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Proving that $\frac{1}{n^2} - \frac{1}{(n+1)^2} \approx \frac{2}{n^3}$ when $n$ is very large This is an example from Mathematical Methods in the Physical Sciences, 3e, by Mary L. Boas.
My question is,
\begin{equation}
\frac{1}{n^2} - \frac{1}{(n+1)^2} \approx \frac{2}{n^3}
\end{equation}
can also be written as,
\begin{equation}\frac{1}{(-n)^2} - \frac{1}{(n+1)^2} \approx \frac{2}{n^3} \end{equation}
so that $\Delta n = dn = -n-(n+1) = -2n-1$.
For $f(n) = 1/n^2$, $f'(n) = -2/n^3$, and
\begin{equation}
df = d(\frac{1}{n^2}) = f'(n)dn
\end{equation}
\begin{equation}
df = \frac{(2)(2n+1)}{n^3}
\end{equation}
Now, for very large $n$, $2n+1 \approx 2n$. Thus, \begin{equation}
df = \frac{4}{n^2}
\end{equation}
But, $4/n^2$ is not approximately equal to $2/n^3$ (required ans.) even if $n$ is very large.
So, please point out my mistake(s).
Thanks in advance (;
|
The claim can be shown with straightforward algebra as b00n heT points out. However, you could also use the mean value theorem, which formalizes your idea.
As for your approach, it's not clear to me why you wrote $\frac{1}{n^2}=\frac{1}{(-n)^2}.$ This is of course true, but this does not make $n=-n$, which is essentially what your subsequent step uses when you claim $\Delta n=-2n-1.$ Herein lies your error.
Let's tie the loose ends on your idea:
Defining $f(x)=1/x^2,$ the mean value theorem tells us
$$\underbrace{f(n+1)-f(n)}_{\Delta f}=f'(n+\epsilon_n)\underbrace{(n+1-n)}_{\Delta n}=f'(n+\epsilon_n),\quad \epsilon_n\in (0,1)\\
\implies \frac{1}{(n+1)^2}-\frac{1}{n^2}=-\frac{2}{(n+\epsilon_n)^3}\\
\implies \frac{1}{n^2}- \frac{1}{(n+1)^2}\sim \frac{2}{n^3},$$
where $g(n)\sim h(n)$ means $g(n)/h(n)\rightarrow 1$ as $n \uparrow \infty$.
Notice how your approach would get this result had you correctly used $\Delta n=1.$
|
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|
How many necklaces can be formed with $6$ identical diamonds and $3$ identical pearls
Find number of ways to make a necklace (or a garland) consisting of $6$ identical diamonds and $3$ identical pearls.
I got the correct answer $7$ by taking different cases but when I applied the formula for arrangement, i.e. $$\frac{(6+3-1)!}{6!3!2}$$ I get a value which is not an integer. I divided by $6!$ and $3!$ because the diamonds and pearls are identical. Further I divided by $2$ since a necklace can be viewed from two sides.
Why am I getting a wrong answer?
|
There are two possibilities here, rotational symmetry (necklace) or
dihedral symmetry (bracelet). This is the OEIS naming convention. For the
first one we have the cycle index of the cyclic group:
$$Z(C_n) = \frac{1}{n} \sum_{d|n} \varphi(d) a_d^{n/d}.$$
For second one we have the cycle index of the dihedral group
$$Z(D_n) =
\frac{1}{2} Z(C_n) +
\begin{cases}
\frac{1}{2} a_1 a_2^{(n-1)/2} & n \text{ odd} \\
\frac{1}{4} \left( a_1^2 a_2^{n/2-1} + a_2^{n/2} \right)
& n \text{ even.}
\end{cases}$$
We get for $n=9$ the cycle indices
$$Z(C_9) = 1/9\,{a_{{1}}}^{9}+2/9\,{a_{{3}}}^{3}+2/3\,a_{{9}}$$
and
$$Z(D_9) =
1/18\,{a_{{1}}}^{9}+1/9\,{a_{{3}}}^{3}+1/3\,a_{{9}}
+1/2\,a_{{1}}{a_{{2}}}^{4}.$$
Then by the Polya Enumeration Theorem we require
$$[D^6 P^3] Z(C_9; D+P)$$
and
$$[D^6 P^3] Z(D_9; D+P).$$
We thus get for the first case
$$\frac{1}{9} {9\choose 6} + \frac{2}{9} {3\choose 2}
= 10$$
and the second one
$$\frac{1}{2} 10 + \frac{1}{2} [D^6 P^3] (D+P) (D^2+P^2)^4
\\ = 5
+ \frac{1}{2} [D^6 P^2] (D^2+P^2)^4
+ \frac{1}{2} [D^5 P^3] (D^2+P^2)^4
\\ = 5 + [D^3 P] (D+P)^4 = 5 + \frac{1}{2} {4\choose 3} = 7.$$
|
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|
A typical inequality: $\frac{8}{27}\leq\frac{(x^2+yz)(y^2+zx)(z^2+xy)}{(xy+yz+zx)^3}$
For $x, y, z\in (0, \infty)$ prove that: $$\frac{8}{27}\leq\frac{(x^2+yz)(y^2+zx)(z^2+xy)}{(xy+yz+zx)^3}.$$
My attempts to apply media inequality or other inequalities have been unsuccessful. In desperation I did the calculations and I have a rather complicated question question that I could not write as the sum of squares:
$$ 0\leq 6x^2y^2z^2 +19(x^3y^3+y^3z^3+z^3x^3)+27(x^4yz+xy^4z+xyz^4)-24(x^3y^2z+x^3yz^2+x^2y^3z+xy^3z^2+x^2yz^3+xy^2z^3).$$
Thanks for any ideas that might help me clarify the issue.
|
Add the following inequalities:
(1) Schur's inequality times $2xyz$:
$$xyz(x(x-y)(x-z)+y(y-z)(y-x)+z(z-x)(z-y))\ge 0$$
That is
$$2(x^4yz+y^4zx+z^4xy)+6x^2y^2z^2\ge 2(x^3y^2z+x^3yz^2+x^2y^3z+xy^3z^2+x^2yz^3+xy^2z^3)$$
(2) $12.5$ times $x^3+y^3-x^2y-xy^2\ge 0$: that is $12.5\sum_{cyc}xyz(x^3+y^3-x^2y-xy^2)\ge 0$. That is,
$$25(x^4yz+y^4zx+z^4xy)\ge 12.5(x^3y^2z+x^3yz^2+x^2y^3z+xy^3z^2+x^2yz^3+xy^2z^3)$$
(3) $9.5$ times $x^3+y^3-x^2y-xy^2\ge 0$: that is $9.5\sum_{cyc}(xy)^3+(yz)^3-(xy)(yz)^2-(xy)^2(yz)\ge 0$. That is,
$$19(x^3y^3+y^3z^3+z^3x^3)\ge 9.5(x^3y^2z+x^3yz^2+x^2y^3z+xy^3z^2+x^2yz^3+xy^2z^3)$$
|
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|
Evaluate $\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{n}} d x$. Latest Edit
By the contributions of the writers, we finally get the closed form for the integral as:
$$\boxed{\int_{0}^{\infty} \frac{\ln x}{(x^{2}+1)^n} d x =-\frac{\pi(2 n-3) ! !}{2^{n}(n-1) !} \sum_{j=1}^{n-1} \frac{1}{2j-1}}$$
I first evaluate $$I_1=\int_{0}^{\infty} \frac{\ln x}{x^{2}+1} d x \stackrel{x\mapsto\frac{1}{x}}{=} -I_1 \Rightarrow I_1= 0.$$
and then start to raise up the power of the denominator
$$I_n=\int_{0}^{\infty} \frac{\ln x}{(x^{2}+1)^n} d x .$$
In order to use differentiation, I introduce a more general integral
$$I_n(a)=\int_{0}^{\infty} \frac{\ln x}{(x^{2}+a)^n} d x. $$
Now we can start with $I_1(a)$. Using $I_1=0$ yields
$$\displaystyle 1_1(a)=\int_{0}^{\infty} \frac{\ln x}{x^{2}+a} d x \stackrel{x\mapsto\frac{x}{a}}{=} \frac{\pi \ln a}{4 \sqrt a} \tag*{}$$
Now we are going to deal with $I_n$ by differentiating it by $(n-1)$ times
$$
\frac{d^{n-1}}{d a^{n-1}} \int_{0}^{\infty} \frac{\ln x}{x^{2}+a} d x=\frac{\pi}{4} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{a}\right)
$$
$$
\int_{0}^{\infty} \ln x\left[\frac{\partial^{n-1}}{\partial a^{n-1}}\left(\frac{1}{x^{2}+a}\right)\right] d x=\frac{\pi}{4} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{\sqrt a}\right)
$$
$$
\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+a\right)^{n}} d x=\frac{(-1)^{n-1} \pi}{4(n-1) !} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{\sqrt{a}}\right)
$$
In particular, when $a=1$, we get a formula for $$
\boxed{\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{n}} d x=\left.\frac{(-1)^{n-1} \pi}{4(n-1)!} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{\sqrt{a}}\right)\right|_{a=1}}
$$
For example, $$
\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{5}} d x=\frac{\pi}{4 \cdot 4 !}(-22)=-\frac{11 \pi}{48}
$$
and
$$
\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{10}} d x=\frac{-\pi}{4(9 !)}\left(\frac{71697105}{256}\right)=-\frac{1593269 \pi}{8257536}
$$
which is check by WA.
MY question
Though a formula for $I_n(a)$ was found, the last derivative is hard and tedious.
Is there any formula for $$\frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{\sqrt{a}}\right)? $$
|
Your integral can be expressed as the derivative of the complete beta function:
From Fubini-Tonelli we can interchange limit/derviative and integral:
$$I = \int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{n}} d x = \int_{0}^{\infty} \frac{\left(\frac{d}{dt}\Big|_{t=0+} x^t\right)}{(x^2+1)^n}dx = \frac{d}{dt}\Big|_{t=0+} \int_{0}^{\infty} \frac{x^t}{(1+x^2)^n} dx $$
Recall the following integral representation of the complete Beta function:
$$ \int_{0}^{\infty} \frac{y^{a-1}}{(1+y)^{a+b}} dx = B(a,b)$$
If we let $x^2 = w$ then
$$ I = \frac{d}{dt}\Big|_{t=0+} \frac{1}{2}\int_{0}^{\infty} \frac{w^{\frac{t-1}{2}}}{(1+w)^n}dw = \frac{d}{dt}\Big|_{t=0+} \frac{1}{2} B\left(\frac{t+1}{2},n-\frac{t+1}{2}\right)$$
Hence
$$\lim_{t \to 0+} \frac{d}{dt} \frac{1}{2}
B\left(\frac{t+1}{2},n-\frac{t+1}{2}\right) = \frac{1}{4}\left[ \psi\left(\frac{1}{2}\right) - \psi\left(n-\frac{1}{2}\right)\right]B\left(\frac{1}{2},n-\frac{1}{2}\right) $$
where $\psi(z)$ is the digamma function.
The right hand side can also be expressed as
$$I= -\frac{1}{4}\left[ H_{n-\frac{3}{2}} + \ln(4)\right]B\left(\frac{1}{2},n-\frac{1}{2}\right) $$
where $H_{x}$ is the Harmonic number
Hence
$$\boxed{\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{n}} d x = \frac{1}{4}\left[ \psi\left(\frac{1}{2}\right) - \psi\left(n-\frac{1}{2}\right)\right]B\left(\frac{1}{2},n-\frac{1}{2}\right) = -\frac{1}{4}\left[ H_{n-\frac{3}{2}} + \ln(4)\right]B\left(\frac{1}{2},n-\frac{1}{2}\right)}$$
Using this formula
$$ \psi\left(\frac{1}{2}-J\right) = \psi\left(\frac{1}{2}+J\right) = -\gamma -\ln(4) + \sum_{j=1}^{J} \frac{2}{2j-1} \quad J=0,1,2...$$
$$\psi\left(\frac{1}{2}\right) = -\gamma -\ln(4)$$
$$\psi\left(n-\frac{1}{2}\right) = -\gamma -\ln(4) + \sum_{j=1}^{n-1} \frac{2}{2j-1} $$
Hence
$$ \left[ \psi\left(\frac{1}{2}\right) - \psi\left(n-\frac{1}{2}\right)\right] = -\sum_{j=1}^{n-1} \frac{2}{2j-1}$$
We have another form:
$$ \boxed{\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{n}} d x = -\frac{1}{4}\sum_{j=1}^{n-1} \frac{2}{2j-1}B\left(\frac{1}{2},n-\frac{1}{2}\right) }$$
Similar to what other users found previously
|
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|
$\text { Show that } \sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=-2 \pi+2 \cos ^{-1} x \text { if }-1 \leq x \leq-\frac{1}{\sqrt{2}} $
Show that
$$\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=-2 \pi+2 \cos ^{-1} x$$ if $-1 \leq x \leq-\frac{1}{\sqrt{2}}.$
I tried solving this question a lot but I’m unable to. My answer comes different.
$\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)$ Putting $x=\cos \theta$
$$
\begin{array}{l}
=\sin ^{-1}\left(2 \cos \theta \sqrt{1-\cos ^{2} \theta}\right) \\
=\sin ^{-1}\left(2 \cos \theta \sqrt{\sin ^{2} \theta}\right) \quad\left(\because 1-\cos ^{2} \theta=\sin ^{2} \theta\right)
\end{array}
$$
$$
=\sin ^{-1}(2 \cos \theta \sin \theta)
$$
$$
=\sin ^{-1}(2 \cos \theta \sin \theta)
$$
$$
=-\sin ^{-1}(\sin 2\theta)
$$
(Using $\sin 2 x=2 \sin x \cos x$)
$=2 \theta \quad$ As $x=\cos \theta$
$=2 \times \cos ^{-1} x$
$=2 \cos ^{-1} x$
EDIT: If any confusion with Q writing
|
Continuing from the substitution in the question, where OP put $x=\cos \theta$. Here I pick $\theta$ to be the particular value $\frac{3\pi}4\le \theta\le \pi$, so that $\theta$ is the principal value $\theta = \cos^{-1}x$.
Using the formula to find general solutions of inverse $\sin$, if given $\sin 2\theta$, $2\theta$ would be in the form
$$2\theta = n\pi + (-1)^n\sin^{-1}(\sin 2\theta)$$
For this case where $\frac{3\pi}2\le 2\theta\le 2\pi$, so $n=2$,
$$\begin{align*}
2\theta &= 2\pi + \sin^{-1}(\sin 2\theta)\\
\sin^{-1}(\sin 2\theta)&=-2\pi + 2\theta\\
\sin^{-1}\left(2x\sqrt{1-x^2}\right) &= -2\pi + 2\cos^{-1}x
\end{align*}$$
The conversion from $\sin 2\theta = 2\cos\theta\sin\theta = 2x\sqrt{1-x^2}$ is what OP already knows.
($0 \le \sin \theta \le 1$ is important here, because that confirms that $\sqrt{\sin^2\theta} = \left|\sin\theta\right| = \sin\theta$.)
|
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|
${ n \choose k}={ n-1 \choose k-1}+{ n-2 \choose k-1}+{ n-3 \choose k-1}+...+{ k-1 \choose k-1}$ i tried to prove this by using the concept of composition
${ n \choose k}={ n-1 \choose k-1}+{ n-2 \choose k-1}+{ n-3 \choose k-1}+...+{ k-1 \choose k-1}$
my attempt:
for explain my idea let's take $ n=6$ and $k =4$
so ${ 6 \choose 4}={ 5 \choose 3}+{ 4 \choose 3}+{ 3 \choose 3}$
${ 6 \choose 4}$is the number of composition of $7$ into $5$ blocks, so it is the number of solution of this equation $a+x+y+z+w=7$ when $x,y,z,w,a \in \mathbb{Z^{*+}}$
let's denote the number of solutions of equation $A$ by $N_s(A)$
so now ,it's easy to see that $N_s(a+x+y+z+w=7)$ =$N_s(x+y+z+w=6)$ + $N_s(x+y+z+w=5)$ +$N_s(x+y+z+w=4)$
so it is ${ 5 \choose 3}+{ 4 \choose 3}+{ 3 \choose 3}$, and that is the RHS
and we can do that with any $n$ and $k$
so the LHS ${ n \choose k}$ = $N_s (\sum_{i=1}^{k+1} x_i=n+1)$
and the RHS ${ n-1 \choose k-1}+{ n-2 \choose k-1}+{ n-3 \choose k-1}+...+{ k-1 \choose k-1}$ = $N_s (\sum_{i=1}^{k} x_i=n)+N_s (\sum_{i=1}^{k} x_i=n-1)+N_s (\sum_{i=1}^{k} x_i=n-2)+...+N_s (\sum_{i=1}^{k} x_i=k)$
and because $N_s (\sum_{i=1}^{k+1} x_i=n+1)=N_s (\sum_{i=1}^{k} x_i=n)+N_s (\sum_{i=1}^{k} x_i=n-1)+N_s (\sum_{i=1}^{k} x_i=n-2)+...+N_s (\sum_{i=1}^{k} x_i=k)$
we can say that :
${ n \choose k}={ n-1 \choose k-1}+{ n-2 \choose k-1}+{ n-3 \choose k-1}+...+{ k-1 \choose k-1}$
Is my attempt correct?
|
Yes, your proof is valid. In words, you counted the number of solutions of the equation
$$x_1 + x_2 + \cdots + x_{k + 1} = n + 1 \tag{1}$$
in two different ways. The left-hand side (LHS) counts the number of solutions directly. The right-hand side (RHS) counts the same thing, depending on the value of $x_{k + 1}$ since you added the number of solutions of each of the equations
$$x_1 + x_2 + \cdots + x_k = n + 1 - x_{k + 1}$$
as $x_{k + 1}$ varies from $1$ to $n + 1 - k$, the set of all possible values $x_{k + 1}$ can assume if equation 1 is an equation in the positive integers.
Alternate Proof: We count $k$-element subsets of the $n$-element set $$\{0, 1, 2, \ldots, n - 1\}$$
in two different ways. The LHS gives a direct count. The RHS counts the number of $k$-element subsets whose largest element is $j$ since if $j$ is the largest element, we must also select $k - 1$ of the $j$ elements smaller than $j$ to be in the $k$-element subset. Since the largest element can vary from $k - 1$ to $n - 1$, we obtain
$$\binom{n}{k} = \sum_{j = k - 1}^{n - 1} \binom{j}{k - 1} = \binom{k - 1}{k - 1} + \cdots + \binom{n - 3}{k - 1} + \binom{n - 2}{k - 1} + \binom{n - 1}{k - 1}$$
|
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|
What is the sum of natural numbers that have $5$ digits or less, and all of the digits are distinct? $1+2+3+\dots+7+8+9+10+12+13+\dots+96+97+98+102+103+104+\dots+985+986+987+1023+1024+1025+\dots+9874+9875+9876+10234+10235+10236+\dots+98763+98764+98765=$
The only thing I can do is to evaluate a (bad) upper bound by evaluating the sum of natural numbers from $1$ to $98765$, that is equal to $98765 \times (98765+1)/2=4,877,311,995$. So the desired answer is less than that.
Any help would be appreciated. Thanks.
|
Allow $0$ as a lead digit, and regard $025$ as different from $25$. Then the average of each digit is $4.5$. Finally, remove numbers whose first digit is $0$, and whose other digits have average $5$.
So there are $10$ one-digit numbers, average $4.5$, $90$ two-digit numbers, average $4.5 ×11$, $720$ three-digit and so on.
Then subtract $9$ two-digit numbers that start with $0$ with average $5$, $72$ three-digit numbers starting with $0$ and average $55$, and so on.
|
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|
Elementary inequalities exercise - how to 'spot' the right sum of squares? I have been working through CJ Bradley's Introduction to Inequalities with a high school student and have been at loss to see how one could stumble upon the solution given for Q3 in Exercise 2c.
Question:
If $ad-bc=1$ prove that $a^2+b^2+c^2+d^2+ac+bd \geq \sqrt{3}$.
Solution:
Make the inequality homogenous by multiplying both sides by $ad-bc=1$. [That seems sensible.] Take everything onto one side so we now want to show
$$a^2+b^2+c^2+d^2+ac+bd -\sqrt{3}(ad-bc) \geq 0 \tag{1}.$$
[That's also a reasonable thing to do. The trouble is coming next...]
Now play around until you notice the left hand side can be written as
$$\frac{1}{4}(2a+c-\sqrt{3}d)^2 + \frac{1}{4}(2b+d+\sqrt{3}c)^2 \tag{2}.$$
[What??]
I played around for a fair while and didn't get to this. I have a suspicion that this question was created by reverse-engineering. What thought processes get you from (1) to (2), without knowing (2) beforehand? How can you get to the solution without pulling a rabbit out of a hat?
|
$a^2+b^2+c^2+d^2+ac+bd -\sqrt{3}(ad-bc) \geq 0 \tag{1}$
Completing the squares is made easier by introducing a factor of $\,2\,$, so
let $\,\lambda = \frac{1}{2}\,$ and $\,\mu = \frac{\sqrt{3}}{2}\,$, then the LHS of $(1)$ can be written as:
$$
a^2+b^2+c^2+d^2+ 2\lambda(ac+bd) - 2\mu(ad-bc)
$$
$$\begin{align}
= \;\;\;& a^2 + 2a(\lambda c - \mu d) \color{red}{+(\lambda c - \mu d)^2-(\lambda c - \mu d)^2}
\\ + \;&b^2+2b(\mu c + \lambda d) \color{blue}{+(\mu c + \lambda d)^2 -(\mu c + \lambda d)^2}
\\ + \;& c^2 + d^2
\end{align}
$$
$$
\require{cancel}
\begin{align}
= \;\;\;& (a+\lambda c - \mu d)^2 + (b + \mu c + \lambda d)^2
\\ - \;& \color{red}{\lambda^2c^2+ \cancel{2 \mu\lambda cd} - \mu^2 d^2} - \color{blue}{\mu^2 c^2 - \cancel{2 \mu\lambda cd} - \lambda^2d^2} + c^2 + d^2
\end{align}
$$
$$
= (a+\lambda c - \mu d)^2 + (b + \mu c + \lambda d)^2 + (1 - \lambda^2 - \mu^2)(c^2+d^2)
$$
The expression reduces to the sum of two squares when $\,\lambda^2+\mu^2=1$, as in OP's question.
|
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|
Existence of limit $\lim_{n\to\infty}\frac{n!}{n^{n+1/2}e^{-n}}$ It is shown that
$$
e^{\frac{11}{12}}n^{n+1/2}e^{-n}< n!<en^{n+1/2}e^{-n}.
$$
The question is, how to conclude from here that the limit
$$
\lim_{n\to\infty}\frac{n!}{n^{n+1/2}e^{-n}}
$$
exists? I suspect that $\frac{n!}{n^{n+1/2}e^{-n}}$ is decreasing, but I can't prove this claim.
Note: I am studying a topic about Stirling's formula, so please avoid mentioning that the limit exists due to that formula.
|
You can prove the monotonicity as follows. Let $a_n = \frac{{n!}}{{n^{n + 1/2} e^{ - n} }}$. Then
$$
\frac{{a_n }}{{a_{n + 1} }} = \left( {1 + \frac{1}{n}} \right)^{n + 1/2} \frac{1}{e} = \exp \left( {\left( {n + \frac{1}{2}} \right)\log \left( {1 + \frac{1}{n}} \right) - 1} \right).
$$
But
\begin{align*}
\left( {n + \frac{1}{2}} \right)\log \left( {1 + \frac{1}{n}} \right) - 1 &= (2n + 1)\frac{1}{2}\log \left( {\frac{{1 + \frac{1}{{2n + 1}}}}{{1 - \frac{1}{{2n + 1}}}}} \right) - 1 \\ & = (2n + 1)\tanh ^{ - 1} \left( {\frac{1}{{2n + 1}}} \right) - 1,
\end{align*}
and $\tanh ^{ - 1} (x) = x + \frac{{x^3 }}{3} + \frac{{x^5 }}{5} + \ldots > x$ for $0<x<1$, i.e.,
$$
\left( {n + \frac{1}{2}} \right)\log \left( {1 + \frac{1}{n}} \right) - 1 > (2n + 1)\frac{1}{{2n + 1}} - 1 = 0.
$$
Consequently,
$$
\frac{{a_n }}{{a_{n + 1} }} > e^0 = 1,
$$
i.e., the sequence $a_n$ is strictly decreasing.
|
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|
Inequality with real exponents I had the following problem on a past homework asignment:
For $x,p \in \mathbb{R}^+$
$$\frac{p}{(1+x)^{p+1}}<\frac{1}{x^p}-\frac{1}{(1+x)^{p}}$$
So, I think one way to show this for $1\leq p$ would be with rules for real exponents for real exponents (since $x^{-1} >-1$):
$$(1+\frac{1}{x})^p \geq 1+p \frac{1}{x}>1+\frac{p}{1+x}$$
and since $(1+x)^p>0$
$$\frac{(1+x)^{p}}{x^p}=\left(\frac{1+x}{x}\right)^p=(1+x^{-1})^p>1+\frac{p}{1+x}$$
$$\iff\frac{1}{x^p}>\frac{1}{(1+x)^{p}}+\frac{p}{(1+x)^{p+1}}$$
$$\iff\frac{p}{(1+x)^{p+1}}<\frac{1}{x^p}-\frac{1}{(1+x)^{p}}$$
For $0<p<1$ this argument doesn't hold since the inequality gets flipped:
$$(1+\frac{1}{x})^p \leq 1+p \frac{1}{x}>1+\frac{p}{1+x}$$
$$\frac{(1+x)^{p}}{x^p}=\left(\frac{1+x}{x}\right)^p=(1+x^{-1})^p\leq 1+px^{-1}>1+\frac{p}{1+x}$$
$$\iff\frac{1}{x^p}\leq\frac{1}{(1+x)^p}+\frac{p}{x(1+x)^p}>\frac{1}{(1+x)^p}+\frac{p}{(1+x)^{p+1}}$$
$$\iff\frac{1}{x^p}-\frac{1}{(1+x)^p}\leq\frac{p}{x(1+x)^p}>\frac{p}{(1+x)^{p+1}}$$
$$\iff\frac{p}{(1+x)^{p+1}}>\frac{1}{x^p}-\frac{1}{(1+x)^{p}}$$
What would I need to do to complete this proof? I tried looking at $x^p(p+x+1)<(x+1)^{p+1}$ as well and got nowhere. Is this 'possible' to prove without the binomial series?
|
Case $0 < p < 1$:
The inequality is written as
$$\frac{1 + x + p}{1 + x} \left(1 - \frac{1}{1 + x}\right)^p < 1.$$
Using Bernoulli inequality
$(1 - u)^r \le 1 - ru$ for all $0 < u < 1$ and $0 < r \le 1$, we have
$$\left(1 - \frac{1}{1 + x}\right)^p \le 1 - \frac{1}{1 + x}p.$$
It suffices to prove that
$$\frac{1 + x + p}{1 + x}\,\frac{1 + x - p}{1 + x} < 1$$
which is clearly true.
|
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|
Prove that $\lim_{n→\infty}\frac{\sqrt{n^2+a^2}}{n}=1$ using "$\varepsilon \to N$" definition how to prove this equation with the "$\varepsilon \to N$" definition? I feel have some trouble.Thanks!
$\lim_{n \to\infty}\frac{\sqrt{n^2+a^2}}{n}=1\\$
I complete my solution:
when $a=0$ it's obviously true. When
$a \neq 0$: $$\begin{align}
\frac{\sqrt{n^2+a^2}}{n}-1&=\frac{\sqrt{n^2+a^2}}{n}-\frac{n}{n}
\\&=\frac{\sqrt{n^2+a^2}-\sqrt{n^2}}{n}\\&=\frac{1}{n}\cdot \frac{a^2}{\sqrt{n^2+a^2}+n}\lt\frac{a^2}{\sqrt{n^2}+n}\cdot\frac{1}{n}\lt\frac{a^2}{n(\sqrt{n^2}+\sqrt{n^2})}=\frac{a^2}{2n}
\\ for\quad \forall \varepsilon \gt0, \quad when\quad \frac{a^2}{2N^2}\lt \varepsilon, \mathit N\gt \frac{a}{\sqrt{2\varepsilon}}+1,|\frac{\sqrt{n^2+a^2}}{n}-1|\lt\varepsilon.
\end{align}$$
|
welcome to MSE.
another point of view may help.
Taylor expansion for $$\sqrt{1+x}=1+\frac{x}2-\frac{x^2}{8}+o(x^3)$$and
$$\frac{\sqrt{n^2+a^2}}{n}=\frac{\sqrt{n^2+a^2}}{\sqrt {n^2}}=\sqrt{1+(\frac{a}{n})^2}\\=1+\frac 12(\frac{a}{n})^2 -\frac 18(\frac{a}{n})^4+...\\\geq 1+\frac 12(\frac{a}{n})^2$$ so
$$|\frac{\sqrt{n^2+a^2}}{n}-1|<\epsilon\\\frac12(\frac{a}{n})^2<\epsilon$$ and ...
|
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|
Derivative of $f(x) = \csc(x)\cot(x)$ using first principles. How to find derivative of $f(x) = \csc(x)\cot(x)$ using First principle of derivative?
I tried the following method.
$f(x) = \csc(x)\cot(x) = \dfrac{\cos(x)}{\sin^2(x)}$
Now using the limit,
$f'(x) = \lim\limits_{h\to0}\dfrac{\dfrac{\cos(x+h)}{\sin^2(x+h)} - \dfrac{\cos(x)}{\sin^2(x)}}{h}$
$f'(x) = \lim\limits_{h\to0}\dfrac{\cos(x+h)\sin^2(x) - \cos(x) \sin^2(x+h)}{h\sin^2(x+h)\sin^2(x)}$
Now, what should I do with this limit? I tried to apply the following identities,
*
*$\cos(A+B)= \cos(A)\cos(B) - \sin(A)\sin(B)$
*$\sin(A+B) = \sin(A) \cos(B) + \cos(A) \sin(B)$
I also tried to change $\sin^2(x) $ into $\dfrac{1 - \cos(2x)}{2}$,
But all of these formulas seem not to work here. Can anyone guide me for the same?
|
If $$f(x) = \csc(x)\cot(x)\\= \dfrac{\cos(x)}{\sin^2(x)},$$ then
$$f'(x) = \lim_{h\to0}\dfrac{\dfrac{\cos(x+h)}{\sin^2(x+h)} - \dfrac{\cos(x)}{\sin^2(x)}}{h}\\
= \lim_{h\to0}\dfrac{\cos(x+h)\sin^2(x) - \cos(x) \sin^2(x+h)}{h\sin^2(x+h)\sin^2(x)}\\
= \lim_{h\to0}\left(\dfrac{\cos(x+h) -\cos(x)}{h}\frac{\sin^2(x)}{\sin^2(x+h)\sin^2(x)} \\-\dfrac{ \sin^2(x+h)-\sin^2(x)}h\frac{\cos(x)}{\sin^2(x+h)\sin^2(x)}\right)\\
= \lim_{h\to0}\dfrac{\cos(x+h) -\cos(x)}{h}\lim_{h\to0}\frac1{\sin^2(x+h)} \\-\lim_{h\to0}\dfrac{ \sin^2(x+h)-\sin^2(x)}h\lim_{h\to0}\frac{\cos(x)}{\sin^2(x+h)\sin^2(x)}\\
= \csc^2(x)\lim_{h\to0}\dfrac{\cos(x+h) -\cos(x)}{h}-\frac{\cos(x)}{\sin^4(x)}\lim_{h\to0}\dfrac{ \sin^2(x+h)-\sin^2(x)}h\\
= \csc^2(x)[\color{red}{-\sin(x)}]-\frac{\cos(x)}{\sin^4(x)}\lim_{h\to0}\left(\left(\sin(x+h)+\sin(x)\right)\dfrac{ \sin(x+h)-\sin(x)}h \right)\\
= -\csc(x)-\frac{\cos(x)}{\sin^4(x)}\lim_{h\to0}\left(\sin(x+h)+\sin(x)\right)\lim_{h\to0}\dfrac{ \sin(x+h)-\sin(x)}h\\
= -\csc(x)-\frac{2\cos(x)}{\sin^3(x)}[\color{red}{\cos(x)}]\\
= -\csc(x)-\frac{2\cos^2(x)}{\sin^3(x)}\\
= -\frac{\sin^2(x)+2\cos^2(x)}{\sin^3(x)}.$$
Answer verified by WolframAlpha.
|
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|
Effective method to solve $\frac{\sqrt[3]{1+x} -\sqrt[3]{1-x}}{\sqrt[3]{1+x} +\sqrt[3]{1-x}} = \frac{x(x^2+3)}{3x^2+1} $ I want to know, is there an easy method to solve below equation $$\frac{\sqrt[3]{1+x} -\sqrt[3]{1-x}}{\sqrt[3]{1+x} +\sqrt[3]{1-x}} = \frac{x(x^2+3)}{3x^2+1} $$ I tried it by plotting and find the solution $x=0,1,-1$ then I tried to solve it by sustitution, multiplying by $(\sqrt[3]{(1+x)^2}+\sqrt[3]{1+x} \sqrt[3]{1-x} +\sqrt[3]{(1-x)^2})$ and so on ...
but I got stuck in solving it normally.
I am thankful if anyone can show me the clue.
Implicit: when I plot both sides , I feel if we name $$f(x)=\frac{(\sqrt[3]{1+x} -\sqrt[3]{1-x})}{(\sqrt[3]{1+x} +\sqrt[3]{1-x})}$$ then $$f^{-1}= \frac{x(x^2+3)}{3x^2+1} $$ so It is an effective way to solve $$f(x)=f^{-1}(x)=x$$
|
I think about it again , $$f(x)=\frac{(\sqrt[3]{1+x} -\sqrt[3]{1-x})}{(\sqrt[3]{1+x} +\sqrt[3]{1-x})}=\\\frac{(\sqrt[3]{\frac{1+x}{1-x}} -1)}{(\sqrt[3]\frac{1+x}{1-x} +1)}$$so $$y=\frac{(\sqrt[3]{\frac{1+x}{1-x}} -1)}{(\sqrt[3]\frac{1+x}{1-x} +1)}=\frac{a+1}{a-1}\\\to a=\frac{y+1}{y-1}\\\to \sqrt[3]{\frac{1+x}{1-x}}=\frac{y+1}{y-1}\\\to x=y\\x=\frac{x(x^2+3)}{3x^2+1}
\\ \to x=0,1,-1$$
|
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|
Trigonometric and exponential integral $\int _0^{\pi }\frac{\cos \left(a\sin x\right)}{1+a\cos x}e^{a\cos x}dx$ How can we prove this Integral relation?
$$\int _0^{\pi }\frac{\cos \left(a\sin x\right)}{1+a\cos x}e^{a\cos x}dx=\frac{\pi }{e}\cdot \frac{e^{\sqrt{1-a^2}}}{\sqrt{1-a^2}}$$
where $\text{ }a\in(-1,1)$.
|
$$I(a)=\int _0^{\pi }\frac{\cos \left(a\sin x\right)}{1+a\cos x}e^{a\cos x}dx=\frac{1}{2}\Re\int _{-\pi}^{\pi }\frac{e^{i \left(a\sin x\right)}}{1+a\cos x}e^{a\cos x}dx=\frac{1}{2}\Re\int _{-\pi}^{\pi }\frac{e^{ ae^{ix}}}{1+a\cos x}dx$$
$$=\Re\int _{-\pi}^{\pi }\frac{e^{ ae^{ix}}e^{ix}}{ae^{2ix}+2e^{ix}+a}dx=\Re\,(-i)\oint_{|z|=1}\frac{e^{az}}{az^2+2z+a}dz$$
The zeros of the denominator are $z_{1,2}=-\frac{1}{a}\pm\sqrt{\frac{1}{a^2}-1}$; only one pole ($\,z_1=-\frac{1}{a}+\sqrt{\frac{1}{a^2}-1}\,\,$) lies inside the closed contour $|z|=1$.
Therefore,
$$I(a)=\Re \operatorname {Rez}_{z=z_1}\,2\pi i\,(-i)\frac{e^{az}}{az^2+2z+a}=\pi\frac{e^{a\big(-\frac{1}{a}+\sqrt{\frac{1}{a^2}-1}\big)}}{a\sqrt{\frac{1}{a^2}-1}}=\frac{\pi}{e}\frac{e^\sqrt{1-a^2}}{\sqrt{1-a^2}}$$
|
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|
Is it hard to tackle the integral $\int_{0}^{\infty} \frac{x^{2}}{\left(1+x^{4}\right)^{2}} d x?$ Putting $ \displaystyle=4, \alpha=2, n=2 $ in my post,
$$\int_{0}^{\infty} \frac{x^{n}}{\left(1+x^{m}\right)^{\alpha}}dx=\frac{\pi}{m(\alpha-1) !} \csc\frac{(n+1) \pi}{m}\prod_{k=1}^{\alpha-1}\left(\alpha-k-\frac{n+1}{m}\right)$$
we can conclude that
$$
\begin{aligned}
I &=\frac{\pi}{4(2-1) !} \csc \frac{3 \pi}{4}\left(1-\frac{3}{4}\right) \\
&=\frac{\sqrt{2} \pi}{16}
\end{aligned}
$$
Question:
Is there any other method?
Your suggestion and alternative are warmly welcome.
|
Integrate both sides of the equation $\left(\frac{x^3}{1+x^4}\right)’= \frac{4x^2}{(1+x^4)^2}-\frac{x^2}{1+x^4}$
to obtain
$$\int_{0}^{\infty} \frac{x^{2}}{(1+x^{4})^{2}} d x
= \frac14\int_{0}^{\infty}
\frac{x^{2}}{1+x^{4}} d x=\frac14 \cdot \frac\pi{2\sqrt2}=\frac\pi{8\sqrt2}
$$
|
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|
Solve $y’’ – 4y’ + 5y = 4e^{2x}\sin(x)$ using $\mathcal D$ operator Hello – I am working through the following question and get stuck at step 6. Could someone please advise in simple terms which I can hopefully understand. Thanks
$$y'' – 4y' + 5y = 4e^{2x}\sin(x)$$
Step one – Order equation so that differential operator is in front of the RHS of the equation
$\newcommand{\D}{\mathcal D}$
$$1 = \frac 1 {\D^2 – 4\D + 5} \cdot 4e^{2x}\sin(x)$$
Step two – move constant and exponential in front of the $\D$ operator
$$1 = 4e^{2x}\cdot \frac 1 {\D^2 – 4\D + 5}\cdot \sin(x)$$
Step three – calculate $a$
Because of $e^{2x}$, $a = 2$, and because of $\sin(x)$, $a = 2 + i$.
Step four – insert $a$ into the $\D$ operator and then calculate to see if it equals zero
\begin{align}
1 &= 4e^{2x}\cdot \frac 1 {(2+i)^2 – 4(2+i) + 5}\cdot \sin(x)
\\
&= 4e^{2x}\cdot \frac 1 {(4+4i+4-8-4i+5)}\cdot\sin(x)
\\
&= 4e^{2x}\cdot \frac 1 {(0)}\cdot \sin(x)
\end{align}
Step 5 – because there is a zero, note $a = 2$ therefore make it $\D+2$
\begin{align}
1 &= 4e^{2x}\cdot \frac 1 {(\D + 2)^2 – 4(\D + 2) + 5}\cdot \sin(x)
\\
&= 4e^{2x}\cdot\frac 1 {\D^2 + 1}\cdot\sin(x)
\end{align}
What do I do for step 6? Please explain in simply terms and assume my calculus knowledge is low.
|
$e^{ix}=\cos(x)+i\sin(x)$
$\sin(x)= Im(e^{ix})$
$$
{1 \over D^2 + 1} \ \sin x = {1 \over D^2 + 1} \ \ Im(e^{ix})=Im{1 \over D^2 + 1} \ \ e^{ix}=Im{1 \over 2D} \ \ xe^{ix} =x Im{1 \over 2i} \ \ e^{ix}=x Im{1 \over 2i} \ \ (\cos(x)+i\sin(x))=\frac{x}{2} Im\ \ (-i\cos(x)+\sin(x))
= \left[ -{x \over 2} \cos x \right]
$$
|
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Find all non-similar solutions of matrix equation
Find all unique $($not conjugate to each other by an element of $GL_2(\mathbb{Z}))$ matrices $A \in M_2(\mathbb{Z})$, such that $A ^ 2 - 4A - I = 0$, where $I$ is the identity matrix.
From the equation it can be easily seen that matrix is a solution iff it has trace equal to $4$ and determinant equal to $-1$. The main problem is to find all conjugacy classes. Since $\mathbb{Z}$ is not a field, Jordan, Frobenius and Smith normal forms are not applicable. I also tried to brute-force the problem and solve the equation $CA = BC$, where $A$ and $B$ are two distinct solutions and $C \in GL_2(\mathbb{Z})$, but this boils down to the system of 3 diophantine equations with 4 variables, one of which is not even linear (as in this paper, page 3) and I have no idea how to check existence of solutions for that.
Another approach that I tried (again, without luck) was to decompose $A$ and $B$ into a product of $GL_2(\mathbb{Z})$ generators and check that they differ by a cyclic permutation (as listed here, point 6) but I was not able to do that in general for this family of matrices.
I have also found a theorem (in this presentation by Svetlana Katok, slide 10), which is very cryptic to me, but I think it is impossible to use in practice.
I feel that the answer is $\pmatrix{1 & 2 \\ 2 & 3}$ and $\pmatrix{0 & 1 \\ 1 & 4}$ because I could not find any counterexamples.
Is there any simple way to find all conjugacy classes?
|
Your guess is correct. Here is an elementary proof. For $b\neq 0$, let
$$
F(a,b)=\begin{pmatrix} a && b \\ -\frac{a^2-4a-1}{b} && 4-a \end{pmatrix} \tag{1}
$$
Then the initial matrix $A$ is necessarily of the form $F(a,b)$ for some integers $a$ and $b$ (such that $b$ divides $a^2-4a-1$). All the conjugates of $A$ will also be of this form.
The first step is then to find, when $b$ is large enough, a suitable conjugate $F(a',b')$ of $F(a,b)$ with $|b'| \lt |b|$ ; this will allow us to argue by induction on $|b|$.
Let $t\in{\mathbb Z}$ and
$$ P=\begin{pmatrix} 1 && 1 \\ -(t+1) && -t \end{pmatrix} \in SL_2({\mathbb Z})\tag{2} $$.
Then we have (where $\ldots$ represent terms that we need not compute)
$$
\begin{array}{lcl}
P^{-1}AP &=& \begin{pmatrix} -t && -1 \\ \ldots && \ldots \end{pmatrix}
\begin{pmatrix} a && b \\ -\frac{a^2-4a-1}{b} && 4-a \end{pmatrix}
\begin{pmatrix} 1 && 1 \\ -(t+1) && -t \end{pmatrix} \\
&=& \begin{pmatrix} -t && -1 \\ \ldots && \ldots \end{pmatrix}
\begin{pmatrix} \ldots && a-bt \\ \ldots && -\frac{a^2-4a-1}{b}+(a-4)t \end{pmatrix} \\
&=&
\begin{pmatrix} \ldots && -t(a-bt)+\frac{a^2-4a-1}{b}-(a-4)t \\ \ldots && \ldots \end{pmatrix} \\
\end{array}\tag{3}$$
We have thus shown that $P^{-1}F(a,b)P$ is of the form $F(a',b')$ where
$$
\begin{array}{lcl}
b' &=& -t(a-bt)+\frac{a^2-4a-1}{b}-(a-4)t \\
&=& \frac{1}{b}\bigg(-bt(a-bt)+(a^2-4a-1)-(a-4)bt\bigg) \\
&=& \frac{1}{b}\bigg(-bt(a-bt)+(a^2-4a)-(a-4)bt-1\bigg) \\
&=& \frac{1}{b}\bigg(-bt(a-bt)+(a-4)(a-bt)-1\bigg) \\
&=& \frac{1}{b}\bigg((a-bt-4)(a-bt)-1\bigg) \\
&=& \frac{1}{b}\bigg((a-bt-2)^2-5\bigg) \\
\end{array}\tag{4}$$
We can certainly find an integer $t$ such that $|a-2-bt|\leq \frac{|b|}{2}$, and
we then deduce from (4) that $|b'|\leq \frac{\max\bigg(5,\big|\frac{b^2}{4}-5\big|\bigg)}{|b|}$. When
$|b| \geq 7$, we have $\max(5,\big|\frac{b^2}{4}-5\big|) \leq b^2$ and hence $|b'| \leq |b|$.
If we look a little closer, we realize that we can extend this argument : for $|b|\geq 3$ we can still find a $t$ making $|b'|\lt |b|$. Indeed, notice first that since $a^2-4a-1=0$ has no integral solutions modulo $3$, $|b|$ cannot equal $3$ or $6$. So we are left with $|b|=4$ or $5$.
When $|b|=4$, $a$ must be odd ; we can write $a=4s\pm 1$ for some integer $s$.
If $a=4s-1$, then $a-4(s-1)-2=1$ ; taking $t=sign(b)(s-1)$ therefore yields $b'=-1$ in (4).
If $a=4s+1$, then $a-4(s-1)-2=3$ ; taking $t=sign(b)(s-1)$ therefore yields $b'=1$ in (4).
When $|b|=5$, $a$ must be of the form $5s+2$ for some integer $s$, so that $a-5(s-1)-2=5$ ; taking $t=sign(b)(s-1)$ therefore yields $b'=4$ in (4).
We are now left with the case $|b|=1$ or $2$, which is the base case of the induction, and here we apply a more direct argument :
When $|b|=1$, so that $b=\varepsilon$ with $\varepsilon \in \lbrace \pm 1\rbrace$, we can write
$$
F(a,b)=Q\begin{pmatrix} 0 && 1 \\ 1 && 4 \end{pmatrix}Q^{-1}, \ \textrm{where} \ Q=\begin{pmatrix} 0 && 1 \\ \varepsilon && \varepsilon(4-a) \end{pmatrix}. \tag{5}
$$
When $|b|=2$, so that $b=2\varepsilon$ with $\varepsilon \in \lbrace \pm 1\rbrace$, we can write
$$
F(a,b)=Q\begin{pmatrix} 1 && 2 \\ 2 && 3 \end{pmatrix}Q^{-1}, \ \textrm{where} \ Q=\begin{pmatrix} 0 && 1 \\ \varepsilon && \varepsilon\frac{3-a}{2} \end{pmatrix}. \tag{6}
$$
This finishes the proof.
|
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|
Does the following definite integral exist? I encounter a problem in which I would need to deal with the folloing definite integral
$$I(t)=\int_{0}^{\infty} \mathrm dp \frac{p^2}{\omega^5} \sin^2\left(\frac{\omega t}{2}\right)$$
in which $$\omega=\sqrt{m^2+p^2}$$
where m is just some positive constant, t is also a parameter. But I would like to obtain the value of $I(t)$ when t is small. My approach to this problem is the following
$$I(t)=I(0)+I(0)'t+\frac{I(0)''}{2}t^2...$$
$$\frac{dI}{dt}|_{t=0}=0$$
However
$$I(0)''\propto \int dp \frac{p^2}{\omega^3}$$
which diverge like $\ln(p)$
Can anyone help me to find the expansion of I(t) in terms of t?
|
I prefer to add another answer since based on a different approach.
$$I=\int_0^{\infty } \frac{p^2 \sin ^2\left(\frac{t}{2} \sqrt{m^2+p^2} \right)}{\left(m^2+p^2\right)^{5/2}} \, dp=\frac12\int_0^{\infty } \frac{p^2}{ \left(m^2+p^2\right)^{5/2}}\, dp-\frac12\int_0^{\infty } \frac{p^2 \cos \left(t\sqrt{m^2+p^2} \right)}{ \left(m^2+p^2\right)^{5/2}} \, dp$$
Let $p=m q$
$$2m ^2\,I=\int_0^{\infty }\frac{q^2}{\left(q^2+1\right)^{5/2}}\,dq-\int_0^{\infty }\frac{q^2 \cos \left(m t\sqrt{q^2+1}
\right)}{\left(q^2+1\right)^{5/2}}\,dq$$
$$\frac 13-2m^2\,I=\int_0^{\infty }\frac{q^2 \cos \left(mt \sqrt{q^2+1}
\right)}{\left(q^2+1\right)^{5/2}}\,dq$$
$$mt \sqrt{q^2+1}=x \implies q=\frac{\sqrt{x^2-m^2 t^2}}{m t}\implies dq=\frac{x}{m t \sqrt{x^2-m^2 t^2}}\,dx$$
$$\int_0^{\infty }\frac{q^2 \cos \left(mt \sqrt{q^2+1}
\right)}{\left(q^2+1\right)^{5/2}}\,dq=(mt)^2 \int_{mt}^\infty \sqrt{x^2-m^2 t^2}\,\,\frac{\cos (x) }{x^4}\,dx$$
$$\frac{1-6m^2\,I}{(mt)^2}=\int_{mt}^\infty \sqrt{x^2-m^2 t^2}\,\,\frac{\cos (x) }{x^4}\,dx$$
$$\sqrt{x^2-m^2 t^2}=\sum_{n=0}^\infty (-1)^n \binom{\frac{1}{2}}{n} (m t)^{2 n}\, x^{1-2 n}$$
$$\frac{1-6m^2\,I}{(mt)^2}=\sum_{n=0}^\infty (-1)^n \binom{\frac{1}{2}}{n} (m t)^{2 n}\int_{mt}^\infty \frac {\cos (x)}{x^{2 n+3} }\,dx$$
$$2 \left(6 m^2\,I-1\right)=\sum_{n=0}^\infty (-1)^n \binom{\frac{1}{2}}{n}\Big[E_{2 n+3}(+i m t)+E_{2 n+3}(-i m t)\Big]$$
|
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|
How to find the inverse function of the function $f$, defined as $f(x)=\frac{x}{1-x^2}$ for $-1 \lt x \lt 1$ I want to find the inverse function of the function $f$, defined as $f(x)=\frac{x}{1-x^2}$ for $-1 \lt x \lt 1$.
Letting $y=f(x)$, we first note that $x=0 \iff y=0$. Next, we consider the case when $y \neq 0$:
$y=\frac{x}{1-x^2} \implies-x^2y+y-x=0 \implies x^2+\frac{x}{y}-1=0 \implies \left(x+\frac{1}{2y}\right)^2-\left(1+\frac{1}{4y^2} \right)=0$
, which can be simplified to:
$\left|x+\frac{1}{2y} \right|=\sqrt{1+\frac{1}{4y^2}} \quad (*)$
If $x+\frac{1}{2y}\gt 0$, then we have: $g_1(y)=x=\sqrt{1+\frac{1}{4y^2}}-\frac{1}{2y}$
If $x+\frac{1}{2y} \lt 0$, then we have: $g_2(y)=x=(-1)\cdot\sqrt{1+\frac{1}{4y^2}}-\frac{1}{2y}$
In its current state, the collection of $(y,x)$ that satisfies $(*)$ does not represent a function...because for a single $y$, there is more than one $x$ that satisfies $(*)$.
Without making reference to graphing software, how does one determine the appropriate inverse for $f$?
Is there a standard procedure? One might argue that I should just try $g_1 \circ f$, $f \circ g_1$, $g_2 \circ f$, and $f \circ g_2$ and see which yield the $I$...but it seems to me like this doesn't necessarily exhaust all possibilities. For example, what if the true inverse is chopped up and pasted together versions of $g_1$ and $g_2$, so as to avoid the doubled values. i.e. let $g_3(x) = \begin{cases}g_1(x) &\text{ if } x \in (a,b)\\ g_2(x) &\text{ if }x \notin (a,b)\end{cases}$
|
To find the inverse function, you have to solve for $x$ the equation $y=\frac{x}{1-x^2}\Leftrightarrow yx^2+x-y=0$ with the restriction $-1<x<1$. When $y=0$, we get $x=0$. When $y\ne 0$, this is a quadratic equation with roots $$x=\frac{-1\pm\sqrt{1+4y^2}}{2y}.$$
Now you have to examine the two roots and determine whether they lie in the interval $(-1,1)$. For $y>0$, you can verify that
$$0<\frac{-1+\sqrt{1+4y^2}}{2y}<1,\quad\quad\quad
\frac{-1-\sqrt{1+4y^2}}{2y}<-1 $$
so $x=\frac{-1+\sqrt{1+4y^2}}{2y}$ is the correct root and the other one must be discarded. Similarly, when $y<0$, we have
$$-1<\frac{-1+\sqrt{1+4y^2}}{2y}<0,\quad\quad\quad
\frac{-1-\sqrt{1+4y^2}}{2y}>1$$
so again we discard the second root and keep the other one. Putting everything together, the inverse function is
$$f^{-1}(x)=
\begin{cases}
0, & x=0\\
\frac{-1+\sqrt{1+4x^2}}{2x}, & x\ne 0\end{cases} $$
|
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|
maximum of reciprocals of integers Suppose you have n integers, such that $\sum_{i=1}^n \frac{1}{a_i}<1$. What is the maximum value of this sum?
The answers for $n=1,2,3,4$ are $1-\frac{1}{2}, 1-\frac{1}{6}, 1-\frac{1}{42}, 1-\frac{1}{42*43}$(checked by hands).
Could it be anyhow proven that this sequence is exactly connected with https://oeis.org/A007018?
|
Claim:
$\max \{ \sum_{i=1}^n \frac{1}{a_i} \} = 1- \frac{1}{A_n}$ , where $A_n=(A_{n-1})^2+A_{n-1}$ with $A_0=1$.
Proof:
Let the claim be true for some $n=k$. That is,
$\max \{ \sum_{i=1}^k \frac{1}{a_i} \} = 1- \frac{1}{A_k}$ .
Now we need to find $$\max \{ \sum_{i=1}^{k+1} \frac{1}{a_i} \} $$.
We already know
$\max \{ \sum_{i=1}^k \frac{1}{a_i} \} = 1- \frac{1}{A_k}$,
$\implies$ to get $\max \{ \sum_{i=1}^{k+1} \frac{1}{a_i} \}$, we need to find $\max \{ 1-\frac{1}{A_k} +\frac{1}{a_{k+1}} \}$ .
Notice that, if $a_{k+1} \le A_k$ then $\frac{1}{a_{k+1}} \ge \frac{1}{A_k} \implies 1 - \frac{1}{A_k} +\frac{1}{a_{k+1}} \ge 1$, contradiction.
$\implies a_{k+1}>A_k$. For maximum, $a_{k+1}=A_k+1$
$\implies \max \{ \sum_{i=1}^{k+1} \frac{1}{a_i} \}=\max \{ 1-\frac{1}{A_k} +\frac{1}{a_{k+1}} \}=1-\frac{1}{A_k} +\frac{1}{A_k+1}=1-\frac{1}{A_k(A_k+1)}=1-\frac{1}{A_{k+1}}$.
Therefore, our claim is true by induction.
|
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|
Trigonometric integral inequality Let us consider the integral equation
\begin{equation}
f(x)=\lambda \int_{0}^{\pi} \cos (x-y) f(y) \hspace{1mm}dy+g(x), \quad x \in[0, \pi],
\end{equation}
where $f$ is an unknown function on $[0, \pi]$, $g(x)$ is a given continuous function on $[0, \pi]$ and $\lambda$ is a given real constant. Prove that the equation has a unique solution $f \in C[0, \pi]$ for each $\lambda \neq \frac{2}{\pi}$.
|
Since
$$
\cos (x-y) =\cos x \cos y + \sin x \sin y,
$$
we have that the equation in question is equivalent to
$$
f(x)=\lambda \cos x \int_0^\pi f(y)\cos y \hspace{1mm}dy + \lambda \sin x \int_0^\pi f(y)\sin y \hspace{1mm}dy + g(x).
$$
We see that $f$ is the function of the form
\begin{equation}
f (x) =a\cos x + b \sin x +g (x)
\label{fab02_009}
\end{equation}
where
$$
a=\lambda \int_0^\pi f(y)\cos y \hspace{1mm}dy \text{ und } b=\lambda \int_0^\pi f(y)\sin y \hspace{1mm}dy
$$
Adding this to the other equation gives the equation for $a$ and $b:$
$$
\begin{aligned}
a \cos x+b \sin x=& \lambda \cos x \int_{0}^{\pi} f(y) \cos y \hspace{1mm}dy+\lambda \sin x \int_{0}^{\pi} f(y) \sin y \hspace{1mm}\hspace{1mm}dy \\
=& \lambda \cos x \int_{0}^{\pi}(a \cos y+b \sin y+g(y)) \cos y \hspace{1mm}dy \\
&+\lambda \sin x \int_{0}^{\pi}(a \cos y+b \sin y+g(y)) \sin y \hspace{1mm}dy .
\end{aligned}
$$
Let us note:
$$
\begin{aligned}
\int_{0}^{\pi} \cos(y)^{2} \hspace{1mm}dy&=\int_{0}^{\pi} \frac{1+\cos 2 y}{2} \hspace{1mm}dy=\frac{\pi}{2} \\
\int_{0}^{\pi} \sin(y)^{2} \hspace{1mm}dy&=\int_{0}^{\pi} \frac{1-\cos 2 y}{2} \hspace{1mm}dy=\frac{\pi}{2}
\end{aligned}
$$
and
$$
\int_{0}^{\pi} \cos y \sin y \hspace{1mm}dy=\frac{1}{2} \int_{0}^{\pi} \sin 2 y \hspace{1mm}dy=0 .
$$
It follows that:
$$
\begin{aligned}
\int_{0}^{\pi}(a \cos y+b \sin y+g(y)) \cos y \hspace{1mm}dy &=a \int_{0}^{\pi} \cos(y)^{2} \hspace{1mm}dy+b \int_{0}^{\pi} \sin y \cos y \hspace{1mm}dy+\int_{0}^{\pi} g(y) \cos y \hspace{1mm}dy \\
&=a \frac{\pi}{2}+A,
\end{aligned}
$$
where:
$$
A=\int_{0}^{\pi} g(y) \cos y \hspace{1mm}dy
$$
and
$$
\begin{aligned}
\int_{0}^{\pi}(a \cos y+b \sin y+g(y)) \sin y dy &=a \int_{0}^{\pi} \cos y \sin y \hspace{1mm}dy+b \int_{0}^{\pi} \sin(y)^{2} \hspace{1mm}dy+\int_{0}^{\pi} g(y) \sin y \hspace{1mm}dy \\
&=b \frac{\pi}{2}+B
\end{aligned}
$$
where:
$$
B=\int_{0}^{\pi} g(y) \sin y \hspace{1mm}dy.
$$
So we get the equation for $a$ and $b$:
$$
a \cos x+b \sin x=\lambda\left (a \frac{\pi}{2}+A\right) \cos x+\lambda\left (b \frac{\pi}{2}+B\right) \sin x
$$
i. e.
$$
\begin{aligned}
a&=\lambda\left (a \frac{\pi}{2}+A\right) \\
b&=\lambda\left (b \frac{\pi}{2}+B\right).
\end{aligned}
$$
For $\lambda \neq \frac{2}{\pi}$ we get the unique solution:
$$
\begin{aligned}
a&= \frac{\lambda A}{1-\lambda\frac{\pi}{2}} \\
b&=\frac{\lambda B}{1-\lambda\frac{\pi}{2}}.
\end{aligned}
$$
|
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|
How to find $\arg(z)$ and $|z|$? How to find $|z|$ and $\arg (z)$
$z$ is complex number and
$z$ is defined by $$z=\left(\cos\frac\pi5+i\sin\frac\pi5\right)^{15}\cdot(3-3i)^{20}$$
I`ve tried to behave it like $$e^{i15\frac\pi5}\cdot e^{i20\frac\pi4}$$ and got in result = $$e^{i3\pi}\cdot e^{i5\pi}=e^{i8\pi}$$ which gives me $$(-1)^8=1$$ and if $$+z=|z|$$ so $$|z|=1$$ So is it my answer and how to find? $$Arg(z)$$
|
Just remember some basic properies of $\arg$ and absolue value: If a value is given in polar form, then the polar coordinates are obtained. With $r, \varphi\in \Bbb R$ and $r\geqslant 0$:
$$\begin{align*}
|r\cdot e^{i\varphi}| &= r \\
\arg (r\cdot e^{i\varphi}) &\equiv \varphi
\end{align*}$$
Where $\equiv$ means that the value is only determined up to an integer multiple of $2\pi$. Common choices for $\arg$ are $0\leqslant\arg<2\pi$ or $-\pi < \arg\leqslant \pi$.
Moreover, for a product of two complex numbers we have:
$$\begin{align}
|z\cdot w| &= |z|\cdot|w| \\
\arg(z\cdot w) &\equiv \arg z + \arg w \\
\end{align}$$
which implies for $z\neq 0$ and real exponents $p\neq 0$
$$\begin{align}
|z^p| &= |z|^p \\
\arg(z^p) &\equiv p\cdot\arg z\\
\end{align}$$
So the first lines for your $\arg z$ would prepare for easy computation and read:
$$\begin{align}
z &= \left(\cos\frac\pi5+i\sin\frac\pi5\right)^{15}\!\!\cdot\,(3-3i)^{20} \\
&= \left(\exp\frac{\pi i}5\right)^{15}\!\!\cdot\,3^{20}\cdot(1-i)^{20} \\
\end{align}$$
Hence:
$$\begin{align}
\arg z &\equiv 15\cdot \frac\pi5 + 20\cdot\underbrace{\arg(3)}_{\textstyle=0} + 20\cdot\underbrace{\arg(1-i)}_{\textstyle=-\pi/4} \\
&\equiv 3\pi - 5\pi \equiv 0 \mod 2\pi
\end{align}$$
|
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|
Find minimum of $\sqrt{a}+\sqrt{b}+\sqrt{c}$, given that $a,b,c \ge 0$ and $ab+bc+ca+abc=4$
Find the minimum of $\sqrt{a}+\sqrt{b}+\sqrt{c}$, given that $a,b,c \ge 0$ and $ab+bc+ca+abc=4$.
I can prove that $\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1,$ because:
$$ab+bc+ca+abc=4 \\ \implies (a+2)(b+2)(c+2)=(a+2)(b+2)+(b+2)(c+2)+(c+2)(a+2),$$
and $3 \le a+b+c$ (using $\frac{9}{a+b+c+6} \le \frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}$). But I didn't know how to deal with the $\sqrt{a}+\sqrt{b}+\sqrt{c}$.
Can anyone help me? Thank you so much.
|
Here is a sketch of the solution (you need to fill in the details).
One might use the following substitution
$$
a=\frac{2x}{y+z},~b=\frac{2y}{z+x},~c=\frac{2z}{x+y},
$$
where $a,b,c>0$ (why do they exist?). Thus, it remains to find the minimum (or rather infimum) of
$$
F(x,y,z)=\sqrt{\frac{2x}{y+z}}+\sqrt{\frac{2y}{z+x}}+\sqrt{\frac{2z}{x+y}}
$$
for positive $x,y,z$.
If $x=y=t>0$ and $z=0$, then $F(x,y,z)=F(t,t,0)=2\sqrt{2}$, so the desired infimum is at least $2\sqrt{2}$ (note that $z$ is not positive, but the conclusion still holds -- why?).
Finally, to prove that the expression above is always greater than $2\sqrt{2}$ it suffices to observe that
$$
\sqrt{\frac{x}{y+z}}\geq\frac{2x}{x+y+z}.
$$
(Why the last inequality holds? Also, why $F(x,y,z)$ will be strictly greater than $2\sqrt{2}$?)
|
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|
A pattern that leads to regular continued fractions of quadratic irrationals The following expression can be obtained by converting the continued fraction of quadratic irrationals to single fraction.
$$
\sqrt{N} = \frac{b\sqrt{N}+aN}{a\sqrt{N}+b}
$$
The equation holds for any values of $a$ and $b$. However, by calculating from the regular continued fraction, $a$ and $b$ are restricted to unique pair for $N$.
While investigating how to infer these $a$ and $b$ from $N$, I found the following properties under limited conditions. (not proven)
Let $N$ be expressed as follows.
$$
N=p^2\pm q \ \ \ \ (p,q \in \mathbb{N}, 1 < q<p)
$$
When $q$ divides $2p$, we obtain $a$ and $b$ as follows.
$$
a = \frac{2p}{q},\ b=ap\pm1
$$
I would like to know why this is happening. If it is already known, please point it out. Thank you.
concrete example
The regular continued fraction of $\sqrt{7}$ is
$$
\sqrt{7} = 2 + \frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{2+\sqrt{7}}}}}
$$
Return to the single fraction.
$$
\sqrt{7} = \frac{8\sqrt{7}+21}{3\sqrt{7}+8}
$$
$(a,b)=(3,8)$ is uniquely determined for $N=7$.
On the other hand, $7$ can be expressed as follows.
$$
7=p^2-q=3^2-2
$$
Since $q=2$ divides $2p=6$, we can obtain $a$ and $b$ as follows.
$$
a = \frac{6}{2} = 3,\ \ b = 3\cdot3-1 = 8
$$
a,b in N=1~50
|
your matrix is the generator of the (oriented) automorphism group of the quadratic form $x^2 - N y^2.$ Given the smallest positive (nonzero) $u,v$ solving the Pell equation $u^2 - N v^2 = 1,$ your matrix is
$$
A =
\left(
\begin{array}{cc}
u & Nv \\
v & u
\end{array}
\right)
$$
and your fraction is
$$ \frac{u \sqrt N + v N}{v \sqrt N + u} $$
Without matrices, taking integers $x,y,$ we can confirm that
$$ (ux + Nv y)^2 - N(vx + uy)^2 = x^2 - N y^2 $$
This identity is the generalization of "Vieta Jumping," which is the phrase used in contests for automorphisms of any form $f(x,y)=x^2 - kxy + y^2 .$ Let's see, for the general indefinite form $g(x,y) = a x^2 + bxy + c y^2$ with $ \Delta = b^2 - 4ac > 0$ but $\Delta$ not a square, there is an expression for the generating automorphism using nontrivial solutions to $ \sigma^2 - \Delta \tau^2 = 4.$ Also, some forms have automorphisms of determinant $-1;$ in your $x^2 - N y^2 $ such a thing is $(x,y) \mapsto (-x,y) $
|
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|
Find all integers $n$ for which $x^6 + nx^2 − 1$ can be written as a product of two non-constant polynomials with integer coefficients. Find all integers $n$ for which $x^6 + nx^2 − 1$ can be written as a product of two non-constant polynomials with integer coefficients.
I first tried expanding: $(x^3+ax^2+bx+1)(x^3+cx^2+dx-1)$ and equating coefficients.
This yields the following relationships:
$$a+c = 0$$
$$ac+b+d = 0$$
$$-b+d = 0$$
and
$$n = -a+bd+c$$
Is this the right approach? How should I go about doing it?
|
This is not too difficult to do case-by-case. Write $f(x)=x^6+nx^2-1$.
*
*By the rational root theorem, the only possible linear factors are $x\pm1$. They are both factors if and only if $n=0$, when $f(x)=(x-1)(x+1)(x^4+x^2+1)$.
*The OP is well on their way to solving the case of two cubics. As the other steps also use this technique, let us record the fact that $f(x)$ is even. In other words $f(x)=f(-x)$. We combine this with the uniqueness of factorization of polynomials (up to unit factors). When concentrating on the case of a factorization into two irreducible cubics, $f(x)=p(x)q(x)$, we see that without loss of generality we can assume $p$ and $q$ to both be monic. As also $f(x)=p(-x)q(-x)$ we have two possibilities. Either $p(-x)=-p(x)$ or $p(-x)=-q(x)$. The former case is impossible, because then $p(0)=0$, and hence also $0=f(0)=1$, which is a contradiction. So if
$p(x)=x^3+ax^2+bx+c$ then $q(x)=x^3-ax^2+bx-c$. Without loss of generality we can assume that $c=1$ (interchange $p$ and $q$ otherwise). In this case
$$
\begin{aligned}f(x)&=p(x)q(x)\\
&=(x^3+bx)^2-(ax^2+1)^2\\
&=x^6+(2b-a^2)x^4+(b^2-2a)x^2-1.
\end{aligned}
$$
A look at the degree four terms tells us that $a$ must be even, say $a=2k$, and $b=2k^2$. Therfore the coefficient of the quadratic term is
$$n=(2k^2)^2-2\cdot(2k)=4k^4-4k.$$
Here the parameter $k$ can be any integer. The corresponding factorization is
$$f(x)=(x^3+2kx^2+2k^2x+1)(x^3-2kx^2+2k^2x-1).$$
*Let us then consider the case of $f(x)=p(x)q(x)$, where $p(x)$ is an irreducible quadratic factor. As above, we conclude that $p(-x)$ is also a factor. Either $p(x)=p(-x)$ or $q(x)=p(-x)r(x)$ where $r(x)$ is yet another irreducible quadratic factor. As $r(-x)$ must also be a factor, in all cases we have either $p(x)=p(-x)$ or $r(x)=r(-x)$, and can conclude that $f(x)$ has a factor without the linear term. In other words $x^2-m\mid f(x)$ for some $m\in\Bbb{Z}$. Plugging in $x^2=m$ tells us that
$$f(\sqrt m)=m^3+nm-1=0.$$
Therefore
$$m(m^2+n)=1.$$
This allows us to conclude that either $m=1$, $n=0$ (a case covered earlier) or $m=-1$, $n=-2$. The new possibility we got out of this is thus
$$
f(x)=x^6-2x^2-1=(x^2+1)(x^4-x^2-1).
$$
Either $n=-2$ or $n=4k^4-4k$ for some integer $k$.
|
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|
Simplify $\frac{1+\sin\alpha-2\sin^2\left(45^\circ-\frac{\alpha}{2}\right)}{4\cos\frac{\alpha}{2}}$ Simplify $$\dfrac{1+\sin\alpha-2\sin^2\left(45^\circ-\dfrac{\alpha}{2}\right)}{4\cos\dfrac{\alpha}{2}}$$
I am reading the solution of the authors and I really don't see how $$\dfrac{1+\sin\alpha-2\sin^2\left(45^\circ-\dfrac{\alpha}{2}\right)}{4\cos\dfrac{\alpha}{2}}=\dfrac{1+\sin\alpha-(1-\cos(90^\circ-\alpha))}{4\cos\dfrac{\alpha}{2}}$$ Which identity have they used?
|
Only one thing changes, namely: $2\sin(45^\circ - a/2)^2 \rightarrow 1-\cos(90^\circ - a).$
So recall the double angle formula: $\cos(2\theta) = 1 - 2\sin(\theta)^2$
Then by rearrangement $2\sin(\theta)^2 = 1 - \cos(2\theta)$.
Let $\theta = 45^\circ - a/2$, then $2\sin(45^\circ - a/2)^2 = 1-\cos(2\cdot (45^\circ - a/2)) = 1-\cos(90^\circ - a)$.
|
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|
Find the range of $\frac{\sqrt{(x-1)(x+3)}}{x+2}$
Find the range of $\frac{\sqrt{(x-1)(x+3)}}{x+2}$
My Attempt:$$y^2=\frac{x^2+2x-3}{x^2+4x+4}=z\\\implies x^2(z-1)+x(4z-2)+4z+3=0$$
Discriminant greater than equal to zero, so, $$(4z-2)^2-4(z-1)(4z+3)\ge0\\\implies z\le\frac43\\\implies -\frac2{\sqrt3}\le y\le\frac2{\sqrt3}$$
But the answer given is $[-\frac2{\sqrt3},1]$
What am I missing?
|
The problem is that squaring an equation may introduce extraneous solutions. You either have to check each solution (by substituting it into the original equation), or you use equivalent equations throughout the solution process which ensures extraneous solutions won't occur. Taking the second approach, your equation $y=\sqrt{(x-1)(x+3)}/(x+2)$ is equivalent to the system
$$\left\{
\begin{array}{l}
(y^2-1)x^2+2(2y^2-1)x+4y^2+3=0\\
y(x+2)\ge 0\\
x+2\ne 0
\end{array}
\right. $$
The range of $f$ is those $y$ for which the system has a solution. For $y^2-1=0\Leftrightarrow y=\pm1$, the equation is $2x+7=0\Leftrightarrow x=-7/2$. With $y=1$ the first inequality isn't satisfied while $y=-1$ is ok. So $y=-1$ is in the range of $f$ and $y=1$ isn't. For $y^2-1\ne 0$, the quadratic formula gives
$$x=\frac{1-2y^2\pm\sqrt{4-3y^2}}{y^2-1},\quad\quad
y\in\left[-\frac{2}{\sqrt 3},\frac{2}{\sqrt 3}\right]\setminus\{-1,1\} $$
and now you have to solve the inequalities $y(x+2)\ge 0,x+2\ne 0$ with each of the two roots for $x$. That is, you have to solve
$$\frac{y\left(-1-\sqrt{4-3y^2}\right)}{y^2-1}\ge0\\-1-\sqrt{4-3y^2}\ne 0 $$
and
$$\frac{y\left(-1+\sqrt{4-3y^2}\right)}{y^2-1}\ge0\\-1+\sqrt{4-3y^2}\ne 0 $$
The solutions to these are, respectively, $y\in\left[-\frac{2}{\sqrt{3}},-1\right)\cup\left[0,1\right)$ and $y\in\left[-\frac{2}{\sqrt{3}},-1\right)\cup\left(-1,0\right]$. Combining the two sets and also adding $y=-1$ (but not $y=1$), we conclude that the range of $f$ is
$$\left[-\frac{2}{\sqrt{3}},1\right)$$
Edit: For the second pair of inequalities, you can use
$$-1+\sqrt{4-3y^2}
=\frac{\left(-1+\sqrt{4-3y^2}\right)\left(1+\sqrt{4-3y^2}\right)}{1+\sqrt{4-3y^2}}
=\frac{-3(y^2-1)}{1+\sqrt{4-3y^2}} $$
|
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|
The characteristic polynomial of this family of matrices I'm looking at the following family of $n\times n$ matrices. The entries are 0 everywhere except above and below the diagonal. Above it takes values from $1 \to n-1$ and below from $ -n +1 \to -1$. Example when $n=4$:
$\left[
\begin{matrix}
0 & 1 & 0 & 0 \\
-3 & 0 & 2 & 0 \\
0 & -2 & 0 & 3 \\
0 & 0 & -1 & 0 \\
\end{matrix} \right]
$
How can I find the characteristic polynomial of these matrices in general?
|
Let $D=\operatorname{diag}(1,i,i^2,\ldots,i^{n-1})=\operatorname{diag}(1,i,-1,-i,1,i,-1,-i,\ldots)$. Then $D^{-1}AD=iK$ where
$$
K=\pmatrix{
0&1\\
n-1&0&2\\
&n-2&\ddots&\ddots\\
& &\ddots&\ddots&\ddots\\
& & &\ddots&0 &n-1\\
& & & & 1 &0}.
$$
is known as the Kac matrix. The spectrum of $K$ is given by
$$
\{-n+1,\,-n+3,\,-n+5,\ldots,\,n-5,\,n-3,\,n-1\}.
$$
Hence the eigenvalues of $A$ are
$$
\{(-n+1)i,\,(-n+3)i,\,(-n+5)i,\ldots,\,(n-5)i,\,(n-3)i,\,(n-1)i\}
$$
and the characteristic polynomial of $A$ is
\begin{cases}
x&\text{when $n=1$,}\\
[x^2+(n-1)^2][x^2+(n-3)^2]\cdots(x^2+4^2)(x^2+2^2)x&\text{when $n\ge3$ is odd,}\\
[x^2+(n-1)^2][x^2+(n-3)^2]\cdots(x^2+3^2)(x^2+1)&\text{when $n$ is even.}\\
\end{cases}
|
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|
Showing $\int _{0} ^{\pi/4} \frac{\cos^{2022}(x)}{\sin^{2022}(x) + \cos^{2022}(x) } dx \approx \frac{\pi}{4}$
Show that $$\int_{0} ^{\pi/4} \frac{\cos^{2022}(x)}{\sin^{2022}(x) + \cos^{2022}(x) } dx \approx \frac{\pi}{4}$$
My method was this: I tried using $x \to \pi/4-x$ conversion but that doesn't lead to common denominator. Next thing I tried was to take help of approximation as the answer too is an approximation, so I thought as $\cos x >\sin x$ in $(0,\pi/4)$, but didn't got a good reason to neglect the $\sin^{2022}(x)$ in comparison to $\cos^{2022}x$. Can anyone explain how we can do so?
Note: also I think at $x= \pi/4$ we cannot neglect at all since both would be equal, so what in that point, will the integral be not a bit more then?
|
Let
$$I_n := \int_{0} ^{\pi/4} \frac{\cos^n(x)}{\sin^n(x) + \cos^n(x) } \,\mathrm{d}x.$$
We have
$$I_n = \frac{\pi}{4} - \int_{0} ^{\pi/4} \frac{\sin^n(x)}{\sin^n(x) + \cos^n(x) } \,\mathrm{d} x .$$
Clearly, $I_n < \frac{\pi}{4}$.
Also, we have
\begin{align*}
I_n &> \frac{\pi}{4} - \int_{0} ^{\pi/4} \frac{\sin^n(x)}{ \cos^n(x) } \,\mathrm{d} x \\
&= \frac{\pi}{4} - \int_{0} ^{\pi/4} \tan^n x \,\mathrm{d} x \\
&= \frac{\pi}{4} - \int_{0} ^{1} \frac{y^n}{1+y^2} \,\mathrm{d} y \\
&\ge \frac{\pi}{4} - \int_{0} ^{1} y^n(1 - y^2/2) \,\mathrm{d} y\\
&= \frac{\pi}{4} - \frac{n+5}{2(n+1)(n+3)}
\end{align*}
where we have used $\frac{1}{1+y^2} \le 1 - y^2/2$
for all $y\in [0, 1]$.
Thus, we have
$$\frac{\pi}{4} - \frac{n+5}{2(n+1)(n+3)} < I_n < \frac{\pi}{4}.$$
Question: Can we obtain asymptotic expansion of $I_n$?
|
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|
How to transform triple integral $\iiint_\Omega \sqrt{1- \frac{x^2}{a^2}- \frac{y^2}{b^2} - \frac{z^2}{c^2} }\ dx dy dz$ I have stumbled across this triple integral
$$\iiint_\Omega \sqrt{1- \frac{x^2}{a^2}- \frac{y^2}{b^2} - \frac{z^2}{c^2} }\ dx dy dz$$
where
$$\Omega =\left\{(x,y,z)\in{\cal{R}}^3\ \bigg| \ \frac{x^2}{a^2}+ \frac{y^2}{b^2} + \frac{z^2}{c^2} \le1 \right\}
$$
I know that I am supposed to use integral transform in which I have to use substitutions. I would substitute $x/a = u, y/b = v, z/c = w$. I can’t find enough examples on how to solve integrals with 3 variables with integral transform.
I really struggled with this integral. Can you guys help me with it?
|
Make the variable changes
$x= a u$, $ y= b v$, $z= c w$, and then integrate in spherical coordinates
\begin{align}
&\iiint_{\frac{x^2}{a^2}+ \frac{y^2}{b^2}
+ \frac{z^2}{c^2}\le1 }\sqrt{1- \frac{x^2}{a^2}- \frac{y^2}{b^2}
- \frac{z^2}{c^2} }\ dx dy dz \\
=&\ abc \iiint_{u^2 +v^2+w^2\le1 }\sqrt{1- {u^2}- {v^2}
- {w^2}}\ dudv dw \\
= &\ 2\pi \ abc\int_0^\pi \int _0^1 \sqrt{1- {\rho^2}}\ \rho^2 \sin\phi \ d\rho d\phi
=\frac{\pi^2}4 abc
\end{align}
|
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|
Let p be a prime. Prove that $x^4 + 4 y^4 = p$ has integer solutions if and only if $p = 5$. In this case, find the solutions for x and y. I have tried to factor
$$x^4 + 4 y^4 = p$$
using the Sophie Germain's Identity, as someone suggested in the comments, which yields:
$$((x+y)^2 + y^2)((x-y)^2 + y^2) = p$$
Since p is a prime, either
$$(x+y)^2 + y^2 = 1$$
or
$$(x-y)^2 + y^2 = 1$$
I do not know how to solve this for $x,y$. And I do not see how this will proves that p can only be 5. Furthermore, I do not know how to find the integer solutions for the equation:
$$x^4 + 4 y^4 = 5$$
|
Here's a hint:
$$x^{4} + 4y^{4} = (x^2)^2 + (2y^{2})^{2} + 4x^{2}y^{2} - 4x^{2}y^{2} = (x^{2} + 2y^{2})^{2} - (2xy)^{2}.$$ Can you factor this further?
If $x^{4} + 4y^{4} = p$ is prime, what does this say about the factors of $x^{4} + 4y^{2}$?
(Edited to give more details)
So, we have $$x^{4} + 4y^{4} = (x^{2} + 2y^{2})^{2} - (2xy)^{2}.$$ Since this is a difference of squares, we can factor it as
$$(x^{2} + 2y^{2}- 2xy)(x^{2}+2y^{2}+2xy).$$
Note that
$$x^{2} +2y^{2} -2xy = (x^{2}-2xy+y^{2})+y^{2} = (x-y)^{2} + y^{2},$$ and that
$$(x^{2}+2y^{2}+2xy) = (x^{2}+2xy+y^{2})+y^{2} = (x+y)^{2} + y^{2}.$$ Both of these are sums of squares, hence are nonnegative.
So, $x^{2} +2y^{2} -2xy$ and $x^{2}+2y^{2}+2xy$ are nonnegative integers which divide the prime $x^{4} +4y^{4} = p.$ So, we know that one of them must be $1.$
Case 1: Suppose that $x^{2} +2y^{2} -2xy = 1.$ Then, we have $(x-y)^{2} + y^{2} = 1.$ Since $(x-y)^{2}, y^{2}$ are both nonnegative integers, we must either have $(x-y)^{2} = 1$ and $y^{2} = 0,$ or $(x-y)^{2} = 0$ and $y^{2} = 1.$
The first possibility results in the possible solutions $(x, y) = (1, 0), (-1, 0)$, and the second possibility results in the possible solutions $(x, y) = (1, 1), (-1, -1).$
Note that if we substitute $(x, y) = (1, 0)$ or $(x, y) = (-1, 0)$ into $x^4 +4y^{4},$ we get $x^{4} + 4y^{4} = 1,$ which is not a prime. So, we eliminate those two solutions from consideration. On the other hand, if we substitute $(x, y) = (1, 1)$ and $(x, y) = (-1, -1)$, we get $x^{4} + y^{4} = 5.$ So, in the first case, we must have $p = 5,$ and we have the corresponding solutions $(x, y) = (1, 1), (-1, -1).$
Case 2: Now, suppose that $x^{2} +2y^{2} + 2xy = 1.$ Then, we get $(x+y)^{2} + y^{2} = 1.$ Using the same argument as in Case $1$, you get the possible solutions $(x, y) = (1, 0), (-1, 0), (-1, 1), (1, -1).$ (You should work out the details here for yourself!) The first two possibilities were already eliminated. If you substitute the remaining two $(x, y) = (-1, 1), (1, -1)$ into the equation, you get
$$x^{4} +4y^{4} = 5.$$ So, in the second case, we must also have $p =5,$ and we have the corresponding solutions $(x, y) = (1, -1), (-1, 1).$
To conclude, we note that the two cases we discussed above are all the possibilities (since one of $x^{2} +2y^{2} -2xy$ and $x^{2}+2y^{2}+2xy$ must be $1$). Therefore, we must have $p =5,$ and we have the solutions
$$(x, y) = (1, 1), (-1, -1), (1, -1), (-1, 1).$$
|
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|
Let $D,m$ be relatively prime integers with $m$ odd. Then $D \equiv 0,1 \pmod 4$ and $D \equiv b^2 \pmod m$ implies that $D \equiv b^2 \pmod{4m}$ This is from a proof in David A. Cox's Primes of the Form $x^2+ny^2$:
Lemma 2.5 Let $D\equiv 0,1 \bmod 4$ be an integer and $m$ be an odd integer relatively prime to $D$. Then $m$ is properly represented by a primitive form of discriminant $D$ if and only if $D$ is a quadratic residue modulo $m$.
Proof: If $f(x,y)$ properly represents $m$ then by Lemma 2.3, we may assume $f(x,y) = mx^2 + bxy + cy^2$. Thus $D=b^2-4mc$, and $D\equiv b^2 \bmod m$ follows immediately.
$\;\;\;\;$ Conversely, suppose that $D\equiv b^2 \bmod m$. Since $m$ is odd, we can assume that $D$ and $b$ have the same parity (replace $b$ by $b+m$ if necessary), and then $D\equiv 0,1 \bmod 4$ implies that $D\equiv b^2 \bmod 4m$. $[\ldots]$
I cannot see how, in the third paragraph, $D \equiv b^2 \pmod{4m}$ follows. I tried doing examples to get an intuition for why it holds, but I cannot figure out why. Any help is appreciated.
|
I don't think this holds unless we assume $b$ is odd. Let $D=81$ and $m=45$. Then $(D \pmod {45}) =36=6^2$ but $(D \pmod {4×45})=81=36+49$ $=36+m$.
If on the other hand $D \pmod m$ is an odd square $b^2$, then because both $D$ and $m$ are odd, it follows that the equation $D=b^2+km$ must hold for an integer $k$ that must be even. But the fact that $D \pmod 4 = b^2 \pmod 4$ implies that $km$ must in fact be a multiple of $4$, which implies $k$ must be a multiple of $4$. Thus $D=b^2+km$ for some $k$ which is a multiple of $4$, which gives $D = b^2+c(4m)$, where $c$ is the integer such that $k=4c$. In particular, as $D=c(4m)$ for some integral $c$, it follows that $D \pmod{4m} = b^2$ as well.
|
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|
Finding distance between a point and parametric equations This problem is from the parametric and trigonometric coordinate systems of the Art of Problem Solving Precalculus book:
Find the smallest distance between the point $ (1,2,3) $ and a point on the graph of the parametric equations $ x = 2-t, y=4+t, z=3+2t.$
I tried eliminating $ t$ by adding the first two equations together, yielding $ x+y=6$, and then multiplying the first equation by two and adding that to the third, yielding $ 2x+z=7$, and then I added these two equations together, and I got $ 3x+y+z=13$.
My textbook only covered simple 3d graphs using spherical and cylindrical coordinates, graphing cones, cylinders, spheres, and planes, so I didn't know what the graph of $ 3x+y+z=13$ was, however using a 3d graphing calculator I saw that it was a plane. I don't know how to proceed any further, any help would be appreciated.
|
You have the given point $P_0 = (1, 2, 3) $
And the curve (which is a line) is given by
$Q(t) = (2,4,3) + t ( -1, 1 , 2) $
The vector extending from $P_0$ to a point on the line $Q(t) $ is
$V(t) = Q(t) - P_0 = (2, 4, 3) + t (-1, 1, 2) - (1, 2, 3) = (1, 2, 0) + t (-1, 1, 2) $
And you want this vector to be perpendicular to the line, i.e. perpendicular to its direction vector $(-1, 1, 2) $. The condition for perpendicularity is that the dot product between the two vectors is $0$. Hence, we have
$ V(t) \cdot (-1, 1, 2) = 0 $
i.e.
$ \big( (1,2,0) + t (-1, 1, 2) \big) \cdot (-1, 1, 2) = 0 $
And this simplifies to
$ 1 + 6 t = 0 $
So $ t = \dfrac{-1}{6} $
Plug this into $V(t)$,
$V(\dfrac{-1}{6}) = (1,2,0) - \dfrac{1}{6} (-1, 1, 2) = \dfrac{1}{6} (7, 11, -2) $
The magnitude of this vector is the shortest distance
$\| V \| = \dfrac{1}{6} \sqrt{ 7^2 + 11^2 + (-2)^2 } = \dfrac{\sqrt{174}}{6} $
One could verify this result by finding the magnitude of the vector $W$ given by
$ W = \dfrac{(P_0 - (2,4,3) ) \times (-1, 1, 2)}{\| (-1 , 1, 2 ) \|} $
This is equal to
$ W = \dfrac{ (-1, -2, 0) \times (-1, 1, 2) }{\sqrt{6} } $
$ = \dfrac{(-4 , 2, -3 )}{\sqrt{6}} $
The magnitude of $W$ is
$\| W \| = \dfrac{\sqrt{ 29 }}{\sqrt{6}} = \dfrac{\sqrt{29 (6) }}{6} = \dfrac{\sqrt{ 174 }}{6} $
So the results match.
|
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|
Simplifiying the trigonometry equation to show that triangle is isosceles In a triangle $ABC$ , $X$ is a point inside it , its given that Angle $BAX = 10^\circ$ , Angle $ABX = 20^\circ$ , Angle $XAC = 40^\circ$ and Angle $XCA = 30^\circ$, prove that triangle is isosceles.
My method was using trigonometry to arrive at value of the unknown angles, and then showing its isoceles. Lets call Angle $XCB$ to be $\alpha$, then applying sin rule on triangle $ABX$ and $AXC$, and similarly for other two pairs keeping in mind one side is common to them, we get arrive at $\sin(\alpha)\sin 20^\circ\sin40^\circ = \sin(80^\circ -\alpha) \sin 10^\circ \sin 30^\circ$. Now my question is how do we solve such trigonometric equation in a restricted domain to show that only solution to this is $20^\circ$? Implying $AB = BC$.
|
Here is an approach using the Law of Sines and the Law of Cosines, that is not too bad.
$$
\begin{align}
BX
&=AB\,\frac{\sin\left(10^{\large\circ}\right)}{\sin\left(150^{\large\circ}\right)}\tag{1a}\\[6pt]
&=2AB\sin\left(10^{\large\circ}\right)\tag{1b}\\[6pt]
CX
&=AX\,\frac{\sin\left(40^{\large\circ}\right)}{\sin\left(30^{\large\circ}\right)}\tag{2a}\\
&=AB\,\frac{\sin\left(20^{\large\circ}\right)}{\sin\left(150^{\large\circ}\right)}\,\frac{\sin\left(40^{\large\circ}\right)}{\sin\left(30^{\large\circ}\right)}\tag{2b}\\[6pt]
&=4AB\sin\left(20^{\large\circ}\right)\sin\left(40^{\large\circ}\right)\tag{2c}\\[12pt]
BC^2
&=BX^2+CX^2-2BX\,CX\,\cos\left(100^{\large\circ}\right)\tag{3a}\\[6pt]
&=AB^2\left(4\sin^2\left(10^{\large\circ}\right)+16\sin^2\left(20^{\large\circ}\right)\sin^2\left(40^{\large\circ}\right)\right.\\
&\phantom{=AB^2\left(\right.}\left.{}-16\sin\left(10^{\large\circ}\right)\sin\left(20^{\large\circ}\right)\sin\left(40^{\large\circ}\right)\cos\left(100^{\large\circ}\right)\right)\tag{3b}\\[6pt]
&=AB^2\left(\color{#C00}{4\sin^2\left(10^{\large\circ}\right)}+\color{#090}{16\sin^2\left(20^{\large\circ}\right)\sin^2\left(40^{\large\circ}\right)}\right.\\
&\phantom{=AB^2\left(\right.}\left.{}+\color{#00F}{16\sin^2\left(10^{\large\circ}\right)\sin\left(20^{\large\circ}\right)\sin\left(40^{\large\circ}\right)}\right)\tag{3c}\\[6pt]
&=AB^2\left(\color{#C00}{2-2\cos\left(20^{\large\circ}\right)}+\color{#090}{4\cos^2\left(20^{\large\circ}\right)-4\cos\left(20^{\large\circ}\right)+1}\right.\\
&\phantom{=AB^2\left(\right.}\left.\color{#00F}{{}-4\cos^2\left(20^{\large\circ}\right)+6\cos\left(20^{\large\circ}\right)-2}\right)\tag{3d}\\[12pt]
&=AB^2\tag{3e}
\end{align}
$$
Explanation:
$\text{(1a)}$: Law of Sines
$\text{(1b)}$: $\sin\left(150^{\large\circ}\right)=\frac12$
$\text{(2a)}$: Law of Sines
$\text{(2b)}$: Law of Sines
$\text{(2c)}$: $\sin\left(30^{\large\circ}\right)=\sin\left(150^{\large\circ}\right)=\frac12$
$\text{(3a)}$: Law of Cosines
$\text{(3b)}$: apply $\text{(1b)}$ and $\text{(2c)}$
$\text{(3c)}$: $\cos\left(100^{\large\circ}\right)=-\sin\left(10^{\large\circ}\right)$
$\text{(3d)}$: use $\sin(a)\sin(b)=\frac{\cos(a-b)-\cos(a+b)}2$ to get
$\phantom{\text{(3d):}}$ $\sin^2\left(10^{\large\circ}\right)=\frac{1-\cos\left(20^{\large\circ}\right)}2$ and
$\phantom{\text{(3d):}}$ $\sin\left(20^{\large\circ}\right)\sin\left(40^{\large\circ}\right)=\frac{\cos\left(20^{\large\circ}\right)-\frac12}2$
$\text{(3e)}$: simplify
|
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Spivak, Ch. 13 "Integrals", Prob. 2: Show $\int_0^b x^4 dx=\frac{b^5}{5}$ by finding unique number such that $L(f,P)\leq \int_0^b x^4 \leq U(f,P)$? Consider the problem (Spivak, Chapter 13 "Integrals", problem 2) of showing that $\int_0^b x^4 dx=\frac{b^5}{5}$ by finding the unique number $\int_0^b x^4$ such that
$$L(f,P)\leq \int_0^b x^4 \leq U(f,P)$$
where $L(f,P)$ the lower sum of $f$ for partition $P$ on $[0,b]$, and $U(f,P)$ is the upper sum of $f$ for partition $P$ on $[0,b]$.
In the course of solving this problem, there is a step in which one must prove
$$\frac{1}{6n^5}\left [ 6n^5-15n^4+10n^3-n \right ] < 1\tag{1}$$
and
$$\frac{1}{n^5} \left [ n^5+\frac{5}{2}n^4+\frac{5}{3}n^3-\frac{n}{6} \right ]>1\tag{2}$$
My question is how to show these two inequalities are true.
Below is context on the steps leading up to the necessity of proving the inequalities $(1)$ and $(2)$.
We start with an equally spaced partition of $[0,b]$, $P_n=\{t_0,...,t_n\}$ where $t_i-t_{i-1}=\frac{b}{n}$.
In each partition subinterval, we have
$$m_i=\inf\{f(x): t_{i-1}\leq x \leq t_i \}=t_{i-1}^4$$
$$M_i=\sup\{f(x): t_{i-1}\leq x \leq t_i \}=t_{i}^4$$
Let's consider $L(f,P)$ specifically
$$L(f,P)=\sum_{i=1}^n m_i(t_i-t_{i-1})=\sum_{i=1}^n t_{i-1}^4(t_i-t_{i-1})$$
$$=\sum_{i=1}^n \left ( \frac{b(i-1)}{n} \right )^4\cdot \frac{b}{n}$$
$$= \frac{b^5}{n^5}\sum_{i=1}^n (i-1)^4$$
$$= \frac{b^5}{n^5}\sum_{i=0}^{n-1} i^4$$
$$= \frac{b^5}{n^5} \left [ \frac{(n-1)^5}{5} +\frac{(n-1)^4}{2}+ \frac{(n-1)^3}{3} -\frac{n-1}{30} \right ]$$
After a bit of algebra we reach
$$L(f,P)=\frac{b^5}{5}\frac{1}{6n^5}\left [ 6n^5-15n^4+10n^3-n \right ]$$
I'd like to show that $(1)$ is $<1$ so that I can assert that $$L(f,P)<\frac{b^5}{5}\tag{3}$$
Similarly,
$$U(f,P)=\sum_{i=1}^n M_i(t_i-t_{i-1})$$
$$=\sum_{i=1}^n t_i^4 (t_i-t_{i-1})$$
$$=\sum_{i=1}^n \left ( \frac{bi}{n} \right )^4\cdot \frac{b}{n}$$
$$=\frac{b^5}{n^5} \sum_{i=1}^n i^4 $$
$$=\frac{b^5}{n^5} \left [ \frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30} \right ]$$
$$=\frac{b^5}{5} \cdot \frac{1}{n^5} \left [ n^5+\frac{5}{2}n^4+\frac{5}{3}n^3-\frac{n}{6} \right ]$$
I'd like to show that $$\frac{1}{n^5} \left [ n^5+\frac{5}{2}n^4+\frac{5}{3}n^3-\frac{n}{6} \right ] > 1$$
so I can assert that $$U(f,P)>\frac{b^5}{5}\tag{4}$$
Once I can assert $(3)$ and $(4)$, I have
$$L(f,P)<\frac{b^5}{5}<U(f,P)\tag{5}$$
for all partitions $P$. Since It is also true that $U(f,P)-L(f,P)<\epsilon$ for any $\epsilon>0$ (by definition of $f$ being integrable), we can conclude that there is only one number satisfying $(5)$, and since $\int_0^b f$ definitely satisfies it by definition of $f$ being integrable, it must be the unique number. Hence $\int_0^b f=\frac{b^5}{5}$.
|
Inequality (1) is algebraically the same as the inequality
$$
15n^3-10n^2+1>0,\tag{1'}
$$
which you can prove by rewriting the LHS in the form:
$$
15n^3-10n^2+1 = 15n^2(n-\frac23)+1
$$ and noting that $n^2(n-\frac23)$ is positive for all integers $n\ge1$.
Inequality (2) is the algebraically the same as
$$
15n^3+10n^2>1\tag{2'}$$
which is true for all integers $n\ge1$. Indeed, the LHS is at least $25$ whenever $n\ge1$.
|
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|
How to solve long integration by partial fraction decomposition problems faster? Some problems are just too time consuming for short exam times what is the fastest way to solve problems like this one for example $$\int \frac{5x^4-21x^3+40x^2-37x+14}{(x-2)(x^2-2x+2)^2}dx$$
|
With $x=t+1$
\begin{align}
&\ \frac{5x^4-21x^3+40x^2-37x+14}{(x-2)(x^2-2x+2)^2}
= \frac{5t^4-t^3+7t^2+1}{(t-1)(t^2+1)^2}
= \frac{A}{t-1}+ {P(t)}
\end{align}
where $A = \lim_{t\to 1}\frac{5t^4-t^3+7t^2+1}{(t^2+1)^2}=3
$ and
\begin{align}
P(t)= &\ \frac{5t^4-t^3+7t^2+1}{(t-1)(t^2+1)^2}
-\frac{3}{t-1}\\
= &\ \frac{2t^3+t^2+2t+2}{(t^2+1)^2}
=\frac{2t(t^2+1)+(t^2+1)+1}{(t^2+1)^2}
= \frac{2t+1}{t^2+1}+ \frac{1}{(t^2+1)^2}\\
\end{align}
Thus, the integral becomes
$$I=\int \frac3{t-1}+ \frac{2t+1}{t^2+1}+ \frac{1}{(t^2+1)^2}\ dt
$$
|
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|
Integral with Weierstrass substituion, limits gone wrong I'm trying to evaluate the following integral with a Weierstrass substitution:
$$
\int_{\pi/3}^{4\pi/3} \frac{3}{13 + 6\sin x - 5\cos x} \text{d}x
$$
This comes out to about 0.55 when evaluated numerically. I thought it would be possible to rewrite the integral using the substitution $t = \tan\frac{x}{2}$ like so:
$$
\text{Upper limit: }\frac{4\pi}{3} \to \tan\frac{4\pi}{6} \to -\sqrt{3}
$$
$$
\text{Lower limit: }\frac{\pi}{3} \to \tan\frac{\pi}{6} \to \sqrt{3}/3
$$
$$
\int_{\sqrt{3}/3}^{-\sqrt{3}} \frac{3(1+t^2)}{2(3t+1)^2 + 6} \cdot \frac{2}{1 + t^2} \text{d}x
$$
However, evaluating this numerically now gives -1.26, so it's clear that this method is wrong. I am sure that
$$
\frac{3}{13 + 6\sin x - 5\cos x} = \frac{3(1+t^2)}{2(3t+1)^2 + 6}
$$
So something must have gone wrong changing the limits, but I've never come across this problem before. I thought the integral may be improper but plotting $\frac{3}{13 + 6\sin x - 5\cos x}$ on a graph shows it has no asymptotes or any other obvious problems in the range of integration.
Any help would be greatly appreciated :)
|
Take care of the discontinuity at $x=\pi$ with the half angle substitution $t=\tan\frac x2$
\begin{align}
&\int_{\pi/3}^{4\pi/3} \frac{3}{13 + 6\sin x - 5\cos x} dx
\overset{t=\tan\frac x2}=\bigg(\int_{\frac1{\sqrt3}}^\infty +\int_{-\infty}^{-\sqrt3}\bigg)
\frac3{4+6t+9t^2}dt
\end{align}
|
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|
Is there any method other than Feynman’s Integration Technique to find $ \int_{0}^{\frac{\pi}{2}} \ln \left(a \cos ^{2} x+b \sin ^{2} x+c\right) d x?$ We are going to find the formula, by Feynman’s Integration Technique, for
$$\int_{0}^{\frac{\pi}{2}} \ln \left(a \cos ^{2} x+b \sin ^{2} x+c\right) d x,$$
where $a+c$ $\textrm{ and }$ $b+c$ are positive real numbers.
First of all, let’s ‘kill’ the term $\sin x$ and the square of $\cos x$ by identity and double angle formula.
$\displaystyle \begin{aligned}\int_{0}^{\frac{\pi}{2}} \ln \left(a \cos ^{2} x+b \sin ^{2} x+c\right) d x =& \int_{0}^{\frac{\pi}{2}} \ln \left[(a-b) \cos ^{2} x+(b+c)\right] d x \\=& \int_{0}^{\frac{\pi}{2}} \ln \left[(a-b)\left(\frac{1+\cos 2 x}{2}\right)+(b+c)\right] d x \\=& \int_{0}^{\frac{\pi}{2}} \ln \left[\frac{a-b}{2} \cos 2 x+\left(b+c+\frac{a-b}{2}\right)\right] d x \\\stackrel{2x\mapsto x}{=} & \frac{1}{2} \int_{0}^{\pi} \ln \left[\frac{a-b}{2} \cos x+\left(\frac{a+b+2 c}{2}\right)\right] d x\end{aligned}\tag*{} $
By my post, I found, by Feynman’s Integration Technique, that
$\displaystyle \int_{0}^{\pi} \ln (b \cos x+c)=\pi \ln \left(\frac{c+\sqrt{c^{2}-b^{2}}}{2}\right),\tag*{} $
where $\left|\frac{c}{b}\right|\geq 1$.
$\displaystyle I=\frac{\pi}{2} \ln \left[\frac{\frac{a+b+2 c}{2}+\sqrt{\left(\frac{a+b+2 c}{2}\right)^{2}-\left(\frac{a-b}{2}\right)^{2}}}{2}\right]\tag*{} $
Simplifying gives the result
$\displaystyle \begin{aligned}I&=\frac{\pi}{2} \ln \left[\frac{a+b+2 c+2 \sqrt{(a+c)(b+c)}]}{4}\right] \\&=\frac{\pi}{2} \ln \left(\frac{\sqrt{a+c}+\sqrt{b+c}}{2}\right)^{2} \\&=\pi \ln \left(\frac{\sqrt{a+c}+\sqrt{b+c}}{2}\right)\end{aligned}\tag*{} $
|
Use the short-hands $p^2=a+c$, $q^2=b+c$ and $r=\frac{p-q}{p+q}$ to rewrite the integral as
\begin{align}
I=&\int_{0}^{\frac{\pi}{2}} \ln \left(a \cos ^{2} x+b \sin ^{2} x+c\right) d x\\
=& \int_{0}^{\frac{\pi}{2}} \ln \left(p^2 \cos ^{2} x+q^2 \sin ^{2} x\right) d x\\
= & \int_{0}^{\frac{\pi}{2}}\bigg(\ln\frac{(p+q)^2}4+ \ln \left(1+2 r\cos 2 x+r^2 \right) \bigg)\ d x\\
\end{align}
and note that
\begin{align}
&\int_0^{\frac\pi2}\ln( 1 + 2r\cos 2x + r^2) dx
= 2\sum_{k=1}^\infty \frac{(-r)^{k+1}}k \int_0^{\frac\pi2}\cos 2kx\ dx =0
\end{align}
Thus
\begin{align}
I=\int_{0}^{\frac{\pi}{2}}\ln\frac{(p+q)^2}4\ d x=\pi\ln \frac{p+q}2\
\end{align}
|
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|
Evaluate $\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx$. $$\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx$$
Edit : $D> 0$.
My work:
Let $x = D\tan \theta$
$$\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx=\int_{-\infty}^\infty \frac{1}{(D^2\sec^2 \theta)^2}dx$$
$$=\int_{-\infty}^\infty \frac{1}{(D^2\sec^2 \theta)^2}D\sec^2\theta d\theta
= \frac{1}{D^3}\int_{-\infty}^\infty \cos^2 \theta d\theta$$
$$=\frac{1}{D^3}[\frac{\theta}{2} + \frac{\sin {2\theta}}{4} + C]_{-\infty}^\infty$$
Put $\theta = \arctan{\frac{x}{D}}$;
$$=\frac{1}{D^3}[\frac{\arctan{\frac{x}{D}}}{2} + \frac{\sin {(2\arctan{\frac{x}{D})}}}{4} + C]_{-\infty}^\infty$$
We get the integral as $\frac{\pi}{2D^3}$. Since $\lim_{x \to +\infty} \arctan(x) = \frac{\pi}{2}$ and $\lim_{x \to -\infty} \arctan(x) = \frac{-\pi}{2}$
I feel something isn't right here. Can anyone point the mistake please. Thank you very much.
|
I’ll suggest another method, involving Integration By Parts. Let the integral to be found be denoted by J. We start by performing an integration by parts of the function $\frac{1}{x^2+D^2}$:
$$I= \int \frac{1}{x^2+D^2} dx= \frac{x}{x^2+D^2}-\int \left(-\frac{x(2x)}{(x^2+D^2)^2}\right)dx$$ $$= \frac{x}{x^2+D^2}+2 \int\frac{x^2+D^2-D^2}{(x^2+D^2)^2}dx$$$$= \frac{x}{x^2+D^2}+2(I-D^2J)$$ so that we have finally, $$J=\frac{1}{2D^2}\left( \frac{x}{x^2+D^2}+I\right)$$ $$J=\frac{1}{2D^2}\left( \frac{x}{x^2+D^2}+\frac 1D \arctan\left(\frac xD\right)\right)$$ which, on putting the limits, gives $\displaystyle \frac{\pi}{2D^3}$.
|
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|
Distribution of amount of green balls drawn Urn of type A contains 4 green and 3 blue balls. Urn of type B contains 4 blue and 5 green balls.
There are three urns of type A and two urns of type B. We pick one urn at random (out of 5) and we draw 3
balls (with a single draw) out of the chosen urn. Let X denote the number of green balls drawn. Determine
the distribution of X.
I tried the following approach
$
P(X=N)=\frac{3}{5}\binom{3}{N}\frac{\binom{4}{N}\binom{3}{3-N}}{\binom{7}{3}}+\frac{2}{5}\binom{3}{N}\frac{\binom{5}{N}\binom{4}{3-N}}{\binom{7}{3}}
$
where $N \in \{ 0,1,2,3 \}$
However sum of probabilities is greater than 1:
$
\sum_{N=0}^{3} \frac{3}{5}\binom{3}{N}\frac{\binom{4}{N}\binom{3}{3-N}}{\binom{7}{3}}+\frac{2}{5}\binom{3}{N}\frac{\binom{5}{N}\binom{4}{3-N}}{\binom{7}{3}} = \frac{283}{105}
$
I have a problem with understanding how to calculate the probability of obtaining N green balls from a single draw.
|
To better understand the question I think it would be better to consider a random variable $Y$ to the problem. Let $Y$ be the random variable that indicates which urn was chosen. Note that
$$
\begin{array}{ll}
\mathbb{P}(Y=A)=3/5
&
\mathbb{P}(Y=B)=2/5
\\
\mathbb{P}(X=n | Y=A)=\frac{\binom{4}{n}\binom{3}{3-n}}{\binom{7}{3}}
&
\mathbb{P}(X=n | Y=B)=\frac{\binom{5}{n}\binom{4}{3-n}}{\binom{9}{3}}
\end{array}
$$
Now just use the following principle to calculate the required probability.
\begin{align}
\mathbb{P}\Big(X=N\Big) & =\mathbb{P}\Big(X=n| Y=A \Big)\mathbb{P}(Y=A) + \mathbb{P}\Big( X=N|Y=B \Big)\mathbb{P}(Y=B)
\end{align}
|
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|
Solving the ODE $(x^2 y-1)y'-(xy^2-1)=0$ I need to solve $(x^2 y-1)y'-(x y^2-1)=0$.
Not sure how to approach this ODE, would love some help regarding solving it or any useful resources.
|
Too long for a comment
Using @projectilemotion's answer, we can easily solve the cubic equation
$$y^3-3x y^2+\frac{3 (2 C+1) x^2}{2 C}y-\frac {(2 C+1) x^3+2 }{2C}=0$$
For $C=1$, the solution is
$$y=x \left(1-\sqrt{2} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\sqrt{2}\frac{
\left(x^3-1\right)}{x^3}\right)\right)\right)$$
For $C=2$,
$$y=x \left(1-\sinh \left(\frac{1}{3} \sinh ^{-1}\left(2-\frac{2}{x^3}\right)\right)\right)$$
So, trying the substitution $y=x \left(1-e^{z}\right)$, the differential equation becomes
$$\left(x^4 \left(e^{z}-1\right)+x\right) z'+1=0$$ Switch variables
$$\frac{\left(e^z-1\right) x^4+x}{x'}+1=0$$ Let $x=\frac 1{\sqrt[3]t}$
$$t'-3 t-3 e^z+3=0\quad \implies t=c_1 \,e^{3 z}+\frac{1}{2} \left(2-3 e^z\right)$$
Now, back to $x$ and $y$ for the implicit solution.
|
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|
Convergence of $\sum_{n=1}^\infty \frac{\sum_{i=1}^n\frac{1}{ \sqrt i}}{n^2}$ How can I prove the following sequence converges?
$$\sum_{n=1}^\infty \frac{\sum_{i=1}^n\frac{1}{ \sqrt i}}{n^2}$$
I tried everything.
Could not find any candidates for comparison test, and failed to find an upper bound.
Any hints will be appreciated.
|
Using generalized harmonic numbers, you face
$$S_p=\sum_{n=1}^p \frac{H_n^{\left(\frac{1}{2}\right)}}{n^2}$$
For large values of $n$
$$H_n^{\left(\frac{1}{2}\right)}=2\sqrt{n}+\zeta \left(\frac{1}{2}\right)+\frac{1}{2 \sqrt{n}}\Bigg[1-\frac{1}{12 n}+\frac{1}{192 n^3}-\frac{1}{512
n^5}+O\left(\frac{1}{n^6}\right) \Bigg]$$ So, the summation does converge.
Using only the above terms
$$\sum_{n=1}^\infty \frac{H_n^{\left(\frac{1}{2}\right)}}{n^2}=\frac{1}{6} \pi ^2 \zeta \left(\frac{1}{2}\right)+2 \zeta
\left(\frac{3}{2}\right)+\frac{\zeta \left(\frac{5}{2}\right)}{2}-\frac{\zeta
\left(\frac{7}{2}\right)}{24}+\frac{\zeta
\left(\frac{11}{2}\right)}{384}-\frac{\zeta \left(\frac{15}{2}\right)}{1024}$$
which is $3.44805$ while numerically, the infinite summation is $3.44844$
|
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|
Proving $(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$ with various solutions.
$(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$
Solutions in the answers.
$\ \\ \ \\ \ \\ \ \\$
Edit) Since this question is closed, I'll add more contexts for this question.
This identity is called "Brahmagupta-Fibonacci identity", which the comment says.
This identity has a special feature, that the form of the expression maintains from LHS to RHS.
Also, for addition, we can expanse this identity to:
$$ (a^2+nb^2)(c^2+nd^2)=(ac\pm nbd)^2+n(ad\mp bc)^2. $$
or:
$$ X=xz-Cyw, Y=axw+a'yz+BYw. \\ (ax^2+Bxy+a'Cy^2)(a'z^2+Bzw+aCw^2)=aa'X^2+BXY+CY^2 $$
, from the answer of @Will Jagy.
This can be proved by various solutions, for example, just multiplying out this identity, or with trigonometric functions, or with the imaginary number "$i$".
I want you to prove this identity with more solutions.
|
There are two possible factorizations.
Here are both of them.
$\begin{array}\\
(a^2+b^2)(c^2+d^2)
&=a^2c^2+a^2d^2+b^2c^2+b^2d^2\\
&=a^2c^2+b^2d^2+a^2d^2+b^2c^2\\
&=a^2c^2+b^2d^2\pm2a^2b^2c^2d^2+a^2d^2+b^2c^2\mp2a^2b^2c^2d^2\\
&=(ac\pm bd)^2+(ad\mp bc)^2
\qquad\text{The two signs are different}\\
\end{array}
$
|
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|
Question about conic sections and determinant Now I have found questions which were on the same topic on this site..but not quite one like this.
Let there be six points $p_1=(x_1,y_1),..,p_6= (x_6,y_6) $
Why is it, that if $$ det\begin{pmatrix}1 &1 &1 & 1& 1 & 1 \\ x_1 & x_2 & x_3 &x_4 & x_5 & x_6 \\
y_1 & y_2 & y_3 &y_4 & y_5 &y_6 \\ x_1^2 & x_2^2 & x_3^2 &x_4^2 & x_5^2 & x_6^2 \\ x_1y_1 & x_2y_2 & x_3y_3 & x_4y_4 & x_5y_5 & x_6y_6 \\ y_1^2 & y_2^2 & y_3^2 &y_4^2 & y_5^2 &y_6^2
\end{pmatrix}=0$$
All of those six points define a conic section?
I know that the matrix is derived from the general conic section formula $ax^2+bxy+cy^2+dx+ey+f=0$
and that a conic section is defined by five points, but here I stumble for an explaination.
|
Let us consider the following homogeneous system with the matrix you have, but transposed :
$$\begin{pmatrix}1 &1 &1 & 1& 1 & 1 \\ x_1 & x_2 & x_3 &x_4 & x_5 & x_6 \\
y_1 & y_2 & y_3 &y_4 & y_5 &y_6 \\ x_1^2 & x_2^2 & x_3^2 &x_4^2 & x_5^2 & x_6^2 \\ x_1y_1 & x_2y_2 & x_3y_3 & x_4y_4 & x_5y_5 & x_6y_6 \\ y_1^2 & y_2^2 & y_3^2 &y_4^2 & y_5^2 &y_6^2
\end{pmatrix}^T\begin{pmatrix}f\\d\\e\\a\\b\\c \end{pmatrix}=\begin{pmatrix}0\\0\\0\\0\\0\\0 \end{pmatrix}$$
It has a nonzero solution, by hypothesis (each line expresses that a certain point belongs to the conic curve).
But a homogeneous system has a nonzero solution iff its determinant is zero.
Remark: Moreover, the equation of this conic is:
$$\det \begin{pmatrix}1 &1 &1 & 1& 1 & \color{red}{1} \\ x_1 & x_2 & x_3 &x_4 & x_5 & \color{red}{x} \\
y_1 & y_2 & y_3 &y_4 & y_5 &\color{red}{y} \\ x_1^2 & x_2^2 & x_3^2 &x_4^2 & x_5^2 & \color{red}{x^2} \\ x_1y_1 & x_2y_2 & x_3y_3 & x_4y_4 & x_5y_5 & \color{red}{xy} \\ y_1^2 & y_2^2 & y_3^2 &y_4^2 & y_5^2 &\color{red}{y^2}
\end{pmatrix}=0$$
|
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|
Limit of sequence given by $x_{n} = \frac{x_{n-2}+x_{n-3}}{2}$? Let $x_1,x_2,x_3\in\mathbb{R}$ be three distinct real numbers. I am interested in the convergence of the sequence
$$
x_{n} = \frac{x_{n-2}+x_{n-3}}{2},\quad n = 4,5,\ldots
$$
ie,
$$
x_{4} = \frac{x_{2}+x_{1}}{2},\quad x_{5} = \frac{x_{3}+x_{2}}{2},\quad x_{6} = \frac{x_{4}+x_{3}}{2},\quad\cdots
$$
Please note that, although related to, this is not a recursive averaging sequence with two initial values, as described and studied elsewhere --- see, for instance, Smith, Scott G. "Recursive Averaging". The Mathematics Teacher, Vol. 108, No. 7 (March 2015), pp. 553-557.
In the present case, convergence is easily verified by noting that
$$
\big|\,x_{n}-x_{n-1}\,\big| = \big|\frac{x_{n-2}+x_{n-3}}{2}-\frac{x_{n-3}+x_{n-4}}{2}\big| = \big|\,\frac{1}{2} (x_{n-2}-x_{n-4})\,\big| = \big|\,\frac{1}{2^{\,n-4}}\,\big|\cdot\big|\, x_{3}-x_{1}\,\big|
$$
Therefore the sequence is Cauchy and hence it converges in $\mathbb{R}$.
However, I was not able to find the limit to where the sequence converges. Computer experiments carried out with several different sets of initial values suggest that this limit should be a linear combination of the those initial values, as occurs in the case mentioned above. Also, the results of these experiments point to the existence of (how many?) sub-sequences, all converging to the very same limit.
I would appreciate some help in the analysis of this problem.
|
For the linear recurrence $$x_{n} = \frac{x_{n-2}+x_{n-3}}{2}$$ the characteristic polynomial is
$$r^3=\frac {r+1}2 \implies r_1=1 \qquad r_2=-\frac {1+i} 2\qquad r_3=-\frac {1-i} 2$$
and, as usual, $$x_n=c_1 + c_2 r_1^n+c_3 r_2^n$$
Using $x_1=a$, $x_2=b$, $x_3=c$, this would give
$$x_n=A_n \,a+B_n\, b+C_n\, c$$ with
$$A_n=\frac{1}{5} \left(1-(3-i) \left(-\frac{1}{2}-\frac{i}{2}\right)^n-(3+i)
\left(-\frac{1}{2}+\frac{i}{2}\right)^n\right)$$
$$B_n=\frac{1}{5} \left(2-(1+3 i) \left(-\frac{1}{2}-\frac{i}{2}\right)^n-(1-3 i)
\left(-\frac{1}{2}+\frac{i}{2}\right)^n\right)$$
$$C_n=\frac{1}{5} \left(2+(2+i) (-1-i)^n 2^{1-n}+(2-i) (-1+i)^n 2^{1-n}\right)$$
Using Euler's formulae
$$A_n=\frac{1}{5}\left(1+2^{1-\frac{n}{2}} \sin \left(\frac{3 \pi n}{4}\right)-3\ 2^{1-\frac{n}{2}} \cos\left(\frac{3 \pi n}{4}\right) \right)$$
$$B_n=\frac{1}{5}\left(2-3\ 2^{1-\frac{n}{2}} \sin \left(\frac{3 \pi n}{4}\right)-2^{1-\frac{n}{2}} \cos \left(\frac{3 \pi n}{4}\right) \right)$$
$$C_n=\frac{1}{5}\left(2+2^{2-\frac{n}{2}} \sin \left(\frac{3 \pi n}{4}\right)+2^{3-\frac{n}{2}} \cos \left(\frac{3 \pi n}{4}\right) \right)$$ and now, make $n \to \infty$ for a very simple result.
|
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|
A cryptogram Diophantine $\overline{ABCDE}=4 \ \overline{AB}^2+\overline{CDE}^2$ ABCDE is a 5-digit number. AB and CDE are 2-digit and 3-digit numbers respectively.
$\overline{ABCDE}=4 \ \overline{AB}^2+\overline{CDE}^2$
Find all possible values of $\overline{ABCDE}$.
Honestly, I designed this problem just for fun, and I tried everything to solve it. It seems that there is a neat solution around. I will be very happy if you take a look at it.
|
We convert the given equation to the following form.
$$1000(AB)+(CDE)=4(AB)^2+(CDE)^2. $$
Now consider the following quadratic equation.
$$(CDE)^2-(CDE)+4(AB)^2-1000(AB)=0$$
The two roots of this equation are,
$$(CDE)=\dfrac{1\pm \sqrt{1-16(AB)^2+4000(AB)}}{2}.$$
Since $(AB)$ is a 2-digit number, $10\le (AB)\le 99$. For all these values of (AB),
$$\sqrt{1-16(AB)^2+4000(AB)} \gt 1$$.
Therefore, we have,
$$(CDE)=\dfrac{1+ \sqrt{1-16(AB)^2+4000(AB)}}{2}.$$
Now, we have to find out the values of $(AB)$, for which the expression under the square-root sign is a square. They turn out to be,
$$(AB)=19\qquad\text{and}\qquad (AB)=70.$$*
The corresponding values of $(CDE)$ are,
$$(CDE)=133\qquad\text{and}\qquad (CDE)=225.$$
*Here is how to get these values of $\overline{AB}$:
First complete the square in the discriminant. This gives
$1-16(\overline{AB})^2+4000\overline{AB}=250001-(4\overline{AB}-500)^2.$
Thus we must have
$250001=m^2+(4\overline{AB}-500)^2$.
Since $250001=500^2+1^2$ where the squares are relatively prime, the odd number must be a product of $4n+1$ primes only. Trial divisions then yield the factorization
$250001=53^2×89=(7^2+2^2)^2(8^2+5^2).$
We now consider Gaussian-integer products of the form
$(7\pm2i)(7\pm2i)(8\pm5i).$
Each product has real and imaginary parts whose squares sum to $250001$. Wlog we may take the first factor to be specifically $7+2i$, and we find three combinations giving different squares:
$(7+2i)(7+2i)(8+5i)=220+449i$
$(7+2i)(7+2i)(8-5i)=500-i$
$(7+2i)(7-2i)(8\pm5i)=424\pm265i.$
The first product above gives
$250001=220^2+449^2,$
from which $4\overline{AB}-500$ which is a multiple of $4$ must be $\pm220$. We choose the negative possibility to force $\overline{AB}<100$, so we then find
$\color{blue}{\overline{AB}=70.}$
The third product similarly gives the other nonzero value for $\overline{AB}$:
$\color{blue}{\overline{AB}=19.}$
|
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|
Solving the equation $x^2+(\frac{x}{x+1})^2=\frac54$
Solve the equation$$x^2+\left(\frac{x}{x+1}\right)^2=\frac54$$
I noticed that for $0< x$, both $x^2$ and $\left(\dfrac x{x+1}\right)^2$ are increasing functions so their sum is also increasing and it has only one root which is $x=1$ (by inspection). But I'm not sure how to find the root for negative values of $x$.
By multiplying both sides by $(x+1)^2$ we get,
$$x^2(x^2+2x+1)+x^2-\frac54(x^2+2x+1)=0$$
$$4x^4+8x^3+3x^2-10x-5=0 \implies (x-1)(4x^3+12x^2+15x+5)=0$$
But I can't factor the third degree polynomial.
|
Simplify as
$$4x^2[(x+1)^2+1]=5(x+1)^2 \implies 4x^4+8x^3+3x^2-10x-5=0$$
Denoting LHS as $f(x)$, note that $f(1)=0 ~\& ~f(-1/2)=0.$ This means that by remainder theorem $$f(x) ~\text{is divisible by}~ (2x+1)(x-1)=2x^2-x-1=g(x)$$
dividing $f(x)$ by $g(x)$ we see that $$f(x)=(2x^2-x-1)(2x^2+5x+5)$$
By solving $2x^2+5x+5=0$ we get $$x=\frac{-5\pm \sqrt{-15}}{4}=\frac{-5\pm i\sqrt{15}}{4}.$$
Finally, the equation has 4 roots $$x=-1/2,1,\frac{-5\pm i\sqrt{15}}{4}.$$
|
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|
Correct version of an inequality Does any one knows the correct version of the inequality:
Let $a_1,\cdots,a_n\in \mathbb R$ and $b_1,\cdots,b_n\in \mathbb R$ with $\sum_{j=1}^n b_j=0$. Then
$$
\left(\sum_{j=1}^n a_jb_j \right)^2\leq \left(\sum_{j=1}^n a_j^2-\left(\sum_{j=1}^n a_j\right)^2\right)\sum_{j=1}^n b_j^2\,,
$$
proposed in https://link.springer.com/content/pdf/10.1007%2F978-1-4939-1926-0_2.pdf Exercise 2.51, page 23?
Indeed, the statement as such is incorrect. e.g. taking $n=2$ and $a_1=a_2=1$, the statement yields
$$
0\leq 2-2^2.
$$
|
I believe the issue is that rather than sums, those should be means. That is,
$$
\left[\frac{1}{n}\sum_{j=1}^na_j^2 - \left(\frac{1}{n}\sum_{j=1}^na_j\right)^2\right]\left(\frac{1}{n}\sum_{j=1}^n b_j^2\right)\ge\left(\frac{1}{n}\sum_{j=1}^na_jb_j\right)^2
$$
This can also be written more compactly in vector notation:
$$
\left[\frac{|\mathbf{a}|^2}{n}-\left(\frac{\mathbf{a}\cdot\mathbf{1}}{n}\right)^2\right]\frac{|\mathbf{b}|^2}{n}\ge \left(\frac{\mathbf{a}\cdot\mathbf{b}}{n}\right)^2,
$$
where $\mathbf{1}$ is the vector of all $1$'s We now use the well-known statistics identity $\langle (x - \langle x\rangle)^2\rangle = \langle x^2\rangle - \langle x\rangle ^2$ to deduce that $|\mathbf{a}|^2/n - (\mathbf{a}\cdot \mathbf{1}/n)^2 = |\mathbf{a}-(\mathbf{a}\cdot\mathbf{1}/n)\mathbf{1}|^2/n$. Then we apply Cauchy-Schwartz to the LHS:
\begin{multline}
\left[\frac{|\mathbf{a}|^2}{n}-\left(\frac{\mathbf{a}\cdot\mathbf{1}}{n}\right)^2\right]\frac{|\mathbf{b}|^2}{n} = \frac{1}{n^2}\left|\mathbf{a}-\left(\frac{\mathbf{a}\cdot\mathbf{1}}{n}\right)\mathbf{1}\right|^2|\mathbf{b}|^2\\\ge\frac{1}{n^2}\left[\mathbf{a}\cdot\mathbf{b}-\frac{(\mathbf{a}\cdot\mathbf{1})(\mathbf{1}\cdot\mathbf{b})}{n}\right]^2 = \left(\frac{\mathbf{a}\cdot\mathbf{b}}{n}\right)^2,
\end{multline}
where the last equality comes from $\sum_{j=1}^n b_j = 0$, i.e., $\mathbf{1}\cdot\mathbf{b} = 0$.
|
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|
$x^{10}+x^{11}+\dots+x^{20}$ divided by $x^3+x$. Remainder? Question:
If $x^{10}+x^{11}+\dots+x^{20}$ is divided by $x^3+x$, then what is the remainder?
Options: (A) $x\qquad\quad$ (B)$-x\qquad\quad$ (C)$x^2\qquad\quad$ (D)$-x^2$
In these types of questions generally I follow the following approach:
Since divisor is cubic so the remainder must be a constant/linear/quadratic expression.
$\Rightarrow F(x)=(x^3+x)Q(x)+ax^2+bx+c$
For $x=0$, we get $c=0$
But since $x^3+x$ has no other roots so I can't find $a$ and $b$.
Please help.
Answer:
Option (B)
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$\overbrace{\!\!\bmod \color{c00}{x^2\!+\!1}}^{\color{#c00}{\large x^2\ \equiv\ -1}\!\!}\!:\ f\color{#0af}{\!+\!x^9} =\! (x^9\!+\!x^{13}\!+\!x^{17})\!\!\overbrace{(1\!+\!x\!+\!x^2\!+\!x^3)}^{\large ((\color{#c00}{x^2})^2-1)/(x-1)=\color{#0a0}0}\!\!\equiv\color{#0a0}0\,$ so $\,f \equiv \color{#0af}{-x^9} = -x(\color{#c00}{x^2})^4\! \equiv -x$
|
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|
Find $q,r$ if $q^2+r=2000$ Let $a,b \in \mathbb{N}$ such that when $a^2+b^2$ is divided by $a+b$, the quotient is $q$ and remainder is $r$ such that $q^2+r=2000$. Find $q,r$
My try:
Obviously $a \ne b$ for if we have $r=0$ $\implies$ $q=\sqrt{2000} \notin \mathbb{N}$
Without loss of generality, let $a>b$ and $a-b=c$
Then we have
$$a^2+b^2=(a-b)(a+b)+2b^2$$
So $q=a-b, r=2b^2$
Using $q^2+r=2000$
$$c^2+2b^2=2000$$
But i guess we cannot find two natural numbers $c,b$ satisfying above equation.
Any help?
|
From
$$a^2+b^2=q(a+b)+r\ ,\qquad q^2+r=2000$$
we get
$$q^2-(a+b)q+(a^2+b^2-2000)=0\ .$$
For this to have real solutions we need
$$(a+b)^2\ge 4(a^2+b^2-2000)$$
which simplifies to
$$8000\ge 3a^2-2ab+3b^2=2a^2+2b^2+(a-b)^2\ .$$
Therefore $8000>2a^2,\,2b^2$ and so
$$a,b\le63\ .$$
However the largest square below $2000$ is $44^2$. If $q\ne44$ then
$$r=2000-q^2\ge2000-43^2=151\ .$$
This is impossible if we require $r<a+b$. So the only possibility is $q=44$, $r=64$ and
$$a^2+b^2=44(a+b)+64\ .$$
Completing the square,
$$(a-22)^2+(b-22)^2=1032\ .$$
Now $3$ is a factor of $1032$, and $3\equiv3\pmod4$, and $3^2\not\mid1032$, so $1032$ is not a sum of two squares and there is no solution.
|
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|
Evaluating $ \frac{24}{\pi}\int_0^\sqrt 2\frac{2-x^2}{(2+x^2)\sqrt{4+x^4}}\,\mathrm dx$ I recently came across a problem on definite integration and couldn't solve it despite my efforts. It goes as
$$
\frac{24}{\pi}\int_0^\sqrt 2\frac{2-x^2}{(2+x^2)\sqrt{4+x^4}}\,\mathrm dx
$$
The $2-x^2$ term and the upper limit along with the $2+x^2$ term at the bottom motivated me to substitute $x=2^{1/2}\tan a$. However subsequent steps proceeded to a stage I couldnt simplify.
I tried other substitutions but was unable to proceed
Any help to solve this problem would be appreciated. Please feel free to share your first thoughts and intuition as well.
|
Integrate as follows
\begin{align}&\ \frac{24}{\pi}\int_0^\sqrt 2\frac{2-x^2}{(2+x^2)\sqrt{4+x^4}}\,dx\\
=&\ \frac{24}{\pi}\int_0^\sqrt2\frac{\frac{2-x^2}{(2+x^2)^2}}{\sqrt{1-\frac{4x^2}{(2+x^2)^2}}}\,
\overset{y=\frac{2x}{2+x^2}}{dx}
=\frac{12}{\pi}\int_0^{\frac1{\sqrt2}}\frac{dy}{\sqrt{1-y^2}}
= \frac{12}{\pi}\cdot\frac\pi{4}=3\\
\end{align}
|
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|
Determine the power series representation of the function $f(x)=\sqrt{(4+x)^3}$ and indicate the radius of convergence I want to find a representation of the function mentioned above, so I took into account that:
$$f(x)=\sqrt{(4+x)^3}=8\left(1+\frac{x}{4}\right)^{\frac{3}{2}}$$ and developing the binomial series for $\left(1+\frac{x}{4}\right)^{\frac{3}{2}}$ we have to
$$\left(1+\frac{x}{4}\right)^{\frac{3}{2}}=1+\frac{3}{2}\left(\frac{x}{4}\right)+\frac{\frac{3}{4}}{2!}\left(\frac{x}{4}\right)^2+\frac{\frac{-3}{8}}{3!}\left(\frac{x}{4}\right)^3+\cdot\cdot\cdot$$
Now, to obtain the terms of the original series simply multiply by 8; now, how can I express the function as a series, since I have only managed to express that as a sum of terms
Any help is appreciated
|
According to the binomial series expansion we obtain
\begin{align*}
\left(1+\frac{x}{4}\right)^{\frac{3}{2}}&=1+\frac{3}{2}\left(\frac{x}{4}\right)+\frac{\frac{3}{4}}{2!}\left(\frac{x}{4}\right)^2+\frac{\frac{-3}{8}}{3!}\left(\frac{x}{4}\right)^3+\cdots\\
&=1+\binom{3/2}{1}\frac{x}{4}+\binom{3/2}{2}\left(\frac{x}{4}\right)^2+\binom{3/2}{3}\left(\frac{x}{4}\right)^3+\cdots\\
&\,\,\color{blue}{=\sum_{j=0}^\infty\binom{3/2}{j}\left(\frac{x}{4}\right)^j}
\end{align*}
|
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|
What's the shortest path to simplify $\tan{85°} \tan{35°} \tan{15°}+\tan{55°} \tan{5°} \tan{75°}+2\cot{50°}$? We have this expression:
$$\tan{85°} \cdot \tan{35°} \cdot \tan{15°}+\tan{55°} \cdot \tan{5°} \cdot \tan{75°}+2\cot{50°}$$
I want this expression to be in its simplest form.
I fed this into the Symbolab calculator but it gave a very long answer. According to my own information, I wrote it like this:
$$\cot{5°} \cdot \cot{55°} \cdot \cot{75°}+\tan{55°} \cdot \tan{5°} \cdot \tan{75°} + 2\cot{50°}=\\ \\ \frac{\cos{5°} \cdot \cos{55°} \cdot \cos{75°}}{\sin{5°} \cdot \sin{55°} \cdot \sin{75°}}+\frac{\sin{5°} \cdot \sin{55°} \cdot \sin{75°}}{\cos{5°} \cdot \cos{55°} \cdot \cos{75°}}+2\frac{\cos{50°}}{\sin{50°}}$$
And now, if: $$a=\cos{5°} \cdot \cos{55°} \cdot \cos{75°}\\b=\sin{5°} \cdot \sin{55°} \cdot \sin{75°}\\c=\cos{50°}\\d=\sin{50°}$$
It will be very long so i show my final way with the variables $a,b,c,d$:
$$\frac{a}{b}+\frac{b}{a}+2\frac{c}{d}=\frac{a^{2}d+b^{2}d+2abd}{abd}$$
but as you see, it is also very long. I am looking for the simplest form for this expression.
|
Consider the identity $\tan(x)\cdot\tan(60°-x)\cdot\tan(60°+x)=\tan(3x)$. We now have
$$\tan5°\cdot\tan55°\cdot\tan65°=\tan15°\implies\tan5°\cdot\tan55°\cdot\tan75°=\cot65°$$
and
$$\tan25°\cdot\tan35°\cdot\tan85°=\tan75°\implies\tan15°\cdot\tan35°\cdot\tan85°=\cot25°$$
The expression now is simplified into $$\tan25°+\frac1{\tan25°}+\frac2{\tan50°}=\frac{\tan^225°+1}{\tan25°}+\frac{2}{\tan50°}$$
Using the fact that $\sin(2x)=\frac{2\tan(x)}{1+\tan^2(x)}$, we simplify as
$$\frac{\tan^225°+1}{\tan25°}+\frac{2}{\tan50°}=\frac2{\sin50°}+\frac{2\cos50°}{\sin50°}$$
From the double angle formulae,
$$\frac{2(1+\cos50°)}{\sin50°}=2\cdot\frac{2\cos^225°}{2\sin25°\cos25°}=2\tan65°$$
Therefore,$$\boxed{\tan{85°} \cdot \tan{35°} \cdot \tan{15°}+\tan{55°} \cdot \tan{5°} \cdot \tan{75°}+2\cot{50°}=2\tan65°}$$
|
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|
Sum of all real roots of $\frac{{3{x^2} - 9x + 17}}{{{x^2} + 3x + 10}} = \frac{{5{x^2} - 7x + 19}}{{3{x^2} + 5x + 12}}$?
The sum of real roots of $\dfrac{{3{x^2} - 9x + 17}}{{{x^2} + 3x + 10}} = \dfrac{{5{x^2} - 7x + 19}}{{3{x^2} + 5x + 12}}$ is____
How do I proceed with this type of problem.
Cross multiplying and segregating the coefficient is a cumbersome process
My putting in desmos.com my answer is $«6»$.
|
Suggestion (Hint):
Use the well-known rule:
$$\frac ab=\frac cd \iff \frac {a-b}{b}=\frac {c-d}{d}$$
Then observe that: $$a-b=c-d$$
|
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|
What periodic function does $\frac{\Gamma(x+1)}{2^x \Gamma(\frac{x}{2}+1)^2}$ tend to? If one graphs $$ y= \frac{\Gamma(x+1)}{2^x \Gamma(\frac{x}{2}+1)^2} = \frac{x!}{2^x \left( \left( \frac{x}{2}\right)!\right)^2} $$
One immediately notices that as $x \rightarrow -\infty$ the function tends to some periodic function that looks an awful lot like $\tan(x)$. Is there a way to make this idea precise and state what that function is? I was thinking something like, there exists a periodic function $f(x)$ such that $f(x) - y(x) \rightarrow 0$ as $x \rightarrow -\infty$.
See this image:
Which was produced using desmos.com/calculator
|
For negative values of $x$, $$y= \frac{\Gamma(x+1)}{2^x\, \Gamma(\frac{x}{2}+1)^2} $$ is closer and closer to
$$\sin ^2\left(\frac{\pi x}{2}\right) \csc (\pi (x+1))=-\frac{1}{2} \tan \left(\frac{\pi x}{2}\right)$$ This comes from Euler's reflection formula of the gamma function.
Edit
As @GregMartin commented, I forgot the next term of the expansion. So, give all credit to him.
Using more terms, for negative values of $x$
$$y=- \sqrt{-\frac{2}{\pi x}} \tan \left(\frac{\pi x}{2}\right)\Bigg[ 1-\frac{1}{4 x}+\frac{1}{32 x^2}+O\left(\frac{1}{x^3}\right)\Bigg]$$
It could even be better if the series expansion is made a simple Padé approximant such as
$$\frac {64 x^2-8 x+11 }{64 x^2+8 x+11 }+O\left(\frac{1}{x^5}\right)$$
In order to check how good (or bad) is the approximation, I computed
$$I_n=\int_{-2n-0.99}^{-2n+0.99} \Bigg[\frac{\Gamma(x+1)}{2^x\, \Gamma(\frac{x}{2}+1)^2}+ \sqrt{-\frac{2}{\pi x}}\, \tan \left(\frac{\pi x}{2}\right)\,\,\frac {64 x^2-8 x+11 }{64 x^2+8 x+11 }\Bigg]^2\, dx$$
$$\left(
\begin{array}{cc}
n & I_n \\
10 & 5.371 \times 10^{-16} \\
9 & 1.766\times 10^{-15} \\
8 & 6.733\times 10^{-15} \\
7 & 3.105\times 10^{-14} \\
6 & 1.846\times 10^{-13} \\
5 & 1.567\times 10^{-12} \\
4 & 2.271\times 10^{-11} \\
3 & 7.961\times 10^{-10} \\
2 & 1.562\times 10^{-07}
\end{array}
\right)$$
What could be interesting is to see if we can approximate the roots of
$$\frac{\Gamma(x+1)}{2^x\, \Gamma(\frac{x}{2}+1)^2}=k $$
|
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|
How many methods are there to find $\int_{0}^{\pi} \frac{d \theta}{a+b \cos \theta}, \textrm{ where }a>b>0$ When I deal with this simple integral, I found there are several methods. Now I share one of them.
Letting $x\mapsto \pi-x $ yields $$I=\int_{0}^{\pi} \frac{d \theta}{a+b \cos \theta}=\int_{0}^{\pi} \frac{d \theta}{a-b \cos \theta}$$
Adding two versions together yields
$$
\begin{aligned}
2 I &=2 a \int_{0}^{\pi} \frac{d \theta}{a^{2}-b^{2} \cos ^{2} \theta} \\
&=4 a\int_0^{\frac{\pi}{2}} \frac{\sec ^{2} \theta}{a^{2} \sec ^{2} \theta-b^{2}} d \theta \quad \textrm{( By symmetry )}\\
&=4 a{\int_{0}^{\infty}} \frac{d t}{\left(a^{2}-b^{2}\right)+a^{2} t^{2}} \\
&=\frac{4}{\sqrt{a^{2}-b^{2}}}\left[\tan^{-1} \left(\frac{at}{\sqrt{a^{2}-b^{2}}}\right)\right]_{0}^{\infty} \\
&=\frac{2 \pi}{\sqrt{a^{2}-b^{2}}}
\end{aligned}
$$
We can now conclude that $$
\boxed{I=\frac{\pi}{\sqrt{a^{2}-b^{2}}}}
$$
Are there any other methods to deal with the integral?
|
Here's an interesting geometric approach I read in one of Jack D' Aurizio's answers (and have never forgotten). It's not as efficient as using a Weierstrass substitution but I believe the geometric interpretation itself merits some sort of recognition.
For a closed and continuous curve $r(\theta)$ where $0\leq\theta\leq2\pi$, then the area of the region bounded by $r$ can be expressed as
$$A=\frac 12\int\limits_0^{2\pi}r^2(\theta)\,\mathrm d\theta$$
In this problem's case, we can transform the integral into the form above by observing that
$$I=\int\limits_0^{\pi}\frac {\mathrm d\theta}{\left(a+b\cos\theta\right)^2}=\frac 12\int\limits_0^{2\pi}\frac {\mathrm d\theta}{(a+b\cos\theta)^2}$$
Therefore, when
$$r=\frac 1{a+b\cos\theta}$$
the integral $I$ describes the area of an ellipse, which is equal to $\pi$ times the product of the semi-axis. Converting to rectangular coordinates, then we get
$$\frac {\left(a^2-b^2\right)^2}{a^2}\left(x+\frac b{a^2-b^2}\right)^2+\left(a^2-b^2\right)y^2=1$$
Thus, the area is then
$$\int\limits_0^{\pi}\frac {\mathrm d\theta}{(a+b\cos\theta)^2}=\pi\cdot\frac a{a^2-b^2}\cdot\frac 1{\sqrt{a^2-b^2}}=\frac {\pi a}{\left(a^2-b^2\right)^{3/2}}$$
Now, integrate both sides with respect to $a$ to see that
$$\int\limits_0^{\pi}\frac {\mathrm d\theta}{a+b\cos\theta}\color{blue}{=\frac {\pi}{\sqrt{a^2-b^2}}}$$
|
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Interesting integral $\int_{0}^{\frac{\pi}{4}} \frac{x^{2}}{(x \sin x+\cos x)^{2}} d x$? Noting that
$\displaystyle d(x \sin x+\cos x)=x \cos xdx,\tag*{} $
we have
$\displaystyle \int_{0}^{\frac{\pi}{4}} \frac{x^{2}}{(x \sin x+\cos x)^{2}} d x=-\int_{0}^{\frac{\pi}{4}} \frac{x}{\cos x} d\left(\frac{1}{x \sin x+\cos x}\right)\tag*{} $
Integration by parts gives
$\displaystyle \begin{aligned}I &=-\left[\frac{x}{\cos x(x \sin x+\cos x)}\right]_{0}^{\frac{\pi}{4}}+\int_{0}^{\frac{\pi}{4}} \frac{\cos x+x \sin x}{\cos ^{2} x} \cdot \frac{1}{x \sin x+\cos x} d x \\&=-\frac{\frac{\pi}{4}}{\frac{1}{\sqrt{2}}\left(\frac{\pi}{4} \cdot \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)}+\int_{0}^{\frac{\pi}{4}} \sec ^{2} x d x \\&=-\frac{2 \pi}{\pi+4}+[\tan x]_{0}^{\frac{\pi}{4}} \\&=\frac{4-\pi}{4+\pi}\end{aligned}\tag*{} $
Is there any other method?
|
\begin{align}
&\int_{0}^{\frac{\pi}{4}} \frac{x^{2}}{(x \sin x+\cos x)^{2}} d x\\
=& \int_{0}^{\frac{\pi}{4}} \frac{x^{2}\sec^2x}{(1+x \tan x)^{2}} d x
= \int_{0}^{\frac{\pi}{4}}
\frac{1}{(1+x \tan x)^{2}}+ \frac{x^{2}\sec^2x -1}{(1+x \tan x)^{2}} \ d x\\
=& \ \left( \frac1{x+\cot x} - \frac{x}{1+x\tan x}\right)_0^{\frac\pi4}= \frac{4-\pi}{4+\pi}
\end{align}
|
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|
Find the limit of $\frac{1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}}{1+\frac15+\frac{1}{5^2}+...+\frac{1}{5^n}}$ Find the limit $$\lim_{n\to\infty}\dfrac{1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}}{1+\frac15+\frac{1}{5^2}+...+\frac{1}{5^n}}$$
For the numerator we have $1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}=\dfrac{1-\left(\frac{1}{3}\right)^{n+1}}{1-\frac13}=\dfrac32-\dfrac12\dfrac{1}{3^n}.$
By analogy, $1+\frac15+\frac{1}{5^2}+...+\frac{1}{5^n}=\dfrac54-\dfrac14.\dfrac{1}{5^n}$
So we have $$\lim_{n\to\infty}\dfrac{1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}}{1+\frac15+\frac{1}{5^2}+...+\frac{1}{5^n}}=\lim_{n\to\infty}{\dfrac{\frac32-\frac12.\frac{1}{3^n}}{\frac54-\frac14.\frac{1}{5^n}}}=\lim_{n\to\infty}{\left(\dfrac{3^{n+1}-1}{2.3^n}\cdot\dfrac{4.5^n}{5^{n+1}-1}\right)}$$ What am I supposed to do next?
My initial mistake was that I thought $1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}=\dfrac{1-\left(\frac13\right)^n}{1-\frac13}$. How could we actually see (and show) that the terms in the sum(s) are not $n$, but $n+1$?
|
You were very close:
$$\lim_{n\to\infty}\dfrac{1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}}{1+\frac15+\frac{1}{5^2}+...+\frac{1}{5^n}}=\lim_{n\to\infty}{\dfrac{\frac32-\frac12.\frac{1}{3^n}}{\frac54-\frac14.\frac{1}{5^n}}}$$
as you correctly derived.
If you simply use $\lim_{n\to\infty}\frac{1}{3^n}=\lim_{n\to\infty}\frac{1}{5^n}=0$ in that expression, you are left with
$$\lim_{n\to\infty}{\dfrac{\frac32-\frac12.\frac{1}{3^n}}{\frac54-\frac14.\frac{1}{5^n}}} = \frac{\frac32-0}{\frac54-0} = \frac65$$
|
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|
Solve the differential equation: $6 \cos^2(x) \dfrac{dy}{dx} -y \sin(x)+2y^4 \sin^3(x)=0$ I have the following differential equation before me:
$6 \cos^2(x) \dfrac{dy}{dx} -y \sin(x)+2y^4 \sin^3(x)=0$
I tried solving it by reducing to Bernoulli form of first order differential equation.
I divided both sides of the equation by $6 \cos^2(x)y^4$ to get:
$\dfrac{1}{y^4} \dfrac{dy}{dx}-\dfrac{1}{y^3} \dfrac{\sin(x)}{6\cos^2(x)}=\dfrac{-\sin^3(x)}{3\cos^2(x)}$
Then I took $\dfrac{1}{y^3}=z$
so that $\dfrac{-3}{y^4}\dfrac{dy}{dx}=\dfrac{dz
}{dx}$
This gave me:
$\dfrac{dz}{dx}+\dfrac{\sin(x)}{2\cos^2(x)}z=\dfrac{\sin^3(x)}{\cos^2(x)}$
This is first order linear differential equation of the type
$ \dfrac{dz}{dx}+P(x)z=Q(x)$
Integrating Factor (IF) is given by:
$IF=e^{\int P(x) \,dx}= e^{\int \dfrac{\sin(x)}{2\cos^2(x)}\,dx}= e^{\dfrac{1}{2\cos(x)}}$
Next, solution is given by:
$z.IF= \int IF.Q(x)$
or
$z.e^{\dfrac{1}{2\cos(x)}}=\int e^{\dfrac{1}{2\cos(x)}}.\dfrac{\sin^3(x)}{\cos^2(x)} \,dx$
This is the step where I falter as I find myself unable to tackle the integral on the RHS of the above equation.
You are requested to help me with the evaluation of this integral or come up with another way of tackling this differential equation. Any help will be highly appreciated.
|
As @abcdefu already did, after $y=\frac{1}{\sqrt[3]{u}}$, we have
$$2 \cos ^2(x)\, u'+ \sin (x)\, u=2 \sin ^3(x)$$
The solution of the homegeneous is simple
$$u=C\,e^{-\frac{\sec (x)}{2}}$$ Variation of parameters leads to
$$\cos ^2(x) e^{-\frac{\sec (x)}{2}} C'=\sin ^3(x) \implies C'=\sin (x) \tan ^2(x) \,e^{\frac{\sec (x)}{2}}$$
Now, using $\sec(x)=2t$
$$C'=2 e^t \left(1-\frac{1}{4 t^2}\right) \implies C=\frac{1}{2} \left(e^t \left(\frac{1}{t}+4\right)-\text{Ei}(t)\right)+\text{Constant}$$
|
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|
If $x^2+2(\alpha-1)x-\alpha+7=0$ has distinct negative solutions... Let $\displaystyle{ \alpha }$ be real such that the equation $\displaystyle{ x^2+2(\alpha-1)x-\alpha+7=0 }$ has two different real negative solutions. Then
*
*$ \ \displaystyle{ \alpha<-2 }$ ;
*$ \ \displaystyle{ 3<\alpha<7 }$ ;
*it is impossible ;
*none of (a)-(c).
$$$$
I have done the following :
The value $\alpha$ is real and such that the equation $x^2+2(\alpha-1)x-\alpha+7=0$ has two different real negative solutions.
The solutions of the equation are given from the quadratic formula \begin{align*}x_{1,2}&=\frac{-2(\alpha-1)\pm \sqrt{[2(\alpha-1)]^2-4\cdot 1\cdot (-\alpha+7)}}{2}\\ & =\frac{-2(\alpha-1)\pm \sqrt{4(\alpha^2-2\alpha-1)-4\cdot (-\alpha+7)}}{2}\\ & =-(\alpha-1)\pm \sqrt{2(\alpha^2-2\alpha-1)-2\cdot (-\alpha+7)} \\ & =-(\alpha-1)\pm \sqrt{2\alpha^2-4\alpha-2+2\alpha-14} \\ & =-(\alpha-1)\pm \sqrt{2\alpha^2-2\alpha-16}\end{align*}
So that we have two different solutions the discriminant must be non-zero.
So that we have two negative solutions, it must hold $-(\alpha-1)\pm \sqrt{2\alpha^2-2\alpha-16}<0$.
So that we have real solutions the expression under the square root must be non negative.
The expression under the square root has the sign of the coefficient of $x^2$, i.e. positive, outside the roots.
We have that \begin{equation*}2\alpha^2-2\alpha-16=0 \Rightarrow \alpha_{1,2}=\frac{1}{2}\pm \frac{\sqrt{33}}{2}\end{equation*}
So we have that the expression under the square root if $\alpha<\frac{1}{2}- \frac{\sqrt{33}}{2}$ and if $\alpha>\frac{1}{2}+ \frac{\sqrt{33}}{2}$.
Is my attempt correct so far? Now do we check if the first two intervals of $\alpha$ can hold for all these conditions? Or how do we continue? Or is there a better way to solve that exercise ?
|
The quadratic polynomial $ \ x^2 + 2(\alpha-1)x - \alpha+7 \ \ $ in "vertex form" is $$ \ ( \ x \ + \ [\alpha - 1] \ )^2 \ + \ (7 - \alpha) - (\alpha - 1)^2 \ \ = \ \ ( \ x \ + \ [\alpha - 1] \ )^2 \ - \ ( \ \alpha^2 \ - \ \alpha \ - \ 6 \ ) $$
$$ = \ \ ( \ x \ + \ [\alpha - 1] \ )^2 \ - \ ( \ \alpha \ + \ 2 \ )·( \ \alpha \ - \ 3 \ ) \ \ . $$
So the polynomial has distinct real zeroes only for $ \ \alpha \ < \ -2 \ $ or $ \ \alpha \ > \ 3 \ \ . \ $ Since the vertex must have a negative $ \ x-$coordinate if the two real zeroes $ \ ( \ x-$intercepts) are to be negative, we are only concerned with the case for $ \ \alpha \ > \ 3 \ \ . $ [For $ \ \alpha \ = \ 3 \ \ , \ $ the polynomial becomes $ \ x^2 + 4x + 4 \ = \ ( x + 2 )^2 \ \ . \ ] $
As $ \ \alpha \ $ increases, the two zeroes "spread" farther from the (increasingly negative) vertex, with the larger zero becoming less and less negative. At $ \ \alpha \ = \ 7 \ \ , \ $ the "constant term" becomes zero (the polynomial is $ \ x^2 + 12x \ ) \ \ , \ $ with one of the zeroes now being $ \ x \ = \ 0 \ \ . \ $ Thus the permissible values are $ \ 3 \ < \ \alpha \ < \ 7 \ \ . $
|
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|
Prove $\cot\frac{\theta}{2} - 2\cot\theta = \tan\frac{\theta}{2}$ This is a math exercise I am doing for my year 10 math class. Basically I would like to show how I would prove that $\cot\frac{\theta}{2} - 2\cot\theta = \tan\frac{\theta}{2}$
I was able to expand the LHS and was able to use fractions to get that $$\cot\frac{\theta}{2} - 2\cot\theta = \tan\frac{\theta}{2} = \frac{\cos\frac{\theta}{2} - 2\cos\theta}{\sin\theta}$$
I do not know what I would do from here.
Any help will be appreciated. Thanks :)
|
$\begin{align}\frac{1}{2}(\cot\frac{\theta}{2} - \tan\frac{\theta}{2})&=\frac{1}{2}\frac{1-\tan^2\frac{\theta}{2}}{\tan\frac{\theta}{2}}\\& =\frac{1}{\frac{2\tan\frac{\theta}{2}}{1-\tan^2\frac{\theta}{2}}}\\&=\frac{1}{\tan\theta} \\&=\cot\theta\end{align}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $\cos^2\left(x\right)\left(2+\cos 2x\right)-2 \cos^2\left(x+\frac{\sin(2x)}{4}\right)\ge0$ Show that for all $x\in\mathbb{R},$
\begin{equation}
\cos^2\left(x\right)\left(2+\cos 2x\right)-2 \cos^2\left(x+\frac{\sin 2x}{4}\right)\ge0
\end{equation}
I have checked this numerically and this is correct but I cannot show analytically. I have used the following useful formulas but had no luck:
\begin{gather}
\sin(2x)=2\sin (x)\cos (x),\newline
\cos (2x)=2\cos^2(x)-1,\newline
\sin^2(x)=1-\cos^2(x).
\end{gather}
I guess there is some bounding trick that I am missing. Any help is appreciated
|
If you expand the first term on the LHS, and add the second term on the LHS to both sides, you'll get
$2\cos^2(x) + \cos^2(x)\cos(2x) \geq 2\cos^2(x+\frac{\sin(2x)}{4})$
Now, using the identity $\cos(2x)+1 = 2\cos^2(x)$, you get:
$\cos(2x)+1 + \frac{\cos(2x)+1}{2}\cos(2x) \geq \cos(2x+\frac{\sin(2x)}{2})+1$
Subtract $1$ from both sides, and write the LHS by taking the expression into the paranthesis of $\cos(2x)$, you get:
$\cos(2x)(\frac{\cos(2x)+3}{2}) \geq \cos(2x + \frac{\sin(2x)}{2})$
Substitute $y=2x$ at this point, and it is easy to see that for each $y \in \mathbb{R}$ that satisfies the equation, $y + 2\pi k$ and $-y$ also satisfies the equation for $k \in \mathbb{Z}$. Thus, it is enough to consider the inequality for $y \in [0,\pi]$.
$\frac{\cos(y)+3}{2} \geq 1$ for all $y$; thus, if $\cos(y) \geq \cos(y + \frac{\sin(y)}{2})$ for all $y$, the inequality holds.
It can be seen that the equality $\cos(y) = \cos(y + \frac{\sin(y)}{2})$ holds when either $\frac{\sin(y)}{2}=0$ or $2\pi- y = y + \frac{\sin(y)}{2}$
The first case holds when $y\in \{0,\pi\}$ and by simple manipulation and checking the derivatives, the only solution for the second case is $y = \pi$ (of course for $y \in [0,\pi]$).
Thus, as there is no zero for $0\leq y \leq \pi$, either:
$\forall y; \cos(y) \geq \cos(y + \frac{\sin(y)}{2})$
or
$\forall y; \cos(y) \leq \cos(y + \frac{\sin(y)}{2})$
For $y = \pi/2, \cos(\pi/2) > \cos(\pi/2 + \frac{1}{2\sqrt{2}}) \Rightarrow \forall y; \cos(y) \geq \cos(y + \frac{\sin(y)}{2})$
Hence, $\forall y\in \mathbb{R}$;
$\cos(2x)(\frac{\cos(2x)+3}{2}) \geq \cos(2x + \frac{\sin(2x)}{2})$
$\blacksquare$
|
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|
Proof of Implicit Differentiation (showing a statement is true)
Prove that if $x^2+y^2-2y\sqrt{1+x^2} = 0$, then $dy/dx = x/\sqrt{1+x^2}$.
Whilst I have implicitly differentiated in terms of x in order to derive that
$$dy/dx = (-x+2xy/\sqrt{1+x^2})/(y\sqrt{1+x^2}-1-x^2)$$
however I am unsure as to what my next steps are. I believe that I will need to rearrange the original supposition in order to achieve the proof however I do require some help as to how can I do this as I cannot see what would possibly cancel out.
Additionally, does anyone have any particularly handy methods/techniques to be able to complete proofs of this format of question more easily? I do understand that there is not one singular technique that can be used when completing this proofs but is there any strategy that minimizes the number of dead-ends that I come across in my working?
Thanks
|
We find from the given relation $ \ x^2 + y^2 \ = \ 2y\sqrt{1+x^2} \ $ that
$$ 2x \ + \ 2y·\frac{dy}{dx} \ \ = \ \ 2·\frac{dy}{dx}·\sqrt{1+x^2} \ + \ 2y·\frac12·\frac{1}{\sqrt{1+x^2}}·2x \ $$
$$ \Rightarrow \ \ ( \ y \ - \ \sqrt{1+x^2} \ ) ·\frac{dy}{dx} \ \ = \ \ y· \frac{1}{\sqrt{1+x^2}}· x \ - \ x \ \ \Rightarrow \ \ \frac{dy}{dx} \ \ = \ \ \frac{-x \ + \ \frac{xy}{\sqrt{1+x^2}}}{y \ - \ \sqrt{1+x^2} } \ \ . $$
Replacing (at least for the present) the radical using the original equation $ ( \ \sqrt{1+x^2} \ = \ \frac{x^2 + y^2}{2y} \ ) \ $ , we obtain
$$ \frac{dy}{dx} \ \ = \ \ \frac{ xy·\frac{2y}{x^2 + y^2} \ - \ x}{y \ - \ \frac{x^2 + y^2}{2y} } \ \ = \ \ \frac{ x·\left( \ 2y^2 \ - \ [x^2 + y^2] \ \right)}{y·(x^2 + y^2) \ - \ \frac{(x^2 + y^2)^2}{2y} } \ \ = \ \ \frac{ x· ( y^2 \ - \ x^2 )·2y}{2y^2·(x^2 + y^2) \ - \ (x^2 + y^2)^2 } $$
$$ = \ \ \frac{ x· ( y^2 \ - \ x^2 )·2y}{(2y^2 \ - \ x^2 \ - \ y^2)·(x^2 + y^2) } \ \ = \ \ \frac{ x· ( y^2 \ - \ x^2 )·2y}{( y^2 \ - \ x^2 )·(x^2 + y^2) } \ \ = \ \ x· \frac{ 2y}{ x^2 + y^2 }
$$
$$= \ \ x· \frac{1}{ \sqrt{1+x^2} } \ \ , $$
again using the original relation. (So it appears there may be an error in your expression for the implicit derivative.)
|
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Suppose the number n = 90k is the smallest integer that has exactly ninety divisors (all positive integers) which includes 1 and n. Find k Suppose the number n = 90k is the smallest integer that has exactly ninety divisors (all
positive integers) which includes 1 and n. Find k.
I know this question has been answered, but I am wondering if there is an elementary way to answer this using modular arithmetic and/or Euler's totient function?
|
We have, $N=90k=2\cdot3^2\cdot5\cdot(k)$
Number of divisors of a number is product of the $(\text{powers}+1)$ of each of the it's prime factors.
Since we want exactly $90$ divisors, so the product would look like:
$$\begin {align} 90 &=2\cdot3^2\cdot5 \\ &= 2\cdot3\cdot3\cdot5 \tag A \\ &= 6\cdot3\cdot5 \tag B \\ &= 2\cdot9\cdot5 \tag C \\ &= 2\cdot3\cdot15 \tag D\end{align}$$
Note: $(A),(B),(C),(D)$ limit our prime factors to 3 or 4 only.
Case (A):
$N=ab^2c^2d^4$
Here, we need to assign lowest prime number to the factor with highest power.
Thus, $a=7,b=5,c=3,d=2$
Thus, $N=25200$
Case (B):
$N=a^5b^2c^4$
Here, we need to assign lowest prime number to the factor with highest power.
Thus, $a=2,b=5,c=3$
Thus, $N=64800$
Case (C):
$N=ab^8c^4$
Here, we need to assign lowest prime number to the factor with highest power.
Thus, $a=5,b=2,c=3$
Thus, $N=103680$
Case (D):
$N=ab^2c^{14}$
Here, we need to assign lowest prime number to the factor with highest power.
Thus, $a=5,b=3,c=2$
Thus, $N=737280$
So, the smallest number satisfying the given conditions is $25200$.
$\therefore 90k=25200$
$$\Rightarrow \boxed {k=280}$$
|
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|
The sum of $n$ terms in $1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots$ I am just confused, considering we can take $1 \cdot 2$ as the first term then we get the $n$th term as $n(n+1)$ so the sum of $n$ terms would be $\frac{n(n+1)(2n+1)}{6}$ + $\frac{n(n+1)}{2}$ but let's assume $0 \cdot 1$ as the first term then the $n$th term becomes $(n-1) \cdot n$ and so it's summation becomes $\frac{n(n+1) (2n+1)}{6} - \frac{n(n+1)}{2}$ but it's the same summation as above but the results are different what am I doing wrong?
|
$S_1=1⋅2+2⋅3+3⋅4+\dots$
$T_n=n(n+1)$
$S_1=\frac{n(n+1)(2n+1)}{6}\bf+\frac{n(n+1)}{2}$
$S_2=1⋅2+2⋅3+3⋅4+\dots$
$T_n=(n-1)n$
$S_2=\frac{n(n+1)(2n+1)}{6}\bf-\frac{n(n+1)}{2}$
As the bold part indicates, the summation isn't the same.
Let's put $\bf n=1$:
$S_1=1⋅2=2$
$S_1=\frac{1(2)(3)}{6}\bf+\frac{1(2)}{2}=1+1=2$
$S_2=0⋅1=0$
$S_2=\frac{1(2)(3)}{6}\bf-\frac{1(2)}{2}=1-1=0$
Let's put $\bf n=1$:
$S_1=1⋅2+2⋅3=8$
$S_1=\frac{2(3)(5)}{6}\bf+\frac{2(3)}{2}=5+3=8$
$S_2=0⋅1+1⋅2=2$
$S_2=\frac{2(3)(5)}{6}\bf-\frac{2(3)}{2}=5-3=2$
So, what you have done is totally correct and verifies.
Whenever there is confusion like such that you have, working out for small values of $n$ which we can ourselves calculate helps to make things clearer.
|
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|
I can't come up with the intended solution to $\frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab>18$ I was going trough some easy algebra problems when I encountered
$$
\frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab>18.
$$
As you can see the problem is easily solvable with AM > GM
I fairly quickly came up with this solution:
$$
\frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab
= \frac{a^2}{2a^5b^5} + \frac{b^2}{2a^5b^5} + \frac{81a^2b^2}{12} + \frac{81a^2b^2}{12} + \frac{81a^2b^2}{12} + \frac{9ab}{2} + \frac{9ab}{2}
$$
and using AM-GM
$$
\frac{a^2}{2a^5b^5} + \frac{b^2}{2a^5b^5} + \frac{81a^2b^2}{12} + \frac{81a^2b^2}{12} + \frac{81a^2b^2}{12} + \frac{9ab}{2} + \frac{9ab}{2} \ge 7\sqrt[7]{\frac{a^{10}b^{10}9^8}{a^{10}b^{10}2^{10}3^3}} \approx 20
$$
(Yes I was able to do that by hand and later check with my calculator)
I am not sure that this is the intended solution though
|
Let functions $A$ and $G$ denote the arithmetic and geometric means of their arguments. Then:
$$x := \frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab$$
$$= A(\frac{a^2}{a^5b^5}, \frac{b^2}{a^5b^5}) + \frac{81a^2b^2}{4} + 9ab$$
$$= A(\frac{1}{a^3b^5}, \frac{1}{a^5b^3}) + \frac{81a^2b^2}{4} + 9ab$$
$$\ge G(\frac{1}{a^3b^5}, \frac{1}{a^5b^3}) + \frac{81a^2b^2}{4} + 9ab$$
$$= \frac{1}{a^4b^4} + \frac{81a^2b^2}{4} + 9ab$$
Note that $a$ and $b$ now only occur in same-exponent pairs. So, for convenience, let $p = ab$.
$$x \ge \frac{1}{p^4} + \frac{81p^2}{4} + 9p$$
$$=A(\frac{2}{p^4}, \frac{81p^2}{2}) + 9p$$
$$\ge G(\frac{2}{p^4}, \frac{81p^2}{2}) + 9p$$
$$= \frac{9}{p} + 9p$$
If $p \ge 2$, then $x > 9p \ge 18$.
If $p \ge \frac{1}{2}$, then $x > \frac{9}{p} \ge 18$.
Now, let's consider $\frac{1}{2} < p < 2$. Then $\frac{1}{2} < \frac{1}{p} < 2$ as well. So,
$$\frac{9}{2} < \frac{9}{p} < 18$$
$$\frac{9}{2} < 9p < 18$$
Adding these gives:
$$18 < \frac{9}{p} + 9p < 36$$
So $x \ge \frac{9}{p} + 9p > 18$.
Finally, just in case anyone nitpicks strict versus non-strict inequality at the boundaries between cases:
$$p = \frac{1}{2} \implies \frac{9}{p} + 9p = 22.5 > 18$$
$$p = 2 \implies \frac{9}{p} + 9p = 22.5 > 18$$
Combining the cases of $0 < p < \frac{1}{2}$, $p = \frac{1}{2}$, $\frac{1}{2} < p < 2$, $p = 2$, and $p > 2$; we can conclude that $\forall p > 0$ (and so $\forall a,b > 0$), $x > 18$, Q.E.D.
|
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|
Arc length of curve stuck with integration Data from exercise
$$y=\frac{4}{3}x^2+2\\
x\in[-1,1]$$
Formula for length of curve
$$L=\int_a^{b}\sqrt{1+(f(x)')^2}\ dx$$
So far i have
$$y'=\frac{8}{3}x$$
$$\int_{-1}^{1}\sqrt{1+\frac{64}{9}x^2}\ dx$$
Substition
$$t^2=\frac{64}{9}x^2$$
$$t=\frac{8}{3}x$$
$$\frac{3}{8}dt=dx$$
So i have
$$\int_{-\frac{8}{3}}^{\frac{8}{3}}\sqrt{1+t^2}\ dt$$
I this point i dont have a clue how to integrate this
|
Let $x = \sinh \theta \implies$
$$
\begin{align}
I &= \int \sqrt{1+x^2}dx \\
&= \int \sqrt{1+\sinh^2 \theta}\cosh \theta d \theta \\
&= \int \cosh^2 \theta d \theta \\
&= \frac{1}{2}\int 1 + \cosh 2 \theta d \theta \\
&= \frac{1}{2}\theta+\frac{1}{4}\sinh 2 \theta + c\\
&= \frac{1}{2}\sinh^{-1}x+\frac{1}{2}x\sqrt{1+x^2}+c\\
&=\frac{1}{2}\ln(x + \sqrt{1+x^2})+\frac{1}{2}x\sqrt{1+x^2}+c
\end{align}$$
|
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|
Solving $\cos3x=-\sin3x$, for $x \in [0,2\pi]$ For my first step, I would choose to either divide both sides or add sin3x to both sides. My professor in Pre-Calc told me I shouldn't divide any trig functions from either side. Despite this, I can only find a solution this way.
$$\frac{\cos3x}{-\sin3x}=1$$
$$-\tan3x=1$$
$$\tan3x=-1$$
Which would lead me to my method of solving trig equations which is:
$$-\tan\,\epsilon\,Q2\quad-\tan\,\epsilon\,Q4$$
$$\tan x=1 \, at\,\frac{\pi}{4}$$
$$Q2:\pi-\frac{\pi}{4}=\frac{3\pi}{4}$$
$$Q4:2\pi-\frac{\pi}{4}=\frac{5\pi}{4}$$
However, this is only for $\tan x=-1$, my professor told me to apply the "modifications of x" after.
$$3x=\frac{3\pi}{4}=\frac{3\pi}{4}\div3=\frac{\pi}{4}$$
$$3x=\frac{5\pi}{4}=\frac{5\pi}{4}\div3=\frac{5\pi}{12}$$
Is this the correct way to solve this equation? I believe dividing out $\sin3x$ leads to domain or range errors. However, I tried solving by adding $\sin3x$ to both sides first but could not figure out how to isolate the trig functions. I will clarify anything needed in the comments.
|
though ur answer is not wrong, as long as u show that $\sin(3x) \neq 0$
a better way to solve would be
$\cos( 3x) + \sin(3x) = 0$
so $\sqrt{2}\sin(3x + \frac{\pi}{4}) = 0$
let y= $3x + \frac{\pi}{4}$
$ y \in [\frac{\pi}{4}, 6\pi+\frac{\pi}{4}]$
$\sin(y) =0$
so $ y \in \{\pi , 2\pi , 3\pi , 4\pi , 5\pi , 6\pi\} $
so $x = \frac{\pi}{4}$ or $\frac{7\pi}{12}$ or $\frac{11\pi}{12}$ or $\frac{5\pi}{4}$ or $\frac{19\pi}{12}$ or $\frac{23\pi}{12}$ or $\frac{9\pi}{4}$
|
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|
Solve for all $x,y\in N$ and $x^3-3^y = 2015$ Question:
Solve for all $x,y\in N$ and $x^3-3^y = 2015$
So far I have tried $\mod(7)$ and $\mod(8)$ but I am not sure if I am in the right path or not I will be happy if someone can toss me a hint.
|
Use Metin's idea of "reject solution of too large" and
someone's idea of mod $13$ -> $3\mid m$, I reach a solution.
Consider mod $13$, then
$3^y\equiv1,3,9 \pmod {13}$
$x^3\equiv1,5,8,12 \pmod {13}$
Thus, $3^y\equiv1 \pmod {13}, x^3\equiv1 \pmod {13}, 3\mid y$
Let $y=3z, k=3^z$, so $x>k>0$
Then $x^3-3^y = x^3-(3^z)^3 = x^3-k^3 = (x-k)(x^2+xk+k^2) = 2015 $
$x^2+xk+k^2$ is factor of $2015$ and thus bounded above, we can have $x<45$. Then the rest is check and reject by computer. (By hand is tough)
Furthermore, consider $x^3-27^z=2015$ with mod $7$, we have $7\mid x$. (Consider all outcomes as above). With $x\equiv2 \pmod{6}$ by Alexander. We can reject all $x<45$ except $14$.
|
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|
Is there a hypergeometric function form of the quadratic formula that converges for all a,b,c? After reading this question about a general formula for roots, I was curious about the simple case of $N=2$.
As is well known to school children, if $ ax^2 + bx + c = 0 $, then the roots are $ \frac{-b \pm \sqrt{b^2-4ac}}{2a} $
Shuffling the binomial series around yields $$ -\frac{c}{b} \sum_{k=0}^{\infty} \binom{-\frac{1}{2}+k}{k} \left( \frac{4ac}{b^2} \right)^k \frac{1}{(k+1)} = -\frac{b}{2a} + \frac{ \sqrt{b^2-4ac}}{2a} $$ but this doesn't converge for simple cases such as $ a=1,b=1,c=-1 $ since the radius of convergence is $<\frac{1}{4}$.
Is there a hypergeometric series that converges for all $a,b,c$ (where there are roots of course)?
|
$$S_n=-\frac{c}{b} \sum_{k=0}^{n} \binom{-\frac{1}{2}+k}{k} \left( \frac{4ac}{b^2} \right)^k \frac{1}{(k+1)} $$
$$S_n=-\frac{b \left(1-\sqrt{1-\frac{4 a c}{b^2}}\right)}{2 a}+\frac{c }{b
}\,\frac{\binom{n+\frac{1}{2}}{n+1}}{n+2 }\,\left(\frac{4a c}{b^2}\right)^{n+1}\, _2F_1\left(1,\frac{2n+3}{2};n+3;\frac{4 a c}{b^2}\right)$$ where appears the Gaussian hypergeometric function.
|
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|
How to proceed with a definite integral of this form? How do I find the definite integral of this sort of function?
$$ \int_m^n \frac{\sin(a \sin(\sqrt{(b \lfloor x\rfloor^2 + b d + 1)}))}{a \sin(\sqrt{(b \lfloor x\rfloor^2 + b d + 1)})} dx $$
This function is actually a form of the $ \operatorname{sinc} $ function except it has a floor variable nested inside it, and also takes the sine of another sine function.
Note: a & b = 10^k (a & b are higher order powers of 10)
|
\begin{eqnarray}
\int_m^n \frac{\sin(a \sin(\sqrt{(b \lfloor x\rfloor^2 + b d + 1)}))}{a \sin(\sqrt{(b \lfloor x\rfloor^2 + b d + 1)})} dx&=&\sum_{k=m}^n\int_k^{k+1} \frac{\sin(a \sin(\sqrt{(b k^2 + b d + 1)}))}{a \sin(\sqrt{(b k^2 + b d + 1)})} dx\\
&=&\sum_{k=m}^n \frac{\sin(a \sin(\sqrt{(b k^2 + b d + 1)}))}{a \sin(\sqrt{(b k^2 + b d + 1)})}
\end{eqnarray}
|
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|
Sum of $\sum_0^\infty \frac1{n^4+a^4}$ How to find the following series as pretty closed form by $a$?
$$S=\sum_0^\infty \frac1{n^4+a^4}$$
I first considered applying Herglotz trick, simply because the expressions are similar. So I changed it like this...
$$2S-a^{-4}=\sum_{-\infty}^\infty \frac1{n^4+a^4}$$
However, the attempt failed to find such an appropriate function like $\pi\cot\pi z$ in this post.
Next I found this post and used Fourier transform in a similar way, and the result was a nightmare!
How on earth can I calculate the value of this series?
|
We start with partial fractions,
$$\frac{1}{n^4+a^4} = \frac{i}{2a^2}\left(\frac{1}{n^2 + i a^2} - \frac{1}{n^2 - i a^2}\right) \tag 1$$
And then note that (Series expansion of $\coth x$ using the Fourier transform)
$$\coth x = \frac{1}{x} + 2 \sum_{n=1}^\infty \frac{x}{x^2+\pi^2n^2}$$
which we rearrange to give:
$$\sum_{n=1}^\infty \frac{1}{\left(\frac{x}{\pi}\right)^2+n^2} = \frac{\pi^2}{2x}\left(\coth x - \frac{1}{x} \right) $$
$$\sum_{n=0}^\infty \frac{1}{\left(\frac{x}{\pi}\right)^2+n^2} = \frac{\pi^2}{x^2}+\frac{\pi^2}{2x}\left(\coth x - \frac{1}{x} \right) =\frac{\pi^2}{2x^2}\left(x \coth x + 1 \right) \tag 2$$
and so
$$\sum_{n=0}^\infty \frac{1}{n^4+a^4} = \frac{i}{2a^2}\sum_{n=0}^\infty \left(\frac{1}{n^2 + i a^2} - \frac{1}{n^2 - i a^2}\right) $$
$$ = \frac{i}{2a^2}\frac{\pi^2}{2ia^2\pi^2}\left(\sqrt{i}a \pi \coth \sqrt{i}a - \sqrt{-i}a \pi \coth \sqrt{-i}a\right) $$
$$ = \frac{\pi}{4a^3}\left(e^{i\pi/4} \coth \sqrt{i}a - e^{3i\pi/4} \coth \sqrt{-i}a\right) $$
which I'm sure can be further simplified but I will leave there for now.
|
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|
Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$
Problem: Find the all possible values of $a$, such that
$$4x^2-2ax+a^2-5a+4>0$$
holds $\forall x\in (0,2)$.
My work:
First, I rewrote the given inequality as follows:
$$
\begin{aligned}f(x)&=\left(2x-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0\end{aligned}
$$
Then, we have
$$
\begin{aligned}
0<x<2\\
\implies -\frac a2<2x-\frac a2<4-\frac a2\end{aligned}
$$
Case $-1:\,\,\,a≥0 \wedge 4-\frac a2≤0$.
This leads,
$$
\begin{aligned}\frac {a^2}{4}>\left(2x-\frac a2\right)^2>\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4>f(x)>\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
For $f(x)>0$, it is enough to take $\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≤0$.
Case $-2:\,\,\,a≤0 \wedge 4-\frac a2≥0$.
We have:
$$
\begin{aligned}\frac {a^2}{4}<\left(2x-\frac a2\right)^2<\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4<f(x)<\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
Similarly, for $f(x)>0$, it is enough to take $\frac{a^2}{4}+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≤0\wedge 4-\frac a2≥0$.
Case $-3:\,\,\,a≥0 \wedge 4-\frac a2≥0$.
This case implies, $\left(2x-\frac a2\right)^2≥0$. This means, $f(x)≥\frac {3a^2}{4}-5a+4$
Thus, for $f(x)>0$, it is enough to take $\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≥0$.
Finally, we have to combine all the solution sets we get.
I haven't done the calculation, because I want to make sure that the method I use is correct. Do you see any flaws in the method?
|
I want to bring the most straightforward way to calculate this problem systematically. Also, you can check the attached graph for better representation. I write the quadratic equation $f_x(a)=4x^2-2ax+a^2-5a+4$ w.r.t. $a$ as an unknown parameter. We have:
$f_x(a)=a^2-(2x+5)a+(4x^2+4)>0$.
So, $a$ must be greater than the bigger root of $f_x(a)=0$ or less than the smaller root, which are:
$f_x(a)=0 \rightarrow a=x+2.5\pm\sqrt{-3x^2+5x+2.25}=g^\pm(x)$ for $x\in(0,2)$.
So, we have $a>g^+(x)$ or $a<g^-(x)$ for $\forall x \in (0,2)$, i.e.:
$a>\max_{x\in(0,2)}g^+(x)$ or $a<\min_{x\in(0,2)}g^-(x)$. We have:
$dg^+(x)/dx=0\rightarrow 1+\frac{-6x+5}{2\sqrt{-3x^2+5x+2.25}}=0\rightarrow -12x^2+20x+9=36x^2-60x+25$
$\rightarrow 3x^{2}-5x+1=0\rightarrow x=\frac{5+\sqrt{13}}{6}\rightarrow g^+(\frac{5+\sqrt{13}}{6})=\frac{2}{3}(5+\sqrt{13})$
Similarly, we have:
$dg^-(x)/dx=0\rightarrow 1-\frac{-6x+5}{2\sqrt{-3x^2+5x+2.25}}=0 \rightarrow x=\frac{5-\sqrt{13}}{6} \rightarrow g^-(\frac{8}{5+\sqrt{13}})=\frac{8}{5+\sqrt{13}}$
So, $a$ range having $f_x(a)=4x^2-2ax+a^2-5a+4>0$ are:
$a>\frac{2}{3}(5+\sqrt{13}) \approx5.7370$ or $a<\frac{8}{5+\sqrt{13}} \approx0.9296$
|
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|
How to solve this group of equations? $$\dfrac{yz(y+z-x)}{x+y+z}=p \tag1$$
$$\dfrac{xz(x+z-y)}{x+y+z}=q \tag2$$
$$\dfrac{xy(x+y-z)}{x+y+z}=r \tag3$$
This group of equations comes from the problem "if the distances of incenter of an triangle to its vertices are $\sqrt{p}$, $\sqrt{q}$ and $\sqrt{r}$. Find the length of sides of the triangle". By applying barycentric coordinates, it is easy to get the group of equations above, with $x$, $y$ and $z$ as three sides of the triangle.
I have no idea how to solve this group of equations now. Any hints? Thanks!
|
Rearranging, we get
$$p+\frac{2xyz}{x+y+z}=yz$$
$$q+\frac{2xyz}{x+y+z}=xz$$
$$r+\frac{2xyz}{x+y+z}=xy$$
if we multiply two of these equations and divide by the third,
$$\implies x=\sqrt{\frac{(q+a)(r+a)}{(p+a)}}$$
$$\implies y=\sqrt{\frac{(r+a)(p+a)}{(q+a)}}$$
$$\implies z=\sqrt{\frac{(p+a)(q+a)}{(r+a)}}$$
where $a=2xyz/(x+y+z)$ must satisfy a cubic equation
$$a^3-(pq+qr+rp)a-2pqr=0$$
The cubic comes from substituting the expressions for $x,y,z$ into $a=\frac{2xyz}{x+y+z}$ and rearranging.
$$a=\frac{2\sqrt{\frac{(q+a)(r+a)}{(p+a)}}\sqrt{\frac{(r+a)(p+a)}{(q+a)}}\sqrt{\frac{(p+a)(q+a)}{(r+a)}}}{\sqrt{\frac{(q+a)(r+a)}{(p+a)}}+\sqrt{\frac{(r+a)(p+a)}{(q+a)}}+\sqrt{\frac{(p+a)(q+a)}{(r+a)}}}$$
$$=\frac{2\sqrt{(q+a)(r+a)}\sqrt{(r+a)(p+a)}\sqrt{(p+a)(q+a)}}{(q+a)(r+a)+(r+a)(p+a)+(p+a)(q+a)}$$
$$=\frac{2(p+a)(q+a)(r+a)}{(q+a)(r+a)+(r+a)(p+a)+(p+a)(q+a)}$$
So
$$3a^3+2(p+q+r)a^2+(pq+qr+rp)a=2a^3+2(p+q+r)a^2+2(pq+qr+rp)a+2pqr$$
|
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|
$P(x)=x^2+ax+b$ and $Q(x)=x^2+a_1x+b_1$ and $P(x)\cdot Q(x)$ has at most one real root then prove that the claims given below are true.
If $P(x)=x^2+ax+b$ and $Q(x)=x^2+a_1x+b_1$ where $a,b,a_1,b_1\in\mathbb{R}$ and equation $P(x)\cdot Q(x)=0$ has at most one real root then prove that the claims given below are true.
$1.$ $(1+a+b)(1+a_1+b_1)>0$
$2.$ $(1+a+b)>0$
$3.$ $\displaystyle\frac{1+a+b}{1+a_1+b_1}>0$
My work:
Let $P(x)\cdot Q(x)=f(x)$ then $f(x)$ has $0$ real roots as it can't have only $1$ real root.
$$f(x)=x^4+x^3(a+a_1)+x^2(aa_1+b+b_1)+x(ab_1+ba_1)+bb_1$$ Now, since the leading coefficient of $f(x)$ is $+ve$ it means that $f(x)>0\:\:\forall x\in\mathbb{R}$ $\implies f(1)>0$
$$f(1)=1+(a+a_1)+(aa_1+b+b_1)+(ab_1+ba_1)+bb_1$$ $$=(1+a+b)(1+a_1+b_1)>0$$
Since $f(x)$ has $0$ real roots, it means that both $P(x)$ and $Q(x)$ have $0$ real roots $\implies$ both $P(1)>0$ and $Q(1)>0$ as both of their leading coefficients are positive.
Or individually,
$$
\begin{align}
(1+a+b)>0\\
(1+a_1+b_1)>0\\
\end{align}$$ It $\implies$ $$\frac{1+a+b}{1+a_1+b_1}>0$$
Is my solution correct$?$
|
When it says both have at most one real root, it means $a^2-4b\le0$ and $a_1^2-4b_1\le0$.
Also, both inequalities cannot be equality, because they will have two real roots, or if both inequalities are equality, then $a=a_1$ and $b=b_1$.
Then $a+b+1\ge a+a^2/4+1=(a/2+1)^2\ge 0$
BUT
Because it can be $a=a_1=-2$ and $b=b_1=1$ with the only root of $1$, then it is a counterexample for all claims (1,2,3) where they are respectively 0, 0, and ambiguous.
|
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|
Solve the polynomial $x^6+x^5+x^4-2x^3-x^2+1=0$ in exact form
Try to reduce the degree of the polynomial $$P(x)=x^6+x^5+x^4-2x^3-x^2+1$$ by algebraic ways and find the possible solution method to $P(x)=0$.
The source of the problem comes from a non-english algebra precalculus workbook. It is a workbook for those who want to be more interested in mathematics.
WolframAlpha couldn't solve the polynomial in exact form. Only approximate solutions are possible.
Rational root theorem shows that there are no rational roots. So, first method failed.
But, I tried different ways.
$$\frac {P(x)}{x^2}=x^3+x^2+x-2-\frac 1x+\frac 1{x^3}=\bigg(x^3+\frac 1{x^3}\bigg)+\bigg(x+\frac 1x\bigg)+x^2-2=0$$
or
$$\frac {P(x)}{x^2}=x^4+x^3+x^2-2x-1+\frac {1}{x^2}=\bigg(x^2+\frac 1{x^2}\bigg)+(x^4+x^3-2x-1)$$
I think that the trick $x+\frac 1x=t$ doesn't work here.
At the end I tried factoring the Polynomial but here are $2$ possibilities to do it
$$P(x)=(x^2+ax+b)(x^4+cx^2+dx+e)$$
or
$$P(x)=(x^3+ax^2+bx+c)
(x^3+dx^2+ex+f)$$
But which of these methods will work, it is not clear.
|
The trick $x+\frac1x=t$ requires the polynomial to be a palindrome, so that is clearly not the case here. However that's not the only substitution possible to simply. Here, lets try something similar to what you have already done, persisting a bit more:
$
\begin{align}
\frac{P(x)}{x^4} &= x^2+x+1-\frac2x-\frac1{x^2}+\frac1{x^4} \\
&= \left(x^2-\frac2x+\frac1{x^4}\right)+\left(x-\frac1{x^2} \right)+1\\
&= \left(x-\frac1{x^2} \right)^2+\left(x-\frac1{x^2} \right)+1\\
&=u^2+u+1 \tag{1}
\end{align}
$
where $u = x-\dfrac1{x^2} \tag{2}$
Now clearly $u$ is solvable as a quadratic from $(1)$, and for both values of $u$, the equation $(2)$ gives a cubic to solve for $x$. As both quadratics and cubics admit solutions in exact algebraic form (though using cumbersome radicals perhaps), we are done.
|
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|
Compute the area of Quadrilateral $ABCD$ As title suggests, the question is to solve for the area of the given convex quadrilateral, with two equal sides, a side length of 2 units and some angles:
I have solved the problem with a synthetic geometric approach involving some angle chasing. However, I believe my solution (which I will post in a comment below since I don't want to clutter the question) is a little messy and not efficient. Are there any better ways to do this? Geometric and/or trigonometric approaches are all welcomed!
EDIT: I have posted my solution below!
|
Let the intersection of the two diagonals be $O$.
And let $ AO = x , OC = y, AD = z $
Then it follows from the law of sines that
$ DO = a x \hspace{25pt}$ where $ a = \dfrac{\sin(105^\circ)}{\sin(30^\circ) } $
and
$ OB = b y \hspace{25pt}$ where $ b = \dfrac{\sin(75^\circ)}{\sin(60^\circ)} $
Apply the law of cosines to $\triangle AOD$, $\triangle OBD$, $\triangle ABC $ gives us the following three quadratic equations:
$ z^2 = x^2 (1 + a^2 - 2 a \cos(45^\circ) )$
$ z^2 = y^2 (1 + b^2 - 2 b (\cos(45^\circ) )$
$ z^2 + (x + y)^2 - 2 z (x + y) \cos(75^\circ) = 2^2 = 4 $
Using a quadratic system solver (for example from Wolframalpha.com), we get:
$ x = \sqrt{3} - 1 $.
$ y = 3 - \sqrt{3} $.
$ z = \sqrt{2} x = \sqrt{6} - \sqrt{2} $
Now the area is given by
$ \text{Area} = \dfrac{1}{2} \sin(45^\circ) \left( a x^2 + b y^2 + xy (a + b) \right) $
Evaluating the above expression, yields
$\text{Area} = 2$
|
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"url": "https://math.stackexchange.com/questions/4536997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
find two primes less than $500$ that divide $20^{22}+1$
Find two primes less than $500$ that divide $20^{22}+1$.
Note that $20^2+1=401$ divides the required number (since for any integer a, if k is odd, then $a+1$ divides $a^k+1$). Also, any prime dividing the given number must be congruent to $1$ modulo $4$ since $-1$ would be a quadratic residue modulo any such prime. Primes like $17$ and $13$ don't work. $20\equiv 7\mod 13, 20\equiv 3\mod 17,$ and in both cases $20$ is a primitive root with respect to the corresponding modulus. So it might be that we'll need to find a large prime. $a^k+1 = (a+1)(a^{(k-1)} - a^{(k-2)}+\cdots + 1),$ where $a = 20^2, k=11.$ I'm not sure if there's some way to factor $a^{(k-1)} -a^{(k-2)}+\cdots + 1.$
|
Suppose a prime $p$ (necessarily odd) divides $20^{22} + 1$. Then $20^{22} \equiv -1 \bmod p$ so $20^{44} \equiv 1 \bmod p$ and it follows that the order of $20 \bmod p$ divides $44$ but does not divide $22$. This means it must be divisible by $4$, so must be either $4$ or $44$. If $20^4 \equiv 1 \bmod p$ then $20^{22} \equiv 20^2 \equiv -1 \bmod p$ so we conclude that $p \mid 401$ and hence $p = 401$. Otherwise the order of $20 \bmod p$ is $44$, from which it follows that $p \equiv 1 \bmod 44$.
The smallest such prime is $p = 89$, and we have
$$\begin{align*} 20^{22} &= 400^{11} \equiv 44^{11} \\
& \equiv \left( \frac{89 - 1}{2} \right)^{11} \equiv \left( - \frac{1}{2} \right)^{11} \\
& \equiv - \frac{1}{32 \cdot 64} \equiv \frac{1}{32 \cdot 25} \\
& \equiv \frac{1}{800} \equiv - \frac{1}{90} \\
& \equiv - 1 \bmod 89 \end{align*} $$
so $p = 89$ works. This is a slightly annoying computation, though, so I don't know if it was what was intended. There were only three primes to check, namely $p = 89, 353, 397$ although if $89$ hadn't worked generating the rest of this list would've been slightly annoying and checking it would've been slightly annoying too, I think.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
The limit $\lim_{n\to \infty}\frac{e+e^{\frac{1}{2}}+e^{\frac{1}{3}}+\ldots+e^{\frac{1}{n}}}{n}$ $$\lim_{n\to \infty}\frac{e+e^{\frac{1}{2}}+e^{\frac{1}{3}}+\ldots+e^{\frac{1}{n}}}{n}$$ is equal to
(a) $0$
(b) $1$
(c) $e$
(d) none of these
How should I deal with such limits? What I know $e >1$, so $$e+e^{\frac{1}{2}}+e^{\frac{1}{3}}+\ldots+e^{\frac{1}{n}}>1+1+\ldots+1$$ ($n$ times), so $$\frac{e+e^{\frac{1}{2}}+e^{\frac{1}{3}}+\ldots+e^{\frac{1}{n}}}{n}>\frac{n}{n}=1.$$
So this limit should not be zero. How to deal further.
Thanks in advance!
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Let's take @Gary:'s hint but in a pedestrian way:
First of all, we have Bernoulli's inequality
$$(1 + x)^n > 1 + n x$$
for $x > 0$, and $n > 1$. Therefore we conclude
$$1+ x > ( 1 + n x)^{\frac{1}{n}}$$
or
$$1 + \frac{y}{n} > (1+y)^{\frac{1}{n}}$$
for $y> 0$ and $n> 1$ natural. In particular
$$1 + \frac{e-1}{n} > e^{\frac{1}{n}} >1$$
for $n > 1$ (the RHS since $e>1$). Now sum all these inequalities and get
$$1 + (e-1) \frac{1+\frac{1}{2} + \cdots + \frac{1}{n}}{n} > \frac{e + e^{\frac{1}{n}} + \cdots +e^{\frac{1}{n}}}{n} > 1$$
Now we only have to show that
$$\frac{1+\frac{1}{2} + \cdots + \frac{1}{n}}{n}$$ has limit $0$. How to proceed? Let's show for instance that for $n$ large enough the terms are $\le \frac{1}{7}$. I claim that this happens for $n\ge 14^2$. Indeed, the expression can be written as
$$\frac{1+ \frac{1}{2} +\cdots + \frac{1}{14} }{n} + \frac{\frac{1}{15} + \cdots + \frac{1}{n}}{n}$$
Now, the first term has in the numerator $14$ terms $\le 1$, so the first fraction $\le \frac{14}{14^2} = \frac{1}{14}$. The second fractions has in the numerator $n-14$ terms $< \frac{1}{14}$, and the denominator $n$. We conclude that the second fraction $< \frac{1}{14}$. Now add them up.
Now, instead of $\frac{1}{7}$, we could reason with $\frac{1}{m}$.
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"timestamp": "2023-03-29T00:00:00",
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|
Expectation of $E(XY^2)$ where $X$ and $Y$ are random events related to drawing balls from a vase. The Question:
a vase has 2 white and 2 black balls in it, we remove balls from the vase randomly one by one without returning them. $X=$ the number of balls that we removed until both white balls were removed (incl the last one), $Y=$ the number of balls we removed until we removed the first black one (including the black one). find $E(XY^2)$
My Process so far:
I constructed a shared measure chart:
https://imgur.com/a/Vmb5DHG
which I then used to calculate the expectation directly.
$$\begin{aligned} E(XY^2) &= \sum_{x,y} P_{X,Y}(x,y) \cdot xy^2 \\ &= \tfrac{2}{6}\cdot 3 \cdot 1^2 + \tfrac{1}{6} \cdot 4 \cdot 1^2 + \tfrac{1}{6}\cdot 2 \cdot 3^2 \\ &=1+2/3+2 +8/3+3 = 9\tfrac13\end{aligned}$$
The right answer is $9.5$ - So my question is this:
where am I wrong? I’m not used to creating charts - is there a different way of solving the problem?
any help is really appreciated!
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There are $6$ equiprobable permutations of the set $\{w,w,b,b\}$, and for each one, we compute the values of $X$ and $Y$ and $XY^2$ as follows:
$$\begin{array}{c|c|c|c}
\text{Outcome} & \color{red}{X} & \color{blue}{Y} & XY^2 \\
\hline
(w,\color{red}{w},\color{blue}{b},b) & 2 & 3 & 18 \\
(w,\color{blue}{b},\color{red}{w},b) & 3 & 2 & 12 \\
(w,\color{blue}{b},b,\color{red}{w}) & 4 & 2 & 16 \\
(\color{blue}{b},w,\color{red}{w},b) & 3 & 1 & 3 \\
(\color{blue}{b},w,b,\color{red}{w}) & 4 & 1 & 4 \\
(\color{blue}{b},b,w,\color{red}{w}) & 4 & 1 & 4 \\
\end{array}$$
I have color coded the respective ball draws that correspond to $X$ and $Y$.
Again, since each of these outcomes is equiprobable, the desired expectation is simply
$$\operatorname{E}[XY^2] = \frac{18+12+16+3+4+4}{6} = 9.5.$$
In the general case where there are $n$ balls of each color (so $2n$ in total), there are $\binom{2n}{n}$ equiprobable elementary outcomes. Let the random variable $X$ denote the number of balls needed to obtain the final white ball, and $Y$ denote the number of balls needed to obtain the first black ball.
Suppose we observe $(X,Y) = (x,y)$. This means the last occurrence of a white ball is in position $x$ and the first occurrence of a black ball is in position $y$. This in turn implies that all of the balls prior to position $y$ must be white, and all of the balls following position $x$ must be black. The order and color of all of the balls in between positions $x$ and $y$ may be freely chosen. So for example, such an outcome might look like this:
$$(w, \ldots, w, \color{blue}{b}, \ldots, \color{red}{w}, b, \ldots, b)$$
where again I have colored the relevant positions of the white and black balls. Notice however, there is a unique exception, which is the outcome
$$(w, \ldots, \color{red}{w}, \color{blue}{b}, \ldots, b),$$
corresponding to the outcome in which the first $n$ balls are white and the last $n$ balls are black. This is the only possible way to have $X < Y$, since if any $b$ precedes the final $w$, then we would have $Y < X$.
So exception aside, we may assume $x > y$ and for each permissible outcome $(X,Y) = (x,y)$, the color of the balls in positions $(1, \ldots, y)$ and $(x, \ldots, 2n)$ are fixed. The colors in positions $(y+1, \ldots, x-1)$ are free within the constraint that there are $n-y$ white balls to choose from, and $x-n-1$ black balls to choose from. So this means there are $$\binom{x-y-1}{n-y}$$ outcomes where $X = x$ and $Y = y$.
Hence we have $$\operatorname{E}[XY^2] = \binom{2n}{n}^{-1} \left( n(n+1)^2 + \sum_{y=1}^n \sum_{x=n+1}^{2n} xy^2 \binom{x-y-1}{n-y} \right),$$ where the additional $n(n+1)^2$ term arises from the single exceptional case where $x = n$ and $y = n+1$ described above.
The evaluation of this sum for $n = 2$ yields $9.5$ as explicitly enumerated above. For $n = 3$, we get $\operatorname{E}[XY^2] = 18.3$, which can also be checked by enumeration.
A closed form for general $n$ is
$$\operatorname{E}[XY^2] = 12n - 28 + \frac{1}{n+1} + \frac{48}{n+2} + \frac{(2n+3)}{2 \binom{2n-1}{n}}.$$ However, as the evaluation of the above double sum is beyond the scope of this question, I have not included it.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is there a closed form expression for $\sum_{1\le i_1 \le i_2 \le \dots \le i_k \le n} i_1 i_2 \dots i_k$? These sums showed up in a probability problem I was working on. They're not quite the Stirling numbers of the first kind since it's possible to have e.g. $i_1 = i_2$. Denoting the sum by $(k\mid n)$ we have the recurrence relation
$(k\mid n) = n(k-1\mid n) + (k\mid n-1)$
I thought maybe the average term of the sum would be similar to the average product over a random list of $k$ numbers from $1$ to $n$. But it seems to always be a few times greater.
I'm beginning to despair of there being a closed form expression for $(k\mid n)$ in terms of factorials and powers. Does anyone know how to analyze the sum further and perhaps find a good approximation?
|
We get from first principles the generating function where we mark
multisets containing some number of instances of $q$ giving the factor
being contributed and the exponent of $z$ how often it occurs:
$$[z^k] \prod_{q=1}^n (1+qz+q^2z^2+q^3z^3+\cdots)
= [z^k] \prod_{q=1}^n \frac{1}{1-qz}
\\ = \frac{1}{n!} [z^k] \prod_{q=1}^n \frac{1}{1/q-z}
= (-1)^n \frac{1}{n!} [z^k] \prod_{q=1}^n \frac{1}{z-1/q}.$$
Now using partial fractions by residues we have
$$(-1)^n \frac{1}{n!} [z^k]
\sum_{p=1}^n \frac{1}{z-1/p}
\prod_{q=1}^{p-1} \frac{1}{1/p-1/q}
\prod_{q=p+1}^n \frac{1}{1/p-1/q}
\\ = (-1)^n \frac{1}{n!} [z^k]
\sum_{p=1}^n \frac{1}{z-1/p}
\prod_{q=1}^{p-1} \frac{pq}{q-p}
\prod_{q=p+1}^n \frac{pq}{q-p}
\\ = (-1)^n \frac{1}{n!} [z^k]
\sum_{p=1}^n \frac{p^{n-1}}{z-1/p}
\prod_{q=1}^{p-1} \frac{q}{q-p}
\prod_{q=p+1}^n \frac{q}{q-p}
\\ = (-1)^n [z^k]
\sum_{p=1}^n \frac{p^{n-2}}{z-1/p}
\prod_{q=1}^{p-1} \frac{1}{q-p}
\prod_{q=p+1}^n \frac{1}{q-p}
\\ = (-1)^n [z^k]
\sum_{p=1}^n \frac{p^{n-2}}{z-1/p}
\frac{(-1)^{p-1}}{(p-1)!} \frac{1}{(n-p)!}
\\ = (-1)^n \frac{1}{n!} [z^k]
\sum_{p=1}^n \frac{p^{n-1}}{z-1/p} {n\choose p} (-1)^{p-1}
\\ = (-1)^n \frac{1}{n!} [z^k]
\sum_{p=1}^n \frac{p^{n}}{pz-1} {n\choose p} (-1)^{p-1}
\\ = (-1)^n \frac{1}{n!} [z^k]
\sum_{p=1}^n \frac{p^{n}}{1-pz} {n\choose p} (-1)^p
\\ = (-1)^n \frac{1}{n!}
\sum_{p=1}^n {n\choose p} (-1)^p p^{n+k}.$$
We may lower $p$ to zero and obtain
$$\frac{1}{n!}
\sum_{p=0}^n {n\choose p} (-1)^{n-p} p^{n+k}
= {n+k \brace n}.$$
|
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"url": "https://math.stackexchange.com/questions/4544787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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|
Angle Between Two Vectors Discrepancy Here is something that is bugging me: Consider the vectors
$$ \mathbf{v} = 2\mathbf{i} - 4\mathbf{j} + 4\mathbf{k} \qquad \text{ and } \qquad \mathbf{w} = 2\mathbf{j} - \mathbf{k}. $$
We have two different formulas for the angle $\theta$ between these vectors:
$$ \theta = \cos^{-1}\left(\frac{\mathbf{v} \cdot \mathbf{w}}{|\mathbf{v}||\mathbf{w}|} \right) \qquad \text{ and } \qquad \theta = \sin^{-1}\left(\frac{|\mathbf{v} \times \mathbf{w}|}{|\mathbf{v} || \mathbf{w}|} \right) .$$
Well:
$$|\mathbf{v}| = 6, \qquad |\mathbf{w}| = \sqrt{5}, \qquad \mathbf{v} \cdot \mathbf{w} = -12, \qquad \text{ and } |\mathbf{v} \times \mathbf{w}| = 6. $$
But:
$$\cos^{-1}\left(-\frac{2}{\sqrt{5}} \right) \ne \sin^{-1}\left( \frac{1}{\sqrt{5}} \right). $$
So, what gives?
|
The angle between two vectors should always be a positive number between $0$ and $\pi$ inclusive, so if the cosine of the angle is a negative number, the value should be between $\pi/2$ and $\pi$ inclusive.
Your numbers $-2/\sqrt5$ and $1/\sqrt5$ clearly obey $\cos^2\theta+\sin^2\theta=1$ and thus form a valid sine and cosine pair, so the value for the $\sin^{-1}$ should be $\pi-\theta$.
Physically, the cross product gives the amount to which vector $\mathbf{w}$ is perpendicular to $\mathbf{v}$. This is ambiguous in that four different vectors, $\mathbf{w}$, $-\mathbf{w}$, and two other vectors of the same length but the direction $\pi-\theta$ with respect to $v$, $\mathbf{x}$ or $-\mathbf{x}$, could all have identical lengths, and have the same absolute value of cross product as $\mathbf{v}\times\mathbf{w}$. See image.
|
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|
easy way to calculate the limit $\lim_{x \to 0 } \frac{\cos{x}- (\cos{x})^{\cos{x}}}{1-\cos{x}+\log{\cos{x}}}$ I have been trying to use L'Hôpital over this, but its getting too long, is there a short and elegant solution for this?
The Limit approaches 2 according to wolfram.
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How to solve
$$
\begin{align*}
\lim_{{x} \to {0}} f(x) &= \lim_{{x} \to {0}} \frac{\cos(x) - \cos(x)^{\cos(x)}}{1 - \cos(x) + \log(\cos(x))}\\
\lim_{{x} \to {0}} f(x) &= \frac{\cos(0) - \cos(0)^{\cos(0)}}{1 - \cos(0) + \log(\cos(0))}\\
\lim_{{x} \to {0}} f(x) &= \frac{1 - 1^{1}}{1 - 1 + \log(1)}\\
\lim_{{x} \to {0}} f(x) &= \frac{0}{0} \quad\mid\quad \text{we have to use L'hospital rule}\\
\\
&\text{if we use the rule of L'hospital one time, we'll get the same situation aigan and aigan aka let's take the rule of L'hospital three
times}\\
\\
\lim_{{x} \to {0}} f(x) &= \lim_{{x} \to {0}} \frac{\cos(x) - \cos(x)^{\cos(x)}}{1 - \cos(x) + \log(\cos(x))}\\
\lim_{{x} \to {0}} f(x) &= \lim_{{x} \to {0}} \frac{\frac{\mathrm{d^{3}}}{\mathrm{d}^{3}x} ~ (\cos(x) - \cos(x)^{\cos(x)})}{\frac{\mathrm{d^{3}}}{\mathrm{d}^{3}x} ~ (1 - \cos(x) + \log(\cos(x)))}\\
\\
\lim_{{x} \to {0}} f(x) &= \lim_{{x} \to {0}} \frac{\frac{\mathrm{d}}{\mathrm{d}x} ~ (\cos(x) - \cos(x)^{\cos(x)})}{\frac{\mathrm{d}}{\mathrm{d}x} ~ (1 - \cos(x) + \log(\cos(x)))}\\
\lim_{{x} \to {0}} f(x) &= \lim_{{x} \to {0}} \frac{\cos(x) \cdot (\sin(x) + \log(\cos(x)) \cdot \sin(x))}{\cos(x) - \tan(x)}\\
\lim_{{x} \to {0}} f(x) &= \lim_{{x} \to {0}} \frac{\sin(x) - \cos(x)^{\cos(x)} \cdot -(\sin(x) + \log(\cos(x)) \cdot \sin(x))}{\cos(x) - \frac{\sin(x)}{\cos(x)}}\\
\lim_{{x} \to {0}} f(x) &= \lim_{{x} \to {0}} \frac{\cos(x) \cdot (-1 + \cos(x)^{\cos(x)} + \cos(x)^{\cos(x)} \cdot \log(\cos(x)))}{\cos(x) - 1}\\
\lim_{{x} \to {0}} f(x) &= \lim_{{x} \to {0}} \cos(x) \cdot \lim_{{x} \to {0}} \frac{-1 + \cos(x)^{\cos(x)} + \cos(x)^{\cos(x)} \cdot \log(\cos(x))}{\cos(x) - 1}\\
\lim_{{x} \to {0}} f(x) &= \cos(0) \cdot \lim_{{x} \to {0}} \frac{-1 + \cos(x)^{\cos(x)} + \cos(x)^{\cos(x)} \cdot \log(\cos(x))}{\cos(x) - 1}\\
\lim_{{x} \to {0}} f(x) &= 1 \cdot \lim_{{x} \to {0}} \frac{-1 + \cos(x)^{\cos(x)} + \cos(x)^{\cos(x)} \cdot \log(\cos(x))}{\cos(x) - 1}\\
\lim_{{x} \to {0}} f(x) &= \lim_{{x} \to {0}} \frac{-1 + \cos(x)^{\cos(x)} + \cos(x)^{\cos(x)} \cdot \log(\cos(x))}{\cos(x) - 1}\\
\lim_{{x} \to {0}} f(x) &= \lim_{{x} \to {0}} \frac{-1 + \cos(x)^{\cos(x)}}{\cos(x) - 1} + \lim_{{x} \to {0}} \frac{\cos(x)^{\cos(x)} \cdot \log(\cos(x))}{\cos(x) - 1} \quad\mid\quad \text{since both the numerator and denominator appraoch to 0 } \frac{-1 + \cos(x)^{\cos(x)}}{\cos(x) - 1} \text{ is our canditate}\\
\\
\lim_{{x} \to {0}} f(x) &= \lim_{{x} \to {0}} \frac{\frac{\mathrm{d}}{\mathrm{d}x} ~ (-1 + \cos(x)^{\cos(x)})}{\frac{\mathrm{d}}{\mathrm{d}x} ~ (\cos(x) - 1)}\\
\lim_{{x} \to {0}} f(x) &= \lim_{{x} \to {0}} \cos(x)^{\cos(x)} \cdot (\log(\cos(x)) + 1) + \lim_{{x} \to {0}} \frac{\cos(x)^{\cos(x)} \cdot \log(\cos(x))}{\cos(x) - 1}\\
\lim_{{x} \to {0}} f(x) &= \cos(0)^{\cos(0)} \cdot (\log(\cos(0)) + 1) + \lim_{{x} \to {0}} \frac{\cos(x)^{\cos(x)} \cdot \log(\cos(x))}{\cos(x) - 1}\\
\lim_{{x} \to {0}} f(x) &= 1^1 \cdot (\log(1) + 1) + \lim_{{x} \to {0}} \frac{\cos(x)^{\cos(x)} \cdot \log(\cos(x))}{\cos(x) - 1}\\
\lim_{{x} \to {0}} f(x) &= 1 \cdot (0 + 1) + \lim_{{x} \to {0}} \frac{\cos(x)^{\cos(x)} \cdot \log(\cos(x))}{\cos(x) - 1}\\
\lim_{{x} \to {0}} f(x) &= 1 + \lim_{{x} \to {0}} \frac{\cos(x)^{\cos(x)} \cdot \log(\cos(x))}{\cos(x) - 1}\\
\lim_{{x} \to {0}} f(x) &= 1 + \lim_{{x} \to {0}} \cos(x)^{\cos(x)} \cdot \lim_{{x} \to {0}} \frac{\log(\cos(x))}{\cos(x) - 1}\\
\lim_{{x} \to {0}} f(x) &= 1 + 1^{\cos(x)} \cdot 1 \frac{\log(\cos(x))}{\cos(x) - 1}\\
\lim_{{x} \to {0}} f(x) &= 1 + 1^{\cos(x)} \frac{\log(\cos(x))}{\cos(x) - 1}\\
\\
\lim_{{x} \to {0}} f(x) &= \lim_{{x} \to {0}} \frac{\frac{\mathrm{d}}{\mathrm{d}x} ~ \log(\cos(x)))}{\frac{\mathrm{d}}{\mathrm{d}x} ~ (\cos(x) - 1)} + 1\\
\lim_{{x} \to {0}} f(x) &= \lim_{{x} \to {0}} \frac{-\tan(x)}{-\sin(x)} + 1\\
\lim_{{x} \to {0}} f(x) &= \lim_{{x} \to {0}} \frac{\tan(x)}{\sin(x)} + 1\\
\lim_{{x} \to {0}} f(x) &= \lim_{{x} \to {0}} \frac{\frac{\sin(x)}{\cos(x)}}{\sin(x)} + 1\\
\lim_{{x} \to {0}} f(x) &= \lim_{{x} \to {0}} \frac{1}{\cos(x)} + 1\\
\lim_{{x} \to {0}} f(x) &= \frac{1}{\cos(0)} + 1\\
\lim_{{x} \to {0}} f(x) &= \frac{1}{1} + 1\\
\lim_{{x} \to {0}} f(x) &= 1 + 1\\
\lim_{{x} \to {0}} f(x) &= 2\\
\end{align*}
$$
Are there still questions?^^
|
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|
Halmos Finite-Dimensional Vector Spaces Sec. 36 Ex. 5 on invertible transformations The exercise
If $A, B, C, D$ are linear transformations (all on the same vector space), and if both $A+B$ and $A-B$ are invertible, then there exist linear transformations $X$ and $Y$ such that
$$AX + BY = C$$
$$BX + AY = D$$
What I tried
I found these values for $X$ and $Y$, which I believe make the equations work. $Y$ is identical to $X$ except for a minus sign in the middle instead of a plus.
$$X = \frac{1}{2} (A+B)^{-1} (C+D) + \frac{1}{2} (A-B)^{-1} (C-D)$$
$$Y = \frac{1}{2} (A+B)^{-1} (C+D) - \frac{1}{2} (A-B)^{-1} (C-D)$$
However, I'm concerned, first, that these expressions involve $\frac{1}{2}$ (i.e. $(1+1)^{-1}$) so they do not work if the field has characteristic 2. Second, they're just a bit unwieldy. Is there a better solution?
|
We can apply Gaussian elimination reasonably well. The equations:
$$AX + BY = C \tag{1}$$
$$BX + AY = D \tag{2}$$
are equivalent to:
$$AX + BY = C \tag{1}$$
$$(A + B)X + (A + B)Y = C + D. \tag{3}$$
As you can see, $(3) = (1) + (2)$. We can recover $(2)$ by computing $(3) - (1)$. As $A + B$ is invertible, $(3)$ is equivalent to
$$X + Y = (A + B)^{-1}(C + D), \tag{4}$$
which implies (not necessarily equivalent to)
$$AX + AY = A(A + B)^{-1}(C + D). \tag{5}$$
Together, $(5)$ and $(1)$ imply $(5) - (1)$:
$$(A - B)Y = A(A + B)^{-1}(C + D) - C, \tag{6}$$
which is equivalent to
$$Y = (A - B)^{-1}(A(A + B)^{-1}(C + D) - C). \tag{8}$$
No (scalar) division is necessary, so there's no assumption about the characteristic of the field! If we instead take $B(4) - (1)$, we get the only possible solution for $X$:
$$X = (B - A)^{-1}(B(A + B)^{-1}(C + D) - C). \tag{9}$$
We now need to check these formulas. Substituting $(8)$ and $(9)$ into the left hand side of $(1)$,
\begin{align*}
AX + BY &= A(B - A)^{-1}(B(A + B)^{-1}(C + D) - C) + B(A - B)^{-1}(A(A + B)^{-1}(C + D) - C) \\
&= (B(A - B)^{-1}A - A(A - B)^{-1}B)(A + B)^{-1}(C + D) + (A - B)(A - B)^{-1}C \\
&= (B(A - B)^{-1}A - (A - B)(A - B)^{-1}B - B(A - B)^{-1}B)(A + B)^{-1}(C + D) + C \\
&= (B(A - B)^{-1}(A - B) - B)(A + B)^{-1}(C + D) + C \\
&= 0(A + B)^{-1}(C + D) + C = C.
\end{align*}
Substituting $(8)$ and $(9)$ into the left hand side of $(2)$,
\begin{align*}
AY + BX &= A(A - B)^{-1}(A(A + B)^{-1}(C + D) - C) + B(B - A)^{-1}(B(A + B)^{-1}(C + D) - C) \\
&= (A(A - B)^{-1}A - B(A - B)^{-1}B)(A + B)^{-1}(C + D) - C \\
&= (A(A - B)^{-1}A + (A - B)(A - B)^{-1}B - A(A - B)^{-1}B)(A + B)^{-1}(C + D) - C \\
&= (A(A - B)^{-1}(A - B) + B)(A + B)^{-1}(C + D) - C \\
&= (A + B)(A + B)^{-1}(C + D) - C \\
&= C + D - C = D.
\end{align*}
If you follow/believe my argument so far, these are the only possible solutions to the equation. Thus, my solution and your solution must agree when $\operatorname{char} F \neq 2$, though it seems to be painful to show directly. But, my solution is well-defined without having to divide scalars, so it works over a general field.
|
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|
Laurent series expansion of $\frac{1}{z(z-2)^2}$ at $z=2$ Im struggling a bit with Laurent series expansions, and I am stuck on the following exercise:
Find a Laurent series expansion with center $z=2$ of $f(z)=\frac{1}{z(z-2)^2}$
My solution so far is not a lot:
I've found that there will be one Laurent series at the punctured disk $D(2,2)$ and one $A=(2,2,\infty)$
Starting with the punctured disk $D(2,2)$ I know I want the polynomial to have the form $\frac{1}{1-c*z}$ since we are expanding in our "inner" area.
The part im stuck on is how I should get $\frac{1}{z(z-2)^2}$ = $\frac{1}{z}\frac{1} {(z-2)^2}$ to $\frac{1}{1-c*z}$ to able to write as a series.
Can anyone help me out please?
Thanks!
|
Using partial fraction expansion, we can write
$$\begin{align}
\frac1{z(z-2)^2}=\frac1{4z}-\frac1{4(z-2)}+\frac1{2(z-2)^2}
\end{align}$$
To obtain the Laurent series for $|z-2|<2$, we write
$$\begin{align}
\frac1{z(z-2)^2}&=\frac1{4(z-2+2)}-\frac1{4(z-2)}+\frac1{2(z-2)^2}\\\\
&=\frac18 \frac1{1+\frac{z-2}2}-\frac1{4(z-2)}+\frac1{2(z-2)^2}\\\\
&=\frac1{2(z-2)^2}-\frac1{4(z-2)}+\frac18\sum_{n=0}^\infty (-1)^n\left(\frac{z-2}{2}\right)^n
\end{align}$$
And to obtain the Laurent series for $|z-2|>2$ we write
$$\begin{align}
\frac1{z(z-2)^2}&=\frac1{4(z-2+2)}-\frac1{4(z-2)}+\frac1{2(z-2)^2}\\\\
&=\frac1{4(z-2)} \frac1{1+\frac2{z-2}}-\frac1{4(z-2)}+\frac1{2(z-2)^2}\\\\
&=\frac1{2(z-2)^2}-\frac1{4(z-2)}+\frac1{8}\sum_{n=1}^\infty (-1)^{n-1}\left(\frac{2}{z-2}\right)^n\\\\
&=\frac1{8}\sum_{n=3}^\infty (-1)^{n-1}\left(\frac{2}{z-2}\right)^n
\end{align}$$
|
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|
Find the total number of roots of $(x^2+x+1)^2+2=(x^2+x+1)(x^2-2x-6)$, belonging to $(-2,4)$.
Find the total number of roots of $(x^2+x+1)^2+2=(x^2+x+1)(x^2-2x-6)$, belonging to $(-2,4)$.
My Attempt:
On rearranging, I get, $(x^2+x+1)(3x+7)+2=0$
Or, $3x^3+10x^2+10x+9=0$
Derivative of the cubic is $9x^2+20x+10$
It is zero at minus zero point something and minus one point something.
So, even at local minima, the cubic is positive. It means it would cross x-axis only once.
At $x=-2$, cubic is positive and at $-3$, it is negative.
It means the only root is minus two point something.
Is there any other way to solve this question? Something that doesn't involve calculator? Or maybe something that doesn't involve calculus?
|
Since $ \ x^2 + x + 1 \ = \ \left(x + \frac12 \right)^2 + \frac34 \ > \ 0 \ $ for all real numbers, we might ask under what circumstances $ \ (x^2+x+1)^2+2 \ > \ (x^2+x+1)(x^2-2x-6) \ \ , \ $ since the product on the right side can plainly take on negative values. Dividing out a factor of $ \ (x^2+x+1) \ $ produces
$$ x^2 \ + \ x \ + \ 1 \ + \ \frac{2}{x^2+x+1} \ \ > \ \ x^2 \ - \ 2x \ - \ 6 \ \ \Rightarrow \ \ \frac{2}{x^2+x+1} \ \ > \ \ - 3x \ - \ 7 \ \ . $$
We will not need to solve for the exact interval over which this inequality holds. The left side of the inequality remains positive for all real numbers, while the linear function on the right side is positive only for $ \ x \ < \ -\frac73 \ \ . $ Hence, this inequality is certainly true for $ \ x \ \ge \ -\frac73 \ \ . \ $ This tells us that the intersection of the curves representing the functions on the two sides of the original equation occurs at a value $ \ c \ < \ -\frac73 \ \ $ (in fact, the intersection is at $ \ c \ \approx \ -2.477 \ ) \ , \ $ so there can be no roots of the original equation in the interval $ \ (-2 \ , \ 4) \ \ . \ $
|
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|
Solve the system $ \left\lbrace \begin{array}{ccc} \sqrt{1-x} + \sqrt{1-y} &=& \frac{y-x}{2} \\ x+y &=& 2xy \end{array}\right. $ I have to solve the following system:
$$
\left\lbrace \begin{array}{ccc}
\sqrt{1-x} + \sqrt{1-y} &=& \frac{y-x}{2} \\
x+y &=& 2xy
\end{array}\right.
$$
I've showed that it is equivalent to
$$
\left\lbrace \begin{array}{ccc}
\sqrt{1-x} - \sqrt{1-y} &=& 2 \\
x+y &=& 2xy
\end{array}\right. \text{ or } x=y= 1
$$
And i can't conclude nothing frome this system:
$$
\left\lbrace \begin{array}{ccc}
\sqrt{1-x} - \sqrt{1-y} &=& 2 \\
x+y &=& 2xy
\end{array}\right.
$$
|
$$
\left\lbrace \begin{array}{ccc}
\sqrt{1-x} - \sqrt{1-y} &=& 2 &(1)\\
x+y &=& 2xy &(2)
\end{array}\right.
$$
From 1, get squared both sides two times, replace all (x + y) by 2xy, you will recieve a equation of (xy).
Find the appropriate of (xy) and then solve the (2) equation.
|
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|
Combinatoric identity from a random walking problem I tried to prove the following equation and got stuck:
$$\sum_{k=1}^{n-r+1}\left[\binom{2k-1}{k}-\binom{2k-1}{k+1}\right]\binom{2n-2k+1}{n-k+r}=\binom{2n+1}{n-r}.$$
For whom are interested in its background:
Let $S_n$ be a symmetric random walking on the line. We call it a sign reverse at time $n$ if $S_{n-1}S_{n+1}<0.$ Let $X_n$ be the total number of reverses by the time $2n+1$. Prove that $$\mathbb{P}[X_n=r]=2\mathbb{P}[S_{2n+1}=2r+1]$$
holds for each $r\geq 0.$
The case that $r=0$ is trivial, and I use induction to prove the statement. The rest process was a bit tedious (confident about its correctness) and I finally arrived at the beginning equation.
THX :)
As @Jean Marie commented, the induction starts by conditioning the first time reaching -1, which is exactly the same as meeting the diagonal in a Catalan path, noted that $\mathbb{P}[X_n=r\big\vert S_1=1]=\mathbb{P}[X_n=r].$
|
We seek to verify that
$$\sum_{k=1}^{n-r+1}
\left[{2k-1\choose k} - {2k-1\choose k+1}\right]
{2n-2k+1\choose n-k+r} = {2n+1\choose n-r}.$$
Note that for the square bracket term with $k\ge 1$ we get a Catalan
number, so the LHS becomes
$$\sum_{k=1}^{n-r+1} \frac{1}{k+1} {2k\choose k}
{2n-2k+1\choose n-r+1-k}.$$
This becomes
$$[z^{n-r+1}] (1+z)^{2n+1}
\sum_{k\ge 1} \frac{1}{k+1} {2k\choose k} \frac{z^k}{(1+z)^{2k}}.$$
Here we have extended to infinity due to the coefficient extractor.
Continuing with the OGF of the Catalan numbers,
$$- {2n+1\choose n-r+1} +
[z^{n-r+1}] (1+z)^{2n+1}
\frac{1-\sqrt{1-4z/(1+z)^2}}{2z/(1+z)^2}.$$
The square root term becomes
$$[z^{n-r+1}] (1+z)^{2n+1}
\frac{1+z-\sqrt{(1-z)^2}}{2z/(1+z)}
\\ = [z^{n-r+1}] (1+z)^{2n+1}
\frac{1}{1/(1+z)}
= [z^{n-r+1}] (1+z)^{2n+2}
= {2n+2\choose n-r+1}.$$
Collecting everything we have
$${2n+2\choose n-r+1} - {2n+1\choose n-r+1}
= {2n+1\choose n-r}
\left[\frac{2n+2}{n-r+1} - \frac{n+r+1}{n-r+1}\right]
\\ = {2n+1\choose n-r}$$
as claimed. This last one was Pascal's rule.
|
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|
How can we generalize factorisation of $(a+b)^n-(a^n+b^n)$
How can we generalize factorisation of
$$(a+b)^n-(a^n+b^n)\,?$$
where $n$ is an odd positive integer.
I found the following cases:
$$(a+b)^3-a^3-b^3=3ab(a+b)$$
$$(a+b)^5-a^5-b^5=5 a b (a + b) (a^2 + a b + b^2)$$
$$(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$$
Exapansions and factoring seem terrible. Maybe there is a good way for generalization.
The question is, can we generalize these factorisations?
$$(a+b)^n-(a^n
+b^n)$$
for all odd integer $n$?
|
For any case obviously you have $a$ and $b$ in each term since you've removed both the $a$-only and the $b$-only terms from $(a+b)^n$. Also $a=-b$ is always a root, so you'll always get an $(a+b)$ term. So $ab(a+b)$ is a given for any odd $n$.
The roots of $a^2+ab+b^2$ are $a={1\over2}(-b\pm{\sqrt{b^2-4b^2}})=-b\omega,-b\bar\omega$ where $\omega$ is a complex number such that $\omega^3=1$, so if $n=5$ then if $a=-\omega b$ then $$a^5+b^5=b^5(1-\omega^5)=b^5(1-\omega^2)$$ and $$(a+b)^5=b^5(1-\omega)^5=b^5{\bar\omega}^5=b^5\bar\omega^2$$ since $$1-\omega=\bar\omega,\tag{1}$$
and
$$\omega^2=-\bar\omega.\tag{2}$$
You'll get a similar result if $n$ is coprime with $3$, but you don't get that root for $9$. So for example if $n=11$ you get
$$
11ab(a+b)(a^2+ab+b^2)(a^6+3a^5b+7a^4b^2+9a^3b^3+7a^2b^4+3ab^5+b^6)$$
and if $n=13$ you get
$$
13ab(a+b)(a^2+ab+b^2)^2(a^6+3a^5b+8a^4b^2+11a^3b^3+8a^2b^4+3ab^5+b^6)
$$
(obtained using Wolfram Alpha).
In general, using (1),
$$(b-\omega b)^n=b^n(1-\omega)^n=b^n(\bar\omega)^r
$$
where $n=3q+r$, $0<r<3$, and $q,r\in\mathbb{N}$, and
$$
b^n+(-\omega b)^n=b^n(1-\omega^r)=b^n(\bar\omega)^r
$$
if $r=1,2$, so
$$
(b-\omega b)^n-(b^n+(-\omega b)^n)=0
$$
and $a=-\omega b$ is a solution if $r=1,2$ and hence $a^2+ab+b^2$ is a factor.
Since the complex conjugate of a complex root of a real equation is always a root too, then obviously $a=-\bar\omega b$ is also a root.
|
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|
Evaluating $\int{\sqrt{x^2+4x+13}\, dx}$ I was trying to calculate with $\sinh$ :
$$\begin{align}I&=\int{\sqrt{x^2+4x+13}\, dx}\\&=\int{\sqrt{(x+2)^2+9}\, dx}\end{align}$$
Now with $x+2=3\sinh(u)$, $dx=3\cosh(u)\,du$
$$\begin{align}I&=3\int{\left(\sqrt{9\sinh^2(u)+9}\, \right)\cosh(u)\, du}\\&=9\int{\cosh(u)\sqrt{\sinh^2(u)+1}\, du}\\&=9\int{\cosh^2(u)\,du}\\&=\frac{9}{2}\int{\cosh(2u)+1\, du}\\&=\frac{9}{2}\left(\frac{1}{2}\sinh(2u)+u\right)+C\end{align}$$
How can I rewrite the $\frac{1}{2}\sinh(2u)$ in terms of $x$? I tried with the double angle formula, but this seemingly just complicates things by introducing $\cosh(u)$ also.
|
Since $\frac12\sinh(2u)=\sinh(u)\cosh(u)=\sinh(u)\sqrt{1+\sinh^2(u)}$, you have
\begin{align}
\frac12\sinh(u)&=\frac{x+2}3\sqrt{1+\left(\frac{x+2}3\right)^2}\\
&=\frac19(x+2)\sqrt{x^2+4x+13}.
\end{align}
|
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|
Prove: $\int_{-1}^1 x^n P_n(x) dx = \frac{2^{n+1}n!^2}{(2n+1)!}$ I'm trying to prove:
$$\int_{-1}^1 x^n P_n(x) dx = \frac{2^{n+1}n!^2}{(2n+1)!}$$
My attempt consisted in applying the Rodrigues formula:
$$\int_{-1}^1 x^n P_n(x)dx = \dfrac{1}{2^n n!} \int_{-1}^1 x^n \dfrac{d^n}{dx^n}(x^2-1)^ndx$$
and integrating by parts n times:
$$\int_{-1}^1 x^nP_n(x)dx=\dfrac{1}{2^n n!}\left(x^n \dfrac{d^{n-1}}{dx^{n-1}}(x^2-1)^n \bigg|_{-1}^1 + (-1)^{n} n!\int_{-1}^1 (x^2-1)^ndx\right)$$
The first term vanishes at the extremes. And the remaining integral I solved through substitution:
$$t = x^{2} \rightarrow dx = \dfrac{dt}{2\sqrt{t}}$$
$$\dfrac{1}{2}\int^{1}_{0}(t - 1)^{n}\dfrac{dt}{2\sqrt{t}} = \dfrac{1}{2}\int^{1}_{0}(t - 1)^{n} t^{-1/2} dt$$
Using the beta function:
$$(-1)^{n}\dfrac{1}{2}\int^{1}_{0}(1 - t)^{n} t^{-1/2} dt =\dfrac{(-1)^{n}}{2}\beta(1/2, n + 1) =\dfrac{(-1)^{n}}{2}\dfrac{\Gamma(1/2)\Gamma(n + 1)}{\Gamma(n + 3/2)}$$
I rewrote the gamma functions as
$$\Gamma(1/2) = \sqrt{\pi}$$
$$\Gamma(n + 1) = n!$$
and $$\Gamma(n + 3/2) = \dfrac{(n+1)(2n)!\sqrt{\pi}}{2^{2n}n!}$$
Finally I reached:
$$(-1)^{n} n!\int_{-1}^1 x^{n-1} \dfrac{d^{n-1}}{dx^{n-1}}(x^2-1)^ndx = \dfrac{(-1)^{n}n!2}{2^{n}n!}\dfrac{(-1)^{n}}{2}\dfrac{\sqrt{\pi}2^{2n}n!n!}{(n+1)(2n)!\sqrt{\pi}} = \dfrac{2^{n}n!n!}{(n+1)(2n)!}$$
I tried simplifying furter, but I didn't get anywhere
|
A method based upon the series for $P_{n}(x)$ is seen by the following.
Using
$$ P_{n}(x) = \sum_{k=0}^{\lfloor{n/2}\rfloor} \frac{(-1)^k \, \left(\frac{1}{2}\right)_{n-k} \, (2 x)^{n-2 k}}{k! \, (n-2 k)!} $$
then
\begin{align}
I &= \int_{-1}^{1} x^n \, P_{n}(x) \, dx \\
&= \sum_{k} a_{k}^{n} \, 2^{n-2 k} \, \int_{-1}^{1} x^{2 (n-k)} \, dx \\
&= \sum_{k} a_{k}^{n} \, 2^{n-2k} \frac{1}{2n -2 k + 1} \, \left( 1^{2n -2k +1} - (-1)^{2n -2k +1} \right) \\
&= \sum_{k} \frac{(-1)^k \, \left(\frac{1}{2}\right)_{n-k} \, 2^{n -2k +1}}{k! \, (n -2k)! \, (2n -2k +1) } \\
&= \frac{2^{n+1} \, \left(\frac{1}{2}\right)_{n}}{n! \, (2n+1)} \, {}_{3}F_{2}\left(-\frac{n}{2}, \frac{1-n}{2}, -n-\frac{1}{2}; \frac{1}{2}-n, \frac{1}{2}-n; 1 \right) \\
&= \frac{\binom{2n}{n} \, \left(\frac{1}{2}\right)_{n}}{2^{n-1} \, \left(\frac{3}{2}\right)_{n}} \, {}_{3}F_{2}\left(-\frac{n}{2}, \frac{1-n}{2}, -n-\frac{1}{2}; \frac{1}{2}-n, \frac{1}{2}-n; 1 \right) \\
&= \frac{\binom{2n}{n} \, \left(\frac{1}{2}\right)_{n}}{2^{n-1} \, \left(\frac{3}{2}\right)_{n}} \, \frac{n!}{\binom{2n}{n} \, \left(\frac{1}{2}\right)_{n}} \\
&= \frac{n!}{2^{n-1} \, \left(\frac{3}{2}\right)_{n}} \\
&= \frac{2^{n+1}}{\binom{2n}{n} \, (2n+1)}.
\end{align}
|
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|
Find the volume of a bounded region rotated about the $x$-axis Question
A region is bounded by the $x$-axis, $y=x^2-9$ and $y=x^2-4$, where the region looks similar to a u-shape. The region is then rotated about the $x$-axis. Find the volume of the solid formed.
My Working
We can see that $y=x^2-9$ intersects the $x$-axis at $-3$ and $3$, while $y=x^2-4$ intersects the $x$-axis at $-2$ and $2$. The volume should be
\begin{align}
V&=\pi\left[\int^3_{-3}(x^2-9)^2dx-\int^2_{-2}(x^2-4)^2dx\right]\\
&=2\pi\left[\int^3_0 (x^4-18x^2+81) dx-\int^2_0(x^4-8x^2+16) dx\right]\\
&=2\pi\left\{\left[\frac{x^5}{5}-6x^3+81x\right]^3_0-\left[\frac{x^5}{5}-\frac{8x^3}{3}+16x\right]^2_0\right\}\\
&=2\pi\left(\frac{243}{5}-162+243-\frac{32}{5}+\frac{64}{3}-32\right)\\
&=2\pi\left(49+\frac{209}{5}+\frac{64}{3}\right)\\
&=2\pi\cdot\frac{735+627+320}{15}\\
&=2\pi\cdot\frac{1682}{15}=\frac{3364}{15}\pi
\end{align}
But my teacher said the answer was $$\frac{3376}{15}\pi$$
Could anyone please point out where I got the answer wrong? Thank you!
|
After looking at CyclotomicField's comment, it seems that the teacher's answer is right and that I have just made a silly mistake, as shown below:
\begin{align}
V&=2\pi\left(49+\frac{211}{5}+\frac{64}{3}\right)\\
&=2\pi\cdot\frac{735+633+320}{15}\\
&=2\pi\cdot\frac{1688}{15}=\frac{3376}{15}\pi
\end{align}
|
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|
Find minimum n :$ \prod_{k=1}^n \frac{\arcsin\left( \frac{9k+2}{\sqrt{27k^3+54k^2+36k+8}}\right)}{\arctan \left( \frac{1}{\sqrt{3k+1}} \right)}>2000$ We've got to find the minimum value of n for which
$$ \prod_{k=1}^n \frac{\arcsin\left( \frac{9k+2}{\sqrt{27k^3+54k^2+36k+8}}\right)}{\arctan \left( \frac{1}{\sqrt{3k+1}} \right)}>2000$$
So I just did some simplifications hoping I'd see something:
$$\prod_{k=1}^n \frac{\arcsin\left( \frac{9k+2}{(3k+2)^{\frac{3}{2}}}\right)}{\arcsin \left( \frac{1}{\sqrt{3k+2}} \right)}>2000$$
I'm thinking that maybe it's something like telescopic series and am trying to bring it into those kinds of terms. Or can it be something else? \
Edit: I think I got it
the numerator is just 3 times the denominator isn't it?
|
It just clicked a few minutes after I posted it
We've got:
$$\prod_{k=1}^n \frac{\arcsin\left( \frac{9k+2}{(3k+2)^{\frac{3}{2}}}\right)}{\arcsin \left( \frac{1}{(3k+2)^{\frac{1}{2}}} \right)}>2000$$
let $\sin(\theta)=\frac{1}{(3k+2)^{\frac{1}{2}}}$
now consider
$$\sin(3\theta)=\frac{3}{(3k+2)^{\frac{1}{2}}}-\frac{4}{(3k+2)^{\frac{3}{2}}}$$
notice that the argument of arcsin in the numerator is
$$\frac{9k+2}{(3k+2)^{\frac{3}{2}}}=\frac{3(3k+2)}{(3k+2)^{\frac{3}{2}}}-\frac{4}{(3k+2)^{\frac{3}{2}}}$$
which is just $sin(3\theta)$
so the whole thing simplifies to
$$3^n>2000$$
$$n>6.918$$
so if the question wants the least integral value of n, then the answer becomes 7.
|
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|
Extracting sequence from generating function I was provided the following generating function, and was unsure how to use it. I have never seen an example where the function “involved” itself.
The generating function is
$F(z)^8$
Where
$$F(z)=z+z^6 F(z)^5+z^{11} F(z)^{10}+z^{16} F(z)^{15}+z^{21} F(z)^{20}$$
Any help appreciated
|
Use Lagrange inversion. From the functional equation that $F(z)$ satisfies, we get, by multiplying through by $z$,
$$
zF(z)=z^2\left(1+(zF(z))^5+(zF(z))^{10}+(zF(z))^{15}+(zF(z))^{20}\right).
$$
Thus,
$$
\frac{z}{1+z^5+z^{10}+z^{15}+z^{20}}\circ(zF(z))=z^2,
$$
so
$$
zF(z)=\left(\frac{z}{1+z^5+z^{10}+z^{15}+z^{20}}\right)^{\langle-1\rangle}\circ z^2,
$$
where "$\langle-1\rangle$" denotes the compositional inverse. Thus, $zF(z)$ is an even function, i.e. $zF(z)=H(z^2)$ for the invertible power series $H(z)=\left(\frac{z}{1+z^5+z^{10}+z^{15}+z^{20}}\right)^{\langle-1\rangle}$. This means that $H=H(z)$ satisfies the functional equation $H=z\varphi(H)$ for $\varphi(z)=1+z^5+z^{10}+z^{15}+z^{20}$. Note that $\varphi(0)\ne 0$. We need to find the coefficients of
$$
F(z)^8=\frac{H(z^2)^8}{z^8}=\frac{H(z)^8}{z^4}\circ z^2.
$$
By Lagrange inversion,
$$
[z^n]f(H(z))=\frac{1}{n}[z^{n-1}]\!\left(\varphi(z)^n f'(z)\right),
$$
where $[z^n]$ means the coefficient at $z^n$.
Here, $f(z)=z^8$, so $f'(z)=8z^7$, and we need essentially
\begin{multline*}
[z^n]\frac{H(z)^8}{z^4}=[z^{n+4}]H(z)^8=\frac{1}{n+4}[z^{n+3}]\!\left(8z^7\varphi(z)^{n+4}\right)=\\=\frac{8}{n+4}[z^{n-4}] (1+z^5+z^{10}+z^{15}+z^{20})^{n+4}.
\end{multline*}
And since $1+z^5+z^{10}+z^{15}+z^{20}=(1+z+z^2+z^3+z^4)\circ z^5$, only the terms where $n\equiv 4\pmod 5$ would be nonzero in the $[z^{n-4}] (1+z^5+z^{10}+z^{15}+z^{20})^{n+4}$. And since we are also composing with $z^2$ at the end, $F(z)^8$ will only have terms with powers congruent to $8\!\!\!\mod\!\!10$.
Finally, here is the pentanomial coefficient triangle. Scroll down to example, it's easier to see there.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4576868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Isolating $θ$ in $a \sin^2 \theta + b \sin \theta + c + d \cos^2 \theta + e \cos \theta + f = 0$. I have an equation that follows this form:
$a \sin^2 \theta + b \sin \theta + c + d \cos^2 \theta + e \cos \theta + f = 0$ where $a,b,c,d,e,f \in \mathbb{R}$ and $a\neq 0, d\neq 0$. I am trying to isolate $θ.$
How can this be achieved?
|
You can convert this expression to a $4$-th order polynomial with the Expotential representations of the trigonometric functions and substitution
these with $t$. You can now solve this equation with the polynomial of degree $4$ using a solution formula or equivalent transformation after t and then resubstitute, so that you only have to take the $\ln()$ and divide by $\mathrm{i}$:
$$
\begin{align*}
a \cdot \sin^{2}(\theta) + b \cdot \sin(\theta) + c + d \cdot \cos^{2}(\theta) + e \cdot \cos(\theta) + f &= 0 \quad\mid\quad -(c + f)\\
a \cdot \sin^{2}(\theta) + b \cdot \sin(\theta) + d \cdot \cos^{2}(\theta) + e \cdot \cos(\theta) &= -c - f \quad\mid\quad \sin(x) = \frac{e^{x \cdot \mathrm{i}} - e^{-x \cdot \mathrm{i}}}{2 \cdot \mathrm{i}} \wedge \cos(x) = \frac{e^{x \cdot \mathrm{i}} + e^{-x \cdot \mathrm{i}}}{2}\\
a \cdot \left( \frac{e^{\theta \cdot \mathrm{i}} - e^{-\theta \cdot \mathrm{i}}}{2 \cdot \mathrm{i}} \right)^{2} + b \cdot \frac{e^{\theta \cdot \mathrm{i}} - e^{-\theta \cdot \mathrm{i}}}{2 \cdot \mathrm{i}} + d \cdot \left( \frac{e^{\theta \cdot \mathrm{i}} + e^{-\theta \cdot \mathrm{i}}}{2} \right)^{2} + e \cdot \frac{e^{\theta \cdot \mathrm{i}} + e^{-\theta \cdot \mathrm{i}}}{2} &= -c - f\\
a \cdot \left( -\frac{e^{2 \cdot \theta \cdot \mathrm{i}} + e^{-2 \cdot \theta \cdot \mathrm{i}}}{4} + \frac{1}{2} \right) + b \cdot \frac{e^{\theta \cdot \mathrm{i}} - e^{-\theta \cdot \mathrm{i}}}{2 \cdot \mathrm{i}} + d \cdot \left( \frac{e^{2 \cdot \theta \cdot \mathrm{i}} + e^{-2 \cdot \theta \cdot \mathrm{i}}}{4} + \frac{1}{2} \right) + e \cdot \frac{e^{\theta \cdot \mathrm{i}} + e^{-\theta \cdot \mathrm{i}}}{2} &= -c - f\\
a \cdot -\frac{e^{2 \cdot \theta \cdot \mathrm{i}} + e^{-2 \cdot \theta \cdot \mathrm{i}}}{4} + \frac{a}{2} + b \cdot \frac{e^{\theta \cdot \mathrm{i}} - e^{-\theta \cdot \mathrm{i}}}{2 \cdot \mathrm{i}} + d \cdot \frac{e^{2 \cdot \theta \cdot \mathrm{i}} + e^{-2 \cdot \theta \cdot \mathrm{i}}}{4} + \frac{a}{2} + e \cdot \frac{e^{\theta \cdot \mathrm{i}} + e^{-\theta \cdot \mathrm{i}}}{2} &= -c - f\\
a \cdot -\frac{\left( e^{\theta \cdot \mathrm{i}} \right)^{2} + \left( e^{\theta \cdot \mathrm{i}} \right)^{-2}}{4} + \frac{a}{2} + b \cdot \frac{e^{\theta \cdot \mathrm{i}} - \left( e^{\theta \cdot \mathrm{i}} \right)^{-1}}{2 \cdot \mathrm{i}} + d \cdot \frac{\left( e^{\theta \cdot \mathrm{i}} \right)^{2} + \left( e^{\theta \cdot \mathrm{i}} \right)^{-2}}{4} + \frac{a}{2} + e \cdot \frac{e^{\theta \cdot \mathrm{i}} + \left( e^{\theta \cdot \mathrm{i}} \right)^{-1}}{2} &= -c - f \quad\mid\quad t := e^{\theta \cdot \mathrm{i}}\\
a \cdot -\frac{t^{2} + t^{-2}}{4} + \frac{a}{2} + b \cdot \frac{t - t^{-1}}{2 \cdot \mathrm{i}} + d \cdot \frac{t^{2} + t^{-2}}{4} + \frac{a}{2} + e \cdot \frac{t + t^{-1}}{2} &= -c - f \quad\mid\quad -a\\
a \cdot -\frac{t^{2} + t^{-2}}{4} + b \cdot \frac{t - t^{-1}}{2 \cdot \mathrm{i}} + d \cdot \frac{t^{2} + t^{-2}}{4} + e \cdot \frac{t + t^{-1}}{2} &= -c - f - a \quad\mid\quad -(-c - f - a)\\
-a \cdot \frac{t^{2} + t^{-2}}{4} + b \cdot \frac{t - t^{-1}}{2 \cdot \mathrm{i}} + d \cdot \frac{t^{2} + t^{-2}}{4} + e \cdot \frac{t + t^{-1}}{2} + c + f + a &= 0 \quad\mid\quad \cdot t^{2}\\
-a \cdot \frac{t^{4} + 1}{4} + b \cdot \frac{t^{3} - t}{2 \cdot \mathrm{i}} + d \cdot \frac{t^{4} + 1}{4} + e \cdot \frac{t^{3} + t}{2} + c + f + a &= 0\\
\end{align*}
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4577312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How to solve the differential equation $y'=\frac{y-xy^2}{x+x^2y}$ $$y'=\frac{y-xy^2}{x+x^2y}$$
This is the equation I want to solve.
My idea is to substitute $xy$ with $u,$ $u=xy$ and $\frac{du}{dx}=xy'+y.$
So, the equation becomes
$y'=\frac{y(1-u)}{x(1+u)}.$
What I am struggling with is how to deal with the $x$ and $y$ that are left in the equation.
|
The equation is $$y^{\prime}=\frac{y-x y^2}{x+x^2 y}$$
HINT:
We can re-write it as: $$-\frac{1}{y^2}\frac{dy}{dx}=\frac{\left(x-\frac{1}{y}\right)}{x+x^2 y}$$
Now letting $\frac{1}{y}=t \Rightarrow -\frac{1}{y^2}\frac{dy}{dx}=\frac{dt}{dx}$
$$\Rightarrow \frac{d t}{d x}=\frac{t(x-t)}{x(x+t)}=\frac{\frac{t}{x}\left(1-\frac{t}{x}\right)}{1+\frac{t}{x}}$$
Now use $t=vx \Rightarrow \frac{dt}{dx}=v+x\frac{dv}{dx}$
$$\Rightarrow \quad v+x \frac{d v}{d x}=\frac{v(1-v)}{1+v} \Rightarrow x \frac{d v}{d x}=\frac{-2 v^2}{1+v}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4577530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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|
Prove that $(x^2−1) \bmod 8$ $\in \{ 0 , 3 , 7 \}, \forall x \in \mathbb{Z}$. It must be verified that for all $x \in \mathbb{Z}$ it holds that $x^2 - 1 \bmod{8} \in \{0, 3, 7\}$. First some definitions. Using the following theorem a definition for $\bmod$ is provided:
Theorem. For all $a \in \mathbb{Z}$ and $d \in \mathbb{N}$ unique integers q, r exist satisfying $a = q \cdot d + r \wedge 0 \leq r <d$
The integers $q$, $r$ correspond to the quotient and remainder, respectively, of $a$, these being defined as:
$a = (a / d) \cdot d + a \bmod d \wedge 0 \leq a \bmod d < d$
With $q = (a/d)$ and $r = a \bmod d$. To verify the claim I use induction. I use a brute force approach where I search for all numbers $x$ such that $x^2 - 1 \bmod 8 = 0$, $x^2 - 1 \bmod 8 = 3$ or $x^2 - 1 \bmod 8 = 7$. First I search for all $x$ such that $x^2 - 1 \bmod 8 = 0$. If $x^2 - 1 \bmod 8 = 0$ then 8 is a divisor of $x^2 - 1$, i.e.: $8 \mid x^2 - 1$. Now I search for some other $x$ such that 8 is a divisor of $x^2 - 1$. Let $f(x) = x^2 - 1$ and $8 \mid f(x)$ then $8 \mid f(x + a)$ for some $a \in \mathbb{N}$. I compute this a:
$8 \mid f(x) = 8 \mid x^2 - 1 \implies 8 \mid (x + a)^2 - 1 = 8 \mid x^2 + a^2 + 2ax - 1 = 8 \mid x^2 - 1 + a^2 + 2ax$
The implication holds for e.g.: $a = 4$ since:
$8 \mid x^2 - 1 + a^2 + 2ax \implies 8 \mid x^2 - 1 + 16 + 8x = 8 \mid f(x) \wedge 8 \mid 16 + 8x = true$
So it can be concluded that if $8 \mid f(x)$ then $8 \mid f(x + 4)$. In fact it may be concluded that if $8 \mid f(x)$ then $8 \mid f(x + 4k), \forall x \in \mathbb{Z} \wedge k \in \mathbb{N}$, since:
$8 \mid (x + 4k)^2 - 1 = 8 \mid x^2 - 1 + 16k^2 + 8kx = 8 \mid f(x) \wedge 8 \mid 16k^2 + 8kx$
So it holds that $8 \mid f(x) \implies 8 \mid f(x + 4k)$. By inspection one finds that $8 \mid f(1) = 8 \mid 1^2 - 1 = 8 \mid 0 = true$ and so $f(x) \bmod 8 = 0$ for all $x \in \{1, 5, 9, 13, ...\}$. Since $f(x)$ is symmetric it holds that $f(-x) = f(x)$ and so it follows that $f(x) \bmod 8 = 0$ for all $x \in \{-1, -5, -9, -13, ...\}$. Upon further inspection one finds that $8 \mid f(3) = 8 \mid 3^2 - 1 = 8 \mid 8 = true$ and so it follows that $x^2 - 1 \bmod 8 = 0$ for all $x \in \{3, 7, 11, 15, ...\}$ and all $x \in \{-3, -7, -11, -15, ...\}$
Next I search for all x such that $x^2 - 1 \bmod 8 = 3$, or equivalently $x^2 - 4 \bmod 8 = 0$. Using the same approach I find that $x^2 - 1 \bmod 8 = 3$ for all $x \in \{2, 6, 10, 14, ...\}$ and all $x \in \{-2, -6, -10, -14, ...\}$. Finally $x^2 - 1 \bmod 8 = 7$ for all $x \in \{0, 4, 8, 12, ...\}$ and $x \in \{0, -4, -8, -12, ...\}$. And so one concludes that for all $x \in \mathbb{Z}$ it holds that $x^2 - 1 \bmod{8} \in \{0, 3, 7\}$.
This question comes from a course in discrete mathematics in computer science. I feel that my approach is overkill and that I'm doing something wrong and that there must be some cleverer way of solving the problem. If anyone can help with a better, cleaner, approach, or point out errors, it will be greatly appreciated :).
|
*
*If $x = 4 k$ with $k \in \mathbb{Z}$ then :
$$x^2 - 1 = 16 k^2 - 1 = 16 k^2 - 8 + 7 \equiv 7 mod 8$$
*If $x = 4 k + 1$ with $k \in \mathbb{Z}$ then :
$$x^2 - 1 = 16 k^2 + 8 k + 1 - 1 = 16 k^2 + 8 k \equiv 0 mod 8$$
*If $x = 4 k + 2$ with $k \in \mathbb{Z}$ then :
$$x^2 - 1 = 16 k^2 + 16 k + 4 - 1 = 16 k^2 + 16 k + 3 \equiv 3 mod 8$$
*If $x = 4 k + 3$ with $k \in \mathbb{Z}$ then :
$$x^2 - 1 = 16 k^2 + 24 k + 9 - 1 = 16 k^2 + 24 k + 8 \equiv 0 mod 8$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4584496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Let $f(xy) = f(x)+ f(y) +\frac{x+y-1}{xy} \forall x,y>0$, $f'(1)=2$, $f$ is differentiable . Find $f(x)$. What I have gotten from $f(xy) = f(x)+ f(y) +\frac{x+y-1}{xy}$ is:
*
*$f(1)=-1$
2.$f(x)+f(\frac{1}{x}) =-(x+\frac{1}{x})$
The information I could get from $f'(1)=2$ is just:
$\lim_{h\to 0} { \frac{f(1+h)+1}{h}}=2$
Since the limit is given, my approach was to find $f'(x)$ and then solve the differential equation. But I'm unable to compute the limit $\lim_{h \to 0} {\frac{f(x+h)-f(x)}{h}}$ with the information given. I have tried stuff like converting $f(x)$ to $f(\frac{x}{h}\cdot h)$ and try to use the original equation and convert it into$ f(\frac{x}{h})+ f(h) +\frac{\frac{x}{h}+h-1}{x}$.But it hasn't been of any use to compute the limit. I need to do it in the way of computing the limit and solving the Differential equation.
|
Here is an answer along the lines requested by OP by taking a limit.
I assume the domain is $x > 0 $
I use an alternate, but equivalent, definition of the derivative involving a product to make use of the given property. It is valid since $xh \to x$ as $h \to 1$.
$$
\begin{align*}
f'(x) &= \lim_{h \to 1} \frac{f(xh) - f(x)}{xh-x}\\
&= \lim_{h \to 1} \frac{f(x)+f(h) + \frac{x+h-1}{xh} - f(x)}{x(h-1)}\\
&= \lim_{h \to 1} \frac{f(h) + \frac{x+h-1}{xh}}{x(h-1)}\\
&= \lim_{h \to 1} \frac{f(h) - f(1) + f(1) + \frac{x+h-1}{xh}}{x(h-1)}\\
&= \lim_{h \to 1} \frac{1}{x}\frac{f(h) - f(1)}{h-1} + \frac{-1+ \frac{x+h-1}{xh}}{x(h-1)} \textrm{ since $f(1) = -1$}\\
&= \frac{1}{x} f'(1) + \lim_{h \to 1} \frac{x+h-xh-1}{x^2h(h-1)}\\
&= \frac{2}{x} + \lim_{h \to 1} \frac{(1-x)(h-1)}{x^2h(h-1)}\\
&= \frac{2}{x} + \lim_{h \to 1} \frac{(1-x)}{x^2h}\\
&= \frac{2}{x} + \frac{1}{x^2} - \frac{1}{x}\\
&= \frac{1}{x} + \frac{1}{x^2}
\end{align*}
$$
Hence $f'(x) = x^{-1} + x^{-2}$, so $f(x) = \ln(x) - \frac{1}{x} + C$
Now using $f(1) = -1 $ we obtained $C=0$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4591396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
What is the value of $f(1999)$? Let be a function $f \colon \Bbb R\to \Bbb R$ given from:
$$f(x)=x^3+\sqrt{x^6+1}+\frac{1}{x^3-\sqrt{x^6+1}}$$
What is the value of $f(1999)$?
To me, it immediately seemed strange that it was necessary to calculate $f(1999)$. With a calculator everything would come easy. Since this question concerns high school students I have done the rationalization of $\frac{1}{x^3-\sqrt{x^6+1}}$ and I have found something very interesting. In fact
$$\frac{1}{x^3-\sqrt{x^6+1}}=\frac{1}{x^3-\sqrt{x^6+1}}\cdot\frac{x^3+\sqrt{x^6+1}}{x^3+\sqrt{x^6+1}}=\frac{x^3+\sqrt{x^6+1}}{-1}$$
because $(x^3-\sqrt{x^6+1})(x^3+\sqrt{x^6+1})=-1$
Thus
$$f(x)=x^3+\sqrt{x^6+1}-x^3-\sqrt{x^6+1}=0$$
and $f(1999)=0$ because $f(x)=0, \forall x\in\Bbb R$. Is it correct?
|
Your work seems correct to me. (this is stated in the comments that your work is correct)
The problem statement is essentially equivalent to:
Let $f:\mathbb R\rightarrow \mathbb R$,
$$f(x):=A(x)+B(x)+\frac {1}{A(x)-B(x)}$$
where, $A^2(x)-B^2(x)=-1$.
Find the value of $f(1999).$
$$
\begin{align}f(x):&=A(x)+B(x)+\frac{A(x)+B(x)}{A^2(x)-B^2(x)}\\
&=A(x)+B(x)-\left(A(x)+B(x)\right)\\
&=0~.\end{align}
$$
Thus $f(x)\equiv 0$, $\forall x\in\mathbb R$, since $\operatorname{dom}f=\mathbb R.$
What we do is based on using conjugate.
Note that, when you add fractions, the conjugate will appear by itself. Indeed,
$$
\begin{align}f(x):&=A(x)+B(x)+\frac {1}{A(x)-B(x)}\\
&=\frac {\left(A(x)-B(x)\right)\cdot\left(A(x)+B(x)\right)+1}{A(x)-B(x)}\\
&=\frac{\overbrace{A^2(x)-B^2(x)}^{\color {#c00}{-1}}+1}{A(x)-B(x)}\\
&=0~.
\end{align}
$$
Because, $A^2(x)-B^2(x)=-1$ implies that, $A(x)±B(x)≠0$.
Finally, I want to note that this type of function is called a constant function in mathematics.
|
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"url": "https://math.stackexchange.com/questions/4592316",
"timestamp": "2023-03-29T00:00:00",
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|
Calculation limit: $\lim\limits_{n\to\infty} \sum\limits_{k=0}^n \frac{1}{2^k-n}$
$$\lim_{n\to\infty} \sum_{k=0}^n \frac{1}{2^k-n}$$
My main issue with the sequence $a_k=\frac{1}{2^k-n}$ is that there will always be a value of $k$ for which the sequence will not be properly defined (at least if $n=2^k$ for some value of $k$). Having said that, if we ignore this issue, I am not really sure what might be the proper way to calculate this limit (Sandwich theorem? )
Will be happy to hear some thoughts. Thank you!
|
Consider the subsequence of $n \not= 2^m$. Now it makes sense talking of the sequence and the limit (if it exists). Write $n=2^m + r$ with $r\in(0,2^m)$.
Then $\lfloor\log{n}\rfloor = m$ and $\lceil\log{n}\rceil = m+1$. Notice that $m\to \infty$ as $n \to \infty$, but the same is not true for $r$. So
$$
\sum_{k=0}^n \frac{1}{2^k-n} = \sum_{k=0}^{\lfloor\log{n}\rfloor} \frac{1}{2^k-n} + \sum_{k=\lceil\log{n}\rceil}^n \frac{1}{2^k-n}
$$
The LHS can be rewritten as
\begin{align}
0>\sum_{k=0}^{\lfloor\log{n}\rfloor} \frac{1}{2^k-n}
&=\sum_{k=0}^{\lfloor\log{n}\rfloor}\frac{-1}{n-2^k} \\
&=\sum_{k=0}^{m}\frac{-1}{2^m + r -2^k} \\
&=\sum_{k=0}^{m}\frac{-1}{r + 2^m - 2^k} \\
&=\sum_{k=0}^{m}\frac{-1}{r + \sum_{j=k}^{m-1} 2^j} \\
&= \frac{-1}{r}+\sum_{k=1}^{m-1}\frac{-1}{r + \sum_{j=i}^{m-1} 2^j} \\
\end{align}
However we have $r + \sum_{j=i}^{m-1} 2^j \geq 2^{m-1}$ so $\sum_{k=1}^{m-1}\frac{-1}{r + \sum_{j=i}^{m-1} 2^j}$ is bounded by $\sum_{k=1}^{m-1}\frac{-1}{2^{m-1}} = \frac{m-1}{2^{m-1}}$.
Finally, the LHS is
$$
\sum_{k=0}^{\lfloor\log{n}\rfloor} \frac{1}{2^k-n} \geq \frac{-1}{r} + \frac{-(m-1)}{2^{m-1}} ``\to \frac{-1}{r}"
$$
More formally, we can write $r=r(n)$, $m=m(n)$ and
$$
\lim_{n\to\infty} \text{LHS} = \lim_{n\to\infty} \frac{-1}{r(n)}
$$
The RHS can be written as
\begin{align}
0<\sum_{k=\lceil\log{n}\rceil}^n \frac{1}{2^k-n} &= \sum_{i=0}^{n-\lceil\log{n}\rceil} \frac{1}{2^{\lceil\log{n}\rceil + i}-n} \\
&= \sum_{i=0}^{n-(m+1)} \frac{1}{2^{m+1 + i}-(2^m+r)} \\
&= \frac{1}{2^{m+1}-(2^m+r)} + \sum_{i=1}^{n-(m+1)} \frac{1}{2^{m+1}2^{i}-(2^m+r)} = \cdots
\end{align}
However, $-(2^m+r) \geq -2^{m+1}$, so
\begin{align}
\cdots&\leq \frac{1}{2^{m+1}-2^m-r} + \sum_{i=1}^{n-(m+1)} \frac{1}{2^{m+1}(2^{i}-1)} \\
&=\frac{1}{2^m-r} + \frac{1}{2^{m+1}}\sum_{i=1}^{n-(m+1)} \frac{1}{2^{i}-1} \\
&\leq \frac{1}{2^m-r} + \frac{1}{2^{m+1}}\frac{1}{2} "\to \frac{1}{2^{m}-r}",
\end{align}
or more formally
$$
\lim_{n\to\infty} \text{RHS} = \lim_{n\to\infty} \frac{1}{2^{m(n)}-r(n)}.
$$
We can conclude that the limit doesn't exist. Consider the two subsequences
$\{n_{1,j} = 2^{j-1} - 1\}$ $(m_1 = {j-1}, r_1=2^{j-1}-1)$ and $\{n_{2,j} = 2^j+1\}$ $(m_2=j, r_2=1)$. Then
$$
\lim_{n_1 \to \infty}\text{LHS} = 0, \quad \lim_{n_1 \to \infty}\text{RHS} = -1,
$$
$$
\lim_{n_1 \to \infty} \sum_{k=0}^\infty \frac{1}{2^k-n_1} = -1,
$$
and
$$
\lim_{n_2 \to \infty}\text{LHS} = -1, \quad \lim_{n_2 \to \infty}\text{RHS} = 0,
$$
$$
\lim_{n_2 \to \infty} \sum_{k=0}^\infty \frac{1}{2^k-n_2} = -1.
$$
Therefore the sequence doesn't converge. You can find many other convergent subsequences, for example a sequence converging to $0$ or to $\frac{\pm 1}{d}$ for any $d$.
Essentially the whole sum reduces to only two terms
\begin{align}
\lim_{n\to\infty}\sum_{k=0}^{n}\frac{1}{2^k-n} &= \lim_{n\to\infty}\left(\frac{1}{2^{\lfloor \log{n}\rfloor} - n} + \frac{1}{2^{\lceil \log{n}\rceil} - n}\right) \\
&= \lim_{n\to\infty}\left( \frac{1}{2^m - r} - \frac{1}{r}\right).
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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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.