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General Inequality Problem. The question goes like this :
Let $a,b,c,d$ be positive reals and given that $a+b+c+d=1$.
Prove that: $$6(a^3+b^3+c^3+d^3) \geqslant a^2+b^2+c^2+d^2 + \frac{1}{8}$$
My approach goes like this:
I wrote $a^3+b^3$ as $(a+b)(a^2+b^2-ab)$
Similarly $c^3+d^3$ as $(c+d)(c^2+d^2-cd)$
Although I am not sure if its the correct method, I tried reducing the powers.
Then, I got:
$6(a+b)(a^2+b^2-ab) -a^2-b^2 \geqslant c^2+d^2 - 6(c+d)(c^2+d^2-cd) +\frac{1}{8} $
I tried substituting $c,d$ in terms of $a,b$ but the calculation is lengthier than I expected. Please check if my reasoning is correct, and help me solve this in a shorter method. Thanks!
|
By Chebyshev's inequality,
$\frac{a^2+b^2+c^2+d^2}{4} \geq \frac{a+b+c+d}{4} \cdot \frac{a+b+c+d}{4}$
$\implies a^2+b^2+c^2+d^2 \geq \frac{1}{4}$
Similarly,
$\frac{a^3+b^3+c^3+d^3}{4} \geq \frac{a+b+c+d}{4} \cdot \frac{a^2+b^2+c^2+d^2}{4}$
$\implies 4(a^3+b^3+c^3+d^3) \geq a^2+b^2+c^2+d^2$
$6(a^3+b^3+c^3+d^3) \geq a^2+b^2+c^2+d^2+ \frac{a^2+b^2+c^2+d^2}{2} \geq a^2+b^2+c^2+d^2+ \frac{1}{8}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is it possible to evaluate $\int\frac{dx}{\sqrt{x^2+4}}$ without using trigonometric substitution? The normal approach to evaluate $\int\frac{dx}{\sqrt{x^2+4}}$ is using the substitution $x=2\tan\theta$. But I wonder is is possible to do it without using trigonometric substitution? I tried this approach:
$$\int\frac{dx}{\sqrt{x^2+4}}=\int\frac{xdx}{\sqrt{x^2(x^2+4)}}=\frac12\int\frac{du}{\sqrt{u(u+4)}}\quad\text{where}\quad u=x^2$$
But I can't see a way to evaluate final integral without completing square and using the substitution $u+2=2\sec t$ (which is a trigonometric substitution!)
|
$$
\begin{aligned}
\int \frac{d x}{\sqrt{x^{2}+4}} &\stackrel{y=\frac{1}{x} }{=}-\int \frac{d y}{y \sqrt{1+4 y^{2}}} \\
&=-\frac{1}{4} \int \frac{d (\sqrt{1+4 y^{2}})}{y^{2}} \\
&=-\int \frac{d (\sqrt{1+4 y^{2}})}{\left(\sqrt{1+4 y^{2}}\right)^{2}-1} \\
&=\frac{1}{2} \ln \left|\frac{\sqrt{1+4 y^{2}}+1}{\sqrt{1+4 y^{2}}-1}\right| \\
&=\frac{1}{2} \ln \left|\frac{\sqrt{x^{2}+4}+x}{\sqrt{x^{2}+4}-x}\right|+c \\
&=\ln \left|\sqrt{x^{2}+4}+x\right|+C
\end{aligned}
$$
|
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|
Limiting Sum including Stirling numbers I would like to simplify the following summation including $S(m,n)$, Stirling numbers of the Second kind.
$\begin{gather*}\\
&&
\frac{1}{n^{n+1}}\sum_{k=1}^{n-1}\binom{n}{n-k}S\left( n,n-k\right)
\left( n-k\right) !k\left( 1-\left( \frac{n-2}{n-1}\right) ^{k}\right)
&&
\end{gather*}$
I conjecture (in fact, computations show) that this sum approaches to $\frac{1}{e}(1-e^{-\frac{1}{e}})$ as $n$ approaches to $\infty$ . However, I do not have a proof.
It would be great if this can be simplified for any $n$ (I am not sure whether this is possible) or at least show the limiting result.
|
We get for the sum term
$$\sum_{k=1}^{n-1} {n\choose k} {n\brace n-k} (n-k)!
\times k x^k
\\ = n \sum_{k=1}^n
{n-1\choose k-1} {n\brace n-k} (n-k)! \times x^k
\\ = n x \sum_{k=0}^{n-1}
{n-1\choose k} {n\brace n-1-k} (n-1-k)! \times x^k
\\ = n x \times n! [z^n] \sum_{k=0}^{n-1}
{n-1\choose k} (\exp(z)-1)^{n-1-k} x^k
\\ = n x \times n! [z^n] (\exp(z)-1+x)^{n-1}.$$
We find with $x=1$
$$n \times n! \times [z^n] \exp((n-1)z)
= n \times (n-1)^n.$$
We obtain for $x=(n-2)/(n-1)$
$$n \times \frac{n-2}{n-1} \times
n! [z^n] (\exp(z)-1/(n-1))^{n-1}.$$
Now the coefficient extractor is
$$\frac{1}{2\pi i}
\int_{|z|=1} \frac{\exp((n-1)z)}{z^{n+1}}
(1-\exp(-z)/(n-1))^{n-1} \; dz.$$
With $n$ large this becomes
$$\frac{1}{2\pi i}
\int_{|z|=1} \frac{\exp((n-1)z)}{z^{n+1}}
\exp(-\exp(-z)) \; dz.$$
We now apply the Saddle point Algorithm for Cauchy Coefficient Integrals
from pages 548 and 553 of Analytic Combinatorics by Flajolet and
Sedgewick. (Theorem VIII.3) Our function $f(z)$ is
$$f(z) = (n-1)z-(n+1)\log z -\exp(-z)$$
and
$$f'(z) = (n-1) - (n+1)\frac{1}{z} + \exp(-z)$$
as well as
$$f''(z) = (n+1)\frac{1}{z^2} - \exp(-z).$$
Now let $\zeta_n$ be the positive real root of the saddle point equation
$f'(\zeta_n) = 0$ so that
$$\zeta_n (n-1) - (n+1) + \zeta_n \exp(-\zeta_n) = 0$$
and
$$\zeta_n = \frac{n+1}{n-1+\exp(-\zeta_n)}.$$
This implies
$$\frac{n+1}{n} \lt \zeta_n \lt \frac{n+1}{n-1}$$
so that with $n$ large $\zeta_n \approx 1.$ We get
for the asymptotics from equation 19 on the cited page
$$\frac{\exp(f(\zeta_n))}{\sqrt{2\pi f''(\zeta_n)}}$$
Now $f''(\zeta_n) = (n+1)/\zeta_n^2+(n-1)-(n+1)/\zeta_n$
and this evaluates to
$$\frac{\exp(n-1)\exp(-\exp(-1))}{\sqrt{2\pi (n-1)}}$$
Collecting everything from the two pieces we find
for the first piece
$$\frac{1}{n^{n+1}} \times n\times (n-1)^n
= \left(1-\frac{1}{n}\right)^n \sim \exp(-1)$$
and for the second piece
$$\frac{1}{n^{n+1}} \times n \frac{n-2}{n-1} n! \times
\exp(-1) \frac{\exp(n)\exp(-\exp(-1))}{\sqrt{2\pi (n-1)}}
\\ = \left(1-\frac{1}{n-1}\right)
\exp(-1) \exp(-\exp(-1))
\frac{n!}{n^n} \frac{\exp(n)}{\sqrt{2\pi (n-1)}}
\\ \sim \left(1-\frac{1}{n-1}\right)
\exp(-1) \exp(-\exp(-1))
\frac{n!}{n^n} \frac{\exp(n)}{\sqrt{2\pi n}}.$$
We apply Stirling's formula to conclude and get
$$\left(1-\frac{1}{n-1}\right)
\exp(-1) \exp(-\exp(-1))
\sim \exp(-1) \exp(-\exp(-1)).$$
We join the two pieces to obtain
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{e}\left(1-e^{-1/e}\right)}$$
as proposed.
|
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|
Error with Differential Forms I'm learning a little bit about differential forms in my Complex Analysis course, and I'm not sure if I'm making an error in what follows. I wonder if anyone could offer any insight into this. Thank you.
So, the question is to show that $\alpha = \frac{xdy - ydx}{x^2 + y^2}$ is closed. To show this, I need to show that $d\alpha = 0$, which I can show as follows (according to my understanding).
Let $f = \frac{1}{x^2 + y^2}, g = xdy - y dx$. Then $d \alpha = f \wedge dg + df \wedge g$.
So, calculate $df = \frac{-1}{(x^2+y^2)^2}2x dx + \frac{-1}{(x^2 + y^2)^2}2y dy = \frac{-2}{(x^2+y^2)^2} (x dx + y dy)$ and $dg = 1 dx \wedge dy - 1 dy \wedge dx = 2 dx \wedge dy$.
Then, substituting these in the right places, we get
\begin{align*}
d \alpha &= \left(\frac{1}{x^2 + y^2}\right) \wedge (2 dx \wedge dy) + \left(\frac{-2}{(x^2+y^2)^2} (x dx + y dy)\right) \wedge (xdy - y dx)\\
&= \frac{2}{x^2+y^2} dx \wedge dy - \frac{2}{(x^2+y^2)^2}(x^2 dx \wedge dy - y^2 dy \wedge dx)\\
&= \frac{2}{x^2 + y^2} dx \wedge dy - \frac{2}{x^2 + y^2} dx \wedge dy\\
&= 0.
\end{align*}
This seems to work.
However, when I interchange $f$ and $g$, I seem to run into a problem, and I'm not sure why. Here's what happens. I keep $d \alpha = f \wedge dg + df \wedge g$. However, now I let $f = xdy - ydx$ and $g = \frac{1}{x^2 + y^2}$.
In this case, $df$ and $dg$ are no different (except of course for interchanging them), so we have $df = 2 dx \wedge dy$ and $dg = \frac{-2}{(x^2+y^2)^2} (x dx + y dy)$.
Now, substituting in, we get
\begin{align*}
d \alpha &= (xdy - ydx) \wedge \left(\frac{-2}{(x^2+y^2)^2} (x dx + y dy)\right) + (2 dx \wedge dy) \wedge \left(\frac{1}{x^2 + y^2}\right)\\
&= \frac{-2}{(x^2+y^2)^2} (xdy - ydx) \wedge (xdx + ydy) + \frac{2}{x^2 + y^2} dx \wedge dy\\
&= \frac{-2}{(x^2+y^2)^2}(x^2 dy \wedge dx - y^2 dx \wedge dy)+ \frac{2}{x^2 + y^2} dx \wedge dy\\
&= \frac{-2}{(x^2+y^2)^2}(-x^2 dx \wedge dy - y^2 dx \wedge dy)+ \frac{2}{x^2 + y^2} dx \wedge dy\\
&= \frac{-2}{(x^2+y^2)^2}((-x^2 -y^2)dx \wedge dy)+ \frac{2}{x^2 + y^2} dx \wedge dy\\
&= \frac{-2}{(x^2+y^2)^2}(-(x^2 +y^2)dx \wedge dy)+ \frac{2}{x^2 + y^2} dx \wedge dy\\
&= \frac{2}{x^2+y^2} dx \wedge dy + \frac{2}{x^2+y^2} dx \wedge dy\\
&= \frac{4}{x^2 + y^2}\\
&\neq 0.
\end{align*}
I've done this several times by hand and keep having this issue, so I assume the issue is not a simple calculation error on my part but a real conceptual misunderstanding about how to do these calculations.
It seems like just interchanging which function is $f$ and which is $g$ might change the sign I wind up with, but I don't see how it could change the result from 0 (which is what it's supposed to be) to something nonzero.
Any insights would be appreciated! Please remember that I have very little experience with differential forms.
|
If $\phi$ is a $p$-form and $\psi$ is a $q$-form, then $d(\phi\wedge\psi) = d\phi\wedge\psi + (-1)^p\phi\wedge d\psi$. In the second case, $f$ is a one-form and $g$ is a $0$-form, so $d\alpha = d(f\wedge g) = df\wedge g + (-1)^1f\wedge dg = df\wedge g \color{red}{-} f\wedge dg$.
|
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|
Angle bisected in a parallelogram
$ABCD$ is a parallelogram and $ID =JB$.
the point $P$ is the intersection of $DJ$ and $IB$.
Prove that the angle $X=Y$
$a,b,α,β$ are angles.
What I realized:
$a=b$.
and if we sum up these angles.
$$180=180-(x+α)+180-(y+β)+a $$
$$\iff 180=x+y+α+β-a \iff 360+2a=2(x+y)+2(α+β)$$
Please Help.
|
It is straightforward by sine law,
$\angle PID = 180^0 - \beta, \angle PJB = 180^0 - \alpha, \angle DPI = \angle BPJ = \gamma$
So in $\triangle PID, \frac{\sin \beta}{PD} = \frac{\sin \gamma}{ID}$
Similarly in $\triangle PBJ, \frac{\sin \alpha}{PB} = \frac{\sin \gamma}{BJ}$
As $BJ = ID$, $ PD \sin \alpha = PB \sin \beta$
Now in $\triangle PCD, \frac{\sin X}{PD} = \frac{\sin \alpha}{PC}$
In $\triangle PBC, \frac{\sin Y}{PB} = \frac{\sin \beta}{PC}$
So, $\frac{\sin X}{\sin Y} = \frac{PD \sin \alpha}{PB \sin \beta} = 1$
$\therefore \angle X = \angle Y$
|
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|
Inequality Induction Proof, how should I proceed? I have been trying to prove inequalities using induction to no avail.
For example,
Prove the following:
$\frac{1}{2^{1}}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+...+\frac{n}{2^{n}}<2$
Base Case:
$\frac{1}{2^{1}} = 2$
$\frac{1}{2}<2\;$ Which is true.
We assume, for n=k, that
$\frac{1}{2^{1}}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+...+\frac{k}{2^{k}}<2\;\;\;$ is true.
As induction step with n=k+1, we need to prove that
$\frac{1}{2^{1}}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+...+\frac{k+1}{2^{k+1}}<2\;\;\;$
How should I proceed from here?
|
You may first prove the following equality:
Prove that $\frac 1 {2^1} + \frac 2 {2^2} + \dots + \frac k {2^k} = 2 - \frac {k + 2} {2^k}$.
This can be proved by induction on $k$ (easy exercise).
Once it is proved, the original inequality follows.
|
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|
Show $\log\left(\frac{x-y}{x}\right)-2\sqrt{\frac{x-y}{x}}<\log\left(\frac{x+y}{x}\right)-2\sqrt{\frac{x+y}{x}}$ if $x\geq 5$ and $1\leq y\leq x-2$ Assume that all logarithms are natural.
Let $x$ and $y$ be integers that satisfy $x \geq 5$ and $1 \leq y \leq x-2$. I am trying to show that $$\log\left( \frac{x-y}{x}\right)-2\sqrt{\frac{x-y}{x}} < \log\left( \frac{x+y}{x}\right)-2\sqrt{\frac{x+y}{x}}.$$
This is equivalent to showing that $$\log\left( \frac{x-y}{x+y}\right)<2\left(\sqrt{\frac{x-y}{x}}-\sqrt{\frac{x+y}{x}}\right).$$
My first inclination was to apply the well known inequality, $\log(z)\leq 2(\sqrt{z}-1)$ if $z>0$, to the left side by letting $z=\frac{x-y}{x+y}$, but this did not help me.
I also tried double induction on $(x,y)$ (since $x$ and $y$ are integers) but I could not finish it.
I appreciate any help, thank you.
|
Here's a proof
by comparing the power series.
Eliding some computation,
$\log\left( \frac{1+t}{1-t}\right)
=2\sum_{n=1}^{\infty} \dfrac{t^{2n-1}}{2n-1}
$.
$\sqrt{1+t}-\sqrt{1-t}
=2\sum_{n=1}^{\infty} \binom{1/2}{2n-1}t^{2n-1}
=2\sum_{n=1}^{\infty} \dfrac{1}{2^{4n-3}(2n-1)}\binom{4n-4}{2n-2}t^{2n-1}
$
So we want
$\dfrac{1}{2^{4n-3}(2n-1)}\binom{4n-4}{2n-2}
\lt \dfrac{1}{2n-1}
$
or
$\dfrac{1}{2^{4n-3}}\binom{4n-4}{2n-2}
\lt 1
$
or,
with $2n-2=m$,
$\dfrac{1}{2^{2m+1}}\binom{2m}{m}
\lt 1
$.
But
$(1+1)^{2m}
=2^{2m}
$
and,
for $m \ge 1$,
$(1+1)^{2m}
=\sum_{k=0}^{2m}\binom{2m}{k}
\gt \binom{2m}{m}
$
so
$\dfrac{1}{2^{2m+1}}\binom{2m}{m}
\lt \dfrac12
$.
|
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|
How to proceed $\int \frac{dx}{(x-2) \left(1+\sqrt{7 x-10-x^2}\right)} $
$$I=\int \frac{dx}{(x-2) \left(1+\sqrt{7 x-10-x^2}\right)} $$
I have taken the second term of denominator (in square root) as $(x - 2)t$. But cannot go further. Please suggest how to proceed or any alternate method to solve it.
|
We are given $I=\int \frac{dx}{(x-2) \left(1+\sqrt{7 x-10-x^2}\right)}$
Here we have $7x-10-x^2=(x-2)(5-x)$,
therefore put $\sqrt{7x-10-x^2} =(x-2)t$
on solving and factoring,
we get $x=\frac{\left(2t^2+5\right)}{\left(t^2+1\right)}\ $ therefore
$dx=\frac{-6t}{(t^2+1)^2}$ and $x-2=$ $\frac{(2t^2+5)}{(t^2+1)}-2 =$$\frac{3}{(t^2+1)}$
and $1+\sqrt{7x-10-x^2} = 1+(x-2)t =1+\frac{3t}{t^2+1}$
$=\frac{t^2+3t+1}{t^2+1}$
hence we have $I=\int\frac{t^2+1}{3}.\frac{t^2+1}{t^2+3t+1}.\frac{-6t}{(t^2+1)^2}dt$
$I=-\int\frac{2t}{t^2+3t+1}=-\int\frac{2t+3-3}{t^2+3t+1}dt$
$I=-\int\frac{2t+3}{t^2+3t+1}dt+3\int\frac{dt}{(t+\frac32)^2 -\frac54 }$
$I=ln(t^2+3t+1) -\frac{3}{\sqrt{5}}ln\frac{(2t+3-\sqrt{5})}{(2t+3+\sqrt{5})}$ + C
where $t=\sqrt{\frac{5-x}{x-2}}$.
|
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|
Determining when $(\sin(\theta) - \cos(\theta))(2+\sin(\theta) \cos(\theta)) \leq 2$ Question:
$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \leq 2 \qquad \text{ LHS} \\ $
Answer key:
$\implies \frac{1}{2}(\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \qquad \text{ RHS} $
I verified RHS = LHS. However, to get to the RHS part, they have factored it somehow and I can't figure out an intuitive way to do so. If anybody has a more intuitive solution (does not have to be the same as what is given here), please provide it. From this step onwards we can easily solve the question because:
$\implies (\sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0\\ $
$ \implies (\sin(\theta)-\cos(\theta)-1) \leq0 \because (3 + \sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)) >0 \text{ }\forall \theta \in \mathbb{R} \\ $
Doubt:
I have no idea how the went from the LHS part to the RHS part. Is there an intuitive way to solve $(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \leq 2$? In this solution they have simply said:
$$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) -2 \leq 0 \\ \implies
\frac{1}{2}(\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \\$$
Please provide a logical way to solve it. Thanks in advance.
I'm looking for a no calculator solution where each step is motivated. And prove one factor is always positive or negative and use the other factor to find the region would be appreciated. Although, if there is another method, (as long as its intuitive) that works too
|
$$(\sin(\theta) - \cos(\theta))(2+\sin(\theta) \cos(\theta)) \leq 2$$
Let $x = \cos \theta$ and $y = \sin \theta$. Then we get
\begin{align}
(y - x)(2+xy) &\leq 2 \\
x^2 + y^2 &= 1
\end{align}
We can now argue
\begin{align}
-(y-x)^2 &= -1 + 2xy \\
5 - (y-x)^2 &= 2(2 + xy) \\
\hline
(y-x)(5 - (y-x)^2) &\le 4
\end{align}
Let $z = y-x$ and we find
\begin{align}
z(5 - z^2) &\le 4 \\
z^3 - 5z + 4 \ge 0 \\
(z-1)(z^2+z-4) \ge 0 \\
\hline
z \ge \dfrac{\sqrt{17}-1}{2} \\
-\frac{\sqrt{17}+1}{2} \le z \le 1
\end{align}
The first answer is superfluous. The second needs to be modified to
$$-\sqrt{2} \le z \le 1$$
$$-\sqrt{2} \le \sin \theta - \cos \theta \le 1$$
$$-\sqrt{2} \le \sqrt 2 \sin\left(x - \frac{\pi}{4} \right) \le 1$$
$$-1 \le \sin\left(x - \frac{\pi}{4} \right) \le \frac{1}{\sqrt 2}$$
$$\theta \in \left[(2n-1)\pi, \frac{\pi}{2}(4n+1) \right] \qquad \forall n \in \Bbb Z$$
|
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|
Equality involving cosine and binomial coefficients - check if true for every $n$ While doing some computations with complex numbers I think I proved that
$$(\cos{ \frac{n \cdot \pi}{4} )\cdot 2^{n/2}} = \binom{n}{0} - \binom{n}{2} + \binom{n}{4} - \binom{n}{6} + ... $$
How? Well, these are two different forms of
$$\frac{1}{2} \cdot ((1+i)^n + (1-i)^n)$$
But because of the late hour, I am not sure if I didn't make some mistake.
Is this equality really true for every natural n?
Also, I wonder how can one prove it without using complex numbers. I mean is these some sort of elementary (high-school) proof?
|
Without complex numbers, by induction, assume that for some integer $n\ge 0$,
$$\begin{align*}
2^{n/2}\cos\frac{n\pi}{4} = \binom{n}{0}-\binom n2+\binom{n}{4}-\binom n6+\cdots\\
2^{n/2}\sin\frac{n\pi}{4} = \binom{n}{1}-\binom n3+\binom{n}{5}-\binom n7+\cdots\\
\end{align*}$$
Then by compound angle formulae,
$$\begin{align*}
\cos\frac{(n+1)\pi}{4} &= \cos\frac{n\pi}{4}\cos\frac\pi 4 - \sin\frac{n\pi}{4}\sin\frac\pi 4\\
&= \frac1{\sqrt2} \left(\cos\frac{n\pi}{4} - \sin\frac{n\pi}{4}\right)\\
2^{(n+1)/2}\cos\frac{(n+1)\pi}{4} &= 2^{n/2} \left(\cos\frac{n\pi}{4} - \sin\frac{n\pi}{4}\right)\\
&= \left[\binom{n}{0}-\binom n2+\cdots\right] - \left[\binom{n}{1}-\binom n3+\cdots\right]\\
&= \binom n0 + \left[-\binom n2+\binom n4 -\cdots\right] + \left[-\binom{n}{1}+\binom n3-\cdots\right]\\
&= \binom{n+1}0 - \binom{n+1}2 + \binom{n+1}4-\cdots
\end{align*}$$
Similarly
$$\begin{align*}
\sin\frac{(n+1)\pi}{4} &= \frac1{\sqrt2} \left(\cos\frac{n\pi}{4} + \sin\frac{n\pi}{4}\right)\\
2^{(n+1)/2}\sin\frac{(n+1)\pi}{4} &= \left[\binom{n}{0}-\binom n2+\cdots\right] + \left[\binom{n}{1}-\binom n3+\cdots\right]\\
&= \binom{n+1}{1}-\binom {n+1}3+\binom{n+1}{5}-\cdots
\end{align*}$$
What remains is to prove that the base case when $n=0$ is true.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Question about simplifying $(a+b+c)(ab+bc+ac)=abc$ To simplify this expression, according to math110's answer, we can write LHS as:
$$(a+b+c)(ab+bc+ac)=(a+b)(b+c)(a+c)+abc$$
then $abc$ cancels and we get: $(a+b)(b+c)(a+c)=0$
But my question is, how we can recognize that we have this equality? I have no clue about it. is there any way to quickly realize we can write $(a+b+c)(ab+bc+ac)$ as $(a+b)(b+c)(a+c)+abc$ ?
|
The original equation is
$$\frac1a + \frac1b+ \frac1c = \frac1{a+b+c}$$
We see that $a=-b, b=-c, c=-a$ are trivial solutions to the equation. Thus
$$abc(a+b+c)\left(\frac1a + \frac1b+ \frac1c - \frac1{a+b+c}\right)$$
must contain $(a+b)(b+c)(c+a)$ as a factor; indeed it is equal to $(a+b)(b+c)(c+a)$.
|
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|
The center of the circumcircle lies on a side of a triangle Consider a triangle $ABC$. Let the angle bisector of angle $A$ be $AP,P\in BC$. $BP=16,CP=20$ and the center of the circumcircle of $\triangle ABP$ lies on the segment $AC$. Find $AB$.
$$AB=\dfrac{144\sqrt5}{5}$$
By Triangle-Angle-Bisector Theorem $$\dfrac{BP}{PC}=\dfrac{AB}{AC}=\dfrac{16}{20}=\dfrac{4}{5}\\ \Rightarrow AB=4x, AC=5x.$$ The cosine rule on $ABC$ gives $$BC^2=AB^2+AC^2-2\cdot AB\cdot AC\cdot\cos\alpha \\ \iff 1296=41x^2-40x^2\cos\alpha,$$ where $\measuredangle A=\alpha.$ Is any of this helpful for the solution? Any help would be appreciated. Thank you in advance!
|
$\angle POC = \angle A$, so $\triangle COP \sim \triangle CAB$
Hence $\frac{R}{AB} = \frac{20}{36} = \frac{5}{9}$ ...(i)
$\angle APB = \frac{1}{2} \angle AOB = 90^0 - \angle A$
$AB = 2R \sin \angle APB = 2R \cos A$
So from (i), $\frac{R}{2R \cos A} = \frac{5}{9}$
$\cos A = \frac{9}{10} = 1 - 2 \sin^2{\frac{A}{2}} \implies \sin \frac{A}{2} = \frac{1}{2\sqrt5} $
$16 = 2R \sin \frac{A}{2} \implies R = 16\sqrt5 $
$AB = 2R \cos A = 32\sqrt5 \times \frac{9}{10} = \frac{144}{\sqrt5}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Trigonometric system with different coefficients I have a trigonometric system as below and I want to solve it for $x$.
\begin{gather}
A\cos(x)+B\cos(2x)+C\sin(2x) = D \\
-C\cos(2x)-A\sin(x)+B\sin(2x) = F
\end{gather}
Can anyone suggest a solution?
Regards.
|
Rewrite the two equations as
$$\sqrt{B^2+C^2}\cos(2x-a) = D-A\cos x $$
$$\sqrt{B^2+C^2}\sin(2x-a) = F+A\sin x $$
with $a = \tan^{-1} \frac CB$. Square and add the two equations
$$B^2+C^2= D^2+F^2 +A^2 +2A(F \sin x - D\cos x)\\
= D^2+F^2 +A^2 +2A\sqrt{D^2+F^2} \sin (x-b)$$
with $b= \tan^{-1} \frac DF$. Now, it should be straightforward.
|
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|
Finding area based on only the sides I came across this simple looking question:
The perimeter of a triangle is $42$ cm. One side of a triangle is $8$ cm longer than the smallest side and the third side is $1$ cm less than $3$ times the smallest side. Find the area of the triangle.
At first it seemed easy enough. The smallest side is $7$ (which can be derived from:$\frac{42-8+1}{5}$, where subtracting $8$ takes care of the long side and now we have $2$ units of the shortest side, then adding one takes care of the third side now we have $5$ units of the third side and dividing by $5$ we can get the length of the smallest side, $7$) and with that knowledge the length of the first and third side is $15$ and $20$ respectively.
That's when I got stuck. The question gives us no indication of the angles in the triangle. I thought we could use the Pythagorean theorem to see if the triangle is right-angled, but it's not. Is there a rule we can use to find the area? Yes, I'm familiar with trigonometry and $sin()$, $cos()$ and $tan()$ so you can use those in your answer.
If you don't mind can you give me the rule to finding the area, not the answer to the problem? Because I want to solve this question myself. Many thanks to whoever answer's this!
|
Here is a proof for Heron's formula, as requested.
We begin with the Cosine rule:
$$a^2=b^2+c^2-2bc\cos A\iff\cos A=\frac{b^2+c^2-a^2}{2bc}$$
Recall the identity $$\sin^2x+\cos^2x\equiv 1\iff\sin^2x\equiv1-\cos^2x$$
This means that $$\begin{align}
\sin^2A&=1-\cos^2A\\
&=1-\frac{(b^2+c^2-a^2)^2}{(2bc)^2}\\
&=\frac{(2bc)^2-(b^2+c^2-a^2)^2}{(2bc)^2}\\
&=\frac{(2bc)^2-(a^2-b^2-c^2)^2}{(2bc)^2}\\
&=\frac{(a^2-(b^2-2bc+c^2))((b^2+2bc+c^2)-a^2)}{(2bc)^2}\\
&=\frac{(a^2-(b-c)^2)((b+c)^2-a^2)}{(2bc)^2}\\
&=\frac{(a+b-c)(a-b+c)(a+b+c)(-a+b+c)}{(2bc)^2}
\end{align}$$
So that factors really nicely! (Just for clarification: I have used the formulae $(x+y)^2=x^2+2xy+y^2$ and $x^2-y^2=(x+y)(x-y)$ quite a few times in the manipulations above.)
Now consider the expression for the 'semi-perimeter' $s$:
$$s=\frac{a+b+c}{2}$$
This means that
$$s-b=\frac{a+b+c}{2}-\frac{2b}{2}=\frac{a-b+c}{2}$$
and so on for $a$ and $c$ as well. Thus, we have
$$\begin{align}
\sin^2A&=\frac{2s \times 2(s-a)\times 2(s-b)\times 2(s-c)}{(2bc)^2}\\
&=\frac{4^2s(s-a)(s-b)(s-c)}{(2bc)^2}
\end{align}$$
Now, in a triangle, the $\sin$ of any angle will be positive. This is because any angle $\theta$ in a triangle must satisfy $0<\theta<180^\circ$, and for $0<\theta<180^\circ$ we know that $\sin\theta$ is positive. Hence when we now square root both sides we can (and must) take the positive square root. So we have
$$\begin{align}
\sin A&=\frac{4}{2bc}\sqrt{s(s-a)(s-b)(s-c)}\\
&=\frac{2}{bc}\sqrt{s(s-a)(s-b)(s-c)}\\
\end{align}$$
But wait! We know that the area of a triangle, let's call it $S$, is equal to $\frac{1}{2}bc\sin A$!! This means that
$$\begin{align}
S&=\frac{1}{2}bc\sin A\\
&=\frac{1}{2}bc\times\frac{2}{bc}\sqrt{s(s-a)(s-b)(s-c)}\\
&=\sqrt{s(s-a)(s-b)(s-c)}
\end{align}$$
So we finally have our truly remarkable endpoint:
$$\mathbf{S=\sqrt{s(s-a)(s-b)(s-c)}}$$
as required!!
I hope you enjoyed that as much as I enjoyed writing it! :) If you have any questions please don't hesitate to ask.
|
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|
How many times does the line $y=\frac{1}{4} x$ cut the curve $y=\sin(x)+\cos(2x)$? This is a multiple choice question, where using calculators is not allowed. Candidates have, on average, two minutes to solve the problem.
PROBLEM:
How many times does the line $y=\frac{1}{4} x$ cut the curve $y=\sin(x)+\cos(2x)$?
CHOICES:
(A) $4$ (B) $5$ (C) $6$ (D) $7$ (E) $8$
What I have tried was not feasible:
*
*Using differentiation, I found that the curve is bounded between $y=\frac{9}{8}$ and $y=-2$.
*When $x<-8$, $\frac{1}{4}x<-2$ , and when $x>\frac{9}{2}$, $\frac{1}{4}x>\frac{9}{8}$. So, any solution to the equation $y=\sin(x)+\cos(2x)=\frac{1}{4}x$ must lie in the interval $(-8,\frac{9}{2})$.
I do not know how to proceed from these two points. Even I know, that takes a long time.
Also, I do not know if using the fact that the curve is periodic will really help.
By Desmos, I took a look to the graph, the line cuts the curve $7$ times. So the correct choice is $D$.
Any help on how to solve this would be appreciated. THANKS!
|
The function is periodic with period $2\pi$ and achieves local extrema
$$y' = \cos x - 2\sin 2x = 0 \implies \cos x (1-4\sin x) = 0$$
at
$$\pm\sqrt{1-\cos^2x}+2\cos^2x-1\Bigr|_{\cos x = 0} = 0,-2$$
and
$$\sin x + 1-2\sin^2x\Bigr|_{\sin x = \frac{1}{4}} = \frac{9}{8}$$
which means the only interval we can expect intersections is on $\left[-8,\frac{9}{2}\right]$. The function has four extrema per period and the interval contains
$$\frac{25}{4\pi} = \frac{8\pi+(25-8\pi)}{4\pi} = 2 + \epsilon$$
two whole periods by estimating since $\epsilon < \frac{1}{12}$. $2\cdot 4 = 8$ turn arounds means $7$ intersections between them.
|
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|
How to show that $\lim_{(x,y) \to (0,0)}\frac{x^{2}y^{2}}{\left|x\right|^{3}+\left|y\right|^{3}}=0$ How to show that $$\lim_{(x,y) \to (0,0)}\frac{x^{2}y^{2}}{\left|x\right|^{3}+\left|y\right|^{3}}=0$$
I know that
$$\left|\frac{x^{2}y^{2}}{\left|x\right|^{3}+\left|y\right|^{3}}\right|\le\frac{x^{2}y^{2}}{\left|x\right|^{3}}=\frac{y^{2}}{\left|x\right|}$$
But now we have a problem with $\left|x\right|$ and I cannot find a lower bound other than $0$.
Also$$\left|\frac{x^{2}y^{2}}{\left|x\right|^{3}+\left|y\right|^{3}}\right|\le\frac{x^{2}y^{2}}{\left|x^{3}+y^{3}\right|}=\frac{\left|x+y\right|}{\left|x^{2}-xy+y^{2}\right|}\le\frac{\left|x\right|+\left|y\right|}{\left|x^{2}-xy+y^{2}\right|}$$
But I'm not sure if that helps.
If we show that $\lim_{(x,y) \to (0,0)}\frac{y^{2}}{\left|x\right|}=0$ then we are done.
|
Easier: if $(x,y) = (r\cos\theta,r\sin\theta)$,
$$
\left|\frac{x^2 y^2}{|x|^3+|y|^3}\right|
= \frac{r^4\cos^2\theta\sin^2\theta}{r^3(|\cos\theta|^3 + |\sin\theta|^3)}\le\frac{r}{m} = \frac{\sqrt{x^2 + y^2}}{m},
$$
with
$$m = \min_{\theta\in[0,2\pi]}(|\cos\theta|^3 + |\sin\theta|^3) > 0.$$
(Why $> 0$)?
|
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|
Find the expectation and variance of X A pdf is defined as
\begin{equation}
f(x)=
\begin{cases}
C(x+\frac{3}{2}),\quad0<x<2\\
0,\quad\quad\quad\quad\text{otherwise}
\end{cases}
\end{equation}
*
*Find the value of C.
*Find the expectation and variance of X.
*Find the expectation of random variable $Z=\frac{X}{2X+3}$
What I tried:
*
*Finding C is straightforward as we just need to make sure that $\int^\infty_{-\infty}f(x)dx=1$.
\begin{align*}
1&=\int^2_0C(x+\frac{3}{2})dx\\
&=\left[\frac{C}{2}x^2+\frac{3}{2}Cx\right]^2_0\\
&=\left[\frac{C}{2}(2)^2+\frac{3}{2}C(2)\right]-0\\
&=5C\\
&C=\frac{1}{5}
\end{align*}
*Using $\mathbb{E}[X]=\int^\infty_{-\infty}xf(x)dx$,
\begin{align*}
\mathbb{E}[X]&=\int^2_0x\left(\frac{1}{5}\left(x+\frac{3}{2}\right)\right)dx\\
&=\left[\frac{1}{15}x^3+\frac{3}{20}x^2\right]^2_0\\
&=\left[\frac{1}{15}(2)^3+\frac{3}{20}(2)^2\right]^2_0-0\\
&=\frac{17}{15}
\end{align*}
Variance is $$\text{Var}[x]=\mathbb{E}[X^2]-\mathbb{E}[X]^2=\mathbb{E}[X(X-1)]-\mathbb{E}[X]-\mathbb{E}[x]^2$$
$$\mathbb{E}[X(X-1)]=\int^2_0x(x-1)(\frac{x}{5}+\frac{3}{10})dx=\frac{7}{15}$$
Then, Var$[X]=\frac{7}{15}-\frac{17}{15}-(\frac{17}{15})^2=-\frac{439}{225}$
My concern is that the variance is negative, which should not be the case. Can I get some pointers on where I went wrong? Thank you.
*This one is simple. I just need to find $\int^2_0\frac{x}{2x+3}f(x)dx$
|
points 1 is ok
the expectation is ok
As variance is concerned
$$E[X^2]=\int_0^2 x^2f(x)dx=\frac{8}{5}$$
thus
$$V[X]=\frac{8}{5}-\left(\frac{17}{15}\right)^2=\frac{71}{225}$$
point 3, easy but incorrect
$$E[Z]=\int_0^2 \frac{x}{2x+3}f(x)dx$$
|
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|
Is there a mistake to the answer given to this problem: Find the matrix of the reflection through the line y=−2x/3. This is the answer given:
I think this answer is not correct because the line given by y=-2x/3 makes an angle that is below the x axis, so the order is incorrect right? We need to start with a counterclockwise rotation first and end with the clockwise rotation, but here, we started with the clockwise rotation first.
|
Here the slope of the line = $m = tan \gamma = -\frac{2}{3}$.
$\therefore sin \gamma = \frac{m}{\sqrt{1+m^2}},\; cos \gamma = \frac{1}{\sqrt{1+m^2}}$.
The transformation matrix for reflection around the line is given by the product of 3 matrices (rotate by angle -$\gamma$ to make it horizontal, reflect w.r.t the horizontal line and rotate back by angle $\gamma$)
\begin{align}T&=T_{-\gamma}.R_0.T_{\gamma}\\
&=\begin{pmatrix}cos \gamma & -sin\gamma\\ sin\gamma & cos \gamma\end{pmatrix}
\begin{pmatrix}1&0\\ 0 & -1\end{pmatrix}
\begin{pmatrix}cos \gamma & sin\gamma\\ -sin\gamma & cos \gamma\end{pmatrix} \\
&=\begin{pmatrix}\frac{1}{\sqrt{1+m^2}} & -\frac{m}{\sqrt{1+m^2}}\\ \frac{m}{\sqrt{1+m^2}} & \frac{1}{\sqrt{1+m^2}}\end{pmatrix}
\begin{pmatrix}1&0\\ 0 & -1\end{pmatrix}
\begin{pmatrix}\frac{1}{\sqrt{1+m^2}} & \frac{m}{\sqrt{1+m^2}}\\ -\frac{m}{\sqrt{1+m^2}} & \frac{1}{\sqrt{1+m^2}}\end{pmatrix} \\
&=\begin{pmatrix}\frac{1}{\sqrt{1+m^2}} & -\frac{m}{\sqrt{1+m^2}}\\ \frac{m}{\sqrt{1+m^2}} & \frac{1}{\sqrt{1+m^2}}\end{pmatrix}
\begin{pmatrix}\frac{1}{\sqrt{1+m^2}} & \frac{m}{\sqrt{1+m^2}}\\ \frac{m}{\sqrt{1+m^2}} & -\frac{m}{\sqrt{1+m^2}}\end{pmatrix} \\
&=\frac{1}{1 + m^2}\begin{pmatrix}1-m^2&2m\\2m&m^2-1\end{pmatrix} \\
&=\begin{pmatrix}\frac{5}{13}&-\frac{12}{13}\\ -\frac{12}{13}&-\frac{5}{13}\end{pmatrix}
\end{align}
The following example in R shows how the black point (1,1) is reflected w.r.t. the life to new location (red point)
x <- seq(-10,10,0.01)
y <- -2*x/3
m <- -2/3
T <- matrix(c((1-m^2)/(1+m^2),2*m/(1+m^2),2*m/(1+m^2),(m^2-1)/(1+m^2)), nrow=2, byrow=T)
p <- c(1,2)
p1 <- T %*% p
plot(x,y, type='l')
points(p[1],p[2],pch=19)
points(p1[1],p1[2],pch=19, col='red')
|
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|
Prove that the least upper bound of the sequence $\left\{ \frac{2^n - 1}{2^{n-1}} \right\}$ is $2$ It can be easily verified that the sequence $\left\{ \dfrac{2^n -1} {2^{n-1}} \right\}$ is strictly positive, the task now is to prove that 2 is an upper bound.
Proof:
Suppose for the sake of contradiction that 2 is not the least upper bound of the sequence. Let $x_n$ be an element of the sequence. Using the density of the real numbers in combination with the fact that the sequence is strictly increasing we can write the following:
$$ x_i < 2 < x_j , \quad j>i $$
That is, we can find a $j \in \mathbb{N}$ such that:
\begin{align}
0 &< x_j - 2 \\
&<\dfrac{2^j - 1}{2^{j-1}} - 2 \\
&< \dfrac{2^j - 1 - 2(2^{j-1})}{2^{j-1}} \\
&< (-1)\dfrac{1}{2^{j-1}} \\
\end{align}
So we arrive at a contradiction.
I just want to check to see if my approach makes sense to this problem.
|
Your answer is okay but it's ludicrously complicated.
Why not just say $\frac {2^n -1}{2^{n-1}} < \frac {2^n}{2^{n-1}} = 2$?
That's it. You are done.
But is that all of the question?
Are you sure you weren't also being as to prove that $2$ is the least upper bound.
In that case if $x < 2$ you must prove there exists a $x < \frac {2^n-1}{2^{n-1} } < 2$ no matter how close $x$ is to $2$.
$x < \frac {2^n-1}{2^{n-1} } = \frac {2^n}{2^{n-1}} - \frac 1{2^{n-1}} = 2-\frac 1{2^{n-1}} < 2 \iff$
$x-2 < -\frac 1{2^{n-1}} < 0\iff$
$0 < \frac 1{2^{n-1}} < 2-x \iff$
$x < 2$ and $2^{n-1} > \frac 1{2-x}$.
If we know that $2^m$ is unbounded then we know that there is an $m = n-1$ so that $2^{n-1} > \frac 1{2-x} $ for any $x < 2$.
If $\{2^m\}$ were bounded above then $\sup 2^m$ would exist. And $\sup 2^m -1$ wouldn't be an upper bound so there would be a $k$ so that $\sup 2^m -1 < 2^k < \sup 2^m$ So $2\sup 2^m - 2 < 2^{k+1}$. But as $\sup 2^m > 2=2^2$ as must have $2\sup 2^m - 2 > \sup 2^m$ and so $\sup 2^m < 2^{k+1}$ which is a contradiction.
|
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|
Evaluate $\frac{4}{\sin^2 20^\circ} - \frac{4}{\sin^2 40^\circ} + 64\sin^2 20^\circ$ Evaluate the following expression:$$\frac{4}{\sin^2 20^\circ} - \frac{4}{\sin^2 40^\circ} + 64\sin^2 20^\circ$$
I tried combining the whole into a single fraction and using double-angle identity, product/sum to sum/product, but it didn't work.
|
Per the identity
$\sin3t =3\sin t -4\sin^3t$,
it is straightforward to verify that $\sin20$, $\sin40$, $-\sin80$ are the roots of
$4x^3-3x+\frac{\sqrt3}2=0$, which leads to
$$\sin20\sin40\sin80=\frac{\sqrt3}8$$
Then
\begin{align}
& \frac{4}{\sin^2 20^\circ} - \frac{4}{\sin^2 40^\circ} + 64\sin^2 20^\circ\\
=& \frac{64}3\left(4\sin^240\sin^280 -4\sin^220\sin^280+3\sin^220 \right)\\
=& \frac{64}3\left((1-\cos80)(1+\cos20)-(1-\cos40)(1+\cos20) + \frac32(1-\cos40)\right)\\
=& \frac{64}3\left(\frac32 - \cos80 -\frac12\cos40
-\cos20\cos80 +\cos20\cos40\right)\\
=& 32 + \frac{64}3\left(- \cos80 -\frac12\cos40
-\frac12(\cos60-\cos80) +\frac12(\cos20+\cos60)\right)\\
=& 32 + \frac{32}3\left(\cos 20- \cos40 -\cos80
\right) \\
= &32 + \frac{32}3\left(\cos 20- 2\cos60\cos20
\right)\\
=& 32
\end{align}
|
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|
Find $ \lim_{r \to 1^{-}} \int_{-\pi}^{\pi} (\frac{1+2r^2}{1-r^2\cos2\theta})^{1/3}d\theta$ Evaluate $$ \lim_{r \to 1^{-}} \int_{-\pi}^{\pi} \left[\frac{1+2r^2}{1-r^2\cos\left(2\theta\right)}\right]^{1/3}{\rm d}\theta $$
Question- Can I take the limit inside the integral?
My try-
$$I= \lim_{r \to 1^{-}} \int_{-\pi}^{\pi} \left(\frac{1+2r^2}{1-r^2\cos2\theta} \right)^{1/3}\, d\theta $$
$$ I= \int_{-\pi}^{\pi} \lim_{r \to 1^{-}} (\frac{1+2r^2}{1-r^2\cos2\theta})^{1/3}d\theta $$
$$ I=3^{1/3} \int_{-\pi}^{\pi} \frac{1}{(1-\cos2\theta)^{1/3}}d\theta $$
$$ I= 3^{1/3}2\int_{0}^{\pi} \frac{1}{(1-\cos2\theta)^{1/3}} d\theta $$
$$ I= 3^{1/3}4\int_{0}^{\pi/2} \frac{1}{(1-\cos2\theta)^{1/3}} d\theta $$
$$ I= (3/2)^{1/3}4\int_0^{\pi/2}
\sin^{-2/3}\theta \ d\theta $$
$$I= 4(3/2)^{1/3} \frac{\Gamma(1/6)\Gamma(1/2) }{\Gamma(2/3)}. $$
|
This is not an unswer, just a hint and too long for a comment: The integral above may be transformed with $\theta = \frac{\arccos(y)}{2}$
$$2\, \left(1+2\, r^2\right)^{1/3} \int_{-1}^1 \left(1-r^2 y\right)^{-1/3} \left(1-y^2\right)^{-\frac{1}{2}} \, dy$$
Next we can split the integral in two parts with $p=-y$ in the first one:
$$2\,\left(1+2\, r^2\right)^{1/3} \left(\int_0^1 \left(1+r^2 p\right)^{-1/3} \left(1-p^2\right)^{-\frac{1}{2}} \, dp + \int_0^1 \left(1-r^2 y\right)^{-1/3} \left(1-y^2\right)^{-\frac{1}{2}} \, dp \right) $$
and evaluate them in form of hypergeometric functions. For this step e.g. we can express the integrand in form of MeijerG - Function. The first integral will be transformed with $q = p^2$ apply the formular Mathematical Functions Site and simplify:
$$I_1 = \frac{\left(1+2\, r^2\right)^{1/3}}{\gamma\left[\frac{1}{3}\right]} \int_0^1 (1-q)^{-\frac{1}{2}} q^{-\frac{1}{2}} \,G_{1,1}^{1,1}\left( \sqrt{q}\, r^2\left\vert
\begin{array}{c}
\frac{2}{3}\\
0%
\end{array}%
\right. \right)\,
\, dq$$
$$ = \left(1+2 \,r^2\right)^{1/3} \left(\pi \,_{2}F_{1}\left[\frac{1}{6},\frac{2}{3},1,r^4\right]-\frac{2}{3} r^2 \,_{3}F_{2}\left[\left\{\frac{2}{3},1,\frac{7}{6}\right\},\
\left\{\frac{3}{2},\frac{3}{2}\right\},r^4\right]\right)$$
The second integral $I_2$ is performed in the same way. The limit $r \to 1$ Mathematical Functions Site and Mathematical Functions Site exists e.g. for $I_1$
$$\lim_{r \to 1}\,I_1=\frac{\frac{3 \pi\, \gamma\
\left[\frac{1}{6}\right]}{\gamma\left[\frac{1}{3}\right]\, \gamma\left[\frac{5}{6}\right]}-2\,_{3}F_{2}\left[\left\{\frac{2}{3},1,\frac{7}{6}\right\},\left\{\frac{3}{2},\frac{3}{2}\right\},1\right]}{3^{2/3}}$$
and for $I_2$ in the same way. The sum of $I = I_1 + I_2$
delivers just the half of your answer:
$$I= \frac{2^{1/3}\, 3^{5/6}\, \gamma\left[\frac{1}{3}\right]^2}{\gamma\left[\frac{2}{3}\right]} $$.
Therefore, inspite of a calculation error of factor 2, @user1055 is right, the limit may be also calculated inside the integral.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $a_n = \frac{\Gamma(\frac{n}{2}+\frac{1}{4})}{\Gamma(\frac{n}{2}+\frac{3}{4})}$ converges to zero According to WolframAlpha, the following sequence
\begin{align*}
a_n = \frac{\Gamma(\frac{n}{2}+\frac{1}{4})}{\Gamma(\frac{n}{2}+\frac{3}{4})}
\end{align*}
seems to converges to zero. However, how can I prove this ? The difficulty I encounter is this "one quarter of an integer" in the Gamma functions.
|
$$a_n = \frac{\Gamma(\frac{n}{2}+\frac{1}{4})}{\Gamma(\frac{n}{2}+\frac{3}{4})}\implies \log(a_n)=\log \left(\Gamma \left(\frac{n}{2}+\frac{1}{4}\right)\right)-\log \left(\Gamma \left(\frac{n}{2}+\frac{3}{4}\right)\right)$$ Now, using Stirling approximation
$$\log (\Gamma (p))=p (\log (p)-1)+\frac{1}{2} \log \left(\frac{2 \pi }{p}\right)+\frac{1}{12
p}-\frac{1}{360 p^3}+O\left(\frac{1}{p^5}\right)$$ apply it twice and continue with Taylor expansion to obtain
$$\log(a_n)=-\frac{1}{2} \log \left(\frac{n}{2}\right)-\frac{1}{16 n^2}+\frac{5}{128
n^4}+O\left(\frac{1}{n^5}\right)$$ Continuing with Taylor
$$a_n=e^{\log(a_n)}=\sqrt{\frac{2}{n}} \exp\Big[-\frac{1}{16 n^2}+\frac{5}{128
n^4} \Big]$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Calculate $2 C_{n}^{k-1} + 3 C_{n}^{k-2} + 4 C_{n}^{k-3} + ... + (k + 1)C_{n}^{0}$ Are there any easy ways to calculate this?
$$ \sum_{i=1}^{k} (i + 1) \cdot C_{2k}^{k-i} = 2 C_{2k}^{k-1} + 3 C_{2k}^{k-2} + 4 C_{2k}^{k-3} + ... + (k + 1) C_{2k}^{0} $$
I tried to "turn" the expression:
$$ \sum_{i=1}^{k} (i + 1) \cdot C_{2k}^{k-i} = n \cdot 2^{n-2} - \frac{1}{2} \sum_{i=1}^{k} (2n -i) \cdot C_{2k}^{k-i} $$
But, it doesn't help me.
Thanks for the help!
|
Consider the formula provided by this answer:
$\sum_{0\leq j\leq n/2}\binom{n}{j}x^j = (x+1)^n - \binom{n}{\left\lfloor\frac{n}{2}\right\rfloor + 1} x^{\left\lfloor\frac{n}{2}\right\rfloor + 1} {_2F_1}\left(1,\left\lfloor\frac{n}{2}\right\rfloor - n + 1;\left\lfloor \frac{n}{2}\right\rfloor + 2;-x\right),$
where $_2F_1$ is the hypergeometric function.
Let $f(x)$ be the function above with $n=2k$, that is
$$f(x) = \sum_{j=0}^{k}\binom{2k}j x^j = (x+1)^{2k} - \binom{2k}{k + 1} x^{k + 1} {_2F_1}\left(1,1-k;k + 2;-x\right).$$
Now write $(x+1)^{2k} = \sum_{j=0}^{2k}\binom{2k}j x^{j}$, so that
$$\begin{align}
(x+1)^{2k} - f(x)
&= \sum_{j=k+1}^{2k}\binom{2k}j x^j
\\&= \sum_{i=1}^{k}\binom{2k}{k+i} x^{k+i}
= \sum_{i=1}^{k}\binom{2k}{k-i} x^{k+i}
\end{align}$$
With $h(x) = (x+1)^{2k} - f(x) = \binom{2k}{k + 1} x^{k + 1} {_2F_1}\left(1,1-k;k + 2;-x\right)$, we hence have
$$x^{1-k}h(x) = \sum_{i=1}^{k}\binom{2k}{k-i} x^{i+1}.$$
Differentiating we get
$$(1-k)x^{-k}h(x) + x^{1-k}h'(x) = \sum_{i=1}^{k}(i+1)\binom{2k}{k-i} x^i,$$
so that your sum can be obtained by evaluating the LHS at $x=1$:
$$(1-k)h(1) + h'(1).$$
This yields a complicated expression in terms of hypergeometric functions:
$$\begin{align}
&&(1-k)&&\binom{2k}{k+1}&&{_2F_1}\left(1,1-k;k + 2;-1\right)
\\+\,\,\,&&(k+1)&&\binom{2k}{k+1}&&{_2F_1}\left(1,1-k;k + 2;-1\right)
\\+\,\,\,&&\frac{k-1}{k+2}&&\binom{2k}{k+1}&&{_2F_1}\left(2,2-k;k + 3;-1\right)
\end{align}
\\= \binom{2k}{k+1}\left(2{_2F_1}\left(1,1-k;k + 2;-1\right) + \frac{k-1}{k+2}{_2F_1}\left(2,2-k;k + 3;-1\right)\right).$$
You can check on WolframAlpha that the formula provided above and the expression in the opening question yield the same results.
|
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|
Given $x^{p^2} = 1, x^p = y^p, yxy^{-1}=x^{p+1}$, show that $(yx^{-1})^p=1$
Suppose $p$ is an odd prime. Given $x^{p^2} = 1, x^p = y^p, yxy^{-1}=x^{p+1}$, show that $(yx^{-1})^p=1$, where $1$ is the identity.
I know that the conjugate of power is the power of conjugate, so $yx^{k}y^{-1} = (x^{p+1})^{k}$, and used this to obtain $yx^{-1} = x^{-p-1} y$, which will allow me to switch $y$ and $x^{-1}$ in each $yx^{-1}$. However, I still cannot figure out where to go from here. Any help is appreciated.
|
$yx^k = x^{k(p+1)}y$, so $y^a x^k = x^{k(p+1)^a} y^a$, so:
$$\begin{array}{rcl}
(yx^{-1})^p
&=& y x^{-1} y x^{-1} y x^{-1} \cdots yx^{-1} \\
&=& x^{-(p+1)}y^2 x^{-1} y x^{-1} \cdots yx^{-1} \\
&=& x^{-(p+1)-(p+1)^2} y^3 x^{-1} \cdots yx^{-1} \\
&=& \cdots \\
&=& x^{-(p+1)-(p+1)^2-\cdots-(p+1)^{p-1}} y^p \\
&=& x^{-(p+1)-(2p+1)-\cdots-((p-1)p+1)} x^p \\
&=& x^{-\frac{(p-1)p}2 p - p} x^p \\
&=& x^{-\frac{(p-1)}2 p^2} \\
&=& 1 \\
\end{array}$$
where $x^{(p+1)^n} = x^{1+np+O(p^2)} = x^{1+np}$ because $x^{p^2}=1$.
|
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|
Simplifying the sum of a product of multinomial coefficients From the multinomial theorem the following holds
$$
\sum_{k_1 + k_2 + \ldots + k_m=n} {n \choose k_1, k_2, \ldots, k_m} = m^n
$$
I have the following sum
$$
\sum_{\substack{k_1 + k_2 + \ldots + k_m &=& B \\ g_1 + g_2 + \ldots + g_m &=& W}} {B \choose k_1, k_2, \ldots k_m} \cdot {W \choose g_1, g_2, \ldots, g_m}
$$
where $B + W = n$, and $\forall_i: g_i = x - k_i$, where $x = \frac{n}{m}$, and $m\mid n$ and $k_i \geq 1$
Can this sum be simplified like the original one?
|
We assume positive integers $m,n\geq 2$. Since $m\mid n$ we know that $n=Nm$ is a multiple of $m$ with $N\geq 1$. This way we can write $W=Nm-B$ and $x=N$. We show by induction of $m$ that for all $N\geq 1, m\leq B\leq Nm$ the following is valid:
\begin{align*}
\color{blue}{\sum_{\substack{k_1+\cdots+k_m=B\\k_1,\ldots,k_m\geq 0}}
\binom{B}{k_1,\ldots,k_m}\binom{mN-B}{N-k_1,\ldots,N-k_m}=\prod_{q=2}^m\binom{qN}{N}}\tag{1}
\end{align*}
Base step: $m=2$
\begin{align*}
\color{blue}{\sum_{{k_1+k_2=B}\atop{k_1,k_2\geq 0}}}
&\color{blue}{\binom{B}{k_1,k_2}\binom{2N-B}{N-k_1,N-k_2}}\\
&=\sum_{k_1=0}^B\binom{B}{k_1}\binom{2N-B}{N-k_1}\tag{2}\\
&=\sum_{k_1=0}^B\binom{B}{k_1}[z^{N-k_1}](1+z)^{2N-B}\tag{3}\\
&=[z^N](1+z)^{2N-B}\sum_{k_1=0}^B\binom{B}{k_1}z^{k_1}\tag{4}\\
&=[z^N](1+z)^{2N-B}(1+z)^{B}\\
&=[z^N](1+z)^{2N}\\
&\,\,\color{blue}{=\binom{2N}{N}}\tag{5}
\end{align*}
in accordance with (1).
Comment:
*
*In (2) we write the multinomial as binomial coefficients and substitute $k_2 = B-k_1$.
*In (3) we use the coefficient of operator $[z^N]$ to denote the coefficient of $z^N$ of a series.
*In (4) we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
*In (5) we select the coefficient of $z^{N}$.
Induction hypothesis: $m=M$
We assume the validity of
\begin{align*}
\color{blue}{\sum_{{k_1+\cdots+k_M=B}\atop{k_1,\ldots,k_M\geq 0}}
\binom{B}{k_1,\ldots,k_M}\binom{MN-B}{N-k_1,\ldots,N-k_M}=\prod_{q=2}^M\binom{qN}{N}}\tag{6}
\end{align*}
Induction step: $m=M+1$
We have to show
\begin{align*}
\color{blue}{\sum_{\substack{k_1+\cdots+k_{M+1}=B\\k_1,\ldots,k_{M+1}\geq 0}}
\binom{B}{k_1,\ldots,k_{M+1}}\binom{(M+1)N-B}{N-k_1,\ldots,N-k_{M+1}}=\prod_{q=2}^{M+1}\binom{qN}{N}}\tag{7}
\end{align*}
We obtain for $M\geq 2$:
\begin{align*}
&\color{blue}{\sum_{\substack{k_1+\cdots+k_{M+1}=B\\k_1,\ldots,k_{M+1}\geq 0}}}\color{blue}{\binom{B}{k_1,\ldots,k_{M+1}}\binom{(M+1)N-B}{N-k_1,\ldots,n-k_{M+1}}}\\
&\qquad=\sum_{{k_1+\cdots+k_{M+1}=B}\atop{k_1,\ldots,k_{M+1}\geq 0}}\frac{B!}{k_1!\cdots k_{M+1}!}\,\frac{\left((M+1)N-B\right)!}{\left(N-k_1\right)!\cdots\left(N-k_{M+1}\right)!}\tag{8}\\
&\qquad=\sum_{k_{M+1}=0}^B\frac{B!}{k_{M+1}!\left(B-k_{M+1}\right)!}\,
\frac{\left((M+1)N-B\right)!}{\left(N-k_{M+1}\right)!\left(MN-\left(B-k_{M+1}\right)\right)!}\\
&\qquad\quad\cdot\sum_{{k_1+\cdots+k_M=B-k_{M+1}}\atop{k_1,\ldots,k_{M}\geq 0}}\frac{\left(B-k_{M+1}\right)!}{k_1!\cdots k_{M}!}\,\frac{\left(MN-\left(B-k_{M+1}\right)\right)!}{\left(N-k_1\right)!\cdots\left(N-k_{M}\right)!}\tag{9}\\
&\qquad=\sum_{k_{M+1}=0}^B\binom{B}{k_{M+1}}\,
\binom{(M+1)N-B}{N-k_{M+1}}\\
&\qquad\quad\cdot\sum_{{k_1+\cdots+k_M=B-k_{M+1}}\atop{k_1,\ldots,k_{M}\geq 0}}
\binom{B-k_{M+1}}{k_1,\ldots, k_{M}}\,\binom{MN-\left(B-k_{M+1}\right)}{N-k_1,\ldots,N-k_{M}}\tag{10}\\
&\qquad=\prod_{q=2}^M\binom{qN}{N}\sum_{k_{M+1}=0}^B\binom{B}{k_{M+1}}
[z^{N-k_{M+1}}](1+z)^{(M+1)N-B}\tag{11}\\
&\qquad=[z^{N}](1+z)^{(M+1)N-B}\prod_{q=2}^M\binom{qN}{N}\sum_{k_{M+1}=0}^B\binom{B}{k_{M+1}}z^{k_{M+1}}\\
&\qquad=[z^{N}](1+z)^{(M+1)N-B}\prod_{q=2}^M\binom{qN}{N}(1+z)^B\\
&\qquad=[z^{N}](1+z)^{(M+1)N}\prod_{q=2}^M\binom{qN}{N}\\
&\qquad=\binom{(M+1)N}{N}\prod_{q=2}^M\binom{qN}{N}\\
&\qquad\,\,\color{blue}{=\prod_{q=2}^{M+1}\binom{qN}{N}}
\end{align*}
in accordance with (7) and the claim follows.
Comment:
*
*In (8) we write the multinomial coefficients using factorials.
*In (9) we separate the summation of the summand $k_{M+1}$ as preparation to apply the induction hypothesis. We expand with $\left(B-k_{M+1}\right)!$ and $\left(MN-\left(B-k_{M+1}\right)\right)!$ to write the expression with binomial and multinomial coefficients.
*In (10) we write the expression using binomial and multinomial coefficients.
*In (11) we apply the induction hypothesis. We also use the coefficient of operator of $[z^N]$ to denote the coefficient of $z^N$ of a series.
*In the following lines we use the same techniques as in the base step.
|
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|
Avoiding brute force: determining when a specific polynomial in $\mathbb{Q}[x]$ is an integer for any integer $x$ I have to prove that $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer for any $n$. I solved this by brute-force, exhausting all the possibilities methods. I was wondering if there was a way to solve this at-a-glance with some sort of theory? Because
Although my answer is correct, I was apparently supposed to use some theory to answer this question.
I started by putting everything on common denominator and factoring out $n$:
$$\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n = \frac{n(3n^4+5n^2+7)}{15}$$
then I proceeded to plug in $\pm1 ,\pm2, \pm3 ,..., \pm7$ into $3n^4+5n^2+7 \pmod {3,5,\text{ or }15}$
$$3(\pm 1)^4 +5(\pm 1)^2 +7 \equiv 15 \equiv 0 \pmod{15}$$
$$3(\pm 2)^4 +5(\pm 2)^2 +7 \equiv 75 \equiv 5\cdot 15 \equiv 0 \pmod{15}$$
$$3(\pm 3)^4 +5(\pm 3)^2 +7 \equiv 295 \equiv 0 \pmod{5} \text{ while } n \equiv 0 \pmod 3$$
$$3(\pm 4)^4 +5(\pm 4)^2 +7 \equiv 855 \equiv 57\cdot 15\equiv 0 \pmod{15}$$
$$3(\pm 5)^4 +5(\pm 5)^2 +7 \equiv 2007 \equiv 0 \pmod{3}\text{ while } n \equiv 0 \pmod 5$$
$$3(\pm 6)^4 +5(\pm 6)^2 +7 \equiv 4075 \equiv 0 \pmod{5}\text{ while } n \equiv 0 \pmod 3$$
$$3(\pm 7)^4 +5(\pm 7)^2 +7 \equiv 7455 \equiv 497 \cdot 15\equiv 0 \pmod{15}$$
to conclude that if $n\equiv 0 \pmod {15}$ then $\frac{n}{15}$ is an integer from which $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer,
and if $n \not\equiv 0 \pmod {15}$ then either $3n^4+5n^2+7 \equiv 0 \pmod{15}$ or $n \equiv 0 \pmod 3$ and $ 3n^4+5n^2+7 \equiv 0 \pmod{5}$ or $n \equiv 0 \pmod 5$ and $ 3n^4+5n^2+7 \equiv 0 \pmod{3}$ from which the statement is clearly true.
Is there a less brute-force-ish way of concluding this? Is there some theory I should be using that would cause me to not be excessively lengthy in calculation if I were given different, larger numbers than $15$?
|
The question asks how to prove a polynomial
$\,f(n)\,$ takes only integer values for any
integer $\,n\,.$ If the polynomial is of
degree $0$, then it is a constant and if
that constant is an integer we are done.
If the polynomial is of degree $1$, then if
any two consecutive values,
$\,f(k), f(k+1)\,$ are integers,
then it is an arithmetic progression and
we are done. In general, for any degree
$\,d\,$ polynomial $\,f(n),\,$ it is
sufficient to verify that
$$f(k),f(k+1), \dots,f(k+d-1) $$ are all integers for some
integer $\,k,\,$ which implies that the polynomial takes on only integer values for all integer $\,n.\,$
A proof of this can be done by using a
difference table to find an explicit
expression for the polynomial as a sum of
binomial coefficients. For example, using
the particular polynomial in the question,
$$ f(n) := \frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n $$
the forward difference table is:
$$ \begin{matrix}
\Delta^6f(n) &&&&&&& 0 \\
\Delta^5f(n) &&&&&& 24 && 24 \\
\Delta^4f(n) &&&&& 48 && 72 && 96 \\
\Delta^3f(n) &&&& 32 && 80 && 152 && 248 \\
\Delta^2f(n) &&& 8 && 40 && 120 && 272 && 520\\
\Delta f(n) && 1 && 9 && 49 && 169 && 441 && 961 \\
f(n) & 0 && 1 && 10 && 59 && 228 && 669 && 1630 \\
n & 0 && 1 && 2 && 3 && 4 && 5 && 6
\end{matrix} $$
where the 5th differences are constant as expected for a
5th degree polynomial. Thus,
$$ f(n) = 0{n \choose 0} +1{n \choose 1} + 8{n \choose 2}
+ 32{n \choose 3} + 48{n \choose 4} + 24{n \choose 5}$$
where the coefficient of $\,{n\choose k}=\Delta^kf(0),\,$
the $k$th difference of $\,f\,$ at zero.
The Wikipedia article finite difference
explains some of this theory. The forward differences of $\,f\,$ are
$$ \Delta f(n) := f(n+1)-f(n),
\:\: \Delta^2 f(n) := \Delta f(n+1) - \Delta f(n), \:\: \dots. $$
In this particular case, the 1st backward difference is the
OEIS sequence A058031
$$ a_n := (n^2-n+1)^2 = \nabla f(n) := f(n)-f(n-1).$$
Thus the polynomial is the partial sums of an integer sequence
and therefore is an integer sequence itself.
|
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|
Remainder When Divided By 70 : Remainder Theorem Problem If $x$ = $16^3$ + $17^3$ + $18^3$ + $19^3$ , then $x$ divided by $70$ leaves a remainder of?
I tried to solve this problem by using the remainder theorem which states that remainder of $$Rem[\frac{a+b+c+....}{x}] = Rem[\frac{a}{x}] + Rem[\frac{b}{x}] + Rem[\frac{c}{x}] + .... $$ where 'Rem' means 'Remainder Of'.
Using this same logic I can write $$Rem[\frac{16^3+17^3+18^3+19^3}{70}] = Rem[\frac{16^3}{70}] + Rem[\frac{17^3}{70}] + Rem[\frac{18^3}{70}] + Rem[\frac{19^3}{70}] = 18 +13+11+69 = 111 \Rightarrow Rem[\frac{111}{70}] = 41$$
So my answer comes up as 41. Can someone please explain to me what I might be doing wrong or what I have understood wrong?
|
To determine the remainder of some computation, $C, \pmod{70}$, simply determine the remainder of $C, \pmod{2}, \pmod{5}$, and $\pmod{7}.$ Then employ Number Theory methods to solve the resultant congruencies.
Let $C = 16^3 + 17^3 + 18^3 + 19^3.$
Since the second and fourth terms of $C$ (only) are odd, it is immediate that $C \equiv 0\pmod{2}.$
You can prove, by binomial expansion, that if $r,s,k,n$ are all positive integers, then $(r+ns)^k \equiv r^k \pmod{n}.$
Therefore, the problem immediately reduces to the twin problems of
computing
$$D = 1^3 + 2^3 + 3^3 + 4^3 \pmod{5}$$
and computing
$$E = 2^3 + 3^3 + 4^3 + 5^3 \pmod{7}.$$
The problem permits shortcuts, by recognizing that if
$a \equiv (-b) \pmod{n},$ then $a^3 \equiv (-b)^3 = (-1)^3b^3 = -(b^3) \pmod{n}.$
Using this approach, note that when applying $\pmod{5}$ congruencies against $D$,
$4 = (-1),$ and $3 = (-2).$
Similarly, when applying $\pmod{7}$ congruencies against $E$
$5 = (-2),$ and $4 = (-3).$
Therefore, it is immediate that $D \equiv 0\pmod{5}$ and $E\equiv 0\pmod{7}$.
Therefore, it is immediate that $C$ is a multiple of $2$, $5$, and $7$.
|
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|
Find $\binom{80}{40}\bmod 2000$
Find $\displaystyle\binom{80}{40}\bmod 2000$.
So far, I've found that $\displaystyle\binom{80}{40}$ is divisible by $2^2$ and $5^1$, so the answer isn't $0$. Usually, with smaller numbers, I would split the $\bmod 2000$ into $2^4$ and $5^3$, find the answer through brute force for each of those, then use the Chinese Remainder theorem. But $80$ and $40$ are too big for this.
|
Since $${80 \choose 40} = \frac{80\cdot 79 \cdot 78 \cdot 77 \cdot ...\cdot 41}{40 \cdot 39 \cdot 38 \cdot 37 \cdot ... \cdot 1}$$ we can cancel the terms in the denominator with terms in the numerator leaving a factor of $2$, so $2^{40}$ is a factor of the answer. It only remains to see how many factors of 5 remain. $75, 65, 55$, and $45$ are all factors, so at least $5^4$ is a factor. As a result, the expression is definitely a multiple of $2000$, so modulo $2000$ the answer is $0$.
EDIT: A few minutes later I realized my mistake, and returned to find several people pointing it out: only the numbers down to 21 in the denominator cancel. That still leaves $$\frac{2 \cdot 79 \cdot 2 \cdot 77 \cdot 2 \cdot 75 \cdot...\cdot 2 \cdot 41}{ 20 \cdot 19 \cdot 18 \cdot...\cdot 3 \cdot 2 \cdot 1}$$ which gives $20$ factors of $2$ in the numerator and $10 + 5 + 2 + 1 = 18$ factors of $2$ in the denominator, so only $2^2$ is a factor. For the $5's$, we have $5$ factors in the numerator and $4$ in the denominator, so it is a multiple of $5$ but not $5^2$.
So now we have $$\frac{79 \cdot 77 \cdot 3 \cdot 73 \cdot...\cdot 43 \cdot 41}{ 19 \cdot 9 \cdot 17 \cdot 3 \cdot 7...\cdot 6 \cdot 3}$$. Continuing to cancel factors, we find $3^{8}$ in the numerator and $3^8$ in the denominator, and so forth. We end up computing ${80 \choose 40}$ and then can find the answer by inspection.
|
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|
Given area of parallelogram, find smallest distance from origin to furthest point (complex numbers)
On the complex plane, the parallelogram formed by the points 0, $z,$
$\frac{1}{z},$ and $z + \frac{1}{z}$ has area $\frac{35}{37}.$ If the
real part of $z$ is positive, let $d$ be the smallest possible value
of $\left| z + \frac{1}{z} \right|.$ Compute $d^2.$
I begin by letting $z=x+yi$ and $\frac{1}{z}=\frac{x-yi}{x^2+y^2}$. Then, I applied the cross product to get that the
$$\text{Area of Parallelogram} = |z| \left|\frac{1}{z}\right| \sin(\theta) . $$
This implies that
$$ \frac{35}{37}=\left(\sqrt{x^2+y^2}\right) \cdot \left(\frac{1}{\sqrt{x^2+y^2}}\right) \sin(\theta) \Longrightarrow \frac{35}{37} = \sin(\theta) , $$
where $\theta$ is the angle between the $\frac{1}{z}$ and $z$. Subsequently, $\sin\left(\frac{\theta}{2}\right)=\frac{5}{\sqrt{74}}$ meaning that $z$ is in the form $x+\frac{5x}{7}i$. The problem reduced to an optimization problem, minimizing
$$ \left(x+\frac{x}{x^2+y^2}\right)^2+\left(y-\frac{y}{x^2+y^2}\right)^2=\left(x+\frac{x}{\frac{x\sqrt{74}}{7}}\right)^2+\left(\frac{5x}{7}-\frac{\frac{5x}{7}}{\frac{x\sqrt{74}}{7}}\right)^2 . $$
I eventually got the answer to be $\left(\frac{35}{37}\right)^2=\frac{1225}{1369}$. However, this answer is wrong. I am not sure where I messed up. Where did I go wrong?
|
Let $a$ and $b$ two $2$-dimensional vectors satisfying $||a||\cdot||b||=1$. Then (via Gram) the square of area of the parallelogram spanned by these vectors is $(35/37)^2=||a||^2\cdot||b||^2-\langle a,b\rangle^2=1-\langle a,b\rangle^2$ from where $\langle a,b\rangle=\pm12/37$.
Now
$$\begin{align}
d^2=||a+b||^2&=||a||^2+||b||^2+2\langle a,b\rangle\\
&=||a||^2+\frac{1}{||a||^2}\pm\frac{12}{37}\\
&=\left(||a||+\frac{1}{||a||}\right)^2-2
\pm\frac{12}{37}\\
&\geq 4-2
\pm\frac{12}{37}\\
&=2
\pm\frac{12}{37},
\end{align}$$
as for all reals $x$ we know $-2<x+1/x<2$.
|
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|
Binomial coefficients identity : $\sum_{k=1}^{n-m+1} k\binom{n-k+1}{m}=\binom{n+2}{m+2}$ For any positive integer m&n.$n\ge m$ , let $\left( {\begin{array}{*{20}{c}}
n\\
m
\end{array}} \right) = {}^n{C_m}$. Prove that $\left( {\begin{array}{*{20}{c}}
n\\
m
\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}
{n - 1}\\
m
\end{array}} \right) + 3\left( {\begin{array}{*{20}{c}}
{n - 2}\\
m
\end{array}} \right) + .. + \left( {n - m + 1} \right)\left( {\begin{array}{*{20}{c}}
m\\
m
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{n + 2}\\
{m + 2}
\end{array}} \right)$
My approach is as follow
$n=4, m=2$
$\left( {\begin{array}{*{20}{c}}
4\\
2
\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}
3\\
2
\end{array}} \right) + 3\left( {\begin{array}{*{20}{c}}
2\\
2
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
6\\
4
\end{array}} \right) = 15$
$6 + 6 + 3 = 15 \Rightarrow LHS = RHS$
But not able to solve it via property
|
Starting from
$$\sum_{k=1}^{n-m+1} k {n-k+1\choose m}$$
we re-write as
$$\sum_{k=1}^{n-m+1} k {n-k+1\choose n-m+1-k}
= [z^{n-m+1}] (1+z)^{n+1}
\sum_{k=1}^{n-m+1} k \frac{z^k}{(1+z)^k}$$
Here the coefficient extractor enforces the upper limit of the sum and
we obtain
$$[z^{n-m+1}] (1+z)^{n+1}
\sum_{k\ge 1} k \frac{z^k}{(1+z)^k}
\\ = [z^{n-m+1}] (1+z)^{n+1}
\frac{z/(1+z)}{(1-z/(1+z))^2}
= [z^{n-m+1}] (1+z)^{n+1}
z (1+z)
\\ = [z^{n-m}] (1+z)^{n+2} = {n+2\choose n-m}
= {n+2\choose m+2}$$
as claimed.
|
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|
How to find angle at centroid of triangle by its edge lengths? I need to write a program that takes length of a triangle's edges and calculates the angle $\angle APB$ ($P$ is the centroid of the triangle). Thanks for any help or clue.
|
$$
PA^2=\frac {2b^2+2c^2-a^2}9,\quad PB^2=\frac {2a^2+2c^2-b^2}9\\
\implies \cos(\angle APB)=\frac {PA^2+PB^2-c^2}{2\cdot PA \cdot PB}
=\frac {a^2+b^2-5c^2}{2\sqrt{(2b^2+2c^2-a^2)(2a^2+2c^2-b^2)}}.
$$
|
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|
What is the intergal of $\int {\dfrac{1}{(x^2+9)^2}}\,dx$ with recursive formula? I used partial integration and I got this:
${\dfrac{x}{(x^2+9)^2}} + 4 $$\int {\dfrac{x^2}{(x^2+9)^3}}\,dx$
According to the recursice formula I should continue like this:
${\dfrac{x}{(x^2+9)^2}} + 4 $$\int {\dfrac{x^2 +9-9}{(x^2+9)^3}}\,dx$
${\dfrac{x}{(x^2+9)^2}} + 4 $$\int {\dfrac{x^2 +9-9}{(x^2+9)^3}}\,dx$ - 4$\cdot 9 $$\int {\dfrac{1}{(x^2+9)^3}}\,dx$
And then end up with something like that:
${\dfrac{1}{2 \cdot 9} \cdot \dfrac{1}{x^2+9} + \dfrac{1}{2 \cdot 27}} arctg{\dfrac{x}{3}}\,$ + C
But I'm sure that I'm doing this wrong or just I'm too confused?
|
You have found the integral of $(x^2+9)^{-3}$ in terms of the integral of $(x^2+9)^{-2}$. So, instead, integrate $(x^2+9)^{-1}$ by parts
|
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|
$p_1,p_2$ roots of $x^2+3x+5=0$, then the value of the expression $M=\frac{p_1^2+5p_1+7}{p_1^2+7p_1+5}+\frac{p_2^2+5p_2+7}{p_2^2+7p_2+5}$ If $p_1,p_2$ are the roots of the equation $x^2+3x+5=0$, then the value of the expression $M=\frac{p_1^2+5p_1+7}{p_1^2+7p_1+5}+\frac{p_2^2+5p_2+7}{p_2^2+7p_2+5}$
I solved the question as follows:
Initially I tried to solve it by finding what the roots are but since $b^2-4ac=9-20<0$, a real solution doesn't exist. Henceforth I attempted to solve it using Vieta's formulas, by saying that $p_1+p_2=-\frac{a_{n-1}}{a_n}=-\frac{3}{1}=-3$ and similarly $p_1*p_2=5$ and then $M=\frac{(p_1^2+5p_1+7)(p_2^2+7p_2+5)+(p_2^2+5p_2+7)(p_1^2+7p_1+5)}{(p_1^2+7p_1+5)(p_2^2+7p_2+5)}$
and eventually after a lot more calculations:
$M=\frac{7}{10}$
I find this method extremely laborious and I believe that there must exist a faster and easier method of solving this question. Could you please explain an easier and faster method?
|
Since $p_1,p_2$ are roots of $x^2+3x+5=0$, then $p_1^2+3p_1+5=p_2^2+3p2+5=0$ and we can simplify the expression to
$$
\frac{2p_1+2}{4p_1}+\frac{2p_2+2}{4p_2}=\frac{p_1+1}{2p_1}+\frac{p_2+1}{2p_2}=\frac{p_1p_2+p_2+p_1p_2+p_1}{2p_1p_2}=1+\frac{p_1+p_2}{2p_1p_2}
$$
Now $p_1+p_2=-3$ and $p_1p_2=5$ so $M=\frac{7}{10}$
|
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|
Calculating the limit of a geometrical construction Suppose I have a rectangle ABCD, with AD larger than AB. I trace its diagonal AC, then the perpendicular to AC that goes through B. It intersects AC in $B_1$. I then trace the perpendicular to AD that goes through $B_1$, which intersects AD in $A_1$, and the perpendicular to CD that goes through $B_1$, which intersects CD in $C_1$.
We iterate the process to make $B_2$, $A_2$, $C_2$...
So on, so forth.
Clearly, $\lim_{n \to +\infty}$$C_n$ is D. But $\lim_{n \to +\infty}A_n$ is not D, and is instead somewhere on AD.
Question : Given the initial size AD and AB, with AD>AB, what's the length of $\lim_{x \to +\infty}A_n$D?
I think that if the initial lengths are $\{a,b\}$, the lengths on the next step are $$\left\{\frac{a^3}{a^2+b^2},\frac{b^3}{a^2+b^2}\right\},$$ but I have no idea where to go from there...
|
The lengths of the next step you obtained are correct.
Now consider the ratio between the initial height and width, i.e. write $b = ka, k>1$.
The new lengths are given by:
$$\frac {a^3}{a^2+ b^2} = \frac {a^3}{a^2 + k^2a^2} = \frac a{1+k^2}$$
$$\frac {b^3}{a^2+b^2} = \frac {k^3a^3}{a^2+b^2}$$
hence the new ratio of height and width is $k^3$. We can also write
$$\frac {b^3}{a^2 + b^2} = \frac {bk^2a^2}{a^2+k^2a^2} = \frac b{1+k^{-2}}$$
and so a recurrence relation becomes apparent. For the height, we are looking for:
$$\frac a{(1+k^2)(1+(k^3)^2)(1+((k^3)^3)^2)\dots} = \frac a{\prod_{n=0}^\infty (1+k^{2\cdot 3^n})}$$
Since $k > 1$, the infinite product diverges, so the height tends to zero as expected.
As for the width, the opposite occurs: we are looking for
$$\frac b{(1+k^{-2})(1+(k^3)^{-2})(1+((k^3)^3)^{-2})\dots} = \frac b{\prod_{n=0}^\infty (1+k^{-2\cdot 3^n})}$$
To show that the infinite product converges, we can check the equivalent series
$$\sum_{n=0}^\infty k^{-2\cdot 3^n} < \sum_{n=0}^\infty k^{-n} = \frac1{1-k^{-1}}$$
which converges as $0 < k^{-1} < 1$. It can also be expressed as the infinite sum:
$$\prod_{n=0}^\infty (1+k^{-2\cdot 3^n}) = 1 + k^{-2} + k^{-6} + k^{-8} + k^{-18} + \dots = \sum_{n \in S} k^{-n}$$
where $S$ is the set of all nonnegative integers that does not contain $1$ in their base $3$ representation (https://oeis.org/A005823).
A trivial upper bound is $\displaystyle \sum_{n = 0}^\infty k^{-2n} = \frac 1{1-k^{-2}} = \frac {k^2}{k^2-1}$, which shows that the limit of the width cannot be less than $\dfrac {b(k^2-1)}{k^2} = b\left(1-\dfrac1{k^2}\right)$. I am however unable to think of a closed form for the infinite product or the infinite sum.
|
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|
Determine whether the series $\sum_{n=1}^\infty \ln\left(\cos \frac{1}{n}\right)$ is convergent. Determine whether the series $\sum_{n=1}^\infty \ln\left(\cos \frac{1}{n}\right)$ is convergent.
Attempt:
Notice that the $p-$ series $\sum_{n=1}^\infty \frac{1}{n^2}$ with $p=2$ is convergent. Then, we have
\begin{align*}
\lim\limits_{n \to \infty} \frac{\ln\left(\cos \frac{1}{n} \right)}{\frac{1}{n^2}} &= -\frac{1}{2} \cdot \lim\limits_{n \to \infty} \frac{\sin \frac{1}{n}}{\frac{1}{n}} \cdot \frac{1}{\cos \frac{1}{n}} \\
&= -\frac{1}{2}.
\end{align*}
Hence, the limit is equal to $-\frac{1}{2} \ne 0$. Since the $p-$ series is convergent with $p=2$, by the limit comparison test, the series is convergent.
Am I true?
|
Alternative way using Taylor series:
$$\ln\left(\cos \frac{1}{n}\right) = \ln\left(1-O\left(\frac{1}{n^2}\right) \right) = O\left(\frac{1}{n^2}\right)$$
|
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|
Determine the value of K so the function have only one point of one horizontal point of tangency
Consider the polynomial function define by
$ f(x) = x^3 + x^2 + kx - 1 $
Determine the value of k so f has only on point of horizontal
tangency.
So I know that $ f^{'}(x) = 3x^2 + 2x + k $
And I should consider $ f^{'}(x) = 0 $ but how I find a k that makes the $ f^{'}(x) = 3x^2 + 2x + k $ have only one root?
I also thought that if $ k = -x^2 $ then the function would have only one point of horizontal tangency. But this does not works because this not equal the answer. ( Though I think it shouldwork too )
Btw the answer is $ \frac{1}{3} $ And I still can't figure why.
|
Continuing where you left off:
We know that the horizontal tangent line is found when $$f'(x) = 3x^2 + 2x + k =0$$
So we need to find the value of $k$ which only gives us $1$ root for this polynomial. Let's us use the quadratic formula to solve for the roots.
$$x=\frac{-2\pm \sqrt{2^2-4(3)(k)}}{2(3)}\\
x=\frac{-2\pm \sqrt{4-4(3)(k)}}{6}\\
x=\frac{-2\pm \sqrt{4(1-3k)}}{6}\\
x=\frac{-2\pm 2\sqrt{1-3k}}{6}\\
x=\frac{-1\pm\sqrt{1-3k}}{3}$$
Now, the only way that $x$ can have one solution is if $$\sqrt{1-3k}=0\\1-3k=0\\ 3k=1\\ \therefore k=\frac{1}{3}$$
Plugging $k$ back in to solve for our single root of $x$, $$x=\frac{-1\pm\sqrt{1-3\left(\frac{1}{3}\right)}}{3}\\
x=\frac{-1\pm\sqrt{1-1}}{3}\\
x=\frac{-1}{3}$$
The only way $x$ can have a single root is if $k=\frac{1}{3}$.
|
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|
Finding a specific polynomial function, with conditions.
Find a polynomial such that $f(x) = a + bx + cx^2 +dx^3 + ex^4$ such
that $f(1)=1, f(2) = -1, f(-1)=5, f(3) = -59, f(-2) = -29$
Any hints on how to approach this? I was thinking about plugging in the f value and then the result in order to build a matrix like this, but I am not sure if that is what I am supposed to do here.
$\displaystyle \begin{pmatrix}
a & b & c & d & e & 1\\
a & -b & -c & -d & -e & -1\\
a & -b & -c & -d & -e & 5\\
a & 3b & 3c & 3d & 3e & -59\\
a & -2b & -2c & -2d & -2e & -29
\end{pmatrix}$
does this make any sense?
|
$f(1)=1, f(2)=-1,f(-1)=5,f(3)=-59,f(-2)=-29$
Provided a linear $5\times 5$ sysytem of equation as $MV=U$
where $$M=\begin{pmatrix} 1 & 1 &1 &1 &1 \\ 1 & 2 & 4 & 8 & 16 \\ 1 & -1 & 1 &-1 & 1\\ 1 & 3 & 9 & 27 & 81 \\ 1 & -2 & 4 & -8 & 16 \end{pmatrix}, V=\begin{pmatrix} a \\ b\\ c\\ d\\ e \end{pmatrix}, U =\begin{pmatrix} 1 \\ 2 \\ -1 \\ 5 \\ -59 \\ -29 \end{pmatrix}.$$ Solving this equation by LinearSolve of Mathematica we get get $$V=\begin{pmatrix} 1 \\ -5 \\ 4 \\ 3 \\ -2 \end{pmatrix}$$
Finally, we get $f(x)=1-5x+4x^2+3x^3-2x^4.$
Mathe matica Command: LinearSolve[m,u]
|
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|
Calculate the Limit of Double Sum Compute
\begin{equation}
L=\lim _{n \rightarrow \infty}\frac{1}{n} \sum_{a=1}^n \sum_{b=1}^n \frac{a}{a^2+b^2 }.
\end{equation}
My attempt:
Define \begin{equation} f(n,m)= \frac{1}{n} \sum_{a=1}^n \frac{1}{m}\sum_{b=1}^m \frac{\frac{a}{n}}{(\frac{a}{n})^2+(\frac{b}{m})^2 } = \frac{1}{m}\sum_{b=1}^m \frac{1}{n} \sum_{a=1}^n \frac{\frac{a}{n}}{(\frac{a}{n})^2+(\frac{b}{m})^2 }
\end{equation}
Now for any $\epsilon >0$ there exists some $B>0$ such that for any $n,m \in \mathbb{N}$ and $n,m\geq B$:
\begin{equation}
|f(n,m)-f(n,n)|<\epsilon
\end{equation}
Thus, we have \begin{equation}
L=\lim _{n \rightarrow \infty} f(n,n)=\lim _{m \rightarrow \infty} \lim _{n \rightarrow \infty} f(n,m)= \lim _{m \rightarrow \infty} \frac{1}{m}\sum_{b=1}^m \int_{0}^{1}\frac{x}{x^2+(\frac{b}{m})^2}dx=\\ \frac{1}{2}\lim _{m \rightarrow \infty} \frac{1}{m}\sum_{b=1}^m \ln\frac{1+(\frac{b}{m})^2}{(\frac{b}{m})^2}=\frac{1}{2}\int_{0}^{1}\ln\frac{1+x^2}{x^2}dx=\frac{2\ln2 +\pi}{4}
\end{equation}
|
While this approach arrives at the correct value for the limit it is not easily justified.
At first glance,
$$f(n,n) = \frac{1}{n}\sum_{a=1}^n\sum_{b=1}^n\frac{a}{a^2 + b^2} = \frac{1}{n^2}\sum_{a=1}^n\sum_{b=1}^n\frac{a/n}{(a/n)^2 + (b/n)^2} $$
is a Riemann sum for the integral of $g:(x,y) \mapsto \frac{x}{x^2+ y^2}$ over $[0,1]\times[0,1]$. However, the integrand is unbounded around $(0,0)$ and is not Riemann integrable. Instead it must be considered as an improper integral which can be evaluated as an iterated integral
$$I = \int_0^1 \left(\int_0^1 \frac{x}{x^2 + y^2} \, dy\right )dx = \frac{2 \log 2 + \pi}{4}$$
Note that the inner integral is properly Riemann and we can write
$$I = \int_0^1 \lim_{m \to \infty} \left(\frac{1}{m}\sum_{b=1}^m \frac{x}{x^2 + (b/m)^2} \right)\, dx$$
At this point it would be nice to switch the integral and the limit and apply a second Riemann sum to obtain
$$I = \lim_{m \to \infty} \frac{1}{m}\sum_{b=1}^m \int_0^1\frac{x}{x^2 + (b/m)^2} \, dx = \lim_{m \to \infty} \left(\frac{1}{m}\lim_{n \to \infty} \frac{1}{n}\sum_{a=1}^n\sum_{b=1}^m \frac{a/n}{(a/n)^2 + (b/m)^2}\right)$$
However justifying that switch is not obvious. For example, we see immediately that
$$\left|\frac{1}{m}\sum_{b=1}^m \frac{x}{x^2 + (b/m)^2} \right| \leqslant \frac{1}{m}\sum_{b=1}^m \frac{x}{x^2 } = \frac{1}{x},$$
but the dominating function $x \mapsto 1/x$ is not integrable on $[0,1]$.
Furthermore, there remains finding the justification for $\lim_{n\to \infty} f(n,n) = \lim_{m\to \infty} \lim_{n\to \infty} f(n,m)$.
A easily justified approach is to apply the Stolz-Cesaro theorem to get
$$\lim_{n \to \infty} f(n,n) = \lim_{n \to \infty} \frac{\sum_{a=1}^{n+1}\sum_{b=1}^{n+1}\frac{a}{a^2 + b^2} - \sum_{a=1}^n\sum_{b=1}^n\frac{a}{a^2 + b^2} }{n+1 - n} \\ = \lim_{n \to \infty} \left(\frac{n+1}{(n+1)^2 + (n+1)^2}+ \sum_{a=1}^n \frac{a}{a^2 + (n+1)^2} + \sum_{b=1}^n \frac{n+1}{(n+1)^2 + b^2} \right) \\ = \lim_{n\to \infty} \sum_{a=1}^{n} \frac{a + n+1}{a^2 + (n+1)^2} = \lim_{n\to \infty} \frac{1}{n+1}\sum_{a=1}^{n} \frac{1 + \frac{a}{n+1}}{1 + \left(\frac{a}{n+1}\right)^2}\\ = \int_0^1 \frac{1+x}{1+x^2} \, dx = \frac{2 \log 2 + \pi}{4}$$
|
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|
Sketch the curve $\frac{ax + b}{cx + d}$ I am working through a pure maths book out of interest. I am stuck on this problem:
If $a$, $b$, $c$ and $d$ are positive, sketch the curve $y = \frac{ax + b}{cx + d}$ when
i) $ad - bc \lt 0$
ii) $ad - bc = 0$
iii) $ad - bc \gt 0$
I can work out the basics. Thus
$x$-intercept: $f(x) = 0 \implies \frac{ax + b}{cx + d} = 0 \implies x = \frac{-b}{a}$
so intercept is $[\frac{-b}{a}, 0]$
$y$-intercept: $f(0) = \frac{b}{d}$ at point $[0,\frac{b}{d}]$
horizontal asymptote: $\frac{a}{c}$
vertical asymptote: $cx + d = 0 \implies x = \frac{-d}{c}$
I can also see that $\frac{ax + b}{cx + d} = \frac{a}{c}\left(1 + \frac{\frac{b}{a}-\frac{d}{c}}{x + \frac{d}{c}}\right)$
but I cannot put all this together to see how one could sketch the graphs except in one case: where $ad -bc = 0$ because then $\frac{b}{a}-\frac{d}{c} = 0$ and the graph will be a horizontal line equal to $\frac{a}{c}$
I think not knowing the actual magnitudes makes it difficult for me to visualise.
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From your last observation, $$\frac{ax+b}{cx+d} =\frac 1c \left( a +\frac{bc-ad}{cx+d} \right)$$
You can think of this as a series of transformations applied to the graph of $y=\frac 1x$.
Starting from it, first shift to the left by $d$. That gives $\frac {1}{x+d}$. Then squish parallel to the $x$ axis by a factor of $c$. This gives $\frac{1}{cx+d}$.
Now, scale along the $y$ axis by a factor of $bc-ad$ (here, the sign comes into play: if $bc-ad\lt 0$, you have to flip the graph over). Then, move the graph up by $a$ units and finally, squish down by a factor of $c$.
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|
Unusual Constant appearing for definite integral I’m learning integration and have come across a question in a textbook. I arrived to the same answer as the textbook itself, but I know it is not correct. It starts as part a) show that $$\frac{d}{dx} tan^{-1} \left(\frac{3tan(x)}{2} \right)= \frac{6}{5sin^2(x)+4}$$ Hence, find area bounded by curve from $ \dfrac{1}{5sin^2(x) + 4}$ from $x=0$ to $x=7$. Clearly using part A to integrate and substituting 7 and 0, I arrive to 0.153 square units, in line with the textbook. Yet, by using desmos to graph the curve, using lower rectangles, minimum area has to be $0.111111 \times 7 = 0.77777$ units squared. I put the integral on integral calculator and it yielded $0.153 + \dfrac{\pi}{3}$, without explaining where this constant came from. I am unfamiliar with constants in definite integrals, and struggled to find anything online to answer this. Thanks.
(Note: I am not asking for homework help, Im just asking about how to find constants in these definite integrals, particularly in a question like this, which the textbook itself overlooked. Thank you. Any help is appreciated.)
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This happens because $\tan x$ is not defined at the points $x = \frac{\pi}{2}, \frac{3\pi}{2} $ which lie in $(0,7)$ so $\tan^{-1} \left(\frac{3\tan(x)}{2} \right)$ becomes discontinuous at these points.
This is why you have to split the interval into sub-intervals $(0,\frac{\pi}{2}), (\frac{\pi}{2}, \frac{3\pi}{2}), (\frac{3\pi}{2},7) $ while substituting the limits.
$$ \Delta = \dfrac16 \left[ \tan^{-1} \left(\frac{3\tan(x)}{2} \right)_0^\frac{\pi^-}{2} + \tan^{-1} \left(\frac{3\tan(x)}{2} \right)_\frac{\pi^+}{2}^\frac{3\pi^-}{2} + \tan^{-1} \left(\frac{3\tan(x)}{2} \right)_\frac{3\pi^+}{2}^7 \right] $$
$$ \Delta = \dfrac16 \left[ \dfrac{\pi}{2} - 0+ \dfrac{\pi}{2} - \left(- \dfrac{\pi}{2} \right) + \tan^{-1}\left(\frac{3\tan7}{2} \right) - \left(- \dfrac{\pi}{2} \right) \right] $$
$$ \Delta = \dfrac{\pi}{3}+\dfrac16 \tan^{-1}\left(\frac{3\tan7}{2} \right) $$
|
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Calculating limit of a square root function, in terms of N and epsilon I've been trying to solve this question:
Prove directly from the definition of a limit, in terms of $N$ and $\varepsilon$:
$$\lim_{n\to\infty} \sqrt{\frac{4n+1}{n}} =2$$
I'm not supposed to use any known properties and use the definition directly.
So, in order to prove this statement, we need to prove that for every $\varepsilon > 0$, there is a natural $N$, which for every $n>N$, the equation $\left|\sqrt{\frac{4n+1}{n}} - 2\right| < \varepsilon$
And since $\sqrt {\frac{4n+1}{n}} = \sqrt {4+\frac{1}{n}} > 2$ for every $n$, we can claim:
$$\sqrt{4+\frac{1}{n}} - 2 < \varepsilon$$
$$=\sqrt{4+\frac{1}{n}} - \sqrt 4 < \varepsilon$$
$$= \frac{\left(\sqrt{4+\frac{1}{n}} - \sqrt4\right)\left(\sqrt{4+\frac{1}{n}} + \sqrt4\right)}{\left(\sqrt{4+\frac{1}{n}} + \sqrt4\right)}< \varepsilon$$
$$=\frac{4+\frac{1}{n}-4}{\sqrt{4+\frac{1}{n}}+\sqrt4} < \varepsilon$$
$$= \frac{1}{n} \cdot \frac{1}{\sqrt{4+\frac{1}{n}}+\sqrt4}$$
I'm not sure where to go on from here in order to find an $N$. I'd appreciate some help, thank you!
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Alternative approach
$\sqrt{4 + 1/n} < 2 + \epsilon = \sqrt{4 + 4\epsilon + \epsilon^2}.$
Therefore, you have to choose $n$ such that
$(4 + 1/n) < (4 + 4\epsilon + \epsilon^2)$.
This means that $(1/n) < 4\epsilon + \epsilon^2.$
This means that $n > \frac{1}{\epsilon} \times
\frac{1}{4 + \epsilon}.$
Since $\frac{1}{4 + \epsilon} < 1$, you can simply specify:
$N > \frac{1}{\epsilon} > \frac{1}{\epsilon} \times
\frac{1}{4 + \epsilon}.$
Then, any positive integer $n > N$ will also be satisfactory.
|
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|
Representing all rational numbers between $\dfrac{1}{2}$ and $1$ How do I show that
$$\dfrac{2^{\left\lfloor\frac12 a_1\right\rfloor} + 2^{\left\lfloor\frac12 a_2\right\rfloor} + \ldots + 2^{\left\lfloor\frac12 a_n \right\rfloor}}{2^{\left\lceil\frac12 a_1\right\rceil} + 2^{\left\lceil\frac12 a_2\right\rceil} + \ldots + 2^{\left\lceil\frac12 a_n \right\rceil}}$$
represents all rationals between $\dfrac{1}{2}$ and $1$, where $a_1, a_2, \ldots, a_n \in \mathbb{N}$?
I have tried representing few rationals such as $\dfrac{7}{8}$ and $\dfrac{5}{8}$ and it seems like we can represent all of them. However I'm not completely sure why and don't have a rigorous proof.
For example
$$\dfrac{5}{8} = \dfrac{2^{\left\lfloor\frac12 3\right\rfloor} + 2^{\left\lfloor\frac12 4\right\rfloor} + 2^{\left\lfloor\frac12 5 \right\rfloor}}{2^{\left\lceil\frac12 3\right\rceil} + 2^{\left\lceil\frac12 4\right\rceil} + 2^{\left\lceil\frac12 5 \right\rceil}}$$
$$\dfrac{7}{8} = \dfrac{2^{\left\lfloor\frac12 1\right\rfloor} + 2^{\left\lfloor\frac12 2\right\rfloor} + 2^{\left\lfloor\frac12 4 \right\rfloor}}{2^{\left\lceil\frac12 1\right\rceil} + 2^{\left\lceil\frac12 2\right\rceil} + 2^{\left\lceil\frac12 4 \right\rceil}}$$
Any help would be highly appreciated. Thanks
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There is an argument using the Farey tree. Briefly, the Farey algorithm fills in rational numbers between $x = \frac{p}{q}$ and $y = \frac{r}{s}$ by
$\frac{p+r}{q+s}$ (and so on recursively) [What some have called the Bart Simpson addition of fractions). The basic fact is that if you start with $\frac12$ and $1,$ the next number you generate is $\frac{1+1}{1+2} = \frac23,$ the next two are $\frac12 + \frac23$ and $\frac23 +1$ (where + is the Bart Simpson sum), this procedure will generate all the rational numbers in the interval $[\frac12, 1].$
It is not hard to see that numbers as in the OP behave nicely with respect to the Bart Simpson addition: if $x$ and $y$ both have an $a_i,$ then their Bart Simpson sum has $a_i+2.$ It also needs to be checked that $\frac12$ and $1$ can be represented, but that is easy. So, to summarize, the bart simpson sum of two numbers having the required representation also has the required representation, which gives both a proof and an algorithm to construct the representation.
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General term for series $0,2,12,70, 408...$ Given the curves $y=\sqrt{2x^2+1}$ and $x=\sqrt{2y^2+1}$, the first line (line with smallest y coordinate) which intersects the curves with integer coordinates is $x+y=1.$ This line intersects the curves at $(0,1)$ and $(1,0)$. Along with this it intersects the curves at $(-2,3)$ and $(3,-2)$. If we use the larger integer coordinates and construct another line $x+y = 2+3=5$ and intersect it with the curves we can find the next integer coordinates which are $(2,3)$ and $(3,2)$, but along with this we find another two points of intersection $(-12,17)$ and $(12,-17).$ If we repeat this process and find the intersection of the curves with $x+y=12+17=29$ we can find the next two integer coordinates: $70,99$. If we continue this process we can iteratively find the next integer coordinates.
So the question at hand is to find the $n^{th}$ largest integer intersecting points on these curves or in other words... the $n^{th}$ largest point of the curve $y=\sqrt{2x^2+1}$ whose both coordinates are integers. I believe by reiterating this line process we can find all integer coordinates, i.e. there are no integer coordinates that can't be found by reiterating this process. We won't skip any. (Feel free to disprove this, this is just my intuition.) Anyways, the main question to generate the $n^{th}$ term in this series, i.e. finding a general term, one the doesn't require redrawing the next line to find the next point.
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I would select the smaller absolute value with each pair of coordinates, thus $0,2,12,70,408,...$.
Suppose you have a line $x+y=a_n+b_n$ which intersects $y^2-2x^2=1$ at $(a_n,b_n)$ and thus intersects $x^2-2y^2=1$ at $(b_n,a_n)$. The other intersection with $y^2-2x^2=1$ is determined by
$(a_n+b_n-x)^2-2x^2=1$
$-x^2+2(a_n+b_n)x+(a_n+b_n)^2-1=0$
The roots of this equation add up to $2(a_n+b_n)$ and by hypothesis one of them is $a_n$, so the other root defined as the next $x$ coordinate on the $y^2-2x^2=1$ hyperbola will be
$a_{n+1}=a_n+2b_n$ Eq. 1
Similarly we can seek the next $x$ coordinate on $x^2-2y^2=1$. That quadratic equation will be
be
$x^2-2(a_n+b_n-x)^2=1$
$-x^2+4(a_n+b_n)x-2(a_n+b_n)^2-1=0$
With one root at $b_n$ and the two roots adding up to $4(a_n+b_n)$ we identify
$b_{n+1}=4a_n+3b_n$ Eq. 2
To get a recursion containing only $a_n$ terms, first increment Eq. 1 by one term to get
$a_{n+2}=a_{n+1}+2b_{n+1}$ Eq. 3
Then solve Eq. 3 for $b_{n+1}$ and Eq. 1 for $b_n$, and substitute these inte Eq. 2. You end up with
$\color{blue}{a_{n+2}=6a_{n+1}-a_n}$
Using your initial $x$ coordinates on $y^2-2x^2+1$ you can put in the initial conditions $a_0=1,a_1=2$ and get the sequence
$0,2,12,70,408,2378,...$
matching your subsequent $x$ coordinates on $y^2-2x^2=1$.
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|
Rolling a die twice in either order with not mutual exclusive events I am reading the following problem:
If a single die is rolled twice, find the probability of rolling an
odd number and a number greater than $4$ in any either order
My solution:
Probability of rolling an odd number = $$\frac{3}{6}$$
Probability of rolling a number greater than $4$ = $$\frac{2}{6}$$
Since the events are not mutually exclusive ($5$ is counted twice) the probabilty of throwing an odd number followed by a number greater than $4$:
$$\frac{3}{6}\cdot \frac{2}{6} - \frac{1}{36} = \frac{5}{36}$$
The probablity of throwing a number greater than $4$ followed by an odd number is: $$\frac{2}{6}\cdot \frac{3}{6} - \frac{1}{36} = \frac{5}{36}$$
Therefore the my final solution is $$\frac{5}{36} + \frac{5}{36} = \frac{10}{36}$$
But it is wrong, as the solution states that it should be $$\frac{11}{36}$$
What am I doing wrong here?
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The outcome potentially counted twice is $(5,5)$, not $5$.
The number of pairs $(a,b)$ where $a$ is odd and $b>4$ is $3\times2=6$. This includes $(5,5)$.
The number of pairs $(a,b)$ where $a>4$ and $b$ is odd is $2\times3=6$. This includes $(5,5)$.
The sum of these is $6+6=12$, but $(5,5)$ was counted twice; there are actually only $11$ pairs in the union of these two sets.
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Trigonometric integral $\int^\pi_0\frac{d\theta}{1+a^2\sin^2\theta}$ The function
$f(x_1,x_2)=\frac{e^{-\sqrt{x^2_1 + x^2_2}}}{\sqrt{x^2_1+x^2_2}}$, $(x_1,x_2)\neq(0,0)$ is integrable in $\mathbb{R}^2\setminus\{(0,0\})$ as $f>0$ and
\begin{align}
\int_{\mathbb{R}^2}f(x_1,x_2)\,dx_1dx_2=2\pi\int^\infty_0e^{-r}\,dr
\end{align}
I am interested in its Fourier transform. Being radial, we have that
\begin{align}
\widehat{f}(t_1,t_2)&=\widehat{f}\big(0,\sqrt{t^2_1+t^2_2}\big)=\int_{(0,\infty)\times(-\pi,\pi)}e^{-r}e^{-2\pi i r\sqrt{t^2_1+t^2_2}\sin\theta}\,drd\theta\\
&=\int^\pi_{-\pi} \frac{d\theta}{1+ 2\pi i \sqrt{t^2_1+t^2_2}\sin\theta}=\int^\pi_{-\pi} \frac{1-2\pi i \sqrt{t^2_1+t^2_2}\sin\theta}{1+ 4\pi^2 (t^2_1+t^2_2)\sin^2\theta}\,d\theta\\
&=2\int^\pi_0\frac{d\theta}{1+ 4\pi^2 (t^2_1+t^2_2)\sin^2\theta}
\end{align}
Question:
At this point, the last integral escapes me. I would like to see if there is either a more direct way to estimate the Fourier transform of $f$, or whether the integral
\begin{align}
\int^\pi_0\frac{d\theta}{1+a^2\sin^2\theta}\tag{1}\label{one}
\end{align}
can be further evaluated.
Edit: As pointed out by @Koro, it turns out that \eqref{one} is rather trivial:
\begin{align}
\int^\pi_0\frac{d\theta}{1+a^2\sin^2\theta}&=2\int^{\pi/2}_0\frac{d\theta}{1+a^2-a^2\cos^2\theta}=2\int^{\pi/2}_0\frac{\sec^2\theta}{\sec^2\theta(1 +a^2)-a^2}\,d\theta\\
&=2\int^{\pi/2}_0\frac{\sec^2\theta}{1+(1+a^2)\tan^2\theta}\,d\theta
\stackrel{u=\tan\theta}{=}2\int^{\infty}_0\frac{du}{1+(1+a^2)u^2}\\
&=\frac{\pi}{\sqrt{1+a^2}}
\end{align}
Hence, the Fourier transform of $f$ is
$$\widehat{f}(t_1,t_2)=\frac{2\pi}{\sqrt{1+4\pi^2(t^2_1+t^2_2)}}$$
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Let $I=\int^\pi_0\frac{d\theta}{1+a^2\sin^2\theta}=2\int_0^{\pi/2} \frac{d\theta}{1+a^2\sin^2\theta}$ Noting that $\csc^2\theta=1+\cot^2\theta$, divide numerator and denominator by $\sin^2\theta$ to get:
$\frac I2=\int_0^{\pi/2} \frac{\csc^2\theta d\theta}{1+\cot^2\theta+a^2}$. Now substitue $t=\cot \theta$ so that $dt=-\csc^2\theta d\theta$ and change the limits accordingly to get
$\frac I2=\int_0^\infty\frac{dt}{t^2+(1+a^2)}$
Can you take it from here?
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|
Show that there is a simultaneously tangent line to both the curves $y = e^x$ and $y = \ln x$ I found the derivatives of both of the curves but I'm having trouble on how to move on from here:
$y = e^x\implies y' = e^x$
$y = \ln x\implies y' = \frac{1}{x}$
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Given \begin{align} y = e^x\implies y' = e^x$, $y = \ln x\implies y' = \frac{1}{x}\end{align}
Tangent line to $y = e^x$ at x=a
\begin{align} y=e^ax+e^a(1-a)\end{align}
Tangent line to $y=lnx$ at x=b\begin{align} y=\frac{1}{b}x+\ln b-1\end{align}
When the two lines become one line:
\begin{align} e^a=\frac{1}{b}, e^a(1-a)=\ln b-1\end{align}
Solve the above equations about a and b, \begin{align}b \approx 0.124, a\approx ln(\frac{1}{0.214}) \\b\approx 4.681, a\approx ln( \frac{1}{4.681}) \end{align}
The common tangent lines \begin{align} y= \frac{1}{0.214}x+\ln(0.214)-1 \\y= \frac{1}{4.681}x+\ln(4.681)-1 \end{align}
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The equation ${4 \choose k}=6$ Find the solution of the equation $${4 \choose k}=6.$$
So $k$ must be a natural number $(n\in \mathbb{N})$ and we can find that when $k=1 \rightarrow {4\choose 1}=4$ and when $k=2\rightarrow {4\choose2}=6$, so the solution of the equation is actually $k=2$. Can we solve the problem without calculating the exact value of ${4\choose k}$? I tried: $$\dfrac{V_4^k}{k!}=6$$ but don't see how to solve this. Thank you in advance!
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You can write $n\choose{k}$ = $\frac{n!}{(n-k)!k!}$. Substituting $n = 4$ into there, we get $\frac{24}{(4-k)!k!}$ = 6. Multiplying both sides by $(4-k)!k!$, we get $24 = 6(4-k)!(k!)$. Dividing both sides by $6$, we get $4 = (4 - k)!(k!)$. Realizing $4 = 2^2$, we can assume that $k! = 2$ and $4 - k = 2$.
The only value of $k$ such that $k! = 2$ is $2$.
Hence, $\boxed{k = 2}$
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How should I check that $6^{\log_{10} x} + 8^{\log_{10} x} = x $ does not have other solutions? The question is to solve the equation $6^{\log_{10} x} + 8^{\log_{10} x} = x $
I know one of the solutions is $x=100$ using Pythagorean triples but I can't show that this is the only solution.
I'm looking for an idea without using differentiation!
any hints would be appreciated. tnx
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Solving for $x$ in $6^{\log_{10} x} + 8^{\log_{10} x} = x$
is equivalent to solving for $y$ in $6^y + 8^y = 10^y$ where $y = \log_{10} x.$
I have an easier time visualizing this equation with $y$ than the one with $x,$ so I would prefer to solve for $y$.
You can also write this $6^{y-2} 6^2 + 8^{y-2} 8^2 = 10^{y-2} 10^2.$
You know you have a solution for $y = 2,$ because $6^2 + 8^2 = 10^2.$
This means you can also write other statements that are always true, such as
$6^{y-2} 6^2 + 6^{y-2} 8^2 = 6^{y-2} 10^2$
or $8^{y-2} 6^2 + 8^{y-2} 8^2 = 8^{y-2} 10^2.$
For $y > 2,$ you have $6^{y-2} < 8^{y-2} < 10^{y-2}.$
For $y < 2,$ you have $6^{y-2} > 8^{y-2} > 10^{y-2}.$
Now you just need to find a way to compare $6^{y-2} 6^2 + 8^{y-2} 8^2$
to $10^{y-2} 10^2$ in each of those two cases.
It might even help to make a further substitution, such as $t = y - 2,$
so now you're trying to solve for $t$ in $6^t 6^2 + 8^t 8^2 = 10^t 10^2,$
you know $t=0$ is a solution, and you want to show there are no solutions for
$t > 0$ or for $t < 0.$ Just like it is easier to think about $6^y$ than
$6^{\log_{10} x}$ (at least for me; your mileage may vary),
it may be easier to think about $6^t$ than $6^{y-2}.$
|
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|
Integral including an Incomplete Gamma function Does anyone have an idea of solving following integral
$$ I =\int_{0}^{\infty}\frac{x\Gamma \left(a,b x\right)}{(x^2+1)}\,dx\,,$$
where $a,b>0$ are positive real values?
Mathematica gives an answer, but I do not know how it can be derived by using integral tables.
\begin{align}
I=&\frac{1}{{2 a (a+1)}}\biggl(\pi (a+1) b^a \csc \left(\frac{\pi a}{2}\right) \, _1F_2\left(\frac{a}{2};\frac{1}{2},\frac{a}{2}+1;-\frac{b^2}{4}\right) \\ &+a b \left((a+1) b \Gamma (a-2) \, _2F_3\left(1,1;2,\frac{3}{2}-\frac{a}{2},2-\frac{a}{2};-\frac{b^2}{4}\right)-\pi b^a \sec \left(\frac{\pi a}{2}\right) \, _1F_2\left(\frac{a}{2}+\frac{1}{2};\frac{3}{2},\frac{a}{2}+\frac{3}{2};-\frac{b^2}{4}\right)\right)+2 \Gamma (a+2) (\psi ^{(0)}(a)-\log (b))\biggr)
\end{align}
Eq. 2.10.1.3 in Integrals and Series [Vol 2 - Spl Functions] - A. Prudnikov would be one of the closet answer, but I could not match parameters well.
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Assuming that $a$ is a positive integer and $b>0$, considering
$$I_a =\int_{0}^{\infty}\frac{x}{x^2+1}\,\Gamma \left(a,b x\right)\,dx$$ there is only one which is simple
$$I_1=\frac{\pi}{2} \sin (b)-(\text{Ci}(b) \cos (b)+\text{Si}(b) \sin (b))$$ where appear the since and cosine integral functions.
The problem is that, for $a>1$, it seems that the result can only be expressed in terms of the Meijer G function.
The case of $a=2$ is
$$I_2=\frac {1}{\sqrt \pi}\,G_{1,3}^{3,1}\left(\frac{b^2}{4}|
\begin{array}{c}
0 \\
0,0,\frac{3}{2}
\end{array}
\right)$$
For $a>2$, we have
$$I_a=\frac {2^{a-2}}{\sqrt \pi}\, G_{2,4}^{4,1}\left(\frac{b^2}{4}|
\begin{array}{c}
0,1 \\
0,0,\frac a 2,\frac{a+1}{2}
\end{array}
\right)$$
Edit
As written in comments, for rational values of $a$ we have quite nasty results. As an example, for $a=\frac 32$, the result given by a CAS is
$$2I_{\frac 32}=-\sqrt{\pi } \left(2 b^2 \,
_2F_3\left(1,1;\frac{3}{4},\frac{5}{4},2;-\frac{b^2}{4}\right)+\log (4b)+\gamma
-2\right)+\pi ^{3/2} \left(C\left( \sqrt{\frac{2b}{\pi
}}\right)-S\left( \sqrt{\frac{2b}{\pi }}\right)\right)+ \pi \sqrt{2b}
(\sin (b)-\cos (b))$$
|
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|
Exact Diff. Equation $ \frac{d y}{d x}=\frac{x y^{2}-\cos x \sin x}{y\left(1-x^{2}\right)} $ Solve:$$
\frac{d y}{d x}=\frac{x y^{2}-\cos x \sin x}{y\left(1-x^{2}\right)}.$$
$${(xy^2 -\cos x\sin x)dx+y(1-x^2)dy=0}.$$
So the first term, $(xy^2 -\cos x\sin x)dx$ is $=M(x,y)$ and the second, $ y(1-x^{2} )dy$ term is $=N(x,y)$.
If $
\frac {\partial {M}}{\partial y}=-2xy$ ; $\frac {\partial N}{\partial x}=-2xy$ there is an exact solution
$$ f(x,y)=\int M(x,y)dx +g(x) = \frac{x^2y^2}{2}+\frac{cos^2x}{2}+g(x)$$
$$\frac{\partial f}{\partial y}=N(x,y)=y(1-x^2)$$
$$f(x,y)=\frac{y^2}{2}(1-x^2)+g(x)$$
$$\frac{\partial f}{\partial x}=-xy^2+g'(x)=-xy^2-\cos x\sin x; g'(x)=-\cos x \sin x$$
$$g(x)= \int -\cos x\sin x dx =-\frac{cos^2x}{2} $$
$$f(x,y) = -\frac{cos^2x}{2} + \frac{y^2(1-x^2)}{2}$$
Am I using the terms properly and is my solution correct?
|
$$\frac{d y}{d x}=\frac{x y^{2}-\cos x \sin x}{y\left(1-x^{2}\right)}.$$
$$y(1-x^2){d y}=x y^{2}dx-\cos x \sin xdx$$
$$dy^2-2x^2yd y-2xy^2dx=2\cos x d\cos x$$
$$dy^2-(x^2d y^2+y^2dx^2)=2\cos x d\cos x$$
$$y^2-x^2 y^2-\cos^2 x =C$$
Your solution looks correct to me.
|
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|
If $a$,$b$ and $y$ are the roots of $3x^3+8x^2-1=0$ find $(b+1/y)(y+1/a)(a+1/b)$ If a b and y are the roots of $3x^3+8x^2-1=0$ find $(b+1/y)(y+1/a)(a+1/b)$
This is what I have done so far, but apparently it is incorrect. I want to know why.
$(b+1/y)(y+1/a)(a+1/b)$
$(by+1/y)(ay+1/a)(ab+1/b)$
$(aby^2+1/ay)(ab+1/b)$
which is equal to
$a^2b^2y^2 + 1/aby$
Using Vieta's formula I get:
$aby = -d/a$
$aby = 1/3$
Subbing in original:
$(1/9 + 1)/1/3$
which is...
10/27
The answer is supposedly 2/3, want to know what I did wrong.
|
You have a few mistakes:
*
*$(b+\frac1y)(y+\frac1a)(a+\frac1b)\neq a^2b^2y^2+\frac1{aby}$, as you write above
*$(b+\frac1y)(y+\frac1a)(a+\frac1b)\neq \frac{a^2b^2y^2+1}{aby}$, as you apply the formula you wrote
The actual solution is:
$$(b+\frac1y)(y+\frac1a)(a+\frac1b)=\frac1{aby}(yb+1)(ay+1)(ab+1)=\frac1{aby}(aby^2+ay+yb+1)(ab+1)=\frac1{aby}(a^2b^2y^2+a^2by+ab^2y+aby^2+ab+ay+yb+1)=aby+a+b+y+\frac{ab+ay+yb}{aby}+\frac1{aby}$$
Now, using Vieta's Formulas, we have that $aby=\frac{1}{3},ab+ay+yb=0,a+b+y=-\frac{8}{3}$. This gives $(b+\frac1y)(y+\frac1a)(a+\frac1b)=\frac{1}{3}-\frac{8}{3}+0+3=\frac23$
|
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|
Prove $\iint_{x^{2}+y^{2} \leq 1} e^{x^{2}+y^{2}} d x d y \leq\left[\int_{-\frac{\sqrt{\pi}}{2}}^{\frac{\sqrt{\pi}}{2}} e^{x^{2}} d x\right]^{2} $ The formula we need to prove is
$$
\displaystyle \iint_{x^{2}+y^{2} \leq 1} e^{x^{2}+y^{2}} d x d y \leq\left[\int_{-\frac{\sqrt{\pi}}{2}}^{\frac{\sqrt{\pi}}{2}} e^{x^{2}} d x\right]^{2}
$$
As I already done the left side
$$
\iint_{x^{2}+y^{2} \leq 1} e^{x^{2}+y^{2}} d x d y = \int_{-\pi}^{\pi} \mathrm{d}\varphi \int_{0}^{1} e^{r^2}r\mathrm{d}r = 2\pi \int_{0}^{1} e^{r^2}r\mathrm{d}r = \left.\pi e^{r^2}\right|_{0}^{1} = \pi(e-1)
$$
I still have no idea about the right side.
|
Integral on the left is the integration over the $\color{red}{red}$ circle, and the integral on the right is the integration over the $\color{blue}{blue}$ square (see the picture below).
We note that the area of the circle is $S_\circ = \pi\cdot 1^2 = \pi$ and the area of the square is $S_\square = (\sqrt \pi)^2 = \pi$. This means the $\color{green}{green}$ and $\color{orange}{orange}$ areas are equal too.
However, note that $$e^{\color{green}{x^2+y^2}} \le e^{\color{orange}{x^2+y^2}} $$ since $\color{green}{x^2+y^2} \le 1$ and $\color{orange}{x^2+y^2} \ge 1$ and therefore
$$\begin{align}
\iint\limits_{x^2+y^2\le 1} e^{x^2+y^2} dxdy
&\le \iint\limits_{\text{square}} e^{x^2+y^2}dxdy \\
&= \int_{-\sqrt\pi/2}^{\sqrt\pi/2}e^{x^2}dx\int_{-\sqrt\pi/2}^{\sqrt\pi/2}e^{y^2}dy \\
&= \left(\int_{-\sqrt\pi/2}^{\sqrt\pi/2}e^{x^2}dx\right)^2
\end{align}$$
$\qquad\qquad\qquad$
|
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|
How to expand this expression in series? From Poisson's first Memoire sur la distribution d'electricite (1812), $\S$39. Suppose $b$ is very small relative to $c-a$ where $a,b,c$ are positive real scalars. Consider the equations
\begin{align*}
af(x)+\frac{b^2}{c-ax} F\left(\frac{b}{c-ax}\right)&=h,\\
bF(x)+\frac{a^2}{c-bx}f\left(\frac{a}{c-bx}\right)&=g;
\end{align*}
for arbitrary functions $f(x)$ and $F(x)$ (representing the "number of electric molecules" on a surface in the original memoir), for $x\in[-1,1]$. The variables $g$ and $h$ are arbitrary constants. With the assumption that $b$ is very smaller relative to $c-a$, we may neglect the cubes of $\frac{b}{c}$.
Given the above, Poisson says that the expansion of $bF(x)$ (presumably about $x=\frac{a}{c}$) takes the form:
\begin{align*}
bF&(x)=g-\frac{a^2}{c}f\left(\frac{a}{c}\right)-\left[\frac{a^2}{c}f\left(\frac{a}{c}\right)+\frac{a^3}{c^2}f'\left(\frac{a}{c}\right)\right]\frac{bx}{c}\\
&-\left[\frac{a^2}{c}f\left(\frac{a}{c}\right)+\frac{2a^3}{c^2} f'\left(\frac{a}{c}\right)+\frac{a^4}{2c^3}f''\left(\frac{a}{c}\right)\right]\frac{b^2x}{c^2};
\end{align*}
Some notes before I go on: Poisson used the notation $fx$ for function $f$, which I have assumed is intended to represent the form $f(x)$. There may be other results related to the above series expansion earlier (some time after $\S$15) but I haven't found anything.
Now, the presence of derivatives brings Taylor series to mind, but expanding about $x=\frac{c}{a}$ gives
$$ bF(x) \approx g-\frac{a^2}{c-bx}\left[ f\left(\frac{c}{a}\right)+f'\left(\frac{c}{a}\right) \left(x-\frac{c}{a}\right)+\frac{1}{2} f''\left(\frac{c}{a}\right) \left(x-\frac{c}{a}\right)^2 \right] $$
Which is far different.
Question: How do you go from the above equations to the expansion for $bF(x)$ given?
For context, I'm writing a translation of the memoir (which is available here). The closest to useful result I've found is this equation:
$$
af\left(\frac{x}{a}\right)-\frac{a^2b}{c^2-b^2-cx}f\left(\frac{ac-ax}{c^2-b^2-cx}\right)=h-\frac{gb}{c-x}\quad (3)
$$
An equation which Poisson says must be integrated, in general, to solve for $f(x)$, which can then be used to solve $F(x)$. But it's not clear still what is happening above.
|
The expansion is around $x=0$.
Then $$bF(x)=g-\frac{a^2}{c-bx}f\left(\frac a{c-bx}\right)$$
Expanding in Taylor series at $x=0$, the constant term is $$g-\frac{a^2}cf\left(\frac ac\right)$$
The linear term is given by the first derivative:
$$\left(g-\frac{a^2}{c-bx}f\left(\frac a{c-bx}\right)\right)'=\frac{a^2}{(c-bx)^2}(-b)f\left(\frac a{c-bx}\right)-\frac{a^2}{c-bx}f'\left(\frac a{c-bx}\right)\left(\frac a{c-bx}\right)'\\=-\frac{a^2b}{(c-bx)^2}f\left(\frac a{c-bx}\right)-\frac{a^3b}{(c-bx)^3}f'\left(\frac a{c-bx}\right)$$
Taking the value at $x=0$ you get $$\left(g-\frac{a^2}{c-bx}f\left(\frac a{c-bx}\right)\right)'_{x=0}=-\frac{a^2b}{c^2}f\left(\frac a{c}\right)-\frac{a^3b}{c^3}f'\left(\frac a{c}\right)$$
You can do the same for the next derivative. Up to the first order, $$bF(x)=g-\frac{a^2}cf\left(\frac ac\right)-\left[\frac{a^2b}{c^2}f\left(\frac a{c}\right)+\frac{a^3b}{c^3}f'\left(\frac a{c}\right)\right]x\\=g-\frac{a^2}cf\left(\frac ac\right)-\left[\frac{a^2}{c}f\left(\frac a{c}\right)+\frac{a^3}{c^2}f'\left(\frac a{c}\right)\right]\frac{bx}c$$
|
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|
Extraneous solutions without performing squaring operation
Solve in $[0, 2\pi]$ $$\sec x+\tan x=2\cos x$$
My mind boggled while solving the trigonometric equation in two different ways:
Method $1.$
Assuming $x\ne \frac{\pi}{2},\frac{3\pi}{2}$
We have:
$$\sec x+\tan x=2\cos x$$
$\implies$
$$\sec x-\tan x=\frac{1}{2\cos x}$$
Adding both we get
$$2\sec x=2\cos x+\frac{1}{2\cos x}$$
$\implies$
$$\frac{3}{2\cos x}=2\cos x$$
$\implies$
$$\cos x=\frac{\pm \sqrt{3}}{2}$$
which gives four solutions in $[0,2\pi]$ which are $\frac{\pi}{6},\frac{11\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6}$ but only two of them which are $\frac{\pi}{6}, \frac{5\pi}{6}$ satisfying the original equation.
Method $2.$
Writing the given equation as
$$\frac{1}{\cos x}+\frac{\sin x}{\cos x}=2\cos x $$
Then we get
$$1+\sin x=2(1-\sin^2 x)$$
$\implies$
$$2\sin^2 x+\sin x-1=0$$
$\implies$
$$\sin x=-1, \sin x=\frac{1}{2}$$
Since $x\ne \frac{\pi}{2},\frac{3\pi}{2}$ we can ignore $\sin x=-1$. Thus we have
$\sin x=\frac{1}{2}$ which gives the two solutions $\frac{\pi}{6}, \frac{5\pi}{6}$
My question is why am i getting two extraneous solutions in Method $1$ even though they are in the domain of $\sec x$ and $\tan x$?
|
Hint
Subtract the two equation to find $$2\tan x=2\cos x-\dfrac1{2\cos x}$$
As $\cos x\ne0,$
multiply out both sides by $2\cos x,$ to find $$4\sin x=4\cos^2x-1=4(1-\sin^2x)-1$$
Your solution needs to satisfy this equation as well
|
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|
Modulus function: finding set of values. I didn't learn this in my syllabus but it is in some past papers. I need a small explanation for how to find the possible values for p to give 0, 1 or 2 roots.
Here is an example question: $f(x)=6-|x+2|$ with roots (-8,0) and (4,0)
Find the set of values for p for which the equation $f(x)=px+5$ has 0 solutions or 1 solution or 2 solutions.
I tried to equate both versions of the modulus to the second definition of the function but it didn't really get me anywhere.
|
I think what you mean it to find the values for which $6-|x+2| = px + 5$ have $0, 1$ or $2$ solutions. The solutions, not roots, are when $6-|x+2|$ and $px+5$ both equal the same number which might not and probably will not be $0$. $6-|x+2|$ has two roots; $x=4$ and $x = -8$ and $px +5$ will have one root if $p\ne 0$ and that is $x =-\frac 5p$. But we are not looking for when they are both $0$ (which can only happen if $-\frac 5p =-8$ or if $-\frac 5p = 4$) but when they are both equal to .... something.
If $6-|x+2| = px + 5$ then we have two possible cases and two equations to solve.
*
*$x+2 \ge 0$. That is $x\ge -2$. Then
$6 -(x+2) = px + 5$
$4 -x = px + 5$
$-1 = px + x= x(p+1)$
$x = -\frac 1{p+1}$. But this can only be a solution if a) $p+1\ne 0$ that is $p \ne -1$ and b) $x \ge -2$. And if $x =-\frac 1{p+1} \ge -2$ then
$\frac 1{p+1} \le 2$
If $p< -1$ then $\frac 1{p+1} < 0 < 2$ so this will always be true.
But if $p > -1$ then we need $0< \frac 1{p+1} < 2$ or $1 < 2(p+1)$ so $p>-\frac 12$.
$x =-\frac 1{p+1}$ will be one solution if $p < -1$ or $p>-\frac 12$.
*$x+2 < 0$. That is $x < -2$. then
$6 -(-(x+2)) = px + 5$
$6 + x+2 = px+5$
$8 + x= px + 5$
$3 = px -x= x(p-1)$
$x = \frac 3{p-1}$
But this can only be a solution if a) $p-1\ne 0$ or $p =1$. and $b) x = \frac 3{p+1} < -2$.
If $p > 1$ then $\frac 3{p-1} > 0 > -2$ and this can never happen.
If $p < 1$ then we have $\frac 3{p-1} < 0$ and we need
$\frac 3{p-1} < -2$
$3 > -2(p-1)= -2p +2$
$1 > -2p$
$p > -\frac 12$.
$x = \frac 3{p-1}$ will be one solution if $-\frac 12 < p < 1$.
Putting those two results together:
if $p < -1$ then $x=-\frac 1{p+1}$ will be the only solution.
if $-1 \le p \le -\frac12$ then there will be no solutions.
if $-\frac 12 < p < 1$ there there will be two solutions. $x =-\frac 1{p+1}$ and $x = \frac 3{p-1}$.
if $x \ge 1$ then $x=-\frac 1{p+1}$ will be the only solution.
|
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|
Prove that $(n+1)^n < n^{n+1}$ for all $n>3$
Prove that $(n+1)^n < n^{n+1}$ for all $n>3$
At $n=4$, $$5^4<4^5$$ which is indeed true.
By mathematical induction, we need to prove that $$(n+2)^{n+1} < (n+1)^{n+2}$$
$$\implies (n+2)^n\times (n+2) < (n+1)^n\times(n+1)^2 $$
I am not getting how to proceed further than this. Any hints or help would be highly appreciated! Thanks.
|
Here is a non-inductive proof that involves a nice application of the AM-GM inequality.
It suffices for us to prove the following inequality:
\begin{align}
\forall \ & n>3, \left(1+\dfrac{1}{n}\right)^n <4 \\
& \iff \left(\dfrac{n+1}{n}\right)^n <4 \\
& \iff \left(\dfrac{n}{n+1}\right)^n >\dfrac{1}{4} \\
& \iff \dfrac{n}{n+1} > \sqrt[^n]{\dfrac{1}{4}}
\end{align}
But
\begin{align}
\sqrt[^n]{\dfrac{1}{4}} &= \sqrt[^n]{\dfrac{1}{2} \cdot \dfrac{1}{2} \cdot 1^{n-2}} \\
& \leq \dfrac{\dfrac{1}{2}+ \dfrac{1}{2}+n-2}{n} (\text{By the AM-GM Inequality}) \\
& = \dfrac{n-1}{n}.
\end{align}
It remains to show that $\dfrac{n-1}{n} < \dfrac{n}{n+1}$, which is obvious since $n^2-1 < n^2$.
|
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|
Computing the generator and error polynomials of a $\left[15, 7, 5\right]$ BCH code My question: Let $\alpha$ be a root of $1+x+x^4 \in \text{F}_{2}\left[x\right]$ and let C be a narrow-sense BCH code with length 15, and designed distance 5.
Find the generator polynomial of $C$ and determine the error position of $100000111000000$.
My work so far:
If $\alpha$ is a root of $1+x+x^4 \in \text{F}_{2}\left[x\right]$ and $C$ is a narrow-sense BCH code with parameters with length 15 and designed distance 5, then a generator polynomial is
$g\left(x\right)=\text{lcm}\left(1+x+x^4\right)\left(1+x+x^2+x^3+x^4\right)=1+x^4+x^6+x^7+x^8$
and we also noted that this code will be able to correct 2 errors, from the equation for designed distance $d=2\delta +1$, and because the degree of $g\left(x\right)$ is $8$, we will have a $\left[15, 7, 5\right]$ narrow-sense BCH code.
Now in order to decode the received word $w\left(x\right)=\left(w_{0}+w_{1}x+...+w_{n-1}x^{n-1}\right)=100000111000000$, first we check
$w\left(x\right)H^{T}=0$, where $H$ is the parity check matrix given by
\begin{bmatrix}1 & \alpha & \alpha^{2} &. . . & \alpha^{14} \\1 & \alpha^{3} & \alpha^{9}&...&\alpha^{12} \end{bmatrix}
and if $w\left(x\right)H^{T}=0$ we conclude that there is no error. If $w\left(x\right)H^{T} \neq 0$, next we calculate the syndromes $\left(s_{0}, s_{1}, ..., s_{\delta-2}\right)=\left(w_{0}, w_{1}, ..., w_{n-1}\right)H^{T}$, where $s_{i}=e\left(\alpha^{i+1}\right)$ is the error locator polynomial.
Because we have a two error pattern, we will need to find the error polynomial to decode $c\left(x\right)=w\left(x\right)-e\left(x\right)$.
We do this in the following way.
First we consider the system of equations
\begin{align*}
\alpha_{1} + \alpha_{2} &= \,s_{1} \\
\alpha_{1}^{3}+\alpha_{2}^{3} &= \, s_{2} \\
\end{align*}
and noting $$\alpha_{1}^{3}+\alpha_{2}^{3}=\underbrace{\left(\alpha_{1}+\alpha_{2}\right)}_{s_{1}} \underbrace{\left(\alpha_{1}^{2}+\alpha_{1}\alpha_{2}+\alpha_{2}^{2}\right)}_{s^{2}_{1}+\alpha_{1}\alpha_{2}}$$
Then, rewriting the two equations for $\alpha_{1}$ and $\alpha_{2}$,
\begin{align*}
\alpha_{1} + \alpha_{2} &= \,s_{1} \\
\alpha_{1}^{3} \cdot \alpha_{2}^{3} &= \, s_{2}s_{1}^{-1}+s_{1}^{2}=\frac{s_{2}+s_{1}^{3}}{s_{1}} \\
\end{align*}
and finally we construct a quadratic equation whose roots are $\alpha_{1}$ and $\alpha_{2}$,
$\left(x-\alpha_{1}\right)\left(x-\alpha_{2}\right)=x^2-\left(\alpha_{1}+\alpha_{2}\right)+\alpha_{1}\alpha_{2}=x^2+s_{1}x+\frac{s_{2}+s_{1}^{3}}{s_{1}}$.
What is giving me trouble:
So after all of this my question is really where to go from here. Do I simply solve the quadratic equations for the roots of $x^2+s_{1}x+\frac{s_{2}+s_{1}^{3}}{s_{1}}$, or is there another detail I am missing? As always, thanks for the help.
|
After taking the comments above into consideration, I believe I have found the solution.
So, using the transpose of $H$,
$$H^{T} = \begin{bmatrix}
1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 0 & 0 & 1 & 1 \\
0 & 0 & 0 & 1 & 0 & 1 & 0 & 1 \\
1 & 1 & 0 & 0 & 1 & 1 & 1 & 1 \\
0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 1 & 0 & 0 & 0 & 1 \\
1 & 1 & 0 & 1 & 0 & 0 & 1 & 1 \\
1 & 0 & 1 & 0 & 0 & 1 & 0 & 1 \\
0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 0 & 1 & 0 & 0 & 0 \\
0 & 1 & 1 & 1 & 0 & 0 & 0 & 1 \\
1 & 1 & 1 & 1 & 0 & 0 & 1 & 1 \\
1 & 0 & 1 & 1 & 0 & 1 & 0 & 1 \\
1 & 0 & 0 & 1 & 1 & 1 & 1 & 1
\end{bmatrix} $$,
and setting $w\left(x\right)=\begin{bmatrix}
1 & 0 & 0 & 0 &0 &0& 1& 1& 1& 0& 0& 0& 0& 0& 0\end{bmatrix}$
then $$w\left(x\right)H^{T}=\begin{bmatrix}
1 & 1 & 0 & 0 & 1 & 1 & 1 & 1
\end{bmatrix}^T=\begin{bmatrix}
1\\
1\\
0\\
0\\
1\\
1\\
1\\
1 \end{bmatrix}=\begin{bmatrix}
\alpha^{4}\\
\alpha^{12}\\
\end{bmatrix}=\begin{bmatrix}
s_{1}\\
s_{2}\\
\end{bmatrix}$$
Now solving for $x^2+s_{1}x+\frac{s_{2}+s_{1}^{3}}{{s_{1}}}=0$,
$$x^2+\alpha^{4}x+\frac{\alpha^{12}+\left(\alpha^{4}\right)^{3}}{\alpha^{4}}=0$$
$$\implies x^2+\alpha^{4}+2\alpha^{8}=x^2+\alpha^{4}$$
is the error check polynomial.
We can check this with $$s\left(x\right)=\left(x-\alpha^{2}\right)\left(x-\alpha^{2}\right)=x^2+\alpha^4$$
So taking this into consideration we can conclude that the errors were in the fourth and twelfth positions.
|
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How to solve $x^4-2x^3-x^2+2x+1=0$? How to solve $x^4-2x^3-x^2+2x+1=0$?
Answer given is: $$\frac{1+\sqrt5}{2}$$
I tried solving it by taking common factors:
$$x^3(x-2)-x(x-2)+1=0 $$ $$x(x-2)(x^2-1)+1=0 $$ $$(x+1)(x)(x-1)(x-2)+1=0$$
But it's not leading me anywhere.
|
obten la raiz negativa de la siguiente funcion con 4 cifras decimales.
$2x^4-2x^3+x^2+3x-4=0 \quad\quad$ newton raphson
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integrating $\int \sqrt{2me-mkr^2-\frac{1}{2}m br^4 - \frac{a^2}{r^2}} \,dr$ I was trying to find Hamiltons principle function, $S$, for the Hamiltonian:
$$ H = \frac{1}{2} m \left( P_{r}^2 + \frac{P_{\theta}^2}{r^2} \right) + \frac{1}{2}kr^2 + \frac{1}{4} b r^4$$
After considering the form $S = f(r, \theta) - Et = R(r) + \Theta(\theta) - Et$ and identifying the canonical momenta with the partial derivatives of $S$ I got to:
$$ \left(\frac{dR}{dr} \right)^2 r^2 + mkr^4 + \frac{1}{2}mbr^6 - 2mEr^2 = -\left(\frac{d \Theta}{d \theta} \right)^2 $$
Each side must be a constant, $-a^2$, and so:
$$ \frac{dR}{dr} = \sqrt{2mE-mkr^2-\frac{1}{2}m br^4 - \frac{a^2}{r^2}} $$
which leads to:
$$ R(r)= \int \sqrt{2mE-mkr^2-\frac{1}{2}m br^4 - \frac{a^2}{r^2}} \,dr$$
I've tried to find a substitution of variables to make it simpler but to no luck. How can I solve this?
|
$$\sqrt{2mE-mkr^2-\frac12mbr^4-\frac{a^2}{r^2}}\,dr=\frac1r\sqrt{2mEr^2-mkr^4-\frac12mbr^6-a^2}\,dr$$
Now with the substitution $u=r^2\Rightarrow dr=\frac{du}{2r}$ so:
$$\frac{1}{2u}\sqrt{2mEu-mku^2-\frac12mbu^3-a^2}du$$
so we have reduced the order of the polynomial inside the square root. However, integrals of this form do not have "nice" closed-form solutions and so you are probably going to end up with a special function/series to represent this
|
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|
Evaluating $\int_0^B \frac{1}{A^2 + (B^2-x^2)^2}\,dx$ I am trying to solve the following integral:
$$\int_0^B \frac{1}{A^2 + (B^2-x^2)^2}\,dx$$
However the polynomial have imaginary roots:
$$-(-i A + B^2)^{1/2},\; (-i A + B^2)^{1/2},\; -(i A + B^2)^{1/2},\;(i A + B^2)^{1/2}$$
I am only interested on the solution from $0$ to $B$. How should I proceed?
|
Using the variable change $x\to Bx$, one can immediately reduce to the case $B=1$. If $A$ is real, we have:
\begin{align*}
\int_0^1\frac1{A^2+(1-x^2)^2}dx&=\frac1{2Ai}\int_0^1\left[\frac1{(x^2-1)-Ai}-\frac1{(x^2-1)+Ai}\right]dx\\
&=\frac1A\Im\left(\int_0^1\frac1{(x^2-1)-Ai}dx\right)\\
&=\frac1A\Im\left(\frac1{2\sqrt{1+Ai}}\int_0^1\left[\frac1{x-\sqrt{1+Ai}}-\frac1{x+\sqrt{1+Ai}}\right]dx\right),
\end{align*}
where $\Im$ denotes the imaginary part. This then just reduces to the sum of a few logarithms.
Alternatively, from the second line we may immediately obtain:
\begin{align*}
\frac1A\Im\left(\int_0^1\frac1{(x^2-1)-Ai}dx\right)&=\frac1A\Im\left(\frac1{\sqrt{1+Ai}}\arctan\left(\frac 1{\sqrt{1+Ai}}\right)\right).\\
\end{align*}
|
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|
Does the following converge to 0? Does the following converge to $0$ for $\theta>-1/2$?
$$\lim_{n\rightarrow\infty}\frac{\sum_{i=1}^ni^{3\theta}}{\left(\sum_{i=1}^ni^{2\theta}\right)^{3/2}}$$
I'd like to use the comparison test but no idea what to compare it to. If $\theta=1$, based on code I wrote it seems to be going to 0 but would like to know how to use math.
|
Rearranging terms we have that
$$\frac{\sum_{i=1}^ni^{3\theta}}{\left(\sum_{i=1}^ni^{2\theta}\right)^{\frac{3}{2}}}\cdot\frac{\frac{1}{n^{3\theta}}}{\frac{1}{n^{3\theta}}}\cdot\frac{\frac{1}{n^{\frac{3}{2}}}}{\frac{1}{n^{\frac{3}{2}}}} = \frac{\sum_{i=1}^n\left(\frac{i}{n}\right)^{3\theta}\frac{1}{n}}{\left(\sum_{i=1}^n\left(\frac{i}{n}\right)^{2\theta}\frac{1}{n}\right)^{\frac{3}{2}}}\cdot\frac{1}{\sqrt{n}}$$
The terms on the left converge to
$$\frac{\sum_{i=1}^n\left(\frac{i}{n}\right)^{3\theta}\frac{1}{n}}{\left(\sum_{i=1}^n\left(\frac{i}{n}\right)^{2\theta}\frac{1}{n}\right)^{\frac{3}{2}}} \longrightarrow\frac{\int_0^1x^{3\theta}dx}{\left(\int_0^1x^{2\theta}dx\right)^{\frac{3}{2}}} = \frac{(1+2\theta)^{\frac{3}{2}}}{1+3\theta}$$
if and only if $\theta>-\frac{1}{3}$, not just $-\frac{1}{2}$. Thus the overall sequence converges to $0$ in that case. If $-\frac{1}{2}<\theta\leq-\frac{1}{3}$, the denominator Riemann sum converges, and the numerator terms
$$N=\frac{1}{\sqrt{n}}\sum_{i=1}^n\left(\frac{i}{n}\right)^{3\theta}\frac{1}{n} = \frac{1}{\sqrt{n}}\sum_{i=1}^n\sqrt{\frac{n}{i}}\cdot\left(\frac{i}{n}\right)^{3\theta+\frac{1}{2}}\frac{1}{n}$$
$$\implies \frac{1}{\sqrt{n}}\sum_{i=1}^n\left(\frac{i}{n}\right)^{3\theta+\frac{1}{2}}\frac{1}{n} < N < \sum_{i=1}^n\left(\frac{i}{n}\right)^{3\theta+\frac{1}{2}}\frac{1}{n}$$
$$\longrightarrow 0 < N < \int_0^1x^{3\theta+\frac{1}{2}}dx = \frac{2}{6\theta+3}$$
by the same logic above. Thus the entire expression will always converge when $\theta > -\frac{1}{2}$, and only possibly to something nonzero when $-\frac{1}{2} < \theta \leq -\frac{1}{3}$.
|
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|
Which of the following is not the correct quadratic equation? If $3/2$ and $4$ are the two roots of a quadratic equation, then which one of the following is not the correct quadratic equation?
(A) $2x^2-11x+6=0$,
(B) $6x^2-33x+18=0$,
(C) $-10x^2-55x-30=0$,
(D) $4x^2-41x+24=0$.
I tried using the analogy $(x-3/2)(x-4)$ but it came to be of no use. My textbook tells me to interpret this as $k(x-3/2)(x-4)$. The equation comes out to have a constant ($c$ term) $+12$. However the options tell a different story altogether. A slight advice about the same will be most welcome.
|
Your textbook is suggesting a uselessly complicated method.
The sum of the roots is $11/2$ and their product is $6$. Dividing by the leading coefficient should make the equation in the form
$$
x^2-\dfrac{11}{2}x+6=0
$$
and the form of this equation is unique. Now we have
(A) $x^2-\dfrac{11}{2}x+3=0$ (divide by $2$)
(B) $x^2-\dfrac{11}{2}x+3=0$ (divide by $6$)
(C) $x^2-\dfrac{11}{2}x+3=0$ (divide by $-10$)
(D) $x^2-\dfrac{41}{4}x+6=0$ (divide by $4$)
So the answer is that none of the proposed equations has roots $3/2$ and $4$.
|
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|
Here why is it wrong to differentiate both sides and put $x=2$ to find $g'(2)$? Here why is it wrong to differentiate both sides and put $x=2$ to find $g'(2)$?
If $\displaystyle I = \int \frac{x-1}{(x+1)\sqrt{x^3+x^2+x}} \,\mathrm dx = g(x) + c$, then $\left\lfloor \dfrac{1}{g'(2)} \right\rfloor = \cdots$,
(where $\lfloor \cdot \rfloor$ represents the greatest integer function)
To me it seemed like it's an identity and so it should be safe to differentiate both sides.
|
Your approach is correct.The formula given clearly indicates that $g(x)$ is the indefinite integral of the function:
$$ \frac{x-1}{(x+1)\sqrt{x^3+x^2+x}} $$
So differentiating $g(x)$ gives us back that function of $x$, and evaluating it at $x=2$ after differentiation is correct.
We should get $g'(2) = \frac{1}{3\sqrt{14}}$, and the greatest integer function applied to the reciprocal gives us $\lfloor 3\sqrt{14} \rfloor = 11$.
Addendum :
$$\begin{align}
11=\sqrt{121}&<\sqrt{126}=\sqrt{2\cdot3^2\cdot7}=3\sqrt{14}=\\
&=\sqrt{126}<\sqrt{144}=12
\end{align}$$
$\text{hence ,}$
$$11<3\sqrt{14}<12\;.$$
|
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|
Integrating $\int_{0}^{\infty}\frac{\cos(x)}{(1+x^2)^2}dx$ without the residue theorem? Consider the integral
$$\int_{0}^{\infty}\frac{\cos(x)}{(1+x^2)^2}dx$$
Are there any other ways to compute this integral besides using the residue theorem?
Edit: Thank you all, kind people, for the answers. I had this on an exam and solved it with the residue theorem and was just wondering what other ways are possible and which of those do not involve complex integration.
|
Let us define integral with parameter:
\begin{align}
I(t) = \int_0 ^\infty \frac{\cos(xt)}{(1+x^2)^2}dt.
\end{align}
Taking the derivative of I:
\begin{align}
\frac{dI}{dt} = \frac{d}{dt} \int_0 ^\infty \frac{\cos(xt)}{(1+x^2)^2}dt
=\int_0 ^\infty \frac{\partial}{\partial t} \frac{\cos(xt)}{(1+x^2)^2}dt= - \int_0^\infty \frac{x\sin(xt)}{(1+x^2)^2}dt.
\end{align}
Taking the second derivative:
\begin{align}
\frac{d^2I}{dt^2} = -\int_0^\infty \frac{x^2\cos(xt)}{(1+x^2)^2}dt =
-\int_0^\infty \frac{(1+x^2-1)\cos(xt)}{(1+x^2)^2}dt =
-\int_0^\infty \frac{\cos(xt)}{(1+x^2)}dt + \int_0^\infty \frac{\cos(xt)}{(1+x^2)^2}dt
\end{align}
From here we get:
\begin{align}
I’’(t) = -\int_0^\infty \frac{\cos(tx)}{1+x^2} dt + I(t)
\end{align}
It is well known that(you can derive it similar to this metdod):
\begin{equation}
\int_0^\infty \frac{\cos(tx)}{1+x^2} dt = \frac{\pi}{2} e^{-t}
\end{equation}
So we get differential equation:
\begin{align}
I’’(t) - I(t) = -\frac{\pi}{2} e^{-t}
\end{align}
Th general solution is of the form is:
\begin{equation}
I(t) = Ae^{-t} + Be^{t}
\end{equation}
And paticular solution will be of the form $I(t) = Cte^{-t}$ and the we calculate $I’’(t) - I(t)$ to get $C = \frac{\pi}{4}$.
So final form is :
\begin{align}
I(t) = Ae^{-t} + Be^t + \frac{\pi}{4}e^{-t} t
\end{align}
To get $A$ and $B$ we calculate $I(0)$ and $I’(0)$:
\begin{align}
I(0) = \int_0^\infty \frac{1}{(1+x^2)^2} dt
\end{align}
U-sub: $u=\tan{\theta}, du = \frac{1}{\cos^2{\theta}} d\theta$
We get:
\begin{equation}
\int_0^{\frac{\pi}{2}} \frac{1}{(1+\tan^2{\theta})^2} \frac{1}{\cos^2(\theta)} dt = \int_0^{\frac{\pi}{2}} \cos^2(\theta) d\theta = \frac{\pi}{4}
\end{equation}
It is obvious that $I’(0) = 0$. After pluging in $I(t)$ and calculating $A$ and $B$ we get:
\begin{equation}
I(t) = \frac{\pi}{4}(xe^{-x} + e^{-x}).
\end{equation}
Our answer is : $I(1) = \frac{\pi}{4}(2e^{-1}) = \frac{\pi}{2e}.$
|
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|
Summation over $a+b+c=5$ Let $a,b,c$ be positive integers. Compute
$$\sum_{a+b+c=5} (6-a)(6-b)(6-c).$$
The first thing I notice is symmetry, so that I can permute $3!=6$ ways, but i'm not really sure how that works with the condition $a+b+c=10.$ The other method is to fix $a$, but that is reall time-consuming and unfeasible if say $a+b+c=20.$ Is there a clever method to evaluate this sum?
I would like to have a generalized method, please.
|
By using generating functions, we show that for any non-negative integer $n$,
$$\begin{align}\sum_{a+b+c=n} (6-a)(6-b)(6-c)
&=[x^n]\left(\sum_{k=1}^{\infty}(6-k)x^k\right)^3\\
&=[x^n]\frac{(x(5-6x))^3}{(1-x)^6}\\
&=[x^n]\frac{125x^3-450x^4+540x^5-216x^6}{(1-x)^6}\\
&=125\binom{n+2}{5}-450\binom{n+1}{5}+540\binom{n}{5}-216\binom{n-1}{5}\\
&=-\frac{(n-2)(n-1)(n^3-87n^2+2072n-12960)}{120}.
\end{align}$$
Therefore for $n=5,10,20$ we obtain $465$, $-36$, $-4788$ respectively.
|
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|
Study of $\sum_{n=1}^{\infty}n^\alpha(\sinh{\frac{1}{n}}-\log{(n+1)}+\log{n}+\cos{\frac{1}{n}}-1)$ For $\alpha\in \mathbb{R}$ I want to study the behaviour of
$$\sum_{n=1}^{\infty}n^\alpha(\sinh{\frac{1}{n}}-\log{(n+1)}+\log{n}+\cos{\frac{1}{n}}-1)$$
I have thought to apply the asymptotic test but in order to be able to apply this I have to check if this is a series of positive term, so: how can I prove this?
Now:
$1)\sinh\frac{1}{n}=\frac{1}{n}+\frac{1}{6n^3}+o(\frac{1}{n^4})$
$2) \cos{\frac{1}{n}}=1-\frac{1}{2n^2}+o(\frac{1}{n^3})$
$3)\log({\frac{n}{n+1}})=-\log(1+{\frac{1}{n}})=-\frac{1}{n}+\frac{1}{2n^2}-\frac{1}{3n^3}+o(\frac{1}{n^3})$
And so:
$$n^\alpha(\sinh{\frac{1}{n}}-\log{(n+1)}+\log{n}+\cos{\frac{1}{n}}-1)\\ =n^\alpha(\frac{1}{n}+\frac{1}{6n^3}+o(\frac{1}{n^4})-\frac{1}{n}+\frac{1}{2n^2}-\frac{1}{3n^3}+o(\frac{1}{n^3})-\frac{1}{2n^2}+o(\frac{1}{n^3}))\\ =n^\alpha(\frac{-1}{6n^3}+o(\frac{1}{n^3}))=-\frac{1}{6n^{3-\alpha}}+o(\frac{1}{n^{3-\alpha}})$$
So the series converges when $3-\alpha<1$ and diverges when $3-\alpha\geq 1$.
Am I right?
For the first question I can say as suggested in an answer that $n^\alpha(\sinh{\frac{1}{n}}-\log{(n+1)}+\log{n}+\cos{\frac{1}{n}}-1)=-\frac{1}{6n^{3-\alpha}}+o(\frac{1}{n^{3-\alpha}})<0$, so the general term of the series is eventually negative. Am I right?
|
We have $$
n^\alpha\left(\frac{1}{n^2+n}-\frac{2n+1}{2(n^2(n+1))}+o\left(\frac{1}{n^2}\right)\right)=\frac{n^\alpha}{n(n+1)}\left(-\frac1{2n}+o\left(\frac{1}{n^2}\right)\right)<0$$ for $n$ sufficiently large. It's just as good to have all the terms negative as positive, of course, so you can proceed.
Perhaps I ought to add that I haven't checked your calculations in detail.
|
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Calculate $\lim_{x\rightarrow0}\frac{(e^{\sin x}+ \sin x)^{\frac{1}{\sin x}}-(e^{\tan x}+ \tan x)^{\frac{1}{\tan x}}}{x^3}$ How to calculate the following limit?
$$\lim_{x\rightarrow0}\frac{(e^{\sin x}+ \sin x)^{\frac{1}{\sin x}}-(e^{\tan x}+ \tan x)^{\frac{1}{\tan x}}}{x^3}$$
I thought of L'Hopital's rule, Taylor expansion, and limit the form of $e^x$, but the presence of $\sin x$ and $\tan x$ make it hard to apply them. Could anyone give me a hint?
|
Hint 1: Consider the Taylor series of $f(\sin(x)) - f(\tan(x))$ for a general function $f$. What information do you need about $(e^x + x)^{1/x}$ to solve the problem?
Hint 2:
The information can be gotten from the Taylor series for $\ln(e^x + x)/x$, which is easier to find.
Full solution:
You can verify by substitution of Taylor series that if we have a function $f(x) = \sum_{i=0}^\infty a_i x^i$, then $f(\sin x) - f(\tan x) = -f'(0) x^3/2 + O(x^4)$. So we simply need to find the derivative of $f(x) = (e^x + x)^{1/x}$ at $0$. Since we only need the first derivative, it'll be easier to look at $\ln(f(x)) = \ln(e^x + x)/x$ instead. The numerator can be easily expanded in power series, and we get $$\frac{\ln(e^x + x)}{x} = \frac{1}{x}\left(2x - \frac{3}{2} x^2 + \frac{11}{6} x^3 + O(x^4)\right) = 2 - \frac{3}{2}x + \frac{11}{6}x^2 + O(x^3)$$ So we have $\ln(f(0)) = 2$ and $\ln(f(x))'|_0 = -3/2$. Using $\ln(f(x))' = f'(x)/f(x)$, we have $f'(0) = -3e^2/2$. Thus, \begin{multline}\lim_{x\rightarrow 0} \frac{(e^{\sin x} - \sin x)^{1/\sin x} - (e^{\tan x} - \tan x)^{1/\tan x}}{x^3} \\= \lim_{x\rightarrow 0}\frac{1}{x^3}\left[\left(-\frac{1}{2}\right)\left(-\frac{3e^2}{2}\right)x^3 + O(x^4)\right) = \frac{3e^2}{4} \end{multline}
|
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|
The sum of three positive integers is $20$. Find the probability that they form the sides of a triangle.
The sum of three positive integers is $20$. Find the probability that they form the sides of a triangle.
Let $a,b,c$ be there positive integers. So, $a+b+c=20$
Total number of solutions would be $^{19}C_2=171$
For $a,b,c$ to form a triangle $0\lt a,b,c\le9$
So, number of such solutions= coefficient of $x^{20}$ in $(x+x^2+x^3+...+x^9)^3=$ coefficient of $x^{17}$ in $(1+x+x^2+...+x^8)^3=$ coefficient of $x^{17}$ in $(1-x^9)^3(1-x)^{-3}=1\times^{19}C_2-3\times^{10}C_2=36$
So, the required probability is $\frac{36}{171}$ but the answer given is $\frac8{33}$
In the hint, they have written total combinations of $a,b,c=\frac{144}{6}+\frac{27}{3}=33$, and favorable combinations of $a,b,c=\frac{12}{3}+\frac{24}{6}=8$
I think they have split $171$ as $144+27$ and $36$ as $24+12$ but why? and why to divide them with $6$ or $3$?
|
Because in the case of a triangle's sides, order has no meaning. The set $\{7,8,5\}$, representing $a,b$ and $c$ forms the same triangle as $\{8,5,7\}$ and likewise. So, if $a,b,c$ are distinct, the $6$ possible ordered pairs give only $1$ combination, hence division by $6$. If two of $a,b,c$ are equal, then $3$ different ordered pairs are possible, which give only one combination, hence division by $3$.
|
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|
I'm facing a problem in differentiating an infinite square root function. How can I differentiate this function?
$y$ = $\displaystyle\frac{x}{x+\frac{\sqrt{x}}{x+\frac{\sqrt{x}}{x+....}}}$
I think I should break this function into an implicit function:
$\displaystyle\frac{y}{\sqrt{x}}=\frac{\sqrt{x}}{x+\frac{y}{\sqrt{x}}}$
this is the equation I get and was able to differentiate it but all the options are in $x$'s. When I differentiate the above function I got this
$\displaystyle\frac{dy}{dx}=-\frac{y(1+2\sqrt{x})}{\sqrt{x}(\sqrt{x}+x)}$
but since the options are all in $x$, I didn't get any solution.
The solution to the derivative of the above function is $\displaystyle\frac{1}{2x\sqrt{x}}$
|
You have to first solve for the function and then differentiate. You already know that
$\frac{y}{\sqrt{x}} = \frac{\sqrt{x}}{x + \frac{y}{\sqrt{x}}}$
so let $z = \frac{y}{\sqrt{x}}$, then we have
$z = \frac{\sqrt{x}}{x + z}$
Then $z^2 + xz = \sqrt{x}$. By the quadratic formula, $z = \frac{-x \pm \sqrt{x^2 + 4 \sqrt{x}}}{2}$. We know that $y$ is positive, hence $z$ is positive, hence $z = \frac{-x + \sqrt{x^2 + 4 \sqrt{x}}}{2}$. Then $y = \sqrt{x} \frac{-x + \sqrt{x^2 + 4 \sqrt{x}}}{2}$.
From here, it's a straightforward-ish derivative. But the answer you were provided is incorrect.
|
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|
Find b to make the three numbers in AP For a positive number '$b$', the value of '$b$' for which the numbers $3^x + 3^{-x} , b ,9^x+ 9^{-x}$ are in A.P. can be : (A) 1 (B) 2 (C) 3 (D) 5
Hello! This question has multiple correct answers and belongs to the topic Arithmetic Progression(Sequence and series). I was able to get the answer b=2. As we know that any number of the form $y + y^{-1}$ can take all values in the set = $(- ∞,-2]∪[2,∞)$. By taking $3^x + 3^{-x} = 2$, I got a constant AP = $2,2,2$. The answers are option (B),(C) and (D). Please help me reach the full solution.
|
\begin{align}
\frac{3^x+3^{-x}+9^x + 9^{-x}}{2} &= \frac{3^x+3^{-x}+(3^x+3^{-x})^2-2}{2}\\
&= -1 + \frac{(3^x+3^{-x})+(3^x+3^{-x})^2}{2}
\end{align}
We know that the range of $3^{x}+3^{-x}$ is $[2, \infty)$
Hence the range of $(3^x+3^{-x})+(3^x+3^{-x})^2$ is $[6, \infty)$.
Hence $b$ can attain any number at least $\frac{6}{2}-1=2$.
|
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Finding expected total number of die rolls Question
Ann and Bob take turns to roll a fair six-sided die. The game ends after a six or three consecutive fives come up, with the winner being the last person who threw the die. Ann will go first.
$(a)\quad$ Find the probability that Ann will win.
$(b)\quad$ Find the expected total number of rolls.
My working
Let $(A, 0)$ denote the state in which it is $A$'s turn and the prior toss was not a $5$ or a $6$, $(A, 5)$ the state in which it is $A$'s turn and the prior toss was a $5$ and $(A, 55)$ the state in which it is $A$'s turn and the prior two tosses were a $5$. The states for $B$ are defined similarly and note also that $(A, 0)$ is the starting state.
Now, for any state $S$, let $P(S)$ denote the probability that $A$ will eventually win, given that we are now in state $S$ and we have the following relationships:
$$\begin{aligned}
P(A, 0) & = \frac 1 6 + \frac 1 6 P(B, 5) + \frac 4 6 P(B, 0)
\\[5 mm] P(A, 5) & = \frac 1 6 + \frac 1 6 P(B, 55) + \frac 4 6 P(B, 0)
\\[5 mm] P(A, 55) & = \frac 2 6 + \frac 4 6 P(B, 0)
\\[5 mm] P(B, 0) & = \frac 1 6 P(A, 5) + \frac 4 6 P(A, 0)
\\[5 mm] P(B, 5) & = \frac 1 6 P(A, 55) + \frac 4 6 P(A, 0)
\\[5 mm] P(B, 55) & = \frac 4 6 P(A, 0)
\end{aligned}$$
The system of linear equations above can be easily solved to give $P(A, 0) = \frac {93} {170}$, which is the correct answer for $(a)$.
However, I am not sure how to approach $(b)$, whose answer is $\frac {129} {22}$. Any intuitive suggestions will be greatly appreciated :)
Edit
Following some hints Joe posted in an answer, I managed to solve $(b)$ :)
Let $X$ be the number of dice rolls it takes for the game to stop, $A$ be the event that a $5$ is first rolled, $B$ the event that a $6$ is first rolled and $C$ the event that neither a $5$ nor a $6$ is first rolled.
By the law of iterated expectation, we have
$$\begin{aligned}
\mathbb{E}(X) & = \mathbb{E}(X \mid A) \mathbb{P}(A) + \mathbb{E}(X \mid B) \mathbb{P}(B) + \mathbb{E}(X \mid C) \mathbb{P}(C)
\\[5 mm] & = (1)\left(\frac 1 6\right) + \mathbb{E}(X \mid B) \mathbb{P}(B) + [\mathbb{E}(X) + 1]\left(\frac 4 6\right)
\end{aligned}$$
To find $\mathbb{E}(X \mid B) \mathbb{P}(B)$, I chose to consider different cases.
*
*The game ends with three $5$s or two $5$s and a $6$, each case happening with probability $\left(\frac 1 6\right)^3$ and taking three turns to end.
*The game ends with one $5$ and one $6$, which happens with probability $\left(\frac 1 6\right)^2$ and takes two turns to end.
We must also not forget the cases where the game does not end (immediately).
*Two $5$s are thrown, followed by neither a $5$ nor a $6$, which happens with probability $\left(\frac 1 6\right)^2\left(\frac 4 6\right)$ and takes $\mathbb{E}(X + 3)$ turns to end.
*One $5$ is thrown, followed by neither a $5$ nor a $6$, which happens with probability $\left(\frac 1 6\right)\left(\frac 4 6\right)$ and takes $\mathbb{E}(X + 2)$ turns to end.
These four cases, when added up, will give
$$\begin{aligned}
\mathbb{E}(X \mid B) \mathbb{P}(B) & = \left(\frac 1 6\right)^3(3)(2) + \left(\frac 1 6\right)^2(2) + \left(\frac 1 6\right)^2\left(\frac 4 6\right)[\mathbb{E}(X + 3)] + \left(\frac 1 6\right)\left(\frac 4 6\right)[\mathbb{E}(X + 2)]
\\[5 mm] & = \frac 1 {12} + \frac 1 {54} \mathbb{E}(X + 3) + \frac 1 9 \mathbb{E}(X + 2)
\end{aligned}$$
Thus,
$$\begin{aligned}
\mathbb{E}(X) & = (1)\left(\frac 1 6\right) + \mathbb{E}(X \mid B) \mathbb{P}(B) + [\mathbb{E}(X) + 1]\left(\frac 4 6\right)
\\[5 mm] & = \frac 1 6 + \frac 1 {12} + \frac 1 {54} \mathbb{E}(X + 3) + \frac 1 9 \mathbb{E}(X + 2) + \frac 4 6 \mathbb{E}(X + 1)
\\[5 mm] & = \frac 1 4 + \frac 1 {54} \mathbb{E}(X + 3) + \frac 1 9 \mathbb{E}(X + 2) + \frac 4 6 \mathbb{E}(X + 1)
\\[5 mm] & = \frac {43} {36} + \frac {43} {54} \mathbb{E}(X)
\\[5 mm] \implies \mathbb{E}(X) & = \frac {129} {22}
\end{aligned}$$
|
If $x$ is the expected number of rolls, $y$ is the expected number of rolls from $S(5)$ and $z$ is the expected number of rolls from $S(55)$, then we have
$\left
\{\begin{array}
{l}x = 1 + \dfrac{1}{6} \cdot y + \dfrac{4}{6} \cdot x \\ y = 1 + \dfrac{4}{6} \cdot x + \dfrac{1}{6} \cdot z \\
z = 1 + \dfrac{4}{6} \cdot x
\end{array}
\right.$
Solving, $x = \dfrac{129}{22}$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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|
$a^2\phi=2$ , what is the value of $2a(\phi+1)$? I reduced a problem to this:
We have $a^2\phi=2$ where $a>0$ what is the value of $2a(\phi+1)$ ?
$1)2\sqrt{2\sqrt5+4}\qquad\qquad2)2\sqrt{\sqrt5+4}\qquad\qquad3)2\sqrt{2\sqrt5+1}\qquad\qquad4)2\sqrt{\sqrt5+1}$
Where $\phi$ is golden ratio ($\frac{1+\sqrt5}2)$.
This is a problem from a timed exam, so I should solve it quickly. Here I used $\phi^2=\phi+1$ several times to get the answer:
$$2a(\phi+1)=2a\phi^2=\sqrt{4a^2\phi^4}=\sqrt{8\phi^3}=\sqrt{8\phi(\phi+1)}=\sqrt{8(2\phi+1)}=2\sqrt{2(\sqrt5+2)}$$
Hence the answer is first choice. although the method I used is quick, but are there other approaches to get the answer (preferably) faster ?
|
Just solve for $a$ and plug it in.
Assuming $a\ge 0$ we have
$a^2 \phi = 2$
$a^2 = \frac 2 {\phi}$
$a= \sqrt{\frac 2\phi } = \frac {\sqrt 2}{\sqrt \phi}$.
So
$2a(\phi +1)= 2\frac {\sqrt 2}{\sqrt \phi} (\phi+1)=2\sqrt 2(\sqrt \phi + \frac 1{\sqrt \phi})=....$
Okay here we can use $\phi^2 = \phi + 1$.
$2\sqrt 2\frac {\phi+1}{\sqrt \phi}= 2\sqrt 2\frac {\phi^2}{\phi^{\frac 12}}= 2\sqrt 2\phi^{\frac 32}=2\sqrt 2\phi \sqrt \phi=$
and given the choices
$2\sqrt{2\phi^2 \phi} = 2\sqrt{2(\phi+1)\phi}=2\sqrt {2\phi^2 + 2\phi}=$
$2\sqrt{2(\phi + 1) + 2\phi}=2\sqrt {4\phi + 2}=2\sqrt{2(1+\sqrt 5) + 2}=$
$2\sqrt{2\sqrt 5 + 4}$.
.....
I guess that wasn't that much quick. But it was more directed.
I wonder if we can put $\phi^2 = \phi + 1$ if we can find a similar expresion for $\sqrt \phi$....
$\sqrt \phi = \sqrt{\phi^2 - 1} = \sqrt{(\phi -1)(\phi + 1)}=$
$\sqrt{\frac {(\sqrt 5-1)(\sqrt 5+3)}4}$...
Yeah... that could do it
$2a(\phi + 1) = 2\frac {\sqrt 2}{\sqrt{\frac {(\sqrt 5-1)(\sqrt 5+3)}4}}= 2\frac {2\sqrt 2}{\sqrt{(\sqrt 5-1)(\sqrt 5+3)} }=...$
That'll get you there but I would say it was easier....
|
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|
Find $ \int \frac{\sec x(2+\sec x)}{(1+2\sec x)^2}\,dx$
Find $\displaystyle \int \frac{\sec x(2+\sec x)}{(1+2\sec x)^2}\,dx$
I couldn't find my answer so I looked up the solution which is as follows
$$\displaystyle \begin{align}&\int \frac{\sec x(2+\sec x)}{(1+2\sec x)^2}\,dx\\&= \int \frac{2\cos x+1}{(2+\cos x)^2}\,dx\\&= \int \frac{2\cos x+\cos^2 x+\sin^2 x}{(2+\cos x)^2}\,dx\\&=\int \frac{\cos x}{2+\cos x}\,dx+\int \frac{\sin^2 x}{(2+\cos x)^2}\,dx\\&=\frac{1}{2+\cos x}\cdot(\sin x)-\int \sin x\cdot\frac{\sin x}{(2+\cos x)^2}\,dx+\int \frac{\sin^2 x}{(2+\cos x)^2}\,dx\\&=\frac{\sin x}{2+\cos x}+C\end{align}$$
I was stuck with $\displaystyle\int \frac{2\cos x+1}{(2+\cos x)^2}\,dx$ and didn't think of replacing $1$ with $\cos^2 x+\sin^2 x$. I want to know the motivation behind this because the reason of this substitution wasn't very obvious to me at first and it only became clear to me in the second last step. Other ways to solve this are also welcome.
|
Analogous of how integration by parts comes from the product rule, from the quotient rule of differentiation:
$$\begin{align*}
\frac uv &= \int \frac {du}{v} - \int\frac{u\ dv}{v^2}\\
\int\frac{u\ dv}{v^2} &= -\frac uv + \int \frac {du}{v}\\
\end{align*}$$
Or directly deriving from integration by parts:
$$
\int \frac{u\ dv}{v^2} = \int u\ d\left(-\frac1v\right) = -\frac{u}{v} + \int\frac{du}{v}
$$
This might have motivated breaking down $1$ into $\cos^2 x + \sin^2 x$, to apply the quotient rule backwards,
$$\begin{align*}
I &= \int \frac{\cos x}{2+\cos x} dx + \int \frac{\sin^2 x}{(2+\cos x)^2}dx\\
&= \int\frac{d(\sin x)}{2+\cos x} - \int \frac{\sin x\ d(2+\cos x)}{(2+\cos x)^2}
\end{align*}$$
But the exact break down of the numerator was not obvious to me. Though by this special case of integration by parts via the quotient rule, let
$$\begin{align*}
v &= 2 + \cos x &
dv &= -\sin x\ dx\\
u &= -(2\cot x +\csc x)&
du &= (2\csc^2x + \csc x \cot x) dx
\end{align*}$$
$$\begin{align*}
\int \frac{2\cos x + 1}{(2+\cos x)^2} dx
&= \int\frac{-(2\cot x + \csc x)(-\sin x\ dx)}{(2+\cos x)^2}\\
&= \frac{2\cot x + \csc x}{2+\cos x} + \int \frac{2\csc^2 x + \csc x \cot x}{2+\cos x} dx\\
&= \frac{2\cot x + \csc x}{2+\cos x} + \int \csc^2 x\cdot \frac{2+\cos x}{2+\cos x} dx\\
&= \frac{2\cot x + \csc x}{2+\cos x} + \int \csc^2 x\ dx\\
&= \frac{2\cot x + \csc x}{2+\cos x} - \cot x + C\\
&= \frac{(2\cot x + \csc x) - \cot x(2+\cos x)}{2+\cos x} + C\\
&= \frac{\csc x - \cot x\cos x}{2+\cos x} + C\\
&= \frac{\sin x (\csc^2 x - \cot^2 x)}{2+\cos x} + C\\
&= \frac{\sin x}{2+\cos x} + C\\
\end{align*}$$
Alternatively, without multiplying the starting numerator and denominator by $\cos^2 x$, let
$$\begin{align*}
v &= 1 + 2\sec x &
dv &= 2\sec x \tan x\ dx\\
u &= \frac12(2\cot x + \csc x)&
du &= -\frac12(2\csc^2x + \csc x \cot x) dx
\end{align*}$$
$$\begin{align*}
\int \frac{\sec x(2+\sec x)}{(1+ 2\sec x)^2} dx
&= \int\frac{\frac12(2\cot x + \csc x)(2\sec x \tan x\ dx)}{(1+2\sec x)^2}\\
&= -\frac{\frac12(2\cot x+\csc x)}{1+2\sec x} + \int\frac{-\frac12 (2\csc^2 x + \csc x \cot x)}{1+2\sec x} dx\\
&= -\frac{\frac12(2\cot x+\csc x)}{1+2\sec x} +\frac12 \int(-\csc x \cot x)\cdot\frac{2\sec x + 1}{1+2\sec x} dx\\
&= -\frac{\frac12(2\cot x+\csc x)}{1+2\sec x} +\frac12 \int(-\csc x \cot x)\ dx\\
&= -\frac{\frac12(2\cot x+\csc x)}{1+2\sec x} +\frac12 \csc x + C\\
&= \frac{-\cot x-\frac12\csc x + \frac12\csc x(1+2\sec x)}{1+2\sec x} + C\\
&= \frac{-\cot x+ \csc x\sec x}{1+2\sec x} + C\\
&= \frac{\tan x(-\cot^2 x+ \csc^2 x)}{1+2\sec x} + C\\
&= \frac{\tan x}{1+2\sec x} + C\\
\end{align*}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Let $(m,n)$ be the pair of integers satisfying $2(8n^3 + m^3) + 6(m^2 - 6n^2) + 3(2m + 9n) = 437.$ Find the sum of all possible values of $mn$.
Let $(m,n)$ be the pair of integers satisfying $2(8n^3 + m^3) + 6(m^2 - 6n^2) + 3(2m + 9n) = 437.$ Find the sum of all possible values of $mn$.
What I Tried:- Given this expression I had no idea to claim something out of it, because it looks like we need to $437$ in terms of sums of $2,3$ (the $6$ does not matter much here). I could break $(8n^3 + m^3)$, but it didn't help much.
Instead I simplified the whole expression and got :-
$$2m^3 + 6m^2 + 6m + 16n^3 - 36n^2 + 27n = 437.$$
$$\rightarrow 2m^3 + 6m^2 + 6m + 16n^3 - 36n^2 = 437 - 27n.$$
From here, $2 \vert (437 - 27n)$, so $n$ is odd. Let $n = (2k + 1)$. Substitute this back to the equation to get :-
$$64k^3 + 24k^2 + 3k + m^3 + 3m^2 + 3m = 215.$$
Now I am stuck here, it is given $(k,m)$ are integers, so they can range to negative values also. So I was not able to make an upper bound on $k$, and I have no other idea what I can claim from here.
Can anyone help me? Thank You.
|
Edit: This is similar Moko19's answer (written in paralel but posted too late...), but I keep it here as it describes process how to get the solutions for the $a^3+b^3=1729$.
Notice that $m^3+3m^2+3m=(m+1)^3-1$, and similarly $$64k^3 + 24k^2 + 3k=\frac{1}{8}((8k)^3+3(8k)^2+3(8k))=\frac{1}{8}((8k+1)^3-1).$$
And so
$$
\frac{1}{8}((8k+1)^3-1)+(m+1)^3-1=215
$$
which in turn after simplification gives
$$
(8k+1)^3+(2m+2)^3=1729.
$$
Let $a=8k+1=4n-3$ and $b=2m+2$. Now notice that $a^3+b^3=(a+b)(a^2-ab+b^2)=1729$ and also $a^2-ab+b^2>0$, which implies $a+b>0$ (regardless of whether any of $a,b$ is negative). Finally, we need to only check finitely many cases, namely for each positive divisor $d \mid 1729 = 7\cdot 13\cdot 19$ (so $d\in \{1, 7, 13, 19, 91, 133, 247, 1729\}$), let $a+b=d$ and $a^2-ab+b^2=\frac{1729}{d}$ , which simplifies to quadratic (diophantine) equation in one variable.
For example, for $d=13$ we have $a+b=13$ and $a^2-ab+b^2=133$, and hence substituting first into the second and cancel $3$, we have $a^2-13a+12=0$. This solves to $a=12$ or $a=1$, applying the original substitution backwards we get $n=1,m=5$. Same way, we find that only other value that works is $d=19$, leading to $n=3,m=4$.
Bonus: You might recognize a famous Ramanujan's taxicab number 1729, which is the smallest natural number expressible as a sum of positive(!) two cubes in exactly two different ways: $$1729=1^3+12^3=9^3+10^3.$$
Which also gives us a solution (although we still need to be careful about the negative cubes).
|
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|
Find all $f : \mathbb R\to\mathbb R$ such that $f(f(x+y)) = f(x+y) + f(x)f(y) -xy$ Find all $f : \mathbb R\to\mathbb R$ such that
$$f(f(x+y)) = f(x+y) + f(x)f(y) -xy$$
for all reals $x,y.$ (Belarusian Mathematical Olympiad-1995)
My answer:
Consider $f(0) = c. ...(i)$
Let $x,y = 0$ at first.
$$f(f(x+y)) = f(x+y) + f(x)f(y) -xy$$
$$\implies f(f(0+0)) = f(0+0) + f(0)f(0) -0 \times 0$$
$$\implies f(f(0)) = f(0) + f(0)f(0) + 0$$
$$\implies f(c) = c + c \times c + 0$$
$$\implies f(c) = c + c^2$$
$$\implies f(c) - c = c^2 ...(ii)$$
Now, let $y = -x.$
Therefore,
$$f(f(x+y)) = f(x+y) + f(x)f(y) -xy$$
$$\implies f(f(x-x)) = f(x-x) + f(x)f(y) -(x\times -x)$$
$$\implies f(f(0)) = f(0) + f(x)f(-x) + x^2$$
$$\implies f(c) = c + f(x)f(-x) + x^2$$
$$\implies f(c)-c = f(x)f(-x) + x^2$$
From $..(ii),$
Therefore, $$c^2 = f(x)f(-x) + x^2$$
Now, let $x = c$
$$Therefore, c^2 = f(x)f(-x) + c^2$$
$$\implies c^2 - c^2 = f(c)f(-c)$$
$$\implies 0 = f(c)f(-c)$$
Hence,
$f(c) = 0 ...(iii)$
$f(-c) = 0 ...(iv)$
Multiplying $(i) and (iii),$
$$f(0) \times f(c) = c \times 0$$
$$\implies f(0) \times f(c) = 0$$
$$\implies f(0) = 0$$
$$\implies c = 0$$
Similarly, Multiplying $(i) and (iv)$
$\implies c = 0$
Hence $c = 0.$
But $f(c) = 0,$
Therefore, $f(c) = c$
I can't go further than that can anyone help me out.
|
$$f(f(x+y))=f(x+y)+f(x)f(y)-xy$$
As mentioned above, I have assumed $f(0)=c$ for some real $c$.
*
*$(x,y)\equiv(c,-c)$
$$f(c)=c+f(c)f(-c)+c^2$$
*$(x,y)\equiv(0,0)$
$$f(c)=c+c^2$$
The above equations imply that $f(c)f(-c)=0$. Hence, there exists some real $k$ such that $f(k)=0$.
*$(x,y)\equiv(k,0)$
$$f(f(k+0))=f(k+0)+f(k)f(0)-k\cdot0\implies c=0\implies f(0)=0 $$
*$(x,y)\equiv(x,0)$
$$f(f(x))=f(x)\implies f(f(x+y))=f(x+y)\implies f(x)=x\cdot \frac{y}{f(y)}$$
If $f(x)=0$ for all real $x$, we have, $xy=0$ for all reals,which is a contradiction. Hence there exists some real $p$ such that $f(p)=q$ and $q\ne 0$.
*$(x,y)\equiv(x,p)$
$$f(f(x+p))=f(x+p)\implies f\left((x+p)\cdot\frac{p}{q}\right)=(x+p)\cdot\frac{p}{q} \implies p=q$$
Therefore, $\boxed{f(x)=x}$ for all real $x$. Substitute this back into the original F.E to verify the result.
|
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|
Proving $\int_a^{b}\frac{\ln (1+x)}{a^2b^2(1+x)^2+({a+b})^2}\ dx=\frac{\ln (a+b)-\ln (ab)}{ab(a+b)}\left(\frac{\pi}{2}-2\tan^{-1}\frac{a}{b}\right)$
How can I prove this?
$$\int_a^{b}\frac{\ln (1+x)}{a^2b^2(1+x)^2+({a+b})^2}\ dx=\frac{\ln (a+b)-\ln (ab)}{ab(a+b)}\left(\frac{\pi}{2}-2\tan^{-1}\frac{a}{b}\right)$$
Here is my attempt
$$ \frac{1}{a^2b^2}\int_a^{b}\frac{\ln (1+x)}{(1+x)^2+(\frac{a+b}{ab})^2}\ dx$$
Applying the substitution $1+x=t \left (\dfrac{a+b}{ab}\right )$
$$ \frac{\frac{a+b}{ab}}{a^2b^2(\frac{a+b}{ab})^2}\int_{\frac{ab(a+1)}{a+b}}^{\frac{ab(b+1)}{a+b}}\frac{\ln (t)+\ln \left (\frac{a+b}{ab}\right )}{t^2+1}\ dt\\\\$$
$$ \frac{1}{ab(a+b)}\int_{\frac{ab(a+1)}{a+b}}^{\frac{ab(b+1)}{a+b}}\frac{\ln (t)+\ln \left (\frac{a+b}{ab}\right )}{t^2+1}\ dt\\\\$$
Again applying substitution $t\ =\dfrac{1}{u}$
$$ \frac{-1}{ab(a+b)}\int_{\frac{a+b}{ab(a+1)}}^{\frac{a+b}{ab(b+1)}}\frac{-\ln (u)+\ln \left (\frac{a+b}{ab}\right )}{u^2+1}\ du\\\\$$
|
Repeating your steps for the antiderivarive ( assuming $a>0$ and $b >0$)$$\int\frac{\log (1+x)}{a^2b^2(1+x)^2+({a+b})}\, dx=\frac 1{a^2\,b^2}\int\frac{\log (1+x)}{(1+x)^2+k^2}\, dx$$ with $k=\sqrt{\frac {a+b}{a^2\,b^2}}$ Now, let $r$ and $s$ be the complex roots of the denominator and use
$$\frac{1}{(1+x)^2+k^2}=\frac 1{(x-r)(x-s)}=\frac 1{r-s}\Big[\frac{1}{x-r}-\frac{1}{x-s} \Big]$$ So, we face two integrals
$$I_t=\int \frac{\log(1+x)}{x-t}\,dx=\text{Li}_2\left(\frac{x+1}{t+1}\right)+\log (x+1) \log
\left(\frac{t-x}{t+1}\right)$$ Thsi would lead to much complex formulae for the definite integral (in which will remain logarithms, polylogarithms (with complex arguments) and arctangents). If I am not mistaken, $\log(a+b)$ would never appear.
|
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|
Show that $u-v=\sqrt5$ given that $u-v>0$ given that $u=b+b^4$ and $v=b^2+b^3$ We're also given that $b$ is a root of $z^5-1=0$ $b^4+b^3+b^2+b+1=0$
If $u=b+b^4$ and $v=b^2+b^3$, show that
i) $u+v=uv=-1$
ii) $u-v=\sqrt5$ given that $u-v>0$
I managed to do part i):
Plugging in $u$ and $v$: $(b+b^4)+(b^2+b^3)=-1$ (using $b^4+b^3+b^2+b+1=0$)
$uv=(b+b^4)(b^2+b^3)$
$=(b^3+b^4+b^6+b^7)$
$=b^3(1+b+b3+b4)$
$=b^3(-b^2)$
$=-b^5$
$=-1^5$
$=-1$ (using $z^5=1$)
For ii)
$u-v=(b+b^4)-(b^2+b3)>0$
$(b+b^4)-(-b-b^4-1)>0$
$2b+2b^4+1>0$
$b+b^4>\frac{-1}{2}$
Although it's a dead-end after that.
What's the general idea for going about solving part ii)?
|
You can compute $(u-v)^2=(u+v)^2-4uv$ simply from what you already know.
You have the elementary symmetric polynomials and $(u-v)^2$ is symmetric, so it will have an expression in terms of $uv$ and $u+v$. The computation is equivalent to solving the quadratic with roots $u$ and $v$, but is more straightforward - you don't need to go via the quadratic.
|
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|
How can you quickly use the graph of $3x-6y=4$ to produce the graph of $3x-6y=10$? To start with, I am asked to graph the lines $3x-2y=k$ for $k=1,2,3$ and $4$.
I find that the equations produce different parallel lines.
For each line where $k=2,3$ and $4$, the coordinates points are $0.5k$ below the coordinates points of the line $3x-2y=1$.
If we let $x=0$, we can find the $y$-intercept of the lines.
\begin{align*}
-2y&=k\\
y &=\frac{k}{-2}
\end{align*}
and if we increase $k$ by $1$ we decrease $y$ by $0.5$.
\begin{align*}
-2y&=k+1\\
y &=\frac{k}{-2}-\frac{1}{2}
\end{align*}
I'm not certain the above logic suffices in quickly using the graph of $3x -6y = 4$ to produce the graph of $3x-6y=10$.
\begin{align*}
-6y&=4+6\\
y&=-\frac{2}{3}-1 \\
y&=-\frac{5}{3}
\end{align*}
Can my answer be improved?
|
Great question. Let's start with a couple of general lines here:
$$ ax + by = k_1 , $$
$$ ax + by = k_2 . $$
As you note, these lines are parallel because their slopes are the same:
$$ ax + by = k_1 \Rightarrow y = -\frac{a}{b}x + \frac{k_1}{b} , $$
$$ ax + by = k_1 \Rightarrow y = -\frac{a}{b}x + \frac{k_2}{b} . $$
And again, you're right to focus on the $x$ and $y$-intercepts.
Compare the $y$-intercepts from the slope-intercept form above. Your approach was to look at the difference, which is perfectly valid:
$$ \frac{k_2}{b} - \frac{k_1}{b} = \frac{k_2 - k_1}{b} . $$
Note, the $y$-intercept increases by $\frac{1}{b}$ times the increase in the constant. A similar pattern holds for the $x$-intercept. If you plug in $y=0$, you'll find the $x$-intercepts are, respectively, $\frac{k_1}{a}$ and $\frac{k_2}{a}$. Their difference is then:
$$ \frac{k_2}{a} - \frac{k_1}{a} = \frac{k_2 - k_1}{a} .$$
So, generally, as the constant increases by $\Delta k = k_2 - k_1$, the intercepts increase by the same amount, scaled down by their respective coefficients. The $x$-intercept increases by $\frac{1}{a}*\Delta k$, and the $y$-intercept increases by $\frac{1}{b}*\Delta k$.
In this case, every time you increase $k$ by 1, you increase the $x$-intercept by $\frac{1}{a} = \frac{1}{3}$ and you increase the $y$-intercept by $\frac{1}{b} = -\frac{1}{6}$, i.e., you decrease the $y$-intercept by $\frac{1}{6}$.
You can also look at what happens to the intercepts multiplicatively. Let's look at the ratio of the $y$-intercepts:
$$ \frac{k_2}{b} \bigg/ \frac{k_1}{b} = \frac{k_2}{b} * \frac{b}{k_1} = \frac{k_2}{k_1} . $$
Similarly, the ratio of the $x$-intercepts is
$$ \frac{k_2}{a} \bigg/ \frac{k_1}{a} = \frac{k_2}{a} * \frac{a}{k_1} = \frac{k_2}{k_1} . $$
So, as the constant scales up by some factor $\kappa = \frac{k_2}{k_1} $, both the $x$ and $y$ intercepts also get scaled up by that same factor.
You can use either approach, looking at the difference $k_2 - k_1$ or the ratio $k_2 / k_1$, depending on which is easier with the intercepts. You just have to calculate the intercepts once.
|
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|
Prove all $2\times2$ real matrices with eigenvalues $1$ and $-1$ can be represented as My exercise asks:
Prove that all $2\times2$ real matrices with eigenvalues $\lambda_1=1$ and $\lambda_2=-1$ can be represented as
\begin{equation}
\begin{bmatrix}
\cos\theta & a\sin\theta \\
\frac{1}{a}\sin\theta & -\cos\theta
\end{bmatrix}
\end{equation}
Starting like
\begin{equation}
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
\end{equation}
Using the fact that $\operatorname{Tr}(A) = a+d =\lambda_1+\lambda_2 = 0 \Rightarrow a=-d$. And using the fact that $\det A = ad-bc=\lambda_1\lambda_2=-1 \Rightarrow-a^2-bc=-1$
\begin{equation}
a^2+bc=1
\end{equation}
How can I complete the proof?
|
$a,b,c,d$ are fixed numbers and we have to show that they can be represented as $\cos\theta, k\sin\theta, \frac1k \sin\theta, -\cos\theta$
Let $\cos \theta=a \implies d = -a = -\cos \theta$
This can be done because there always exists a $\theta \in \mathbb{C}$ such that $\cos\theta = a$
Now,
$$ a^2 + bc=1$$
$$ bc = 1-a^2 = \sin^2 \theta$$
You will always be able to choose a $k$ such that $b=k\sin\theta$
This is because
$$ \sin^2\theta + \cos^2\theta=1$$
$$ \dfrac{b^2}{k^2} + a^2 = 1 $$
$$ k = \pm\dfrac{b}{\sqrt{1-a^2}} $$
This $k$ will also satisfy $c = \dfrac1k \sin\theta$ because $c = \dfrac{\sin^2\theta}{b}$
That completes the proof.
|
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|
Simplifying a summation with binomial coefficients Simplify the sum
$$
f(k,q)=\sum_{i=0}^k\binom{2i}{i+q}\binom{2k-2i}{k-i}
$$
where $k,q$ are given integers satisfying $0<q\le k$.
I tried to simplify it combinatorially. For a lattice path $p$ using steps (1,1) and (1,-1) and starts from the origin, define $g_q(p)$ to be the number of intersections between $p$ and the line $y=2q$. One can notice that $f(k,q)$ calculates the sum $g_q(p)$ for all lattice paths from the origin to (2k,2q). If we enumerate the rightmost intersection between the path and the line $y=2q$, and denote its $x$ position by $2t$, then we have
$$
f(k,q)=\sum_{t=0}^k\left(\binom{2t}{t+q}-2\binom{2t-1}{t+q}\right)4^{k-t}
$$
However this does not make the problem easier. How should I proceed? Any help is appreciated!
|
Supposing we seek to simplify
$$\sum_{j=0}^k {2j\choose j+q} {2k-2j\choose k-j}.$$
where $0\le q\le k.$ This is
$$[z^k] (1+z)^{2k}
\sum_{j=0}^k {2j\choose j+q} \frac{z^j}{(1+z)^{2j}}.$$
Here the coefficient extractor enforces the upper limit of the sum
and we find
$$[z^k] (1+z)^{2k}
\sum_{j\ge 0} {2j\choose j+q} \frac{z^j}{(1+z)^{2j}}.$$
At this point we see that we will require residues and complex
integration and continue with
$$\frac{1}{2\pi i}
\int_{|z|=\varepsilon}
\frac{(1+z)^{2k}}{z^{k+1}}
\frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{q+1}}
\sum_{j\ge 0} \frac{(1+w)^{2j}}{w^j} \frac{z^j}{(1+z)^{2j}}
\; dw \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\varepsilon}
\frac{(1+z)^{2k}}{z^{k+1}}
\frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{q+1}}
\frac{1}{1-z(1+w)^2/w/(1+z)^2}
\; dw \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\varepsilon}
\frac{(1+z)^{2k+2}}{z^{k+1}}
\frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{q}}
\frac{1}{w(1+z)^2-z(1+w)^2}
\; dw \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\varepsilon}
\frac{(1+z)^{2k+2}}{z^{k+1}}
\frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{q}}
\frac{1}{(w-z)(1-wz)}
\; dw \; dz.$$
For the geometric series to converge we must have
$|z(1+w)^2/w/(1+z)^2| \lt 1$ or $|z/(1+z)^2| \lt |w/(1+w)^2|.$ This
requires $\varepsilon/(1-\varepsilon)^2 \lt \gamma/(1+\gamma)^2.$ We
will also require $w=z$ to be inside the contour for $w$ so we need
$\varepsilon \lt \gamma.$ With $\varepsilon \ll 1$ and $\gamma \ll 1$
we may take $\varepsilon = \gamma^2$ for the latter inquality. We then
get for the inquality from the geometric series
$\gamma^2/(1-\gamma^2)^2 \lt \gamma / (1+\gamma)^2$ or $\gamma \lt
(1-\gamma^2)^2/(1+\gamma)^2$ or $\gamma \lt (1-\gamma)^2.$ This holds
for $\gamma\lt 1-1/\varphi$ with $\varphi$ the golden mean.
Now we have the pole at zero and the one at $w=z$ inside the
contour in $w$. This means we can evaluate the integral by using the
fact that residues sum to zero, taking minus the residue at $w=1/z$
and minus the residue at infinity, which is zero by inspection,
however. (The pole at $w=1/z$ has modulus $1/\varepsilon$ and is
outside the contour.) Computing minus the residue at $w=1/z$ we write
$$- \frac{1}{2\pi i}
\int_{|z|=\varepsilon}
\frac{(1+z)^{2k+2}}{z^{k+2}}
\frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{q}}
\frac{1}{(w-z)(w-1/z)}
\; dw \; dz.$$
With the sign change we obtain
$$\frac{1}{2\pi i}
\int_{|z|=\varepsilon}
\frac{(1+z)^{2k+2}}{z^{k+2}}
z^q \frac{1}{1/z-z} \; dz
= \frac{1}{2\pi i}
\int_{|z|=\varepsilon}
\frac{(1+z)^{2k+2}}{z^{k-q+1}}
\frac{1}{1-z^2} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\varepsilon}
\frac{(1+z)^{2k+1}}{z^{k-q+1}}
\frac{1}{1-z} \; dz.$$
This is zero when $q\gt k$ and otherwise
$$\sum_{j=0}^{k-q} {2k+1\choose j}
= \sum_{j=0}^k {2k+1\choose j}
- \sum_{j=k-q+1}^k {2k+1\choose j}$$
or alternatively
$$\bbox[5px,border:2px solid #00A000]{
4^k - \sum_{j=k-q+1}^k {2k+1\choose j}}$$
which is a closed form term plus a sum of $q$ terms. E.g. with $q=0$
we obtain $4^k$ and with $q=1,$ $4^k - {2k+1\choose k}$. For $q=2$ we
have $4^k - {2k+1\choose k-1} - {2k+1\choose k}$ and so on.
|
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|
If $2\sin\theta+\cos\theta=\sqrt3$, what is the value of $\tan^2\theta+4\tan\theta$? If $2\sin\theta+\cos\theta=\sqrt3$, what is the value of $\tan^2\theta+4\tan\theta$ ?
$1)1\qquad\qquad2)2\qquad\qquad3)3\qquad\qquad4)5$
First I tried plugging in some values for $\theta$ like $0,\frac{\pi}4,\frac{\pi}3,...$ but neither of these known angles worked. But by doing it I realized that for $\theta=\frac{\pi}4+k\pi\quad$, $\tan^2\theta+4\tan\theta=5\quad$ and $2\sin\theta+\cos\theta\neq\sqrt3$ Hence the fourth choice is wrong.
Also tried to expanding,
$$\tan^2\theta+4\tan\theta=\dfrac{\sin^2\theta}{\cos^2\theta}+\dfrac{4\sin\theta}{\cos\theta}=\dfrac{\sin^2\theta+4\sin\theta\cos\theta}{\cos^2\theta}$$But can't continue even writing $4\sin\theta\cos\theta=2\sin2\theta$ doesn't help.
|
Simply square the constraint and you get
$4 \sin^2(x)+4\sin(x)\cos(x)+\cos^2(x)=3$
then double angle and trig identity
$3\sin^2(x)+2\sin(2x)=2$
furthermore this is equivalent to
$2\sin(2x) = 3\cos^2(x)-1$
and with your definition of the left hand side
$\tan^2(x)+4\tan(x)=\frac{\sin^2(x)+2\sin(2x)}{\cos^2(x)}$
we can insert the definition of $2\sin(2x)$ we got from the constraint and we get
$\tan^2(x)+4\tan(x)=\frac{\sin^2(x)+3\cos^2(x)-1}{\cos^2(x)}=\frac{2\cos^2(x)}{\cos^2(x)}=2$
I hope that helps and I hope I didn't make a mistake
;)
|
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|
Why does the triangle inequality seem to be false when rewritten based on $|x| = \sqrt{x^{2}}$? The triangle inequality is $$|x + y| \leq |x| + |y|.$$
Also, we know that $|x| = \sqrt{x^{2}}$. Then,
\begin{align*}
\sqrt{x + y} &\leq \sqrt{x} + \sqrt{y} \\
x + y &\leq x + y + 2\sqrt{xy} \\
0 &\leq 2\sqrt{xy} \\\\
\sqrt{xy} &\geq 0
\end{align*}
Now, we can see that the inequality $|x + y| \le |x| + |y|$ holds for real $x$ and $y$, but $\sqrt{xy} \geq 0$ does not for $x < 0, y > 0$ or $x > 0, y < 0$.
What seems to be the problem? Is it the statement $|x| = \sqrt{x^{2}}$ or is it much more than that?
|
The statement you have written using $|x|=\sqrt {x^2}$ is wrong. The correct statement would be: $$\sqrt {(x+y)^2}\leq \sqrt {x^2}+\sqrt {y^2}$$ which, indeed, does hold for all real $x,y$, as can be verified by squaring and simplifying.
|
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|
Finding the remainder of a polynomial divided by $x^4+x^2+1$ if remainders when dividing by $x^2+x+1$, $x^2-x+1$ are $-x+1$, $3x+5$.
Find the remainder of $f$ divided by $g(x)=x^4+x^2+1$ if the remainder of $f$ divided by $h_1 (x)=x^2+x+1$ is $-x+1$ and the remainder of $f$ divided by $h_2(x)=x^2-x+1$ is $3x+5$.
My attempt was to write $f(x)=(x^4+x^2+1)q(x)+Ax^3+Bx^2+Cx+D=(x^2+x+1)(x^2-x+1)q(x)+Ax^3+Bx^2+Cx+D$ and then factor out $x^2+x+1$ to get $f(x)=(x^2+x+1)[(x^2-x+1)q(x)+B]+Ax^3+(C-B)x+(D-B)$, then do the same for $x^2-x+1$ and then use that along with the known remainder of $f$ divided by $x^2+x+1$ and $x^2-x+1$ to obtain $A$, $B$, $C$, $D$. However, $Ax^3$ is in the way so I don't know how to proceed nor do I have any other ideas to start with.
|
Hint:
As the roots of $x^2+x+1=0$ are $w,w^2$ where $w$ is a complex cube root of unity,
the roots of $x^2-x+1=0$ are $-w,-w^2$
we can write $$f(x)$$
$$=p(x)(x^2+x+1)(x^2-x+1)+A(x-w)(x-w^2)(x+w)+B(x-w)(x-w^2)(x+w^2)+C(x-w^2)(x+w)(x+w^2)+D(x-w)(x+w)(x+w^2)$$
$$=p(x)(x^2+x+1)(x^2-x+1)+(x^2-x+1)(c(x-w^2)+d(x-w))+\cdots$$
$$-w+1=f(w)=(-2w)c(w-w^2)\implies2c=-w$$
Similarly, we can find $a,b,d$
|
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|
$\min\left(\sqrt{ka+1}+\sqrt{kb+1}+\sqrt{kc+1}\right)$ Let $k>0~$ be fixed. Find
$$\min\left(\sqrt{ka+1}+\sqrt{kb+1}+\sqrt{kc+1}\right)$$
over all $c, b, a\geq0~\text{satisfying}~ab+bc+ac=a+b+c>0~.$
==================
In case we have 2 positive numbers $a+b=ab$ results that $a, b > 1$ and $a= \dfrac{b}{b-1}$,
so the minimum of the expression $ab$ is 4 because $\dfrac{b}{b-1} \cdot b \geqslant 4$ --- equality when $a=b=2$
Then $(\sqrt{ka+1} + \sqrt{kb+1})^2 = kab+2+ 2\sqrt{k^2ab + kab +1} ~~~$ (I've used $a+b = ab$)
so the minimum is reached when $ab$ is minimal, that is 4 so we obtain $4k+2 + 2\sqrt{4k^2 + 4k + 1} = 4(2k+1)$
That is the minimum searched is $2\sqrt{2k+1}$.
In case of 3 numbers I wonder if an idea is to raise to power 2 and proceed in some similar way...
|
Sketch of my solution:
Set WLOG $a\leq b\leq c.$ We denote $bc=p^2$ and $b+c=2s.$ Clearly, $s\geq p\geq1.$ By AM-GM and CBS,we get:
$$\sqrt{ka+1}+\sqrt{kb+1}+\sqrt{kc+1}\geq\sqrt{ka+1}+2\cdot\sqrt{\sqrt{kb+1}\cdot\sqrt{kc+1}}\geq\sqrt{ka+1}+2\cdot\sqrt{kp+1}.$$
Case 1: $p>2.$ We have:
$$E(a,b,c,k)\geq \sqrt{ka+1}+2\cdot\sqrt{kp+1}>1+2\cdot\sqrt{2k+1}.$$
Case 2: $p\in[1,2].$ According to BMO Problem 2/2019, $a\geq\frac{2-p}p,$ with equality iff $a=b=c=1$ or $a=0,b=c=2.$ Thus,
$$E(a,b,c,k)\geq \sqrt{ka+1}+2\cdot\sqrt{kp+1}\geq\sqrt{k\cdot\frac{2-p}p+1}+2\cdot\sqrt{kp+1}.$$
We now take the first derivative of the function $f(p):=\sqrt{k\cdot\frac{2-p}p+1}+2\cdot\sqrt{kp+1}$ and trivially get:
$$\min E(a,b,c,k)=\min_{k>0}\left\{3\cdot\sqrt{k+1},1+2\cdot\sqrt{2k+1}\right\}.$$
|
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|
Help for $\int _0^1\int _0^z\int _0^y\frac{1}{\left(1-x^2\right)\left(1+y^2\right)\left(1+z^2\right)}\:\mathrm{d}x\:\mathrm{d}y\:\mathrm{d}z$ With the help of programs I have been able to conjecture
$$\int _0^1\int _0^z\int _0^y\frac{1}{\left(1-x^2\right)\left(1+y^2\right)\left(1+z^2\right)}\:\mathrm{d}x\:\mathrm{d}y\:\mathrm{d}z=\frac{\pi }{8}G-\frac{7}{32}\zeta \left(3\right)$$
But is there a simple way to prove this?
What I have got thus far is:
$$\int _0^1\int _0^z\int _0^y\frac{1}{\left(1-x^2\right)\left(1+y^2\right)\left(1+z^2\right)}\:\mathrm{d}x\:\mathrm{d}y\:\mathrm{d}z=-\frac{1}{2}\int _0^1\frac{1}{1+z^2}\int _0^z\frac{\ln \left(\frac{1-y}{1+y}\right)}{1+y^2}\:\mathrm{d}y\:\mathrm{d}z$$
$$=-\frac{1}{2}\int _0^1\frac{\ln \left(\frac{1-y}{1+y}\right)}{1+y^2}\int _y^1\frac{1}{1+z^2}\:\mathrm{d}z\:\mathrm{d}y=-\frac{1}{2}\int _0^1\frac{\left(\frac{\pi }{4}-\arctan \left(y\right)\right)\ln \left(\frac{1-y}{1+y}\right)}{1+y^2}\:\mathrm{d}y$$
But I'm not sure how to advance from here, is the current path I'm taking correct?
|
The last integral, after the substitution $$x=\frac\pi4-\arctan y=\arctan\frac{1-y}{1+y},$$ becomes $\int_0^{\pi/4}x\ln\tan x\,dx$ which is evaluated here (or here).
|
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|
Proving that $f(t)=\frac{n^2}{2}\cdot t^{n-4}(1-t^2)\left(t^2-\frac{n-3}{n}\right)$ is bounded above by $1$, for $n\geq6$ and $t\in[0,1]$ I have a problem that looks like a typical problem of maximizing functions in a compact interval. However, I am not being able to prove the bound I need.
Let $n\geq 6$ be an integer number. Consider the function: $$f(t) = \frac{n^2}{2} \cdot t^{n-4}(1-t^2) \left(t^2 - \frac{n-3}{n}\right) $$
Prove that for all $t\in [0,1]$ it holds $f(t) \leq 1$.
The points where the derivative $f'$ is zero are very ugly expressions.
By maximizing the factor $t^{n-4}(1-t^2)$ and using that the last factor is at most $\frac{3}{n}$ it is possible to deduce that for $n\geq 6$ it is $f(t) \leq \frac{3}{2}$ (in fact, it is possible to bound it in the limit by $\frac{3}{e}\approx 1.1036...$ but that is far from $1$.
I have checked that the claim is true for several random values of $n$ (in fact, I think that the bound can be reduced to something like $0.61...$ for $n>20$ say).
|
In case a less-clever estimate by separation of cases is useful:
Our goal is to show
$$
\frac{n^{2}}{2}\, t^{n-4}(1 - t^{2}) \left(t^{2} - \frac{n-3}{n}\right)
= \frac{n^{2}}{2}\, t^{n-4}(1 - t^{2}) \left(\frac{3}{n} - (1 - t^{2}\right)
\leq 1
$$
for $0 \leq t \leq 1$.
Write $n = 4 + m$ for $m \geq 2$ and $u = 1 - t^{2}$. The desired inequality becomes
$$
\tfrac{1}{2}(4 + m)^{2}\, (1 - u)^{m/2}\, u(\tfrac{3}{4 + m} - u) \leq 1
$$
for $0 \leq u \leq 1$. Put $c = \frac{3}{4 + m}$ and $v = \frac{u}{c}$, so the desired inequality reads
$$
\tfrac{9}{2}(1 - c v)^{m/2} (v - v^{2}) \leq 1
$$
for $0 \leq v \leq \frac{1}{c}$.
We have $v - v^{2} = \frac{2}{9} - (v - \frac{1}{3})(v - \frac{2}{3}) \leq \frac{2}{9}$ unless $\frac{1}{3} \leq v \leq \frac{2}{3}$. Since $0 \leq 1 - cv \leq 1$, the inequality in this case follows at once.
Otherwise (if $\frac{1}{3} \leq v \leq \frac{2}{3}$), we have $v - v^{2} = \frac{1}{4} - (v - \frac{1}{2})^{2} \leq \frac{1}{4}$ (for all real $v$), while
$$
(1 - cv)^{m/2} \leq (1 - \tfrac{1}{3}c)^{m/2} = (1 - \tfrac{1}{4 + m})^{m/2}.
$$
The right-hand side decreases with $m$, and for $m = 2$ is equal to $\frac{5}{6} < \frac{8}{9}$, so the inequality holds in this case as well.
|
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|
Integer solution of $abc=a+b+c+2$. Let $a,b,c$ be integers greater than $1$. I am trying to prove that $$abc\geq a+b+c+2$$
with equality if and only if $a=b=c=2$.
I can prove the inequality by using the fact that $ab\geq a+b$. Since $ab\geq 4$ and $c\geq 2$, it follows that $abc\geq 4c$ and $abc\geq 2ab$. Therefore $abc\geq ab+2c\geq ab+c+c\geq a+b+c+c\geq a+b+c+2$.
The main problem I face is to justify that $$abc=a+b+c+2\implies a=b=c=2.$$
My idea is to assume that $a>2$ and try to get a contradiction.
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Let $a=x+2$, $b=y+2$ and $c=z+2$.
Thus, $x\geq0$, $y\geq0$, $z\geq0$ and $$(x+2)(y+2)(z+2)=x+y+z+8$$ or
$$xyz+2(xy+xz+yz)+3(x+y+z)=0,$$ which gives $x=y=z=0$ and $a=b=c=2$.
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"timestamp": "2023-03-29T00:00:00",
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|
Prove if number is rational or irrational I've been asked to prove if $\frac{\sqrt{3+\sqrt5}}{\sqrt{2} + \sqrt {10}}$ is a rational number. I've tried a proof as follows:
Suppose the number is rational, so it can be written as the quotient of 2 numbers $a$ and $b$
\begin{align*}
\frac{\sqrt{3+\sqrt{5}}}{\sqrt{2} + \sqrt{10}} = \frac{a}{b} \\
\frac{3+\sqrt{5}}{12 + 4\sqrt{5}} = \frac{a^2}{b^2} \\
\frac{3+\sqrt{5}}{4(3+\sqrt{5})} = \frac{a^2}{b^2} \\
\frac{1}{4} = \frac{a^2}{b^2}
\end{align*}
And because we get $\frac{1}{4}$ which is rational, we can conclude that the proof is right and there aren't any contradictions. Hence $\frac{\sqrt{3+\sqrt{5}}}{\sqrt{2} + \sqrt{5}}$ is a rational number.
I guess my proof lacks something and I don't feel it's complete yet. I would appreciate any recommendations on how to improve my answer. Thanks in advance.
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$$\frac{\sqrt{3+\sqrt5}}{\sqrt{2} + \sqrt {10}} =\frac{\sqrt{6+2\sqrt5}}{\sqrt{2}(\sqrt{2} + \sqrt {10})} = \frac{\sqrt{5}+1}{2+2\sqrt{5}}=\frac{1}{2}$$
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Find the values of $a$ for which the function $f(x)=\sin (2x)-8(a+1) \sin x+(4a^2+8a-14)x$ increases for all $x\in R$.
Find the values of $a$ for which the function $f(x)=\sin (2x)-8(a+1) \sin x+(4a^2+8a-14)x$ increases for all $x\in R$ and has no critical points.
For increasing function, $$f'(x)\gt0\\2\cos(2x) - 8(a+1)\cos x + (4a^2 + 8a -14)\gt0$$
Using $\cos(2x)=2\cos^2x-1$, $$4\cos^2x-8(a+1)\cos x+4a^2+8a-16\gt0\\\cos^2x-2(a+1)\cos x+a^2+2a-4\gt0\\(\cos x-(a+1))^2-5\gt0\\(\cos x-a-1+\sqrt5)(\cos x-a-1-\sqrt5)\gt0\\\cos x\lt a+1-\sqrt5 \text{ or }\cos x\gt a+1+\sqrt5$$
Now, I know that $\cos x$ lies between $-1$ and $1$. Not able to use that to solve the inequality.
Also, the same question has been discussed here, but the approach is different. Also, the final answer posted there is not correct.
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You are almost done! Each of the two inequalities has to be true for all $x \in \mathbb{R}$.
Let us consider the first situation:
$$\text{cos} \ x<a+1-\sqrt{5} , \ (\forall) \ x \in \mathbb{R}$$
Since the maximum value of $\text{cos} \ x$ is $1$ we get that $$1<a+1-\sqrt{5} \iff \sqrt{5} <a$$
For the second case:
$$\text{cos} \ x>a+1+\sqrt{5} , \ (\forall) \ x\in \mathbb{R}$$
Since the minimum value of $\text{cos} \ x$ is $-1$ we get that $$-1>a+1+\sqrt{5} \iff -2-\sqrt{5} >a$$
Thus, $$a \in (-\infty,-2-\sqrt{5}) \cup (\sqrt{5}, \infty)$$
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Patterns in indefinite integral (1) So far, I've found out this:
$\int \arcsin x \mathrm d x = x \arcsin x + \sqrt{1-x^2}+C$
$\int x\arcsin x \mathrm d x = \dfrac{x^2}{2}\arcsin x +\dfrac {x}{4}\sqrt{1-x^2} -\dfrac{1}{4}\arcsin x + C$
$\int x^2 \arcsin x \mathrm d x = \dfrac {x^3}{3}\arcsin x+\dfrac{x^2}{9}\sqrt{1-x^2}+\dfrac{2}{9}\sqrt{1-x^2} + C$
$\int x^3 \arcsin x \mathrm d x = \dfrac{x^4}{4}\arcsin x+\dfrac{x^3}{4\times 4}\sqrt{1-x^2}+\dfrac{3x}{4\times 4\times 2}\sqrt{1-x^2}-\dfrac{3\times 1}{4\times 4\times 2}\arcsin x + C$
$\int x^4 \arcsin x \mathrm d x = \dfrac {x^5}{5}\arcsin x +\dfrac {x^4}{5 \times 5}\sqrt{1-x^2}+\dfrac{4x^2}{5\times 5\times 3}\sqrt{1-x^2} + \dfrac{4\times2}{5\times 5\times 3}\sqrt{1-x^2}+C$
$\int x^5 \arcsin x \mathrm d x =
\dfrac {x^6}{6}\arcsin x
+\dfrac {x^5}{6\times6}\sqrt{1-x^2}
+\dfrac{5x^3}{6\times6\times4}\sqrt{1-x^2}
+\dfrac{5\times3x}{6\times6\times4\times2}\sqrt{1-x^2}
-\dfrac{5\times3\times1}{6\times6\times4\times2}\arcsin x + C$
There seems to be a pattern. It seems like for any positive whole number $n$,
$\int x^n \arcsin x \mathrm d x = \mathrm{(polynomial)}\arcsin x + \mathrm{(polynomial)}\sqrt{1-x^2} + C$
So, what is the pattern? What should the above polynomial be?
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Note: please help with formatting
Here's what I found:
$\int x^n \arcsin x \mathrm d x$
$= \dfrac{x^{n+1}}{n+1}\arcsin x - \int \dfrac{x^{n+1}}{(n+1)\sqrt{1-x^2}}\mathrm d x$
$= \dfrac{x^{n+1}}{n+1}\arcsin x - \dfrac{1}{n+1}\int \dfrac{x^{n+1}}{\sqrt{1-x^2}}\mathrm d x$
[ let x = sin u ]
$= \dfrac{x^{n+1}}{n+1}\arcsin x - \dfrac{1}{n+1}\int \dfrac{(\sin u)^{n+1}}{\sqrt{1-(\sin u)^2}}\mathrm d (\sin u)$
$= \dfrac{x^{n+1}}{n+1}\arcsin x - \dfrac{1}{n+1}\int (\sin u)^{n+1}\mathrm d u$
[ see integration chart ]
$= \dfrac{x^{n+1}}{n+1}\arcsin x - \dfrac{1}{n+1} \left( -\dfrac{1}{n+1}(\sin u)^n\cos u +\dfrac{n}{n+1}\int (\sin u)^{n-1}\mathrm d u \right)$
$= \cdots$
[ sin u = x ]
$= \dfrac{x^{n+1}}{n+1}\arcsin x
+ \dfrac{x^n}{(n+1)^2}\sqrt{1-x^2}
+ \dfrac{n x^{n-2}}{(n+1)^2(n-1)}\sqrt{1-x^2}\\
+ \dfrac{n}{(n+1)^2}\cdot
\dfrac{(n-2)}{(n-1)}\cdot
\dfrac{x^{n-4}}{(n-3)}\sqrt{1-x^2} + \cdots \\
+ \dfrac{n}{(n+1)^2}\cdot
\dfrac{(n-2)}{(n-1)}\cdot
\dfrac{x^{n-4}}{(n-3)}\cdots
\dfrac{5}{6}\cdot
\dfrac{3}{4}\cdot
\dfrac{x^1}{2}\sqrt{1-x^2}
- \dfrac{n}{(n+1)^2}\cdot
\dfrac{(n-2)}{(n-1)}\cdot
\dfrac{x^{n-4}}{(n-3)}\cdots
\dfrac{5}{6}\cdot
\dfrac{3}{4}\cdot
\dfrac{1}{2}\arcsin x + C$ [... when n is odd]
Or
$\dfrac{x^{n+1}}{n+1}\arcsin x
+ \dfrac{x^n}{(n+1)^2}\sqrt{1-x^2}
+ \dfrac{n x^{n-2}}{(n+1)^2(n-1)}\sqrt{1-x^2}\\
+ \dfrac{n}{(n+1)^2}\cdot
\dfrac{(n-2)}{(n-1)}\cdot
\dfrac{x^{n-4}}{(n-3)}\sqrt{1-x^2} + \cdots \\
+ \dfrac{n}{(n+1)^2}\cdot
\dfrac{(n-2)}{(n-1)}\cdot
\dfrac{x^{n-4}}{(n-3)}\cdots
\dfrac{4}{5}\cdot
\dfrac{2}{3}\cdot
\sqrt {1-x^2} + C$ [...when n is even]
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Point inside a right angled triangle A right angled triangle $ABC$ $(\measuredangle C=90^\circ)$ is given with $\measuredangle BAC=\alpha$. The point $O$ lies inside the triangle $ABC$ such that $\measuredangle OAB=\measuredangle OBC=\measuredangle OCA=\varphi$. Show that $\tan\varphi=\sin\alpha\cdot\cos\alpha.$
We can write the RHS of the equality that we are supposed to prove as $$\sin\alpha\cdot\cos\alpha=\dfrac{a}{c}\cdot\dfrac{b}{c}=\dfrac{ab}{c^2}$$
So we can try to show the equality $$\tan\varphi=\dfrac{ab}{c^2}$$
To express $\tan\varphi$ in some way, all I can think of is to include $\measuredangle\varphi$ in a right triangle and then use the definition of tangent of an acute angle. For this purpose, let's draw perps from $O$ to the sides of the triangle $ABC$. Their foots are $H, H_1, H_2$ on $AB,BC$ and $AC$, respectively. Then we have $$\tan\varphi=\dfrac{OH}{AH}=\dfrac{OH_1}{BH_1}=\dfrac{OH_2}{CH_2}$$
This seems pointless. Thank you in advance!
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Let the hypotenuse $AB = 1$. Then $BC=\sin\alpha$.
Consider $\triangle OAB$. $\angle OAB = \varphi$ and $\angle ABO = (90^\circ-\alpha) - \varphi$, so $\angle BOA = 90^\circ + \alpha$. By the laws of sine,
$$\begin{align*}
\frac{OB}{\sin \angle OAB} &= \frac{AB}{\sin\angle BOA}\\
OB &= \frac{1\cdot\sin \varphi}{\sin (90^\circ + \alpha)}\\
&= \frac{\sin \varphi}{\cos \alpha}\\
\end{align*}$$
Then consider $\triangle OBC$. $\angle OBC = \varphi$ and $\angle BCO = 90^\circ - \varphi$, so $\angle COB = 90^\circ$.
$$\begin{align*}
\cos \angle OBC = \cos\varphi &= \frac{OB}{BC}\\
\cos\varphi &= \frac{\sin\varphi}{\cos \alpha} \cdot \frac{1}{\sin\alpha}\\
\sin\alpha\cos\alpha &= \frac{\sin\varphi}{\cos \varphi}\\
\tan \varphi&= \sin\alpha \cos \alpha
\end{align*}$$
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A question involving the use of complex numbers.
Evaluate $2^{n-1}\left(\cos \theta -\cos(\frac{\pi}{n})\right)\left(\cos \theta -\cos(\frac{2\pi}{n})\right)...\left(\cos \theta -\cos(\frac{(n-1)\pi}{n})\right)$
In the above question the terms $\cos(\frac{\pi}{n}),\cos(\frac{2\pi}{n}),...$ are the real part of a complex number given by $$z=e^{i\frac{k\pi}{n}}$$
where $k = 1,2,...,n-1 $.
If we expand the above problem then we would have to compute $\Sigma_0^{n-1}\cos(\frac{k\pi}{n}),...$ and also the summation of other terms.It will obvious not work.
The author has given a hint to solve the above question.
Hint: Use the expansion for $\frac{x^{2n}-1}{x^2-1}$ to solve the problem.
I literally have no idea how to proceed further and solve it. I don't know in what way we can use the hint to solve the question as I cannot see or establish any clear connection between them.
Any help will be appreciated.
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Let $z^{2n}=1$
$$\implies z^{2n}-1=(z-z_1)(z-z_2)(z-z_3)....(z-z_{2n})$$
where $z_1,z_2...z_{2n}$ are the roots of unity
Since $z_k^{2n}=e^{2k\pi i}$,
$$z_k=e^{i(\frac{k\pi}{n})}=\cos\bigg(\frac{k\pi}{n}\bigg)+i\sin\bigg(\frac{k\pi}{n}\bigg)$$
Notice that
$$z_{2n-k}=\cos\bigg(\frac{(2n-k)\pi}{n}\bigg)+i\sin\bigg(\frac{(2n-k)\pi}{n}\bigg)$$
$$=\cos\bigg(\frac{2n\pi-k\pi}{n}\bigg)+i\sin\bigg(\frac{2n\pi-k\pi}{n}\bigg)$$
$$=\cos\bigg(2\pi-\frac{k\pi}{n}\bigg)+i\sin\bigg(2\pi-\frac{k\pi}{n}\bigg)$$
$$\cos\bigg(\frac{k\pi}{n}\bigg)-i\sin\bigg(\frac{k\pi}{n}\bigg)=\bar{z_k}$$
Also,
$$z_n=\cos(\pi)+i\sin(\pi)=-1$$
and
$$z_{2n}=\cos(2\pi)+i\sin(2\pi)=1$$
Grouping all the conjugates together and then multiplying them,
$$z^{2n}-1=(z-z_1)(z-z_{2n-1})(z-z_2)(z-z_{2n-2})....(z_n)(z_{2n-n})$$
$$=(z-z_1)(z-\bar{z_1})(z-z_2)(z-\bar{z_2})....(z-1)(z+1)$$
$$=\big[z^2-z(z_1+\bar{z_1})+|z_1^2|\big]\big[z^2-z(z_2+\bar{z_2})+|z_2^2|\big]....(z^2-1)$$
Since $|z_k|^2=1$ and $(z_k+\bar{z_k})=2Re(z_k)=2\cos(\frac{k\pi}{n})$,
$$z^{2n}-1=\big[z^2+1-2z\cos\bigg(\frac{\pi}{n}\bigg)\big]\big[z^2+1-2z\cos\bigg(\frac{2\pi}{n}\bigg)\big]....(z^2-1)$$
$$\implies \frac{z^{2n}-1}{z^2-1}=\big[z^2+1-2z\cos\bigg(\frac{\pi}{n}\bigg)\big]\big[z^2+1-2z\cos\bigg(\frac{2\pi}{n}\bigg)\big]....$$
Dividing both sides by $z^{n-1}$,
$$\frac{z^n-\frac{1}{z^n}}{z-\frac{1}{z}}=\big[z+\frac{1}{z}-2\cos\bigg(\frac{\pi}{n}\bigg)\big]\big[z+\frac{1}{z}-2\cos\bigg(\frac{2\pi}{n}\bigg)\big]....$$
Substituting $z=\cos(\theta)+i\sin(\theta)$
$\implies \frac{1}{z}=\cos(\theta)-i\sin(\theta)$
Also, $z^n=\cos(n\theta)+i\sin(n\theta)$ $\implies\frac{1}{z^n}=\cos(n\theta)-i\sin(n\theta)$
$$\implies\frac{z^n-\frac{1}{z^n}}{z-\frac{1}{z}}= \frac{2i\sin(n\theta)}{2i\sin(\theta)}=\bigg[2\cos(\theta)-2\cos\bigg(\frac{\pi}{n}\bigg)\bigg]\bigg[2\cos(\theta)-2\cos\bigg(\frac{2\pi}{n}\bigg)\bigg].....$$
$$\implies 2^{n-1}\prod_{k=1}^{n-1}[\cos(\theta)-\cos\big(\frac{k\pi}{n}\big)]=\fbox{$\frac{\sin(n\theta)}{\sin(\theta)}$}$$
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Given $a,b,c$ are sides of a triangle, Prove that :- $\frac{1}{3} \leq \frac{a^2+b^2+c^2}{(a+b+c)^2} < \frac{1}{2}$
Given $a,b,c$ are sides of a triangle, Prove that :-
$$\frac{1}{3} \leq \frac{a^2+b^2+c^2}{(a+b+c)^2} < \frac{1}{2}$$
What I Tried:- I was able to solve the left hand side inequality. From RMS-AM Inequality on $a,b,c$ :-
$$\sqrt{\frac{a^2+b^2+c^2}{3}} \geq \frac{a+b+c}{3}$$
$$\rightarrow \frac{a^2+b^2+c^2}{3} \geq \frac{(a+b+c)^2}{9}$$
$$\rightarrow \frac{1}{3} \leq \frac{a^2+b^2+c^2}{(a+b+c)^2}$$
I got no progress for the second part. I also have a clue, that $a,b,c$ are sides of a triangle, which I have not used yet. So maybe that should be used somehow, but I am not getting it.
Can anyone help me? Thank You.
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Note that the sum of any two sides is greater then the third side, which shows $$0 < (a+b-c)(a-b+c) = a^2 - (b-c)^2.$$
Similarly $$\begin{eqnarray}
a^2 &>& (b-c)^2\\
b^2 &>& (a-c)^2\\
c^2 &>& (a-b)^2
\end{eqnarray}.$$ Summing these lines:
$$a^2+b^2+c^2 > 2(a^2+b^2+c^2 - ab - ac - bc)\\ = 3(a^2+b^2+c^2) - (a+b+c)^2.$$
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solve the equation: $\cos (z) = 1 + i$ , $z \in ℂ$. I want to solve the equation: $\cos (z) = 1 + i$ , $z \in ℂ$.
I started by saying $\cos (z) = \frac {e^{iz} + e^{-iz}}{2}$.
I am therefore led to solve:
$$e^{iz} + e^{-iz} = 2 + 2i $$ this implies that $$e^{2iz}+1 =(2 + 2i) e^{iz}$$
By setting $x = e^{iz}$
Consider the equation:
$$x^2-(2 + 2i) x + 1 = 0$$
which is equivalent to solving the equation $$ x^2-2 \sqrt{2} e^{i \frac {\pi}{4}} x + 1 = 0$$
then $$ (x- \sqrt{2} e^{i \frac{\pi}{4} })^2 + 1 + 2i = 0$$
We set $$w = x- \sqrt {2} e^{i \frac{\pi}{4}}$$ so $$w^2 = -1-2i$$ But I can't seem to continue. An idea please.
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Try to use the complex definition of the inverse cosine and its analytic extension to simplify:
$$\cos^{-1}(z)=\frac\pi 2+i \ln\left(iz+\sqrt{1+z}\sqrt{1-z}\right)+2k\pi\Bbb Z$$
$$\mathop \implies^\text{z=1+i}\cos^{-1}(1+i)=\frac\pi2+i\ln\left(i(1+i)+\sqrt{1-(1+i)^2}\right)=\frac\pi2+i\ln\left(i-1+\sqrt{1-2i}\right)$$
To take a radical of a complex number, use De Moivre’s theorem and $n\in \Bbb Z$:
$$\sqrt{1-2i}=\sqrt[4]{1^2+(-2)^2}e^{\frac12i\left(\tan^{-1}\left(-\frac21\right)+2\pi n\right)}=\sqrt[4]5\cos\left(\frac12 \tan^{-1}(2)\right) + \sqrt[4]5i\sin\left(\frac12 \tan^{-1}(2)\right)=\pm\sqrt[4]5 \sqrt{\frac12+\frac{\sqrt{5}}{10}}\pm\sqrt[4]5\sqrt{\frac{2}{5+\sqrt5}}i$$
Putting it all together gives the general solution of the following. The signs can be chosen to be +,- or -,+:
$$\cos^{-1}(1+i)=i\ln\left(\pm\sqrt[4]5 \sqrt{\frac12+\frac{\sqrt{5}}{10}}-1\mp\sqrt[4]5\sqrt{\frac{2}{5+\sqrt5}}i+i\right)+ \frac\pi2+ 2\pi k, k\in \Bbb Z.$$
Proof of result for a case. Please correct me and give me feedback!
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$\lim_{x \to \frac{\pi}{2}} \biggl|\frac{1}{(1+\sin x)\log(\sin x)}+\frac{1}{\cos^2x} \biggl| $
$$\lim_{x \to \frac{\pi}{2}} \biggl|\frac{1}{(1+\sin x)\log(\sin x)}+\frac{1}{\cos^2x} \biggl| $$
My Approach:
$$\lim_{x \to \frac{\pi}{2}} \biggl|\frac{1}{(1+\sin x)\log(\sin x)}+\frac{1}{(1-\sin x)(1+\sin x)} \biggl|$$
$$\lim_{x \to \frac{\pi}{2}} \frac{1}{2}\biggl|\frac{1}{\log(\sin x)}+\frac{1}{(1-\sin x)} \biggl| $$
$$\lim_{x \to \frac{\pi}{2}} \frac{1}{2}\biggl|\frac{(\sin x - 1)}{(\sin x-1)\log(1+(\sin x - 1))}+\frac{1}{(1-\sin x)} \biggl|$$
I could not processed further.
By Applying L'Hôpital's Rule I get answer $0$ but given answer is $\frac{1}{4}$
Also L'Hôpital's Rule is going lengthy.
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Let me use $\ln$ in place of $\log$ and get from your third line
$$\frac{1}{\ln(\sin x)}+\frac{1}{(1-\sin x)} =\frac{1-\sin x+\ln(\sin x)}{(1-\sin x)\ln(\sin x)} = \\=
\frac{1-\sin x+\sin x -1+(\sin x-1)^2\frac{1}{2}-(\sin x - 1)^3\frac{1}{3}+\cdots}{-(\sin x - 1)^2}\to -\frac{1}{2}$$
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Finding the sum of the series $1^2 + 2 × 2^2 + 3^2 + 2 × 4^2 + 5^2 + 2 × 6^2 + . . . + 2(n − 1)^2 + n^2 ,$ The following is a question which has been bugging me for quite a while,
Find the sum of the series
$$1^2 + 2 × 2^2 + 3^2 + 2 × 4^2 + 5^2 + 2 × 6^2 + . . . + 2(n − 1)^2 + n^2 ,$$
Where $n$ is odd
I started by denoting this entire series with $S$, from there it is apparent that $S$ infact consists of two lesser series I shall call;
*
*$S_A$ The sum of the squares of odd natural numbers up till $n^2$
*$S_B$ Twice the sum of the squares of even natural numbers up till $(n-1)^2$
I resolved this would be easier to tackle by noting that
$$S_A + \frac{1}{2}S_B = \sum_{r=1}^{n} r^2$$
Which is just the sum of the squares of the first $n$ natural numbers
$$\therefore S_A +\frac{1}{2}S_B=\frac{n}{6}(n+1)(2n+1)$$
Leaving only the value of $\frac{1}{2}S_B$ to be found, this is where I am currently facing difficulty as I am unsure on whether my working is correct;
For the sum of the squares of the first n even natural numbers;
$$2^2 + 4^2 .... (2n)^2=2^2\sum_{r=1}^{n} r^2$$
$$\implies \frac{2}{3}n(n+1)(2n+1) $$
Hence the sum of the first $n-1$ even natural numbers should be
$$ \frac{2}{3}n(n-1)(2n-1)$$
And
$$S= \frac{n}{6}(n+1)(2n+1) + \frac{2}{3}n(n-1)(2n-1)$$
$$\therefore S= \frac{1}{6}n(10n^2 -9n + 5n) $$
However the correct answer is
$$\frac{1}{2}n^2(n+1)$$
Where has my working gone wrong and how would I arrive at the correct answer?
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The given sum is equal to
$$\begin{align}
1^2 + 2 × 2^2 + 3^2 + &2 × 4^2 + 5^2 + 2 × 6^2 + \dots + 2(n − 1)^2 + n^2\\
&=1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + \dots + (n − 1)^2 + n^2\\
&\qquad\;+2^2 + 4^2 + 6^2 + \dots + (n − 1)^2\\
&=\sum_{k=1}^n k^2+\sum_{k=1}^{m}(2k)^2=\sum_{k=1}^n k^2+4\sum_{k=1}^{m}k^2\\
&=\frac{n(n+1)(2n+1)}{6}+4\frac{m(m+1)(2m+1)}{6}\\&=\frac{n(n+1)(2n+1)}{6}+\frac{(n-1)(n+1)n}{6}\\
&=\frac{n(n+1)(2n+1+n-1)}{6}=\frac{n^2(n+1)}{2}
\end{align}$$
where $m=\frac{n-1}{2}$. Note that in the sum involving the even squares there are $m$ terms (not $(n-1)$).
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"url": "https://math.stackexchange.com/questions/4225909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Are there infinitely many positive integer solutions to $(xz+1)(yz+1)=P(z)$? Let $P(z)\equiv 1($ mod $ \ z) $ be a polynomial of degree $n>3$ with integer coefficients. Are there infinitely many positive integers $x, y, z$ such that $(xz+1)(yz+1)=P(z)$?
If $P(z) = a_nz^n+1$, it has be proven that the Diophantine equation has infinitely many solutions in positive integers $x, y, z$ Prove that the diophantine equation $(xz+1)(yz+1)=az^{k}+1$ has infinitely many solutions in positive integers..
From experiment, it appears the assertion is true for all polynomials $P(z)\equiv 1$(mod$ \ z)$ of degree $n>3$. How do we go about proving this?
Note if $n=3$, it has been shown that the Diophantine equation has a finite number of solutions
https://mathoverflow.net/questions/392002/is-xz1-a-proper-divisor-of-a-3z3a-2z2a-1z1-finitely-often/392018#392018
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I figured out all solutions in positive integers $\ x, y , z \ $ to $(xz+1)(yz+1)=z^4+z^3+z^2+z+1$.
First, define some sequences as follows:
$r_m = m^2+m-1, \ \ \ m=1, 2, \ldots $
$q_m = (r_m+2)^2-2,\ \ \ m=1, 2, \ldots $
$a_m = m+1, \ \ \ m=1,2,\ldots $
$b_m = m^5 + 3m^4 + 5m^3 + 4m^2 +m, \ \ \ m=1,2,\ldots $
$c_m = m^3 + 2m^2 +2m, \ \ \ m=1,2,\ldots $
$e_m = m^3 + m^2 +m+1, \ \ \ m=1,2,\ldots $
$g_m = m^5 + 2m^4 + 3m^3 + 3m^2 +m, \ \ \ m=1,2,\ldots $
And for a particular $m$, define sequences $A,B,C,E,F,G$ as follows;
$A_1 = a_m$, $A_2 = b_m$, $A_n = q_mA_{n-1} - A_{n-2} - r_m, \ \ \ n = 3, 4 , \dots$
$B_n = A_{n+1}, \ \ \ n = 1, 2, \ldots $
$C_1 = c_m$, $C_2 = q_mC_1+r_m$, $C_n = q_mC_{n-1} - C_{n-2} + r_m, \ \ \ n = 3, 4 , \dots$
$E_1 = e_m$, $E_2 = q_mE_1-r_m$, $E_n = q_mE_{n-1} - E_{n-2} - r_m, \ \ \ n = 3, 4 , \dots$
$F_n = E_{n+1}, \ \ \ n = 1, 2, \ldots $
$G_1 = g_m$, $G_2 = q_mG_1+ r_m - m$, $G_n = q_mG_{n-1} - G_{n-2} + r_m, \ \ \ n = 3, 4 , \dots$
All positive integer solutions $(x,y,z)$ are given by $(A_n, B_n, C_n)$, $(E_n, F_n, G_n), n = 1, 2, \dots $.
NB. This is simply an observation. A proof is required to show that $(A_n, B_n, C_n)$, $(E_n, F_n, G_n)$ are indeed solutions for every $n$ and that these are the only solutions in positive integers.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/4226419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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|
Trigonometric equality proof $ \cos^2(\omega t) + \cos^2(\omega t + \delta) = \sin^2\delta + 2\cos(\omega t + \delta)\cos(\omega t)\cos(\delta)$ Looking to prove
$x = A\cos(\omega t)\\
y = A\cos(\omega t + \delta)
\\
\\$
YIELDS
$x^2-2xy\cos(\delta)+y^2=A^2\sin^2(\delta)$
Specifically we're trying to express the equations without any reference to $t$. If it helps, we can look in terms of ${x^2 + y^2 \over {A^2}}$
and just prove
$$ \cos^2(\omega t) + \cos^2(\omega t + \delta) = \sin^2\delta + 2\cos(\omega t + \delta)\cos(\omega t)\cos(\delta)$$
I know the angle sum identity $$\cos(\omega t + \delta) = \cos(\omega t) \cos(\delta) - \sin(\omega t)\sin(\delta)$$
$$ \cos^2(\omega t) + \cos^2(\omega t + \delta) \\ =\cos^2(\omega t) + [\cos(\omega t )\cos(\delta) - \sin(\omega t)\sin(\delta)]^2\\
=\cos^2(\omega t) + \cos^2(\omega t )\cos^2(\delta)+ \sin^2(\omega t) \sin^2(\delta)-2\sin(\omega t)\sin(\delta)\cos(\omega t )\cos(\delta) $$
But I can't get anywhere after that. I suspect that there is some other trig identity I either don't know or am overlooking somewhere.
This is from a physics textbook talking about two dimensional oscillation with the same $\omega$ but offset by a $\delta$ phase angle. It's not a problem i'm just trying to follow along the text and fill in the author's gaps.
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Alternative approach, similar to Deepak's answer.
$\underline{\text{intermediate results}}$
Lemma 1: $~~\cos(a+b) \cos(b-a) = \cos^2(a) - \sin^2(b).$
Proof:
$(\cos a \cos b - \sin a \sin b) \times (\cos a \cos b + \sin a \sin b)$
$= (\cos^2 a ~\cos^2 b) - (\sin^2 a ~\sin^2 b)$
$= [(\cos^2 a)(1 - \sin^2 b)] - [(1 - \cos^2 a)(\sin^2 b)]$
$= \cos^2 a - (\cos^2 a ~\sin^2 b) - \sin^2 b + (\cos^2 a ~\sin^2 b).$
Lemma 2: $~~\cos(a + b) + \cos(b - a) = 2(\cos a)(\cos b).$
This is routinely shown via the angle sum formulas.
Let $a = \omega t, ~b = \delta.$
Then the problem is to prove that
$$\cos^2 a + \cos^2 (a + b) ~=~ \sin^2 b + 2\cos(a+b)\cos(a)\cos(b). \tag1 $$
Using Lemma 2, the RHS of (1) above may be re-expressed as
$= \sin^2 b +
~~\left\{2 \cos(a + b) \left(\dfrac{1}{2}\right) [\cos(a + b) + \cos (b - a)]\right\}$
$= \sin^2 b + \cos^2 (a + b) + \cos(a + b)\cos (b - a).$
Using Lemma 1, the RHS may therefore be re-expressed as:
$\sin^2 b + \cos^2 (a + b) + \cos^2 a - \sin^2 b.$
Therefore, the RHS of (1) equals the LHS of (1).
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/4226986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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|
How to solve $\int_0^\infty \ln(x)\left(e^{-(x+c)^2}+e^{-(x-c)^2}\right)dx$? Mathematica is telling me
$$\int_0^\infty \ln(x)\left(e^{-(x+c)^2}+e^{-(x-c)^2}\right)dx=-\frac{\sqrt\pi}2\left(\gamma+\ln4+f'(0)\right),$$
where
$$f(a)={}_1F_1(a;1/2;-c^2)$$
is the confluent hypergeometric function. I already know how to derive the special case $c=0$,
$$2\int_0^\infty \ln(x)\,e^{-x^2}dx=-\frac{\sqrt\pi}2\left(\gamma+\ln4\right)$$
so I am curious how to arrive at the extra $f'(0)$ for $c>0$. I'm not familiar with hypergeometric functions, so would appreciate an answer assuming minimal background knowledge of them.
An alternative way to write the solution to the integral would be
$$\int_0^\infty \ln(x)\left(e^{-(x+c)^2}+e^{-(x-c)^2}\right)dx=-\frac{\sqrt\pi}2\left(\gamma+\ln4+{{}_1F_1}^{(1,0,0)}(0;1/2;-c^2)\right),$$
where the superscripts denote that a derivative has been taken with respect to the first argument. Not sure if people prefer this notation to what I wrote above.
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A ridiculous approach compared to the simple and elegant solution given by @WillG.
Hoping to tecognize some patterns, I used a series expansion of
$$f(x)=e^{-(x+c)^2}+e^{-(x-c)^2}$$ around $c=0$ which gives
$$f(x)=e ^{-x^2}\, \sum_{n=0}^\infty \frac{P_n(x^2)}{n!}c^{2n}$$ and
$$\int_0^\infty x^{2k}\,e ^{-x^2}\,\log(x)\,dx=\frac{1}{4} \Gamma \left(n+\frac{1}{2}\right) \psi \left(n+\frac{1}{2}\right)$$
Summing from $n=0$ to $n=12$ the result is
$$I=-\frac{ \sqrt{\pi }}{2} \sqrt{\pi } (\gamma +\log (4))+\sqrt \pi \Big[c^2-\frac{c^4}{3}+\frac{4 c^6}{45}-\frac{2 c^8}{105}+\frac{16
c^{10}}{4725}-\frac{16 c^{12}}{31185}\Big]$$ At this point, we recognize the pattern since the series expansion of Dawson's function is
$$F(c)=c-\frac{2 c^3}{3}+\frac{4 c^5}{15}-\frac{8 c^7}{105}+\frac{16 c^9}{945}-\frac{32
c^{11}}{10395}+O\left(c^{13}\right)$$
$$\int_0^c F(t)\,dt=\frac12 \Bigg[c^2-\frac{c^4}{3}+\frac{4 c^6}{45}-\frac{2 c^8}{105}+\frac{16
c^{10}}{4725}-\frac{16 c^{12}}{31185}\Bigg]+O\left(c^{14}\right)$$
In fact
$$\int_0^c F(t)\,dt=\frac{1}{2} c^2 \, _2F_2\left(1,1;\frac{3}{2},2;-c^2\right)$$
Edit
Notice that, if we write
$$\int_0^c F(t)\,dt=\sum_{n=0}^p (-1)^n\frac{\sqrt{\pi } \, c^{2 n+2}}{4 (n+1) \Gamma \left(n+\frac{3}{2}\right)}+\sum_{n=p+1}^\infty (-1)^n\frac{\sqrt{\pi } \, c^{2 n+2}}{4 (n+1) \Gamma \left(n+\frac{3}{2}\right)}$$ we know in advance $p$ such that
$$R_p=\frac{\sqrt{\pi }\, c^{2 p+4}}{4 (p+2) \Gamma \left(p+\frac{5}{2}\right)} \leq \epsilon$$.
It is given by
$$p \sim \left\lceil c^2\, e^{1+W(t)} \right\rceil -3 \qquad \text{with} \qquad t=-\frac{\log \left(32 c^4 \epsilon ^2\right)}{2 e c^2}$$
Trying for $c=2$ and $\epsilon=10^{-20}$, this gives $p=33$. Checking
$$R_{32}=7.63\times 10^{-20} > 10^{-20}\qquad \text{while} \qquad R_{33}=8.59\times 10^{-21} < 10^{-20}$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/4227738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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|
I'm stuck in derivatives for finding Curvature of radius I know the formula How can we find the radius of curvature for this equation
$$(x^2+y^2)^2=a^2(x^2-y^2)$$
Here I know the formula but I just want to find y' and y''
I'm stuck in it
here is what know
$$\delta=\frac{(1+y'^2)^\frac{3}{2}}{y''}$$
i got y'
$$y'=\frac{x}{y}\frac{a-2\sqrt{(x^2-y^2)}}{a+2\sqrt{(x^2-y^2)}}$$
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It is easier to do this in polar coordinates,
$(x^2+y^2)^2=a^2(x^2-y^2)$
Using $x = r \cos\theta, y = r \sin\theta$,
$r^4 = a^2 r^2 \cos2\theta \implies r^2 = a^2 \cos2\theta$
$r_{\theta} = - \cfrac{a^2 \sin2\theta}{r} \implies r_{\theta}^2 = \cfrac{a^4 - r^4}{r^2}$
$r_{\theta \theta} = \cfrac{a^2 \sin2\theta}{r^2} r_{\theta} - \cfrac{2 a^2 \cos2\theta}{r} = - \cfrac{r_{\theta}^2}{r} - 2 r$
Radius of curvature, $R = \cfrac{(r^2 + r_{\theta}^2)^{3/2}}{|r^2 + 2 r_{\theta}^2 - r r_{\theta \theta}| } = \cfrac{\sqrt{r^2 + r_{\theta}^2}}{3}$
$ = \cfrac{a^2}{3 r} = \cfrac{a^2}{3 \sqrt{x^2+y^2}}$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/4228218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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