sqrt{ab}$$
It's given that $A$ and $B$ are two-digit numbers such that the tens digit of $A$ is the same as the ones digit of $B$, and the tens digit of $B$ is the same as the ones digit of $A$.
So $A = 10x + y\;\,$and$\;B = 10y + x$.
Also given is $A\ne B$.
What is $a$ and $b$?
My teacher gave us the answer without explaining it as: $a = 98$ and $b = 32$, which makes $A = 65$ and $B=56$.
My question is: How do you prove this? I know $98 = 2\cdot 7^2$ and $32 = 2^5$, but I don't understand how to find this specific answer.
|
Rearrange
$B=\sqrt{ab}$
to get
$B^2/a=b.$
Substitute in to get
$2A=a+B^2/a.$
Rearrange and solve:
$a^2-2Aa+B^2=0$
$(a-A)^2-A^2+B^2=0$
$a=A\pm\sqrt{A^2-B^2}.$
So we need $A^2-B^2=C^2$ for some $C$.
$(10x+y)^2-(10y+x)^2=C^2$
$((10x+y)-(10y+x))((10x+y)+(10y+x))=C^2$
$(9x-9y)(11x+11y)=C^2$
$3^211(x-y)(x+y)=C^2$
So we need $11(x-y)(x+y)$ to be a square for digits $x$ and $y$. Clearly we need one of the factors to be $\pm 11$ and the other to be $\pm 1$. Then since $x-y$ is smaller than $x+y$ we need $x-y=1$, $x+y=11$. Hence $x=6$ and $y=5$.
|
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|
Prove $\sin(\alpha -\beta)+\sin(\alpha-\gamma)+\sin(\beta-\gamma)=4\cos\frac{\alpha-\beta}{2}\sin\frac{\alpha-\gamma}{2}\cos\frac{\beta-\gamma}{2}$ Here is a problem from Gelfand's Trigonometry:
Let $\alpha, \beta, \gamma$ be any angle, show that $$\sin(\alpha -\beta)+\sin(\alpha-\gamma)+\sin(\beta-\gamma)=4\cos\left(\frac{\alpha-\beta}{2}\right)\sin\left(\frac{\alpha-\gamma}{2}\right)\cos\left(\frac{\beta-\gamma}{2}\right).$$
I have tried to worked through this problem but cannot complete it. If I let $A= \alpha -\beta$, $B=\beta-\gamma$ and $C= \beta-\gamma$, and $A+B+C=\pi$ (now $A$, $B$ and $C$ are angles of a triangle), then I could prove the equality. But without this condition, I am stuck.
Could you show me how to complete this exercise?
|
$$
\begin{align}
\color{#C00}{\sin(x)+\sin(y)}+\color{#090}{\sin(x+y)}
&=\color{#C00}{2\sin\left(\frac{x+y}2\right)\cos\left(\frac{x-y}2\right)}+\color{#090}{2\sin\left(\frac{x+y}2\right)\cos\left(\frac{x+y}2\right)}\\
&=2\sin\left(\frac{x+y}2\right)\left[\cos\left(\frac{x-y}2\right)+\cos\left(\frac{x+y}2\right)\right]\\
%&=2\sin\left(\frac{x+y}2\right)\,\color{#00F}{2\cos\left(\frac x2\right)\cos\left(\frac y2\right)}\\
%&=4\sin\left(\frac{x+y}2\right)\cos\left(\frac x2\right)\cos\left(\frac y2\right)
\end{align}
$$
Finish off by using the formula for the cosine of a sum/difference, then set $x=\alpha-\beta$ and $y=\beta-\gamma$.
|
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|
Is there a fast way to prove a tridiagonal matrix is positive definite? I' m trying to prove that
$$A=\begin{pmatrix}
4 & 2 & 0 & 0 & 0 \\
2 & 5 & 2 & 0 & 0 \\
0 & 2 & 5 & 2 & 0 \\
0 & 0 & 2 & 5 & 2 \\
0 & 0 & 0 & 2 & 5 \\
\end{pmatrix}$$
admits a Cholesky decomposition.
$A$ is symmetric, so it admits a Cholesky decomposition iff it is positive definite. The only methods I know for checking this are:
*
*$X^tAX > 0, \quad \forall X \in \mathbb{K}^n- \{0\}$.
*If $\lambda$ is an eigenvalue of $A$, then $\lambda>0.$
I have failed to prove it using 1 and 2 is taking me so much time. Is there any easier way to do this, given that $A$ is tridiagonal?
|
Notice $A$ can be rewritten as a sum of 5 matrices.
$$A =
\left[\begin{smallmatrix}
2 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 3
\end{smallmatrix}\right] +
\left[\begin{smallmatrix}
2 & 2 & 0 & 0 & 0\\
2 & 2 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0
\end{smallmatrix}\right] +
\left[\begin{smallmatrix}
0 & 0 & 0 & 0 & 0\\
0 & 2 & 2 & 0 & 0\\
0 & 2 & 2 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0
\end{smallmatrix}\right] +
\left[\begin{smallmatrix}
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 2 & 2 & 0\\
0 & 0 & 2 & 2 & 0\\
0 & 0 & 0 & 0 & 0
\end{smallmatrix}\right] +
\left[\begin{smallmatrix}
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 2 & 2\\
0 & 0 & 0 & 2 & 2
\end{smallmatrix}\right]
$$
The first matrix
is diagonal with positive entries on diagonals, so it is positive definite.
The remaining four matrices are clearly positive semi-definite. Being a sum
of a positive definite matrix with a bunch of positive semi-definite matrices,
$A$ itself is positive definite.
|
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|
Simplifying $\operatorname{tanh}(\operatorname{arsinh}(x))$ So I am trying to simplify $\tanh(\operatorname{arsinh}(x))$ to $\frac{x}{\sqrt{1+x^2}}$
In general,
$$\tanh(x) = \frac{e^x-e^{-x}}{e^x+e^{-x}}$$
and $$\operatorname{arsinh}(x)= \ln(x+\sqrt{x^2-1})$$
therefore
\begin{align}\tanh(\operatorname{arsinh}(x)) & =\frac{e^{\ln(x+\sqrt{x^2-1})}-e^{\ln(\frac{1}{x+\sqrt{x^2-1}})}}{e^{\ln(x+\sqrt{x^2-1})}+e^{\ln(\frac{1}{x+\sqrt{x^2-1}})}}\\
&=\frac{x+\sqrt{x^2-1}-\frac{1}{x+\sqrt{x^2-1}}}{x+\sqrt{x^2-1}+\frac{1}{x+\sqrt{x^2-1}}} \\
&=\frac{(x+\sqrt{x^2-1})^2-1}{(x+\sqrt{x^2-1})^2+1} \\
&=\frac{x^2+2x\sqrt{x^2-1}+x^2-1-1}{x^2+2x\sqrt{x^2-1}+x^2-1+1} \\
&=\frac{x^2+x\sqrt{x^2-1}-1}{x^2+x\sqrt{x^2-1}} \\
&=\frac{x^2+x\sqrt{x^2-1}}{x^2+x\sqrt{x^2-1}}-\frac{1}{x^2+x\sqrt{x^2-1}} \\
&=1-\frac{1}{x(x+\sqrt{x^2-1})} \\
&=1-\frac{x-\sqrt{x^2-1}}{x(x+\sqrt{x^2-1})(x-\sqrt{x^2-1})} \\
&=1-\frac{x-\sqrt{x^2-1}}{x(x^2-x^2+1)} \\
&=1-\frac{x-\sqrt{x^2-1}}{x} \\
&=\frac{\sqrt{x^2-1}}{x} \\
&\ne \frac{x}{\sqrt{1+x^2}} \end{align}
I'm struggling to get my answer into the required form.
|
Your route is fine, but you have just made a mistake from the beginning, observe that
$$
\operatorname{arsinh} x =\ln \left ( x + \sqrt{x^2 \color{red}{+} 1} \right )\ne \ln(x+\sqrt{x^2-1}).
$$
|
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"url": "https://math.stackexchange.com/questions/2777721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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|
Finding the residue of $\frac{1}{z^3 \sin z}$ at z = 0 Given the function $f(z) = \frac{1}{z^3 \sin{(z)}}$, what is the residue of $f(z)$ at $z = 0$?
I want to find $b_1$ from the Laurent expansion. So I did the following:
\begin{align*}
\frac{1}{z^3 \sin{(z)}}
&= ( \dots \frac{b_5}{z^5} + \frac{b_4}{z^4} + \frac{b_3}{z^3} + \frac{b_2}{z^2} + \frac{b_1}{z} + a_0 + a_1z + \dots )\\
1& = \Big ( \dots \frac{b_5}{z^5} + \frac{b_4}{z^4} + \frac{b_3}{z^3} + \frac{b_2}{z^2} + \frac{b_1}{z} + a_0 + a_1z + \dots \Big ) \cdot \Big ( z^3 \sin{(z)} \Big )\\
&= \Big ( \dots \frac{b_5}{z^2} + \frac{b_4}{z} + b_3 + b_2z + b_1z^2 + a_0z^3 + a_1z^4 + \dots \Big ) \cdot \Big ( \sin{(z)} \Big )\\
&= \Big ( \dots \frac{b_5}{z^2} + \frac{b_4}{z} + b_3 + b_2z + b_1z^2 + a_0z^3 + a_1z^4 + \dots \Big ) \cdot \Big ( z - \frac{z^3}{3!} + \frac{z^5}{5!} - \dots \Big )\\
\end{align*}
After some more thought...
Is it true to say that because f(z) has a pole of order 4 at $z=0$ that our $b_n$'s only go out to the 4th term? Meaning there are no $b_5$, $b_6$, etc like how wrote previously. That is,
$\frac{1}{z^3 \sin{(z)}}
= \Big ( \frac{b_4}{z^4} + \frac{b_3}{z^3} + \frac{b_2}{z^2} + \frac{b_1}{z} + a_0 + a_1z + \dots \Big )\\$
followed by
\begin{align*}
1
&= \Big ( \frac{b_4}{z} + b_3 + b_2z + b_1z^2 + a_0 + \dots \Big ) \cdot \Big ( z - \frac{z^3}{3!} + \frac{z^5}{5!} - \dots\Big )\\
\end{align*}
Which then when multiplying out $b_1z^2$ with each term from sin(z)'s Laurent expansion will never yield a $\frac{1}{z}$ term, concluding that the coefficient $b_1 = 0$?
|
I came up with the solution my professor was looking for and figured I would share it here for both my own clarification and for anyone else in the future who wants to find the solution via a Laurent expansion.
For our function $f(z) = \frac{1}{z^3 \sin{(z)}}$, we can solve for the residue of $f$ at $z = 0$ by doing the following:
We know that this is a pole of order 4, as $z^3$ has order 3 at $z=0$ and $\sin{(z)}$ has a pole of order 1 (simple pole) at $z=0$. So our general Laurent series for $$\frac{1}{z^3 \sin{(z)}} = \frac{b_4}{z^4} + \frac{b_3}{z^3} + \frac{b_2}{z^2} + \frac{b_1}{z} + a_0 + \dots$$ We want to solve for $b_1$, which is our residue.
We can do some algebra and some known Laurent expansions to get
\begin{align*}
\frac{1}{z^3 \sin{(z)}} &= \frac{b_4}{z^4} + \frac{b_3}{z^3} + \frac{b_2}{z^2} + \frac{b_1}{z} + a_0 + \dots\\
1 &= \Big ( \frac{b_4}{z^4} + \frac{b_3}{z^3} + \frac{b_2}{z^2} + \frac{b_1}{z} + a_0 + \dots \Big ) \cdot \Big (z^3 \sin{(z}) \Big )\\
1 &= \Big ( \frac{b_4}{z} + b_3 + b_2z + b_1z^2 + a_0z^3 + \dots \Big ) \cdot \Big (\sin{(z}) \Big )\\
1 &= \Big ( \frac{b_4}{z} + b_3 + b_2z + b_1z^2 + a_0z^3 + \dots \Big ) \cdot \Big (z - \frac{z^3}{3!} + \frac{z^5}{5!} - \dots \Big)\\
\end{align*}
Gathering all the constant terms from the left and right sides of the equation, we have
\begin{align*}
1 &= \frac{b_4}{z} \cdot z\\
& = b_4
\end{align*}
So $b_4 = 1$.
Similarly, gathering all the z terms from the left and right sides, we have
\begin{align*}
0z &= b_3 \cdot z\\
0 &= b_3z
\end{align*}
So $b_3 = 0$.
Remembering that $b_4 = 1$ and gathering all the $z^2$ terms from the left and right sides, we have
\begin{align*}
0z^2 &= \frac{b_4}{z} \cdot \Big (-\frac{z^3}{3!} \Big) + b_2z \cdot z\\
0 &= -\frac{b_4}{3!}z^2 +b_2z^2\\
0 &= \Big (-\frac{1}{3!} + b_2 \Big )z^2\\
\implies b_2&= \frac{1}{3!}
\end{align*}
So $b_2 = \frac{1}{6}$.
Remembering that $b_3 = 0$ and gathering all the $z^3$ terms, we have
\begin{align*}
0z^3 &= -\frac{b_3}{3!}z^3 + b_1z^3\\
\implies 0 &= -\frac{0}{3!} + b_1\\
\implies b_1 &= 0
\end{align*}
Thus the residue of $f(x) = \frac{1}{z^3 \sin{(z)}} = b_1 = 0$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Power series of $f(x)=\ln (x^2+4)$ I am supposed to find a power series representation of
$$f(x)=\ln\left(x^{2}+4\right).$$
Then, I am to graph it and observe what happens as $n$ increases. My attempt at a solution:
$$\ln\left(x^2+4\right) = \int \frac{1}{x^2+4}\,dx = \frac{1}{4}\int \frac{1}{\frac{x^2}{4}+1}\,dx = \frac{1}{4}\int \frac{1}{1-\left(-\frac{x^2}{4}\right)}\,dx.$$
Now that it is in the $\frac{1}{1-x}$ format, the power series representation is $\sum_{n=0}^{\infty} \int \frac{1}{4}\left(-\frac{x^2}{4}\right)^n$. The first terms are
$$\frac{1}{4} \left(x-\frac{x^5}{80}+\frac{x^7}{448}-\frac{x^9}{2304}+\cdots\right).$$
But as I graph these, they look nothing like the graph of $f(x)=\ln(x^2 +4)$. I am not sure if I turned the revised formula into a sum correctly.
|
By starting with $g(x) = \ln(1 + x)$ it can be determined that $g^{(n)}(x) = (-1)^{n-1} \, (n-1)! (1+x)^{-(n+1)}$ for $n \geq 1$ along with $g(0) = \ln(1 + 0) = 0$ and $g^{(n)}(0) = (-1)^{n-1} \, (n-1)!$. From this it is developed
$$\ln(1 + x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{n}}{n}.$$
Now, as Mark Viola has pointed out in his comment, it can be seen that
\begin{align}
\ln(x^2 + 4) &= \ln\left( 4 \, \left(1 + \left(\frac{x}{2}\right)^{2} \right) \right) = \ln(4) + \ln\left(1 + \left(\frac{x}{2}\right)^{2} \right) \\
&= 2 \, \ln(2) + \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{2 n}}{4^{n} \, n} \\
&= 2 \, \ln(2) + \frac{x^{2}}{4} - \frac{x^{4}}{32} + \frac{x^{6}}{192} - \mathcal{O}(x^{8})
\end{align}
In terms of the Proposer's work:
With
\begin{align}
\ln\left(x^2+4\right) &= \int \frac{2 \, x}{x^2+4}\,dx = \frac{1}{2} \, \int \frac{x}{\frac{x^2}{4}+1}\,dx + c_{0} \\
&= \frac{1}{2} \, \int \frac{x}{1-\left(-\frac{x^2}{4}\right)}\,dx + c_{0} \\
&= \frac{1}{2} \, \sum_{n=0}^{\infty} \, \int x \, \left(-\frac{x^2}{4}\right)^n + c_{0} \\
&= \frac{1}{2} \, \sum_{n=0}^{\infty} \frac{(-1)^{n} \, x^{2n+2}}{4^{n} \, (2n+2)} + c_{0} \\
&= \frac{1}{4} \, \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{2n}}{4^{n-1} \, n} + c_{0} \\
&= \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{2n}}{4^{n} \, n} + c_{0}.
\end{align}
The value of $c_{0}$, an essential constant of integration may be obtained by setting $x = 0$ and yields $c_{0} = \ln(4)$. This then yields the result
$$\ln(x^2 + 4) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{2n}}{4^{n} \, n} + 2 \, \ln 2. $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integrate $\sin^{-1}\frac{2x}{1+x^2}$
Integrate $\sin^{-1}\frac{2x}{1+x^2}$
The solution is given in my reference as: $2x\tan^{-1}x-\log(1+x^2)+C$.
But, is it a complete solution ?
My Attempt
$$
\int 2\tan^{-1}x \, dx=\int \tan^{-1}x \cdot 2\,dx=\tan^{-1}x\int2\,dx-\int\frac{1}{1+x^2}\int2\,dx\cdot dx\\
=\tan^{-1}x \cdot2x-\int\frac{2x}{1+x^2}\,dx=2x\tan^{-1}x-\log(1+x^2)+C
$$
$$
2\tan^{-1}x=\begin{cases}\sin^{-1}\frac{2x}{1+x^2}\text{ if }|x|\leq{1}\\
\pi-\sin^{-1}\frac{2x}{1+x^2}\text{ if }|x|>{1}\text{ and }x>0\\
-\pi-\sin^{-1}\frac{2x}{1+x^2}\text{ if }|x|>{1}\text{ and }x<0
\end{cases}\\
\sin^{-1}\frac{2x}{1+x^2}=\begin{cases}2\tan^{-1}x\text{ if }|x|\leq{1}\\
\pi-2\tan^{-1}x\text{ if }|x|>{1}\text{ and }x>0\\
-\pi-2\tan^{-1}x\text{ if }|x|>{1}\text{ and }x<0
\end{cases}
$$
$$
\int\sin^{-1}\frac{2x}{1+x^2}\,dx=\begin{cases}\int2\tan^{-1}x\,dx&\text{ if } |x|\leq{1}\\\int\pi\, dx-\int2\tan^{-1}x\,dx&\text{ if }|x|>{1}&\text{ and } x>0\\-\int\pi \,dx-\int2\tan^{-1}x\,dx&\text{ if }|x|>{1}&\text{ and } x<0\end{cases}=\begin{cases}\color{red}{2x\tan^{-1}x-\log(1+x^2)+C\text{ if } |x|\leq{1}}\\\pi x-2x\tan^{-1}x+\log(1+x^2)+C&\text{ if }|x|>{1}&\text{ and } x>0\\-\pi x-2x\tan^{-1}x+\log(1+x^2)+C&\text{ if }|x|>{1}&\text{ and } x<0\end{cases}$$
So don't we have two more cases for our solution rather than that is given in my reference, right ?
|
The "identity"
$$\arcsin\left(2x\over1+x^2\right)=2\arctan x$$
does not hold for all $x$. You can see this by comparing the ranges of the two sides. The arcsine function has $[-\pi/2,\pi/2]$ as its range, but twice the arctangent function has $(-\pi,\pi)$ as its range. In particular, if $x\gt1$, then $2\arctan x\gt\pi/2$, and likewise if $x\lt-1$ then $2\arctan x\lt-\pi/2$ (since $\arctan1=\pi/4$).
The correct identity is
$$\arcsin\left(2x\over1+x^2\right)=
\begin{cases}
\pi-2\arctan x\qquad\text{if }x\ge1\\
2\arctan x\qquad\quad\text{ }\text{ if }-1\le x\le1\\
-\pi-2\arctan x\quad\text{ if }x\le-1
\end{cases}$$
(Note, I've intentionally overlapped the intervals at $x=\pm1$ to stress the agreement of $\pi-2\arctan1=2\arctan1$ and $2\arctan(-1)=-\pi-2\arctan(-1)$.) The OP's observations regarding the integral follow.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Sum of $1+4\epsilon +9\epsilon^2 +16\epsilon^3+...+2018^2 \epsilon^{2017}$ I have this problem from a college exam: Let $\epsilon$ be a non-real root of unity of order 2018, find the sum$$S=1+4\epsilon +9\epsilon^2 +16\epsilon^3+...+2018^2 \epsilon^{2017}$$Here is my try. First I considered $$S_1=\sum_{k=0}^{2018} x^k=\frac{1-x^{2019}}{1-x}$$ Now I derivate one time then I multiply again by x to get: $$\sum_{k=0}^{2018} kx^k=\frac{x-x^{2020}-2019x^{2019}}{(1-x)^2}$$ And now I must derivate one more time and set $x=\epsilon$ in order to get the desired sum: $$S=\sum_{k=0}^{2018} k^2\epsilon^{k-1}=\frac{2019^2\epsilon^{2018}-2020\epsilon^{2020}-(2019^2-3\cdot2019-1)\epsilon^{2019}-\epsilon-1}{(1-\epsilon)^3}$$ And my final simplification to the numerator was:$$\epsilon(2019^2-2021)-2018\epsilon^2-2019^2-1$$ Now there were 5 answers given, and not a single one was even close to this one. Out of luck because only 2 answer had $1-\epsilon$ in the denominator I have choosen the correct one, which was: $$S=\frac{2018(2018\epsilon-2020)}{(1-\epsilon)^2}$$ Can you help me to get that answer?
|
Using the operator $\delta = x \, \frac{d}{dx}$ leads to
\begin{align}
\sum_{k=0}^{n} x^{k} &= \frac{1 - x^{n+1}}{1-x} \\
\sum_{k=0}^{n} k \, x^{k} &= \delta \left(\frac{1 - x^{n+1}}{1-x} \right) = \frac{x - (n+1) x^{n+1} + n x^{n+2}}{(1-x)^2} \\
\sum_{k=0}^{n} k^{2} \, x^{k} &= \delta^{2} \left(\frac{1 - x^{n+1}}{1-x} \right) = \frac{x(1 + x - (n+1)^2 x^n + (2n^2 +2n -1) x^{n+1} - n^2 x^{n+2})}{(1-x)^{3}}
\end{align}
Letting $x^{n} =1$ gives
\begin{align}
\sum_{k=1}^{n} k^{2} \, x^{k-1} &= \frac{1 + x - (n+1)^2 + (2n^2 +2n -1) x - n^2 x^2}{(1-x)^{3}} \\
&= - \frac{n \, (n+2 + 2 (n+1) x + n x^2)}{(1-x)^3}
\end{align}
Now let $n=2018$.
|
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"url": "https://math.stackexchange.com/questions/2782393",
"timestamp": "2023-03-29T00:00:00",
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|
Laurent series of $f(z)=\frac{z-1}{z(z^3-1)}$ for $0<|z|<1$ and $|z|>1$. I need to find the Laurent series of $f(z)=\frac{z-1}{z(z^3-1)}$ for $0<|z|<1$ and $|z|>1$.
I believe I am nearly there but I have a confusion.
Notice $f(z)=\frac{z^2-1}{1-z^3}+\frac{1}{z}$ so if $0<|z|<1$ we have
$$f(z)=\frac{1}{z}+(z^2-1)\sum_{n=0}^{\infty}z^{3n}=\frac{1}{z}+\sum_{n=0}^{\infty}(-z^{3n}+z^{3n+2})$$
First question: Is there any better way to present this series or is this in general acceptable?
Now for $|z|>1$, $f$ is entire so it should have a Taylor Series, however I am getting this:
$$f(z)=\frac{1}{z}+\left(\frac{1+z^2}{z^3}\right)\left(\frac{1}{1-\frac{1}{z^3}}\right)=\frac{1}{z}+\sum_{n=1}^{\infty}\left(z^{-3n}+z^{-3n+2}\right)$$
and this is not a Taylor series...
Edit $1$: For the case $0<|z|<1$ I have tried a different approach and got
$$f(z)=\frac{1-z}{z}\left(\frac{1}{1-z^3}\right)=\sum_{n=0}^{\infty}(z^{3n-1}-z^{3n})$$
It is the same series but a much cleaner approach
|
If $|z|<1$, then\begin{align}\frac{z-1}{z^3-1}&=\frac{1-z}{1-z^3}\\&=(1-z)(1+z^3+z^6+z^9+\cdots)\\&=1-z+z^3-z^4+z^6-z^7+\cdots\end{align}and therefore$$f(z)=z^{-1}-1+z^2-z^3+z^5-z^6+\cdots$$If $|z|>1$, then\begin{align}\frac{z-1}{z^3-1}&=\frac{1-z}{1-z^3}\\&=-(1-z)(z^{-3}+z^{-6}+z^{-9}+\cdots)\\&=z^{-2}-z^{-3}+z^{-5}-z^{-6}+\cdots\end{align}and therefore$$f(z)=z^{-3}-z^{-4}+z^{-6}-z^{-7}+\cdots$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2782936",
"timestamp": "2023-03-29T00:00:00",
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|
Find the values of $(2\sin x-1)(\cos x+1)=0$ How many different value of x from 0° to 180° for the equation $(2\sin x-1)(\cos x+1) = 0$?
The solution shows that one of these is true:
$\sin x = \frac12$ and thus $x = 30^\circ$ or $120^\circ$
$\cos x = -1$ and thus $x = 180^\circ$
Question: Inserting the $\arcsin$ of $1/2$ will yield to $30°$, how do I get $120^\circ$? and what is that $120^\circ$, why is there $2$ value but when you substitute $\frac12$ as $x$, you'll only get $1$ value which is the $30^\circ$?
Also, when I do it inversely: $\sin(30^\circ)$ will result to 1/2 which is true as $\arcsin$ of $1/2$ is $30^\circ$. But when you do $\sin(120^\circ)$, it will be $\frac{\sqrt{3}}{2}$, and when you calculate the $\arcsin$ of $\frac{\sqrt{3}}{2}$, it will result to $60^\circ$ and not $120^\circ$. Why?
|
If $ab = 0$, then either $a=0$ or $b=0$.
Given $(2\sin x - 1)(\cos x + 1) = 0$, we can separate these and solve separately.
So... $\ \ \ \ \ 2\sin x -1 = 0$
$\Rightarrow \sin x = \frac12 \Rightarrow x = 30^\circ, 150^\circ$
We get those two values because $\sin x$ is symmetrical across the $y$-axis.
Also... $\ \ \ \ \ \cos x + 1 = 0$
$\Rightarrow \cos x = -1 \Rightarrow x = 180^\circ$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Testing $\sum\limits_{k=1}^∞(\frac{k+1}k)^{k^2}3^{-k}$ for convergence and absolute convergence
Test $$\sum_{k=1}^{\infty}\left(\frac{k+1}{k}\right)^{k^2}3^{-k}$$ for convergence and absolute convergence.
We apply the ratio test for $\displaystyle \sum_{k=1}^{\infty}\left|\left(\frac{k+1}{k}\right)^{k^2}3^{-k}\right|$:
$$
\left|\frac{\left(\dfrac{k+2}{k+1}\right)^{(k+1)^2}3^{-(k+1)}}{\left(\dfrac{k+1}{k}\right)^{k^2}3^{-k}}\right|=\left|\frac{\left(\left(\dfrac{k+2}{k+1}\right)^{k}\right)^2\left(\dfrac{k+2}{k+1}\right)^{2k+1}3^{-k}\dfrac{1}{3}}{\left(\dfrac{k+1}{k}\right)^{k^2}3^{-k}}\right|→\frac{e^2\cdot e^2\cdot\dfrac13}{e^2}=\frac{1}{3}e^2.
$$
Since $\dfrac{1}{3}e^2 \geq 1$, the series diverges. Is this correct? I feel like I made a mistake somewhere that I cannot pin down.
|
Yet another way:
\begin{align*}
\frac{a_{k+1}}{a_{k}} & =\frac{1}{3}\left(1+\frac{1}{k}\right)^{-k^{2}}\left(1+\frac{1}{1+k}\right)^{(1+k)^{2}}\\
& =\frac{1}{3}\left(1+\frac{1}{k}\right)^{-k^{2}}\left(1+\frac{1}{1+k}\right)^{k^{2}+k}\left(1+\frac{1}{1+k}\right)^{k+1}.
\end{align*}
Now, if
$$
b_{k}\equiv\left(1+\frac{1}{k}\right)^{-k^{2}}\left(1+\frac{1}{1+k}\right)^{k^{2}+k}
$$
has limit one, we can conclude $a_{k+1}/a_{k}\rightarrow e/3$. Take
logs to get
$$
\log b_{k}=k\left[\left(1+k\right)\log\left(1+\frac{1}{1+k}\right)-k\log\left(1+\frac{1}{k}\right)\right].
$$
As in Claude Leibovici's answer, expand around $\infty$ to get
$$
\log b_{k}=k\left(\frac{1}{2k^{2}}+O\left(\frac{1}{k^{3}}\right)\right)\rightarrow0
$$
so that $b_{k}=e^{\log b_k}\rightarrow e^0=1$, as desired.
|
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"url": "https://math.stackexchange.com/questions/2789545",
"timestamp": "2023-03-29T00:00:00",
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|
Solve the indefinite integral $\int \frac{1}{x^{1/2}+x^{1/3}} dx$
Solve $\int \frac{1}{x^{1/2}+x^{1/3}} dx$
My Attempt
$$
\int \frac{1}{x^{1/2}+x^{1/3}} dx=\int\frac{dx}{x^{1/2}(1+x^{-1/6})}
$$
Put $t=x^{1/2}\implies dt=\frac{dx}{2.x^{1/2}}\implies\frac{dx}{x^{1/2}}=2dt$
$$
\int \frac{1}{x^{1/2}+x^{1/3}} dx=2\int\frac{dt}{1+\frac{1}{t^{1/3}}}
$$
What is the easiest method to solve this indefinite integral and how do I choose a proper substitution in such problems ?
|
$$\int \frac{1}{\sqrt{x}+\sqrt[3]{x}}dx=6\left(\frac{(x^{\frac{1}{6}}+1)^3}{3}-\frac{3(x^{\frac{1}{6}}+1)^2}{2}+3(x^{\frac{1}{6}}+1)-\ln \left|x^{\frac{1}{6}}+1\right|\right)+k, \,k\in \mathbb Z$$
Proof:
If I take this substitution $t=x^{\frac{1}{6}}$ we have:
$$\int \frac{1}{\sqrt{x}+\sqrt[3]{x}}dx=\int \frac{6t^3}{t+1}dt$$
With another substitution $v=t+1$ we have:
$$\int \frac{6t^3}{t+1}dt=6\cdot \int \:\frac{(v-1)^3}{v}dv=6\left(\int \:v^2dv-\int \:3vdv+\int \:3dv-\int \frac{1}{v}dv\right)=$$
Hence
$$=6\left(\frac{v^3}{3}-\frac{3v^2}{2}+3v-\ln \left|v\right|\right)+k=*$$
and considering the two substitutions we have:
$$*=6\left(\frac{(x^{\frac{1}{6}}+1)^3}{3}-\frac{3(x^{\frac{1}{6}}+1)^2}{2}+3(x^{\frac{1}{6}}+1)-\ln \left|x^{\frac{1}{6}}+1\right|\right)+k, \,k\in \mathbb Z$$.
|
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"url": "https://math.stackexchange.com/questions/2790218",
"timestamp": "2023-03-29T00:00:00",
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|
Can I Systematically Determine the Cardinality of a Finite Set? Is there any way you can, in a systematically manner, determine the cardinality of a finite set in which the elements are given by a specific expression?
For example, let $A$ be a set $\ A=\{1,2,3,4,5,6\}$ and $B$ be a set defined by:
$$B=\biggl\{\frac{a-b}{a+b} : a,b\in A\biggl\}$$
My question is thus, can I somehow determine the cardinality of the set without having to calculate and enumerate each and every element?
|
It's really dependent on how you built the set. The thing that can make things rather tricky is when things in the set are equal only one thing is counted. So usually inequalities are easy but the exact number you'd have to eleminate all equivalent elements.
For example, let $A$ be a set $\ A=\{1,2,3,4,5,6\}$ and $B$ be a set
defined by: $$B=\biggl\{\frac{a-b}{a+b} : a,b\in A\biggl\}$$
Since $|A|=6$ and you are taking two elements in $A$ you get the easy inequality $|B| \le 6^2=36$.
Since $B$ has at least $1$ element $1 \le |B|$
Enumerateing through the values and evaluating them in the function $\frac{a-b}{a+b}$ i get the following table
$\begin{array}{rrrrrr}
0 & \frac{1}{3} & \frac{1}{2} & \frac{3}{5} &
\frac{2}{3} & \frac{5}{7} \\
-\frac{1}{3} & 0 & \frac{1}{5} & \frac{1}{3} &
\frac{3}{7} & \frac{1}{2} \\
-\frac{1}{2} & -\frac{1}{5} & 0 & \frac{1}{7} &
\frac{1}{4} & \frac{1}{3} \\
-\frac{3}{5} & -\frac{1}{3} & -\frac{1}{7} & 0 &
\frac{1}{9} & \frac{1}{5} \\
-\frac{2}{3} & -\frac{3}{7} & -\frac{1}{4} & -\frac{1}{9}
& 0 & \frac{1}{11} \\
-\frac{5}{7} & -\frac{1}{2} & -\frac{1}{3} & -\frac{1}{5}
& -\frac{1}{11} & 0
\end{array}$
So sorting out all the duplicates:
$$B=\left\{0, \pm\frac{1}{2}, \pm\frac{1}{3}, \pm\frac{2}{3}, \pm\frac{1}{4}, \pm\frac{1}{5}, \pm\frac{3}{5}, \pm\frac{1}{7}, \pm\frac{3}{7}, \pm\frac{5}{7}, \pm\frac{1}{9}, \pm\frac{1}{11}\right\}$$
So it looks like in your example $|B|=23$
|
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|
Evaluating $\int_0^1 \frac{x-x^2}{\sin \pi x} dx = \frac{7 \zeta(3)}{\pi^3}.$ I tried to use the series for $\sin \pi x$ and maybe find something related to $\zeta(3)$, but didn't work. I'm guessing this integral needs more than the little calculus that I know.
\begin{equation}
\int_0^1 \frac{x-x^2}{\sin \pi x} dx = \frac{7 \zeta(3)}{\pi^3}.
\end{equation}
|
An elementary evaluation:
$$\int_0^1 \frac{x-x^2}{\sin \pi x} dx \overset{ibp}
=\int_0^1 \frac{2x-1}\pi\ln \tan \frac{\pi x}2 dx
\overset{t=\tan^2\frac{\pi x}2}=
\frac2{\pi^3}\int_0^\infty \frac{\ln t\tan^{-1}\sqrt t}{\sqrt t(1+t)}dt
$$
Let $J(a)=\int_0^\infty \frac{\ln t\tan^{-1}a\sqrt t}{\sqrt t(1+t)}dt$
$$J’(a)= \int_0^\infty \frac{\ln t \>dt}{(1+t)(1+a^2t)}
\overset{y=\frac1{a^2t}}=- \ln a\int_0^\infty \frac{ dy}{(1+y)(1+a^2y)}= \frac{2\ln^2a}{1-a^2}
$$
Then
$$I = \frac2{\pi^3}J(1)= \frac2{\pi^3}\int_0^1 J’(a)da
=\frac4{\pi^3}\int_0^1 \frac{\ln^2a}{1-a^2} da
= \frac4{\pi^3}\frac{7\zeta(3)}4
=\frac{7 \zeta(3)}{\pi^3}
$$
|
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|
Inequality in number theory Prove that:
(i)$5<5^{\frac{1}{2}}+5^{\frac{1}{3}}+5^{\frac{1}{4}}$
(ii)$8>8^{\frac{1}{2}}+8^{\frac{1}{3}}+8^{\frac{1}{4}}$
(iii)$n>n^{\frac{1}{2}}+n^{\frac{1}{3}}+n^{\frac{1}{4}}$ for all integer $n\geq9$
Can raising both sides to exponent $12$ help
|
(i) $5<2.2+1.7+1.4<5^{1/2}+5^{1/3}+5^{1/4}$.
(ii) $8>2.9+2+1.7>8^{1/2}+8^{1/3}+8^{1/4}$.
(iii) $n\ge3\sqrt n>n^{1/2}+n^{1/3}+n^{1/4}$ because $n \ge 9$.
|
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|
A problem on series . I found this problem on series in my school textbook.
If one of the two series below is 2,what is the value of the other series?
The series are:
$$(1+x+x^{2}+x^{3}+\cdots+x^{n})$$
and the other one is
$$(x+2x^{2}+3x^{3}+\cdots+nx^{n})$$
The four options are:
A) 1
B) 2
C) 3
D) 4
I did it in the following way
We can prove
$$ (1+x+x^{2}+x^{3}+x^{4}+\cdots+x^{n})=\frac{1}{1-x}$$
and differentiating both sides of the series we get ,
$$(x)\frac{d}{dx} (1+x+x^{2}+x^{3}+x^{4}+\cdots+x^{n})=
(x)\frac{d}{dx}\left(\frac{1}{1-x}\right)$$
So we can say $$(x+2x^{2}+3x^{3}+4x^{4}+\cdots+nx^{n})=\frac{x}{(1-x)^{2}}$$
So assuming the second series to be equal to 2, we get,
$$\frac{x}{(1-x)^{2}}=2 $$
solving which we get
$$x=\frac{5\pm\sqrt{25-16}}{4}$$
Thus the rooots of $x$ are $2 , 0.5$.
Thus putting it into the equation of the first series $$\frac{1}{1-x}$$ for x=2, we get $$\frac{1}{(1-x)}=-2$$ which is not there in the option.So IT WILL CERTAINLY NOT BE THE ANSWER .
Thus our assumption as to the second series being equals to two was wrong .
Thus now we CAN say that the first series is equal to 2.
Thus, $$\frac{1}{1-x}=2$$ and solving for $x$ we get $$x=0.5$$
Plugging in this value into the second series we get ,
$$(0.5+2(0.5)^{2}+3(0.5)^{3}+4(0.5)^{4}+\cdots+n(0.5)^{n})=\frac{0.5}{(1-0.5)^{2}} $$
We get the series being equal to $$\frac{0.5}{(0.5)^{2}}$$ which we find is INDEED EQUAL TO 2
Plz suggest any better way in which i can do the problem.
THANX WITH ALL MY HEART TO ANYONE FOR THEIR SUGGESTIONS!!
|
In case the sum is infinite, that is, you're considering $$1+x+x^2+\dots\tag{1}$$
Assume that $(1)$ is equal to $2$, so that $$x+x^2+x^3+\dots=1\tag{2}$$ Then look at: $$\begin{align}x+x^2+x^3+\dots\\x^2+x^3+\dots\\x^3+\dots\\\\x+2x^2+3x^3+\dots\tag{3}\end{align}$$
The desired sum $(3)$ is equal to $$x(1+x+x^2+\dots)+x^2(1+x+x^2+\dots)+\dots$$ Using $(2)$, this last sum is equal to $$2x+2x^2+2x^3+\dots=2(x+x^2+x^3+\dots)=2\cdot 1=2$$
|
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|
Prove that $\frac{1}{\sqrt {a^2 - b^2}} = \frac{1/a}{\sqrt {1 - \frac{b^2}{a^2}}}$ Recently I have been reading physics book and saw interesting equation, like this:
$$\frac{1}{\sqrt {a^2 - b^2}} = \frac{1/a}{\sqrt {1 - \frac{b^2}{a^2}}}$$
But I still don't understand how to get the right part of the equation from left part, and I ask for the explaining of this. Thanks a lot!
|
Divide numerator and denominator by $a$ (where $a>0$). The denominator becomes
$$\frac{1}{a}\sqrt{a^2-b^2}=\sqrt{\frac{1}{a^2}}\times\sqrt{a^2-b^2}=\sqrt{1-\frac{b^2}{a^2}}.$$
|
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Find the Eigenvalues and its Eigen vectors. Consider the $ \ n \times n \ $ matrix $$ \begin{pmatrix} a & -1 & & & & \\ -1 & a & -1 & & & \\ & -1 & a & -1 & & & \\ & & -1 & a & -1 & \\ & & ..... & .... & .... \\ & & && a & -1 \\ & & & &-1 & a \end{pmatrix} $$
Find the Eigenvalues and its Eigen vectors.
Also determine the values of $ \ a \ $ for which the matrix is positive definite.
Answer:
For our convenience consider the $ \ 3 \times 3 \ $ matrix as follows
$$ A=\begin{pmatrix} a & -1 & \\ -1 & a & -1 \\ & -1 & a \end{pmatrix} $$
The blank positions must be filled with $ \ 0 \ $. Thus,
$$ A=\begin{pmatrix} a & -1 & 0 \\ -1 & a & -1 \\ 0& -1 & a \end{pmatrix} $$
Let $ \ \lambda \ $ be the Eigen value of matrix $ \ A \ $ , then
$ |A-\lambda I |=0 \\ \Rightarrow \begin{vmatrix} a-\lambda & -1 &0 \\ -1 & a-\lambda & 0 \\ 0 & -1 & a-\lambda \end{vmatrix} =0 \\ \Rightarrow (a-\lambda)^3-(a-\lambda)=0 \\ \Rightarrow (a-\lambda) [(a-\lambda)^2-1)=0 \\ \Rightarrow \lambda=a, \ a-1, \ a+1 $
For $ \ 2 \times 2 \ $ such matrix we have
$$ A'=\begin{pmatrix} a & -1 \\ -1 & a \end{pmatrix} $$
The Eigen values of $ \ A' \ $ are $ \ a-1 , \ a+1 \ $
For $ \ 4 \times 4 \ $ such that
the eigen values are $ \ a-1, a-1 , a+1, a+1 \ $
Thus in general the eigen vlaues are ($ \ if \ n=even \ $ )
$ a-1, a-1, ...... \frac{n}{2} \ times \ \\ a+1 , a+1, ........ \frac{n}{2} \ times \ $
If $ n=odd \ $ , then the eigen values of the $ \ n \times n \ $ matrix are
$ a-1 , a-1, ................. \frac{n-1}{2} \ times \\ a+1,a+1,..............\frac{n-1}{2} \ times \\ and \ \ a \ \ $
Am I right ?
But how to find the eigen vectors ?
|
You're not quite right about the eigenvalues for the $3\times 3$ matrix since you're missing a $-1$ in the $(2,3)$ position.
For the higher dimensional cases let $A_n$ be the $n\times n$ matrix with $a-\lambda$ along the diagonal and $-1$s above and below the diagonal. Then $A_n$ has the form
\begin{pmatrix}
a-\lambda & -1 & 0 & \cdots\\
-1 & & & \\
0 & & A_{n-1}&\\
\vdots & & &
\end{pmatrix}
From this you can see that $\det A_n = (a-\lambda)\det A_{n-1} - \det A_{n-2}$. Considering the case $n=1$ the only eigenvalue is $a$. Then if $n$ is odd we can then see that $a$ is always an eigenvalue.
This isn't a complete answer but I hope it helps!
|
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|
Help finding the limit of the following function as x tends to 0 $$
\lim_{x\to0}\left(\sqrt{\frac{1}{x} + \sqrt{\frac{1}{x} + \sqrt{\frac{1}{x}}}} - \sqrt{\frac{1}{x} - \sqrt{\frac{1}{x} + \sqrt{\frac{1}{x}}}}\,\right)
$$
From Demidovich 5000 problems in mathematical analysis
|
You can try with $x=t^2$ (the domain is $x>0$, we assume also $t>0$), so your function becomes
$$
\sqrt{\frac{1}{t^2} + \sqrt{\frac{1}{t^2} + \sqrt{\frac{1}{t^2}}}} - \sqrt{\frac{1}{t^2} - \sqrt{\frac{1}{t^2} + \sqrt{\frac{1}{t^2}}}} =
\frac{\sqrt{1+\sqrt{t^2+t^3}}-\sqrt{1-\sqrt{t^2+t^3}}}{t}
$$
and your limit is reduced to computing a derivative.
You can simplify the task observing that the function is also writable as
$$
\frac{2\sqrt{t^2+t^3}}{t}\frac{1}{\sqrt{1+\sqrt{t^2+t^3}}+\sqrt{1-\sqrt{t^2+t^3}}}
$$
and the limit is essentially immediate.
|
{
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Integral $\int_0^1 \frac{x^n}{x^2-x+1} dx$ Greetings I desire to find a closed form for $$I=\int_0^1 \frac{x^n}{x^2-x+1} dx$$ My try was to use $x-\frac{1}{2}=t$ to get $$\int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{(t+\frac{1}{2})^n}{t^2+\frac{3}{4}}dt$$ then $\frac{\sqrt 3}{2}\tan u=t$ $$I=\frac{2}{\sqrt 3}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\left(\frac{\sqrt3}{2}\tan u+\frac{1}{2}\right)^ndu$$ I dont know how to takle this down. I am thinking that this can be evaluated using Contour integration because I saw other similarly integrals evaluated that way, but I dont have so much knowdeledge about it. Could you help me find a closed form for this integral?
|
Sketch: Write $I_n = \int_0^1 \frac{x^n}{x^2-x+1}\,dx$, then consider $f(t) = \sum_{n=0}^\infty I_n t^n$:
\begin{align}\sum_{n=0}^\infty I_n t^n&=\sum_{n=0}^\infty \int_0^1 \frac{(xt)^n}{x^2-x+1}\,dx
\\&=\int_0^1\frac{1}{1-xt}\frac{1}{x^2-x+1}\,dx\\
&=-\frac{\sqrt{3} \pi (t-2) + 9t\log(1-t)}{9t^2-9t+9}\end{align}
by partial fractions, and the swap of the limit and sum is justified easily since they are non-negative, so Fubini applies. Then
$$f(t) = -\frac{\frac{\pi}{3 \sqrt{3}}(t-2)+t \log(1-t)}{t^2-t+1}$$
is a generating function for $I_n$, so Taylor expansion gives the integrals each as a finite sum of the form $q + \frac{\pi}{3 \sqrt{3}} a$ for rational $q$ and $a \in \{\pm 1, \pm 2\}$.
|
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|
How to solve the problem on number theory Find the number of positive integer pairs $x,y$ such that
$$xy+\dfrac{(x^3+y^3)}3=2007.$$
I solved the question by using factorization and further checking possible values of $x$ and $y$. But it was very lengthy as I had to check many cases for $x$ and $y$. Is there any possible other method?
|
I show here that the only correct answer is, as it maintain @fleablood and @farruhota in their answers.
$$\frac{x^3+y^3}{3}=\frac{(x+y)^3}{3}-xy(x+y)\Rightarrow xy+\frac{(x+y)^3}{3}-xy(x+y)=2007$$ It follows $x+y\equiv 0\pmod3$.
The cubic $$xy+\frac{x^3+y^3}{3}=2007$$ is symmetric respect to the diagonal $y=x$ then if the line $x + y = 3a$ cuts the cubic at the point $(x, y)$ it also cuts it at the point $(y, x)$. The system
$$\begin{cases}x+y=3a\\xy+\dfrac{x^3+y^3}{3}=2007\end{cases}$$ has a solving equation of the second degree whose solution $(x,y)$ (and the symmetric $(y,x))$ is$$\begin{cases}x=\dfrac{3(-3a^2+a+X)}{2-6a}\\y=\dfrac{3(3a^2-+Xa+X)}{2-6a}\end{cases}\quad\text{ where } X=\sqrt{-3a^4-2a^3+a^2+2676a-892}$$ It follows that there are quite few values of $a$ giving real roots. In the attached graph of the cubic, only the black lines $x + y = 21,24,27$ (corresponding to $a=7,8,9$) cut the curve in real positive points. For $a\le6$ and $a\ge10$ does not. For the line $x+y=18$ the red points of intersection are not in the first quadrant (approximately $(18.204,-0.204)$).
Finally we have just three candidates (and the symmetric) produced by the lines $x+y=21,24,27$ but only the line $x+y=21$ agrees with the requirement of positive integers for the intersection. In this line we have marked all the points of integer coordinates of the first quadrant (in red those of intersection with the cubic). These three candidates are
$$(18,3)\quad\text{ and approximately }\quad(17.56,6.44),(16.162,10.838).$$
Thus the only solutions are given by $\{18,3\}$.
|
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Prove that $\tan\frac{\alpha}{2}=\pm\tan\frac{\beta}{2}\tan\frac{\gamma}{2}$ based on given condition
If $$\cos\alpha=\cos\beta\cos\phi=\cos\gamma\cos\theta$$
and $$\sin\alpha=2\sin\frac{\phi}{2}\sin\frac{\theta}{2}$$
then prove that $$\tan\frac{\alpha}{2}=\pm\tan\frac{\beta}{2}\tan\frac{\gamma}{2}$$
My approach:
$$\frac{\cos \alpha}{\cos\beta}=\cos\phi$$
$$=1-2\sin^2\frac{\phi}{2}$$
$$\Rightarrow 2\sin^2\frac{\phi}{2}=1-\frac{\cos\alpha}{\cos\beta}$$
and(in the similar fashion)
$$\Rightarrow 2\sin^2\frac{\theta}{2}=1-\frac{\cos\alpha}{\cos\gamma}$$
Please help me to proceed.
|
$$\sin^2\alpha=2\sin^2\dfrac\phi2\cdot2\sin^2\dfrac\theta2$$
$$=\left(1-\frac{\cos\alpha}{\cos\beta}\right)\left(1-\frac{\cos\alpha}{\cos\gamma}\right)$$
$$\cos\beta\cos\gamma(1-\cos^2\alpha)=(\cos\beta-\cos\alpha)(\cos\gamma-\cos\alpha)$$
$$-\cos^2\alpha\cos\beta\cos\gamma=-\cos\alpha\cos\beta-\cos\alpha\cos\gamma+\cos^2\alpha$$
Assuming $\cos\alpha\ne0,$
$$-\cos\alpha\cos\beta\cos\gamma=-\cos\beta-\cos\gamma+\cos\alpha$$
$$\iff\cos\alpha(1+\cos\beta\cos\gamma)=\cos\beta+\cos\gamma$$
$$\dfrac{\cos\beta+\cos\gamma}{1+\cos\beta\cos\gamma}=\dfrac{\cos\alpha}1$$
Apply Componendo and Dividendo and use $$\cos2x=\dfrac{1-\tan^2x}{1+\tan^2x}$$
|
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|
Find a coefficient of $x^{12}$ in $(x + x^{2} +x^{3} +x^{4})^{5}$ Find a coefficient of $x^{12}$ in $$(x + x^{2} +x^{3} +x^{4})^{5}$$
Using formula for the sum of a geometric series I have turned it into $$x^5(1-x^4)(1-x)^{-4} $$
So now I see that in $(1-x^4)(1-x)^{-4}$ we need to find coefficient of $x^7$ as we actually have $x^5$, am I right?
If so I unfortunately still don't know what I should do as my next step to calculate this coefficient.
|
you can have a look at this post too : Coefficient of $x^{25}$
$$(x+x^2+x^3+x^4)^5=\sum\limits_{i+j+k+l=5} \binom{5}{i,j,k,l}x^{i+2j+3k+4l}$$
The coefficient $x^{12}$ is obtained when $i+2j+3k+4l=12$.
$\begin{array}{cccc|c}
i & j & k & l & \dfrac {5!}{i!j!k!l!}\\\hline
0 & 3 & 2 & 0 & 10\\
0 & 4 & 0 & 1 & 5\\
1 & 1 & 3 & 0 & 20\\\
1 & 2 & 1 & 1 & 60\\
2 & 0 & 2 & 1 & 30\\
2 & 1 & 0 & 2 & 30\\\hline
&&&&155\end{array}$
|
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|
Cool way of finding $\cos\left(\frac{\pi}{5}\right)$ while trying to solve a problem, I stumbled upon a way of finding $\cos\left(\frac{\pi}{5}\right)$ using identities and the cubic formula. Is it possible to find other values of sine or cosine in a similar way ?
Consider
$$\cos\left(\frac{\pi}{5}\right) - \cos\left(\frac{2\pi}{5}\right).$$
Using the difference of cosines identity, we have
$$\cos\left(\frac{\pi}{5}\right) - \cos\left(\frac{2\pi}{5}\right) = -2\sin\left(\frac{3\pi}{10}\right)\sin\left(-\frac{\pi}{10}\right).$$
Now we change the RHS using the identity $\sin(x) = \cos\left(\frac{\pi}{2}-x\right)$ and the fact that $\sin(x)$ is odd.
$$-2\sin\left(\frac{3\pi}{10}\right)\sin\left(-\frac{\pi}{10}\right) = 2\cos\left(\frac{\pi}{5}\right) \cos\left(\frac{2\pi}{5}\right)$$
So,
$$\cos\left(\frac{\pi}{5}\right) - \cos\left(\frac{2\pi}{5}\right) = 2\cos\left(\frac{\pi}{5}\right) \cos\left(\frac{2\pi}{5}\right)$$
Now we make use of the identity $\cos(2x)=2\cos^2(x)-1$.
$$\cos\left(\frac{\pi}{5}\right) - 2\cos^2\left(\frac{\pi}{5}\right)+1 = 2\cos\left(\frac{\pi}{5}\right) \left(2\cos^2\left(\frac{\pi}{5}\right)-1\right)$$
Let $y=\cos\left(\frac{\pi}{5}\right)$ and we have
$$y-y^2+1=2y(2y^2-1)$$
$$4y^3+2y^2-3y-1=0$$
which has the correct solution
$$ y=\frac{\sqrt{5}+1}{4} =\cos\left(\dfrac{\pi}{5}\right) $$
One of the roots is also $\sin\left(\dfrac{\pi}{10}\right)$ which I'm guessing is because you end up with the same cubic if you apply the above to sin too.
|
Using complex numbers we can derive $\sin(\frac{\pi}{5})$ and $\cos(\frac{\pi}{5})$
$$ z^5=-1 \implies z^4-z^3+z^2-z+1=0 \implies z^2+\frac{1}{z^2}-z-\frac{1}{z}+1=0 $$
By substitution $t=z+\frac{1}{z}$ we get:
$$ t^2-t-1=0 \implies t=\frac{1\pm\sqrt{5}}{2} $$
And because $\cos(\frac{\pi}{5})$ is positive, we may consider just $t=\frac{1+\sqrt{5}}{2}$
$$ z^2-z\frac{1+\sqrt{5}}{2}+1=0 \implies z=\frac{1+\sqrt{5}}{4}\pm i\frac{\sqrt{10-2\sqrt{5}}}{4} $$
And we get that $\cos(\frac{\pi}{5}) = \frac{1+\sqrt{5}}{4} $ and $\sin(\frac{\pi}{5})= \frac{\sqrt{10-2\sqrt{5}}}{4}$
|
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|
differentiation under the integral sign $\int_{0}^{\pi}\ln\left(1-2r\cos x +r^2 \right)dx$ Using differentiation with respect to the parameter,show that for $|r|<1$
$$\mathbf{F}(r)=\int_{0}^{\pi}\ln\left(1-2r\cos x +r^2 \right)dx =0$$
my attempt is
$$\mathbf{F}'(r)=\int_{0}^{\pi}{-2\cos x + 2r \over1-2r\cos x +r^2}dx$$
let $u=\tan{x \over2}$ then $\cos x = {1-u^2 \over 1+u^2} $ and $dx={2du \over 1+u^2}$
Now
$$\mathbf{F}'(r)=\int_{0}^{\infty}{-2{1-u^2 \over 1+u^2} + 2r \over1-2r{1-u^2 \over 1+u^2} +r^2}{2du \over 1+u^2}$$
$$\mathbf{F}'(r)= 4\int_{0}^{\infty}{r+ru^2+u^2-1 \over 1+u^2-2r+2ru^2+r^2}{du \over 1+u^2}$$
but I do not know how to continue,
I tried with partial fractions but it becomes too tedious.
|
A slightly different approach.
It is clear that $F'(0)=0$. We assume $r\neq 0$ from now on. Notice that for $|r|<1$ one has:
\begin{align}
\sum_{n=0}^\infty r^n \cos(nx)& = \frac{1-r\cos(x)}{1-2r\cos(x)+r^2}\\\sum_{n=0}^\infty r^n \sin(nx)&= \frac{r\sin(x)}{1-2r\cos(x)+r^2}
\end{align}
One can use that to write $F'(r)$ as:
\begin{align}
F'(r) = \int^\pi_0\left[ \frac{2}{r}\sum_{n=0}^\infty r^n \cos(nx) + 2\left( 1-\frac 1 {r^2}\right)\sum_{n=0}^\infty r^n \frac{\sin(nx)}{\sin(x)} \right]\,dx
\end{align}
Both series converge uniformly on $x\in [0,\pi]$ so one can interchange the order of integration and summation to obtain:
\begin{align}
F'(r) = \frac{2}{r}\sum_{n=0}^\infty r^n \int^\pi_0\cos(nx)\,dx + 2\left( 1-\frac 1 {r^2}\right)\sum_{n=0}^\infty r^n \int^\pi_0\frac{\sin(nx)}{\sin(x)} \,dx
\end{align}
Notice that \begin{align}\int^\pi_0 \cos(nx) = \begin{cases}
\pi & \text{ if } n =0\\
0 & \text{ if } n\neq 0
\end{cases}
\end{align}
and:
\begin{align}\int^\pi_0 \frac{\sin(nx)}{\sin(x)} = \begin{cases}
\pi & \text{ if } n \text{ is odd} \\
0 & \text{ if } n \text{ is even}
\end{cases}
\end{align}
So:
\begin{align*}
F'(r) = \frac {2 \pi} r + 2 \left( 1- \frac 1{r^2}\right) \sum_{n=0}^\infty \pi r^{2n+1} =\frac {2 \pi} r + 2\pi\left( 1 -\frac 1{ r^2}\right) \frac{r}{1-r^2} = 0
\end{align*}
This implies that $F(r)=C$ for some $C\in\mathbb R$. Set $r=0$ to find that $C=0$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\frac{dy}{dx}+\sec^2(\frac{\pi}{4}-x)=0$
If $y=\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$, prove that $\frac{dy}{dx}+\sec^2(\frac{\pi}{4}-x)=0$
My attempts:
Attempt 1:
$y=\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$
$\implies y=\sqrt{\frac{\sin^2 x+\cos^2 x-2\sin x \cos x}{\sin^2 x+\cos^2 x+2\sin x \cos x}}$
$\implies y=\sqrt{\frac{(\sin x -\cos x)^2}{(\sin x+\cos x)^2}}$
$\implies y=\frac{\sin x -\cos x}{\sin x +\cos x}$
$\therefore \frac{dy}{dx}=\frac{(\sin x +\cos x)(\cos x+\sin x)-(\sin x -\cos x)(\cos x -\sin x)}{(\sin x +\cos x)^2}$
$\implies \frac{dy}{dx}=\frac{(\sin x+\cos x)^2+(\sin x -\cos x)^2}{(\sin x+\cos x)^2}$
$\implies \frac{dy}{dx}=\frac{2(\sin^2 x+\cos^2 x)}{(\cos(\frac{\pi}{2}-x)+\cos x)^2}$
$\implies \frac{dy}{dx}=\frac{2}{(2\cos(\frac{\pi}{4})\cos(\frac{\pi}{4}-x))^2}$
$\implies \frac{dy}{dx}=\frac{2}{4\cos^2(\frac{\pi}{4})\cos^2(\frac{\pi}{4}-x)}$
$\implies \frac{dy}{dx}=\frac{2}{4\times\frac{1}{2}\cos^2(\frac{\pi}{4}-x)}$
$\implies \frac{dy}{dx}=\sec^2(\frac{\pi}{4}-x)$
$\implies \frac{dy}{dx}-\sec^2(\frac{\pi}{4}-x)=0$
But i have to prove $\frac{dy}{dx}+\sec^2(\frac{\pi}{4}-x)=0$
Attempt 2:
$y=\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$
$\implies y=\sqrt{\frac{\cos^2 x+\sin^2 x-2\cos x \sin x}{\cos^2 x+\sin^2 x+2\cos x \sin x}}$
$\implies y=\sqrt{\frac{(\cos x -\sin x)^2}{(\cos x+\sin x)^2}}$
$\implies y=\frac{\cos x -\sin x}{\cos x +\sin x}$
$\therefore \frac{dy}{dx}=\frac{(\cos x +\sin x)(-\sin x-\cos x)-(\cos x -\sin x)(-\sin x +\cos x)}{(\cos x +\sin x)^2}$
$\implies \frac{dy}{dx}=\frac{-(\cos x+\sin x)^2+(\cos x -\sin x)^2}{(\cos x+\sin x)^2}$
$\implies \frac{dy}{dx}=\frac{(\cos x -\sin x)^2-(\cos x+\sin x)^2}{(\cos x+\sin x)^2}$
$\implies \frac{dy}{dx}=\frac{-4\sin x\cos x}{(\cos x+\cos(\frac{\pi}{2}-x))^2}$
$\implies \frac{dy}{dx}=\frac{-4\sin x\cos x}{(2\cos(\frac{\pi}{4})\cos(x-\frac{\pi}{4}))^2}$
$\implies \frac{dy}{dx}=\frac{-4\sin x\cos x}{4\cos^2(\frac{\pi}{4})\cos^2(x-\frac{\pi}{4})}$
$\implies \frac{dy}{dx}=\frac{-4\sin x\cos x}{4\times \frac{1}{2}\cos^2(x-\frac{\pi}{4})}$
$\implies \frac{dy}{dx}=\frac{-2\sin x\cos x}{\cos^2(x-\frac{\pi}{4})}$
Attempt 3:
$y=\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$
$\implies y=\sqrt{\frac{(1-\sin 2x)(1-\sin 2x)}{(1+\sin 2x)(1-\sin 2x)}}$
$\implies y=\frac{1-\sin 2x}{\cos 2x}$
$\therefore \frac{dy}{dx}=\frac{\cos 2x(-2\cos 2x)-(1-\sin 2x)(-2\sin 2x)}{\cos^2 2x}$
$\implies \frac{dy}{dx}=\frac{-2\cos^2 2x+2\sin 2x(1-\sin 2x)}{\cos^2 2x}$
My questions:
(i) Why does Attempt 2 give a result different from that given by Attempt 1.
(ii) How do i prove the result?
|
The statement that you are trying to prove is not true for every $x$. If it was, we would always have $\frac{\mathrm dy}{\mathrm dx}\leqslant0$ and therefore $y$ would be decreasing. But it is increasing in $\left[\frac\pi4,\frac\pi2\right]$, for instance.
The statement holds in $\left(-\frac\pi4,\frac\pi4\right]$. Yours first attempt proves it, as long as you correct this mistake: what you deduce from the equality$$y=\sqrt{\frac{(\sin x-\cos x)^2}{(\sin x+\cos x)^2}}$$is that$$y=\frac{-\sin x+\cos x}{\sin x+\cos x},$$since $y$ is a square root and therefore $y\geqslant0$.
|
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|
$1^n-3^n-6^n+8^n$ is divisible by $10$ Prove that $1^n-3^n-6^n+8^n$ is divisible by $10$ for all $n\in\mathbb{N}$
It is divisible by $2$ and $5$ if we rearrange it will it be enough
$(1^n -3^n)$ and $(6^n -8^n)$ is divisible by $2$.
And
$(1^n-6^n)$ and $(8^n-3^n)$ is divisible by $5$.
Hence $\gcd(2,5)$ is $1$ and it is divisible by $2\cdot5=10$.
Is it correct?
|
Perhaps another method to notice is that, by Euler's Generalization of Fermat's Little Theorem, we have that:
$$1^n-3^n-6^n+8^n \equiv 1^{n \mod \phi(10)} - 3^{n \mod \varphi(10)} - 6^{n \mod \varphi(10)} + 8^{n \mod \varphi(10)} \mod 10$$
Since $\varphi(10) = 4$, you really only need to check the cases of $n = 0, 1, 2$ and $3$. For $n = 0$, you have:
$$1^0-3^0-6^0+8^0 \equiv 1 - 1 - 1 + 1 \equiv 0 \mod 10$$
For $n = 1$:
$$1^1-3^1-6^1+8^1 \equiv 1 - 3 - 6 + 8 \equiv 0 \mod 10$$
For $n = 2$:
$$1^2-3^2-6^2+8^2 \equiv 1 - 9 - 36 + 64 \equiv 20 \equiv 0 \mod 10$$
For $n = 3$:
$$1^3-3^3-6^3+8^3 \equiv 1 - 27 - 216 + 512 \equiv 270 \equiv 0 \mod 10$$
Since all the other exponents leading forward will be one of these cases, you are done. It is not the most elegant way to prove this, however, it is fairly direct.
|
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|
Find out if polynomial has inverse in quotient ring Find out if polynomial $x^3-x^2+x-1$ has inverse element in quotient ring ${\displaystyle \mathbb {Z} }_{11}/(x^4+3x^3-3x^2-4x-1)$, if yes find this inverse.
I know that $x^3-x^2+x-1$ has inverse element when $gcd(x^4+3x^3-3x^2-4x-1;x^3-x^2+x-1)$~$1$. I also know that I can use Euclidean algorithm. But I'm not sure about my result, I have got result that the g$cd$~$5$ it means that $x^3-x^2+x-1$ does not have inverse. Am I correct?
Thanks for any help!
|
The extended Euclidean algorithm over $\mathbb Q$ gives:
$$
232 = (49 x^3 + 168 x^2 - 75 x - 195)(x^3-x^2+x-1) +(-49 x^2 + 28 x - 37)(x^4+3x^3-3x^2-4x-1)
$$
Mod $11$, this becomes
$$
1 \equiv (5 x^3 + 3 x^2 - 9 x - 8)(x^3-x^2+x-1) \bmod (x^4+3x^3-3x^2-4x-1)
$$
and so the inverse is $5 x^3 + 3 x^2 - 9 x - 8 \equiv 5x^3+3x^2+2x+3$.
|
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|
Integral $\int_0^{\pi} \frac{\cos(2018x)}{5-4\cos{x}}dx$ I wish to evaluate $$I(2018)=\int_{0}^{\pi}\frac{\cos(2018x)}{5-4\cos x} dx$$ Considering $$X=I(k)+iJ(k)=\int_{-\pi}^{\pi}\frac{\cos{kx}}{5-4\cos x} dx +i\int_{-\pi}^{\pi}\frac{\sin{kx}}{5-4\cos x} dx=\int_{-\pi}^{\pi}\frac{e^{ikx}}{5-4\cos x} dx$$ let us substitute $$e^{ix}=z\rightarrow dx=\frac{dz}{iz} \, ,|z|=1$$ Due to Euler's formula we can rewrite $$\cos x=\frac{z^2+1}{2z}$$ $$X=\oint_{|z|=1} \frac{z^k}{5-4\frac{z^2+1}{2z}}\frac{dz}{iz}=\frac{1}{i}\oint_{|z|=1} \frac{z^k}{-2z^2+5z-2}dz$$ $$-2z^2+5z-2=-\frac{1}{2}((2z)^2-5(2z)+4)=-\frac{1}{2}(2z-4)(2z-1)=-2(z-2)(z-\frac{1}{2})$$ Now let us notice that in our contour $|z|=1\,$ only the pole $z_2=\frac{1}{2}$ is found. Thus the integral we seek to evaluate is $$\frac{1}{i} \cdot 2\pi i \, \text{Res} (f(z),z_2)$$ where $f(z)=\frac{z^k}{-2(z-2)(z-\frac{1}{2})}$ $$X=2\pi \lim_{z\to z_2} (z-z_2)\frac{z^k}{-2(z-2)(z-z_2)}=\frac{2}{3}\pi \frac{1}{2^k}$$ therefore $$I(k)=\Re (X) =\frac{2\pi}{3}\frac{1}{2^k}$$ And $$\int_{0}^{\pi}\frac{\cos(2018 x)}{5-4\cos x} dx=\frac{\pi}{3}\cdot\frac{1}{2^{2018}}.$$ Now someone told me that the answer is $0$ and I am wrong (also wolfram gives $0$ as an answer). Could you please clarify? Or maybe give another solution to this integral if it's $0$ or another answer?
|
I am going to show the general case:
$$
I_n=\int_{0}^{\pi} \frac{\cos nx}{5-4 \cos x} d x= \frac{\pi}{3 \cdot 2^{n}}
$$
using contour integration on the unit circle $|z|=1$.
Noting $I_n= \frac{1}{2} \int_{0}^{2\pi} \frac{\cos nx}{5-4 \cos x} dx=\frac{1}{2} Re(J_n)$, where
$$J_n=\int_{0}^{2\pi} \frac{e^{nxi}}{5-4 \cos x} dx$$
Let $ z=e^{xi}, $ then $ dz=i z dx.$
$$
\begin{aligned}
J _n &=\oint_{|z \mid=1} \frac{z^{n}}{5-4\left(\frac{z+z^{-1}}{2}\right)} \frac{d z}{z i} \\
&=i \oint_{|z|=1} \frac{ z^{n}}{(2 z-1)(z-2)} d z
\end{aligned}
$$
The integrand has a simple pole within $|z|=1$ at $\frac{1}{2} $, and the residue is
\begin{aligned}
\operatorname{Res}\left(\frac{z^{n}}{(2 z-1)(z-2)}, z=\frac{1}{2}\right)=& \lim _{z \rightarrow \frac{1}{2}}\left(z-\frac{1}{2}\right) \cdot \frac{z^{n}}{(2 z-1)(z-2)} \\
=& -\frac{1}{3 \cdot 2^{n}}\end{aligned}
$$
J_{n}=i \cdot 2 \pi i \cdot\left(-\frac{1}{3 \cdot 2^{n}}\right)=\frac{\pi}{3 \cdot 2^{n-1}}
$$
Plugging $J_n$ back yields the result.
$$
\boxed{I_{n}=\frac{1}{2} Re\left(J_{n}\right)=\frac{\pi}{3 \cdot 2^{n}}}
$$
As a bonus, $$\int_{0}^{\pi} \frac{\sin nx}{5-4 \cos x} d x=0$$
|
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|
How do you read $\pm$? Why does $|x|=3$ imply $x=\pm 3$? Q1) How is this symbol read as $±$?
"Plus AND Minus" or Plus OR Minus"?
All this time I've been reading it as Plus-Minus.
I asked my math teacher and he said it's the former one. Then my question is if it's "Plus AND Minus" then when we take square roots on both sides of an equation, like,
$$x^2 = 9 \implies x= ± 3 $$
How can we say $x$ equals $2$ values? It should be $+3$ or $-3$ right?
Q2) My teacher said, $$|x| ≠ ± x$$ but $$|x| = 5 \implies x = ±5$$
Wait... what? What is going on here? I think I lack the basic math knowledge here.
EDIT: I forgot to mention another thing.
He says, $$\sqrt{9} = 3$$
And then he says, $$\sqrt{x^2}= |x|$$ but wouldn't that mean of $$\sqrt{9} = |3|$$ & $$ |3| = ± 3$$ Didn't he just contradict himself?
Thanks :)
|
The symbol $\pm$ should be read plus or minus.
Background
By convention, if $x \geq 0$, then $\sqrt{x}$ is the principal square root of $x$, meaning the nonnegative square root of $x$. Hence, $\sqrt{9} = 3$. If we want to denote the negative square root, we write $-\sqrt{x}$. Thus, the negative square root of $9$ is expressed in the form $-\sqrt{9} = -3$. Notice that the principal square root of a number is uniquely defined.
The absolute value of a real number $x$, denoted $|x|$, is its distance from $0$ on the real number line. Since distances are nonnegative, the absolute value of a number is never negative. If $x \geq 0$, then $|x| = x$. For instance, $|5| = 5$ since $5$ is five units from $0$ and $|0| = 0$ since $0$ is zero units from $0$. If $x < 0$, then $|x| = -x$. For instance, $|-4| = 4$ since $-4$ is four units from $0$. Consequently, we can write a piecewise definition of the absolute value as follows:
$$|x| = \begin{cases}
x & \text{if $x \geq 0$}\\
-x & \text{if $x < 0$}
\end{cases}$$
Observe that
\begin{align*}
\sqrt{5^2} & = \sqrt{25} = 5 = |5|\\
\sqrt{0^2} & = \sqrt{0} = 0 = |0|\\
\sqrt{(-4)^2} & = \sqrt{16} = 4 = -(-4) = |-4|
\end{align*}
Observe that if $x \geq 0$, then $\sqrt{x^2} = x = |x|$. If $x < 0$, then $\sqrt{x^2} = -x = |x|$. Therefore, we obtain the alternative definition of absolute value:
$$\sqrt{x^2} = |x|$$
Solve the equation $x^2 = 9$.
\begin{align*}
x^2 & = 9\\
\sqrt{x^2} & = \sqrt{9} && \text{take the principal square root of each side of the equation}\\
|x| & = 3 && \text{evaluate the principal square roots}
\end{align*}
The equation $|x| = 3$ means that $x$ is a real number whose distance from $0$ on the real number line is $3$. There are two such numbers. They are $3$ and $-3$. However, $x$ cannot simultaneously be both $3$ and $-3$. Hence, we say that $x = 3$ or $-3$. Our solution set is $S = \{3, -3\}$. Thus, this equation has two distinct solutions that happen to be opposite in sign. When this occurs, the solution is sometimes expressed in the form
$$x = \pm 3$$
to indicate that if $x$ is a solution of the equation, then $x = 3$ or $x = -3$. To put is another way, either $3$ or $-3$ could be substituted for $x$ to make the equation $x^2 = 9$ true.
Clearly, it does not make sense to say $x = 3$ and $x = -3$ since $3 \neq -3$. Consequently, it also does not make sense to read the symbol $\pm$ as plus and minus.
The symbol $\pm$ indicates that the symbol may be substituted by either the positive or negative values of the variable (see this definition). Therefore, it does not make sense to write
$$|x| = \pm x$$
since there is only a single value which may be substituted for $|x|$. If $x \geq 0$, that value is $x$; if $x < 0$, that value is $-x$.
|
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|
Probability that all three shirts were worn Mr. Jones has three shirts: Red, Green and White. Each day he picks randomly a red shirt with probability of $\frac{1}{2}$, green with probability of $\frac{1}{3}$ and white with probability of $\frac{1}{6}$.
What is the probability that he wears all three shirts after 6 days?
My try:
$$\binom{6}{1}\binom{5}{1}\binom{4}{1}\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{6}\cdot[\binom{3}{2}(\frac{1}{2})^2\cdot[\frac{1}{3}+\frac{1}{6}]+[\binom{3}{2}(\frac{1}{3})^2\cdot[\frac{1}{2}+\frac{1}{6}]+[\binom{3}{2}(\frac{1}{6})^2\cdot[\frac{1}{2}+\frac{1}{3}]+\binom{3}{1}\binom{2}{1}\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{6}]$$
First, we pick possessions for red, green and white. Then, we multiply it by the other combinations possible of the other three possessions left, and summing it up (I didn't count all possibilities). The problem is the answer is bigger than 1, so it must be wrong. Also, I guess there is more elegant way.
Thanks in advance.
|
The alternative problem is this: one has 3 shirts of red color, 2 of green and 1 of white of which he chooses arbitrarily. What is the probability that he puts on all of the 3 colors within 3 days?$$P=\dfrac{N}{N_{\text{tot}}}\\N=\binom{3}{1}\binom{2}{1}\binom{1}{1}=6\\N_{\text{tot}}=\dfrac{6!}{1!2!3!}=60\\P=0.1$$
|
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|
Limit of sequence as $n$ goes to infinity with $n$ in every term I am studying for an exam in introductory analysis and came across this exercise in a practice exam:
Determine the limit of $(\frac{1}{n^2}+\frac{2}{n^2-1}+\frac{3}{n^2-2}+ \cdots + \frac{n}{n^2-n+1})$ as $n$ goes to $+\infty$.
I know that every single fraction in the sequence goes to $0$ so by the addition rule I thought maybe the entire sequence adds to $0$? My intuition tells me I'm wrong since there are infinite terms.
I'm not sure where else to start.
Any hints would be appreciated. Thanks!
|
We have,
$(\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+ \cdots + \frac{n}{n^2})\le(\frac{1}{n^2}+\frac{2}{n^2-1}+\frac{3}{n^2-2}+ \cdots + \frac{n}{n^2-n+1})\le(\frac{1}{n^2-n}+\frac{2}{n^2-n}+\frac{3}{n^2-n}+ \cdots + \frac{n}{n^2-n})\ \forall \ n \in N$
$\Rightarrow a_n= \frac{(n)(n+1)}{2(n^2)} \le (\frac{1}{n^2}+\frac{2}{n^2-1}+\frac{3}{n^2-2}+ \cdots + \frac{n}{n^2-n+1}) \le \frac{(n)(n+1)}{2(n^2-n)} =b_n $
By L-Hopital's rule on the upper and lower bouding sequences we have $\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}b_n=\frac{1}{2}$ and therefore by Squeeze theorem we get
$\lim_{n\rightarrow\infty}(\frac{1}{n^2}+\frac{2}{n^2-1}+\frac{3}{n^2-2}+ \cdots + \frac{n}{n^2-n+1})=\frac{1}{2}$
|
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|
If $0⩽x⩽y⩽z⩽w⩽u$ and $x+y+z+w+u=1$, prove $xw+wz+zy+yu+ux⩽\frac15$
If $0⩽x⩽y⩽z⩽w⩽u$ and $x+y+z+w+u=1$, prove$$
xw+wz+zy+yu+ux⩽\frac15.
$$
I have tried using AM-GM, rearrangement, and Cauchy-Schwarz inequalities, but I always end up with squared terms. For example, applying AM-GM to each pair directly gives$$
x^2+y^2+z^2+w^2+u^2 ⩾ xw + wz + zy + yu + ux,
$$
but I cannot seem to continue from here or use $x + y + z + w + u = 1$. Other inequalities like Chebyshev's rely on the multiplied pairs to be in order from least to greatest or vice versa, so I am stuck here.
|
Let $y=x+a$, $z=x+a+b$, $w=x+a+b+c$ and $u=x+a+b+c+d$.
Thus, $a$, $b$, $c$ and $d$ are non-negatives and we need to prove that:
$$(x+y+z+u+w)^2\geq5(xw+wz+zy+yu+ux)$$ or
$$(5x+4a+3b+2c+d)^2\geq5(x(x+a+b+c)+(x+a+b+c)(x+a+b)+(x+a+b)(x+a)+(x+a)(x+a+b+c+d)+(x+a+b+c+d)x)$$ or
$$(5x+4a+3b+2c+d)^2\geq5(5x^2+(8a+6b+4c+2d)x+(a+b+c)(a+b)+(a+b)a+a(a+b+c+d))$$ or
$$(4a+3b+2c+d)^2\geq5((a+b+c)(a+b)+(a+b)a+a(a+b+c+d)),$$
which is obvious.
Done!
|
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|
Finding $C_n=4C_{n-1}+6n-1$ 2.$$\begin{cases}
C_n=4C_{n-1}+6n-1\\
C_0=2\\
\end{cases} $$
$\begin{cases}
C_n=D_n+dn+e\\
D_n=\lambda D_{n-1}\end{cases}$
$D_n+dn+e=4C_{n-1}+6n-1$
But $C_{n-1}=D_{n-1}+d(n-1)+e$
So $D_n+dn+e=4(D_{n-1}+d(n-1)+e)+6n-1$
$D_n+dn+e=4D_{n-1}+4dn-4d+4e+6n-1$
But for $\lambda=4$ we get $4D_{n-1}+dn+e=4D_{n-1}+4dn-4d+4e+6n-1$ and
$0=-dn-e+4dn-4d+4e+6n-1$ or $0=(3d+6)n+3e-4d-1$
So $d=-2,e=1.5$
$C_n=D_n+dn+e$ So $C_n=D_n-2n+1.5$ And $C_0=D_0+1.5$ So $2=D_0+1.5$ And $D_0=-0.5$
$D_n=4D_{n-1}$
So
$C_n=D_n+dn+e=4^n-2n+1.5$ which is $O(4^n)$
Is it correct?
|
We start by solving the homogenous part: letting $x^n=C_n$ gives the characteristic equation $$x^n=4x^{n-1}\Rightarrow x=4,$$ which gives $$C_n=A(4^n).$$ Now, for the the inhomogenous part, let us guess that the solution is of the form $$p(n)=d_0+d_1n,$$ substituting into the recurrence gives $$d_0+d_1n=4(d_0+d_1(n-1))+6n-1,$$ which, by comparing coefficients gives $$d_0=-\frac{7}{3},\quad d_1=-2.$$ The final solution is obtained by adding the homogenous and inhomogenous parts; and thus $$C_n=A(4^n)-\frac{7}{3}-2n.$$ Finally, we may find $A$ by imposing the initial condition $$2=A-\frac{7}{3}\Rightarrow A=\frac{13}{3}.$$ This therefore gives $$C_n=\frac{13}{3}(4^n)-\frac{7}{3}-2n.$$
|
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|
Consider this system of congruence equations. \begin{cases}
4x \equiv 14 \pmod m \\
3x \equiv 2 \pmod 5
\end{cases}
I want to prove that for $m \in 4\mathbb{Z}$ there are no solutions(1). Moreover, I want to determine all m for which I have solutions(2).
First of all, the second equation is equivalent to $ x \equiv 4$ (mod 5).
If $m$ and $5$ are coprime, the Chinese remainder theorem states that I have solution, so I pick $m$ and $5$ not coprime. In this case, $m = 5t$ for some $t \in \mathbb{Z}$.
I can write:
$$ 4x = 14 + (5t)c$$
$$ 4x = 4 + 5(2 + tc)$$
$$ 4x = 4$$
$$ x \equiv 1 mod 5$$
However, I also have that $ x \equiv 4 \pmod 5$, so there are no solution. In the proof I did not use the fact that $m$ is a multiple of $4$, so I think the answer for (2) is that we have solution only for $(m,a) = 1$. Is that right?
|
If $m=4n$, then
\begin{align}
4x \equiv 14 \pmod m
&\implies 4x \equiv 14 \pmod{4n} \\
&\implies 4n \mid 14-4x \\
&\implies 2 \mid 7-2x \\
&\implies 2 \mid 7
\end{align}
which is false.
$3x \equiv 2 \pmod 5 \implies x \equiv 4 \pmod 5 \implies x = 4 + 5u$ for some integer, $u$.
\begin{align}
4x \equiv 14 \pmod m
&\implies 4(4 + 5u) \equiv 14 \pmod m \\
&\implies 16 + 20u \equiv 14 \pmod m \\
&\implies 20u \equiv -2 \pmod m \\
&\implies m \mid 2(10u+1)
\end{align}
If $u=0$, we get $x=4$ and $m \mid 2$
If $u=1$ we get $x=9$ and $m \mid 22$
$\dots$
So $x=4+5u$ and $m \mid 2(10u+1)$
Check
$3x = 12 + 15u \equiv 2 \pmod 5$
$4x - 14 = 16 + 20u - 14 = 20u + 2 = 2(10u+1)$
So $m \mid 2(10u+1) \implies m \mid 4x-14 \implies 4x \equiv 14 \pmod m$.
|
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|
Find $\lim\limits_{n→∞}\frac{3^n(\overbrace{\sin(\sin(\cdots\sin(x)))}^{n\text{ times}}+(\sqrt2\cos x+2)^n+2^n\cos x}{3^n+\sin x(\sqrt2\cos x+2)^n}$
Let$$f(x)= \lim_{n \to \infty}\dfrac{3^n(\overbrace{\sin(\sin(...\sin(x)))}^{\text{n times}}+(\sqrt 2 \cos x+2)^n+2^n\cos x}{3^n+\sin x(\sqrt 2\cos x+2)^n}.$$
If $l = \lim\limits_{x\to \frac{\pi}{4}^+}f(x)$ and $m = \lim\limits_{x\to \frac{\pi}{4}^-}f(x)$, find the value of $l^2+m^2$.
Attempt:
I do not see any reason why left hand and right limits should differ.
Dividing by $3^n$ we get:
$$\lim_{\substack{n \to \infty\\x \to \frac{\pi}{4}}}\dfrac{(\overbrace{\sin(\sin(...\sin(x)))}^{n\text{ times}}+\dfrac{(\sqrt 2 \cos x+2)^n}{3^n}+\dfrac{2^n\cos x}{3^n}}{1+\dfrac{\sin x(\sqrt 2\cos x+2)^n}{3^n}} \equiv \dfrac{0+1+0}{1+\dfrac{1}{\sqrt 2}},$$
so $l=m = \dfrac{\sqrt 2}{\sqrt 2+1}$. But that does not give the right answer of $l^2+m^2$.
The answer is:
$2$
Where have I gone wrong? What is the method to solve it?
|
Keep in mind that you are first taking the limit with respect to $n$ and only then taking the limit with respect to $x$. So in particular, you care about the behavior of $\dfrac{(\sqrt 2 \cos x+2)^n}{3^n}$ for $x<\pi/4$ for the limit from below and for $x>\pi/4$ for the limit from above.
When $x<\pi/4$, the terms with $\dfrac{(\sqrt 2 \cos x+2)^n}{3^n}$ get large as $n$ gets large and so they dominate both the numerator and denominator in the limit and you get $f(x)=1/\sin x$ for all $x<\pi/4$ and thus $m=\sqrt{2}$. On the other hand, for $x>\pi/4$ the terms with $\dfrac{(\sqrt 2 \cos x+2)^n}{3^n}$ go to $0$ as $n$ gets large and so $f(x)$ is just $0$, which gives $l=0$.
|
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|
Solve the differential equation $x^5\frac{d^2y}{dx^2}+3x^3\frac{dy}{dx}+(3-6x)x^2y=x^4+2x-5$ Solve the differential equation $$x^5\frac{d^2y}{dx^2}+3x^3\frac{dy}{dx}+(3-6x)x^2y=x^4+2x-5$$
In the solution of the above the question the author is trying to convert it into an exact differential equation by multiplying $x^m$ and comparing the given equation to $$P_{0}y''+P_{1}y'+P_{2}y=\phi(x)$$
To make it exact it tried to $P_{2}-P_{1}'+P_{0}''=0$ how is he getting to this equation ?
Also further on using this equation a value of $m$ was derived and substituted , then a first integral was made.
My question is what exactly is first integral and how is it relevant in solving ordinary differential equation . If someone can guide me to a good source or link I would be thankful
Also the method shown above is it a standard method of solving ODE ? if yes could you also please let me know the name / link of a source from where I can read this method.
|
Hint:
$x^5\dfrac{d^2y}{dx^2}+3x^3\dfrac{dy}{dx}+(3-6x)x^2y=x^4+2x-5$
$x^3\dfrac{d^2y}{dx^2}+3x\dfrac{dy}{dx}-(6x-3)y=x^2+\dfrac{2}{x}-\dfrac{5}{x^2}$
Let $y=x^nu$ ,
Then $\dfrac{dy}{dx}=x^n\dfrac{du}{dx}+nx^{n-1}u$
$\dfrac{d^2y}{dx^2}=x^n\dfrac{d^2u}{dx^2}+nx^{n-1}\dfrac{du}{dx}+nx^{n-1}\dfrac{du}{dx}+n(n-1)x^{n-2}u=x^n\dfrac{d^2u}{dx^2}+2nx^{n-1}\dfrac{du}{dx}+n(n-1)x^{n-2}u$
$\therefore x^3\left(x^n\dfrac{d^2u}{dx^2}+2nx^{n-1}\dfrac{du}{dx}+n(n-1)x^{n-2}u\right)+3x\left(x^n\dfrac{du}{dx}+nx^{n-1}u\right)-(6x-3)x^nu=x^2+\dfrac{2}{x}-\dfrac{5}{x^2}$
$x^{n+3}\dfrac{d^2u}{dx^2}+2nx^{n+2}\dfrac{du}{dx}+n(n-1)x^{n+1}u+3x^{n+1}\dfrac{du}{dx}+3nx^nu-(6x-3)x^nu=x^2+\dfrac{2}{x}-\dfrac{5}{x^2}$
$x^{n+3}\dfrac{d^2u}{dx^2}+x^{n+1}(2nx+3)\dfrac{du}{dx}+x^n((n(n-1)-6)x+3n+3)u=x^2+\dfrac{2}{x}-\dfrac{5}{x^2}$
Choose $n=3$ , the ODE becomes
$x^6\dfrac{d^2u}{dx^2}+x^4(6x+3)\dfrac{du}{dx}+12x^3u=x^2+\dfrac{2}{x}-\dfrac{5}{x^2}$
$x^3\dfrac{d^2u}{dx^2}+x(6x+3)\dfrac{du}{dx}+12u=\dfrac{1}{x}+\dfrac{2}{x^4}-\dfrac{5}{x^5}$
Let $t=\dfrac{1}{x}$ ,
Then $\dfrac{du}{dx}=\dfrac{du}{dt}\dfrac{dt}{dx}=-\dfrac{1}{x^2}\dfrac{du}{dt}=-t^2\dfrac{du}{dt}$
$\dfrac{d^2u}{dx^2}=\dfrac{d}{dx}\left(-t^2\dfrac{du}{dt}\right)=\dfrac{d}{dt}\left(-t^2\dfrac{du}{dt}\right)\dfrac{dt}{dx}=\left(-t^2\dfrac{d^2u}{dt^2}-2t\dfrac{du}{dt}\right)\left(-\dfrac{1}{x^2}\right)=\left(-t^2\dfrac{d^2u}{dt^2}-2t\dfrac{du}{dt}\right)(-t^2)=t^4\dfrac{d^2u}{dt^2}+2t^3\dfrac{du}{dt}$
$\therefore\dfrac{1}{t^3}\left(t^4\dfrac{d^2u}{dt^2}+2t^3\dfrac{du}{dt}\right)-\dfrac{1}{t}\left(\dfrac{6}{t}+3\right)t^2\dfrac{du}{dt}+12u=t+2t^4-5t^5$
$t\dfrac{d^2u}{dt^2}-(3t+4)\dfrac{du}{dt}+12u=t+2t^4-5t^5$
For $t\dfrac{d^2u}{dt^2}-(3t+4)\dfrac{du}{dt}+12u=0$ ,
It has the trivial solution $u=e^{3t}$ .
Then solve it by reduction of order.
|
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|
Proving the product of four consecutive integers, plus one, is a square I need some help with a Proof:
Let $m\in\mathbb{Z}$. Prove that if $m$ is the product of four consecutive integers, then $m+1$ is a perfect square.
I tried a direct proof where I said:
Assume $m$ is the product of four consecutive integers.
If $m$ is the product of four consecutive integers, then write $m=x(x+1)(x+2)(x+3)$ where $x$ is an integer.
Then $m=x(x+1)(x+2)(x+3)=x^4+6x^3+11x^2 +6x$.
Adding $1$ to both sides gives us:
$m+1=x^4+6x^3+11x^2+6x+1$.
I'm unsure how to proceed. I know I'm supposed to show $m$ is a perfect square, so I should somehow show that $m+1=a^2$ for some $a\in\mathbb{Z}$, but at this point, I can't alter the right hand side of the equation to get anything viable.
|
Empirically:
Consider the function
$$p(n):=\sqrt{n(n+1)(n+2)(n+3)+1}.$$
For $n=0,1,2,3,\cdots$ we have $p(n)=1,5,11,19,29,\cdots$ a sequence with constant second order differences ($2$), and we can postulate the polynomial
$$n^2+3n+1$$
(because $p(0)=1$, the coefficient of $n^2$ must be $1$, and $p(n)-n^2-1=0,3,6,9,\cdots$)
Now the identity
$$n(n+3)(n+1)(n+2)=(n^2+3n)(n^2+3n+2)
\\=(n^2+3n+1-1)(n^2+3n+1+1)
\\=(n^2+3n+1)^2-1.$$
becomes apparent.
Another approach is by bringing more symmetry and shifting the variable by $3/2$.
$$\sqrt{\left(m-\frac32\right)\left(m-\frac12\right)\left(m+\frac12\right)\left(m+\frac32\right)+1}
=\sqrt{\left(m^2-\frac94\right)\left(m^2-\frac14\right)+1}
=\sqrt{m^4-\frac52m^2+\frac{25}{16}}=m^2-\frac54,
$$
which is
$$n^2+2\frac32n+\left(\frac32\right)^2-\frac54.$$
|
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|
Proving $x^2+x+1\gt0$ I was doing a question recently, and it came down to proving that $x^2+x+1\gt0$. There are of course many different methods for proving it, and I want to ask the people here for as many ways as you can think of.
My methods:
*
*$x^2+x+1=(x+\frac12)^2+\frac34$, which is always greater than $0$.
*Let it be $0$ for some $x=k$. Then $x^2+x+1=0$ has a real solution. But since $1^2\not\gt4$, this has no real solution. Therefore it is more than $0$.
|
Here a rather geometric way:
$$y = x^2+x+1 = x(x+1) + 1$$
So, $y = x^2+x+1$ is the parabola $y=x(x+1)$ shiftet by $1$ upwards.
$y=x(x+1)$ has its vertex at $x_V = -\frac{1}{2} \Rightarrow y_V = -\frac{1}{4}$
So, the vertex of $y= x(x+1) + 1$ is also at $x_V = -\frac{1}{2}$ with a minimum value of $y_{min}= -\frac{1}{4}+1 = \frac{3}{4}>0$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Facing difficulty in working $\int_{0}^{1}\frac{\arctan\left(\frac{ax}{1-x}\right)}{\sqrt{1-x}}\frac{dx}{x^{3/2}}$ I would like to evaluate this integral,$$\int_{0}^{1}\frac{\arctan\left(\frac{ax}{1-x}\right)}{\sqrt{1-x}}\frac{dx}{x^{3/2}}\tag1$$
This is the approach I will take:
We can begin with a sub: $y=\sqrt{x}$, $dx=2\sqrt{x}dy$
$$-2\int_{0}^{1}\frac{\arctan\left(\frac{ay^2}{y^2-1}\right)}{\sqrt{1-y^2}}\frac{dy}{y^2}\tag2$$
Not sure, but we can try integration by parts: $u=\arctan\left(\frac{ay^2}{y^2-1}\right)$, $$du=\frac{2ay}{y^2-1}-\frac{2ay^3}{(y^2-1)^2}\times \frac{1}{\frac{a^2y^4}{(y^2-1)^2}+1}dy$$
$dv=\frac{1}{y^2\sqrt{1-y^2}}dy$, $$v=-\frac{\sqrt{1-y^2}}{y}$$
$$2a\int_{0}^{1}\frac{(1-y^2)^{3/2}}{(y^2-1)+a^2y^4}+2a\int_{0}^{1}\frac{y^2(1-y^2)^{1/2}}{(y^2-1)^2+a^2y^4}dy=2a\left(I+J\right)\tag3$$
Integral I: Making another sub: $y=\sin(u)$, $u=\arcsin(y)$, $dy=\cos(u) du$
It is too much to write everything down, finally got to:
$$I=\int_{0}^{\pi/2}\frac{cos^4(u)}{a^2\sin^4(u)+\cos^4(u)}du$$
Using trig identities we can rewrite
$$I=\int_{0}^{\pi/2}\sec^2(u)\frac{du}{(1+\tan^2(u))(1+a^2\tan^4(u))}$$
Make another sub: $s=\tan(u)$, $du=\frac{1}{\sec^2(s)}ds$
$$I=\int_{0}^{\infty}\frac{ds}{(1+s^2)(1+a^2s^4)}$$
Using partial fraction decomp:
$$I=\frac{\pi}{2(a^2+1)}-\frac{a^2}{a^2+1}\int_{0}^{\infty}\frac{s^2-1}{a^2s^4+1}ds$$
Integral J: Making another sub: $y=\sin(v)$, $v=\arcsin(y)$, $dy=\cos(v) dv$
$$J=\int_{0}^{\pi/2}\frac{\cos^2(v)\sin^2(v)}{(a^2+1)\sin^4(v)-2\sin^2(v)+1}$$
Using trig identities to rewrite
$$J=\int_{0}^{\pi/2}\sec^2(v)\cdot \frac{\tan^2(v)}{(1+\tan^2(v))(a^2\tan^4(v)+1)}$$
Make another sub: $t=\tan(v)$, $dv=\frac{1}{\sec^2(v)}dt$
$$J=\int_{0}^{\infty}\frac{t^2}{(1+t^2)(1+a^2t^4)}dt$$
Using partial fraction decomp:
$$J=\frac{1}{1+a^2}\color{red}{\int_{0}^{\infty}\frac{a^2t^2+1}{a^2t^4+1}dt}-\frac{\pi}{2(1+a^2)}$$
The red integral it is definitely way out of my reach!
$$\int_{0}^{\infty}\frac{a^2t^2+1}{a^2t^4+1}dt=\int_{0}^{\infty}\frac{t^2}{t^4+a^{-2}}dt+\int_{0}^{\infty}\frac{dt}{a^2t^4+1}$$
The above approach seem to be not helping in evaluating the question.
After simplification I got to:
$$2a(I+J)=\int_{0}^{1}\frac{\arctan\left(\frac{ax}{1-x}\right)}{\sqrt{1-x}}\frac{dx}{x^{3/2}}
=\frac{1}{1+a^2}\int_{0}^{\infty}\frac{a^2+1}{a^2t^4+1}dt$$
If my work so far it is correct, then I am shruggle in solving this integral
$$\int_{0}^{\infty}\frac{1}{a^2t^4+1}dt=\int_{0}^{\infty}\frac{dt}{(at^2-i)(at^2+i)}dt$$
|
$\newcommand{\Im}{\operatorname{Im}}$I don't think your last integral is correct. You have made a mistake somewhere which I could not find.
Assume wolog $a\geq 0$ (because the case $a<0$ follows from the case $a>0$ by noticing that the arctangent is an odd function). Let the original integral be $F(a)$. Set $u=\frac{x}{1-x}$ to get:
$$F(a)= \int^\infty_0 \frac{\arctan(au)}{u^{3/2}}\,du$$
Integrate by parts to get:
$$F(a)= 2\int^\infty_0 \frac{a}{\sqrt[]{u} (1+a^2u^2)}\,du$$
Now set $t=a u$ to get:
$$F(a) = 2\sqrt[]{a} \int^\infty_0 \frac{dt}{\sqrt[]{t}(1+t^2)}$$This integral can be easily found using Residue Theorem or many other methods. For instance you can set $z=\sqrt[]{t}$ to get:
$$F(a) = 4 \sqrt[]{a} \int^\infty_0 \frac{1}{1+z^4}\,dz = 2\sqrt[]{a} \int^\infty_{-\infty} \frac{1}{1+z^4}\,dz = 2\sqrt[]{a} \ \ \Im \left[ \int^\infty_{-\infty} \frac{1}{z^2-i}\,dz \right]$$
Using the Residue Theorem by enclosing the pole at $e^{i\pi/4}$ we conclude that:
$$F(a) = 2\sqrt[]{a} \Im\left[ \pi i e^{-i\pi/4}\right] = \sqrt[]{2a}\ \pi$$
That was for $a>0$. In general $$F(a) = \operatorname{sign} (a) \sqrt[]{2|a|}\ \pi$$
|
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|
Solve the equation $X^2+X=\text{a given matrix}$ I want to solve the quadratic matrix equation
$$X^2+X=\begin{pmatrix}1&1\\1&1\end{pmatrix}$$
If I put $X$ in the form
$$X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
then I find complicated equations. Is there a simple way to tackle the problem without using diagonalization?
|
Write $J=\pmatrix{1&1\\1&1}$. The equation $X^2+X=J$ implies $XJ=JX$. The matrices satisfying $XJ=JX$ are those of the form $X=aI+bJ$. Then $X^2=a^2I+(2ab+2b^2)J$, so we get
$$(a^2+a)I+(b+2ab+2b^2)J=J.$$
Therefore $a^2+a=0$, entailing $a\in\{0,-1\}$ and $b+2ab+2b^2=1$,
which gives a quadratic equation for each of the two possible $a$-values.
|
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.