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Compute $\lim \limits_{n \to \infty}\left(\frac {\sqrt[n]a + \sqrt[n]b}2\right)^{n} ~~~ (a, b>0)$
$$\lim_{n \to \infty}\left(\frac {\sqrt[n]a + \sqrt[n]b}2\right)^{n} ~~~ (a, b>0)$$
I extended its domain and applied L'Hopital's rule to get the answer $\sqrt{ab}$.
However, is it possible to avoid using L'Hopital's rule here? (I tried for a few times, but failed.)
Or any hint maybe?
|
If $a=b$, then it is trivial.So assume without loss of generality. $x = \frac ab>1.$
Then it is the same as:
$$\left(\dfrac{\sqrt[n]{x}+1}{2}\right)^n\to\sqrt{x}$$.For simplicity, denote $\sqrt[2n]{x} = y >1.$ Then, we have $$\left(\dfrac{y^2+1}{2}\right)^n - y^n = y^n\left[\left(1+\dfrac{(y-1)^2}{2y}\right)^n-1\right].$$
However, notice that $y^n=\sqrt{x},\quad y\to 1,\quad\dfrac{(y-1)^2}{2y} <(\sqrt[2n]{1+(x-1)}-1)^2 <\frac{(x-1)^2}{4n^2} $ as $n\to 0.$
$$\left(\dfrac{y^2+1}{2}\right)^n - y^n = y^n\left[\left(1+\dfrac{(y-1)^2}{2y}\right)^n-1\right] < \sqrt{x} \left[\left(1+\frac{(x-1)^2}{4n^2}\right)^n-1\right] = $$
$$ = \sqrt{x}\left[\left(\left(1+\frac{(x-1)^2}{4n^2}\right)^{n^2}\right)^{\frac 1n}-1\right] < \sqrt{x}(e^{(x-1)^2/4n}-1)\to 0$$
as $n\to 0.$
Here, we used the fact that $(1+a/m)^m<e^a.$
|
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|
Complex number problem $i \tan (\theta)$ proof.
Given $z = \cos (\theta) + i \sin (\theta)$,
prove $\dfrac{z^{2}-1}{z^{2}+1} = i \tan(\theta)$
I know $|z|=1$ so its locus is a circle of radius $1$; and so $z^{2}$ is also on the circle but with argument $2\theta$; and $z^{2}+1$ has argument $\theta$ (isosceles triangle) so it lies on a line through the origin and $z$.
$z^{2}-1$, $z^{2}$, and $z^{2}-1$ all lie on a horizontal line $i \sin (\theta)$.
On the Argand diagram I can show $z^{2}+1$ and $z^{2}-1$ are perpendicular so the result follows.
Can anyone give an algebraic proof?
|
For $z = \cos \theta + i \sin \theta$ you have
$$
\frac{z^2-1}{z^2+1} = \frac{ \cos^2 \theta - \sin^2 \theta + 2i \cos\theta \sin \theta -1}{\cos^2 \theta - \sin^2 \theta + 2i \cos\theta \sin \theta +1 } \, .
$$
Now substitute $1 = \cos^2 \theta + \sin^2 \theta $ in both numerator and denominator, and collect terms:
$$
\frac{-2 \sin^2 \theta + 2 i \cos\theta \sin\theta}{2 \cos^2\theta + 2 i \cos\theta \sin\theta} =
\frac{- \sin\theta + i \cos \theta}{\cos \theta + i \sin\theta} \cdot \frac{\sin\theta }{\cos\theta} \, .
$$
Finally convince yourself that the first factor is equal to $i$, and
you are done.
|
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|
If $\frac{2x + 6} {(x + 2)^2}- \frac{2} {x + 2} = \frac{a} {(x + 2)^2}$, then what is $a$? $$\frac{2x + 6} {(x + 2)^2}- \frac{2} {x + 2}$$
The expression above is equivalent to $$\frac{a} {(x + 2)^2}$$
where $a$ is a positive constant and $x \neq -2$.
What's the value of $a$?
|
We have
$$2x+\frac{6}{(x+2)^2}-\frac{2}{x+2}=\frac{a}{(x+2)^2}$$
$$2x(x+2)^2+6-2(x+2)=a$$
$$2x^3+4x^2+4x+6-2x-2=a$$
$$2x^3+4x^2+2x+4=a$$
|
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|
Mean of two numbers by infinite sequences Consider two numbers $a$ and $b$, and the following sequence alternating between even and odd positions:
$$
a+2b+3a+4b+5a+6b\ldots,
$$
If we ''normalize''
$$
\frac{a+2b+3a+4b+\ldots}{1+2+3+4+\ldots},
$$
it turns out this ratio approaches the mean value of $a$ and $b$: $(a+b)/2$. In general
$$
\frac{a+2^n b+3^n a+4^n b+\ldots}{1^n+2^n+3^n+4^n+\ldots}=\frac{a+b}{2}
$$
for $n\geq1$. However if we use exponential functions instead powers:
$$
\frac{m^1 a+m^2 b+m^3 a+m^4 b+\ldots}{m^1 +m^2 +m^3 +m^4 +\ldots}
$$
for some $m>1$, this ratio oscillates and does not approach any number.
Could someone explain why convergence to the mean is obtained by using sequences of powers, and the ratio diverges for sequences of exponentials?
|
Some thoughts; not sure if this fully answers "why" but an attempt:
The difference in coefficients of $a$ and $b$ goes roughly like the last coefficient term (whichever one's ahead is ahead by around that much). In the polynomial case this difference becomes small relative to the sums; in the geometric case, this stays large.
This is because last term of a geometric sequence with ratio $m > 1$, no matter how long, is a constant fraction of its sum. But for a polynomial sequence, eventually the last term becomes a vanishingly small fraction of the sum.
More formally: consider a sequence of positive reals $c_1, c_2, \dots$ (may be geometric, polynomial, whatever). We're interested in
$$\frac{c_1a + c_2b + c_3a + \dots}{c_1 + c_2 + c_3 + \dots}$$
Or more precisely, let the partial sum $s_k = \sum_{1 \le i \le k} c_i$. Let the sum of odd terms $o_k = \sum_{1 \le 2i-1 \le k} c_{2i-1}$ and the sum of even terms $e_k = \sum_{1 \le 2i \le k} c_{2i} = s_k - o_k$. We're interested in
$$\lim_{k \rightarrow \infty} \frac{o_k a + e_k b}{s_k}$$
Now we can observe this converges for all $a, b$ if and only if $\frac{o_k}{s_k}$ converges.
So the question is: for what kinds of sequences does $\frac{o_k}{s_k}$ converge?
Don't have a nice full answer to this.
But we're more interested in the case when $c_i$ increases it seems, so let's restrict ourselves to that case. Hereafter $c_i$ are increasing.
In this case, we can say the following (the formal version of the above)
Claim:
(i) if $\frac{c_k}{s_k} \rightarrow 0$ then $\frac{o_k}{s_k} \rightarrow \frac{1}{2}$.
(ii) if $\frac{o_k}{s_k} \rightarrow L$ then $L = \frac{1}{2}$ and $\frac{c_k}{s_k} \rightarrow 0$.
Proof of claim:
Observe first that $\frac{o_k}{s_k} \rightarrow L$ if and only if $\frac{o_k - e_k}{s_k} \rightarrow 2L-1$. In particular $\frac{o_k}{s_k} \rightarrow \frac{1}{2} \Leftrightarrow \frac{o_k - e_k}{s_k} \rightarrow 0$
(i) Assume $\frac{c_k}{s_k} \rightarrow 0$. Because the $c_i$ are increasing and positive: when $k$ is odd, $0 \le o_k - e_k \le c_k$. When $k$ is even $0 \le e_k - o_k \le c_k$. So $\left|\frac{o_k - e_k}{s_k}\right| \le \frac{c_k}{s_k}$ for all $k$, hence $\left|\frac{o_k - e_k}{s_k}\right| \rightarrow 0$.
(ii) Assume $\frac{o_k}{s_k} \rightarrow L$. We have $\frac{o_k - e_k}{s_k} \rightarrow 2L-1$. But as in (i) we have $o_k - e_k \ge 0$ for odd $k$ and $o_k - e_k \le 0$ for even $k$. So we must have $2L-1 = 0 \Rightarrow L = 1/2$.
Now let's try to upper bound $\frac{c_k}{s_k}$ in terms of $|\frac{o_k - e_k}{s_k}|$.
For any odd $k$, $(e_{k-1} - o_{k-1}) + (o_k - e_k) = o_k - o_{k-1} = c_k$. Both those terms are positive, so we have that either $\frac{c_k}{2} \le e_{k-1} - o_{k-1} = |o_{k-1} - e_{k-1}|$ or $\frac{c_k}{2} \le o_k - e_k = |o_k - e_k|$. This also holds for even $k$ and can be shown similarly.
So we have
$$c_k \le 2\left(\max \{|o_{k-1} - e_{k-1}|, |o_k - e_k|\}\right)$$
Divide both sides by $s_k$, and we see that the RHS goes to 0 and hence $\frac{c_k}{s_k} \rightarrow 0$ $\square$.
Applied to polynomial case: so we have $c_1, c_2, \dots = 1^n, 2^n, \dots$. Let's lower bound $s_k$ by some multiple of $k^{n+1}$. Very roughly, note that for say $k \ge 4$,
$$s_k = \sum_{i=1}^k i^n \ge \sum_{i \ge k/2} i^n \ge \sum_{i \ge k/2} \left(\frac{k}{2}\right)^n \ge \lfloor \frac{k}{2} \rfloor \left(\frac{k}{2}\right)^n \ge \frac{k}{4}\left(\frac{k}{2}\right)^n = \frac{k^{n+1}}{2^{n+2}}$$
And so $\frac{c_k}{s_k} \le \frac{2^{n+2}}{k} \rightarrow 0$ as $k \rightarrow \infty$.
From the claim, this means $\frac{o_k}{s_k} \rightarrow \frac{1}{2}$ and so the sequence of fractions $\frac{o_ka + e_kb}{s_k}$ approaches $\frac{a+b}{2}$ as expected.
Geometric case: By the formula for sum of geometric series, we can show that $\frac{c_k}{s_k} = \frac{m^{k-1}}{(m^{k} - 1)/(m - 1)} \ge \frac{m-1}{m}$; hence it cannot go to 0. By the claim, the series of fractions diverges.
[More specifically I think it can be shown that $\frac{o_k}{s_k}$ oscillates between $\frac{1}{m+1}$ and slightly larger than $\frac{m}{m+1}$, using the facts that for even $k$, $e_k = mo_k$ and for odd $k$, $o_k = me_k + c_1$]
|
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|
Find $\sin81^\circ$ given $\sin18^\circ=\dfrac{\sqrt{5}-1}{4}$
If $\sin18^\circ=\dfrac{\sqrt{5}-1}{4}$, then $\sin81^\circ$ is equal to
\begin{align}
&a)\quad\frac{\sqrt{5}+1}{4}\\
&b)\quad\frac{\sqrt{3+\sqrt{5}}+\sqrt{5-\sqrt{5}}}{4}\\
&c)\quad\frac{\sqrt{10+2\sqrt{5}}}{4}
\end{align}
$$
\cos18^\circ=\sqrt{1-\frac{6-2\sqrt{5}}{16}}=\frac{\sqrt{10+2\sqrt{5}}}{4}
$$
$$
\sin81^\circ=\cos9^\circ=\sqrt{\frac{1+\cos18^\circ}{2}}=\sqrt{\frac{4+\sqrt{10+2\sqrt{5}}}{8}}=\frac{\sqrt{8+2\sqrt{10+2\sqrt{5}}}}{4}
$$
How do I proceed further and find the solution ?
Or from here, at least can I just identify the solution as $\sin81^\circ=\dfrac{\sqrt{3+\sqrt{5}}+\sqrt{5-\sqrt{5}}}{4}$ from the options given ?
|
If you suspect (b) is the answer, just square it. You'll get
$$
\frac1{16}\left(3 +\sqrt 5+2\sqrt{(3+\sqrt 5)(5-\sqrt 5)} + 5 - \sqrt 5\right)
$$
which simplifies to
$$
\frac1{16}\left(8 + 2\sqrt{10+2\sqrt 5}\right),
$$
the square of your answer. Your answer must then agree with (b), since both answers are positive (as is $\sin 81^\circ$).
|
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|
Proving $\lim_{n\rightarrow-\infty}\frac{3x^2+x}{2x^2+1}=\frac{3}{2}$ Using Definition Definition of Limit of Function As n $\rightarrow\infty$: Let $g:\mathbb{R}\rightarrow\mathbb{R}$ be a function and let $B\in\mathbb{R}$. If for all $\epsilon>0$, there exists $N>0$ such that $x<-N\Rightarrow |g(x)-B|<\epsilon$, we write $\lim_{n\rightarrow\infty}g(x)=B$.
I want to prove $\lim_{n\rightarrow-\infty}\frac{3x^2+x}{2x^2+1}=\frac{3}{2}$.
Here is my attempt.
We must find the value $-N<0$ such that: $x<-N\Rightarrow |g(x)-\frac{3}{2}|=|\frac{3x^2+x}{2x^2+1}-\frac{3}{2}|=|\frac{3x^2+x}{2x^2+1}-\frac{3(x^2+\frac{1}{2})}{2(x^2+\frac{1}{2})}|=|\frac{x-\frac{3}{2}}{2x^2+1}|<\epsilon$.
When $N>\frac{3}{2}$, we have: $-x>N\Rightarrow |\frac{-x-\frac{3}{2}}{2x^2+1}|=\frac{-x-\frac{3}{2}}{2x^2+1}<\frac{-x-\frac{3}{2}}{2x^2}<\frac{-x}{2x^2}=\frac{-1}{2x}<\epsilon\Rightarrow -x>\frac{1}{2\epsilon}$. So, choose $N>\frac{1}{2\epsilon}$.
Now, this is the bit where I'm having some issues.
Then, for all $x<-N<\min\{-\frac{1}{2\epsilon},-\frac{3}{2}\}$, we have $|\frac{x-\frac{3}{2}}{2x^2+1}|=\frac{-x+\frac{3}{2}}{2x^2+1}<\frac{-x+\frac{3}{2}}{2x^2}=\frac{\frac{1}{2\epsilon}+\frac{3\epsilon}{2\epsilon}}{\frac{9}{2}}$... I can't quite get this inequality to work the way I need it to.
Any help appreciated.
|
You want the $x^2$ in the bottom to become dominant. And the negatives are confusing things. Bout $x < -\frac 32$ so $-x \ge \frac 32 > 0$ so
$\frac {-x + \frac 32}{2x^2} = \frac {1 + \frac 3{-2x}}{-2x}$
Replace $y = -x$ and $y > \max(\frac 32, \frac 1{2\epsilon})$ and $\frac 1y < \min (\frac 23, 2\epsilon)$.
$\frac {1+ \frac 3{2y}}{2y} < \frac 1y\frac {1 + \frac 32*\frac 1y}2< (2\epsilon)\frac {1 + \frac 32\frac 32}2= 2\epsilon$
So ... some minor arithmetic error slipped in. But it's minor and fixable. If we chose $N> \max (\frac 32, \frac 1\epsilon)$ it'd probably work.
|
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|
Show that $\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)\cdots\left(1+\frac{1}{n^3}\right) < 3$ I have this problem which says that for any positive integer $n$, $n \neq 0$ the following inequality is true: $$\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)\cdots\left(1+\frac{1}{n^3}\right) < 3$$
This problem was given to me in a lecture about induction but any kind of solution would be nice.And also I'm in 10th grade :)
|
By induction we can prove the stronger
$$\prod_{k=1}^n\left(1+\frac{1}{k^3}\right)<3\left(1-\frac{1}{n}\right)$$
indeed
1. base cases: by inspection the inequality is satisfied for for
$n=1,2, 3$
2. induction step:
*
*assume true that (Ind. Hyp.): $\prod_{k=1}^n\left(1+\frac{1}{k^3}\right)<3\left(1-\frac{1}{n}\right)$
*we want to prove that: $\prod_{k=1}^{n+1}\left(1+\frac{1}{k^3}\right)<3\left(1-\frac{1}{n+1}\right)$
then we have
$$\prod_{k=1}^{n+1}\left(1+\frac{1}{k^3}\right)=\prod_{k=1}^n\left(1+\frac{1}{k^3}\right) \cdot \left(1+\frac{1}{(n+1)^3}\right)<$$
$$\stackrel{Ind. Hyp.}<3\left(1-\frac{1}{n}\right)\left(1+\frac{1}{(n+1)^3}\right)\stackrel{?}<3\left(1-\frac{1}{n+1}\right)$$
thus we need to show that
$$3\left(1-\frac{1}{n}\right)\left(1+\frac{1}{(n+1)^3}\right)\stackrel{?}<3\left(1-\frac{1}{n+1}\right)$$
which is true indeed
$$1+\frac{1}{{n+1}^3}-\frac1n-\frac{1}{n(n+1)^3}\stackrel{?}<1-\frac{1}{n+1}$$
$$n-(n+1)^3-1\stackrel{?}<-n(n+1)^2$$
$$n-n^3-3n^2-3n-1-1\stackrel{?}<-n^3-2n^2-n$$
$$n^2+n+2\stackrel{?}>0$$
|
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|
Integrating $\int_{0}^{π/4} \frac{\sin x+\cos x}{\sin^4 x+\cos^2 x} dx$ Here is the problem
$$\int_{0}^{π/4} \frac{\sin x+\cos x}{\sin^4 x+\cos^2 x} dx$$
I have tried diving by $\cos^2 x$ and using partial fractions. Also substituting $\tan$ formulas or separating and partial fraction mess up the limits for me.
So can I get a full solution with answer please?
|
Use your trig identities to create a $u$-substitution. We have
$$
[\sin(x) + \cos(x)]dx = -d[\cos(x) - \sin(x)]
$$
and
$$
\sin(x)^4 + \cos(x)^2 = 1 - \sin(x)^2\cos(x)^2 = 1 - \frac{1}{4}\left(1 - [\cos(x) - \sin(x)]^2\right)^2,
$$
so
$$
\int_0^{\pi/4}\frac{\sin(x)+\cos(x)}{\sin(x)^4 + \cos(x)^2}dx = 4\int_0^1\frac{du}{4 - (1-u^2)^2} = 4\int_0^1\frac{du}{(3-u^2)(1+u^2)}
$$
and I'm guessing you know where to go from here.
Note that if you didn't know those trig identities for the denominator, you could still use $u = \cos(x) -\sin(x) = \sqrt{2}\cos\left(x+\frac{\pi}{4}\right)$, then plug in $x = \cos^{-1}\left(\frac{u}{\sqrt{2}}\right) - \frac{\pi}{4}$ and do the algebra. It's not fun, but it gets there in the end.
|
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|
I would really appreciate some help solving this induction problem!!
$3.$ for all $n\ge1$, $\displaystyle\sum_{i=1}^n(2i)^2=\frac{2n(2n+1)(2n+2)}{6}$
I have
$$\frac{2n(2n+1)(2n+2)(12(n+1)^2)}6= \frac{2n(2n+1)2(n+1)12(n+1)(n+1)}6.$$
I think I need to do something with the $(n+1)$.
However, I'm not sure where to go from here. I know the end goal is:
$$\frac{2(n+1)(2(n+1)+1)(2(n+1)+2)}6.$$
Any help is appreciated!
|
I'm not sure where you got
$$
\frac{2n(2n+1)(2n+2)(12(n+1)^2)}{6}
$$
from. It should be
$$
\frac{2n(2n+1)(2n+2)\color{red}{+}12(n+1)^2}{6}
$$
and with this fix you should be able to finish up.
However, it's better if you do some simplifications, to get rid of useless factors.
The induction step consists in writing
$$
\sum_{i=1}^{n+1}(2i)^2=
\sum_{i=1}^{n}(2i)^2+(2(n+1))^2
$$
and substituting the summation with the formula provided by the induction hypothesis, so to get
$$
\frac{2n(2n+1)(2n+2)}{6}+4(n+1)^2
$$
Your goal is to prove that this equals
$$
\frac{2(n+1)(2(n+1)+1)(2(n+1)+2)}{6}=
\frac{2(n+1)(2n+3)(n+2)}{3}
$$
which you obtain from the formula to prove by changing $n$ into $n+1$, which you seem to have done right.
Consider
\begin{align}
\frac{2n(2n+1)(2n+2)}{6}+4(n+1)^2
&=\frac{2n(2n+1)(n+1)}{3}+4(n+1)^2 \\[4px]
&=\frac{2}{3}(n+1)\bigl(n(2n+1)+6(n+1)\bigr) \\[4px]
&=\frac{2}{3}(n+1)(2n^2+7n+6)
\end{align}
Since $2n^2+7n+6=(n+2)(2n+3)$, you're done.
|
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|
Trigonometric Identities: Given that $2\cos(3a)=\cos(a)$ find $\cos(2a)$
Given that $2\cos(3a)=\cos(a)$ find $\cos(2a)$.
$2\cos(3a)=\cos(a)$
I converted $\cos(2a)$ into $\cos^2(a)-\sin^2(a)$
Then I tried plugging in. I know this is not right, but I have no clue how to solve this. Hints please?
edit:
Because I got that $\cos(2a) = 4\cos^2(3a)-1$
|
I like to split $\color{blue}{\cos(3 a) = \cos(a+2a) = \cos(a) \cos(2 a) -\sin(a) \sin(2 a)}$. This makes the expression
$$ 2 \left( \cos(a) \cos(2 a) -\sin(a) \sin(2 a) \right) = \cos(a) $$
or (re-arrange and divide by $2\cos(a)$)
$$ \frac{2 \cos(a) \cos(2 a)}{2 \cos(a)} = \frac{\cos(a)}{2 \cos(a)} + \frac{2\sin(a) \sin(2 a)}{2 \cos(a) }$$
of course $\color{blue}{ \sin(2 a) = 2\sin(a)\cos(a)}$
$$ \cos(2 a) = \tfrac{1}{2} + \frac{ 2\sin(a) \left( 2 \sin(a) \cos(a) \right)}{2 \cos(a)} = \tfrac{1}{2} + 2 \sin^2(a) $$
Now use $\color{blue}{\sin^2(a) = \tfrac{1}{2} - \tfrac{1}{2} \cos(2 a)}$ to make the above
$$ \cos(2 a) = \tfrac{1}{2} + 2 \left(\tfrac{1}{2} - \tfrac{1}{2} \cos(2 a)\right) = \tfrac{3}{2} - \cos(2 a) $$
The above is solved for
$$ \cos(2 a) = \tfrac{3}{4} $$
|
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|
Examining whether $\sum \limits_{n=0}^\infty\frac{(-1)^{n+1}}{5n+1}$ is convergent, absolute convergent or divergent Everything in red is edited
To show, that the series is convergent we show at first, that $\color{red}{\lim \limits_{n \to \infty} \left(\dfrac{1}{5n+1}\right)}=0$.
$\color{red}{\lim \limits_{n \to \infty} \left(\dfrac{1}{\underbrace{5n+1}_{1/\infty}}\right)=\lim \limits_{n \to \infty} \left(\dfrac{1}{n}\right)}=0 \implies a_n>0$
Leibniz criterion $\sum \limits_{n=0}^\infty (-1)^{n+1}\cdot \underbrace{\dfrac{1}{5n+1}}_{a_n}$
We still need to show, that $a_n$ is monotonic decreasing:
\begin{align}
a_{n}&\ge a_{n+1}\\
\color{red}{\frac1{5n+1}}&\color{red}{\ge\frac1{5n+6}\iff 5n+6\ge5n+1\iff 6\ge 1\;\checkmark}
\end{align} $\implies$ monotonic decreasing.
$\implies$ The series is convergent.
To prove, that the series is absolute convergent, we show that $\sum \limits_{n=0}^\infty \left|\dfrac{(-1)^{n+1}}{5n+1}\right|$ is converging.
\begin{align}
\sum \limits_{n=0}^\infty \left|\dfrac{(-1)^{n+1}}{5n+1}\right|&=\sum \limits_{n=0}^\infty \dfrac{\mid (-1)^{n+1}\mid }{\mid 5n+1\mid}\\
&=\sum \limits_{n=0}^\infty \frac{1}{5n+1}\\
&\ge\color{red}{\sum \limits_{n=0}^\infty \frac{1}{5n+5}}\\
&=\color{red}{\frac15\sum \limits_{n=0}^\infty \frac{1}{n+1}}\\
&=\color{red}{\frac15\sum \limits_{n=1}^\infty \frac{1}{n}}\\
&\implies \text{harmonic series} \implies divergent
\end{align}
$\sum \limits_{n=0}^\infty \dfrac{(-1)^{n+1}}{5n+1}$ is convergent but not absolute convergent. $_\blacksquare$
|
You first part is fine for the second that step is wrong
$$\sum \limits_{n=0}^\infty \frac{1}{5n+1}
\color{red}{=\sum \limits_{n=1}^\infty \frac{1}{5n}}$$
we can simply refer directly to limit comparison test with $\sum \frac 1n$ and conclude for divergence indeed
$$\frac{\frac{1}{5n+1}}{\frac1n}=\frac{n}{5n+1}\to \frac15$$
or as an alternative
$$\sum \limits_{n=0}^\infty \frac{1}{5n+1}\ge \sum \limits_{n=0}^\infty \frac{1}{5n+5}=\frac15\sum \limits_{n=0}^\infty \frac{1}{n+1}=\frac15\sum \limits_{n=1}^\infty \frac{1}{n}$$
|
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|
Evaluate $\lim_{x\to 0}\frac{e^{(x+1)^{1/x}}-(x+1)^{e/x}}{x^2}$ By L'Hôpital's rule, it gets $$\lim_{x\to 0}-\frac{\left(e^{(x+1)^{1/x}} (x+1)^{1/x}-e (x+1)^{e/x}\right) ((x+1) \log (x+1)-x)}{2x^3(1+x)},$$which becomes more complicated.
What could I do then?
|
We have that
$$(x+1)^{\frac{1}{x}}=e^{\frac1x \log (1+x)}=e^{1-\frac12 x+\frac13x^2+o(x^2)}=e \cdot e^{-\frac12 x+\frac13x^2+o(x^2)}=e\left(1-\frac 1 2 x+\frac 1 3 x^2 +\frac12\left(-\frac12 x+\frac13x^2\right)^2+ o(x^2)\right)
=e\left(1-\frac 1 2 x+\frac{11}{24}x^2+o(x^2)\right)$$
$$e^{{(x+1)}^{\frac{1}{x}}}=e^{e}e^{-\frac e 2 x+\frac{11e}{24}x^2+o(x^2)}=e^{e}\left(1-\frac e 2 x+\frac{11e}{24}x^2+\frac{e^2}{8}x^2+o(x^2)\right)$$
and
$$(x+1)^{\frac{e}{x}}=e^{\frac e x \log (1+x)}=e^{e-\frac e 2 x+\frac e3x^2+o(x^2)}=e^ee^{-\frac e 2 x+\frac e3x^2+o(x^2)}=e^e\left(1-\frac e 2 x+\frac e3x^2+\frac {e^2}{8}x^2+o(x^2)\right)$$
and thus
$$\dfrac{e^{(x+1)^{\frac{1}{x}}}-(x+1)^{\frac{e}{x}}}{x^2}=\frac{e^{e}\left(1-\frac e 2 x+\frac{11e}{24}x^2+\frac{e^2}{8}x^2+o(x^2)\right)-e^e\left(1-\frac e 2 x+\frac e3x^2+\frac {e^2}{8}x^2+o(x^2)\right)}{x^2}=$$
$$=\frac{\frac{11e^{e+1}}{24}x^2-\frac{e^{e+1}}{3}x^2+o(x^2)}{x^2}=\frac{\frac{e^{e+1}}{8}x^2+o(x^2)}{x^2}=\frac{e^{e+1}}{8}+o(1)\to \frac{e^{e+1}}{8}$$
|
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|
Calculating coefficient I have a generating function,
$$ \frac{(1-x^7)^6}{(1-x)^6} $$
and I want to calculate the coefficient of $x^{26}$
Solution for this is,
$$ {26+5 \choose 5} - 6{19+5 \choose 5} + 15{12+5 \choose 5} - 20{5+5 \choose 5} $$
Is there formula for this? If there is, what is called?
If there is no formula, how can I calculate it?
Thanks!
|
Use the Negative binomial series:
$$\begin{align}[x^{26}]\frac{(1-x^7)^6}{(1-x)^6}&=[x^{26}](1-x^7)^6\cdot (1-x)^{-6}=\\
&=[x^{26}]\sum_{i=0}^6 {6\choose i}(-x^7)^i\cdot \sum_{j=0}^{\infty} {-6\choose j}(-x)^j=\\
&=[x^{26}]\sum_{i=0}^6 {6\choose i}(-x^7)^i\cdot \sum_{j=0}^{\infty} {6+j-1\choose j}x^j=\\
&=[x^{26}]\left[{6\choose 0}(-x^7)^0\cdot {6+26-1\choose 26}x^{26}+\\
\qquad \qquad {6\choose 1}(-x^7)^1\cdot {6+19-1\choose 19}x^{19}+\\
\qquad \qquad {6\choose 2}(-x^7)^2\cdot {6+12-1\choose 12}x^{12}+\\
\qquad \qquad {6\choose 3}(-x^7)^3\cdot {6+5-1\choose 5}x^{5}+\right]=\\
&={6\choose 0}{31\choose 26}-{6\choose 1}{24\choose 19}+{6\choose 2}{17\choose 12}-{6\choose 3}{10\choose 5}.\end{align}$$
|
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|
Prove $\left(1-\frac{1}{n^2}\right)^n\times\left(1+\frac 1n\right)<1$ for every natural $n > 0$ by induction Consider this inequality
$$\left(1-\dfrac{1}{n^2}\right)^n\times\left(1+\dfrac 1n\right)<1$$
which is meant to be valid for any nonzero natural number $n$.
It is asked to prove it by induction. I haven't made any significant progress after many tries.
Any advice is welcome.
Note :
A way I encountered which looks nicer and more straightforward is to notice that
$1-\dfrac{1}{n^2}<1-\dfrac{1}{n^2+k}$ for $k\in \{1,\cdots,n\}$
Multiplying side by side those $n$ inequalities, a nice telescoping takes place to give exaclty the sought result.
|
The inequality
$\left(1-\dfrac{1}{n^2}\right)^n\cdot\left(1+\dfrac 1n\right)
=\left(1-\dfrac{1}{n}\right)^n\left(1+\dfrac{1}{n}\right)^{n+1}
=\dfrac{\left(1+\dfrac{1}{n}\right)^{n+1}}{\left(1+\dfrac{1}{n-1}\right)^{n}}<1$
is equivalent to the sequence
$\left\{\left(1+\dfrac{1}{n}\right)^{n+1}\right\}$
is strictly decreasing.
So we will show that：sequence
$\left\{\left(1+\dfrac{1}{n}\right)^{n+1}\right\}$
is strictly decreasing.
By AG-MG inequality
$\dfrac{1}{\left(1+\dfrac{1}{n}\right)^{n+1}}=\left(\dfrac{n}{n+1}\right)^{n+1}
=1\cdot\dfrac{n}{n+1}\cdot\dfrac{n}{n+1}\cdots\dfrac{n}{n+1}
<\left(\dfrac{n+1}{n+2}\right)^{n+2}$
$=\left(\dfrac{n+1}{n+2}\right)^{n+2}=\dfrac{1}{\left(1+\dfrac{1}{n+1}\right)^{n+2}}.$
Hence $$\left(1+\dfrac{1}{n}\right)^{n+1}>\left(1+\dfrac{1}{n+1}\right)^{n+2}.$$
|
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|
Integral using polar coordinates Let $X=\{(x,y)\in\mathbf{R}^2\mid x^2+y^2\leqslant 1,x,y\geqslant 0 \}$. Calculate $\int_X xye^{x^2+y^2}\,dx\,dy$.
By using polar coordinates, I get $=\int_0^{\pi/2}\int_0^1e \cos\theta\sin\theta\,dr\,d\theta=\int_0^{\pi/2}\frac{1}{2}e\sin 2\theta\,d\theta=\left.-\frac{1}{4}e\cos 2\theta\right\vert_0^{\pi/2}=\frac{1}{4}e+\frac{1}{4}e=\frac{1}{2}e $.
By rewrting, I get
$=\int_0^1 \int_{0}^{\sqrt{1-y^2}} xye^{x^2+y^2}\,dx\,dy=\int_0^1 ye^{y^2}\left(\frac{1}{2}e^{x^2}\right)_0^{\sqrt{1-y^2}}\,dy=\int_0^1 \frac{1}{2}ey-\frac{1}{2}ye^{y^2}\,dy=\left.\frac{1}{4}ey^2-\frac{1}{4}e^{y^2}\right\vert_0^1=\frac{1}{4}e-\frac{1}{4}e+\frac{1}{4}e=\frac{1}{4}e$.
The code in Mathematica
f[x] := x*y*Exp[x^2 + y^2]
Integrate[f[x] Boole[x^2 + y^2 < 1], {x, 0, 1}, {y, 0, 1}]
gives me $\frac{1}{4}$.
Who can tell me which of these three answers is correct and where are my mistakes?
|
Your polar integral is wrong; it should be
$$\int_0^{\pi/2}\int_0^1r^2\cos\theta\sin\theta\cdot e^{r^2}r\,dr\,d\theta$$
where the $r^2\cos\theta\sin\theta$ comes from rewriting $xy$ and the extra $r$ factor is the Jacobian of the transformation to polar coordinates. Continuing the evaluation, we get
$$=\int_0^{\pi/2}\cos\theta\sin\theta\int_0^1r^3e^{r^2}\,dr\,d\theta$$
$$=\int_0^{\pi/2}\cos\theta\sin\theta[(r^2-1)e^{r^2}/2]_0^1\,d\theta$$
$$=\int_0^{\pi/2}\frac12\cos\theta\sin\theta\,d\theta$$
$$=\frac14\int_0^{\pi/2}\sin2\theta\,d\theta$$
$$=\frac14[-1/2\cdot\cos2\theta]_0^{\pi/2}=\frac14$$
agreeing with Mathematica.
|
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|
How to solve this nonlinear equation? This is the system of equations:
$$xy + z = -30\\
yz + x = 30\\
zx + y = -18$$
I have also done some work: if we add the first two equations we'll get:
$$xy + z + yz + x = 0\\
\implies y (x+z) + z + x = 0\\
\implies (x+z)(y+1) = 0$$
Now, a product equals zero if at least one of the factors equals zero.
Thus we have:
$$y+1=0 \implies y=-1$$
and
$$x+z = 0 \implies x = -z \ \textrm{ or }\ z = -x$$
Now if we substitute this into the equations we get:
$$xy + z = -30 \implies x(-1)+(-x) = -30 \implies x=15$$
We have:
$$x = 15 \ \textrm{ and }\ y = 1$$
But then the third equation makes no sense.
$$zx + y = -18$$
How can I solve this and is there a strategy that does not use numerical algorithms?
Thanks!
|
As you observe, from the first two equations, $y=-1$ or $z=-x$. If $y=-1$, then we have two equations to solve for $x$ and $z$, namely, $x-z=30$ and $xz=-17$. Therefore, $x$ and $-z$ are the roots of the quadratic polynomial $t^2-30t+17$, whereby it follows that $$(x,y,z)=(15-4\sqrt{13},-1,-15-4\sqrt{13})$$
and
$$(x,y,z)=(15+4\sqrt{13},-1,-15+4\sqrt{13})$$
are solutions.
If $z=-x$, then we have two equations to solve for $x$ and $y$, i.e., $x(y-1)=-30$ and $x^2-y=18$. Thus,
$$x^2+\frac{30}{x}=x^2-(y-1)=(x^2-y)+1=19\,.$$
That is, $$(x+5)(x-2)(x-3)=x^3-19x+30=0\,.$$
Thus,
$$(x,y,z)=(-5,7,5)\,,$$
$$(x,y,z)=(2,-14,-2)\,,$$
and $$(x,y,z)=(3,-9,-3)$$
are the solutions in this case.
|
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|
How to calculate $\sum_{n=0}^{\infty}C_n\left(\frac{1}{2}\right)^{2n+1}\left(\frac{n+1}{2n+1}\right)$ Does the following sum equal 1 (or some amount less than 1)?
$$S\equiv\sum_{n=0}^{\infty}C_n\left(\frac{1}{2}\right)^{\!2n+1}\!\!\left(\frac{n+1}{2n+1}\right)=\sum_{n=0}^{\infty}\left(\frac{(2n)!}{(n+1)!\cdot n!}\right)\left(\frac{1}{2}\right)^{\!2n+1}\!\!\left(\frac{n+1}{2n+1}\right)$$
where $C_n$ is the $n$th Catalan number.
The first 100 sums yields .7573;
The first 1000 sums yields .7765;
The first 10000 sums yields .7826;
The first 100000 sums yields .7845. It is not clear to me if $S=1$ or $S<1$.
I know the following:
$$\sum_{n=0}^{\infty}C_n\left(\frac{1}{2}\right)^{\!2n+1}=1$$
|
Simplifying a bit the notation, since $C_n=\frac{1}{n+1}\binom{2n}{n}$, we want to compute
$$ S = \frac{1}{2}\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}\int_{0}^{1}x^{2n}\,dx \tag{1}$$
and we may easily recognize the Maclaurin series of $\frac{1}{\sqrt{1-x^2}}$, leading to:
$$ S = \frac{1}{2}\int_{0}^{1}\frac{dx}{\sqrt{1-x^2}}=\frac{\arcsin 1}{2}=\color{red}{\frac{\pi}{4}}.\tag{2}$$
Notice that $\frac{1}{4^n}\binom{2n}{n}\leq\frac{1}{\sqrt{\pi n}}$ implies, through creative telescoping,
$$ S\leq \frac{1}{2}\left(1+\frac{1}{\sqrt{\pi}}\sum_{n\geq 1}\frac{1}{(2n+1)\sqrt{n}}\right)\stackrel{\text{CT}}{\leq}\frac{1}{2}\left(1+\frac{1}{\sqrt{\pi}}\sum_{n\geq 1}\frac{1}{\sqrt{n-\frac{1}{6}}}-\frac{1}{\sqrt{n+\frac{5}{6}}}\right)$$
or $S\leq \frac{1}{2}\left(1+\sqrt{\frac{6}{5\pi}}\right)$, from which it follows that $\pi<3+\frac{2}{9}$.
|
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|
For which values of $a$ is $f$ primitivable?
Let $a \in \mathbb{R}$ and $p,q$ be natural numbers with $p \geq q+2.$ For which values of $a$ is the function
$$f(x) =
\begin{cases}
\frac{1}{x}\sin \frac{1}{x^p}\sin \frac{1}{x^q}, &x \neq 0 \\
a, &x=0
\end{cases}
$$
primitivable?
I noticed that $\frac{1}{x}\sin \frac{1}{x^p}\sin \frac{1}{x^q}$ doesn't have an elementary antiderivative, so I tried to obtain it from the derivative of some function and work this from there. But those powers in the denominator are really getting in the way of any attempt.
I also thought of using the formula $\sin a \sin b = \frac{1}{2}(\cos(a-b)-\cos(a+b))$ and so I split the function like this:
$$f(x)=
\frac{1}{2}\begin{cases}
\frac{1}{x}\cos(\frac{1}{x^p}-\frac{1}{x^q}), &x \neq 0 \\
a, &x=0
\end{cases} -
\frac{1}{2}\begin{cases}
\frac{1}{x}\cos(\frac{1}{x^p}+\frac{1}{x^q}), &x \neq 0 \\
a, &x=0
\end{cases}
$$
and this definitely looks more promising, but I don't know how to proceed.
|
Take $a=0$. Consider the function
$$
G(x):=x^{p}\cos\frac{1}{x^{p}}\sin\frac{1}{x^{q}}.
$$
Then for $x\neq0$,
\begin{align*}
G^{\prime}(x) & =px^{p-1}\cos\frac{1}{x^{p}}\sin\frac{1}{x^{q}}-px^{-1}%
\sin\frac{1}{x^{p}}\sin\frac{1}{x^{q}}\\
& -qx^{p-q-1}\cos\frac{1}{x^{p}}\cos\frac{1}{x^{q}}\\
& =:h(x)-pf(x),
\end{align*}
while for $x=0$,
$$
\frac{G(x)-G(0)}{x-0}=x^{p-1}\cos\frac{1}{x^{p}}\sin\frac{1}{x^{q}}%
\rightarrow0
$$
since $p\geq2$. Now the function
$$
h(x)=px^{p-1}\cos\frac{1}{x^{p}}\sin\frac{1}{x^{q}}-qx^{p-q-1}\cos\frac
{1}{x^{p}}\cos\frac{1}{x^{q}}%
$$
is continuous at $x=0$ provided we set $h(0):=0$, since $p-q-1\geq1$ and so
$$
\lim_{x\rightarrow0}h(x)=0.
$$
It follows that $h$ has an antiderivative $H(x):=\int_{0}^{x}h(t)\,dt$. It
follows that$$
-pf(x)=G^{\prime}(x)-H^{\prime}(x)
$$
and so the function given for $x\in\lbrack-1,1]$ by$$
F(x):=-\frac{1}{p}G(x)+\frac{1}{p}H(x)
$$
has derivative $F^{\prime}(x)=-\frac{1}{p}G^{\prime}(x)+\frac{1}{p}H^{\prime
}(x)=f(x)$.
For $x\geq1$ define
$$
F(x):=F(1)+\int_{1}^{x}f(t)\,dt.
$$
Note that $f$ is integrable in $[1,\infty)$ since $\sin t\sim t$ for
$t\rightarrow0$ and so as $x\rightarrow\infty$
we have $$
\frac{1}{x}\sin\frac{1}{x^{p}}\sin\frac{1}{x^{q}}\sim\frac{1}{x^{1+p+q}}
$$
and $1+p+q\geq3$. Then $F^{\prime}(x)=f(x)$ for all $x\geq1$.
Similarly, for $x\leq-1$ define
$$
F(x):=F(-1)+\int_{-1}^{x}f(t)\,dt.
$$
|
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|
Inverse of tridiagonal Toeplitz matrix Consider the following tridiagonal Toeplitz matrix. Let $n$ be even.
$${A_{n \times n}} = \left[ {\begin{array}{*{20}{c}}
{0}&{1}&{}&{}&{}\\
{1}&{0}&{1}&{}&{}\\
{}&{1}&{\ddots}&{\ddots}&{}\\
{}&{}&{\ddots}&{\ddots}&{1}\\
{}&{}&{}&{1}&{0}
\end{array}} \right]$$
What is the inverse $A^{-1}$?
Clearly, $A^{-1}$ is symmetric.
I look for a proof of the following conjecture that $A^{-1}$ is given as follows:
If $A_{i, j}^{-1}$ such that $j$ is odd and $i =1+j + 2 m$ with $m\in \cal N_0$, then $A_{i, j}^{-1} = (-1)^m$. From which follows by symmetry:
If $A_{i, j}^{-1}$ such that $j$ is even and $i =-1+j - 2 m$ with $m\in \cal N_0$, then $A_{i, j}^{-1} = (-1)^m$.
All other $A_{i, j}^{-1} = 0$.
Here is an example, computed with Matlab, for $n=10$ which shows the structure:
$${A_{10 \times 10}^{-1}} = \left[ {\begin{array}{*{20}{r}}
0 & 1 & 0 & -1 & 0 & 1 & 0 & -1 & 0 & 1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & -1 & 0 & 1 & 0 & -1 \\
-1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & -1 & 0 & 1 \\
1 & 0 & -1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & -1 \\
-1 & 0 & 1 & 0 & -1 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
1 & 0 & -1 & 0 & 1 & 0 & -1 & 0 & 1 & 0
\end{array}} \right]$$
|
I found an algorithm that I hope it help you. Let the inverse be$$A^{-1}=\begin{bmatrix}c_1&c_2&\cdots&c_n\end{bmatrix}$$In other words, $c_i$s are the columns of $A^{-1}$. We build the $c_i$s recursively as following:
$$c_{2}=e_1\\c_{2k+2}=-c_{2k}+e_{2k+1}\qquad,\qquad 1\le k\le {n\over 2}-1$$also $$c_{n-1}=e_n\\c_{2k-1}=e_{2k}-c_{2k+1}\qquad,\qquad 1\le k\le {n\over 2}-1$$where $e_i$s are the columns of $I_n$
|
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|
How does one solve $\sin x-\sqrt{3}\ \cos x=1$? I thought this one up, but I am not sure how to solve it. Here is my attempt:
$$\sin x-\sqrt{3}\ \cos x=1$$
$$(\sin x-\sqrt{3}\ \cos x)^2=1$$
$$\sin^2x-2\sqrt{3}\sin x\cos x\ +3\cos^2x=1$$
$$1-2\sqrt{3}\sin x\cos x\ +2\cos^2x=1$$
$$2\cos^2x-2\sqrt{3}\sin x\cos x=0$$
$$2\cos x(\cos x-\sqrt{3}\sin x)=0$$
$2\cos x=0\Rightarrow x\in \{\frac{\pi }2(2n-1):n\in\Bbb Z\}$
But how do I solve $$\cos x-\sqrt{3}\sin x=0$$
|
Hint :
$$\cos x - \sqrt{3}\sin x = 0 \Leftrightarrow \frac{\sin x}{\cos x} = \frac{\sqrt{3}}{3} \Leftrightarrow \tan x = \frac{\sqrt{3}}{3}$$
Note : You can divide by $\cos x$, since if the case was $\cos x =0$, it would be $\sin x = \pm 1$ and thus the equation would yield $\pm \sqrt{3} \neq 0$, thus no problems in the final solution, as the $\cos$ zeros are no part of it.
|
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|
How to show $\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n}<2$ using induction I´d like to show that $$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n}+\frac{n+1}{2^{n+1}}<2$$ using the fact that $$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n}<2$$
I guess the answer use transitivity of natural numbers and the inequality $$n+1<2^n$$ In this case, i can use it because i suppose it for natural n such that n>1. I'd like to have something like $$....+(n+1)<2(2^{n+1})$$ because you can divide both sides by $2^{n+1}$, and so, you get $$...+\frac{n+1}{2^{n+1}}<2$$
I´ve noticed that $$2^n+n+1<2^n+2^n=2^{n+1}<2(2^{n+1})$$ and it leads me to think $$...+n+1<2^n+n+1$$ i´ve tried dividing both sides ,of the inequality i´m assuming as a fact, by $2^n/2$ in order to get $2^n$
in the right hand of the inequality and then, add $n+1$ to both sides: $$(\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n})\frac{2^n}{2}+n+1<2(\frac{2^n}{2})+n+1=2^n+n+1$$ but at this part i get stuck because if i continue it doesn't lead me to what i´d like to show.
Please, help me. I hope someone here have already proved this before, because i don't find anything similar to this at any part of the web.
|
If you can drop the method by induction you may also proceed as follows:
For $|x| < 1$ you have
$$\frac{1}{1-x}= \sum_{n=0}^{\infty} x^n \Rightarrow \frac{1}{(1-x)^2}= \sum_{n=1}^{\infty} n \cdot x^{n-1} \Rightarrow \color{blue}{\frac{x}{(1-x)^2}= \sum_{n=1}^{\infty} n \cdot x^{n}}$$
For $x=\frac{1}{2}$ you have
$$\color{blue}{\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \cdots} = \frac{\frac{1}{2}}{\left(1-\frac{1}{2}\right)^2} \color{blue}{= 2} \Rightarrow \boxed{\sum_{n=1}^N \frac{n}{2^n}< 2} \mbox{ for all } N \in \mathbb{N}$$
|
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Prove that number $1^{2005}+2^{2005}+\cdots+n^{2005}$ is not divisible by number $n+2$ Prove that number $1^{2005}+2^{2005}+\cdots+n^{2005}$ is not divisible by number $n+2$ for every $n\in \mathbb N$
I have solution $2(1^{2005}+2^{2005}+ \cdots +n^{2005})=2+(2^{2005}+n^{2005})+(3^{2005}+(n-1)^{2005})+\cdots+(n^{2005}+2^{2005})=2+(n+2)M$, but I do not know how he get that result and i try to do on another way. I know that $1+2+\cdots+n=\frac{n(n+1)}{2}$and that $1+2+\cdots+n| 1^{2005}+2^{2005}+ \cdots +n^{2005}$.
So $\frac{n(n+1)}{2}|1^{2005}+2^{2005}+ \cdots +n^{2005}.$ that mean $n(n+1)|2(1^{2005}+2^{2005}+ \cdots +n^{2005})$ so $2(1^{2005}+2^{2005}+ \cdots +n^{2005})=M_1n(n+1)$ now if $n+2|1^{2005}+2^{2005}+ \cdots +n^{2005}$ that mean that $1^{2005}+2^{2005}+ \cdots +n^{2005}=M_2(n+2)+r$, where $r=0$ but I need to show that is true or not so I do this.
$2(M_2(n+2)+r)=M_1n(n+1)$ , but I stuck here so maybe this is not good option to show that $r \not =0,$ do you have some idea?
|
The solution in your statement is using the following fact:
Let $k$ be an odd natural number. Then $x+y$ divides $x^k+y^k$.
The simplest way to get this is perhaps to view this as a polynomial in $x$ and then if we substitute $x=-y$, we get $x^k+y^k=x^k+(-x)^k=0$.
In your question each of the term: $(m^{2005}+(n-m+2)^{2005})$ will be divisible by $m+(n-m+2)=n+2$. This shows that
\begin{align*}
2(1^{2005}+2^{2005}+ \cdots +n^{2005}) &=2+(2^{2005}+n^{2005})+(3^{2005}+(n-1)^{2005})+\cdots+(n^{2005}+2^{2005})\\
&=2+A_2(n+2)+A_3(n+2)+\dotsb+A_n(n+2)\\
&=2+(n+2)[A_2+\dotsb+A_n]\\
&=2+(n+2)M
\end{align*}
|
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|
Can you prove this equality? In their book An Introduction to Optimization, on the chapter on gradient algorithms, to prepare for discussing convergence properties of the descent methods, authors Chong and Zak have following:
$f(x) = \frac{1}{2}x^TQx-b^Tx$, where $Q$ is symmetric and $Q>0$.
...
$V(x)=f(x)+\frac{1}{2}(x^*)^TQx^*=\frac{1}{2}(x-x^*)^T Q (x-x^*)$, where $x^*$ is the solution point obtained by solving $Qx=b$, that is, $x^*=Q^{-1}b$
I could not follow the equation $f(x)+\frac{1}{2}(x^*)^TQx^*=\frac{1}{2}(x-x^*)^T Q (x-x^*)$. Can you prove it?
|
\begin{align}
V(x)=&f(x)+\frac{1}{2}(x^*)^TQx^* \\
=&\frac12 x^TQx-b^Tx+ \frac{1}{2}(x^*)^TQx^* \\
=&\frac12 x^TQx-(Qx^*)^Tx+ \frac{1}{2}(x^*)^TQx^* \\
=&\frac12 x^TQx-(x^*)^TQx+ \frac{1}{2}(x^*)^TQx^* \quad(\because Q^T=Q)\\
=&\frac12 x^TQx-\frac12(x^*)^TQx-\frac12(x^*)^TQx+ \frac{1}{2}(x^*)^TQx^*\\
=&\frac12 x^TQx-\frac12x^TQx^*-\frac12(x^*)^TQx+ \frac{1}{2}(x^*)^TQx^* \quad(\because Q^T=Q)\\
=& \frac{1}{2}x^T Q (x-x^*) - \frac{1}{2}(x^*)^T Q (x-x^*) \\
=&\frac{1}{2}(x-x^*)^T Q (x-x^*)
\end{align}
|
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|
System of equations $a(1 - b^2) = b(1 -c^2) = c(1 -d^2) = d(1 - a^2)$ Given a positive real number $t$, find the number of real solutions $a, b, c, d$ of the system
$$a(1 - b^2) = b(1 -c^2) = c(1 -d^2) = d(1 - a^2) = t$$
I have a solution
Let $f(x)=\frac t{1-x^2}$ and we get $f(f(f(f(a))))=a$ and so we are looking for the number of fixed points of $f(f(f(f(x))))$
1) fixed points of $f(x)$
Any fixed point $u$ of $f(x)$ gives a solution $(u,u,u,u)$.
These fixed points are solution of $x^3-x+t=0$ and so :
If $t\in(0,\frac 2{3\sqrt 3})$ : three such solutions
If $t=\frac 2{3\sqrt 3})$ : two such solutions
If $t\in(\frac 2{3\sqrt 3},+\infty)$ : one such solution
2) fixed points of $f(f(x))$ which are not fixed points of $f(x)$
Any fixed point $u$ of $f(f(x))$ which is not fixed point of $f(x)$ gives a solution $(u,f(u),u,f(u))$
These fixed points are solution of $x^2-tx-1=0$ and so :
Always two solutions to this quadratic giving the same solutions to the original equation (just a circular permutation)
3) fixed points of $f(f(f(f(x))))$ which are neither fixed points of $f(x)$, neither fixed points of $f(f(x))$
These are solutions of a degree $12$ or degree $11$ ugly polynomial :
$(t^4-3t^2+1)x^{12}-t^3x^{11}+(-7t^4+18t^2-6)x^{10}$
$+(-t^5+6t^3)x^{9}+(26t^4-48t^2+15)x^{8}+(6t^5-14t^3)x^{7}$
$+(t^6-54t^4+72t^2-20)x^{6}+(-14t^5+16t^3)x^{5}+(-7t^6+64t^4-63t^2+15)x^{4}$
$+(-t^7+14t^5-9t^3)x^{3}+(11t^6-41t^4+30t^2-6)x^{2}+(2t^7-5t^5+2t^3)x^{1}$
$+(t^8-6t^6+11t^4-6t^2+1)$
And I dont know how to determine the number of real roots of this polynomial :
For certain values (for example $t=\frac 38$), this polynomial has $12$ real roots and so $3$ new solutions of original equation (leading to $3+1+3=7$ solutions to the equation (counting as one all circular permutations))
For certain values (for example $t=10$), this polynomial has no real roots and so only $2$ solutions for the equation.
There is certainly a clever transformation for the equation $f(f(f(f(x))))=x$ allowing to count the roots depending on $t$ easily.
But I did not see it up to now.
|
If you graph the function $x(1-y^2)=t$, you get a clear picture of what the solutions are.
Assume $1<t$, then you can have infinite negative solutions as per the path
shown. You cannot have a positive value for any one of the unknowns.
Only if $0<t<<1$ you may find infinite positive solutions starting with a high value for $a$.
|
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|
Find largest eigenvalue of specific nonnegative matrix I look for the largest eigenvalue of the following matrix (or at least a small upper bound).
The only thing I know is that the eigenvalue is smaller than 1 and converges to $\cos(\frac{\pi}{2n+1})$ with growing n.
In general, it is very hard to compute the characteristic polynomial to calculate the eigenvalue and that's why I hope for an easier way.
Has anyone some ideas?
The dimension of the matrix is $n \times n$.
$A = \begin{bmatrix}
\frac{1}{2-\frac{1}{n+1}}& \frac{1}{2} & 0 & 0 & \dots & 0 \\
\frac{1}{2} & 0 & \frac{1}{2} & 0 & \dots & 0 \\
0 & \frac{1}{2} & 0 & \frac{1}{2} & \dots & 0 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
0 & 0 & 0 &0 & \frac{1}{2} & 0
\end{bmatrix}$
|
An asymptotic expansion for large $n$ can be obtained as follows. Expanding $A - \lambda I$ by the first row and expanding one of the resulting matrices by the first column gives
$$\det(A - \lambda I) =
\left( \frac {n + 1} {2 n + 1} - \lambda \right) \det \tilde A_{n - 1} -
\frac 1 4 \det \tilde A_{n - 2},$$
where $\tilde A_n$ is a tridiagonal Toeplitz matrix with $-\lambda$ on the main diagonal and $1/2$ on the two adjacent diagonals. The determinant of $\tilde A_n$ is $2^{-n} U_n(-\lambda)$, where $U_n$ is the Chebyshev polynomial of the second kind. Since $U_n(-\cos \theta) = (-1)^n \csc \theta \,\sin \,(n + 1) \theta$, $\det(A - \lambda I) = 0$ reduces to
$$\cot n \theta =
\left( \frac {2 n + 2} {2 n + 1} - \cos \theta \right) \csc \theta.$$
Taking $\theta = \alpha/n$ and expanding both sides of the equation into series, we obtain $\cot \alpha = 1/(2 \alpha) + O(1/n)$, determining $\alpha$. Then
$$\lambda_{\max} = \cos \theta \sim
1 -\frac {\alpha^2} {2 n^2},$$
where $\alpha$ is the smallest positive root of $\cot \alpha = 1/(2 \alpha)$. The next terms can be obtained in the same manner.
|
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|
$x$ intercept problem How would I find the $x$ intercept of $x^5-x^3+2=0$? I haven’t learned about things like synthetic division or any theorems, just algebraic manipulations.
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My answer is not totally correct but I hope it helps a little :
Consider $f(x)=x^5-x^3+2$ . we have $f'(x)=5x^4-3x^2$ and $f'(x)=0$ if $x=0$ or $x=\pm \sqrt{\frac{3}{5}}$ and it is also possible to determine the sign of $f'(x)$ for all other $x$.
Note that $5x^4-3x^2>0$ if $5x^4>3x^2$ if $x^2>3/5$ . that is, $x >+ \sqrt{\frac{3}{5}}$ or $x <- \sqrt{\frac{3}{5}}$. Also $5x^4-3x^2<0$ if $- \sqrt{\frac{3}{5}}<x<+ \sqrt{\frac{3}{5}}$.
Thus $f$ is increasing for $x <- \sqrt{\frac{3}{5}}$, decreasing between $- \sqrt{\frac{3}{5}}$ and $+ \sqrt{\frac{3}{5}}$ and again increasing for $x>+ \sqrt{\frac{3}{5}}$.
For this $f$, finding $x$ for $f(x)=0$ is difficult. But IVT guranteed that such a number is between $-2$ and $-1$, since $f(-2)<0$ and $f(-1)>0$
with this information the graph look like this:(with the help of Desmos)
|
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|
A differentiation/derivative/calculus problem The question is as follows:
$$y=x^2/(x+1)$$
The normal to this curve at $x=1$ meets the $x$-axis at point $M$.
The tangent to the curve at $x=-2$ meets the $y$-axis at point $N$.
Find the area of triangle $MNO$, where $O$ is the origin.
PS- this is not a school h.w so don't worry. And I did try....for a good 100 min...not even joking
EDIT - The drawing is NOT EXACT, it is just to give an idea.
|
$y(x) = \dfrac{x^2}{x + 1} = (x + 1)^{-1}x^2; \tag 1$
$y'(x) = -(x + 1)^{-2}x^2 + 2x(x + 1)^{-1} = -(x + 1)^{-2}x^2 + 2x(x + 1)(x + 1)^{-2}$
$= -(x + 1)^{-2}x^2 + (2x^2 + 2x)(x + 1)^{-2} = (x^2 + 2x)(x + 1)^{-2} = \dfrac{x^2 + 2x}{(x + 1)^2}; \tag 2$
$y(1) = \dfrac{1}{2}; \; y'(1) = \dfrac{3}{4}; \tag 3$
$y(-2) = -4; \; y'(-2) = 0; \tag 4$
the slope of the normal line through $(1, y(1)) = (1, 1/2)$ is then
$m = -\dfrac{1}{y'(1)} = -\dfrac{4}{3}; \tag 5$
the equation of the normal line through $(1, 1/2)$ is thus
$y - \dfrac{1}{2} = -\dfrac{4}{3}(x - 1), \tag 6$
which meets the $x$-axis where $y = 0$:
$-\dfrac{1}{2} = -\dfrac{4}{3}(x - 1) \Longrightarrow x = \dfrac{3}{8} + 1 = \dfrac{11}{8}; \tag 7$
thus,
$M = \left (\dfrac{11}{8}, 0 \right); \tag 8$
likewise, the tangent line through $(-2, y(-2)) = (-2, -4)$ is
$y + 4 = 0(x + 2) = 0, \tag 9$
which intersects the $y$-axis where $x = 0$, with $y$-coordinate given by
$y + 4 = 0 \Longrightarrow y = -4, \tag{10}$
and so
$N = \left (0, -4 \right ); \tag{11}$
the area $A$ of $\triangle MNO$ is thus
$A = \dfrac{1}{2}ON \cdot OM = \dfrac{1}{2} 4 \cdot \dfrac{11}{8} = \dfrac{11}{4}. \tag{12}$
|
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|
Prove that the polynomial is $g(x,y)(x^2 + y^2 -1)^2 + c$ This is from a Brazilian math contest for college students (OBMU):
Let $f(x,y)$ be a polynomial in two real variables such that the polynomials
$$\frac{\partial f}{\partial x}(x,y)$$
$$\frac{\partial f}{\partial y}(x,y)$$
are divisible by $x^2+y^2-1$. Prove that there's a polynomial $g(x,y)$ and a constant $c$ such that
$$f(x,y) = g(x,y)(x^2+y^2 -1)^2 +c$$
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Treating $f$ as a polynomial in $(\mathbb{R}[y])[x]$, there exist polynomials $p(x,y) \in (\mathbb{R}[y])[x], q(y), r(y) \in \mathbb{R}[y]$, such that $f(x,y)=(x^2+y^2-1)p(x,y)+xq(y)+r(y)$.
We have that $\frac{\partial f}{\partial x}(x,y)=(x^2+y^2-1)\frac{\partial p}{\partial x}(x,y)+2xp(x,y)+q(y)$ and $\frac{\partial f}{\partial y}(x,y)=(x^2+y^2-1)\frac{\partial p}{\partial y}(x,y)+2yp(x,y)+xq’(y)+r’(y)$ are divisible by $(x^2+y^2-1)$. Therefore $2xp(x,y)+q(y)$ and $2yp(x,y)+xq’(y)+r’(y)$ are divisible by $(x^2+y^2-1)$
As we did earlier, there exist polynomials $s(x,y) \in (\mathbb{R}[y])[x], t(y),u(y) \in \mathbb{R}[y]$ such that $p(x,y)=(x^2+y^2-1)s(x,y)+xt(y)+u(y)$.
We have that
\begin{align}
2xp(x,y)+q(y)&=2x(x^2+y^2-1)s(x,y)+2x^2t(y)+2xu(y)+q(y) \\
&=(x^2+y^2-1)(2xs(x,y)+2t(y))+x(2u(y))+(q(y)-2(y^2-1)t(y))\\
\end{align}
is divisible by $x^2+y^2-1$. Thus $2u(y)=0,q(y)-2(y^2-1)t(y)=0$. Have $q’(y)=4yt(y)+2(y^2-1)t’(y)$
Next we have
\begin{align}
2yp(x,y)+xq’(y)+r’(y)&=2y(x^2+y^2-1)s(x,y)+2xyt(y)+2yu(y)+xq’(y)+r’(y)\\
&=2y(x^2+y^2-1)s(x,y)+x(2yt(y)+4yt(y)+2(y^2-1)t’(y))+r’(y)\\
\end{align}
is divisible by $x^2+y^2-1$. Thus $6yt(y)+2(y^2-1)t’(y)=0, r’(y)=0$. Thus $r(y)=c$ for some constant $c$.
We shall show $t(y)=0$. Assume on the contrary $t$ is not identically $0$. Let $t$ have degree $n$ with nonzero leading coefficient $a$. Comparing the leading coefficient in $6yt(y)+2(y^2-1)t’(y)=0$, we get $6a=2an$, so $n=3$. Note $t$ is divisible by $y^2-1$, so $t(y)=a(y^2-1)(y+b)$, some $b \in \mathbb{R}$. Substituting $y=0$ in $6yt(y)+2(y^2-1)t’(y)=0$ gives $t’(0)=0$. However $t’(0)=-a$ is nonzero, a contradiction. Thus $t(y)=0$, and so $q(y)=2(y^2-1)t(y)=0$.
Thus
\begin{align}
f(x,y)&=(x^2+y^2-1)p(x,y)+xq(y)+r(y)\\
&=(x^2+y^2-1)^2s(x,y)+x(x^2+y^2-1)t(x,y)+(x^2+y^2-1)u(y)+xq(y)+r(y)\\
&=(x^2+y^2-1)^2s(x,y)+c\\
\end{align}
and we are done.
|
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|
$\lim\limits_{x\to 0} \frac{\tan x - \sin x}{x^3}$? $$\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}$$
Solution
\begin{align}\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}&=\\&=\lim_{x\to 0} \frac{\tan x}{x^3} - \lim_{x\to 0} \frac{\sin x}{x^3}\\
&= \lim_{x\to 0}\frac{\tan x}{x}\lim_{x\to 0} \frac{1}{x^2} -\lim_{x\to 0} \frac{\sin x}{x}\lim_{x\to 0} \frac{1}{x^2}\\&= \lim_{x\to 0} \frac{1}{x^2} -\lim_{x\to 0} \frac{1}{x^2}\\
&= \lim_{x\to 0} \frac{1}{x^2} -\frac{1}{x^2}\\&=0 \end{align}
But the answer is $\dfrac{1}{2}$ by L'Hopital's Rule.
|
Another way of evaluation can be the use of Taylor Maclurin Expansion of $tan x$ and $sin x$.
We have
$$\lim_{x \to 0} \tan x= \frac{x}{1} +\frac{x^{3}}{3} +\frac{2x^{5}}{15} + . . .$$
$$\lim_{x \to 0} \sin x= \frac{x}{1} - \frac{x^{3}}{6} +\frac{x^{5}}{120} + . . .$$
Therefore expression turns to,
$$\lim_{x \to 0} \frac{\frac{x}{1} +\frac{x^{3}}{3} +\frac{2x^{5}}{15} + . . . - (\frac{x}{1} - \frac{x^{3}}{6} +\frac{x^{5}}{120} + . . .)}{x^{3}}$$
Cancel the $x$ and then enforce the limit after dividing the numerator by $x^{3}$ . The expression simplifies to the calculation of sum of $\frac{1}{3}$ and $\frac{1}{6}$ which is $\color{red} {\frac{1}{2}}$
|
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How to calculate $\sum_{n=1}^{\infty}\frac{(2^n+(-1)^n)^2}{11^n} $ How to calculate $\sum_{n=1}^{\infty}\frac{(2^n+(-1)^n)^2}{11^n} $
I know that result $\frac{331}{910}$ because I checked it in Mathematica but I have troubles with calculate that.
$$\sum_{n=1}^{\infty}\frac{(2^n+(-1)^n)^2}{11^n} = ... = $$
$$ \sum_{n=1}^{\infty}(2^n \cdot (\frac{2}{11})^n+2\cdot(\frac{2}{11})^n\cdot(-1)^n + \frac{1}{11^n})$$ But what should be done after...?
|
For $|p|<1$, $\sum_{n=1}^{\infty}p^n=\frac{p}{1-p}$, so
$$\sum_{n=1}^{\infty}\frac{(2^n+(-1)^n)^2}{11^n}=\sum_{n=1}^{\infty}\frac{4^n}{11^n}+2\sum_{n=1}^{\infty}\frac{(-2)^n}{11^n}+\sum_{n=1}^{\infty}\frac{1}{11^n}\\
= \frac{\frac{4}{11}}{1-\frac{4}{11}}-2\frac{\frac{2}{11}}{1+\frac{2}{11}}+\frac{\frac{1}{11}}{1-\frac{1}{11}}\\=\frac{4}{7}-\frac{4}{13}+\frac{1}{10}=\frac{331}{910}$$
|
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Find the integral $\int _{1}^{e} (x \ln x)^2 dx$.
Find : $$\int _{1}^{e} (x \ln x)^2 \;dx.$$
My answer:
I have tried integration by parts with $u = x^2$ and $dv = (\ln x)^2$ but I end up having the same integration another time!
I reversed the role of $u$ & $v$, but it also did not work?
Do you have any suggestions ?
|
Let $u=(\ln x)^2$ and $dv=x^2 \,dx$. Then $du=2(\ln x)\frac{1}{x} \, dx$ and $v=\frac{x^3}{3}$. So
$$I=\int_1^e(x \ln x)^2 \, dx=(\ln x)^2 \frac{x^3}{3}\Big|_{1}^{e}-\frac{2}{3}\int_1^e x^2 \ln x \, dx.$$
Now we will solve the integral on the right side. Call the integral as $J$. For this $u=\ln x$ and $dv=x^2 \, dx$. So $du= \frac{1}{x}\,dx$ and $v=\frac{x^3}{3}$.Then
$$J=(\ln x) \frac{x^3}{3}\Big|_{1}^{e}-\frac{1}{3}\int_1^e x^ 2\, dx=(\ln x) \frac{x^3}{3}\Big|_{1}^{e}-\frac{x^3}{9}\Big|_{1}^{e}=\frac{e^3}{3}-\left(\frac{e^3-1}{9}\right)=\frac{2e^3+1}{9}.$$
So
$$I=\frac{e^3}{3}-\frac{2}{3}\left(\frac{2e^3+1}{9}\right)=\frac{5e^3-2}{27}$$
|
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Let $3\sin x +\cos x =2 $ then $\frac{3\sin x}{4\sin x+3\cos x}=?$ Let $3\sin x +\cos x =2 $ then $\dfrac{3\sin x}{4\sin x+3\cos x}=\,?$
My try :
$$\frac{\frac{3\sin x}{\cos x}}{\frac{4\sin x+3\cos x}{\cos x}}=\frac{3\tan x}{4\tan x+3} =\;?$$
Now we have to find $\tan x$ from $3\sin x +\cos x =2 $ but how?
|
Let $t=\tan(\frac x2)$.
use the identities
$$\sin(x)=\frac{2t}{1+t^2}$$
$$\cos(x)=\frac{1-t^2}{1+t^2},$$
and
$$\tan(x)=\frac{2t}{1-t^2}.$$
thus
$$3\sin(x)+\cos(x)=2\implies$$
$$6t+1-t^2=2(1+t^2) \implies$$
or
$$3t^2-6t+1=0$$
hence
$$\frac{3\tan(x)}{4\tan(x)+3}=$$
$$\frac{6t}{8t+3(1-t^2)}=$$
$$\frac{6t}{2t+4}=3-\frac{6}{t+2}$$
with $$t=1\pm \sqrt{\frac 23}$$
|
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|
Prove $a_n = \sqrt{n+1}-\sqrt{n}$ is monotone $n \ge 0$.
Prove $a_n = \sqrt{n+1}-\sqrt{n}$ is monotone $n \ge 0$.
To be monotone it must be either increasing or decreasing, so:
$a_n \ge a_{n-1}$ or $a_n \le a_{n-1}$
$\sqrt{n+1}-\sqrt{n} \ge? \sqrt{n}-\sqrt{n-1} $
$\sqrt{n+1}+\sqrt{n-1} \ge? \sqrt{n} + \sqrt{n}$
I know that $\sqrt{n+1} \ge \sqrt{n}$ and $\sqrt{n} \ge \sqrt{n-1}$
But I can't make a statement about their sum since they have opposite signs that is one is greater or equal and the other is less or equal.
Can you help me to figure out the solution?
|
$$
\\a_{n+1}\le a_{n}<=>
\\\sqrt{n+1}+\sqrt{n-1}\le2\sqrt{n}<=>
\\(\sqrt{n+1}+\sqrt{n-1})^2\le(2\sqrt{n})^2<=>
\\\sqrt{(n-1)(n+1)}=\sqrt{n^2-1}\le \sqrt{n^2}=n
$$
|
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|
Evaluate the limit of $\lim_{n\to\infty}\frac{1}{n^2}\left(\frac{2}{1}+\frac{9}{2}+\frac{64}{9}+\cdots+\frac{(n+1)^{n}}{n^{n-1}}\right)$ $$\lim_{n\to\infty}\frac{1}{n^2}\left(\frac{2}{1}+\frac{9}{2}+\frac{64}{9}+\cdots+\frac{(n+1)^{n}}{n^{n-1}}\right)$$
My try:
The limit can be written as follows:
$$\lim_{n\to\infty}\left(\frac{1}{n^2}\cdot\sum_{k=1}^{n}\frac{(k+1)^{k}}{k^{k-1}}\right)$$
Evaluate the following series:
$\sum_{k=1}^{\infty}\frac{(k+1)^{k}}{k^{k-1}}$
$\frac{(k+1)^{k}}{k^{k-1}}=k\cdot\frac{(k+1)^{k}}{k^{k}}=k\cdot\left(1+\frac{k+1}{k}-1\right)^k=k\cdot\left(1+\frac{1}{k}\right)^k$
Then:
$\lim_{k\to\infty}\frac{(k+1)^{k}}{k^{k-1}}=\lim_{k\to\infty}k\cdot\left(1+\frac{1}{k}\right)^k=e\cdot\infty\neq0 \Longrightarrow \sum_{k=1}^{\infty}\frac{(k+1)^{k}}{k^{k-1}}$ diverges.
Therefore:
$$\lim_{n\to\infty}\frac{1}{n^2}\left(\frac{2}{1}+\frac{9}{2}+\frac{64}{9}+\cdots+\frac{(n+1)^{n}}{n^{n-1}}\right)=0\cdot\infty$$
What to do next?
|
We have that by Stolz-Cesaro
$$\frac{a_n}{b_n}=\frac{\sum_{k=1}^{n}\frac{(k+1)^{k}}{k^{k-1}}}{n^2}$$
$$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\frac{\frac{(n+2)^{n+1}}{(n+1)^{n}}}{(n+1)^2-n^2}=\frac{(n+2)^{n+1}}{(2n+1)(n+1)^{n}}=\frac{n+2}{2n+1}\left(1+\frac{1}{n+1}\right)^{n} \to \frac e 2$$
|
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|
Evaluate integral $\int_{-2}^0 x^2+x\ dx$ using Riemann Sum Consider the integral $$\int_{-2}^0 x^2+x\ dx.$$
The question says to use Riemann Sum theorem which is $$\sum_{i=1}^nf(x_i)\delta x$$
I know that $\delta x= \frac{-2}{n}$ and that $x_i=-2+(\frac{2}{n}i)$
After i plug everything in I get $$\sum_{i=1}^n\frac{2}{n}\left(-2-\frac{2}{n}i\right)^2+\left(-2-\frac{2}{n}i\right)$$
After completing the square I have $$\frac{2}{n}\sum_{i=1}\left(4-\frac{8}{n}i\right)+\left(\frac{4}{n^2}i^2\right)+\left(-2-\frac{2}{n}i\right)$$
I know that $i=\left(\frac{(n+1)}{2}\right)$ and $i^2=\left(\frac{(n+1)(2n+1)}{6}\right)$
but how do I manipulate the equation so that I can use them?
|
I know that $\delta x= \frac{-2}{n}$
The $\delta x$ should be positive. You should think of it as the length of the corresponding rectangle. In this case it is simplest to take $x_i=-\frac{2i}{n}$. So you have,
\begin{align*}
&\int_{-2}^0 x^2+x\ dx\\
=&\lim_{n\rightarrow+\infty}
\sum_{i=1}^n \left(\left(-\frac{2i}{n}\right)^2+\left(-\frac{2i}{n}\right)\right)\frac{2}{n}\\
=&\lim_{n\rightarrow+\infty}
\sum_{i=1}^n \left(\frac{8i^2}{n^3}-\frac{4i}{n^2}\right)\\
=&\lim_{n\rightarrow+\infty}
\frac{8}{n^3}\left(\sum_{i=1}^n i^2\right)
-\frac{4}{n^2}\left(\sum_{i=1}^n i\right)\\
=&\lim_{n\rightarrow+\infty}
\frac{8}{n^3}\left(
\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}\right)
-\frac{4}{n^2}\left(\frac{1+n}{2}\right)\\
=&\lim_{n\rightarrow+\infty}
\left(
\frac{8}{3}+\frac{8}{2n}+\frac{n}{6n^2}\right)
-\left(\frac{2}{n}+2\right)\\
=& \frac{8}{3}-2
\end{align*}
|
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|
Let $x \in \mathbb{N}$ and let $p$ be a prime divisor of $x^4+x^3+x^2+x+1$, prove that $p = 5$ or $p \equiv 1 \mod{5}$
Let $x \in \mathbb{N}$ and let $p$ be a prime divisor of $x^4+x^3+x^2+x+1$, prove that $p = 5$ or $p \equiv 1 \mod{5}$
Note that $(x^4+x^3+x^2+x+1)(x-1) = x^5 - 1$.
I tried reducing the equation $\mod{5}$, which gave me some information on $x$, but I was not able to utilize this information.
|
*
*Notice that $x^4+x^3+x^2+x+1 = \frac{x^5-1}{x-1}$. Now, $\frac{x^5-1}{x-1} \equiv 0 \pmod p \implies x^5\equiv 1 \pmod p$. It follows that the order of $x$, call it $r$, $\pmod p$ divides $5$. But $5$ is a prime number, hence $r=1,5$.
*We know that The order of any number $\pmod p$ divides $\phi(p)$. If the order is $5$, then $5|(\phi(p)=p-1) \implies p\equiv 1 \pmod 5$. If otherwise, i.e. the order is $1$, then $x \equiv 1 \pmod p$. Therefore, $(x^1)^5-1 \equiv 0 \pmod 5$ implying that $5$ divides $x^5-1$.
|
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|
Mysterious polynomial sequence Can someone identify this polynomial sequence? Is it known in mathematics? I'm interested in various properties of this sequence.
I'd like to find $P(n)$, $n\in \mathbb{Z}^+$
\begin{align}
P(0)&= 1\\
P(1)&= a\\
P(2)&= a^2+b\\
P(3)&= a^3+2ab\\
P(4)&= a^4+3a^2b+b^2\\
P(5)&= a^5+4a^3b+3ab^2\\
P(6)&= a^6+5a^4b+6a^2b^2+b^3\\
P(7)&= a^7+6 a^5 b+10 a^3 b^2+4 a b^3\\
P(8)&= a^8 + 7 a^6 b + 15 a^4 b^2 + 10 a^2 b^3 + b^4\\
P(9)&= a^9 + 8 a^7 b + 21 a^5 b^2 + 20 a^3 b^3 + 5 a b^4\\
P(10)&= a^{10} + 9 a^8 b + 28 a^6 b^2 + 35 a^4 b^3 + 15 a^2 b^4 + b^5
\end{align}
More steps upon request.
I'll be grateful for any hints!
|
Hint. Note that the following recurrence holds: for $n\geq 2$,
$$P(n)=aP(n-1)+bP(n-2).$$
They are related to the Fibonacci polynomials. The wiki page gives a list of properties. For example we have that
$$P(n)=\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n-k}{k}a^{n-2k}b^k.$$
|
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|
How to factor this quadratic expression? A bit confused on how to factor $2x^2 + 5x − 3 = 0$. Firstly, I multiplied $a \cdot c$, so $2(-3)=-6$, however couldn't find two numbers that will add up to $5$. Then, I thought of the factors could be $6(-1)=-6$, and $6+-1=5$. However, shouldn't the $a$ and $c$ values always be multiplied together, (the $(2)(-3)$, rather than $6(-1)$?
When following the factors $-1$ and $6$. I have
$(2x^2-1x)(6x-3)$
$x(2x-1)+3(2x-1)$
Is this correct, if not; what is the best way to solve a leading coefficient when factoring?
|
We need to find two numbers with product $2(-3) = -6$ and sum $5$. You are correct that the numbers are $-1$ and $6$. Splitting the linear term yields
\begin{align*}
2x^2 + 5x - 3 & = 2x^2 - x + 6x - 3 && \text{split the linear term}\\
& = x(2x - 1) + 3(2x - 1) && \text{factor by grouping}\\
& = (x + 3)(2x - 1) && \text{extract the common factor}
\end{align*}
which you can verify by multiplying the factors.
You should not write $(2x^2 - 1x)(6x - 3)$ since
\begin{align*}
(2x^2 - 1x)(6x + 3) & = 2x^2(6x + 3) - 1x(6x + 3)\\
& = 12x^3 + 6x^2 - 6x^2 -3x\\
& = 12x^3 - 3x\\
& \neq 2x^2 + 5x - 3
\end{align*}
Instead, you can write $(2x^2 - 1x) + (6x - 3)$ or $2x^2 - 1x + 6x - 3$.
Also, you should be including equals signs since you are asserting that
$$2x^2 + 5x - 3 = 2x^2 - x + 6x - 3$$
|
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|
How to solve equations of this type My math teacher gave us some questions to practice for the midterm exam tomorrow, and I noticed some of them have the same wired pattern that I don't know if it has a name:
Q1. If: $x - \frac{1}{x} = 3$ then what is $x^2 + \frac{1}{x^2}$ equal to?
The answer for this question is 11, but I don't know how, I thought it should be 9, but my answer was wrong.
Q2. If: $\frac{x}{x + y} = 5$ then what is $\frac{y}{x + y}$ equal to?
The answer for this is -5, I also don't know how.
Q3. If: $x^4 + y^4 = 6 x^2 y^2 \land x\neq y$ then what is $\frac{x^2 + y^2}{x^2 - y^2}$ equal to?
I guess the answer for this was $\sqrt{2}$ but I'm not sure.
Any body can explain how to solve these questions, and questions of the same pattern?
|
Hints.
$1) x-\frac1x=3\implies x^2+\frac1{x^2}-2=9\\2)\frac y{x+y}+\frac x{x+y}=1\\3)x^4+y^4=6x^2y^2\\\implies x^4+y^4+2x^2y^2=(x^2+y^2)^2=8x^2y^2\\\implies x^4+y^4-2x^2y^2=(x^2-y^2)^2=4x^2y^2$
Divide the two and take the square root.
|
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|
I calculated $\sin 75^\circ$ as $\frac{1}{2\sqrt{2}}+\frac{\sqrt{3}}{2\sqrt{2}}$, but the answer is $\frac{\sqrt{2}+\sqrt{6}}{4}$. What went wrong? I calculated the exact value of $\sin 75^\circ$ as follows:
$$\begin{align}
\sin 75^\circ &= \sin(30^\circ + 45^\circ) \\
&=\sin 30^\circ \cos 45^\circ + \cos 30^\circ \sin 45^\circ \\
&=\frac12\cdot\frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2}\cdot\frac{1}{\sqrt{2}} \\
&= \frac{1}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}}
\end{align}$$
The actual answer is
$$\frac{\sqrt{2} + \sqrt{6}}{4}$$
My main confusion is how the textbook answer is completely different from mine, even though if I compute $\sin 30^\circ \cos45^\circ + \cos 30^\circ \sin 45^\circ$, it will be approximately the same value of $\sin 75^\circ$.
I think I'm having difficulty adding and subtracting the radicals. So, if someone can demonstrate to me how they got that answer, it will be helpful. Thanks.
|
They’re the same value. Multiply the numerator and denominator of your answer by $\sqrt 2$ to see why.
$$\frac{1+\sqrt 3}{2\sqrt 2} = \frac{\sqrt 2}{\sqrt 2}\cdot\frac{1+\sqrt 3}{2\sqrt 2} = \frac{\sqrt 2+\sqrt 6}{4}$$
You can also use $\sin 45 = \cos 45 = \frac{\sqrt 2}{2}$ (rationalizing $\frac{1}{\sqrt 2}$) to get the answer more easily.
|
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|
Isn't $\frac{ax+bi}{ax+bi}$ equal to $1$? Isn’t $\dfrac{ax+bi}{ax+bi}$ equal to $1$?
Here, $i=\sqrt{-1}$, & $a$,$b$ & $x$ $\in$ $R$
|
Division in the field of complex numbers is defined in the following manner:
$\frac{ax+bi}{ax+bi}=\frac{(ax+bi)(ax-bi)}{(ax+bi)(ax-bi)}$, so
$\frac{ax+bi}{ax+bi}=\frac{a^2x^2+b^2}{a^2x^2+b^2}$, which is a real number and equal to 1. Hence your intuition is correct. But note that this is not true for $b=0$ and either $a=0$ or $x=0$ or both. Because $\frac{a^2x^2+b^2}{a^2x^2+b^2}$ is not defined when $b=0$ and either $a=0$ or $x=0$ or both.
|
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|
Evaluate $\int_0^1\frac{\ln{(x^2+1)}}{x+1}dx$ Any idea on how to solve the following definite integral?
$$\int_0^1\frac{\ln{(x^2+1)}}{x+1}dx$$
I have tried to parameterize the integral like $\ln{(a^2x^2+1)}$ or $\ln{(x^2+a^2)}$, which don't seem to work.
|
Not a full answer (yet)
I'll try to provide a solution which doesn't use any special functions, only the well known "classic" series.
$$I=\int_0^1 \frac{\ln{(x^2+1)}}{x+1}dx=\sum_{k,n=0}^\infty \frac{(-1)^{k+n}}{k+1} \int_0^1 x^{2(k+1)+n}dx=\sum_{k,n=0}^\infty \frac{(-1)^{k+n}}{(k+1)(2k+n+3)}$$
Let's use partial fractions:
$$\frac{1}{(k+1)(2k+n+3)}=\frac{a}{k+1}+\frac{b}{2k+n+3} \\ 2a+b=0 \\ (n+3)a+b=1 \\ (n+1)a=1 \\ a=\frac{1}{n+1} \\ b=-\frac{2}{n+1}$$
So we have:
$$I=\sum_{k,n=0}^\infty \frac{(-1)^n}{n+1} \frac{(-1)^k}{k+1}-2 \sum_{k,n=0}^\infty \frac{(-1)^n}{n+1} \frac{(-1)^k}{2k+n+3}=\ln^2 2-2\sum_{n=0}^\infty \frac{(-1)^n}{n+1} \sum_{k=0}^\infty \frac{(-1)^k}{2k+n+3}$$
The second series can be represented the following way, separating even and odd terms:
$$\sum_{n=0}^\infty \frac{(-1)^n}{n+1} \sum_{k=0}^\infty \frac{(-1)^k}{2k+n+3}=\sum_{l=0}^\infty \frac{1}{2l+1} \sum_{k=0}^\infty \frac{(-1)^k}{2(k+l)+3}-\sum_{l=0}^\infty \frac{1}{2l+2} \sum_{k=0}^\infty \frac{(-1)^k}{2(k+l)+4}$$
Now let's deal with the inner series separately:
1) $$\sum_{k=0}^\infty \frac{(-1)^k}{2(k+l)+3}=(-1)^{l+1} \sum_{k=l+1}^\infty \frac{(-1)^k}{2k+1}=(-1)^{l+1} \left( \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}- \sum_{k=0}^l \frac{(-1)^k}{2k+1} \right)= \\ =(-1)^{l+1} \left( \frac{\pi}{4}- \sum_{k=0}^l \frac{(-1)^k}{2k+1} \right)$$
The first term is the well known Leibniz series for $\pi$.
2) $$\sum_{k=0}^\infty \frac{(-1)^k}{2(k+l)+4}=\frac{(-1)^{l+1}}{2} \sum_{k=l+1}^\infty \frac{(-1)^k}{k+1}=\frac{(-1)^{l+1}}{2} \left( \sum_{k=0}^\infty \frac{(-1)^k}{k+1}- \sum_{k=0}^{l} \frac{(-1)^k}{k+1} \right)= \\ =\frac{(-1)^{l+1}}{2} \left(\ln 2- \sum_{k=0}^{l} \frac{(-1)^k}{k+1} \right)$$
The first terms of each can now be summed w.r.t. $l$:
$$\sum_{l=0}^\infty \frac{1}{2l+1} (-1)^{l+1} \frac{\pi}{4}=-\frac{\pi^2}{16}$$
$$\frac{1}{2} \sum_{l=0}^\infty \frac{1}{l+1} \frac{(-1)^{l+1}}{2} \ln 2=-\frac{\ln^2 2}{4}$$
At this point we have:
$$I=\frac{\ln^2 2}{2}+\frac{\pi^2}{8}-\sum_{l=0}^\infty (-1)^l \left(\frac{2}{2l+1} \sum_{k=0}^l \frac{(-1)^k}{2k+1}-\frac{1}{2}\frac{1}{l+1} \sum_{k=0}^{l} \frac{(-1)^k}{k+1} \right)$$
Well, I tried. Now it seems that there's at least two similar questions with some good answers, but I'll still leave this attempt here in case it could be finished.
We need to prove that:
$$\sum_{l=0}^\infty (-1)^l \left(\frac{2}{2l+1} \sum_{k=0}^l \frac{(-1)^k}{2k+1}-\frac{1}{2}\frac{1}{l+1} \sum_{k=0}^{l} \frac{(-1)^k}{k+1} \right)=\frac{7 \pi^2}{48}-\frac{\ln^2 2}{4}$$
which seems like a long shot, even though numerically it's true.
Mostly I just wanted to show how the squared logarithm and $\pi$ appear in the answer.
|
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|
Solving a Diophantine Equation dealing with powers of 2 Let x and y be positive integers with no prime factors larger than 5. Find all such $x$ and $y$ which satisfy $$x^2-y^2=2^k$$
where the range of k is $1\leq k\leq2019$.
I managed to solve the equation for general solutions where $(x, y)=(3 \times 2^z, 2^z)$ and $(x, y)=(5 \times 2^z, 3 \times 2^z)$ for non-negative integers for $z$. However, I still have trouble finding the range of values for $z$ that fit the given range of $k$.
|
First, we have the equation:
$$x^2 - y^2 = 2^k$$
We define $\nu_2(n)$ to be the power of $2$ that divides $n$.
Now, assume that $\nu_2(x) \neq \nu_2(y)$. We divide our equation by $2^{2\min(\nu_2(x),\nu_2(y))}$. Then, we will have the LHS as an odd value, as one term would be even, and the other would be odd. Thus, we would have an odd power of $2$, that is, $2^{k-2\min(\nu_2(x),\nu_2(y))} = 1$. This would mean that the difference of two positive perfect squares is $1$. Contradiction.
Let $\nu_2(x) = \nu_2(y) = t$. Then, let $x = 2^t \cdot p$ and $y = 2^t \cdot q$. Here, $p$ and $q$ are odd. Let $l = k-2t$. By dividing by $2^{2t}$ :
$$p^2-q^2 = 2^l \implies (p-q)(p+q) = 2^l \implies p-q = 2^{l_1} \space , \space p+q = 2^{l_2}$$
Again, we can note that if $4 \mid p-q$ and $4 \mid p+q$, then $4 \mid (p-q) + (p+q) \implies 4 \mid 2p \implies 2 \mid p$. However, this is wrong as $p$ is odd. Thus, it is not possible for both of $p+q$ and $p-q$ to be divisible by $4$. Since they are both even and powers of $2$, and $p-q < p+q$, we have $p-q = 2$.
Solving $p-q = 2$ and $p+q = 2^{l-1}$, we get $p = 2^{l-2}+1$ and $q = 2^{l-2}+1$. We are given the condition that $x$ and $y$ have no prime factors greater than $5$. Then, we can note that the only prime factors of $p$ and $q$ are $3$ and $5$. Moreover, as $p-q = 2$, we can have $3$ and $5$ only dividing one of $p$ and $q$. Thus, one of $p$ and $q$ is a power of $3$ and the other is a power of $5$.
Case 1: Power of $5$ is equal to $1$
Here, we have $p > q$ and as the power of $5$ is equal to $1$, we have $q=1$ which would give us $p=3$. Then, we would have the solution:
$$ (x,y,k) = (3 \cdot 2^t , 2^t , 2t+3)$$
Case 2: Power of $5$ is more than $1$
Here, we can note that $p=2^{l-2}+1$ and $q = 2^{l-2}-1$. We have:
$$5 \mid 2^n \pm 1 \implies 2 \mid n$$
Thus, we have $2 \mid l-2$. Then, $3 \mid 2^{l-2}-1$.
Now, we get $p= 2^{l-2}+1 = 5^m \implies 2^{l-2} = 5^m-1$. By lifting the exponent lemma: $$\nu_2(5^m-1) = \nu_2(5-1) + \nu_2(m) = \nu_2(m) + 2 \implies 2^{l-4} \mid m$$
This shows that $m \geqslant 2^{l-4}$. Hence:
$$2^{l-2} = 5^m-1 \geqslant 5^{2^{l-4}}-1$$
which is true only when $l=4$. In that case, we get $p=5$ and $q=3$, which shows:
$$(x,y,k) = (5 \cdot 2^t , 3 \cdot 2^t , 2t+4)$$
Therefore, the only solutions are $(x,y,k) = (3 \cdot 2^t , 2^t , 2t+3)$ and $(x,y,k) = (5 \cdot 2^t , 3 \cdot 2^t , 2t+4)$.
|
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Prove that $a_n \in [0,2)$
Let $(a_n)_{n \in \mathbb{N}}$ be a sequence, with $a_0=0$, $a_{n+1}=\frac{6+a_n}{6-a_n}$.
Prove that $a_n \in [0,2)$ $\forall n \in \mathbb{N}$
Here's what I did:
I tried to prove this by induction:
Base case:
$0 \leq a_0 (=0) < 2$.
Inductive step:
Suppose that $0 \leq a_n < 2$
So $$\begin {split} 0 \leq a_n < 2 &\iff 0 \geq -a_n > -2 \\ &\iff 6 \geq 6-a_n > 6-2 \\ &\iff \frac{1}{6} \geq \frac{1}{6-a_n} > \frac{1}{4} \\ &\iff \frac{a_n}{6}+1 \geq \frac{6+a_n}{6-a_n} > \frac{3}{2} + \frac{a_n}{4} \end{split}$$
To be fair I have no idea if this is going somewhere.
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Let's do it by induction:
$0 \le a_n < 2$
So $6= 6- 0 \ge 6- a_n > 6-2 = 4$ and $\frac 16 \le \frac 1{6-a_n} < \frac 14$
$6 + a_0 \ge 6 > 0$ so
$\frac {6 + a_n}6 \le \frac {6+a_n}{6-a_n} < \frac{6+a_n}4$.
And $6+a_n \ge 6+0 = 6$ and $6+a_n < 6+2 = 8$ so
$0 \le 1 = \frac 66 \le \frac {6+a_n}6 \le \frac {6+a_n}{6-a_n} < \frac {6+a_n}4 < \frac 84 = 2$
So $a_{n+1} = \frac {6+a_n}{6+a_n} \in [0, 2)$. (If fact $a_{n+1} \in [1,2)$).
|
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|
How do I get from $x - x^2 = \frac{1}{4}$ to $x =\frac{1}{2}$? I'm working on a text book problem where I need to sketch the graph of $y = 4x^2 - 4x+1$ by finding where the curve meets the $x$ axis.
To start out I set $y = 0$ then tried to isolate $x$ then,
$4x - 4x^2 = 1$
$x - x^2 = \frac{1}{4}$
From here I want to continue algebraically to reach $x = \frac{1}{2}$ which I know is the solution from plotting the curve using an app and I can see that that makes sense.
However I am missing some concepts which allow me to turn $x - x^2 = \frac{1}{4}$ into $x = \frac{1}{2}$ and wanted some help to get unblocked.
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$$x-x^2=x(1-x)\leq\left(\frac{x+1-x}{2}\right)^2=\frac{1}{4}.$$
The equality occurs for $x=1-x$ only, which gives $x=\frac{1}{2}.$
|
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|
Does the constant $C$ in this solution to a differential equation equal infinity? The problem is $y' = -\frac{1}{t^2} - \frac{1}{t}y + y^2;\ y_p = \frac{1}{t}$. My solution is
$$\begin{align}
y = \frac{1}{t} + B &\implies y' = -\frac{1}{t^2} + B' \\
&\implies -\frac{1}{t^2} - \frac{1}{t}y + y^2 = -\frac{1}{t^2} + B' \\
&\implies -\frac{1}{t^2} - \frac{1}{t} \left(\frac{1}{t} + B \right) + \left(\frac{1}{t} + B\right)^2 = -\frac{1}{t^2} + B' \\
&\implies B' - \frac{1}{t}B = B^2 \\
&\implies L = B^{-1} \\
&\implies L' = -B^{-2} \left(B^2 + \frac{1}{t}B \right) \\
&\implies L' + \frac{1}{tB} = -1 \\
&\implies L' + \frac{1}{t}L = -1 \\
&\implies L_h = \frac{1}{t} \\
&\implies L = \frac{1}{t}\int\frac{-1}{\frac{1}{t}}dt \\
&\implies L = \frac{1}{t} \left(-\frac{1}{2}t^2 + C_{tentative} \right) \\
&\implies L = \frac{C - t^2}{2t} \\
&\implies B = \frac{2t}{C - t^2} \\
&\implies y = \frac{1}{t} + \frac{2t}{C - t^2}
\end{align}$$
Although this solution does not appear equivalent to the answer given in my book, one or two people in chat reviewed this work and could not see anything incorrect. The issue is that the given particular solution, $y_p = \frac{1}{t}$, cannot be obtained by plugging any finite value into the $C$ in my general solution. We could say that the general solution yields the given particular solution when $C =$ infinity, more specifically, when $C =$ some $\aleph$ expression which makes the term $\frac{2t}{C - t^2}$ disappear regardless of $t$, but this feels a bit outside the box for a textbook problem. Is my general solution correct, and if so, how if at all can we derive the given particular solution from it?
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For non-linear equations like this, it is valid to take $C\to\infty$. You can also write $c = \frac{1}{C}$ to get an alternate form
$$ y(t) = \frac{1}{t} + \frac{2ct}{1-ct^2} $$
|
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Is my approach correct to this equation? The problem is the following:
Does $a \in \mathbb{R}$ exist such that $[a + \sqrt{2n + 1}] = [a + \sqrt{2n + 2}]$ for all $n \in \mathbb{N}$? ($[x]$ denotes the whole part of $x$).
Note: I will also use $\{x\}$ to denote the fractional part of $x$. Also that $[x] + \{x\} = x$.
My approach is the following:
There is no such that $a$.
Proof: For the sake of simplicity, let's assume $a = k + \alpha$, $k \in \mathbb{Z}$ and $0 \le \alpha \lt 1$. Because of $[x + k] = [x] + k$ for any $k \in \mathbb{Z}$, and substituting $a = k + \alpha$, our equation $[k + \alpha + \sqrt{2n + 1}] = [k + \alpha + \sqrt{2n + 2}]$ becomes $[\alpha + \sqrt{2n + 1}] = [\alpha + \sqrt{2n + 2}]$. Now our task is to find $\alpha \in [0, 1)$ that satisfies the equation for any $n \in \mathbb{N}$.
First, let's find a formula for $\alpha$ for given $n$. There are two cases:
1. $[\sqrt{2n + 1}] = [\sqrt{2n + 2}] = m$, $m \in \mathbb{N}$. This happens when $2n + 2$ isn't a perfect square $\iff$ $2n + 2 \ne p^2 \iff n \ne \frac {p^2} {2} - 1$, $p \in \mathbb{N}$.
$[\alpha + \sqrt{2n + 1}] = [\alpha + \sqrt{2n + 2}] \iff [\alpha + [\sqrt{2n + 1}] + \{\sqrt{2n + 1}\}] = [\alpha + [\sqrt{2n + 2}] + \{\sqrt{2n + 2}\}] \iff [\alpha + \{\sqrt{2n + 1}\} + m] = [\alpha + \{\sqrt{2n + 2}\} + m] \iff [\alpha + \{\sqrt{2n + 1}\}] = [\alpha + \{\sqrt{2n + 2}\}]$.
From this we get that $\alpha \in [0, 1 - \{\sqrt{2n + 2}\}) \tag {1'}$
Edit #1 (no longer actual beacuse of Edit #2): took out the maximum function from $(1')$ for obvious reason: the fractional part function is monotonically growing on its period, so if $[x] = [y]$, then $\{x\} \lt \{y\}$ only if $x \lt y$.
Edit #2: $(1')$ is only a partial solution for the 1. case, it is indeed a solution if both sides of the latest equation is equal to $0$. But they can both equal to $1$, so we have to consider that $\alpha \in [1 - \{\sqrt{2n + 1}\}, 1)$. 1. case overall:
$\alpha \in [0, 1 - \{\sqrt{2n + 2}\}) \cup [1 - \{\sqrt{2n + 1}\}, 1) \tag 1$
2. $[\sqrt{2n + 1}] \ne [\sqrt{2n + 2}] \implies [\sqrt{2n + 1}] = m, [\sqrt{2n + 2}] = m + 1$, $m \in \mathbb{N}$. This happens when $2n + 2$ is a perfect square $\iff$ $2n + 2 = p^2 \iff n = \frac {p^2} {2} - 1$, $p \in \mathbb{N}$. Now $[\sqrt{2n + 2}] = \sqrt{2n + 2} \implies \{\sqrt{2n + 2}\} = 0$.
$[\alpha + \sqrt{2n + 1}] = [\alpha + \sqrt{2n + 2}] \iff [\alpha + [\sqrt{2n + 1}] + \{\sqrt{2n + 1}\}] = [\alpha + [\sqrt{2n + 2}] + \{\sqrt{2n + 2}\}] \iff [\alpha + \{\sqrt{2n + 1}\}] = [\alpha + \{\sqrt{2n + 2}\}] + 1 \iff [\alpha + \{\sqrt{2n + 1}\}] = [\alpha] + 1 \iff [\alpha + \{\sqrt{2n + 1}\}] = 1$
From this we get that $\alpha = 1 - \{\sqrt{2n + 1}\} \tag 2$
Now we prove the following: (this is where it gets really ambiguous)
There does exists $n_1, n_2 \in \mathbb{N}$, $n_1 \ne \frac {{p_1}^2} {2} - 1$, $n_2 = \frac {{p_2}^2} {2} - 1$, $p_1$, $p_2 \in \mathbb{N} \iff [\sqrt{2n_1 + 1}] = [\sqrt{2n_1 + 2}]$, $[\sqrt{2n_2 + 1}] \ne [\sqrt{2n_2 + 2}]$ such that $ 1 - \{\sqrt{2n_1 + 2}\} \le 1 - \{\sqrt{2n_2 + 1}\} \lt 1 - \{\sqrt{2n_1 + 1}\} \tag 3 \iff$ there is no $\alpha$ satisfying both $(1)$ and $(2)$.
$\implies \{\sqrt{2n_1 + 1}\} \lt \{\sqrt{2n_2 + 1}\} \le \{\sqrt{2n_1 + 2}\} \tag 4$
(this is where it gets really-really-really ambiguous)
If we substitute $n_1 = 3$ and $n_2 = 1$ in $(4)$ we get the following:
$\{\sqrt{7}\} \lt \{\sqrt{3}\} \le \{\sqrt{8}\} \implies \approx 0.64 \lt 0.73 \le 0.82$ wich is indeed true. So we found two exceptions for $n$ and we should be done with the proof.
This part is strongly wrong! I made a mistake. $(4)$ is wrong. In fact, I can't find $n_1$ and $n_2$ satisfying $(3)$. Now I believe that I am wrong and there does exist a desired $a$.
Note: I made a lot of edits. My proof changed since I posted the question. Please take a look at the edits.
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Your idea of looking at $\alpha=\{a\}$ is a good one.
If $\alpha \lt 2-\sqrt 3\approx 0.268$ we choose $n=1$ and note that $\sqrt {2n+1}+\alpha=\sqrt 3+\alpha$ has floor $1$ while $\sqrt {2n+2}=\sqrt 4=2$
If $\alpha \ge 2-\sqrt 3$ we want to choose $n$ so that $$\lfloor\alpha+\sqrt {2n+2}\rfloor = k \gt \lfloor \alpha +\sqrt {2n+1} \rfloor\\
\lfloor \alpha^2+2\alpha\sqrt{2n+2}+2n+2 \rfloor=k^2\gt \lfloor \alpha^2+2\alpha\sqrt{2n+1}+2n+1\rfloor \\
\lfloor \alpha^2+2\alpha\sqrt{2n+2}+1 \rfloor=k^2\gt \lfloor \alpha^2+2\alpha\sqrt{2n+1}\rfloor $$
We note that when $2n$ is a square $\sqrt {2n+1}\approx \sqrt{2n}(1+\frac 1{2n})$ (rounding up the Taylor series) so if we choose $\sqrt{2n} \gt \frac 1\alpha$ and $2n$ a square it will fail.
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$I_{m,n}=\int\frac{x^m}{(ax^2+bx+c)^n}dx$ Reduction Formula I'm having trouble proving the following reduction formula:
If
$$I_{m,n}=\int\frac{x^m}{(ax^2+bx+c)^n}dx$$
then
$$\int\frac{x^m}{(ax^2+bx+c)^n}dx=-\frac{x^{m-1}}{a(2n-m-1)(ax^2+bx+c)^{n-1}}-\frac{b(n-m)}{a(2n-m-1)}I_{m-1,n}+\frac{c(m-1)}{a(2n-m-1)}I_{m-2,n}$$
My attempt went as follows:
$$\int\frac{x^m}{(ax^2+bx+c)^n}dx\space\begin{vmatrix}u=\frac{1}{(ax^2+bx+c)^n}\\du=\frac{-n(2ax+b)}{(ax^2+bx+c)^{n+1}}dx\end{vmatrix}\space dv=x^mdx\quad v=\frac{1}{m+1}x^{m+1}\\\int\frac{x^m}{(ax^2+bx+c)^n}dx=\frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+\bigg(\frac{n}{m+1}\bigg)\int\frac{(2ax+b)x^{m+1}}{(ax^2+bx+c)^{n+1}}dx\\=\frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+\bigg(\frac{n}{m+1}\bigg)\bigg(2a\int\frac{x^{m+2}}{(ax^2+bx+c)^{n+1}}dx\\+b\int\frac{x^{m+1}}{(ax^2+bx+c)^{n+1}}dx\bigg)\\=\frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+\bigg(\frac{2an}{m+1}\bigg)\int\frac{x^{m+2}}{(ax^2+bx+c)^{n+1}}dx\\+\bigg(\frac{bn}{m+1}\bigg)\int\frac{x^{m+1}}{(ax^2+bx+c)^{n+1}}dx\\=\frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+\bigg(\frac{2an}{m+1}\bigg)I_{m+2,n+1}+\bigg(\frac{bn}{m+1}\bigg)I_{m+1,n+1}$$
This is where I'm stuck, any help?
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So this means you are using the integration by parts in the wrong direction or with a wrong choice. In the reduction formula you are trying to achieve, you can notice that $m$ is decreasing while $n$ stays constant. Thus, you need a choice of $u,v$ such that:
$$udv = \dfrac{x^mdx}{f^n(x)},$$
where $f(x) = ax^2+bx+c$ and:
$$vdu = C\dfrac{x^{m-1}dx}{f^n(x)},$$
with $C$ some constant. If you tried to naively take $u = x^m,$ then you will have $v = \int\dfrac{dx}{f^n(x)},$ which you cannot evaluate. Therefore, one needs to improvise somehow. For instance,
$$d\left(\dfrac{1}{g^k(x)}\right) = \dfrac{kg'(x)g^{k-1}(x)dx}{g^{2k}(x)} = k\dfrac{g'(x)dx}{g^{k+1}(x)}$$ for appropriately smooth function $g(x).$ This is where I will leave you with a hint:
$$\dfrac{x^mdx}{f^n(x)} = \dfrac{x^{m-1}}{2a}\cdot\dfrac{d(ax^2+bx+c)}{(ax^2+bx+c)^n} -\dfrac{b}{2a}\cdot\dfrac{x^{m-1}}{(ax^2+bx+c)^n}=\dots $$
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|
Laplace's equation: separation of variables Question:
Let $(r,\theta)$ denote plane polar coordinates. Show that there are countably infinitely many $k \in \Bbb R$ for which
$$\nabla^2 u=0 \qquad 1≤r≤2 \\ ku + \frac{\partial u}{\partial r}=0 \qquad r=1,2$$
has a non-trivial solution of the form $u(r,\theta) = f(r)g(\theta)$.
Attempt:
Writing out the Laplacian in plane polars and plugging in $u(r,\theta) = f(r)g(\theta)$, we get
$$\nabla^2 u = \frac{\partial^2 u}{\partial r^2} + \frac 1r \frac{\partial u}{\partial r}+ \frac{1}{r^2} \frac{\partial u^2}{\partial \theta^2} = g\frac{d^2f}{dr^2} + \frac gr \frac{df}{dr} + \frac{f}{r^2}\frac{d^2g}{d\theta^2}$$
Thus
\begin{align}
\nabla^2 u = 0 & \implies g\frac{d^2f}{dr^2} + \frac gr \frac{df}{dr} + \frac{f}{r^2}\frac{d^2g}{d\theta^2} = 0 \\
& \implies \frac{r^2}{f}\frac{d^2f}{dr^2} + \frac rf \frac{df}{dr} = - \frac 1g \frac{d^2g}{d\theta^2} = \text{constant} : = \lambda
\end{align}
First assuming that $\lambda \neq 0$, I solved the $r$-equation:
$$\frac{r^2}{f}\frac{d^2f}{dr^2} + \frac rf \frac{df}{dr} = \lambda \implies r^2\frac{d^2f}{dr^2} + r\frac{df}{dr}-\lambda f = 0 \implies f(r) = Ar^{\sqrt \lambda} + Br^{-\sqrt \lambda}$$
The boundary condition given is
$$ku + \frac{\partial u}{\partial r} = 0 \implies kfg + g\frac{df}{dr}=0 \implies g(\theta) \big(k f(r) + f'(r)\big)=0 \qquad r=1,2 \; \; , \; \; \theta \in \Bbb R$$
$g(\theta) \equiv 0$ gives only non-trivial solutions, thus we must have $kf(1)+f'(1) = kf(2)f'(2)=0$. That is:
\begin{align*}
& k f(1)+f'(1)=0 \implies k (A+B) + \sqrt\lambda(A-B)=0 \\
& k f(2)+f'(2)=0 \implies k \big(A2^{\sqrt\lambda} + B2^{-\sqrt{\lambda}} \big) + \frac {\sqrt\lambda}{2} \big(A2^{\sqrt\lambda} - B2^{-\sqrt{\lambda}} \big)=0
\end{align*}
In matrix form, this is
$$\begin{pmatrix} k + \sqrt \lambda & k - \sqrt \lambda \\ \Big(k + \frac {\sqrt\lambda}{2}\Big) 2^{\sqrt \lambda} & \Big(k - \frac {\sqrt\lambda}{2}\Big) 2^{-\sqrt \lambda} \end{pmatrix} \begin{pmatrix} A \\ B \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$
and we get a non-trivial solution iff the determinant of the matrix is zero.
However, it seems to me that for every $k \in \Bbb R$ there is a corresponding $\lambda$ that gives a non-trivial solution. Have I done something wrong? Or am I misunderstanding something? Any help would be much appreciated.
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From the periodicity condition, we get a general solution
$$ u_n(r,\theta) = f(r)g(\theta) = (Ar^n + Br^{-n})(C\cos n\theta + D\sin n \theta) $$
where $n=1,2,3,\dots$
Since the boundary conditions are radially symmetric, it follows that $ kf(r) +f'(r) =0 $ on $r=1,2$. Plugging this in
\begin{align}
k(A+B) + n(A-B) = 0 \\
k(A2^n + B2^{-n}) +n(A2^{n-1} -B2^{-n-1})=0
\end{align}
Converting into a matrix equation
$$ \begin{pmatrix} k+n & k-n \\ 2^nk + 2^{n-1}n & 2^{-n}k - 2^{-n-1}n \end{pmatrix}\begin{pmatrix} A \\ B \end{pmatrix} = 0 $$
There's a non-trivial solution for $(A, B)$ if the determinant of the coefficient matrix is zero
$$ (k+n)(2k-n) - 2^{2n}(k-n)(2k+n) = 0 $$
In quadratic form this is
$$ 2(2^{2n}-1)k^2 - n(2^{2n}+1)k - (2^{2n}-1)n^2 = 0 $$
where $\Delta = b^2-4ac = n^2(2^{2n}+1)^2 + 8n^2(2^{2n}-1)^2 > 0$
So there exists two values of $k$ for each corresponding positive integer $n$, thus "countably infinite"
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Repeatedly dividing $360$ by $2$ preserves that the sum of the digits (including decimals) is $9$ Can someone give me a clue on how am I going to prove that this pattern is true or not as I deal with repeated division by 2? I tried dividing 360 by 2 repeatedly, eventually the digits of the result, when added repeatedly, always result to 9 for example:
\begin{align*}
360 ÷ 2 &= 180 \text{, and } 1 + 8 + 0 = 9\\
180 ÷ 2 &= 90 \text{, and } 9 + 0 = 9\\
90 ÷ 2 &= 45 \text{, and } 4 + 5 = 9\\
45 ÷ 2 &= 22.5 \text{, and } 2 + 2 + 5 = 9\\
22.5 ÷ 2 &= 11.25 \text{, and } 1 + 1 + 2 + 5 = 9\\
11.25 ÷ 2 &= 5.625 \text{, and } 5 + 6 + 2 + 5 = 18 \text{, and } 1 + 8 = 9\\
5.625 ÷ 2 &= 2.8125 \text{, and } 2 + 8 + 1 + 2 + 5 = 18 \text{, and } 1 + 8 = 9
\end{align*}
As I continue this pattern I found no error (but not so sure)... please any idea would be of great help!
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1) The sum of the digits of a multiple of $9$ will be a multiple of $9$.
Why? If $N$ is a multiple of $9$ then $N - 9k$ will also be a multiple of $9$ for all integers $k$.
If $N = \sum_{k=0}^n a_k 10^k$ is a multiple of $9$ then $(\sum_{k=0}^n a_k 10^k) - 9(a_1 + 11a_2 + 111a_3 + ..... + 1111.....1111a_n)$ is a multiple of $9$.
And $(\sum_{k=0}^n a_k 10^k) - 9(a_1 + 11a_2 + 111a_3 + ..... + 1111.....1111a_n)=$
$(\sum_{k=0}^n a_k 10^k) - (9a_1 + 99a_2 + 999a_3 + ..... + 9999.....9999a_n)=$
$\sum_{k=0}^n [a_k*10^k - \underbrace{999...9}k*a_k]=$
$\sum_{k=0}^n a_k(10^k - \underbrace{999...9}k)=\sum_{k=0}^n a_k*1 =$ the sum of the digits of $N$.
2) we can extend that decimals:
If $N$ is $9$ times some terminating decimal than the sum of the digits are a multiple of $9$.
Why? Well, its the same as above except $N$ has be divided by a power of $10$. But that only affects the position of the decimal point. It doesn't affect the digits.
3) If $N$ is a $9$ times a terminating decimal than $\frac N2$ is $9$ times a terminating decimal.
If $N = 9\times M$ and $M$ is a terminating decimal then $\frac M2$ is a terminating decimal. Then $\frac N2 = 9\times \frac M2$ is also $9$ times a terminating decimal and the digits add up to a multiple of $9$.
4) Since $360 = 9*40$ we start with a number whose digits add to $9$. Successively dividing by $2$ will result in $9*20$ and $9*10$ and so on, always $9$ times a terminating decimal with the sum of the digits being a multiple of $9$.
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|
Prove $\sum_{k=1}^m \cot^2 k\pi/(2m+1)=m(2m-1)/3$
Prove that $$ \sum_{k=1}^m \cot^2 \frac{k\pi}{2m+1}=\frac{m(2m-1)}{3} $$
I have tried to use $$\sin\left((2m+1)x\right)=
\left(\sin^{2m+1}x\right) \cdot \left(\sum_{j=0}^m (-1)^j \binom{2m+1}{2j+1}\left(\cot^2x\right)^{m-j}\right)$$
and induction without any success. Thanks for any help!
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Further to my last comment and given you used
$$\sin\left((2m+1)x\right)=
\left(\sin^{2m+1}x\right) \left(\sum_{j=0}^m (-1)^j \binom{2m+1}{2j+1}(\cot^2x)^{m-j}\right)=
\left(\sin^{2m+1}x\right) \cdot P\left(\cot^2{x}\right)\tag{1}$$
where
$$P(x)=\sum_{j=0}^m (-1)^j \binom{2m+1}{2j+1}x^{m-j}=\binom{2m+1}{1}x^m-\binom{2m+1}{3}x^{m-1}+...$$
is a polynomial of degree $m$, with (easy to see from $(1)$ since $\sin\left((2m+1)\frac{k\pi}{2m+1}\right)=0$) $\cot^2{\frac{k\pi}{2m+1}}, k=1..m$ as roots. Using Vieta's formulas
$$\sum_{k=1}^m \cot^2{\frac{k\pi}{2m+1}}=-\frac{-\binom{2m+1}{3}}{\binom{2m+1}{1}}=\frac{(2m+1)(2m)(2m-1)}{(2m+1)\cdot 2 \cdot 3}=\frac{m(2m-1)}{3}$$
|
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|
Simplify $\frac{x\sqrt{64y}+4\sqrt{y}}{\sqrt{128y}}$ into $\frac{2\sqrt{2x}+\sqrt{2}}{4}$ I am to simplify $$\frac{x\sqrt{64y}+4\sqrt{y}}{\sqrt{128y}}$$
into $\frac{2\sqrt{2x}+\sqrt{2}}{4}$
I am able to get to $\frac{x+4\sqrt{y}\sqrt{2}}{2}$ but cannot arrive at the provided solution.
Here is my working:
$\frac{x\sqrt{64y}+4\sqrt{y}}{\sqrt{128y}}$ = $\frac{x\sqrt{64y}+4\sqrt{y}}{\sqrt{64y}\sqrt{2}}$ = $\frac{x+4\sqrt{y}}{\sqrt{2}}$ = $\frac{x+4\sqrt{y}}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}}$ = $\frac{x+4\sqrt{y}\sqrt{2}}{2}$
I cannot see how to arrive at $\frac{2\sqrt{2x}+\sqrt{2}}{4}$?
Screen shot of my online textbooks question and answer in case I've typed it incorrectly:
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$$\frac{x \sqrt{64y} + 4 \sqrt{y}}{\sqrt{128 y}}$$
Factor our common factor $\sqrt{y}$ from numerator and denominator.
$$\frac{\sqrt{y}(x \sqrt{64} + 4)}{\sqrt{y}(\sqrt{128})}$$
Notice $\sqrt{64} = \sqrt{8^2} = 8$ and $\sqrt{128} = \sqrt{64 * 2} = \sqrt{8^2 * 2} = 8\sqrt{2}$.
$$\frac{8x + 4}{8\sqrt{2}}$$
Factor out $4$ from numerator and denominator and cancel.
$$\frac{2x + 1}{2\sqrt{2}}$$
Multiply numerator and denominator by $\sqrt{2}$.
$$\frac{2\sqrt{2}x + \sqrt{2}}{4}$$
|
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|
Prove that $N - \lfloor{N/p}\rfloor = \lfloor{\frac{p-1}{p}\left({N + 1}\right)}\rfloor$ for positive $N$ and prime $p$ I am counting the number of positive integers less than or equal to some positive integer $N$ and not divisible by some prime $p$. This gets generalized for $k$ primes where I use the principle of inclusion-exclusion for this result. The simple result for a single prime is $N - \lfloor{\frac{N}{p}}\rfloor$. However, I have noticed by experiment that this is also equal to $\lfloor{\frac{p-1}{p} \left({N + 1}\right)}\rfloor$. I am looking for a proof of this relation if true.
|
let
$$
f(N) = N - \lfloor \frac{N}{p} \rfloor
$$
then
$$
f(N+1) - f(N) = 1 - \bigg(\lfloor \frac{N+1}{p} \rfloor - \lfloor \frac{N}{p} \rfloor \bigg)
$$
thus when $N$ is incremented by 1, $f(N)$ is also incremented by 1 except when $N \equiv_p -1$. in this case $f(N+1) = f(N)$
if now we let
$$
g(N) = \lfloor{\frac{p-1}{p} \left({N + 1}\right)}\rfloor
$$
a little rearrangement gives
$$
g(N) = \lfloor N+1 - \frac{N+1}p \rfloor= N+1 + \lfloor -\frac{N+1}p \rfloor
$$
and
$$
g(N+1) - g(N) = 1 + \bigg(\lfloor -\frac{N+2}{p} \rfloor - \lfloor -\frac{N+1}{p} \rfloor \bigg)
$$
since $N$ is positive we again find that when $N$ increases by 1 the increment of $g$ is 1 except when $N \equiv_p -1$.
since $f(1)=g(1)=1$ the result is demonstrated
|
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|
show this inequality with $\sum_{i=1}^{n}a_{i}=n$
Let $n\ge 3$ be postive number,$a_{i}>0,i=1,2,\cdots,n$,and $\displaystyle\sum_{i=1}^{n}a_{i}=n$,show that
$$a^3_{1}a_{2}+a^3_{2}a_{3}+\cdots+a^3_{n}a_{1}+n\ge 2(a_{1}a_{2}\cdots a_{n-1}+a_{2}a_{3}\cdots a_{n}+a_{n}a_{1}\cdots a_{n-2})$$
it seem can use indution to prove it.when $n=3$,it must prove
$$a^3_{1}a_{2}+a^3_{2}a_{3}+a^3_{3}a_{1}+3\ge 2(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1})$$
it seem use three shcur inequaliy
$$a^3+b^3+c^3+3abc\ge \sum ab(a+b)$$
then we have
$$a^2+b^2+c^2+3(abc)^{2/3}\ge 2(ab+bc+ca)$$
|
The hint.
Prove this inequality for $n=3$ and for $n=4$.
But for all $n\geq5$ by AM-GM we obtain:
$$\sum_{k=1}^na_k^3a_{k+1}\geq n\sqrt[n]{\prod_{k=1}^na_k^4}.$$
Thus, it's enough to prove that:
$$n\sqrt[n]{\prod_{k=1}^na_k^4}+n\geq2\prod_{k=1}^na_k\sum_{k=1}^n\frac{1}{a_k}.$$
Now, let $\prod\limits_{k=1}^na_k=const$ and $f(x)=-\frac{1}{x}.$
Thus, $$g(x)=f'\left(\frac{1}{x}\right)=x^2$$ is strictly convex on $(0,+\infty),$ which says that by Vasc's EV Method, theorem 1.3 (1), $p=0$ from here
https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf
the expression $\sum\limits_{k=1}^nf(x_k)=-\sum\limits_{k=1}^n\frac{1}{a_k}$ gets a minimal value, when $n-1$ variables are equal.
After homogenization it's enough to assume that $a_1=a$ and $a_2=...=a_n=1$ and we need to prove that:
$$n\sqrt[n]{a^4}\left(\frac{a+n-1}{n}\right)^{n-5}+n\left(\frac{a+n-1}{n}\right)^{n-1}\geq2((n-1)a+1),$$
which is true by AM-GM!
Indeed, let $a=x^n$.
Thus, we need to prove that
$$P(x)=x^{n^2-n}+b_1x^{n^2-n-1}+...+b_{n^2-n}\geq0,$$ where only coefficient before $x^n$ is negative.
But the inequality $$n\sqrt[n]{\prod_{k=1}^na_k^4}+n\geq2\prod_{k=1}^na_k\sum_{k=1}^n\frac{1}{a_k}.$$
is symmetric, which says that the polynomial $P$ is divisible by $(x-1)^2.$
Now, we see that the sum of coefficients of $P'(x)$ is equal to zero,
which says that $P(x)\geq0$ by AM-GM.
|
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|
integral with parameter and 2 denominators Hello I want to solve following integral
$\displaystyle\int \frac{x^22t}{(1+x^2)(1+t^2x^2)}\,dx=-\dfrac{2\arctan(tx)-t \arctan(x)}{t^2-1}$
My Problem is that I just do not know how to start solving this problem .
|
Partial fraction decomposition gives$$\frac 1{(1-t^2)(1+t^2x^2)}-\frac 1{(1-t^2)(1+x^2)}=\frac {x^2}{(1+x^2)(1+t^2x^2)}$$Hence, the integral now becomes$$\begin{align*}\mathfrak{I} & =2t\int\mathrm dx\,\left[\frac 1{(1-t^2)(1+t^2x^2)}-\frac 1{(1-t^2)(1+x^2)}\right]\\ & =\frac {2t}{1-t^2}\int\frac {\mathrm dx}{1+t^2x^2}-\frac {2t}{1-t^2}\int\frac {\mathrm dx}{1+x^2}\end{align*}$$Can you continue?
EDIT: The partial fraction decomposition can be derived by assuming the fraction splits into two separate components $$\frac {x^2}{(1+x^2)(1+t^2x^2)}=\frac {Ax+B}{1+x^2}+\frac {Cx+D}{1+t^2x^2}$$Now multiply both sides by $(1+x^2)(1+t^2x^2)$ to clear the fractions completely. Hence$$x^2=(Ax+B)(1+t^2x^2)+(Cx+D)(1+x^2)$$
To find the coefficients, we evaluate the equation first at $x^2=-1$ and then $x^2=-\tfrac 1{t^2}$. The aim of these substitutions is to set one of the products equal to zero and compare the coefficients to each other. By setting $x^2=-1$, then$$\begin{align*}-1 & =(Ax+B)(1-t^2)\\ & =Ax(1-t^2)+B(1-t^2)\end{align*}$$From here, it's evident that $A=0$ because there's no linear term in the left - hand side. Similarly, $B=-\tfrac 1{1-t^2}$. Repeating the process similarly for $x^2=-\tfrac 1{t^2}$ to get that $C=0$ and $D=\tfrac 1{1-t^2}$.$$\frac {x^2}{(1+x^2)(1+t^2x^2)}\color{red}{=\frac 1{(1-t^2)(1+t^2x^2)}-\frac 1{(1-t^2)(1+x^2)}}$$which aligns with what we have above.
|
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|
Find shortest distance from the parabola $y=x^2-9$ to the origin.
Find shortest distance from the parabola $y=x^2-9$ to the origin.
First, I find minima of $\sqrt{x^2+(x^2-9)^2}$, so use derivative and ...
Is have an easier way?
|
$$x^2+(x^2-9)^2=(x^2-9)^2+x^2-9+\frac{1}{4}+\frac{35}{4}=\left(x^2-9+\frac{1}{2}\right)^2+\frac{35}{4}\geq\frac{35}{4}.$$
The equality occurs for $x^2=\frac{17}{2},$ which says that $\frac{\sqrt{35}}{2}$ is a minimal value.
|
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|
Solve differential equation $xyy'=x^4+y^4$ How to find general solution to this differential equation (if it exists):
$$ xyy'=x^4+y^4 ?$$
I do not know how to even approach it since I never dealt with nonlinear equations. Only thing that I notice is that $x$ and $y$ are symetric in equation: $$ y'=\dfrac{x^4+y^4}{xy}$$
so I expect (maybe) some kind of symetric function for y(x).
Thanks for any help.
|
Observing that $y y' = \left(\frac{1}{2}y^2\right)'$ we define the new dependent variable $z := \frac{1}{2} y^2$, $y^4 = 4 z^2$. We then obtain a Riccati equation for $z$: $z' = x^3 + \frac{4}{x} z^2$.
This Riccati equation (like any Riccati equation) can be reduced to a second-order linear ordinary differential equation by writing
\begin{equation}
z = - \frac{x}{4} \frac{u'}{u}, \quad z^2 = \frac{x^2}{16} \frac{(u')^2}{u^2}, \quad z' = - \frac{1}{4} \frac{u'}{u} - \frac{x}{4} \frac{u'' u - (u')^2}{u^2},
\end{equation}
which yields $x^2 u'' + x u' + 4 x^4 u = 0$. With the definition of a new independent variable
\begin{equation}
\xi := x^2, \quad \frac{d}{dx} = 2 \xi^{1/2} \frac{d}{d\xi}, \quad \frac{d^2}{dx^2} = 2 \frac{d}{d\xi} + 4 \xi \frac{d^2}{d\xi^2},
\end{equation}
we obtain the Bessel differential equation
\begin{equation}
\xi^2 \frac{d^2 u}{d \xi^2} + \xi \frac{d u}{d\xi} + \xi^2 u = 0,
\end{equation}
with fundamental solutions $J_0(\xi)$, $Y_0(\xi)$ (zeroth-order Bessel functions).
|
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|
Counting, Probability and Binomial Coefficients If $$P_{2n+2}=\sum_{k=n+2}^{2n+2}{2n+2 \choose k}p^kq^{2n+2-k}$$
and,
$$P_{2n}=\sum_{k=n+1}^{2n}{2n \choose k}p^kq^{2n-k}$$
where $0<p<q<1$ and $q=1-p$
Prove that
$$P_{2n+2}=P_{2n}+{2n \choose n}p^{n+2}q^n-{2n \choose {n+1}}p^{n+1}q^{n+1}$$
$\mathbf {Inspiration:}$ A and B play a series of games where the probability of winning $\mathit p$ for A is kept less than 0.5. However A gets to choose in advance the total no. of plays. To win the game one must score more than half the games . If the total no. of games is to be even, How many plays should A choose?
$\mathbf {Here}$ $P_{2n}$ and $P_{2n+2}$ represents the probability of A winning the play in $2n$ and $2n+2$ games where $2n$ is considered the optimum number of games
|
We consider $(p+q)^2P_{2n}$ instead of $P_{2n}$ (because then our identity will be homogeneous). Then, we will simply compare coefficents of monomials $p^kq^{2n+2-k}$ in both sides of identity.
Note that (from equality $p+q=1$)
$$
P_{2n}=\sum_{k=n+1}^{2n}{2n \choose k}p^kq^{2n-k}=\sum_{k=n+1}^{2n}{2n \choose k}p^kq^{2n-k} (p+q)^2,
$$
which equals
$$
\sum_{k=n+1}^{2n}{2n \choose k}p^kq^{2n-k} (p^2+2pq+q^2)=
\\
\sum_{k=n+1}^{2n}{2n \choose k}p^{k+2}q^{2n-k}+\sum_{k=n+1}^{2n}2{2n \choose k}p^{k+1}q^{2n-k+1}+\sum_{k=n+1}^{2n}{2n \choose k}p^{k}q^{2n-k+2}=
\\
\sum_{k=n+3}^{2n+2}{2n \choose k-2}p^{k}q^{2n+2-k}+\sum_{k=n+2}^{2n+1}2{2n \choose k-1}p^{k}q^{2n+2-k}+\sum_{k=n+1}^{2n}{2n \choose k}p^{k}q^{2n-k+2}.
$$
Therefore, $P_{2n}$ equals
$$
\sum_{k=n+2}^{2n+2}\left({2n \choose k}+2{2n \choose k-1}+{2n \choose k-2}\right)p^kq^{2n+2-k}-\left({2n \choose n}p^{n+2}q^{n}-{2n \choose n+1}p^{n+1}q^{n+1}\right).
$$
We also know that
$$
{2n \choose k}+2{2n \choose k-1}+{2n \choose k-2}=\left({2n \choose k}+{2n \choose k-1}\right)+\left({2n \choose k-1}+{2n \choose k-2}\right)=
\\
{2n+1 \choose k}+{2n+1 \choose k-1}={2n+2 \choose k},
$$
so we obtain
$$
P_{2n}=\sum_{k=n+2}^{2n+2}{2n+2 \choose k}p^kq^{2n+2-k}=P_{2n+2}-\left({2n \choose n}p^{n+2}q^{n}-{2n \choose n+1}p^{n+1}q^{n+1}\right).
$$
Hence,
$$
P_{2n+2}=P_{2n}+{2n \choose n}p^{n+2}q^{n}-{2n \choose n+1}p^{n+1}q^{n+1},
$$
as desired.
|
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|
How to apply CRT to a congruence system with moduli not coprime? $x=1 \pmod 8$
$x=5 \pmod{12}$
8 and 12 are not coprime, I could break it to:
$x=1 \pmod 2$
$x=1 \pmod 4$
and
$x=5 \pmod 3$
$x=5 \pmod 4$
But what are the next steps to solve it? By the way, $x$ should be $17$ not sure how to get that number ...
Thanks in advance.
|
Alternatively:
$$\begin{cases}x\equiv 1\pmod{8}\\ x\equiv 5\pmod{12}\end{cases} \Rightarrow \begin{cases} x=8n+1\\x=12m+5\end{cases} \Rightarrow 8n+1=12m+5 \Rightarrow \\
2n-3m=1 \Rightarrow \begin{cases}n=2+3k\\m=1+2k\end{cases} \Rightarrow \begin{cases} x=8(2+3k)\\ x=12(1+2k)+5\end{cases} \Rightarrow \\
x=24k+17 \Rightarrow x\equiv 17\pmod{24}.$$
|
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|
How to prove this closed formula for Cantor set?
Let $C_0=[0,1]$ and $C_{n+1} = \dfrac{C_n}{3} \bigcup\left(\dfrac{2}{3}+\dfrac{C_n}{3}\right)$.
Theorem: $$C_n=\bigcap_{m=0}^{n}\bigcup_{k=0}^{\left\lfloor \frac{3^{m}}{2}\right\rfloor}\left[\frac{2k}{3^m},\frac{2k+1}{3^m}\right]$$
I have tried to prove this assertion by induction on $n$, but to no avail. I am stuck at inductive step.
Please shed me some light to accomplish the proof. Thank you so much!
My attempt:
The formula is trivially true for $n=0$. Let it hold for $n$.
$$C_{n+1}=\frac{C_n}{3} \cup\left(\frac{2}{3}+\frac{C_n}{3}\right)$$
$$=\left(\frac{1}{3} \bigcap_{m=0}^{n}\bigcup_{k=0}^{\left\lfloor \frac{3^{m}}{2}\right\rfloor}\left[\frac{2k}{3^m},\frac{2k+1}{3^m}\right]\right) \cup \left(\frac{2}{3}+\frac{1}{3} \bigcap_{m=0}^{n}\bigcup_{k=0}^{\left\lfloor \frac{3^{m}}{2}\right\rfloor}\left[\frac{2k}{3^m},\frac{2k+1}{3^m}\right] \right)$$
$$=\left(\bigcap_{m=0}^{n}\bigcup_{k=0}^{\left\lfloor \frac{3^{m}}{2}\right\rfloor}\left[\frac{2k}{3^{m+1}},\frac{2k+1}{3^{m+1}}\right]\right) \cup \left(\bigcap_{m=0}^{n}\bigcup_{k=0}^{\left\lfloor \frac{3^{m}}{2}\right\rfloor}\left[\frac{2k+2.3^m}{3^{m+1}},\frac{2k+2.3^m+1}{3^{m+1}}\right]\right)$$
$$=\left(\bigcap_{m=0}^{n}\bigcup_{k=0}^{\left\lfloor \frac{3^{m}}{2}\right\rfloor}\left[\frac{2k}{3^{m+1}},\frac{2k+1}{3^{m+1}}\right]\right) \cup \left(\bigcap_{m=0}^{n}\bigcup_{k=0}^{\left\lfloor \frac{3^{m}}{2}\right\rfloor}\left[\frac{2(k+3^m)}{3^{m+1}},\frac{2(k+3^m)+1}{3^{m+1}}\right]\right)$$
$$=\left(\bigcap_{m=0}^{n}\bigcup_{k=0}^{\left\lfloor \frac{3^{m}}{2}\right\rfloor}\left[\frac{2k}{3^{m+1}},\frac{2k+1}{3^{m+1}}\right]\right) \cup \left(\bigcap_{m=0}^{n}\bigcup_{k=3^m}^{\left\lfloor \frac{3^{m}}{2}\right\rfloor+3^m}\left[\frac{2k}{3^{m+1}},\frac{2k+1}{3^{m+1}}\right]\right)$$
|
Notice that
$$\bigcap_{m=0}^{n+1}\bigcup_{k=0}^{\lfloor 3^m/2\rfloor}\left[\frac{2k}{3^m},\frac{2k+1}{3^m}\right]=\left(\bigcap_{m=0}^{n}\bigcup_{k=0}^{\lfloor 3^m/2\rfloor}\left[\frac{2k}{3^m},\frac{2k+1}{3^m}\right]\right)\cap \bigcup_{k=0}^{\lfloor 3^{n+1}/2\rfloor}\left[\frac{2k}{3^{n+1}},\frac{2k+1}{3^{n+1}}\right]\\=C_n\cap \bigcup_{k=0}^{\lfloor 3^{n+1}/2\rfloor}\left[\frac{2k}{3^{n+1}},\frac{2k+1}{3^{n+1}}\right]$$
We may notice that if we divide the interval $[0,1]$ into $3^{n+1}$ parts we will get $[0,\frac{1}{3^{n+1}}],\ldots,[\frac{3^{n+1}-1}{3^{n+1}},1]$.
$\bigcup_{k=0}^{\lfloor 3^{n+1}/2\rfloor}\left[\frac{2k}{3^{n+1}},\frac{2k+1}{3^{n+1}}\right]$ is just taking the even parts of the list together, notice that $C_{n+1}\subseteq \bigcup_{k=0}^{\lfloor 3^{n+1}/2\rfloor}\left[\frac{2k}{3^{n+1}},\frac{2k+1}{3^{n+1}}\right]\cap C_n$, now notice that the even intervals of this stage are always the first and last thirds of the even intervals of the previous stages, so if $x\in C_n$, it had to be in the form $0.d_1d_2\ldots_3$ where $i\in \{1,2,\ldots, n-1\}$ implies $d_i\in\{0,2\}$, or in other words $x\in[0.d_1\ldots d_{n-1}0_3,0.d_1\ldots d_{n-1}2_3]$, $d_n$ will be $0$ on the first third and $2$ on the last third, so $C_n \cap \bigcup_{k=0}^{\lfloor 3^{n+1}/2\rfloor}\left[\frac{2k}{3^{n+1}},\frac{2k+1}{3^{n+1}}\right]\subseteq C_{n+1}$
|
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Integration by substitution to take out square root
Find $$\int (x+1)\sqrt{x^2+1}\,dx .$$
In order to not bother with the square root I thought of doing this:
$let$ $ x^2+1=(x+t)^2$ $\Rightarrow$ $x=\frac{1-t^2}{2t}$ $\Rightarrow x-t=\frac{1+t^2}{2t}$ $(1)$
$\Rightarrow$ $t=\sqrt{x^2+1}-x$ $(2)$
And
$dx=-\frac{t^2+1}{2t^2}dt $ $(3)$
and also $x+1=-\frac{t^2-2t-1}{2t}$ $(4)$
Then
$(1),(3),(4)\Rightarrow$$\int (x+1)\sqrt{x^2+1}dx=$ $\int\frac{t^2-2t-1}{2t} \frac{1-t^2}{2t} \frac{t^2+1}{2t}dt=$
$\frac{1}{8} \int \frac{-t^6+2t^5+t^4+t^2-2t-1}{t^3}dt=$
$\frac{1}{8} [-\int t^3 dt+2\int t^2 dt+\int t dt +\int \frac{1}{t} dt-2 \int \frac{1}{t^2} dt- \int \frac{1}{t^3}dt]=$
$\frac{1}{8}[-\frac{t^4}{4}+2\frac{t^3}{3}+\frac{t^2}{2}+lnt+\frac{2}{t}-\frac{1}{2t^2}]$
We do the substitution from $(2)$ and we get the result
You think that's correct?
|
After the rationalizing substitution, I think you made a few errors. It should be
$$\int (x+1)\sqrt{x^2+1}\,dx=\int\frac{t^2-2t-1}{2t}\cdot \frac{t^2+1}{2t}\cdot \frac{t^2+1}{2t^2}dt.$$
Alternative approach by using hyperbolic functions.
Let $x=\sinh(t)$ then $\cosh^2(t)-\sinh^2(t)=1$ and
$$\begin{align*}\int (x+1)\sqrt{x^2+1}\,dx &=\int (\sinh(t)+1)\sqrt{\sinh^2(t)+1}\,\cosh(t) dt \\
&=\int (\sinh(t)+1)\cosh^2(t) dt\\
&=\frac{\cosh^3(t)}{3}+ \frac{\cosh(t)\sinh(t)}{2}+\frac{t}{2}+c\\
&=\frac{(x^2+1)^{3/2}}{3}+\frac{x(x^2+1)^{1/2}}{2}+\frac{\text{arcsinh}(x)}{2}+c
\end{align*}.$$
|
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Find range of $x$ if $\log_5\left(6+\frac{2}{x}\right)+\log_{1/5}\left(1+\frac{x}{10}\right)\leq1$
If $\log_5\left(6+\dfrac{2}{x}\right)+\log_{1/5}\left(1+\dfrac{x}{10}\right)\leq1$, then $x$ lies in _______
My Attempt
$$
\log_5\bigg(6+\dfrac{2}{x}\bigg)+\log_{1/5}\bigg(1+\dfrac{x}{10}\bigg)=\log_5\bigg(6+\dfrac{2}{x}\bigg)-\log_{5}\bigg(1+\dfrac{x}{10}\bigg)\leq1\\
\log_5\frac{(6x+2)10}{x(10+x)}\leq1\implies\frac{(6x+2)10}{x(10+x)}\leq5\\
\frac{4(3x+1)}{x^2+10x}\leq1\\
\implies 12x+4\leq x^2+10x\quad\text{or}\quad12x+4>x^2+10x\\
x^2-2x-4\geq0\quad\text{or}\quad x^2-2x-4<0\implies x\in\mathcal{R}
$$
My reference gives the solution $(-\infty,1-\sqrt{5})\cup(1+\sqrt{5},\infty)$, what is going wrong here ?
|
Writing your inequality in the form $$\frac{\ln\left(6+\frac{2}{x}\right)}{\ln(5)}+\frac{\ln\left(1+\frac{x}{10}\right)}{-\ln(5)}\le 1$$ we get the inequalities
$$6+\frac{2}{x}>0$$ and $$1+\frac{x}{10}>0$$ and
$$\frac{6+\frac{2}{x}}{1+\frac{x}{10}}\le 5$$ we get
$$1-\sqrt{5}\le x<-\frac{1}{3}$$ or $$x\geq 1+\sqrt{5}$$
|
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solution of an algebraic equation with integers Prove that the equation $x^2+y^2+z^2=(x-y)(y-z)(z-x)$ has an infinite number of solutions when $x,y,z$ are integers.
I started from specific cases.
|
Let $r$ be odd, then $(r+2)^2+r^2+(r+1)^2$ is even, so it is a multiple of $((r+2)-r)(r-(r+1))((r+1)-(r+2))=2$. The quotient is $d=(3r^2+6r+5)/2$. Then $x=d(r+2)$, $y=dr$, $z=d(r+1)$ works.
Of course, there are many more. E.g., let $b=a+1$, $c=a+3$. Then $(a-b)(b-c)(c-a)=6$. If $a$ is even, then $a^2+b^2+c^2$ is divisible by 2, and if $a-1$ is divisible by $3$, then so is $a^2+b^2+c^2$, so if $a=6r-2$, then $a^2+b^2+c^2=108r^2-24r+6=(a-b)(b-c)(c-a)s$ where $s=18r^2-4r+1$. So, $x=(6r-2)s$, $y=(6r-1)s$, $z=(6r+1)s$ is a solution for $r=1,2,3,\dots$.
|
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|
Evaluate $\sum_{n=1}^{\infty}\frac{\psi^{''}(n)}{2n-1}$, where $\psi^{''}(n)$ is 2nd derivative of digamma function. Does the following sum have a closed form?
$$\sum_{n=1}^{\infty}\frac{\psi^{"}(n)}{2n-1},$$ where $\psi^{"}(n)$ is 2nd derivative of digamma function.
|
The $n$-th derivate of the digamma function is known as "polygamma function of order $n$-th". Moreover, we have the integral representation
$$\psi^{(m)}(x) = (-1)^{m+1} \int_0^\infty \frac{t^m}{1-e^{-t}} e^{-xt} \, \mathrm{d} t.$$
Using the monotone convergence theorem, we get
$$\label{1} \tag{1}\sum_{n=1}^\infty \frac{\psi''(n)}{2n-1} = - \int_0^\infty \frac{t^2}{1-e^{-t}} \sum_{n=1}^\infty \frac{e^{-nt}}{2n-1} \, \mathrm{d}t. $$
Since
$$\sum_{n=1}^\infty e^{-(n-1/2)t} = \frac{e^{t/2}}{e^t-1},$$
we get by integration
$$ \sum_{n=1}^\infty \frac{e^{-nt}}{2n-1} = \frac{e^{-t/2}}{2} \int_t^\infty \frac{e^{s/2}}{e^s-1} \, \mathrm{d} s.$$
But we have
$$\int_t^\infty \frac{e^{s/2}}{e^s-1} \, \mathrm{d} s = 2 \int_{e^{t/2}}^\infty \frac{1}{x^2-1} \, \mathrm{d} x = 2 \mathrm{arcoth}(e^{t/2}) $$
and therefore
$$\sum_{n=1}^\infty \frac{e^{-nt}}{2n-1} = e^{-t/2} \mathrm{arcoth}(e^{t/2}).$$
Note that we have proven that $$\mathrm{arcoth}(x) = \sum_{n=1}^\infty \frac{x^{-(2n-1)}}{2n-1} \quad \text{for} \quad |x| >1.$$
Thus, we can rewrite \eqref{1} by
$$\tag{2}\label{2}\sum_{n=1}^\infty \frac{\psi''(n)}{2n-1} = - \int_0^\infty \frac{t^2}{1-e^{-t}} e^{-t/2} \mathrm{arcoth}(e^{t/2}) \, \mathrm{d} t.$$
Changing variables we see that
\begin{align}
- \int_0^\infty \frac{t^2}{1-e^{-t}} e^{-t/2} \mathrm{arcoth}(e^{t/2}) \, \mathrm{d} t = - 8 \int_1^\infty \frac{\ln(x)^2}{x^2-1} \mathrm{arcoth}(x) \, \mathrm{d} x.
\end{align}
Using $\mathrm{artanh}(1/x) = \mathrm{coth}(x)$ we can rewrite the integral also by
$$ - 8 \int_1^\infty \frac{\ln(x)^2}{x^2-1} \mathrm{arcoth}(x) \, \mathrm{d} x= 8 \int_0^1 \mathrm{artanh}(y) \frac{\ln(y)^2}{y^2-1} \, \mathrm{d}y.$$
With the help of ComplexYetTrivial's comment we can compute the last integral explicitly.
Using partial integration in the last line we get that
\begin{align}
8 \int_0^1 \mathrm{artanh}(y) \frac{\ln(y)^2}{y^2-1} \, \mathrm{d}y &= \left.-4 \mathrm{artanh}(y)^2 \ln(y)^2 \right|_{y=0}^1 + 8 \int_0\mathrm{artanh}(y)^2 \frac{\ln(y)}{y} \, \mathrm{d} y \\
&= 8 \int_0^1\mathrm{artanh}(y)^2 \frac{\ln(y)}{y} \, \mathrm{d} y.
\end{align}
Since $$\mathrm{artanh}(y) = \frac{1}{2} \left( \ln(y+1)-\ln(1-y) \right) $$
we get that \eqref{2} is equal to
$$2 \int_0^1 \frac{\ln(y+1)^2 \ln(y)}{y} \, \mathrm{d} y - 4 \int_0^1 \frac{\ln(y+1) \ln(1-y) \ln(y)}{y} \, \mathrm{d} y + 2 \int_0^1 \frac{\ln(1-y)^2 \ln(y)}{y} \, \mathrm{d} y $$
The remaining integrals are evaluated here, here and here. In fact, we have
$$2 \int_0^1 \frac{\ln(y+1)^2 \ln(y)}{y} \, \mathrm{d} y = - \zeta(4) = - \frac{\pi^4}{90} $$
and
$$- 4 \int_0^1 \frac{\ln(y+1) \ln(1-y) \ln(y)}{y} \, \mathrm{d} y = \frac{3}{40} \pi^4 - 7 \log(2) \zeta(3) + \frac{\pi^2 \log(2)^2}{3}- \frac{\log(2)^4}{3} - 8 \mathrm{Li}_4(1/2)$$
and also
$$2 \int_0^1 \frac{\ln(1-y)^2 \ln(y)}{y} \, \mathrm{d} y = \frac{\pi^2}{12} - 8 \mathrm{Li}_4(1/2) -7 \log(2) \zeta(3)+ \frac{\pi^2 \log(2)^2}{3} - \frac{\ln(2)^4}{3} $$
Finally, we see that \eqref{2} can be written as
$$\frac{53}{360} \pi^4 - 14 \log(2) \zeta(3) + \frac{2}{3} [\pi^2 -\log(2)^2] \log(2)^2 - 16 \mathrm{Li}_4(1/2).$$
|
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|
Evaluate the limit: $\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9}$ $$\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9} $$
I want to know how to evaluate without using L'Hopital Rule. I'm unable to factorise or simplify it suitably.
|
An elementary approach: "recognize the derivative."
You can rewrite, for $x\notin\{-3,3\}$,
$$
\frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9}
= 1+ \frac{\sqrt{x+6}-3}{x^{2}-9}
= 1+ \frac{\sqrt{x+6}-3}{(x-3)(x+3)}
= 1+ \frac{\sqrt{x+6}-3}{(x-3)(x+3)} \tag{1}
$$
so the question boils down to computing $\lim_{x\to 3} \frac{\sqrt{x+6}-3}{x-3}$ (the rest is "under control").
Now, we can recognize a derivative here: let $f\colon[0,\infty) \to [0,\infty)$ be defined by
$$
f(x) = \sqrt{x+6}, \qquad x\geq 0\,. \tag{2}
$$
Note that $f$ is differentiable, with $f'(x) = \frac{1}{2\sqrt{x+6}}$. Therefore,
$$
\frac{1}{6} = \frac{1}{2\sqrt{9}} = f'(3) = \lim_{x\to 3} \frac{f(x)-f(3)}{x-3} = \lim_{x\to 3} \frac{\sqrt{x+6}-3}{x-3} \tag{3}
$$
giving the desired limit.
Putting it together, from (1) and (3) we get
$$
\begin{align*}
\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9}
&= 1+ \lim_{x\to 3} \frac{\sqrt{x+6}-3}{(x-3)(x+3)}= 1+\frac{1}{6}\lim_{x\to 3} \frac{\sqrt{x+6}-3}{x-3}
\\&= 1+\frac{1}{36} = \boxed{\frac{37}{36}}
\end{align*}$$
|
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Prove that $\sum_{k=0}^n 2^k \binom{n}{k} \binom{n-k}{\lfloor (n-k)/2 \rfloor}=\binom{2n+1}{n}$ Where the thing that looks like a floor function is the floor function. This is an interesting result which I hoped to prove by induction, but ran into trouble applying the inductive hypothesis. The base case is trivial. Here's the progress I made:
Inductive hypothesis: $\sum_{k=0}^m 2^k \binom{m}{k} \binom{m-k}{\lfloor{(m-k)/2}\rfloor}=\binom{2m+1}{m}$ for some $m>0$. Then
\begin{align*}
&\sum_{k=0}^{m+1} 2^k \binom{m+1}{k} \binom{m+1-k}{\lfloor{(m+1-k)/2}\rfloor}\\
=&\sum_{k=0}^m 2^k \binom{m+1}{k} \binom{m+1-k}{\lfloor{(m+1-k)/2}\rfloor}+2^{m+1} \binom{m+1}{m+1} \binom{m+1-m+1}{\lfloor{(m+1-(m+1))/2}\rfloor}\\
=&2^{m+1}+\sum_{k=0}^m 2^k \binom{m+1}{k} \binom{m+1-k}{\lfloor{(m+1-k)/2}\rfloor}
\end{align*}
Let $A$ be the set of all $k\in[0,m]$ such that $m+1-k$ is even. Let $B$ be the set of $k\in[0,m]$ such that $m+1-k$ is odd. Observe that for each $k\in[0,m]$, $k$ is in exactly one of $A$ or $B$. It follows that
\begin{align*}
&\sum_{k=0}^m 2^k \binom{m+1}{k} \binom{m+1-k}{\lfloor{(m+1-k)/2}\rfloor}\\
=&\sum_{k\in A} 2^k \binom{m+1}{k} \binom{m+1-k}{\lfloor{(m+1-k)/2}\rfloor}+\sum_{k\in B} 2^k \binom{m+1}{k} \binom{m+1-k}{\lfloor{(m+1-k)/2}\rfloor}\\
=&\sum_{k\in A} 2^k \binom{m+1}{k} \binom{m+1-k}{(m+1-k)/2}+\sum_{k\in B} 2^k \binom{m+1}{k} \binom{m+1-k}{(m-k)/2}
\end{align*}
Splitting the sum into $A$ and $B$ was a last-ditch attempt to form the expression into something to which I could apply the inductive hypothesis. Did I make a mistake somewhere, is this the wrong approach, or am I just missing the next step?
|
There are $ 2n+1 \choose n $ strings consisting of $n$ ones and $n+1$ zeroes.
Given any such string $S$, let $S_1$ consist of the first $n$ digits of the string $S$ and let $S_2$ consist of the last $n+1$ digits of $S$. Suppose that there $k$ values of $i$ such that the $i^{th}$ element of $S_1$ is different from the $i^{th}$ element of $S_2$. Then there are $n \choose k$ ways to choose which digits $i$ the two strings differ at. There are $2^k$ ways to assign these $k$ elements in $S_1$.
Now consider the number of ways in which the rest of the elements may be chosen. We need to choose the remaining $n-k$ elements of $S_1$ (which are the same as their respective elements in $S_2$) and the last element of $S_2$. The number of zeroes must be one greater than the number of ones. It can be shown that the number of ways to do this is $n-k \choose \lfloor\frac{n-k}{2}\rfloor$.
|
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|
Find solutions of the given equation in the form of power series Find solutions of the given equation in the form of power series
$$y'' +2xy' = 0$$
My approach:
$$2xy' = 2a_1x + 2\times2a_2x^2 + 2\times3a_3x^3 + ... = \sum 2ia_{i}x^{i}$$
$$y'' = 2a_2 + 3\times2a_3x + 4\times3a_4x^2 + ... = \sum (i+1)(i+2)a_{i+2}x^{i}$$
So to have $y'' +2xy' = 0$ we need to have every coefficient equal to $0$.
So
$x^0$ // $2a_2 + 0 = 0 => a_2=0$
$x^1$ // $3\times2a_3 + 2a_1 = 0 => a_3 = -a_1/3$
$x^2$ // $4\times3a_4 + 2\times2a_2=0 => a_4 = -4a_2/12 = 0$
...
So it was easy to observe that every even coefficient is equal to $0$ and every odd one can be written in terms of $a_0$
Afterwards I wanted to find a recursive formula for odd coefficients and this is a problem that I can't solve.
I tried doing it in that way:
$$2n(2n+1)a_{2n+1} + 2(2n-1)a_{2n-1} = 0$$
$$a_{2n+1} = -\frac{2(2n-1)a_{2n-1}}{2n(2n+1)}$$
Now probably it should be somehow easy to write it as a recursive formula in terms of $a_0$, but I can't really see it.
|
Let the solution of the given ODE be $y=\sum c_{s}x^{s}$.
Then this satisfies the equation $y^{''}+2xy^{'}=0$.
Now, $y^{'}=\sum sc_{s}x^{s-1}$ and $y^{''}=\sum s(s-1)c_{s}x^{s-2}$. Substituting these values in the differential equation.
$\sum s(s-1)c_{s}x^{s-2} + 2x \sum sc_{s}x^{s-1}=0$
$=>\sum s(s-1)c_{s}x^{s-2} + 2\sum sc_{s}x^{s}=0$
$=>\sum 0(0-1)c_{0}x^{0} + \sum 1(1-1) c_{1}x^{1} + \sum_{s=2} s(s-1)c_{s}x^{s} + 2\sum 0c_{0}x^{0} + 2\sum_{s=1} sc_{s}x^{s}=0$
$=>\sum_{s=0} (s+1)(s+2)c_{s+2}x^{s} + 2\sum_{s=0} (s+1)c_{s+1}x^{s+1} =0$
Now since the last relation is an identity, therefore
$(s+1)(s+2)c_{s+2}=0$ and $2(s+1)c_{s+1}=0$
$=>(s+1)(s+2)c_{s+2}=2(s+1)c_{s+1}$
$=>c_{s+2}=\frac{2c_{s+1}}{s+2}$
$=>c_{s+1}=\frac{2c_{s}}{s+1}$
Now I think you can see the recurrence relation
|
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What is the value of $\sqrt{49-x^2}+\sqrt{25-x^2}, x \in \mathbb{R}$ if $\sqrt{49-x^2}-\sqrt{25-x^2}=3$ Suppose that real number $x$ satisfies $$\sqrt{49-x^2}-\sqrt{25-x^2}=3$$What is the value of $\sqrt{49-x^2}+\sqrt{25-x^2}$?
This is what I did:
I try to multiply by the conjugate. Its value I believe is technically the solution. $(\sqrt {49-x^2} + \sqrt {25-x^2})(\sqrt {49-x^2} - \sqrt {25-x^2})=24$. Given that $(\sqrt {49-x^2} - \sqrt {25-x^2}) = 3$, $(\sqrt {49-x^2} + \sqrt {25-x^2})(\sqrt {49-x^2} - \sqrt {25-x^2})= 3(\sqrt {49-x^2} + \sqrt {25-x^2}) =24\implies \sqrt {49-x^2} + \sqrt {25-x^2}=8$
My question is that will this method work for similar problems, and is there a faster method?
Thanks!
|
You basically used the formula $$a+b=\frac{a^2-b^2}{a-b}$$
in an indirect way.
The problem gave you $a-b$ and, since $a,b$ are square roots of simple expressions, it is easy to calculate $a^2-b^2$.
And for the same reason, the same trick can be used to calculate
$$\sqrt{\mbox{nice}}\pm \sqrt{\mbox{nice}}$$
whenever when $\sqrt{\mbox{nice}}\mp \sqrt{\mbox{nice}}$ is given.
|
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|
A series question whose indices involves greatest integer function Let $x$ be a real number and $\lfloor{x}\rfloor$ denote the greatest integer less than or equal to $x$. Let $a$ and $n$ be nonnegative integers such that $n\geq a+1$. Prove that
$$ \sum_{k=a}^{n} f(k) - \sum_{k=a}^{n} (-1)^{k} f(k) = 2 \sum_{k=\lfloor{\frac{a+2}{2}}\rfloor}^{\lfloor{\frac{n+1}{2}}\rfloor} f(2k-1).$$
I proved this by considering the situations where $a$ and $n$ takes even or odd values respectively. Are there any other way to show the equation is true? I generally have trouble with series whose indices involves greatest integer function, are there fast techniques to solve them?
|
The index of summation is ranging as
$$
a \le k \le n
$$
Replace it with its even and odd components
$$
k = 2j - i\quad \left| {\,i = 0,1} \right.
$$
Then for the even component it shall be
$$
\eqalign{
& i = 0\quad \Rightarrow \quad a \le 2j \le n\quad \Rightarrow \quad \left\lceil {{a \over 2}} \right\rceil \le j \le \left\lfloor {{n \over 2}} \right\rfloor
\quad \Rightarrow \cr
& \Rightarrow \quad \left\lfloor {{{a + 1} \over 2}} \right\rfloor \le j \le \left\lfloor {{n \over 2}} \right\rfloor \cr}
$$
and for the odd
$$
\eqalign{
& i = 1\quad \Rightarrow \quad a \le 2j - 1 \le n\quad \Rightarrow \quad \left\lceil {{{a + 1} \over 2}} \right\rceil \le j \le
\left\lfloor {{{n + 1} \over 2}} \right\rfloor \quad \Rightarrow \cr
& \Rightarrow \quad \left\lfloor {{{a + 2} \over 2}} \right\rfloor \le j \le \left\lfloor {{{n + 1} \over 2}} \right\rfloor \cr}
$$
Note that for the lower bound we shall employ the ceiling and the floor for the upper. Then we can convert the ceiling to floor.
The sum over $k$ of $f(k)$ can be split into the sum over the even and the odd component
$$
\eqalign{
& \sum\limits_{a\, \le \,k\, \le \;n} {f(k)} = \sum\limits_{a\, \le \,k\, \le \;n} {\left. {f(k)\,} \right|_{\,k = 2j} + \left. {f(k)\,} \right|_{\,k = 2j - 1} } = \cr
& = \sum\limits_{a\, \le \,2j\, \le \;n} {f(2j)} + \sum\limits_{a\, \le \,2j - 1\, \le \;n} {f(2j - 1)} = \cr
& = \sum\limits_{\left\lfloor {{{a + 1} \over 2}} \right\rfloor \le j \le \left\lfloor {{n \over 2}} \right\rfloor } {f(2j)}
+ \sum\limits_{\left\lfloor {{{a + 2} \over 2}} \right\rfloor \le j \le \left\lfloor {{{n + 1} \over 2}} \right\rfloor } {f(2j - 1)} \cr}
$$
and same for that of $(-1)^k f(k)$
$$
\eqalign{
& \sum\limits_{a\, \le \,k\, \le \;n} {\left( { - 1} \right)^{\,k} f(k)}
= \sum\limits_{\left\lfloor {{{a + 1} \over 2}} \right\rfloor \le j \le \left\lfloor {{n \over 2}} \right\rfloor } {\left( { - 1} \right)^{\,2j} f(2j)}
+ \sum\limits_{\left\lfloor {{{a + 2} \over 2}} \right\rfloor \le j \le \left\lfloor {{{n + 1} \over 2}} \right\rfloor } {\left( { - 1} \right)^{\,2j - 1} f(2j - 1)} = \cr
& = \sum\limits_{\left\lfloor {{{a + 1} \over 2}} \right\rfloor \le j \le \left\lfloor {{n \over 2}} \right\rfloor } {f(2j)}
- \sum\limits_{\left\lfloor {{{a + 2} \over 2}} \right\rfloor \le j \le \left\lfloor {{{n + 1} \over 2}} \right\rfloor } {f(2j - 1)} \cr}
$$
Now, just subtract the two above to confirm the thesis.
--- addendum ----
Concerning your request on how to transform ceiling <-> floor, consider that
$$
\eqalign{
& n = 2\left\lceil {{n \over 2}} \right\rceil + 2\left\{ {{n \over 2}} \right\} = 2q + i\quad \left| \matrix{
\;n,q \in Z \hfill \cr
\,i = 0,1 \hfill \cr} \right. \cr
& \left\lceil {{n \over 2}} \right\rceil = \left\lceil {{{2q + i} \over 2}} \right\rceil
= \left\lceil {q + {i \over 2}} \right\rceil = q + \left\lceil {{i \over 2}} \right\rceil = q + i \cr
& \left\lfloor {{{n + 1} \over 2}} \right\rfloor = \left\lfloor {{{2q + 1 + i} \over 2}} \right\rfloor
= \left\lfloor {q + {{1 + i} \over 2}} \right\rfloor = q + \left\lfloor {{{1 + i} \over 2}} \right\rfloor = q + i \cr}
$$
and, in general
$$
\left\lceil {{n \over m}} \right\rceil = \left\lfloor {{{n + m - 1} \over m}} \right\rfloor \quad \left| \matrix{
\;n,m \in Z \hfill \cr
\;1 \le m \hfill \cr} \right.
$$
re. for details to this article.
|
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|
$\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5}=3x$ Problem:
solve equation
$$\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5}=3x$$
I don't look for easy solution (square booth side and things like that...) I look for some tricks for "easy" solution because:
I would like to use substitution, but we have $3x$ and $-3x$, but I can't see it.
Solution:
|
$\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5}=3x$
Or $$(\sqrt{2x^2 + 3x +5}-7) + (\sqrt{2x^2-3x+5}-5)=3x-12$$
Or $$\frac{\left(x-4\right)\left(2x+11\right)}{\sqrt{2x^2+3x+5}+7}+\frac{\left(x-4\right)\left(2x+5\right)}{\sqrt{2x^2-3x+5}+5}=3(x-4)$$
It is not difficult to prove $$\frac{2x+11}{\sqrt{2x^2+3x+5}+7}+\frac{2x+5}{\sqrt{2x^2-3x+5}+5}>3$$
Then you have $x=4$ is a root.
Another way:
Let $(\sqrt{2x^2 + 3x +5} ; \sqrt{2x^2-3x+5})\rightarrow (a;b)$ where $a;b\ge0$
So $$\frac{a^2-b^2}{2}=\frac{2x^2+3x+5-2x^2+3x-5}{2}=3x$$
Then you have a new equation $$a+b=\frac{a^2-b^2}{2}$$
Or $$-\frac{1}{2}(a-b-2)(a+b)=0$$
Can you solve it ?
|
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|
Combinatorics - Limited Replacement A man is selling $15$ different items, and has exactly $3$ of each item. They all cost the same and we have enough to buy $7$ items in total. How many combinations of items can we buy? I know that if he had unlimited of each item, the formula would be ${n+k-1\choose k}$, so ${15+7-1\choose 7}$ or choose ${21\choose 7} = 116280$, but I am not sure how to implement the restriction of only being allowed up to $3$ of each item.
|
If we let $x_j$ denote the number of items of the $j$th type, $1 \leq j \leq 15$, that are selected, then
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 + x_{10} + x_{11} + x_{12} + x_{13} + x_{14} + x_{15} = 7 \tag{1}$$
A particular solution of the equation in the nonnegative integers corresponds to the placement of $15 - 1 = 14$ addition signs in a row of seven ones. For instance,
$$1+ + + + + + 1 + + + + 1 1 1 + + + 1 1 +$$
corresponds to the solution $x_1 = 1$, $x_2 = x_3 = x_4 = x_5 = x_6 = 0$, $x_7 = 1$, $x_8 = x_9 = x_{10} = 0$, $x_{11} = 3$, $x_{12} = x_{13} = 0$, $x_{14} = 2$, $x_{15} = 0$. The number of such solutions is the number of ways we can place $14$ addition signs in a row of seven ones, which is
$$\binom{7 + 15 - 1}{15 - 1} = \binom{21}{14} = \binom{21}{7}$$
as you found.
From these, we wish to subtract those solutions in which one or more variables exceeds $3$ since there are only three items of each type. Observe that it is possible for only of the variables to exceed $3$ since $2 \cdot 4 = 8 > 7$.
Choose which variable exceeds $3$. Suppose it is $x_1$. Then $x_1' = x_1 - 4$ is a nonnegative integer. Substituting $x_1' + 4$ for $x_1$ in equation 1 yields
\begin{align*}
x_1' + 4 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 + x_{10} + x_{11} + x_{12} + x_{13} + x_{14} + x_{15} & = 7\\
x_1' + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 + x_{10} + x_{11} + x_{12} + x_{13} + x_{14} + x_{15} & = 3 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers with
$$\binom{3 + 15 - 1}{15 - 1} = \binom{17}{14} = \binom{17}{7}$$
solutions. Hence, there are
$$\binom{15}{1}\binom{17}{14}$$
solutions that violate the restriction that no variable exceeds $3$.
Hence, the number of admissible selections is
$$\binom{21}{14} - \binom{15}{1}\binom{17}{14}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving $2\sin\theta\cos\theta + \sin\theta = 0$
The question is to solve the following question in the range $-\pi \le \theta \le \pi$
$$2\sin\theta\cos\theta + \sin\theta = 0$$
I missed the obvious sin factorisation so proceeded as below. I see the correct solutions should be $\pm2/3\pi$ and the values when $\sin\theta = 0$. Although I missed the early factorisation I don't know what I'm doing to actually arrive at an incorrect answer:
$$\begin{align}
2\sin\theta\cos\theta + \sin\theta &= 0 \qquad\text{(square)} \tag{1} \\
4\sin^2\theta\cos^2\theta + \sin^2\theta &= 0 \tag{2}\\
4\sin^2\theta(1-\sin^2\theta) + \sin^2\theta &= 0 \tag{3} \\
4\sin^2\theta - 4\sin^4\theta + \sin^2\theta &= 0 \tag{4} \\
5\sin^2\theta - 4\sin^4\theta &= 0 \tag{5}
\end{align}$$
and then solving by substitution/the quadratic equation I get $\sin\theta = \pm\sqrt(5)/2$ and $0$ but as this out of bounds for sin so cannot be the answer.
I know using Symbolab that this solution is correct for the quadratic I've generated, so I must be going wrong somewhere above after missing that factoristion. I feel like it is in the squaring step but not sure what would be wrong here...
Thanks a lot for your help.
|
For $-\pi\leq\theta\leq\pi$:
\begin{align*}
2\sin{\theta}\cos{\theta}+\sin{\theta}&=0 \\
\sin{\theta}\big(2\cos{\theta}+1\big)&=0 \\
\end{align*}
Thus, $\sin{\theta}=0$ or $2\cos{\theta}+1=0$ .
*
*If $\sin{\theta}=0$, then $\theta\in\{-\pi,0,\pi\}$ .
*If $2\cos{\theta}+1=0$, then $\cos{\theta}=-1/2<0$, then $\theta\in(-\pi,-\pi/2)\cup(\pi/2,\pi)$ and thus $\theta\in\{-2\pi/3,2\pi/3\}$
The solution is $\theta\in\bigg\{-\pi,-\dfrac{2\pi}{3},0,\dfrac{2\pi}{3},\pi\bigg\}$ .
|
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|
If $S=1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}\cdots$, then what is $\lfloor S \rfloor$?
If
$$S=1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}\cdots$$
then $$\lfloor S \rfloor = \text{?}$$
What I tried:
I know that
$$S=1+\frac{1}{2^4}+\frac{1}{3^4}+\cdots=\zeta(4)=\frac{\pi^4}{90}\approx 1.1$$
then $\lfloor S \rfloor =1$.
But how do I find with inequality? Please have a look.
|
Note that $$\frac{1}{k^4}=\int_{k-1}^{k}\frac{1}{k^4}\,dx\le \int_{k-1}^k\frac{1}{x^4}\,dx$$
Therefore, $$\sum_{k=2}^n \frac{1}{k^4}\le \sum_{k=2}^n\int_{k-1}^k\frac{1}{x^4}\,dx=\int_1^n \frac{1}{x^4}\,dx=\frac{1}{3}\left[1-\frac{1}{n^3}\right]$$
Taking limit as $n\to\infty$, we get $$\sum\limits_{k=2}^\infty \frac{1}{k^4}\le \frac{1}{3}$$
Hence we can say that $$ 1\le S\le 1+\frac{1}{3}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3097084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
$\sqrt{24ab+25}+\sqrt{24bc+25}+\sqrt{24ca+25}\geq 21$ if $a+b+c=ab+bc+ca$? For $a,b,c>0 $ and $a+b+c=ab+bc+ca$ . Prove or disprove that : $\sqrt{24ab+25}+\sqrt{24bc+25}+\sqrt{24ca+25}\geq 21$
I checked in very many cases. Example :$c=1,
a=2,b=\frac{1}{2}...$ then it’s true, but cannot prove that
My attempts:
I consider function $ f(x)=\sqrt{24x^2+25}$
And $f’(x)=\frac{24x}{\sqrt{24x^2+25}}$,
$f’’(x)=\frac{600}{(24x^2+25)(\sqrt{24x^2+25}}>0$.
So $f(x)+f(y)+f(z)\geq 3f(\frac{x+y+z}{3})$.
But cannot prove that $\sqrt{ab}+\sqrt{bc}
+\sqrt{ca}\geq 3$
|
I consider function $ f(x)=\sqrt{24x^2+25}$
And $f’(x)=\frac{24x}{\sqrt{24x^2+25}}$,
$f’’(x)=\frac{600}{(24x^2+25)\sqrt{24x^2+25}}>0$
So $f(x)+f(y)+f(z)\geq 3f(\frac{x+y+z}{3})$
(inequalyti Jensen’s)
But can not prove that $\sqrt{ab}+\sqrt{bc}
+\sqrt{ca}\geq 3$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3098290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Show that for all $n \in \Bbb{N}^*$, $\frac{1}{n+1} \le u_n$ $$u_n:= \int_{1}^{e}x^{1/3}(1-\ln(x))^n \, \mathrm{d}x$$
*
*I showed that $u_n$ is decreasing.
*I showed that for all $n \in \Bbb{N}^*$, $u_{n+1} = -\dfrac{3}{4} + \dfrac{3}{4}(n+1)u_n$.
*I have to show that for all $n \in \Bbb{N}^*$, $\dfrac{1}{n+1} \le u_n.$
Do you have any idea? I tried to use the fact that the sequence is decreasing but it doesn't help.
|
Taking the $Z-$transform for the equation that you proved, i.e.
\begin{equation}
u_{n+1}=-\frac{3}{4}+\frac{3}{4}(n+1)u_n
\end{equation}
\begin{equation}
zU(z)-zu(0)=-\frac{3}{4}\frac{z}{z-1}-\frac{3z}{4}\frac{\text{dU(z)}}{\text{dt}}+\frac{3}{4}U(z)
\end{equation}
Assuming $u(0)=1$, and after some manipulation of the above equation and taking inverse $Z-$transform, we get,
\begin{equation}
u_n=\frac{1}{4}\Bigg((4c_1+3)\Big(\frac{3}{4}\Big)^n\Gamma(n+1)-4e^\frac{4}{3}E_{-n}\Big(\frac{4}{3}\Big)\Bigg)
\end{equation}
To find $c_1$, we substitute $u_0=1$ and use the asymptotic expansion of $E_n(x)$ viz.,
\begin{equation}
E_n(x)= \frac{e^{-x}}{x}\Bigg[1-\frac{n}{x}+\frac{n(n+1)}{x^2}-...\Bigg]
\end{equation}
\begin{equation}
E_0\Big(\frac{4}{3}\Big)=\frac{3}{4}e^{-\frac{4}{3}}
\end{equation}
Thus, $c_1=1$
Thus,
\begin{equation}
u_n=\frac{1}{4}\Bigg(7\cdot\Big(\frac{3}{4}\Big)^n\Gamma(n+1)-4e^{\frac{4}{3}}E_{-n}\Big(\frac{4}{3}\Big)\Bigg).
\end{equation}
Now, replacing $n$ with $-n$ and $x$ by $\frac{4}{3}$ in the Exponential Integral function $E_n(x)$, we get,
\begin{equation}
u_n=\frac{1}{4}\Bigg(7\cdot\Big(\frac{3}{4}\Big)^nn!-3\displaystyle \sum_{k=0}^{n} \Bigg(\frac{3}{4}\Bigg)^k{}^nP_k\Bigg)
\end{equation}
This can be represented in other form as:
\begin{equation}
y_n=\frac{4^{-n}\Big(7\cdot 3^{n+1}\cdot (n+1)!-4^{n+1}e^{4/3}E_{-n}\Big(\frac{4}{3}\Big)(n+3)\Big)}{3(n+1)}
\end{equation}
where, $ y_n = 4u_n-\frac{4}{n+1}$.
Since this is a strictly increasing function, it is proved that $u_n\ge\frac{1}{n+1}$
Wolframalpha plot for $y_n$
|
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|
How do the chances of winning Game A compare to the chances of winning Game B?
A coin is biased in such a way that on each toss the probability of heads is $\frac{2}{3}$ and the probability of tails is $\frac{1}{3}$. The outcomes of the tosses are independent. A player has the choice of playing Game A or Game B. In Game A she tosses the coin three times and wins if all three outcomes are the same. In Game B she tosses the coin four times and wins if both the outcomes of the first and second tosses are the same and the outcomes of the third and fourth tosses are the same. How do the chances of winning Game A compare to the chances of winning Game B?
A The probability of winning Game A is $\frac{4}{81}$ less than the probability of winning Game B.
B The probability of winning Game A is $\frac{2}{81}$ less than the probability of winning Game B.
C The probabilities are the same.
D The probability of winning Game A is $\frac{2}{81}$ greater than the probability of winning Game B.
E The probability of winning Game A is $\frac{4}{81}$ greater than the probability of winning Game B.
Let H denote getting heads and T denote getting tales. For game A, the probability of winning is either getting HHH or TTT. The first outcome has a probability of $(\frac{2}{3})^3$ and the second has a probability of $(\frac{1}{3})^3$. Adding them together gives P(Winning A) = $\frac{27}{81}$.
For game B, the player can win by getting either HHTT or TTHH. So we can do $(\frac{2}{3})^2 \cdot (\frac{1}{3})^2 \cdot 2$ (since we can just count for one of HHTT and double it to get the other), which gives $\frac{8}{81}$. None of the answers A-E seems to fit what I have, though. Where did I go wrong?
|
For Game A, winning corresponds to getting $3$ heads in a row or $3$ tails in a row. These have probabilities $(2/3)^3$ and $(1/3)^3$, respectively. This means that the probability of winning Game A is, as you correctly computed,
$$\left(\frac{2}{3}\right)^3 + \left(\frac{1}{3}\right)^3 = \frac{1}{3}$$
For Game B, winning corresponds to one of these four combinations: HHHH, TTTT, HHTT, TTHH. These have probabilities $(2/3)^4$, $(1/3)^4$, $(2/3)^2(1/3)^2$ and $(2/3)^2(1/3)^2$, respectively. This means that the probability of winning Game A is $$\left(\frac{2}{3}\right)^4 + \left(\frac{1}{3}\right)^4 + \left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)^2+ \left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)^2 = \frac{16+1+4+4}{81} = \frac{25}{81}$$
The difference between the probability of winning Game A and that of winning Game B is therefore $$\left|\frac{25}{81}-\frac{1}{3}\right| = \boxed{\frac{2}{81}\,}$$
|
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"url": "https://math.stackexchange.com/questions/3099379",
"timestamp": "2023-03-29T00:00:00",
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|
In a right angled $\triangle ABC$, $DE$ and $DF$ are perpendicular to $AB$ and $BC$ respectively. What is the probability of $DE\cdot DF>3$?
In a right angled $\triangle ABC$, $\angle B = 90^\circ$, $\angle C = 15^\circ$ and $|AC| = 7.\;$ Let a point $D$ (Random Point) be taken on $AC$ and then perpendicular lines $DE$ and $DF$ are drawn on $AB$ and $AC$ respectively. What is the probability of $DE\cdot DF >3?$
Attempt:
By trigonometry, I got the length of other two side from the hypotenuse $AC:$
$AB$ $\approx 1.8117$
$BC$ $\approx 6.7614$. And than, I got the equation that
$DE\cdot DF = (6.7614 - DE)\cdot AE\;$ (from the similarity of both $\triangle AED$ and $\triangle DFC$)
Again, from right angled $\triangle AED$,
$\dfrac{AE}{DE} = \tan 15^{\circ}\quad \implies \quad AE = DE\cdot \tan 15^\circ$
Here, I got stuck. I couldn't find a way out to proceed and skip that situation. I became lost and was unable to complete that process. Any kind of help or clue will be greatly helpful for me to step forward.
|
Let $DC = x , DF = o , DE = a $ and $ DA=y $ .
We have :- $$ o = x \sin 15 \tag{1}$$ $$a= y \ cos 15 \tag{2} $$ Multiplying $(1),(2)$ , we get :- $$ o\cdot a = xy \sin 15 \cos 15 = \frac{xy}{2} sin 30 = \frac{xy}{4} > 3 $$ ( $\because 2 \sin \theta \cos \theta = \sin 2\theta $)
Hence , we need $xy = x(7-x) > 12$ or $x^2-7x+12<0$
As this is of the form $ax^2+bx+c$ , and since $a>0$ , the expression is negative between the roots . Therefore, we must have $x \in (3,4) $
$\therefore $ The probability $\frac{4-3}{7} = \frac{1}{7}$
|
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"url": "https://math.stackexchange.com/questions/3099727",
"timestamp": "2023-03-29T00:00:00",
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|
Differentiation uner the integral sign - help me find my mistake This is my integral:
$$I(a)=\int_0^\infty\frac {\ln(a^2+x^2)}{(b^2+x^2)}dx.$$
Taking the first derivative with respect to a:
$$I'(a)=\int_0^\infty \frac {2adx} {(a^2+x^2)(b^2+x^2)}.$$
This is how I did the partial fraction decomposition:
$\frac {2a} {(a^2+x^2)(b^2+x^2)}=\frac {Ax+B} {(a^2+x^2)}+\frac {Cx+D} {(b^2+x^2)}$.
From here I get that $A=C=0$, $B=\frac {2a} {(b^2-a^2)}$ and $D=\frac {-2a} {(b^2-a^2)}$
Is this correct? Because when I try to solve $I'(a)$ using these values for $B$ and $D$ I get a different solution from the textbook?
|
Here you are solving:
\begin{equation}
I(a,b)=\int_0^\infty\frac {\ln\left(a^2+x^2\right)}{\left(b^2+x^2\right)}dx
\end{equation}
The method you have taken is perfectly fine. Here I will employ Feynman's Trick and introduce a new parameter:
\begin{equation}
J(t;a,b)=\int_0^\infty\frac {\ln\left(a^2+tx^2\right)}{\left(b^2+x^2\right)}dx
\end{equation}
We see that $I(a,b) = \lim_{t \rightarrow 1^+} J(t;a,b)$ and that $J(0;a,b)$ is easy to resolve (will do later). Applying Leibniz's Integral Rule we take the derivative with respect to $t$:
\begin{align}
\frac{dJ}{dt} &= \int_0^\infty\frac {x^2}{\left(a^2 + tx^2\right)\left(b^2+x^2\right)}dx = \frac{1}{b^2t - a^2}\int_0^\infty\left[\frac{b^2}{b^2 + x^2} - \frac{a^2}{a^2+tx^2}
\right]dx \\
&= \frac{1}{b^2t - a^2}\left[\left|b\right|\operatorname{arctan}\left(\frac{x}{\left|b\right|} \right) - \frac{\left|a\right|}{\sqrt{t}} \operatorname{arctan}\left(\frac{\sqrt{t}x}{\left|a\right|}\right)
\right]_0^\infty \\
&=\frac{1}{b^2t - a^2}\left[\left|b\right|\cdot \frac{\pi}{2} -\frac{\left|a\right|}{\sqrt{t}}\cdot \frac{\pi}{2}\right] =\frac{\pi}{2\left(b^2t - a^2\right)}\left[\left|b\right|\ -\frac{\left|a\right|}{\sqrt{t}}\right] = \frac{\pi}{2\sqrt{t}\left(b^2t - a^2\right)}\left[\left|b\right|\sqrt{t} -\left|a\right|\right] \\
&=\frac{\pi}{2\sqrt{t}\left(\left|b\right|\sqrt{t} + \left|a\right|\right)}
\end{align}
We now integrate with respect to $t$:
\begin{equation}
J(t;a,b) = \int \frac{\pi}{2\sqrt{t}\left(\left|b\right|\sqrt{t} + \left|a\right|\right)}\:dt
\end{equation}
Here let $t = u^2$
\begin{align}
&\int \frac{\pi}{2\sqrt{t}\left(\left|b\right|\sqrt{t} + \left|a\right|\right)}\:dt = \int \frac{\pi}{2\sqrt{u^2}\left(\left|b\right|\sqrt{u^2} + \left|a\right|\right)}\cdot 2u\:du = \int \frac{\pi}{\left|b\right||u| + \left|a\right|}\:du \\
&= \frac{\pi}{|b|}\ln\left|\left|b\right||u| + \left|a\right| \right| + C = \frac{\pi}{|b|}\ln\left|\left|b\right|\sqrt{t} + \left|a\right| \right| + C
\end{align}
Where $C$ is the constant of integration.To resolve we use $J(0;a,b)$:
\begin{equation}
J(0;a,b) = \int_0^\infty\frac {\ln\left(a^2+0\cdot x^2\right)}{\left(b^2+x^2\right)}dx = \frac{\pi}{|b|}\ln\left|\left|b\right|\sqrt{0} + \left|a\right| \right| + C = \frac{\pi}{|b|}\ln\left|ab \right| + C
\end{equation}
And thus
\begin{align}
\frac{\pi}{|b|}\ln\left|ab \right| + C&= \int_0^\infty\frac {\ln\left(a^2\right)}{\left(b^2+x^2\right)}dx = 2\ln|a| \int_0^\infty\frac {1}{\left(b^2+x^2\right)}dx\\
& = 2\ln|a| \left[\frac{1}{\left|b\right|}\operatorname{arctan\left(\frac{x}{\left|b\right|} \right)}
\right]_0^\infty = 2\ln|a| \cdot \frac{1}{\left|b\right|} \frac{\pi}{2} = \frac{\pi\ln|a|}{|b|}
\end{align}
Thus
\begin{equation}
C = \frac{\pi\ln|a|}{|b|} - \frac{\pi}{|b|}\ln\left|ab \right| = -\frac{\pi}{|b|}\ln|b|
\end{equation}
Thus, we form our solutions for $J(t;a,b)$:
\begin{equation}
J(t;a,b) = \frac{\pi}{|b|}\ln\left|\left|b\right| \sqrt{t}+ \left|a\right| \right| -\frac{\pi}{|b|}\ln|b|
\end{equation}
We now can solve for $I(a,b)$ by applying the limit as above:
\begin{align}
I(a,b) &= \lim_{t \rightarrow 1^+} J(t;a,b) = \lim_{t \rightarrow 1^+}\frac{\pi}{|b|}\ln\left|\left|b\right| \sqrt{t}+ \left|a\right| \right| -\frac{\pi}{|b|}\ln|b| \\
&= \frac{\pi}{|b|}\ln\left|\left|b\right| + \left|a\right| \right| -\frac{\pi}{|b|}\ln|b| =\frac{\pi}{|b|}\ln\left|\frac{\left|b\right| + \left|a\right|}{|b|} \right|
\end{align}
Or
\begin{equation}
\int_0^\infty\frac {\ln\left(a^2+x^2\right)}{\left(b^2+x^2\right)}dx = \frac{\pi}{|b|}\ln\left|\frac{\left|b\right| + \left|a\right|}{|b|} \right|
\end{equation}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Use the definition of limit to show that $\lim_{x \to -1} \frac{x+5}{2x+3}=4$
Use the definition of limit to show that
$\lim_{x \to -1} \frac{x+5}{2x+3}=4$
We have $|\frac{x+ 5}{2x+3} -4|= |\frac{x+5-8x-12}{2x+3}|=|\frac{-7x-7}{2x+3}|=\frac{7}{|2x+3|}|x+1|$
To get a bound on the coefficient of $|x+1|$, we restrict $x$ by the condition $-2<x<0$ [neighborhood of $-1$] . For $x$ in this interval, we have $-1<2x+3<3$, so that
$|\frac{x+ 5}{2x+3} -4|= \frac{7}{|2x+3|}|x+1|<...$
Is that true, please? And I don’t know how can I complete.
|
Yes that's right. From this step on, as $x\to -1$ you can write$$-\epsilon<x+1<\epsilon$$ for some small choice of $\epsilon>0$ therefore $$1-2\epsilon<2x+3<1+2\epsilon$$since for small enough $\epsilon>0$ we have $1-2\epsilon >{1\over 2}$ we can conclude that $$0\le {|1+x|\over |2x+3|}<{\epsilon \over {1\over 2}}=2\epsilon$$which completes the proof.
|
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|
$\sum_{k=1}^{n-1}\left(1-e^{\frac{2\pi ki}{n}}\right)^{-1}$ How can I go about computing
$$
\sum_{k\ =\ 1}^{n - 1}
\left(1 - \mathrm{e}^{\large 2\pi k\mathrm{i}/n}\right)^{-1}\
{\Large ?}
$$
I originally thought that it was supposed to be the reciprocal of the sum, and I ended up with $1/n$, but now I realized that it is the sum of the reciprocals. I've tried using
$$e^{ix}=\cos x+i\sin x,$$
but I didn't get anywhere with that.
|
With $\zeta = \exp(2\pi i/n)$ and
$$f(z) = \frac{1}{1-z} \frac{n/z}{z^n-1}$$
we have for $1\le k\le n-1$
$$\mathrm{Res}_{z=\zeta^k} f(z)
= \frac{1}{1-\zeta^k}$$
so that
$$S = \sum_{k=1}^{n-1} \frac{1}{1-\zeta^k}
= \sum_{k=1}^{n-1} \mathrm{Res}_{z=\zeta^k} f(z).$$
Residues sum to zero and the residue at infinity is zero, so we find
$$S = - \mathrm{Res}_{z=1} f(z) - \mathrm{Res}_{z=0} f(z).$$
For the first one we have
$$- \mathrm{Res}_{z=1} f(z)
= \mathrm{Res}_{z=1} \frac{1}{z-1} \frac{n/z}{z^n-1}
\\ = \mathrm{Res}_{z=1} \frac{1}{(z-1)^2}
\frac{n/z}{1+z+\cdots+z^{n-1}}
\\ = n
\left.\left(\frac{1}{z} \frac{1}{1+z+\cdots+z^{n-1}}
\right)'\right|_{z=1}
\\ = n
\left.\left(- \frac{1}{z^2} \frac{1}{1+z+\cdots+z^{n-1}}
- \frac{1}{z} \frac{(1+\cdots+(n-1) z^{n-2})}
{(1+z+\cdots+z^{n-1})^2}
\right)\right|_{z=1}
\\ = n
\left( - \frac{1}{n} - \frac{1}{n^2} \frac{1}{2} (n-1) n \right)
= -1 - \frac{1}{2} (n-1).$$
The second one is
$$- \mathrm{Res}_{z=0} f(z) = - (1 \times n \times -1) = n.$$
Collecting everything we get
$$\bbox[5px,border:2px solid #00A000]{
S = \frac{1}{2} (n-1).}$$
|
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|
Evaluate the values of $x$ in $\sqrt{2x-5} = \sqrt[3]{6x-15}$
$$\sqrt{2x-5} = \sqrt[3]{6x-15}$$
*
*Evaluate the values of $x$
I believe that there would be an easier approach to this problem because I will have to expand 3th degree binomial as shown below
$$(2x-5)^3 = (6x-15)^2$$
What am I missing?
Regards
|
$$(2x-5)^3 = (6x-15)^2$$
$$(2x-5)^3 = (3(2x-5))^2$$
$$(2x-5)^3 = 9(2x-5)^2$$
This may help :)
Then: $(2x-5)=0\lor (2x-5)=9$
Thus: $x=2\frac{1}{2}\lor x=7$
|
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|
Better proof for $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$ Prove this $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$
My attempt
Proof - by using [axiomdistributive] and [axiommulcommutative]:
$$\begin{split}
&(x+y)(x^2 - xy + y^2)\\
&= (x+y)x^2 - (x+y)xy + (x+y)y^2\\
&= (x^3+x^2y) - (x^2y+xy^2) + (xy^2+y^3)\\
&= x^3 + x^2y - x^2y - xy^2 + xy^2 + y^3\\
&= x^3 + y^3\\
\end{split}$$
Q.E.D.
Question:
Spivak says there is an easy proof that, if I use this other theorem:
$$
x^3 - y^3 = (x-y)(x^2 + xy + y^2)
$$
then, I will also allow me to find out $x^n+y^n$ whenever $n$ is odd.
How to do this? I fail to see how.
|
$$x^3+y^3 = x^3+\color{red}{x^2y-x^2y}+y^3 $$
$$ = x^2(x+y)-y(x^2-y^2)$$
$$ = x^2(x+y)-y(x-y)(x+y)$$
$$ = (x+y)(x^2-y(x-y))$$
$$ = (x+y)(x^2-xy+y^2)$$
We can proceede similary for arbitrary odd $n$.
|
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|
How to prove that $\frac{m^{n+1}}{n+1}<1^n+2^n+\dots+m^n<(1+\frac{1}{m})^{n+1}\frac{m^{n+1}}{n+1} $? The first part of the problem is:
Prove that for all integers $n \ge 1$ and real numbers $t>1$,
$$ (n+1)t^n(t-1)>t^{n+1}-1>(n+1)(t-1)$$
I have done the first part by induction on $n$ for any real $t>1$.
However, I don't know how to do the second part, which is:
Use this to prove that if $m$ and $n$ are positive integers,
$$ \frac{m^{n+1}}{n+1}<1^n+2^n+\dots+m^n<\Big(1+\frac{1}{m}\Big)^{n+1}\frac{m^{n+1}}{n+1} $$
I factorized $t^{n+1}-1$ into $(t-1)(1+t+t^2+\dots+t^n)$ and cancelled $t-1$ to obtain $(n+1)t^n>1+t+t^2+\dots+t^n>n+1$. And have no idea what to do next. This is the only way I could think of in order to get a "sum", but couldn't see any relation between the two sum (if there are any...).
The last part of the question is:
Find
$$ \lim_{m\rightarrow \infty} \frac{1^n+2^n+\dots+m^n}{m^{n+1}}$$
I think I know how to do the last part. It should be divide the inequality by $m^{n+1}$, then apply Squeeze Theorem. I believe the answer is $\frac{1}{n+1}$.
|
Let $S=1^n+2^n+3^n+\cdots+m^n$.
From a quick graph sketch it is clear that:
$$\begin{align}
\int_0^m x^n dx \;\;&<\qquad \qquad \quad S &&< \int_0^m (x+1)^n dx\\
\left[\frac {x^{n+1}}{n+1}\right]_0^m\;\;&<\qquad \qquad \quad S &&<\;\;\left[\frac {(x+1)^{n+1}}{n+1}\right]_0^m\\
\frac {m^{n+1}}{n+1}\;\;&<\qquad \qquad \quad S &&<\;\;\frac {(m+1)^{n+1}}{n+1}-\frac 1{n+1}\\
\frac {m^{n+1}}{n+1}\;\;&<\qquad \qquad \quad S &&<\;\;\left[\left(1+\frac 1m\right)^{m+1}-1\right]\frac {m^{n+1}}{n+1}<\left(1+\frac 1m\right)^{m+1}\frac {m^{n+1}}{n+1}\\
\frac {m^{n+1}}{n+1}\;\;&<\qquad1^n+2^n+3^n+\cdots+m^n &&<\left(1+\frac 1m\right)^{m+1}\frac {m^{n+1}}{n+1}\\
\end{align}$$
|
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|
solve an equation in complex plane $a(5-i)+b=ai-3$, $a$ and $b$ are conjugate complex number.
Find $a$ and $b$.
I have tried several methods to solve it but it stuck.
How find the relationship of a and b in the equation with the complex number?
|
$a(5-i)+b=ai-3$
so
$b=a(2i-5)-3$
but
$b=x+iy$
and
$a=x-iy$
$x+iy=(x-iy)(2i-5)-3=2ix-5x+2y+5iy-3=$
$=(2y-5x-3)+i(2x+5y)$
So you must solve the system
$x=2y-5x-3$
and
$y=2x+5y$
$x=\frac{y}{3}-\frac{1}{2}$
$y=\frac{2y}{3}-1+5y=\frac{17}{3}y-1$
To sum up
$y=\frac{3}{14}$
$x=-\frac{3}{7}$
|
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|
Let a,b and c be the side lengths of triangle ABC respectively...find the greatest value of b*c. Let a,b and c be the side lengths of triangle ABC respectively. If the perimeter of $\Delta$ABC is 7, and that $\cos A=-\frac{1}{8}$, find the greatest value of $b*c$.
This is how I start the solution:
$$a+b+c=7 \implies b+c=7-a,\quad \cos A=-\frac{1}{8}\\
a^2=b^2+c^2-2bc\cdot \cos A\\
\implies a^2=b^2+c^2+\frac{bc}{4} \\
\implies a^2=(7-a)^2-2bc+\frac{bc}{4} \\
\implies bc=4(7-2a)$$
|
$$ a^2=b^2+c^2+\frac{bc}{4}$$
so $$ (7-b-c)^2 = b^2+c^2+bc/4$$
so $$ 49-14b-14c+2bc = bc/4$$
so $$ 7bc= 56(b+c)-196 \geq 112\sqrt{bc}-196$$
Put $ x = \sqrt{bc} \;\;\;\;(\leq {b+c\over 2} < {7\over 2})$ then $$7x^2-112x+196\geq 0$$
so $$x^2-16x+28\geq 0$$
so $$x\in (0,2]\cup[14,\infty)$$
and thus $x_{\max} \leq 2$ (and this is achieved if $b=c=2$)
|
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|
Evaluate : $\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}$
Evaluate: $\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}$
My solution:
\begin{align}
\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}} & = \lim_{x\to -\infty} \frac{\sqrt{4-\frac{2}{x}}}{\sqrt[3]{1+\frac{1}{x^3}}}\\
& = \frac{\sqrt{4-0}}{\sqrt[3]{1+0}} \\
& = 2
\end{align}
Despite the steps I've taken seems plausible to me, the answer is given as $-2$.
Is dividing both the numerator and the denominator by $x$ allowed here? Where am I making a mistake?
|
Keep in mind that it is a limit at $-\infty$. Therefore, the way you should deal with the numerator is$$\sqrt{4x^2-2x}=-x\frac{\sqrt{4x^2-2x}}{-x}=-x\sqrt{4-\frac2x}.$$
|
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|
If $a, b, c, d$ exists s.t. $p=a^2+kb^2$, $pn=c^2+kd^2$, proof that integer $x, y$ such that $n=x^2+ky^2$ exists. Question. $p$ is a prime, $k$ is a given natural number. If $a, b, c, d$ exists s.t. $p=a^2+kb^2$, $pn=c^2+kd^2$, proof that integer $x, y$ such that $n=x^2+ky^2$ exists.
My approach. Let $n=x^2+ky^2$.
$pn=(a^2+kb^2)(x^2+ky^2)=(ax \pm kby)^2+k(ay \mp bx)^2$.
Let $(x_1, y_1)$ the root of $ax+kby=c, -bx+ay=d$.
Let $(x_2, y_2)$ the root of $ax-kby=c, bx+ay=d$.
If one of $(x_1, y_1)$ or $(x_2, y_2)$ is integer pair, the proof will be done.
$$x_1=\frac{ac-kbd}{p}, y_1=\frac{ad+bc}{p}$$
$$x_2=\frac{ac+kbd}{p}, y_2=\frac{ad-bc}{p}$$
So when we proof these three,
*
*$p \mid (ad+bc)(ad-bc)$
*$p \mid ad+bc \Rightarrow p \mid ac-kbd$
*$p \mid ad-bc \Rightarrow p \mid ac+kbd$
The proof will be done.
How should I prove this?
|
feb. 21 2019
LEMMA: with integers, if $v^2 | w^2,$ then $v | w.$
With $b \neq 0,$ since $p$ is a prime rather than a square,
$$ p = a^2 + k b^2 $$
$$ np = c^2 + k d^2 $$
$$ kb^2 = p - a^2 $$
$$ b^2 np = b^2 c^2 + (k b^2) d^2 = b^2 c^2 + (p-a^2)d^2 = (b^2 c^2 - a^2 d^2) + p d^2 $$
Thus
$$ p | (bc-ad)(bc+ad) $$
and $p$ divides at least one of them. Define some
$$ \phi = \pm 1$$ so that
$$ p | bc - \phi ad. $$
We then have an integer $\tau$ with
$$ bc - \phi ad = p\tau, $$
$$ bc + \phi ad = p\tau + 2 \phi a d , $$
$$ (b^2 c^2 - a^2 d^2) = p^2 \tau^2 + 2 \phi adp\tau . $$ Recall
$$ b^2 np = (b^2 c^2 - a^2 d^2) + p d^2 $$
$$ b^2 np = p^2 \tau^2 + 2 \phi adp\tau + p d^2 $$ Divide by $p$ and switch order
$$ b^2 n = d^2 + 2 \phi da\tau + p \tau^2 $$
Add and subtract $a^2 \tau^2, \;$
$$ b^2 n = (d^2 + 2 \phi da\tau + a^2 \tau^2) + (p \tau^2 - a^2 \tau^2) $$
$$ b^2 n = (d + \phi a \tau)^2 + \tau^2 (p-a^2) $$
Recall $p - a^2 = k b^2$
$$ b^2 n = (d + \phi a \tau)^2 + k b^2 \tau^2 $$
Now, $b^2 | (d + \phi a \tau)^2,$ so the LEMMA says $b | d + \phi a \tau.$ We may introduce an integer $\psi$ with
$$ d + \phi a \tau = b \psi, $$ whence
$$ (d + \phi a \tau)^2 = b^2 \psi^2. $$ Then
$$ b^2 n = b^2 \psi^2 + k b^2 \tau^2, $$ finally
$$ n = \psi^2 + k \tau^2. $$
|
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|
Finding the Exact X-Coordinates of the Points on the Graph f(x) for Which the Tangent Line is Parallel to the Line g(x) I cannot seem to solve the following situation:
When $f'(x)$ is equal to the slope of $g(x)$ over the interval of $\frac{\pi}{2}\le x \le \pi $ when
.
$$f'(x)=\frac{2\cos(2x)}{3\sin(2x)^{\frac{2}{3}} }$$
and $g(x)$ is defined by the equation
$$2x-3(6)^\frac{1}{3}y=0$$
.
So, my fundamental problem is with the trig functions in this case. The problem simplifies to:
$$\frac{\cos(2x)}{\sin(2x)^{\frac{2}{3}}}-\frac{1}{6^\frac{1}{3}}=0$$
.
As always, thanks!
|
Since $g^\prime(x)=\dfrac{2}{3\sqrt[3]{6}}$ we wish to find a solution for
$$ \frac{2\cos(2x)}{3\sin(2x)^{\frac{2}{3}} }= \dfrac{2}{3\sqrt[3]{6}}$$
in the interval $\frac{\pi}{2}\le x \le \pi$. Simplifying and cubing both sides gives
\begin{eqnarray}
\frac{\cos^3(2x)}{\sin^2(2x)}&=&\frac{1}{6}\\
6\cos^3(2x)-\sin^2(2x)&=&0\\
6\cos^3(2x)+\cos^2(2x)-1&=&0\\
6u^3+u^2-1&=&0\\
\end{eqnarray}
Can you finish from here, given that the polynomial has a rational zero?
|
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|
Show that sequence is a Cauchy sequence Prove that given sequence $$\langle f_n\rangle =1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{(-1)^{n-1}}{n}$$
is a Cauchy sequence
My attempt :
$|f_{n}-f_{m}|=\Biggl|\dfrac{(-1)^{m}}{m+1}+\dfrac{(-1)^{m+1}}{m+2}\cdots\dots+\dfrac{(-1)^{n-1}}{n}\Biggr|$
using $ m+1>m \implies \dfrac{1}{m+1}<\dfrac{1}{m} $
$|f_{n}-f_{m}|\le \dfrac{1}{m}+\dfrac{1}{m}+\dfrac{1}{m}\cdots\cdots\dfrac{1}{m}$
$|f_{n}-f_{m}|\le\dfrac{n-m}{m}$
I don't know if I am proceeding correctly or if I am, how to proceed further, any hint would be really helpful .
|
If you ignore the signs of the terms,
the result diverges.
So you can't do that.
$f_n
=\sum_{k=1}^n \dfrac{(-1)^k}{k}
$
so,
if $n > m$,
$f_n-f_m
=\sum_{k=m+1}^n \dfrac{(-1)^k}{k}
=\sum_{k=1}^{n-m} \dfrac{(-1)^{k+m}}{k+m}
=(-1)^m\sum_{k=1}^{n-m} \dfrac{(-1)^{k}}{k+m}
$.
If
$n-m$ is even,
so $n-m = 2j$,
then
$\begin{array}\\
f_n-f_m
&=(-1)^m\sum_{k=1}^{2j} \dfrac{(-1)^{k}}{k+m}\\
&=(-1)^m\sum_{k=1}^{j} \left(\dfrac{(-1)^{2k-1}}{2k-1+m}+\dfrac{(-1)^{2k}}{2k+m}\right)\\
&=(-1)^m\sum_{k=1}^{j} (-1)^{2k-1}\left(\dfrac{-1}{2k-1+m}+\dfrac{1}{2k+m}\right)\\
&=(-1)^m\sum_{k=1}^{j} (-1)^{2k-1}\left(\dfrac{(2k-1+m)-(2k+m)}{(2k-1+m)(2k+m)}\right)\\
&=(-1)^{m+1}\sum_{k=1}^{j} \left(\dfrac{-1}{(2k-1+m)(2k+m)}\right)\\
&=(-1)^{m}\sum_{k=1}^{j} \left(\dfrac{1}{(2k-1+m)(2k+m)}\right)\\
\text{so}\\
|f_n-f_m|
&=\sum_{k=1}^{j} \left(\dfrac{1}{(2k-1+m)(2k+m)}\right)\\
&=\sum_{k=1}^{j}\dfrac14 \left(\dfrac{1}{(k-\frac12+\frac{m}{2})(k+\frac{m}{2})}\right)\\
&\lt \dfrac14\sum_{k=1}^{j} \left(\dfrac{1}{(k-1+\frac{m}{2})(k+\frac{m}{2})}\right)
\quad\text{this is the sneaky part}\\
&\lt \dfrac14\sum_{k=1}^{j} \left(\dfrac{1}{k-1+\frac{m}{2}}-\dfrac{1}{k+\frac{m}{2}}\right)\\
&= \dfrac14 \left(\dfrac{1}{\frac{m}{2}}-\dfrac{1}{j+\frac{m}{2}}\right)\\
&= \dfrac12 \left(\dfrac{1}{m}-\dfrac{1}{2j+m}\right)\\
&= \dfrac12 \left(\dfrac{1}{m}-\dfrac{1}{n}\right)\\
&< \dfrac{1}{2m}\\
&\to 0 \text{ as } m \to \infty\\
\end{array}
$
If $n-m$ is odd,
the sum changes
by at most $\frac1{n}$
so it still goes to zero.
|
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|
The intersection of a sphere and the xy-plane is the circle $(x-1)^2+(y-2)^2=5$ and point $(1,2,1)$ is on the sphere The intersection of a sphere and the $xy$-plane is the circle $$(x-1)^2+(y-2)^2=5$$ and point $(1,2,1)$ is on the sphere. Find the center and radius of the sphere.
When the sphere and $xy$-plane intersect than $z=0$.
I tried $(x-1)^2+(y-2)^2+(z+?)^2=5$ but what can I do next?
|
Let $O(a,b,c)$ be center of sphere, $A(1,2,0)$ be the center of intersection circle, $B(1,2+\sqrt{5},0)$ be the point of the sphere and intersection circle, $C(1,2,1)$ be the point on the sphere.
Then the radius of the sphere is $R=OC=OB=AO+AC=AO+1$.
The triangle $AOB$ is right, so:
$$AO^2+AB^2=OB^2 \Rightarrow AO^2+5=(AO+1)^2 \Rightarrow AO=2.$$
Thus:
$$R=AO+1=3;O(1,2,-2).$$
|
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|
Prove that $\ x_{n}=\int_{n}^{2n} \frac{x+a}{x^{3}+2a}dx$ is decreasing Let $\ a>0$ and the sequence $(x_{n})_{n>=0}$ defined by $\ x_{n}=\int_{n}^{2n} \frac{x+a}{x^{3}+2a}dx$. Prove the sequence is monotonically decreasing and $0<x_n<\frac{4+3a}{8}$, for any $n>0$. I've made little progress towards proving that $x_{n}<x_{n-1}$. First, if a function takes the form of the fraction , $f(x)=\frac{x+a}{x^{3}+2a}$, then it's monotonically decreasing. Also, $x_n$ compared to $x_{n-1}$ boils down to comparing $2\frac{x+2a}{x^{3}+16a}$ with $\frac{x+a}{x^{3}+2a}$. From here I have really no idea how to continue. These simple results are the work of some ruminations and I am unable to bring something out of the blue to complete the proof...
|
First, let's note that
$$ \int_n^{2n} \frac{x+a}{x^3+2a}dx \leq \int_n^{2n} \frac{x+a}{x^3}dx.$$
Evaluating the integral on the right, we have
$$ \int_n^{2n} \frac{x+a}{x^3}dx = \int_n^{2n} \frac{1}{x^2}dx + a\int_n^{2n} \frac{1}{x^3}dx.$$
These are easy integrals to evaluate. This gives us that
$$ \int_n^{2n} \frac{x+a}{x^3}dx = \frac{3a + 4n}{8n^2} \leq \frac{3a + 4}{8}$$
for $n \geq 1$. At $n = 0$, we have that the integral is
$$\int_0^0 \frac{x+a}{x^3+2a}dx = 0,$$
and so denoting
$$ I_n = \int_n^{2n} \frac{x+a}{x^3+2a}dx,$$
we get
$$ 0 \leq I_n \leq \frac{3a + 4}{8}$$
for all $a > 0$. Now, we would like to establish
$$ I_{n+1} \leq I_n.$$
Can you take it from here?
|
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|
Why does the domain of convergence change after this integration? So, while trying to derive the Taylor expansion for $\ln(x+1)$ around $x=0$, I started by using the well-known power series:
$$\frac{1}{1-x}=1+x+x^2 +x^3 +... \qquad \text{when } \lvert x \rvert < 1$$
Letting $x \mapsto-x$, we get:
$$\frac{1}{1+x}=1-x+x^2 -x^3 + ... \qquad \text{when } \lvert x \rvert < 1$$
So then perfoming this integration gives us:
$$\begin{align*}
\ln(1+x) = \int_0^x \frac{dt}{1+t} &=\int_0^x (1-t+t^2-t^3+...)dt \\
&= x-\frac{x^2}{2}+\frac{x^3}{3}- \frac{x^4}{4} +... \qquad \text{when } \vert x \rvert<1
\end{align*}$$
However, the real domain of convergence for this is $-1 < x \leq 1$ as $\ln(2) = 1- \frac{1}{2} +\frac{1}{3} - \frac{1}{4}+ ...$
So my question is why did the domain over which the original function converges in change after this integration and is there a way to predict when this happens given some power series, or should one always check the endpoints again?
|
Writing
$$\tag{*}\ln(1+x) = \int_0^x \frac{dt}{1+t} =\int_0^x (1-t+t^2-t^3+...)\,dt \\ = x-\frac{x^2}{2}+\frac{x^3}{3}- \frac{x^4}{4} + \ldots $$
is true for $x < 1$ because the power series converges uniformly on the compact interval $[0,x]$ (within the interval $(-1,1)$ bounded by the radius of convergence) and termwise integration is permissible.
The series $1-t+t^2-t^3+...$ is neither pointwise convergent at $t=1$ nor uniformly convergent on $[0,1)$.
As much as one might like to simply substitute $1$ for $x$ in (*) and conclude that
$$\tag{**} \ln 2 = 1-\frac{1}{2}+\frac{1}{3}- \frac{1}{4} + \ldots ,$$
it cannot be justifed without further analysis, i.e. there are other examples of this type where it would be false.
In this case, however, we can establish that the series on the RHS of (**) is convergent by the alternating series test and by Abel's theorem it follows that
$$\ln 2 = \lim_{x \to 1-} \ln(1+x) = 1-\frac{1}{2}+\frac{1}{3}- \frac{1}{4} + \ldots $$
Again -- knowing only the radius of convergence is not enough information to establish this result.
|
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|
This polynomial is divisible by $(x-2)$ so the remainder is $0$. How do we evaluate the product $ab$? $$P(x) = x^{2a+b-1} +x^{a-2b+5}-2x^{a+b-1}$$
This polynomial is divisible by $(x-2)$ so the remainder is $0$. How do we evaluate the product $ab$?
My attempt:
This polynomial is divisible by $(x-2)$ so the remainder is $0$
$$P(x) = (x-2)Q(x) + 0 \implies P(2) = 0$$
Plugging $x = 2$
$$P(2) = (2)^{2a+b-1}+(2)^{a-2b+5}-2(2)^{a+b-1}$$
Let $2^a = u$ and $2^b = m$
$$0 = \dfrac{1}{2}(u^2m)+32um^{-2}-2\dfrac{1}{2}(um) = \dfrac{1}{2}(u^2m)+32um^{-2}-um$$
Multiplying both sides by $2m^2$
$$0 = u^2m^3+64u-2um^3$$
Factoring $u$
$$0 = u\biggr(um^3+64-2m^3\biggr ) $$
Here we get one solution for $u = 0$. But I'm not sure If I went correctly.
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We continue from the last line, since it has been pointed out in comments that $u\ne0$:
$$0=um^3+64-2m^3=(u-2)m^3+64$$
$$64=(2-u)m^3$$
Given that $u,m$ are powers of 2, $2-u$ is positive, but $u>0$ so $u=1$ and $m=4$, i.e. $a=0,b=2$ and $ab=0$.
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"timestamp": "2023-03-29T00:00:00",
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|
Coefficients of $1,x,x^2$ in $((\cdots (x-2)^2-2)^2-2)^2\cdots )-2)^2$
Finding coefficients of $x^0,x^1,x^2$ in
$((\cdots (x-2)^2-2)^2-2)^2\cdots )-2)^2$
where there are $k$ parenthesis in the left side
Try:
Let $P(x)=((\cdots (x-2)^2-2)^2-2)^2\cdots )-2)^2$
Let we assume
$P(x)=a_{k}+b_{k}x+c_{k}x^2+d_{k}x^3+\cdots\cdots $
For constant term put $x=0$, we have
$$((\cdots (0-2)^2-2)^2-2)^2\cdots )-2)^2=a_{k}$$
Now i did not know how to find coeff. of $1,x,x^2$ in $P(x)$ . Thanks
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Let the $x^2,x,1$ coefficients of the polynomial formed from the expression with $k$ brackets be $a_k,b_k,c_k$ respectively. We have
$$a_1,b_1,c_1=1,-4,4$$
$$a_{k+1},b_{k+1},c_{k+1}=2a_k(c_k-2)+b_k^2,2b_k(c_k-2),(c_k-2)^2$$
We first notice that $c_k=4$ for all $k\ge1$, so we reduce the recurrences left to solve to
$$a_{k+1},b_{k+1}=4a_k+b_k^2,4b_k$$
Solving the recurrence for $b_k$ gives $b_k=-4^k$ and
$$a_{k+1}=4a_k+4^{2k}$$
the solution of which is OEIS A166984:
$$a_k=\frac{16^k-4^k}{12}$$
Thus the three lowest-degree terms of the expression with $k$ are
$$\frac{16^k-4^k}{12}x^2-4^kx+4$$
The $x^3$ coefficient
In the same vein, denote the $x^3$ coefficient for $k$ brackets $d_k$, so
$$d_1=0\qquad d_{k+1}=2(d_k(c_k-2)+a_kb_k)=4d_k-\frac{16^k-4^k}6\cdot4^k$$
$$=4d_k-\frac{64^k-16^k}6$$
This may be solved in a similar manner to $a_k$ to obtain
$$d_k=\frac{-64^k+5\cdot16^k-4\cdot4^k}{360}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$6\sqrt{24} + 7\sqrt{54} - 12\sqrt{6}$ = $21\sqrt{6}$ but I get $207\sqrt{6}$ I'm asked to simplify $6\sqrt{24} + 7\sqrt{54} - 12\sqrt{6}$
The provided solution is $21\sqrt{6}$ but I arrive at a different amount.
Here is my working, trying to understand where I went wrong:
First expression:
$6\sqrt{24}$ = $6\sqrt{4}$ * $6\sqrt{6}$ = $6*2*6\sqrt{6}$ = $72\sqrt{6}$
Second expression:
$7\sqrt{54}$ = $7\sqrt{9} * 7\sqrt{6}$ = $147\sqrt{6}$
Third expression is already the remaining common expression $12\sqrt{6}$.
So: $147\sqrt{6} + 72\sqrt{6} - 12\sqrt{6}$ = $207\sqrt{6}$
Where did I go wrong?
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$$6\sqrt{24} + 7\sqrt{54} - 12\sqrt{6}=6\cdot2\sqrt6+7\cdot3\sqrt6-12\sqrt6=21\sqrt6.$$
I used the following law.
$$a(bc)=(ab)c.$$
For example,
$$6\cdot2\sqrt6=(6\cdot2)\sqrt6=12\sqrt6.$$
|
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"url": "https://math.stackexchange.com/questions/3131572",
"timestamp": "2023-03-29T00:00:00",
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|
Inductive proof of $D_N=-H_n\cdot R_N \cdot H_n $ (Grover Iterator) I am currently working on an inductive proof $D_N=-H_n\cdot R_N \cdot H_n $ (H is the Hadamard matrix for n Bits), the induction assumption (base case) and the induction condition have been done by me, but I can not go further in the Inductive step.
First, there are a few key details to mention:
$$D_N = \begin{pmatrix}-1+\frac{2}{N}&\frac{2}{N}&...&\frac{2}{N}\\\frac{2}{N}&-1+\frac{2}{N}&...&\frac{2}{N}\\\vdots&\vdots&\ddots&\vdots\\\frac{2}{N}&\frac{2}{N}&...&-1+\frac{2}{N}\end{pmatrix}$$
In the proof there is to shown that: $D_N=-H_n\cdot R_N \cdot H_n, \quad R_N=\begin{pmatrix}-1&0&...&0\\0&1&\ddots&\vdots\\\cdots&\ddots&\ddots&0\\0&...&0&1\end{pmatrix}$
These are my ideas:
Induction assumption (base case): $N=2, N=2^n$
$$D_2 = -H\cdot R_2 \cdot H = -\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\\1&-1\end{pmatrix}\cdot \begin{pmatrix}-1&0\\0&1\end{pmatrix}\cdot\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\\1&-1\end{pmatrix} = \begin{pmatrix}0&1\\1&0\end{pmatrix}$$
Thus:
$$D_2=\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}0&1\\1&0\end{pmatrix}$$
Induction condition:
For any but firm $n$ the statement applies.
Inductive step: n+1
$$D_{N+1}=-H_{n+1}R_{N+1}H_{n+1}$$
$H_{n+1}$ is equal to $H_1\otimes H_n=\frac{1}{\sqrt{2}}\begin{pmatrix}H_n&H_n\\H_n&-H_n\end{pmatrix}$
then you would have first:
$$D_{N+1}=-\frac{1}{\sqrt{2}}\begin{pmatrix}H_n&H_n\\H_n&-H_n\end{pmatrix}\cdot R_{N+1}\cdot \frac{1}{\sqrt{2}}\begin{pmatrix}H_n&H_n\\H_n&-H_n\end{pmatrix}$$
Now, however, the question remains open, how to express or rewrite the $D_{N+1}$ and the $R_{N+1}$.
For me, in the induction step already raises the question of whether $N+1$ and $n+1$ is right at all.
I'm stuck with this step. I am grateful for the answers and hope that the question is clear and understandable.
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You have to notice that $N=2^{n-1}$ and you are doing induction on $n$. The induction step is that
$$-H_nR_{2^{n-1}}H_n=D_{2^{n-1}}.$$
Also notice that $R_2^{n}$ can be writen as a block matrix in the following way:
$$ R_{2^n}=\left(\begin{array}{c|c}
R_{2^{n-1}} & 0\\
\hline
0 & I_{2^{n-1}}
\end{array}\right). $$
And so,
$$-H_{n+1}R_{2^n}H_{n+1}=-\frac{1}{2}
\left(\begin{array}{c|c}
H_n & H_n\\
\hline
H_n & -H_n
\end{array}\right)
\left(\begin{array}{c|c}
R_{2^{n-1}} & 0\\
\hline
0 & I_{2^{n-1}}
\end{array}\right)
\left(\begin{array}{c|c}
H_n & H_n\\
\hline
H_n & -H_n
\end{array}\right),
$$
Using the induction step, this product equals
$$-\frac{1}{2}\left(\begin{array}{c|c}
-D_{2^{n-1}} + H_n^2 & -D_{2^{n-1}} - H_n^2\\
\hline
-D_{2^{n-1}} - H_n^2 & -D_{2^{n-1}} + H_n^2
\end{array}\right) =
-\frac{1}{2}\left(\begin{array}{c|c}
-D_{2^{n-1}} + I & -D_{2^{n-1}} - I\\
\hline
-D_{2^{n-1}} - I & -D_{2^{n-1}} + I
\end{array}\right) = D_{2^n}, $$
where the equality $H_n^2=I$ comes from the fact that $H_n$ is symmetric. The last equality is checked easily by computing the diagonal and off-diagonal entries of the blocks.
|
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"timestamp": "2023-03-29T00:00:00",
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|
If nine coins are tossed, what is the probability that the number of heads is even? If nine coins are tossed, what is the probability that the number of heads is even?
So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.
We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$
$n = 9, k = 0$
$$\binom{9}{0}\bigg(\frac{1}{2}\bigg)^0\bigg(\frac{1}{2}\bigg)^{9}$$
$n = 9, k = 2$
$$\binom{9}{2}\bigg(\frac{1}{2}\bigg)^2\bigg(\frac{1}{2}\bigg)^{7}$$
$n = 9, k = 4$
$$\binom{9}{4}\bigg(\frac{1}{2}\bigg)^4\bigg(\frac{1}{2}\bigg)^{5}$$
$n = 9, k = 6$
$$\binom{9}{6}\bigg(\frac{1}{2}\bigg)^6\bigg(\frac{1}{2}\bigg)^{3}$$
$n = 9, k = 8$
$$\binom{9}{8}\bigg(\frac{1}{2}\bigg)^8\bigg(\frac{1}{2}\bigg)^{1}$$
Add all of these up:
$$=.64$$ so there's a 64% chance of probability?
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If there are an even number of heads then there must be an odd number of tails. But heads and tails are symmetrical, so the probability must be $1/2$.
|
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"url": "https://math.stackexchange.com/questions/3134991",
"timestamp": "2023-03-29T00:00:00",
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|
Integral involving incomplete beta function I have the following integral,
$$\int_{0}^1x^{a-1}(1-x)^{b-1}B_x(c,d)dx$$
where $B_x(c,d) = \int_{0}^xt^{c-1}(1-t)^{d-1}dt$ is the incomplete beta function, and $a,b,c,d>0$.
Question: Does this have a closed form?
My attempt:
*
*First, playing around in Wolfram Alpha makes me think that there may be a (simple?) closed form: example 1 and example 2. The second example can also be written as $\int x^2(1-x)B_x(12,2)dx = \frac{1}{4}(B_x(16,2) - x^4B_x(12,2))+\frac{1}{3}(x^3B_x(12,2)-B_x(15,2))$.
*It seems there is a reduction when $a=c$, $b=d$ as
$$\int_0^1 x^{a-1}(1-x)^{b-1}B_x(a,b)dx=\frac{1}{2}\left(\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\right)^2$$
where $\Gamma(a)$ is the gamma function. However, for the case when $a\neq c$, $b\neq d$, things are not quite as clear for me.
*I found a similar question here but it has not been answered.
*I also found this but I'm not sure if it is useful.
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Using the hypergeometric representation of the incomplete Beta function
\begin{equation}
B_x\left( c,d \right)=\frac{x^c}{c}{}_2F_1\left( c,1-d;1+c;x \right)
\end{equation}
the integral can be written as
\begin{align}
I\left( a,b,c,d \right)&=\int_{0}^1x^{a-1}(1-x)^{b-1}B_x(c,d)\,dx\\
&=\frac{1}{c}\int_{0}^1x^{a+c-1}(1-x)^{b-1}{}_2F_1\left( c,1-d;1+c;x \right)\,dx\\
&=\frac{1}{c}\sum_{k=0}^\infty \frac{(c)_k(1-d)_k}{(1+c)_kk!} \int_0^1x^{a+c+k-1}(1-x)^{b-1}\,dx\\
&=\frac{1}{c}\sum_{k=0}^\infty \frac{(c)_k(1-d)_k}{(1+c)_kk!} \frac{\Gamma(b)\Gamma(a+c+k)}{\Gamma(a+b+c+k)}\\
&=\frac{1}{c}\frac{\Gamma(b)\Gamma(a+c)}{\Gamma(a+b+c)}\sum_{k=0}^\infty \frac{(c)_k(1-d)_k}{(1+c)_kk!} \frac{(a+c)_k}{(a+b+c)_k}\\
&=\frac{1}{c}\frac{\Gamma(b)\Gamma(a+c)}{\Gamma(a+b+c)}\,{}_3F_2\left( 1-d,a+c,c;1+c,a+b+c ;1\right)
\end{align}
Using this identity for the generalized hypergeometric function:
\begin{equation}
{}_3F_2\left( a_1,a_2,a_3;b_1,b_2;1 \right)=\frac{\Gamma(b_1)\Gamma(b_1+b_2-a_1-a_2-a_3)}{\Gamma(b_1-a_1)\Gamma(b_1+b_2-a_2-a_3)}{}_3F_2\left( a_1,b_2-a_2,b_2-a_3;b_2,b_1+b_2-a_2-a_3;1 \right)
\end{equation}
Here we choose $a_1=1-d,a_2=a+c,a_3=c,b_1=a+b+c,b_2=1+c$ to obtain
\begin{align}
I\left( a,b,c,d \right)&=\frac{1}{c}\frac{\Gamma(b)\Gamma(a+c)}{\Gamma(a+b+c)}\frac{\Gamma(a+b+c)\Gamma(b+d)}{\Gamma(a+b+c+d-1)\Gamma(b+1)}\,{}_3F_2\left( 1-d,1-a,1;1+c,1+b ;1\right)\\
&=\frac{1}{bc}\frac{\Gamma(a+c)\Gamma(b+d)}{\Gamma(a+b+c+d-1)}\,{}_3F_2\left( 1,1-a,1-d;1+b,1+c ;1\right)\\
&=\frac{a+b+c+d-1}{bc}B(a+c,b+d)\,{}_3F_2\left( 1,1-a,1-d;1+b,1+c ;1\right)
\end{align}
which gives simple results if $a$ or $d$ are positive integer, as the hypergeometric series is finite: as $1-a\le0$ or $1-b\le0$, in the series definition of the hypergeometric function, there are $\operatorname{min}(a,b)$ terms (numerator of the coefficients cancel after that). One can also check the given result when $a=c$ and $b=d$.
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|
Compute $\lim\limits_{x\to \infty} \left(\sqrt{\left(x+\frac{1}{\sqrt x}\right) ^3} - \sqrt{x^3} \right) $ Compute $\lim\limits_{x\to \infty} \left(\sqrt{\left(x+\frac{1}{\sqrt x}\right) ^3} - \sqrt{x^3} \right) $.
I tried multiplying by the conjugate and I get to an ugly expression. What should I do?
|
Hello Taylor my old friend...
$\begin{array}\\
\sqrt{\left(x+\frac{1}{\sqrt x}\right) ^3} - \sqrt{x^3}
&=\left(x+x^{-1/2}\right) ^{3/2} - x^{3/2}\\
&=x^{3/2}\left(\left(1+x^{-3/2}\right) ^{3/2} - 1\right)\\
&=x^{3/2}\left(\left(1+\frac32x^{-3/2}+O(x^{-3})\right) - 1\right)\\
&=x^{3/2}\left(\frac32x^{-3/2}+O(x^{-3})\right)\\
&=\frac32+O(x^{-3/2})\\
\end{array}
$
Actually, of course,
it's the generalized binomial theorem,
but that doesn't scan as well.
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the general formula of $a_{n+1}=\frac{a_n^2+4}{a_{n-1}}$ with $a_1=1$ and $a_2=5$
Find the general formula of $a_{n+1}=\dfrac{a_n^2+4}{a_{n-1}}$ with $a_1=1$, $a_2=5$.
I have tried to write the recursion as a product, make summations, tried to look at patterns but its value grows very fast: $1,5,29,169,985,5741$… So I ran out of ideas.
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FYI,
I noticed that
$\dfrac{a_{n+1}+a_{n-1}}{a_n}=6 \hspace{1cm}\forall n \gt1\tag1$
(unfortunately I could not prove it so far).
Using this recursion formula easy to determine the general formula of $a_{n+1}$.
Searching the general formula in the form
$a_n=c_1 r_1^n+c_2 r_2^n\tag2$
where $r_1, r_2$ are the roots of the following equation:
$r^2-6r+1=0\tag3$
So $\hspace{1cm}$ $r_1=3+2\sqrt2$ $\hspace{0,5cm}$ and $\hspace{0.5cm}$$r_2=3-2\sqrt2. $
$c_1, c_2$ can be determined from the initial values:
$a_1=1=c_1 (3+2\sqrt2)+c_2 (3-2\sqrt2)\tag4$
$a_2=5=c_1 (3+2\sqrt2)^2+c_2 (3-2\sqrt2)^2\tag5$
From (4) and (5) we get:
$c_1=\frac {1}{4+2\sqrt2}$ and $c_2=\frac {3+2\sqrt2}{4+2\sqrt2}$
Finally, based on (2):
$a_{n+1}=\Big(\frac {3+2\sqrt2}{4+2\sqrt2}\Big)^{n+1}+ \Big(\frac {3-2\sqrt2}{4+2\sqrt2}\Big)^{n}\tag5$
I have checked the first 15 items by Excel; the results are matches.
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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.