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If $a^2 + b^2 + c^2 = 2$, find the maximum of $\prod(a^5+b^5)$ Given that $a, b, c > 0$ and $a^2 + b^2 + c^2 = 2$, what is the maximum value of $(a^5 + b^5)(a^5 + c^5)(b^5 + c^5)$?
Normally when I encounter a problem like this, I seem to be able to push through with AM-GM. This one seems a little problematic since I can't seem to make any valid substitutions. I would prefer an elementary solution if there is one, that is without resorting to calculus. Thanks for your help.
|
Let $c=\min\{a,b,c\}$, $a^2+\frac{c^2}{2}=x$ and $b^2+\frac{c^2}{2}=y$.
Hence, $x+y=2$, $xy\leq1$, $\sqrt{x}+\sqrt{y}\leq\sqrt{(1+1)(x+y)}=2$ and
$a^5+c^5\leq\left(a^2+\frac{c^2}{2}\right)^{\frac{5}{2}}=x^{\frac{5}{2}}$,$b^5+c^5\leq\left(b^2+\frac{c^2}{2}\right)^{\frac{5}{2}}=y^{\frac{5}{2}}$ and
$$a^5+b^5\leq\left(a^2+\frac{c^2}{2}\right)^{\frac{5}{2}}+\left(b^2+\frac{c^2}{2}\right)^{\frac{5}{2}}=x^{\frac{5}{2}}+y^{\frac{5}{2}}.$$
Thus,
$$\prod_{cyc}(a^5+b^5)\leq(xy)^{\frac{5}{2}}\left(x^{\frac{5}{2}}+y^{\frac{5}{2}}\right)=(\sqrt{x}+\sqrt{y})(xy)^{\frac{5}{2}}(x^2-\sqrt{x^3y}+xy-\sqrt{xy^3}+y^2)\leq$$
$$\leq2(xy)\cdot(xy)\cdot(xy)^{\frac{1}{2}}(x^2+xy+y^2-2\sqrt{xy})\leq$$
$$\leq2(xy)\cdot(xy)^{\frac{1}{2}}\cdot(xy)^{\frac{1}{2}}(x^2+xy+y^2-2\sqrt{xy})\leq$$
$$\leq2\left(\frac{xy+\sqrt{xy}+\sqrt{xy}+x^2+xy+y^2-2\sqrt{xy}}{4}\right)^4=2\left(\frac{(x+y)^2}{4}\right)^4=2.$$
|
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|
Find the area of the region inside: $r= 6\sin(\theta)$ but outside of $r = 1$ How do we find the area of the region inside $r = 6 \sin(\theta)$, but outside $r = 1$?
So, here's my work thus far:
First off, we know:
$r^2 = x^2 + y^2$
and $\mathrm{sin}(\theta) = y/r$
Therefore,
$r = \sqrt{x^2+y^2}$
and $6\mathrm{\sin}(\theta) = \frac{6y}{r}$
The original equation turns into:
$\sqrt{x^2 + y^2} = 6y/r$ ; or equivalently $x^2 + y^2 = 6y$
Completing the square turns this into
$x^2 + (y-3)^2 = 9$, a circle centered at $(0,3)$ with a radius of $3$
The area of that circle is $9\pi$, so the answer is going to be $9\pi$ - something. I'm having a hard time figuring out the something.
The other circle is $x^2 + y^2 = 1$, a circle centered at the origin with a radius of $1$.
Well, I was thinking, let's find the points of intersection by setting the two equations equal to eachother. So set $x^2 + y^2 - 1 = x^2 + y^2 - 6y$ (because they both = 0).
So cancel out the $x^2$ and $y^2$, so we get $-1 = -6y$ or $y = 1/6$. Plug that back in and find the two x coordinates. Plugging it back into $x^2 + y^2 = 1$, so you get +/- the $\sqrt{35/36}$. So the two intersection points are $(-\sqrt{35/36} , 1/6)$ and $(\sqrt{35/36} , 1/6)$. So I'm thinking, alright, just integrate the following from $-\sqrt{35/36}$ to $\sqrt{35/36}$: $x^2 + y^2 - 1 - (x^2 + y^2 - 6y)dx.$ But I get the integral of $(6y - 1)dx$. I don't understand please help!
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You want the region where $r>1$. That's the same as $\sin\theta > 1/6$. The boundary of that region is where $\sin\theta=1/6$. That happens when $\theta = \arcsin(1/6)$ and when $\theta = \frac\pi 2 - \arcsin(1/6)$. So you want
$$
\int_{\arcsin(1/6)}^{\pi/2-\arcsin(1/6)} \frac{r^2}{2} \, d\theta = \int_{\arcsin(1/6)}^{\pi/2-\arcsin(1/6)} 18\sin^2\theta \, d\theta = \int_{\arcsin(1/6)}^{\pi/2-\arcsin(1/6)} 9 - 9\cos(2\theta)\,d\theta
$$
$$
= \left[ 9\theta - \frac92\sin(2\theta) \right]_{\arcsin(1/6)}^{\pi/2-\arcsin(1/6)}
$$
To find $\sin(2\arcsin\frac 1 6)$ use the double-angle formula for the sine. For that, you want to remember that $\cos(\arcsin x) = \sqrt{1-x^2}$. To find $\sin(2(\frac\pi 2 - \arcsin(1/6))) = \sin(\pi - 2\arcsin(1/6))$, remember that $\sin(\pi - \theta)$ is the same as $\sin\theta$.
And remember that $\cos(\pi-\theta) = -\cos(\theta)$.
|
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|
What is the maximum value of this trigonometric expression What is the maximum value of the expression
$1/(\sin^2 \theta + 3\sin\theta \cos\theta+ 5\cos^2 \theta$).
I tried reducing the expression to $1/(1 + 3\sin\theta$ $\cos\theta + 4\cos^2 \theta)$.
How do I proceed from here?
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There are as usual many approaches. We describe one that uses pure trigonometry (and in particular no calculus), in line with the tag on the question.
Trigonometry can be bypassed in favour of algebra, as we show in the second part of this answer. But it would not be a good idea to omit the trigonometric approach altogether, since in the process we describe an idea that is useful in Physics, Electrical Engineering, and elsewhere.
A solution via trigonometry: In order to have less messy expressions, we will mainly study
$$\sin^2\theta+3\sin\theta\cos\theta+5\cos^2\theta.$$
It looks sensible enough to use $\sin^2\theta=1-\cos^2\theta$ to rewrite our expression as
$$1+3\sin\theta\cos\theta+4\cos^2\theta.$$
Using the trigonometric identities $2\sin\theta\cos\theta=\sin 2\theta$ and $\cos 2\theta=2\cos^2\theta-1$, we can then rewrite our expression as
$$3+\frac{3}{2}\sin 2\theta +2\cos 2\theta.$$
Next comes the idea which is useful elsewhere. We express $\frac{3}{2}\sin2\theta+2\cos 2\theta$ as $k \sin(2\theta+\alpha)$ for suitable $k$ and $\alpha$.
The idea is to use the identity
$$\sin(2\theta +\alpha)=(\sin2\theta)(\cos\alpha)+(\cos2\theta)(\sin\alpha).$$
If our expression is to be equal to $k\sin(2\theta+\alpha)$,
we need to have $(3/2)/k=\cos\alpha$ and $2/k=\sin\alpha$. This means that $k=\sqrt{(3/2)^2+2^2}$. It turns out that $k$ is the nice number $5/2$ (thank you, problem designer). So we are looking at
$$3+\frac{5}{2}\left(\frac{3}{5}\sin 2\theta +\frac{4}{5}\cos 2\theta\right).$$
This is
$$3+\frac{5}{2}\sin(2\theta+\alpha),$$
where $\alpha$ is the angle whose cosine is $3/5$ and whose sine is $4/5$. Note by the way that our expression is always positive, since $\sin(2\theta+\alpha)$ can never be less than $-1$.
The above expression is smallest when $\sin(2\theta+\alpha)=-1$. This is obviously achievable by letting $2\theta=3\pi/2-\alpha$.
So the minimum value of our expression is $1/2$.
The maximum value of the original function is therefore $2$. If we wanted the minimum value of the original function, that is also easily found.
Comment: The method that we used to get to the expression $k\sin(2\theta+\alpha)$ can be used to express $a\sin\phi +b\cos\phi$, where $a$ and $b$ are constants, as $k\sin(\phi + \delta)$, for suitable $k$ and $\delta$.
A solution via algebra: Let $x=\sin\theta$ and $y=\cos\theta$. We want to minimize $x^2+3xy+5y^2$ subject to the condition $x^2+y^2=1$.
To do this we examine the values of $k$ for which the curves $x^2+3xy+5y^2=k$ and $x^2+y^2=1$ kiss. (From the geometry we can see that at the minimum the two curves will be tangent, so have double point of intersection. That will also happen at the maximum.)
For the minimum, exactly one of $x$ and $y$ will be negative, it doesn't matter which. So let $x=\sqrt{1-y^2}$. Substitute for $x$ in $x^2+ 3xy+4y^2=k$. We obtain $4y^2-(k-1)=-3y\sqrt{1+y^2}$. Square both sides and simplify. We arrive at the equation
$$25y^4-(8k+1)y^2 +(k-1)^2=0.$$
For $y^2$ to be a double root of this equation, the discriminant must be $0$. So we obtain the equation $(8k+1)^2=100(k-1)^2$, and thus
$8k+1=\pm10(k-1)$. There is an extraneous root which gives us the maximum. The relevant $k$ is $1/2$. Thus the minimum value of $x^2+3xy+5y^2$ subject to $x^2+y^2=1$ is $1/2$, so the maximum value of the function in the post is $2$.
|
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|
Trigonometric system I would like to solve:
$ x +y+z=\frac{11\pi}{6} $
$ \sin(x)+\sin(y)+\sin(z)= \frac{\sqrt{3}}{2} $
$ \cos(x)+\cos(y)+\cos(z)=\frac{1}{2} $
After eliminating $ z $ I get:
$ 2\sin(x)+2\sin(y)-\cos(x+y)-\sqrt{3}\sin(x+y)=\sqrt{3}\tag{1}$
$ 2\cos(x)+2\cos(y)+\sqrt{3}\cos(x+y)-\sin(x+y)=1\tag{2}$
Also: $\sqrt{3}\times(1)+(2)$ gives $ \cos(x-\frac{\pi}{3})+\cos(y-\frac{\pi}{3})-\sin(x+y)=1 $...
My attempt:
$ \sin(x)+\sin(y)+\sin(z)=\sin(x+y+z+\frac{\pi}{2}) $
$ \sin(x)+\sin(y)+\sin(z)-\sin(x+y+z+\frac{\pi}{2})=0 $
$(1)$: $ \sin(\frac{x+y}{2})\cos(\frac{x-y}{2})-\sin(\frac{x+y}{2}+\frac{\pi}{4})\cos(\frac{x+y+2z}{2}+\frac{\pi}{4})=0 $
$ \cos(x)+\cos(y)+\cos(z)-\cos(x+y+z+\frac{\pi}{2})=0 $
$ (2) $: $ \cos(\frac{x+y}{2})\cos(\frac{x-y}{2})+\sin(\frac{x+y+2z}{2}+\frac{\pi}{4})\sin(\frac{x+y}{2}+\frac{\pi}{4})=0 $
$(1)+(2)$: $ \cos(\frac{x-y}{2})\cos(\frac{x+y}{2}-\frac{\pi}{4})+\sin(\frac{x+y}{2}+\frac{\pi}{4})\sin(\frac{x+y+2z}{2})=0 $
$ \cos(\frac{x+y}{2}-\frac{\pi}{4})(\cos(\frac{x-y}{2})+\sin(\frac{x+y+2z}{2}))=0 $
$ \cos(\frac{x+y}{2}-\frac{\pi}{4})(\sin(\frac{x-y}{2}+\frac{\pi}{2})+\sin(\frac{x+y+2z}{2}))=0 $
$ \cos(\frac{x+y}{2}-\frac{\pi}{4})\sin(\frac{x+z}{2}+\frac{\pi}{4})\cos(\frac{y+z}{2}-\frac{\pi}{4})=0 $
$ x+y=-\frac{\pi}{2} (\mod2\pi) $
$ x+z=-\frac{\pi}{2} (\mod2\pi) $
$ y+z=-\frac{\pi}{2} (\mod2\pi) $
Using only these equations to determine $ x,y,z $:
$ x=y=z=-\frac{\pi}{4} (\mod\pi) $
and the system is not satisfied.
Using $ x+y+z=\frac{11\pi}{6} $ to determine one of the three quantities:
$x=y=-\frac{\pi}{4} (\mod\pi) $, $ z=\frac{\pi}{3} (\mod\pi) $
or
$x=z=-\frac{\pi}{4} (\mod\pi) $, $ y=\frac{\pi}{3} (\mod\pi) $
or
$y=z=-\frac{\pi}{4} (\mod\pi) $, $ x=\frac{\pi}{3} (\mod\pi) $
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Multiply the 2nd equation by $i$ and add the 3rd equation to get $$e^{ix}+e^{iy}+e^{iz}=e^{i\pi/3}$$ Now you can argue geometrically that for three numbers of modulus 1 to sum to a number of modulus 1, one of the summands must be the negative of another summand (and so the third summand must equal the lone number on the other side). From there, it's easy.
If you don't like geometry, you can rewrite as $$e^{ix}+e^{iy}=e^{i\pi/3}-e^{iz}$$ then express the left side as $e^{is}\cos t$ and the right side as $e^{iu}\sin v$ for appropriate $s,t,u,v$, and that should give you enough relations among $x,y,z$ to solve the problem.
If you haven't learned about complex exponentials, the above won't mean much to you - but the upside is that you have some very nice mathematics to look forward to.
|
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|
Proving the identity $\sum_{k=1}^n {k^3} = \big(\sum_{k=1}^n k\big)^2$ without induction I recently proved that
$$\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2$$
using mathematical induction. I'm interested if there's an intuitive explanation, or even a combinatorial interpretation of this property. I would also like to see any other proofs.
|
$f(n)=1^3+2^3+3^3+\cdots+n^3$
$f(n-1)=1^3+2^3+3^3+\cdots+(n-1)^3$
$f(n)-f(n-1)=n^3$
if $g(n)= (1+2+3+4+\cdots+n)^2$ then
$$g(n)-g(n-1)=(1+2+3+4+\cdots+n)^2-(1+2+3+4+\cdots+(n-1))^2$$
using $a^2-b^2=(a+b)(a-b)$
$$\begin{align}g(n)-g(n-1)&=\\
&=[(1+\dots+n)+(1+\dots +(n-1))][(1+\dots+ n)-(1+\dots+(n-1)]\\
&=[2(1+2+3+4+\cdots+(n-1))+n]n\\
&=\left(2\frac{n(n-1)}{2}+n\right)n\\
&==(n(n-1)+n)n\\
&=n^3
\end{align}$$
So $f(n)$ and $g(n)$ are equal.
|
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|
Is this asymptotic equation correct? Is this equation correct?
$$ \frac {1 + \Theta(\frac 1 {2n})} {(1 + \Theta(1/n))^2} = 1 + O(1 / n) $$
I need this equation to prove that
$$ \binom {2n} n = \frac {2 ^ {2n}} {\sqrt {\pi n}} (1 + O(1 / n)) $$
which could be calculated from Stirling's Approximation:
$$ n! = \sqrt {2\pi n} \left(\frac n e\right)^n \left(1 + \Theta({\frac 1 n})\right).$$
|
Yes. One way to see it is to do the long division; i.e., divide the denominator directly into the numerator. The numerator is $1 + \Theta(\frac{1}{2n}) = 1 + \Theta(\frac{1}{n})$, and the denominator is $\left(1 + \Theta(\frac{1}{n})\right)^2 = 1 + \Theta(\frac{1}{n}) + \Theta(\frac{1}{n^2}) = 1 + \Theta(\frac{1}{n})$. I'm not going to attempt to typeset the long division on this forum, but try it, and you'll see that you get $1$ with a remainder of $O(\frac{1}{n})$. So you have
$$\frac {1 + \Theta(\frac{1}{2n})} {\left(1 + \Theta(\frac{1}{n})\right)^2} = \frac{1 + \Theta(\frac{1}{n})}{1 + \Theta(\frac{1}{n})} = 1 + \frac{O(\frac{1}{n})}{1 + \Theta(\frac{1}{n})} = 1 + O\left(\frac{1}{n}\right),$$
since the denominator is $O(1)$.
(Remember that $\Theta(\frac{1}{2n}) = \Theta(\frac{1}{n})$, since the constant $\frac{1}{2}$ doesn't affect the asymptotic order.)
|
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|
Difficult Integral: $\int\frac{x^n}{\sqrt{1+x^2}}dx$ How to calculate this difficult integral: $\int\frac{x^2}{\sqrt{1+x^2}}dx$?
The answer is $\frac{x}{2}\sqrt{x^2\pm{a^2}}\mp\frac{a^2}{2}\log(x+\sqrt{x^2\pm{a^2}})$.
And how about $\int\frac{x^3}{\sqrt{1+x^2}}dx$?
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Recall the hyperbolic functions
$$\cosh t= \frac{e^t + e^{-t}}{2} = \cos(it)$$
and $$\sinh t=\frac{e^t - e^{-t}}{2} = i\sin(-it).$$
Note that $\frac{d}{dt}\sinh t = \cosh t$, $\frac{d}{dt}\cosh t = \sinh t$ and also $\cosh^2 t -\sinh^2 t = 1$.
Making the substitution $\sinh t=x $ we see that
$$\frac{x^n\, dx}{\sqrt{1+x^2}} = \frac{\sinh^n t\, \cosh t\,dt}{\sqrt{1+\sinh^2t}}=
\frac{\sinh^n t\, \cosh t\,dt}{\sqrt{\cosh^2t}}=\sinh^n t\, dt$$
which leads us to $$\int\frac{x^n\, dx}{\sqrt{1+x^2}} = \int \sinh^n t\, dt.$$
To complete the problem, the binomial theorem is useful.
|
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|
Solving Radical Equations $x-7= \sqrt{x-5}$ This the Pre-Calculus Problem:
$x-7= \sqrt{x-5}$
So far I did it like this and I'm not understanding If I did it wrong.
$(x-7)^2=\sqrt{x-5}^2$ - The Square root would cancel, leaving:
$(x-7)^2=x-5$ Then I F.O.I.L'ed the problem.
$(x-7)(x-7)=x-5$
$x^2-7x-7x+14=x-5$
$x^2-14x+14=x-5$
$x^2-14x-x+14=x-x-5$
$x^2-15x+14=-5$
$x^2-15x+14+5=-5+5$
$x^2-15x+19=0$
$(x-1)(x-19)=0$
Now this is where I'm stuck because when I tried to see if I got the right numbers in the parentheses I got this....
$x^2-19x-1x+19=0$
$x^2-20x+19=0$
As you may see I'm doing something bad because I don't get $x^2-15x+19$
Could anyone please help me and tell me what I'm doing wrong?
|
$x-7= \sqrt{x-5}$
$(x-7)^2=\sqrt{x-5}^2$
$(x-7)^2=x-5$
$(x-7)(x-7)=x-5$
$x^2-7x-7x+49=x-5$
$x^2 - 15x + 54 = 0$
$(x - 9)(x - 6) = 0$
$x - 9 = 0 $ or $x - 6 = 0$
$x = 9$ or $x = 6$
Now check for extraneous solutions...
|
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|
How to show a scalar inequality? Let $a_1,\cdots, a_n\ge0$ and $S=a_1+\cdots+a_n$ with $k\ge 2, k\in\mathbb{Z}$. How to show
$$\sum_{i=1}^n(S-a_i)^k\le 2^{k-2}\left(\sum\limits_{i=1}^n a_i ^k+(n-2)S^k\right)?$$
|
There must be a slick way to do this, but here's one way. We will split the problem into different cases; the inequality is quite obvious in each of the cases. (Convince yourself that the cases exhaust all the possibilities. :))
Case 1: $n = 2$. For $n=2$, we have $S = a_1 + a_2$. In this case the inequality is obvious:
$$
(S - a_1)^k + (S - a_2)^k = a_2^k + a_1^k,
$$
which is clearly at most the right hand side. So we will assume $n \geq 3$ from now on.
Case 2: $k = 2$, $n$ arbitrary. For $k=2$, there's equality:
$$
\sum_{i} (S - a_i)^2 = nS^2 - 2S \sum_i a_i + \sum_i a_i^2 = nS^2 - 2S^2 + \sum_i a_i^2= (n-2) S + \sum_i a_i^2,
$$
which is also the right hand side.
Case 3: $k=3, n=3$. In this case, the left hand side is:
$$
\begin{eqnarray*}
&& (a_2+a_3)^3 + (a_3+a_1)^3 + (a_1+a_2)^3 \\
&=& 2(a_1^3+a_2^3+a_3^3) + 3(a_1^2a_2+a_1a_2^2+ a_2^2 a_3+a_2a_3^2+ a_3^2a_1+a_3a_1^2),
\end{eqnarray*}
$$
whereas the right hand side is
$$
2(a_1^3+a_2^3+a_3^3) + 2(a_1+a_2+a_3)^3 .
$$
Expanding the right hand side by the multinomial theorem, it is easy to check that the right hand side dominates the left term-by-term, and hence it is greater than or equal to the left.
Case 4: $k = 3, n \geq 4$. The only remaining case is when $k = 3$ and $n \geq 3$. Collect the terms to the right:
$$
\begin{eqnarray*}
&& 2(\sum_i a_i^3 + (n-2)S^3) - \sum_i (S - a_i)^3
\\ &=& (2n-4)S^3 + 2 \sum_i a_i^3 - n S^3 + 3S^2 \sum_i a_i - 3 S \sum_i a_i^2 + \sum_i a_i^3
\\ &=& (n-4)S^3 + 3 \sum_i a_i^3 + 3S^2 \cdot S - 3 S \sum_i a_i^2
\\ &\geq& 3S^3 + 3 \sum_i a_i^3 - 3 S \sum_i a_i^2
\\ &\geq& 3S^3 - 3 S \sum_i a_i^2
\\ &\geq& 3S^3 - 3 S \left(\sum_i a_i \right)^2
\\ &\geq& 3S^3 - 3 S \cdot S^2 = 0.
\end{eqnarray*}
$$
Case 5: $k \geq 4, n \geq 3$. Similarly, for $k \geq 4$ (and $n \geq 3$), the inequality is again obvious:
$$
\sum_i (S - a_i)^k \leq n S^k \leq 4(n-2) S^k \leq 2^{k-1}(n-2)S^k,
$$
which is clearly smaller than the right hand side. This leaves us with a narrow range of parameters to check the inequality.
|
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|
If $a^3 =a$ for all $a$ in a ring $R$, then $R$ is commutative. Let $R$ be a ring, where $a^{3} = a$ for all $a\in R$. Prove that $R$ must be a commutative ring.
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I happen to have come across a recent set of exercises on many of the small-$n$ cases of Jacobson's Theorem. It also happens that my solution is different than those contained in the link of @lhf above.
So we have that $a^3 = a \quad\forall a \in R$, and so $2a = (a+a)^3 =8a$, thus $6a = 0$.
Now consider the ideals $2R$ and $3R$. The intersection of $2R$ and $3R$ is trivial, as if $a \in 2R \cap 3R$, then $a = 2r = 3s$ for some $r,s$. Thus $3a = 6r = 0 = 6s = 2a$, and so $(3-2)a = a= 0$. So $2R \oplus 3R = R$. Further, if $a \in 2R$, $b \in 3R$, then $ab, ba \in 2R \cap 3R$ and so $ab = ba = 0$. So we only worry about commutativity in each ideal separately.
In $3R$, we have both $a^3 = a$ and $2a = 2 \cdot 3r = 0$ (some $r$). Then $1 + a = (1 + a)^3 = 1 + 3a + 3a^2 + a^3 = 1 + a + a^2 +a = 1 + a^2 \implies a^2 = a$. So what? In that case, we also have $(1 + a) = (1 + a)^2 = 1 + 2a + a^2 = 1 + 2a + a$, and so $2a = 0$ (yes, we have this in our ideal, but this is true in general in Boolean rings). Continuing, $(a + b) = (a + b)^2 = a^2 + ab + ba + b^2 = a + ab + ba + b$, so $ab = -ba = -ba + 2ba = ba$.
For $2R$, we have both $a^3 = a$ and $3a = 0$. Then we have that $a + b = (a + b)^3 = a^3 + a^2b + aba + ab^2 + ba^2 + bab + b^2a + b^3 $$= a + a^2b + aba + ab^2 + ba^2 + bab + b^2a + b$ on the one hand, and $a - b = (a - b)^3 = a^3 - a^2b - aba + ab^2 - ba^2 + bab + b^2a - b^3$ $= a - a^2b - aba + ab^2 - ba^2 + bab + b^2a - b$.
Taking the difference between these, we see $2(a^2b + aba + ba^2) = 0$, and so $a^2 b + aba + ba^2 = 0$. Multiply by $a$, and we get $a^3b + a^2ba + aba^2 = ab + (a^2b + aba)a = ab + (-ba^2)a = ab - ba = 0$. Thus $ab = ba$.
As both ideals commute separately and in products, $R$ commutes in general.
|
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|
Convergence of Series, problems with intermediate steps I have problems with two exercises. I know the answer of those two but both of them have a step which I don't understand.
1) I have to prove that $\prod_{k=2}^n \frac{k^3-1}{k^3+1}$ is convergent.
My first steps were:
$\prod_{k=2}^n \frac{k^3-1}{k^3+1}= \prod_{k=2}^n \frac{k-1}{k+1}\cdot \prod_{k=2}^n \frac{k^2+k+1}{k^2-k+1}=\frac{2(n-1)!}{(n+1)!}\prod_{k=2}^n \frac{k^2+k+1}{k^2-k+1}$
And then I got stuck. The author of the exercise does that:
$\frac{2(n-1)!}{(n+1)!}\cdot\prod_{k=2}^n \frac{k^2+k+1}{k^2-k+1} = \frac{2}{n(n+1)}\cdot\prod_{k=2}^n((k+1)^2-(k+1)+1)\cdot\prod_{k=2}^n\frac{1}{k^2-k+1}=$
$\frac{2}{n(n+1)}\cdot\prod_{k=3}^{n+1}(k^2-k+1)\cdot\prod_{k=2}^n\frac{1}{k^2-k+1}=\frac{2}{n(n+1)}\cdot\frac{(n+1)^2-(n+1)+1}{3}=\frac{2}{3}\cdot\frac{n^2+n+1}{n^2+n}\rightarrow\frac{2}{3}$
I understand what he does, but I don't know how he knew it. How can I use it for other exercises, so does somebody know how to remember this trick and for which kind of exercises this works?
2) I have to prove that $a_n:=\sqrt{n^2+n}-n$ is convergent and that a constant $A$ exist with $|a_n-a|<\frac{A}{n}.$
I proved that $a_n$ is convergent to $\frac{1}{2}$. For finding the constant $A$ the author says
It's easy to see, that: $|a_n -\frac{1}{2}|=|\frac{n}{\sqrt{n^2+n}+n}-\frac{1}{2}|<\frac{1}{8n}$, therefore $A:=\frac{1}{8n}$.
Unfortunately I don't see it. Can somebody please give a hint how he gets $\frac{1}{8}$?
thanks
|
1) Factoring further
$$
\begin{align}
k^3-1&=(k-1)(k^2+k+1)=(k-1)(k+\alpha)(k+\bar{\alpha})\\
k^3+1&=(k+1)(k^2-k+1)=(k+1)(k-\alpha)(k-\bar{\alpha})
\end{align}
$$
where $\alpha+\bar{\alpha}=1$. It is easy to see that $\prod\frac{k-1}{k+1}$ is telescoping. However, when we write $\prod\frac{(k+\alpha)(k+\bar{\alpha})}{(k-\alpha)(k-\bar{\alpha})}=\prod\frac{(k+1-\bar{\alpha})(k+1-\alpha)}{(k-\alpha)(k-\bar{\alpha})}$, it becomes evident that this telescopes, too.
2)
$$
\begin{align}
\left|\frac{n}{\sqrt{n^2+n}+n}-\frac{1}{2}\right|
&=\frac{1}{2}\left|\frac{n-\sqrt{n^2+n}}{n+\sqrt{n^2+n}}\right|\\
&=\frac{1}{2}\frac{n}{(n+\sqrt{n^2+n})^2}\\
&<\frac{1}{2}\frac{n}{4n^2}\\
&=\frac{1}{8n}
\end{align}
$$
|
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|
Finding $a$ and $r$ such that $\lim\limits_{n\to \infty} n^r \cdot \frac12 \cdot \frac34 \cdots \frac{2n-1}{2n}=a$ Find $a,r>0$ such that
$$\lim_{n\to \infty} n^r \cdot \frac12 \cdot \frac34 \cdots \frac{2n-1}{2n}=a$$
I don't have any idea to solve it. How can I solve it?
|
I'll start out from a celebre limit, namely Wallis product that states that:
$$ \frac{\pi}{2} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdot \cdot \cdot $$
Without loss of generality, we consider an even factors number of the limit excepting ${n^r}$, and then by applying Wallis product we get that:
$$ \lim_{n\to \infty}\frac{n^r}{\sqrt{2n+1}} \frac{\sqrt{2}}{\sqrt{\pi}}$$ that obviously gives us $L =\frac{1}{\sqrt{\pi}}$ for $r=\frac{1}{2}$
The proof is complete.
|
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|
Integrate $\int \sin (2x) \cos (2x)\,dx$ I have
$$\int \sin(2x) \cos (2x)\,dx = \frac12 \int \sin(4x)\,dx = -\frac18 \cos(4x),$$
but I also have
$$\int \sin(2x) \cos (2x)\,dx = \frac12 \int \sin 2x \cdot 2 \cos 2x \, dx = \frac14 \sin^2(2x).$$
Which one is correct, and why is the other method wrong?
|
To illustrate the phenomena in your calculation, I've chosen the number $\pi/8$ as the lower boundary.
\begin{align*}
\int_{\frac{\pi}{8}}^{x} \sin(4t) \,\mathrm{d}t &= \int_{\frac{\pi}{8}}^{x} 2 \sin(2t) \cos(2t) \,\mathrm{d}t \\
&= \int_{\frac{\pi}{8}}^{x} \sin(2t) \,\mathrm{d}(\sin(2t)) \\
\left . -\frac{\cos(4t)}{4} \right\rvert_{\frac{\pi}{8}}^{x} &= \left . \frac{\sin^2(2t)}{2} \right\rvert_{\frac{\pi}{8}}^{x} \\
-\frac{\cos(4x)}{4} + \color{red}{\frac{\cos(4\cdot\frac{\pi}{8})}{4}} &=
\frac{\sin^2(2x)}{2} - \color{red}{\frac{\sin^2(2\cdot\frac{\pi}{8})}{2}} \\
-\frac{\cos(4x)}{4} + \color{red}{0} &= \frac{\sin^2(2x)}{2} + \color{red}{\frac{1}{4}}
\end{align*}
|
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|
The $3 = 2$ trick on Google+ I found out this on Google+ yesterday and I was thinking about what's the trick. Can you tell?
How can you prove $3=2$?
This seems to be an anomaly or whatever you call in mathematics. Or maybe I'm just plain dense.
See this illustration:
$$ -6 = -6 $$
$$ 9-15 = 4-10 $$
Adding $\frac{25}{4}$ to both sides:
$$ 9-15+ \frac{25}{4} = 4-10+ \frac{25}{4} $$
Changing the order
$$ 9+\frac{25}{4}-15 = 4+\frac{25}{4}-10 $$
This is just like $a^2 + b^2 - 2a b = (a-b)^2$. Here $a_1 = 3, b_1=\frac{5}{2}$ for L.H.S, and $a_2 =2, b_2=\frac{5}{2}$ for R.H.S. So it can be expressed as follows:
$$ \left(3-\frac{5}{2} \right) \left(3-\frac{5}{2} \right) =
\left(2-\frac{5}{2} \right) \left( 2-\frac{5}{2} \right) $$
Taking positive square root on both sides:
$$ 3 - \frac{5}{2} = 2 - \frac{5}{2} $$
$$ 3 = 2 .$$
I think it's something near the root.
|
HINT $\ $ You erroneously inferred $\rm\ x^2 =\: (-x)^2\ \Rightarrow\ x\: =\: -x\:,\ $ for $\rm\ x\:=\:1/2\:.$
|
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|
Getting the root out of the denominator Alright quick question,
If I have
$$\frac{1}{2^{1/3} - 2},$$
how do i change it to not have the root in the denominator? I know the trick with the root of 2, but this seems a bit more complicated. Thanks in advance :)
|
If you mean
$$\frac{1}{2^{1/3}} - 2,$$
then just multiply the numerator and denominator by $2^{2/3}$; we get
$$\frac{2^{2/3}}{2} - 2.$$
If you mean
$$\frac{1}{2^{1/3} - 2}$$
then you can use the formula
$$(a-b)(a^2+ab+b^2) = a^3-b^3$$
so
$$\frac{1}{2^{1/3}-2} = \frac{2^{2/3} + 2(2^{1/3}) + 4}{(2^{1/3}-2)(2^{2/3} + 2(2^{1/3})+4)} = \frac{2^{2/3} + 2^{4/3} + 4}{2 - 8} = -\frac{2^{2/3} + 2^{4/3} + 4}{6}. $$
|
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|
Why doesn't this method of calculating Gaussian curvature of a sphere work? Suppose we have a sphere defined as $x^2 + y^2 + z^2 = 9$. With radius of 3, we know the Gaussian Curvature should be $\frac{1}{9}$ at every point on the surface.
Since this defines a level set, the gradient gives the normals. So, the normals should be
$$\begin{bmatrix}
2x \\\\
2y \\\\
2z
\end{bmatrix}$$
We can then normalize these normals, giving
$$\begin{bmatrix}
\frac{2x}{\sqrt{4x^2+4y^2+4z^2}} \\\\
\frac{2y}{\sqrt{4x^2+4y^2+4z^2}} \\\\
\frac{2z}{\sqrt{4x^2+4y^2+4z^2}} \\\\
\end{bmatrix}$$
By my understanding, this gives us the surface normals on the unit sphere (the Gauss Map).
According to Wolfram MathWorld's article on Gauss Map, if we take the partial derivatives of the Gauss Map function, and then take the determinant of that matrix, we should get the curvature. The matrix in question is
$$\begin{bmatrix}
\frac{y^2+z^2}{{(x^2+y^2+z^2)}^{\frac{3}{2}}} & \frac{-xy}{{(x^2+y^2+z^2)}^{\frac{3}{2}}} & \frac{-xz}{{(x^2+y^2+z^2)}^{\frac{3}{2}}} \\\\
\frac{-xy}{{(x^2+y^2+z^2)}^{\frac{3}{2}}} & \frac{x^2+z^2}{{(x^2+y^2+z^2)}^{\frac{3}{2}}} & \frac{-yz}{{(x^2+y^2+z^2)}^{\frac{3}{2}}} \\\\
\frac{-xz}{{(x^2+y^2+z^2)}^{\frac{3}{2}}} & \frac{-yz}{{(x^2+y^2+z^2)}^{\frac{3}{2}}} & \frac{x^2+y^2}{{(x^2+y^2+z^2)}^{\frac{3}{2}}}
\end{bmatrix}$$
However, if we pick a point on the sphere, (0,3,0), then the resulting determinant is zero when we plug in that point. Where am I going wrong? The result should be $\frac{1}{9}$.
|
First observe that your normalization simplifies greatly since $x^2 + y^2 + z^2 = 9$. So really the unit normal vector is just $\frac{1}{3} (x,y,z)$. Now the Gauss map takes the sphere to the unit sphere and the differential of this in coordinates should be a two by two matrix since this is a map between surfaces. Locally we can choose any two of $x,y,z$ to serve as coordinates on the sphere. Then the differential of the Gauss map in such coordinates is just 1/3 times the (2x2) identity matrix so the determinant is 1/9.
|
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|
Maximize the distance between a line normal to an ellipse and its center My friend sent me this problem, which (upon Googling) seems to be from a Cornell class (1220?). Anywho.
My advice to him was to parametrize the ellipse (say, in the first quadrant) with
$x = a \cos(t); y = b \sin(t)$, find normal lines, then use the formula for the distance between a point and a line. But then I started wondering:
Is there a better way? (Also, will this method even lead to a solution?!?)
|
The same maximal distance occurs in each quadrant, so we can restrict attention to $t\in[0,\pi/2]$. The tangent vector at $t$ is $(-a\sin t,b\cos t)$. This vector is normal to the line, so we just have to take the scalar product of a unit vector in this direction with the position vector in order to find the distance of the origin from the line:
$$
\begin{eqnarray}
D
&=&
\left\lvert\frac{(-a\sin t,b\cos t)}{\sqrt{a^2\sin^2t+b^2\cos^2t}}\cdot(a\cos t,b\sin t)\right\rvert
\\
&=&
\frac{(a^2-b^2)\sin t\cos t}{\sqrt{a^2\sin^2t+b^2\cos^2t}}\;.
\end{eqnarray}$$
Differentiating with respect to $t$ yields
$$\frac{a^2\sin^4 t-b^2\cos^4t}{\left(a^2\sin^2t+b^2\cos^2t\right)^{3/2}}\;,$$
and setting this to zero yields
$$a^2\sin^4t=b^2\cos^4t\;,$$
$$t=\arctan\sqrt{\frac ba}\;.$$
Using $\cos t=1/\sqrt{1+\tan^2 t}$, we can evaluate $D$ at this parameter:
$$
\begin{eqnarray}
D
&=&
\frac{(a^2-b^2)\sin t\cos t}{\sqrt{a^2\sin^2t+b^2\cos^2t}}
\\
&=&
\frac{(a^2-b^2)\tan t}{\sqrt{a^2\tan^2t+b^2}}\cos t
\\
&=&
\frac{(a^2-b^2)\tan t}{\sqrt{a^2\tan^2t+b^2}}\frac1{\sqrt{1+\tan^2 t}}
\\
&=&
\frac{(a^2-b^2)\sqrt{b/a}}{\sqrt{a^2(b/a)+b^2}}\frac1{\sqrt{1+b/a}}
\\
&=&
\frac{a^2-b^2}{a+b}\;.
\\
&=&
a-b\;.
\end{eqnarray}$$
The result obviously supports your idea that there might be a simpler way to do this.
|
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|
Problem with generating functions and binary recurrences I am considering the following recurrence:
$a_0 = 1$;
$a_1 = 2$
$a_{n} = 2 (a_{n - 1} + a_{n - 2})$
Then I proceeded with the generating function:
$F(x) = \displaystyle\sum_{n = 0}^\infty a_n x^n = 1 + 2x + \displaystyle\sum_{n = 2}^\infty a_{n} x^{n} = 1 + 2x + \displaystyle\sum_{n = 2}^\infty 2x^n(a_{n - 1} + a_{n - 2})$
$F(x) = 1 + 2x + \displaystyle\sum_{n = 2}^{\infty} 2x^{n} a_{n - 1} + \displaystyle\sum_{n = 2}^{\infty} 2x^{n} a_{n - 2}$
$F(x) = 1 + 2x + (2x \displaystyle\sum_{n = 2}^{\infty} x^{n - 1} a_{n - 1}) + (2x^{2} \displaystyle\sum_{n = 2}^{\infty} x^{n - 2} a_{n - 2})$
$F(x) = 1 + 2x + 2x(F(x) - 1) + 2x^{2}F(x)$
$F(x) = \frac{1}{1 - 2x - 2x^{2}}$ Let a, b be the roots of the quadratic.
$F(x) = \frac{1}{(x - a)(x - b)} = \displaystyle\sum_{n = 0}^{\infty} \frac{x^{n}(b^{-1 - n} - a^{-1 - n})}{\sqrt{3}}$
We should then have $a_{n} = \frac{b^{-1 - n} - a^{-1 - n}}{\sqrt{3}}$, but I know that this is false. Where have I gone wrong?
|
A simpler way to handle such is to write the recurrence as:
$$
a_{n + 2} = 2 (a_{n + 1} + a_n)
$$
multiply by $x^n$, sum over $n \ge 0$. Recognize the sums that result:
$$
\frac{A(x) - a_0 -a_1 z}{z^2}
= 2 \left( \frac{A(x) - a_0}{x} + A(x) \right)
$$
Plugging in $a_0 = 1$ and $a_1 = 2$, solving for $A(z)$:
$$
A(z) = \frac{1}{1 - 2 z - 2 z^2}
= \frac{3 + \sqrt{3}}{6} \frac{1}{1 - (1 + \sqrt{3}) x}
- \frac{3 - \sqrt{3}}{6} \frac{1}{1 - (1 - \sqrt{3}) x}
$$
This is just two geometric series:
$$
a_n = \frac{3 + \sqrt{3}}{6} \cdot (1 + \sqrt{3})^n
- \frac{3 - \sqrt{3}}{6} \cdot (1 - \sqrt{3})^n
$$
|
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|
Find all integers $m$ such that $\frac{1}{m}=\frac{1}{\lfloor 2x \rfloor}+\frac{1}{\lfloor 5x \rfloor} $ How would you determine all integers $m$ such that the following is true?
$$\frac{1}{m}=\frac{1}{\lfloor 2x \rfloor}+\frac{1}{\lfloor 5x \rfloor} .$$
Note that $\lfloor \cdot \rfloor$ means the greatest integer function. Also, $x$ must be a positive real number.
|
You can solve it by cases. Let $x=n+r$, where $n$ is an integer and $r=\lfloor x\rfloor$. Then $\lfloor 2x\rfloor = \lfloor 2n+2r\rfloor = 2n+\lfloor 2r\rfloor$ and $\lfloor 5x\rfloor = \lfloor 5n+5r\rfloor = 5n+\lfloor 5r\rfloor$, and the original equation can be written as $$\frac1m = \frac1{2n+\lfloor 2r\rfloor}+\frac1{5n+\lfloor 5r\rfloor}.$$
You know that $0\le r<1$, and you can get exact values for $\lfloor 2r\rfloor$ and $\lfloor 5r\rfloor$ for $r$ in different subintervals of $[0,1)$.
*
*If $0\le r<\frac15$, $\lfloor 2r\rfloor=\lfloor 5r\rfloor=0$.
*If $\frac15\le r <\frac25$, $\lfloor 5r\rfloor = 1$ and $\lfloor 2r\rfloor = 0$.
*If $\frac25\le r<\frac12$, $\lfloor 5r\rfloor = 2$ and $\lfloor 2r\rfloor = 0$.
*If $\frac12\le r<\frac35$, $\lfloor 5r\rfloor = 2$ and $\lfloor 2r\rfloor = 1$.
And there are two more intervals, which I’ll leave to you.
If $0\le r<\frac15$, the equation becomes simply $$\frac1m=\frac1{2n}+\frac1{5n}=\frac{7n}{10n^2}=\frac7{10n};$$ in order for $m$ to be an integer, $n$ must be a multiple of $7$, say $n=7k$, we get $m=10k$. In other words, this case gives us every positive multiple of $10$ as a solution.
If $\frac15\le r <\frac25$, the equation becomes $$\frac1m=\frac1{2n}+\frac1{5n+1}=\frac{7n+1}{10n^2+2n},$$ or $$m=\frac{10n^2+2n}{7n+1}.$$ Divide this out to get $$m = \frac{10}7n+\frac4{49}-\frac4{49(7n+1)} = \frac{70n+4}{49}-\frac4{49(7n+1)},$$ and multiply through by $49$ to get $$49m=70n+4-\frac4{7n+1}$$ or, with a little rearrangement, $$49m-70n-4=\frac4{7n+1}.$$ But this is impossible if $n$ and $m$ are integers, because the lefthand side is an integer, and the righthand side isn’t. Thus, there are no solutions in this case.
If $\frac25\le r<\frac12$, the equation becomes $$\frac1m=\frac1{2n}+\frac1{5n+2}=\frac{7n+2}{10n^2+4n},$$ so $$m=\frac{10n^2+4n}{7n+2}=\frac{10n}7+\frac8{49}-\frac{16}{49(7n+2)},$$ and $$49m=70n+8-\frac{16}{7n+2}.$$ This time there is one value of $n$ that makes the righthand side an integer, namely, $n=2$. Substituting $n=2$ into the last equation, we get $49m=140+8-1=147$, and $m=3$; this is the only solution in this case. (Note that had the righthand side not been a multiple of $49$ when $n=2$, there would have been no solutions in this case.)
You can continue in this fashion through the remaining three cases. There may be an easier approach, but this one is at least systematic and workable.
|
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|
How to prove a fact about the sum of three squares? How would I go about proving the following?
If $a$, $b$, $c$, $n$ are positive integers, then
$a^2+b^2+c^2 \neq 2^nabc$
I tried doing something similar to the proof for Adrien-Marie Legendre's Three Square theorem: $a^2+b^2+c^2=n$ iff there are not integers $k$, and $m$ so that $n=4^k(8m+7)$. It didn't quite work out...
$2^nabc$ is always even. So if $a^2+b^2+c^2 = 2^nabc$, then $a^2+b^2+c^2$ must be even.
That means there is $a_1$, $b_1$, $c_1$ so that
$a = 2a_1$, $b = 2b_1$, and $c = 2c_1$
So
$(2a_1)^2+(2b_1)^2+(2c_1)^2 = 2^nabc \rightarrow 2(2a_1^2+2b_1^2+2c_1^2)= 2^nabc$
and we get $2a_1^2+2b_1^2+2c_1^2= 2^{n-1}abc$
We can continue to do this procedure with $a_2$, $b_2$, $c_2$ then $a_3$, $b_3$, $c_3$ then ... $a_n$, $b_n$, $c_n$.
With $a_n$, $b_n$, $c_n$ we'd get
$2^na_n^2+2^nb_n^2+2^nc_n^2= 2^{n-n}abc=abc$
Since $a_n=2a_{n-1}$ and $a_0=a$,
$a_n = \frac{a}{2^n}$
and we get
$2^n(\frac{a}{2^n})^2+2^n(\frac{b}{2^n})^2+2^n(\frac{c}{2^n})^2=abc$
This just becomes the original equation.
$a^2+b^2+c^2 = 2^nabc$
|
To go from $a^2+b^2+c^2$ even to $a,b,c$ even in this case, you need an argument, although it is true for $n$ strictly positive.
You should try to express your equation for only $a_k$, $b_k$ and $c_k$, then you will see that your argument does not end after $n$ steps. Either, you argue with infinite descent or equivalently, you divide immediately by the largest factor of 2 possible.
|
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|
How to solve $z^4 +6z^2 +25 = 0$ into complex conjugate? How to solve $z^4 +6z^2 +25 = 0$ into complex conjugate?
I started with
$$(z^2 + 3)^2 + 16 = 0$$
$$(z^2 + 3)^2 = - 16$$
$$z^2 + 3 = \pm 4i$$
Is this the way to start solving the equation or am I completely of?
|
You have $z^2=-3\pm4i$. You need the two square roots of $-3+4i$ and the two square roots of $-3-4i$. I can think of two ways to do this. One is polar coordinates:
$$
-3+4i = \sqrt{3^2+4^2}\left(\cos\theta+i\sin\theta\right) = 5\left(\cos\theta+i\sin\theta\right)
$$
where $\theta$ is an angle between $\pi/2$ and $\pi$ whose tangent is $-4/3$. Since $\arctan(-4/3)=-\arctan(4/3)$ is between $-\pi/2$ and $0$, we add $\pi$ to it and get $\theta=\pi-\arctan(4/3)$.
Then the square roots of $-3+4i$ are
$$
\pm\sqrt{5}\left(\cos\left(\frac\theta 2\right)+i\sin\left(\frac\theta 2\right)\right).
$$
And a similar thing works for the square roots of $-3-4i$, and in fact you get
$$
\pm\sqrt{5}\left(\cos\left(\frac\theta 2\right)-i\sin\left(\frac\theta 2\right)\right)
$$
where $\theta$ is the same number as above.
The other method is this. Say $a+bi$ is a square root of $-3+4i$, where $a$ and $b$ are real. Then $(a+bi)^2=-3+4i$. So $a^2-b^2 + 2abi = -3+4i$. Then
$$
\begin{align}
a^2 - b^2 & = -3, \\
2ab & = 4.
\end{align}
$$
If you write $b=2/a$ and substitute that for $b$ in the first equation, you have and equation in $a$ that can be solved.
|
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|
Finding the slope of the tangent line to $\frac{8}{\sqrt{4+3x}}$ at $(4,2)$ In order to find the slope of the tangent line at the point $(4,2)$ belong to the function $\frac{8}{\sqrt{4+3x}}$, I choose the derivative at a given point formula.
$\begin{align*}
\lim_{x \to 4} \frac{f(x)-f(4)}{x-4} &=
\lim_{x \mapsto 4} \frac{1}{x-4} \cdot \left (\frac{8}{\sqrt{4+3x}}-\frac{8}{\sqrt{4+3 \cdot 4}} \right )
\\ \\ & = \lim_{x \to 4} \frac{1}{x-4} \cdot \left ( \frac{8}{\sqrt{4+3x}}-\frac{8}{\sqrt{16}} \right ) \\ \\ & = \lim_{ x \to 4} \frac{1}{x-4} \cdot \left ( \frac{8}{\sqrt{4+3x}}-2\right)
\end{align*}$
But now I can't figure it out, how to end this limit.
I know that the derivative formula for this function is $-\frac{12}{(4+3x)\sqrt{4+3x}}$.
Thanks for the help.
|
Just note that
$$\begin{align}
\frac{8}{\sqrt{4 + 3x}} - 2 &= \frac{8 - 2 \sqrt{4 + 3x}}{\sqrt{4 + 3x}}
\\
&= \frac{8 - 2 \sqrt{4 + 3x}}{\sqrt{4 + 3x}} \cdot \frac{8 + 2\sqrt{4 + 3x}}{8 + 2 \sqrt{4 + 3x}}
\\
&= \frac{64 - 4(4 + 3x)}{\sqrt{4 + 3x}(8 + \sqrt{4 +3x})}.
\end{align}$$
I'll leave the rest up to you.
|
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|
Evaluating $\sum\limits_{n=0}^{20} \frac{(-1)^{n}2^{n+1}}{3^{n}},$ I know the sum of the series $$2 - \frac{4}{3} + \frac{8}{9} - \cdots + \frac{(-1)^{20}2^{21}}{3^{20}}$$ is equal to $$\sum\limits_{n=0}^{20} \frac{(-1)^{n}2^{n+1}}{3^{n}},$$ but I don't know how to calculate the sum without manually entering it into the calculator.
|
$$\sum_{n=1}^{20}\frac{(-1)^n2^{n+1}}{3^n}=2\sum_{n=1}^{20}\left(-\frac{2}{3}\right)^n=2\left(-\frac{2}{3}\right)\frac{\left(-\frac{2}{3}\right)^{20}-1}{\left(-\frac{2}{3}-1\right)}=$$
$$=\left(-\frac{4}{3}\right)\left(-\frac{3}{5}\right)\frac{2^{20}-3^{20}}{3^{20}}=\frac{4}{5}\frac{2^{20}-3^{20}}{3^{20}}$$
|
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|
Expected value and Variance of $Y=\frac{1}{a} X-b$ where $X \sim N(\mu, \sigma^2)$ I absolutely know I am not doing this right. :[
Could I get some input or point back in the right direction?
My work done so far is shown below.
Let $X$ be a normal random variable with parameters $N(\mu, \sigma^2)$. Please find the Expected value and Variance of random variable $Y=\frac{1}{a} X-b$, where $a$ and $b$ are constant values.
My work.
$$
\begin{align*}
E(Y) &= aE(x)-b = \sum_x \Big(\frac{1}{a} x - b \Big) p_x(x) = \frac{1}{a} \sum_x x p_x(x) - b
\\ &= \frac{1}{a} \sum_x x p_x(x) - b \sum_x p_x(x) = a E(x) - b \cdot 1.
\end{align*}
$$
If $a = 0$, then $E(x-b) = E(x)$ and if $b = 0$, then $E(ax)= \frac{1}{a}E(x)$.
$$
\begin{align*}
\mu &= E(X) = \int_{-\infty}^{\infty} x f(x) dx = \int_{0}^{1} x \Big(\frac{1}{a} X - b \Big) dx = \frac{1}{a} \int_0^1 X^2 - xb ~dx
\\ & = \frac{1}{a} \Big( \frac{x^3}{3} - \frac{bx^2}{2} \Big) \text{ from } 1 \text{ to } 0
\\ & = \frac{1}{a} \Big( \frac{1}{3} - \frac{b}{2} \Big) .
\end{align*}
$$
$$
\text{RV } Y = (X - E(x)^2).
$$
$$
\sigma^2 = \operatorname{Var}(x) = E[(x) - E[x])^2]
$$
$$
\operatorname{Var} \Big( \frac{1}{a} X - b \Big) = a^2 \operatorname{Var}(x).
$$
$$
\int_{\Box}^{\Box}\Big( \frac{1}{a} X - b \Big) - \Big( \frac{1}{a} X - b \Big)^2 \ldots
$$
|
Since expectation is linear, you have that $\mathsf{E}(Y)=\mathsf{E}(1/a\:X-b)=1/a\:\mathsf{E}(X)-b$. Furthermore,
$$
\begin{align}
\mathsf{Var}(Y)
&=\mathsf{E}(Y^{\;2})-\mathsf{E}(Y)^2\\
&=\mathsf{E}(1/a^2X^2-2b/aX+b^2)-(1/a\:\mathsf{E}(X)-b)^2\\
&=1/a^2\mathsf{E}(X^2)-2b/a\mathsf{E}(X)+b^2-1/a^2\mathsf{E}(X)^2+2b/a\mathsf{E}(X)-b^2\\
&=1/a^2\mathsf{E}(X^2)-1/a^2\mathsf{E}(X)^2\\
&=1/a^2\mathsf{Var}(X).
\end{align}
$$
|
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|
What is the least positive integer $n$ for which $(-\sqrt{2}+i\sqrt{6})^n$ is an integer?
Compute the least positive integer $n$ for which $(-\sqrt{2}+i\sqrt{6})^n$ will be an integer, where $i$ is the imaginary unit.
I did the binomial expansion and just plugged in numbers for $n$ starting from $1$ to see any pattern. I coudn't find any pattern but I eventually solved the problem to be $n=6$, but is there any easier more practical approach to this problem?
|
$$
\begin{align}
(-\sqrt{2} + i \sqrt{6})^n
& = \left(\sqrt{2} \left(-1 + i \sqrt{3} \right)\right)^n
& = 2^{n/2} \left(-1 + i \sqrt{3} \right)^n\\
& = 2^{n/2} \left( 2 \left( - \frac12 + i \frac{\sqrt{3}}{2} \right) \right)^n
& = 2^{n/2} 2^n \left( - \frac12 + i \frac{\sqrt{3}}{2} \right)^n \\
& = 2^{3n/2} \left( - \frac12 + i \frac{\sqrt{3}}{2} \right)^n
& = 2^{3n/2} \left(\cos \left(\frac{2 \pi}{3} \right) + i \sin \left(\frac{2 \pi}{3} \right)\right)^n
\end{align}
$$
$$(-\sqrt{2} + i \sqrt{6})^n = 2^{3n/2} \left(\cos \left(\frac{2 n\pi}{3} \right) + i \sin \left(\frac{2 n\pi}{3} \right)\right)$$
For $2^{3n/2}$ to be an integer, we need $\frac{3n}{2} \in \mathbb{Z}$. We also need $\sin \left(\frac{2 n\pi}{3} \right) = 0$. This gives us $ \frac{2n}{3} \in \mathbb{Z}$.
Hence, $n = \frac{3k_1}{2} = \frac{2k_2}{3}$ where $k_1,k_2 \in \mathbb{Z}$. Hence, $n = 6k$ where $k \in \mathbb{Z}$.
|
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|
Four kissing circles How can one go about solving the following problem?
Inscribe a circle in an arbitrary triangle. Call it's radius $r_1$.
Inscribe three more circles so that each one is tangent to two sides of
the triangle and the first circle (i.e., each at a different corner). Call the
radii $r_2, r_3, r_4$. Find a relationship between $r_1, r_2, r_3$ and
$r_4$.
The most promising method of attack for me was to consider the isosceles triangles at each corner: the base being the tangent line to the point of intersection of the angle bisector of the triangle and the first circle. But I'm stuck.
Any suggestions much appreciated.
|
I'm using the same notation as in the link mentioned by David Mitra. By similarity we have
$$
\frac{r}{OA}=\frac{r_a}{OA-r-r_a}=\sin\left(\frac{\alpha}{2}\right)
$$
where $\alpha=\angle A$. Then
$$
\frac{r_a}{r} = \frac{1-\sin(\alpha/2)}{1+\sin(\alpha/2)}
= \frac{1-\cos((\beta+\gamma)/2)}{1+\cos((\beta+\gamma)/2)}
= \tan\left(\frac{\beta+\gamma}{4}\right)^2
$$
where we used that $\alpha+\beta+\gamma=\pi$. Similarly
$$
\frac{r_b}{r} = \tan\left(\frac{\alpha+\gamma}{4}\right)^2
\qquad\qquad
\frac{r_c}{r} = \tan\left(\frac{\alpha+\beta}{4}\right)^2
$$
Therefore, $\sqrt{r_ar_b}+\sqrt{r_ar_c}+\sqrt{r_br_c}$ can be written as
$$
\begin{split}
\frac{\sqrt{r_ar_b}+\sqrt{r_ar_c}+\sqrt{r_br_c}}{r}
= \frac{\sin(\rho)\sin(\sigma)\cos(\tau)+\sin(\rho)\cos(\sigma)\sin(\tau)
+\cos(\rho)\sin(\sigma)\sin(\tau)}{\cos(\rho)\cos(\sigma)\cos(\tau)}
\end{split}
$$
where $\rho=(\beta+\gamma)/4$, $\sigma=(\alpha+\gamma)/4$, and $\tau=(\alpha+\beta)/4$. Note that $\rho+\sigma+\gamma=\pi/2$.
Consider $x$, $y$, $z$ such that $x+y+z=\pi/2$. Then
$$
\begin{split}
\sin(x)\sin(y)\cos(z) &= \frac{1}{2}(\cos(x-y)-\cos(x+y))\cos(z)
\\&=
\frac{1}{4}(\cos(x-y+z)+\cos(x-y-z) -\cos(x+y+z)-\cos(x+y-z))
\\&= \frac{1}{4}( \cos\left(\frac{\pi}{2}-2y\right) + \cos\left(-\frac{\pi}{2}+2x\right) - \cos\left(\frac{\pi}{2}\right)-\cos\left(\frac{\pi}{2}-2z\right)
\\&=\frac{1}{4}\left( \sin(2y)+\sin(2x)-\sin(2z)\right)
\end{split}
$$
And similarly
$$
\begin{split}
\cos(x)\cos(y)\cos(z) &= \frac{1}{2}(\cos(x-y)+\cos(x+y))\cos(z)
\\&=
\frac{1}{4}(\cos(x-y+z)+\cos(x-y-z) +\cos(x+y+z)+\cos(x+y-z))
\\&=\frac{1}{4}\left( \sin(2y)+\sin(2x)+\sin(2z)\right)
\end{split}
$$
This means that $(\sqrt{r_ar_b}+\sqrt{r_ar_c}+\sqrt{r_br_c})/r$ is equal to
$$
\frac{\sqrt{r_ar_b}+\sqrt{r_ar_c}+\sqrt{r_br_c}}{r} = \frac{\sin(2\rho)+\sin(2\sigma)+\sin(2\tau)}{\sin(2\rho)+\sin(2\sigma)+\sin(2\tau)}=1
$$
|
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|
Help with a geometry problem The problem says:
A triangle has its lengths in an arithmetic progression,
with difference d. The area of the triangle is t. Find the dimensions.
the solution says:
the notation can be even better if we make it more symmetrical, by making the
side lengths $b − d, b,$ and $b + d$ .
by Heron’s formula we know that $t^2 = s(s − b + d)(s − b)(s − b − d)$ , where $s =
((b − d) + b + (b + d))/2$ is the semi-perimeter;
and after simplification
$$ t^2 = \frac{3b}{2}(\frac{3b}{2} - b + d )(\frac{3b}{2} - b ) (\frac{3b}{2} - b - d ) $$
$$ \implies t^2 = \frac{3b^2(b-2d)(b+2d)}{16} = \frac{3b^2(b^2-4d^2)}{16} $$
then we get $$ 3b^4 − 12d^2b^2 − 16t^2 = 0 $$
and using the quadratic formula :
$$ b^2 = \frac{12d^2 \pm \sqrt{144d^4 + 169t^2} }{6} = 2d^2 \pm \sqrt{4d^4 + \frac{16}{3} t^2} $$
and because b has to be positive , we get
$$ b = \sqrt{2d^2 + \sqrt{4d^4 + \frac{16}{3}t^2}} $$
Which is the part that i have a problem with , my question is :
why should we select only the positive sign solution of the quadratic formula ? is that because $\sqrt{ 4d^4 + \frac{16}{3}t^2} > 2d^2$ which means that the negative sign solution leads to the square root of a negative number which is not valid? why is the positive sign solution is the right solution ?
In other words :
If $\sqrt{ 4d^4 + \frac{16}{3}t^2} > 2d^2$ , how is that ? how can we prove it ?
thank you
|
Yes, you know that $b^2$ and $b$ are both positive real numbers, so you select the positive sign twice while taking square root.
If you want to know why the inequality is true, note that both sides of it are positive, so you can just square both sides to get an equivalent inequality which is easily seen to be correct.
|
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|
Proof of an inequality: $\sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$
Possible Duplicate:
Proving $\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$ with induction
How do I prove the following?
$$\sqrt{n} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \cdots + \dfrac{1}{\sqrt{n}},$$ for all $n \in\mathbb{Z}$, $n\ge 2$.
|
We can even do better:
$$
\sqrt{n}-\sqrt{n-1}=\frac{1}{\sqrt{n}+\sqrt{n-1}}<\frac{1}{2\sqrt{n-1}}\tag{1}
$$
Summing, we get
$$
\sqrt{n}-1<\frac{1}{2}\left(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{n-1}}\right)\tag{2}
$$
So, for $n\ge2$,
$$
\sqrt{n}<1+\frac{1}{2}\left(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{n-1}}\right)\tag{3}
$$
which is a better bound for every $n$.
|
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|
How can we prove the locus is a circle? Given two fixed points A and B, find the locus of the point P, satisfying PA=2PB. Of course we can use Cartesian geometry to find the equation of the curve.
Let the midpoint of A and B be the origin, and the line AB be the x-axis, then the coordinates of A and B is (a,0), (-a,0). Let the coordinates of P be (x,y). Then
$\sqrt{(x-a)^2+y^2}=2\sqrt{(x+a)^2+y^2}$,
we get $(x+\frac{5}{3})^2+y^2=(\frac{4}{3}a)^2$. So the locus is a circle. But how can we solve this problem by pure geometry?
|
Without loss of generality, let the distance between $A$ and $B$ equal 3 (this is for convenience in calculations) and suppose that $\overline{AB}$ is horizontal (this just makes it easier to talk about where things are).
Now, the point $P_1$ that is on $\overline{AB}$ a distance of 2 from $A$ and 1 from $B$ is in the locus, as are points $P_2$ and $P_3$ that are directly above and below $B$, a distance of $\sqrt{3}$ from $B$ and $2\sqrt{3}$ from $A$. Using these three points, we can determine that if the locus were a circle, its center would have to be at a point we'll call $C$, on $\overrightarrow{AB}$ a distance of 4 from $A$ and 1 from $B$ (1 unit past $B$ from $A$).
Next, consider a point $P$ in the locus with $PA=2x$, $PB=x$, and $PC=y$. Apply Stewart's Theorem to get $$\begin{align}
3y^2+(2x)^2&=4(x^2+3) \\
3y^2+4x^2&=4x^2+12 \\
y^2=4 \\
y=2.
\end{align}$$ So, all possible points $P$ in the locus are a distance of 2 from $C$, which means that the locus is a circle of radius 2, centered at $C$.
n.b. This suggests that if $A=(a,0)$ and $B=(-a,0)$, $C=(-2a,0)$ and the radius is $2a$, which would give an equation of $(x+2a)^2+y^2=4a^2$, which is different from what you said you got.
|
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|
Rectifiability of a curve Let $f$ be a function defined on $[0,1]$ by
$$f(x) = { 0, \text{ if } x = 0} $$
$$f(x) = { x \sin \frac 1 x , \text{ if } 0 < x \leq 1} $$
Prove that the curve $\{(x, f(x)) : x \in [0,1]\}$ is not rectifiable.
I'm not sure how to approach this. The general idea seems logical, we're proving that the length of the curve is infinite, but the method seems difficult to find.
|
The arclength $L$ of a curve $(x,f(x))$ from $x=a$ to $x=b$ is defined as
$$L=\int_a^b\sqrt{1+(f'(x))^2}dx.$$
Therefore, in this case, $f(x)=\displaystyle x\sin(\frac{1}{x})$, which implies that
$$(f'(x))^2=\Big[\sin(\frac{1}{x})-\frac{1}{x}\cos(\frac{1}{x})\Big]^2\geq-\frac{2}{x}\sin(\frac{1}{x})\cos(\frac{1}{x})+\frac{1}{x^2}\cos^2(\frac{1}{x}).$$
Hence, if $x\in\displaystyle[\frac{1}{2\pi n+\pi/3},\frac{1}{2\pi n}]$ where $n\in\mathbb{N}$, then
$$(f'(x))^2\geq -2(2\pi n+\frac{\pi}{3})+4\pi^2 n^2\cos^2(2\pi n+\pi/3)=\pi^2n^2-4\pi n-\frac{2\pi}{3}.$$
Therefore, the arclength $L$ of $(x,f(x))$ from $x=0$ to $x=1$ can be estimated as follows:
$$L=\int_0^1\sqrt{1+(f'(x))^2}dx\geq \sum_{n=1}^\infty\int^{\frac{1}{2\pi n}}_{\frac{1}{2\pi n+\pi/3}}\sqrt{1+(f'(x))^2}dx$$
$$\geq \sum_{n=1}^\infty\int^{\frac{1}{2\pi n}}_{\frac{1}{2\pi n+\pi/3}}\sqrt{1+\pi^2n^2-4\pi n-\frac{2\pi}{3}}dx$$
$$=\sum_{n=1}^\infty\sqrt{1+\pi^2n^2-4\pi n-\frac{2\pi}{3}}\cdot\frac{\frac{\pi}{3}}{(2\pi n)(2\pi n+\pi/3)}.$$
It's easy to see that the last series in $n$ diverges to infinity by using limit comparison test with the harmonic series $\displaystyle\sum_{n=1}^\infty\frac{1}{n}$.
|
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|
A quartic diophantine equation Here is the statement: Let $a,b \in \mathbb{Z}$ positive integers such that $a^2=b^4+b^3+b^2+b+1.$ Prove $b=3.$
I've tried is the following: Let $\Sigma=b^4+b^3+b^2+b+1$. If $a\equiv 0\mod 3$, then $a^2\equiv 0 \mod 3$. If $a\equiv 1\mod 3$ or $a\equiv 2\mod 3$, then $a^2\equiv 1\mod 3$. In the other hand, $b\equiv 0\mod 3$ or $b\equiv 2\mod 3$, then $\Sigma\equiv 1\mod 3$ and if $b\equiv 1\mod 3$, $\Sigma\equiv 1\mod 3$. Thus $a=3n$ or $a=3n+2$ for some $n\in\mathbb{Z}$, and $b=3m$ or $b=3m+2$ for some $m\in\mathbb{Z}$. In a similar way, I have prove $a$ cannot be even.
If $a=3n$, for some $n\in \mathbb{Z}$ and $b=3m$, for some $n\in \mathbb{Z}$, we arrive at the following equation $(3n-1)(3n+1)=3m(3³m³+3²m²+3m~1).$ So, $a\equiv 0\mod 3$ or $b\equiv 0\mod 3$ but not both at same time.
I think I should use the Legendre symbol. In this case $(\frac{\Sigma}{3})=0 \text{ or }1$.
|
If $b$ is even then $b^4+b^3+b^2+b+1$ is caught between $$(b^2+(b/2))^2=b^4+b^3+(1/4)b^2$$ and $$b^4+b^3+(9/4)b^2+b+1=(b^2+(b/2)+1)^2$$
If $b$ is odd, have a look at $(b^2+(b-1)/2)^2$ and $(b^2+(b+1)/2)^2$.
|
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|
How to find $\sqrt[3]{8i}$ How do I find the following cube root?
$$\sqrt[3]{8i} = ?$$
I tried by adding $\sqrt[3]{i^3 + 8i + i}$ but that is where my imagination quits.
|
Here is another way that avoids exponential functions
for this problem but is not a method that I would
recommend in general, e.g. to find $\sqrt[10]{8i}$,
for which the method described Andre Nicolas's and
Agusti Roig's answers work much better.
Suppose $a$ and $b$ are real numbers such that $(a+bi)^3 = 8i$. We have
$$\begin{align*}
(a+bi)^3 &= a^3 + 3a^2bi + 3a(bi)^2 + (bi)^3\\
&= (a^3 -3ab^2) + (3a^2b - b^3)i\\
&= 0 + 8i,
\end{align*}$$
and so
$$\begin{align*}
a^3 - 3ab^2 = a(a^2 - 3b^2) &= 0\\
3a^2b - b^3 = b(3a^2 - b^2) &= 8
\end{align*}$$
*
*From the first equation we see that one possible value for $a$ is $0$,
and in this case, the second equation reduces to $-b^3 = 8$ giving $b = -2$.
In other words, one cube root of $8i$ is $-2i$.
*Alternatively, if $a \neq 0$, then from the first equation we see
that it must be that $a^2 - 3b^2 = 0$
so that $a = \pm \sqrt{3}b$. Substituting into the second equation, we
get $b(9b^2 - b^2) = 8b^3 = 8$, that is $b^3 = 1$. Now, $b = 1$ is
the only real number solution to $b^3 = 1$, and so we get that
the other two
cube roots of $8i$ are $\sqrt{3} + i$ and $-\sqrt{3} + i$.
|
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|
Dice Question Probability A fair die is rolled until a $6$ appears. What is the probability that is must be cast more than 5 times?
So this is $1- P(\text{dice has to be cast less than or equal to}\ 5 \ \text{times})$. So this probability is equal to $$ P =\frac{1}{6}+ \frac{5}{6} \frac{1}{6}+\left(\frac{5}{6} \right)^{2} \frac{1}{6} + \cdots + \left(\frac{5}{6} \right)^{4} \frac{1}{6}$$
So just add $-1$ to this?
|
The easiest way to do this is simply to raise the probability that a six is not rolled to the fifth power, as these probabilities are independent, which gives $$P(\text{cast more than five times}) = P(\text{individual roll is not 6})^5 = \left(\frac{5}{6}\right)^5$$
This is equivalent to $1 - P$ where $P$ is what you wrote, as
$$1 - P = 1 - \frac{1}{6}\left(\frac{5}{6} + \cdots + \left(\frac{5}{6}\right)^4\right) = 1 - \frac{1}{6}\frac{1-(5/6)^5}{1 - 5/6} = \frac{1}{6}\frac{(5/6)^5}{1/6} = \left(\frac{5}{6}\right)^5$$
|
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Number of possible triangle for the given triangle. An equilateral triangle has n equally spaced dots on each side. How many triangles can be formed (of any size)?
Analysis
If there are n dots on each side then total no of dots =n(n+1)/2
Number of combination using 3 at a time =${n(n+1)/2 \choose 3}$
but all set of 3 pts wont be ∆ some will be co linear points
like x1,x2,x3 are coliner on line AB , similarly on CD,EF n so on2
n similarly on other side of ∆ BT, GH & so on & AT ,SJ & so on
combination of 3 dots on AB = ${n \choose 3}$
on CD = ${n-1 \choose 3}$
on EF = ${n-2 \choose 3}$
so total = ${n \choose 3}+{n-1 \choose 3}+{n-2 \choose 3}....+{3 \choose 3}$
as there are 3 side
$\therefore \ $ total number of such combination = $3\cdot\left[{n \choose 3}+{n-1 \choose 3}+{n-2 \choose 3}....+{3 \choose 3}\right]$
= $\left[3 \cdot \sum\limits_{k = 3}^n \binom{k}{3} \right]$
$\therefore \ $ total number of triangle
= ${n(n+1)/2 \choose 3}-\left[3 \cdot \sum\limits_{k = 3}^n \binom{k}{3} \right]$
But this is not the answer
as x1,x4,x6,x7 are also collinear
also as the length of triangle increases ,we find more co-linearity at different
angles
|
This question has previously been considered for triangles created by points in a square lattice; see here.
As with the square case, the challenge on the triangular lattice is determining the number of collinear triples. A000938 in OEIS gives the answer for the square lattice as
$$\left(2 \;\sum _{m=2}^n\; \sum _{k=2}^n (n-k+1) (n-m+1) \gcd (k-1,m-1)\right)-\frac{1}{6} n^2 (n^2-1).$$
This can be rearranged as
$$\left(2 \;\sum _{m=2}^n\; \sum _{k=2}^n (n-k+1) (n-m+1) (\gcd (k-1,m-1)-1)\right)+\frac{1}{3} n^2 (n-1)(n-2)$$
where each summand is the number of collinear triples with end-points offset by $(k-1,m-1)$, and the final term is the number of horizontal and vertical triples.
For the triangular lattice, a similar analysis gives the number of collinear triples as
$$
3\; \sum _{k=2}^n \;\sum _{m=2}^k \frac{1}{2} (n-k+1) (n-k+2) (\gcd (k-1,m-1)-1).
$$
After rearranging, we get the total number of triangles in the triangular lattice as
$$
\frac{1}{2} \left(n^2+n+2\right) \binom{n+2}{4}
-\frac{3}{2}\; \sum _{k=2}^n\; \sum _{m=2}^k (n-k+1) (n-k+2) \gcd (k-1,m-1).
$$
The first few values are $0, 1, 17, 105, 407, 1216, 3036, 6696, 13428, 25005, 43861, 73277$. As commented below, this is A194131 in OEIS.
|
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|
How can I solve a complicated induction exercise? (formula for sum of fourth powers) I have an induction exercise:
It is for $n \in \mathbb{N_0}$. Show:
$$ \sum\limits_{k=1}^{n} k^4 = \frac 1 {30} n(n+1)(2n+1)(3n^2+3n-1) $$
As far as I understand it, you have to put in $(n+1)^4$ at the end and you have to resolve it to the old function replacing $n$ with $n+1$. So I have on my paper:
$$ \frac 1 {30} (n+1)((n+1)+1)(2(n+1)+1)(3(n+1)^2+3(n+1)-1) = \frac 1 {30} n(n+1)(2n+1)(3n^2+3n-1) + (n+1)^4 $$
How can I solve this exercise? I probably have to disprove it because it is not true for numbers $> 1$.
All the other examples were with very tiny formulas.
I really tried to find a solution; what should I do?
|
There is one way to more or less predict the coefficients in one of these sums. It is not a use of induction as such, but I think it is worth knowing when you are beginning the study of induction. It is just Pascal's triangle.
$$ \sum_{k=0}^n \left( \begin{array}{c} k \\ 0 \end{array} \right) = \left( \begin{array}{c} n+1 \\ 1 \end{array} \right), $$
$$ \sum_{k=1}^n \left( \begin{array}{c} k \\ 1 \end{array} \right) = \left( \begin{array}{c} n+1 \\ 2 \end{array} \right), $$
$$ \sum_{k=2}^n \left( \begin{array}{c} k \\ 2 \end{array} \right) = \left( \begin{array}{c} n+1 \\ 3 \end{array} \right), $$
$$ \sum_{k=3}^n \left( \begin{array}{c} k \\ 3 \end{array} \right) = \left( \begin{array}{c} n+1 \\ 4 \end{array} \right), $$
$$ \sum_{k=4}^n \left( \begin{array}{c} k \\ 4 \end{array} \right) = \left( \begin{array}{c} n+1 \\ 5 \end{array} \right). $$
So then you need expressions such as
$$ k = \left( \begin{array}{c} k \\ 1 \end{array} \right), $$
$$ k^2 = 2 \; \left( \begin{array}{c} k \\ 2 \end{array} \right) + \left( \begin{array}{c} k \\ 1 \end{array} \right), $$
$$ k^3 = 6 \; \left( \begin{array}{c} k \\ 3 \end{array} \right) + 6 \; \left( \begin{array}{c} k \\ 2 \end{array} \right) + \left( \begin{array}{c} k \\ 1 \end{array} \right), $$
$$ k^4 = 24 \; \left( \begin{array}{c} k \\ 4 \end{array} \right) + 36 \; \left( \begin{array}{c} k \\ 3 \end{array} \right) + 14 \; \left( \begin{array}{c} k \\ 2 \end{array} \right) + \left( \begin{array}{c} k \\ 1 \end{array} \right), $$
$$ k^5 = 120 \; \left( \begin{array}{c} k \\ 5 \end{array} \right) + 240 \; \left( \begin{array}{c} k \\ 4 \end{array} \right) + 150 \; \left( \begin{array}{c} k \\ 3 \end{array} \right) + 30 \; \left( \begin{array}{c} k \\ 2 \end{array} \right) + \left( \begin{array}{c} k \\ 1 \end{array} \right). $$
It helps that these expressions still work if you take
$$ \left( \begin{array}{c} k \\ t \end{array} \right)= \; 0 $$ whenever
$ 0 \leq k < t.$ Indeed, under the same rule, the sums in the first section can all start at $k=0$ instead of the beginning values I typed in.
Well, just for laughs, I worked it through,
$$ \sum_{k=0}^n k^4 = 24 \; \left( \begin{array}{c} n+1 \\ 5 \end{array} \right) + 36 \; \left( \begin{array}{c} n+1 \\ 4 \end{array} \right) + 14 \; \left( \begin{array}{c} n+1 \\ 3 \end{array} \right) + \left( \begin{array}{c} n+1 \\ 2 \end{array} \right). $$ Writing things out, I got a common denominator of 30, then I put everything together into one fraction and got
$$ \sum_{k=0}^n k^4 = \frac{6 n^5 + 15 n^4 + 10 n^3 -n}{30}, $$ where the lack of a constant term was predictable but the lack of an $n^2$ term seems an accident.
Now that I have been through this one, line by line, I can see how absolutely crucial it is that I be able to correctly rewrite one of the formulas in the first section as starting at $k=0,$ as
$$ \sum_{k=0}^n \left( \begin{array}{c} k \\ 4 \end{array} \right) = \left( \begin{array}{c} n+1 \\ 5 \end{array} \right), $$ which is true because we
take
$$ \left( \begin{array}{c} k \\ t \end{array} \right)= \; 0 $$ whenever
$ 0 \leq k < t.$
|
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Calculating this residue (residue theorem) Given $$\int_{\gamma}\frac{1}{(z-a)(z-\frac{1}{a})}dz,$$ and $0<a<1$, where $\gamma(t)=e^{it}$ and $0\le t \le 2\pi$
I am trying to find the residue of$$f(z)=\frac{1}{(z-a)(z-\frac{1}{a})}$$
The answer says $\operatorname{res}(f,\mathbb C)=\frac{1}{(a-\frac{1}{a})}$?
Any help will be appreciated, thanks.
|
$$ f(z)=\frac{1}{(z-a)(z-\frac{1}{a})}=\frac{\frac{1}{a-\frac{1}{a}}}{z-a}+\frac{\frac{-1}{a-\frac{1}{a}}}{z-\frac{1}{a}} $$
$$ \gamma \text{ is unit circle} \hspace{20mm} \text{Poles of Function} \begin{cases} z_1=a &\text{placed inside of unit circle} \\
z_2=\frac{1}{a} &\text{placed outside of unit circle} \end{cases} $$
$$ |a|<1 \rightarrow |\frac{1}{a}|>1 \hspace{10mm} $$
$$ \text{Newton's generalised binomial theorem} $$
$$ (a+b)^n =a^n+\frac{n}{1!}a^{n-1}b+\frac{n(n-1)}{2!}a^{n-2}b^2+...+b^n $$
$$ \frac{\frac{1}{a-\frac{1}{a}}}{z-a}= \frac{1}{a-\frac{1}{a}}(z-a)^{-1}=\frac{1}{a-\frac{1}{a}}(z^{-1}+z^{-2}a+z^{-3}a^2+...)=\frac{1}{a-\frac{1}{a}}\sum_{n-1}^{\infty}z^{-n}a^{n-1}$$
$$ \text{so residue is coefficient of term }z^{-1} \rightarrow\frac{1}{a-\frac{1}{a}} $$
$$ I=\int_{\gamma}\frac{1}{(z-a)(z-\frac{1}{a})}dz=2 \pi i .\operatorname{res}(f,\mathbb C)=\frac{2 \pi i}{(a-\frac{1}{a})} $$
|
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|
Showing $24|(n+1)\implies24|\sigma(n)$ Question:
Show that if $n$ is a positive integer such that $24$ divides into $n + 1$, then $24$ divides the sum of all divisors of $n$ (denoted in number theory by $\sigma(n)$ or $\sigma_1(n)$).
For example if $n = 95$, then $n + 1 = 96 = 4 \times 24$ and the sum of the divisors of $n$ is $$1 + 5 + 19 + 95 = 120 = 5 \times 24.$$ (Note that the number $n$ is included among its divisors.)
|
We're looking for divisors $a, b$ such that $a \times b = n \equiv - 1 \pmod{24}$. In order for this to be possible, $a$ and $b$ must both be coprime to 24. Now note the following:
$a \times b \equiv -1 \pmod{24}$
$a^2 + a \times b \equiv a^2 -1 \pmod{24}$
$a(a + b) \equiv (a-1)(a+1) \pmod{24}$
Since $a$ is coprime to 24, and thus $a$ is odd, $(a-1), (a+1)$ must be even numbers. Furthermore, they are consecutive even numbers and thus at least one of them is also divisible by 4. In addition, as $a$ is coprime to 24, one of $(a-1), (a+1)$ must be divisible by 3 because if $a \equiv 1 \pmod{3}$ then $3|(a-1)$ and if $a \equiv 2 \mod{3}$ then $3|(a+1)$. This means that $4 \times 2 \times 3 = 24 | (a+1)(a-1)$. As $a$ is coprime to 24, this implies that $(a + b) \equiv 0 \pmod{24}$ for all $a, b$.
|
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Proving the inequality: $[x-(x^2)/2 < \ln(1+x) < x]$ , $x>0$ Prove the inequality:
$[x-(x^2)/2 < \ln(1+x) < x]$ , $x>0$
The right side is easy, I used taylor expansion to show that $e^x > 1+x$ since
$e^x = 1 + x + x^2/2 + x^3/3! +\cdots $
The left side I expanded $\ln(1+x)$ to $x-x^2/2 + x^3/3 - x^4/4 +\cdots$
moved $(x-x^2/2)$ to the left and got this:
$$0 < x^3/3 - x^4/4 \cdots $$
The only thing I can think of here is that since the first term is positive and this series goes on forever the first term will always stay positive since the other terms cancel out...
Not sure this is mathematically coherent or even correct.
thanks.
|
For the left inequality, since you know that $x-\frac{x^2}{2}=\ln(1+x)$ when $x=0$ you need only check the derivatives: $\frac{d}{dx}(x-\frac{x^2}{2})=1-x$ while $\frac{d}{dx}(\ln(1+x))=\frac{1}{1+x}$, and so
$$\frac{d}{dx}(\ln(1+x))-\frac{d}{dx}(x-\frac{x^2}{2})=\frac{1}{1+x}-(1-x)=\frac{1-(1+x)(1-x)}{1+x}=\frac{x^2}{(1+x)}$$
which is positive for positive $x$, thus for $x>0$, $\ln(1+x)$ is always growing faster than $x-\frac{x^2}{2}$ so it must be greater than $x-\frac{x^2}{2}$.
|
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differential equation nondevelopable I try to solve this differential equation whose solution seems not to be constructable in power series
$y''+(x+a/x^2+b)y=0$, where $a$ and $b$ are some positive real numbers.
If one can help me please?
|
Another way of solving this equation will be a series expansion in the parameter $b$. We know that for $b=0$ the two independent solutions can be easily found by means of the series expansion method.Those solutions are related to Bessel functions. Let us therefore assume that the whole solution reads:
\begin{equation}
y(x) = \sum\limits_{j=0}^\infty b^j y^{(j)}(x)
\end{equation}
We call the $j=0$ function the basic solution and the $j>0$ functions the corrections of order $j$.
Inserting the ansatz into the ODE we get:
\begin{equation}
y^{(0)}_\pm(x) = \sqrt{x} J_{\pm \frac{1}{3} \sqrt{1-4 a}} \left(\frac{2}{3} x^{\frac{3}{2}}\right) =
\frac{x^\alpha}{{\mathcal A}_\pm} F_{0,1}[1\pm \frac{1}{3} \sqrt{1-4 a};-\frac{x^3}{9}]
\end{equation}
where $\alpha=1/2(1\pm \sqrt{1-4 a})$ and ${\mathcal A}_\pm = 3^{\frac{\pm}{3} \sqrt{1-4 a} } \left(\frac{\pm}{3} \sqrt{1-4 a} \right)!$.
We have:
\begin{equation}
\left[\frac{d^2}{d x^2} + (x + \frac{a}{x^2})\right] y^{(j)}(x) = -y^{(j-1)}(x)
\end{equation}
for $j=1,2,\dots$. The above equation can be solved by means of Greens functions. The solution reads:
\begin{equation}
y^{(j)}(x) = \int\limits_0^x \left(\frac{y^{(0)}_-(x) y^{(0)}_+(\xi) - y^{(0)}_+(x) y^{(0)}_-(\xi)}{{\mathcal W}\left[y^{(0)}_+,y^{(0)}_-\right](\xi)}\right) \cdot (-) y^{(j-1)}(\xi) d \xi
\end{equation}
Here ${\mathcal W}[y^{(0)}_+,y^{(0)}_-]$is the Wronskian. It is a constant as a function of $\xi$ and it reads:
\begin{equation}
{\mathcal W}[y^{(0)}_+,y^{(0)}_-](\xi) = - \frac{\sin\left[\frac{\pi}{3} \sqrt{1- 4 a}\right]}{\frac{\pi}{3}}
\end{equation}
Writing the above equation in a compact way we get:
\begin{equation}
y^{(J)}(x) = \frac{1}{{\mathcal W}^J} \int\limits_0^x {\mathcal K}^{(J)}(x,\xi) \cdot y^{(0)}(\xi) d \xi
\end{equation}
where
\begin{equation}
{\mathcal K}^{(J)}(x,\xi) := \int\limits_\xi^x {\mathcal K}^{(J-1)}(x,\eta) \cdot {\mathcal K}(\eta,\xi) d \eta
\end{equation}
for $J=2,3,\dots$ and
\begin{equation}
{\mathcal K}^{(1)}(x,\xi) := \left| \begin{array}{rr} y_+^{(0)}(x) & y_-^{(0)}(x) \\ y_+^{(0)}(\xi) & y_-^{(0)}(\xi) \end{array} \right|
\end{equation}
Now, in order to construct the higher order corrections we compute a following quantity(to be termed moments):
\begin{equation}
{\mathcal M}_{J,l}(x) := \int\limits_0^x {\mathcal K}^{(J)}(x,\xi) \cdot \xi^{l+\alpha} d \xi
\end{equation}
where $J=1,2,\dots$ and $l=0,1,2,\dots$. Integrating by parts twice we obtain a following recursion relation for the moments:
\begin{equation}
(l+2)(l+2\pm \sqrt{1-4 a}){\mathcal M}_{J,l} + {\mathcal M}_{J,l+3} = -{\mathcal W}\cdot\left({\mathcal M}_{J-1,l+2} 1_{J>1} + \delta_{J,1} x^{l+2+\alpha}\right)
\end{equation}
for $J=1,2,\dots$. The solutions to the recursion relations are pretty straightforward and read:
\begin{eqnarray}
{\mathcal M}_{1,l} &=& {\mathcal W} \cdot x^{l+\alpha-1} \cdot \left\{ F_{1,2}\left[\begin{array}{rr} 1 \\
\frac{l+2}{3} & \frac{l+2\pm\sqrt{1-4 a}}{3}\end{array};-\frac{x^3}{3^2}\right]-1\right\} \\
{\mathcal M}_{J,l} &=& {\mathcal W} \sum\limits_{j=0}^\infty \left(\frac{-1}{3^2}\right)^{j+1} \cdot \frac{{\mathcal M}_{J-1,l+3 j+2}}{\left(\frac{l+2}{3}\right)^{(j+1)} \left(\frac{l+2\pm \sqrt{1-4 a}}{3}\right)^{(j+1)}}
\end{eqnarray}
for $J>1$. Using the expressions for the moments along with the expression for the correction of order $j$ we obtain the first order correction:
\begin{eqnarray}
&& y_\pm^{(1)}(x) =
\frac{x^{\alpha+2}}{(-9)^1 {\mathcal A}_\pm} \left[(-\frac{1}{3})!\right]^2 \left(\pm \frac{\sqrt{1-4 a}}{3}\right)!
\sum\limits_{j=0}^\infty \frac{(-\frac{x^3}{9})^j)}{(\frac{2}{3}+j)!(\frac{2\pm\sqrt{1-4 a}}{3}+j)!}\\
&&\sum\limits_{j_1=0}^j \binom{j_1-\frac{1}{3}}{-\frac{1}{3}} \cdot \binom{j_1-\frac{1}{3} \pm \frac{\sqrt{1-4 a}}{3}}{-\frac{1}{3}}
\end{eqnarray}
Likewise the second order correction reads:
\begin{eqnarray}
&&y_\pm^{(2)}(x) = \frac{x^{\alpha+4}}{(-9)^2 {\mathcal A}_\pm} \left[(-\frac{1}{3})!\right]^4 \left(\pm \frac{\sqrt{1-4 a}}{3}\right)!
\sum\limits_{j=0}^\infty \frac{(-\frac{x^3}{9})^j)}{(\frac{4}{3}+j)!(\frac{4\pm\sqrt{1-4 a}}{3}+j)!}\\
&&\sum\limits_{0\le j_1 \le j_2 \le j}^j
\binom{j_1-\frac{1}{3}}{-\frac{1}{3}} \cdot \binom{j_1-\frac{1}{3} \pm \frac{\sqrt{1-4 a}}{3}}{-\frac{1}{3}}
\binom{j_2+\frac{1}{3}}{-\frac{1}{3}} \cdot \binom{j_2+\frac{1}{3} \pm \frac{\sqrt{1-4 a}}{3}}{-\frac{1}{3}}
\end{eqnarray}
It is now easy to see what is the pattern for all higher order corrections. As such the problem is in principle solved. It would be nice to reduce the multiple sums to single sums though.
|
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|
Lim $x\to 0$ of $\frac{\sin(\pi x)}{\tan(\sqrt{3}x)}$ $$
\lim_{x \to 0} \frac{\sin(\pi x)}{\tan(\sqrt{3} x)}
$$
I need a step by step explanation. Thank you.
|
Hint:
First, let's expand what we are looking at:
$$\begin{align}
\frac{\sin(\pi x)}{\tan(\sqrt{3}x)} = \frac{\sin(\pi x)}{\frac{\sin(\sqrt{3} x)}{\cos(\sqrt{3} x)}} = \sin(\pi x) \cdot \frac{\cos(\sqrt{3}x)}{\sin(\sqrt{3}x)} = \sin(\pi x)\cdot\frac{1}{\sin(\sqrt{3}x)}\cdot \cos(\sqrt{3}x).
\end{align}$$
Next, we want to get each $\sin$ term in the form $\frac{\sin(t)}{t}$. One way to do that is to multiply by "$1$" in a helpful way:
$$\begin{align}
\sin(\pi x)\cdot\frac{1}{\sin(\sqrt{3}x)}\cdot \cos(\sqrt{3}x) &= \sin(\pi x)\frac{\pi x}{\pi x} \cdot \frac{1}{\sin(\sqrt{3}x)} \cdot \frac{\sqrt{3}x}{\sqrt{3} x} \cos(\sqrt{3}x)
\\
&= \frac{\sin(\pi x)}{\pi x} \cdot \frac{\sqrt{3} x}{\sin(\sqrt{3} x)} \cdot \frac{\pi x}{\sqrt{3} x} \cos(\sqrt{3} x).
\end{align}$$
I'll let you take a limit and see what happens.
|
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How to find$\int x^2 \arctan(x) \;dx $ I was solving $\int x^2 \arctan(x) \;dx $
I set $u=x^2$, $dv= \arctan(x)$, so I could get $du=2x$, $v=x\arctan(x)-\frac12\ln(1+x^{2})$.
From $\int x^2 \arctan(x)\;dx = uv - \int v \; du$
I got
$$\begin{align*}&\int x^2 \arctan(x) \; dx =\\
&x^3\arctan(x) -\frac12\ln(1+x^{2})-\int 2x\left[x^{2}\arctan x -\frac12\ln(1+x^2)\right]\;dx \end{align*}$$
and simplified if; then I got
$$3\int x^2\arctan(x) \;dx = x^3 \arctan(x)-\frac12\ln(1+x^2)+\int x\ln(1+x^2) \;dx$$
after that I use $w=\ln(1+x^2) \; dv =dx$ to find $\int x\ln(1+ x^2)$
but I got
$$x^2 \arctan(x)-\int 2x\arctan(x) \; dx$$
If I got $$x^2 \arctan(x)-\int 2x^2 \arctan(x)\;dx$$ instead, it would be easy to solve the question....
How can I solve this question and if you find any my mistake could you post this wall ??
Thank you !
|
We have
$$
\int x^2 \arctan x\;dx.
$$
Let
$$
\begin{align}
u & = \arctan x, \\ \\
du & = \frac{dx}{x^2+1}, \\ \\
dv & = x^2 \;dx, \\ \\
v & = \frac{x^3}{3}.
\end{align}
$$
Then
$$
\begin{align}
\int u\;dv & = uv - \int v\;du = \frac{x^3}{3}\arctan x - \int \frac{x^3\;dx}{3(x^2+1)} = \frac{x^3}{3}\arctan x - \frac 1 3\int \left(x- \frac{x}{x^2+1}\right)\;dx \\ \\
& = \frac{x^3}{3}\arctan x - \frac{x^2}{6} + \frac 16 \log(x^2+1) +C.
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/106827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Steps in evaluating $\int \frac{1}{x^4-c^4} dx$? $$ \int\ \frac {dx} {x^4 - c^4} \\ $$
is equal to (from integral tables)
$$ \frac {1} {4 c^3} \ln \frac {x-c} {x+c} - \frac {1} {2 c^3} \tan^{-1}\frac {x} {c}$$
If I let $\frac{x}{c}=\tan u$, then $dx/c= \sec^2 u du$, and the integral becomes
$$ \frac {1} {c^3} \int\ \frac {\sec ^2u du} {\tan^4u -1} = \frac {1} {c^3} \int\ \frac {\sec ^2u du} {(\tan^2u +1)(\tan^2u -1)} $$
Since $\tan^2u+1=\sec^2u$ the integral becomes
$$ \frac {1} {c^3} \int\ \frac {du} {(\tan^2u -1)} $$
Am I on the right track? What should I do next?
|
You’d be better off writing factoring $x^4-c^4$ as $(x^2-c^2)(x^2+c^2)$ and then as $(x-c)(x+c)(x^2+c^2)$ and decomposing $\frac1{x^4-c^4}$ into partial fractions:
$$\frac1{x^4-c^4}=\frac{A}{x-c}+\frac{B}{x+c}+\frac{Cx+D}{x^2+c^2}\;.$$
Once you’ve done the algebra to find $A,B$, and $C$, the integrations should be pretty straightforward: two logs and a tangent substitution.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $\lim\limits_{(x,y) \to (0,0)} \frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4} $ I'm trying to find $$\lim\limits_{(x,y) \to (0,0)} \frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4} .$$
After I tried couple of algebraic manipulation, I decided to use the polaric method.
I choose $x=r\cos \theta $ , $y=r\sin \theta$, and $r= \sqrt{x^2+y^2}$, so I get
$$\lim\limits_{r \to 0} \frac{e^{-\frac{1}{r^2}}}{r^4\cos^4 \theta+r^4 \sin^4 \theta } $$
What do I do from here?
Thanks a lot!
|
We have for $t>0$ that $e^t\geq\frac{t^3}3$ so
$$\frac{\exp\left(-\frac 1{x^2+y^2}\right)}{x^4+y^4}\leq \frac 3{\frac 1{(x^2+y^2)^3}}\frac 1{x^4+y^4}=\frac{3(x^2+y^2)^3}{x^4+y^4}=3(x^2+3y^2+3x^2+y^2)=12(x^2+y^2),$$
and you are done.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Computing $ \sum_{n=0}^{\infty} (-1)^{n+1}\int_{0}^{1}\frac{t^{3n+3}}{1+t^3}\mathrm{d} t$ I would like to find the exact value of the following series:
$$ \sum_{n=0}^{\infty} (-1)^{n+1}\int_{0}^{1}\frac{t^{3n+3}}{1+t^3}\mathrm{d} t$$
We can easily show that the series converges using the alternating series test:
$$ 0 \leq\int_{0}^{1}\frac{t^{3n+3}}{1+t^3}\mathrm{d} t \leq\frac{1}{3n+4}$$
So $$ \int_{0}^{1}\frac{t^{3n+3}}{1+t^3}\mathrm{d} t \rightarrow_{n\rightarrow\infty} 0$$
And $$ \int_{0}^{1}\frac{t^{3n+6}}{1+t^3}\mathrm{d} t-\int_{0}^{1}\frac{t^{3n+3}}{1+t^3}\mathrm{d} t=\int_{0}^{1}\frac{t^{3n+3}(t^3-1)}{1+t^3}\mathrm{d} t\leq0 $$
So the series converges.
Do you have any idea to compute the sum?
|
We have
$$ \begin{align*}
\sum_{n=0}^{\infty} (-1)^{n+1} \int_{0}^{1} \frac{t^{3(n+1)}}{t^3 + 1} \; dt
& = - \int_{0}^{1} \frac{t^3}{(t^3 + 1)^2} \; dt \\
&= \left[ \frac{t}{3} \frac{1}{t^3 + 1}\right]_{0}^{1} - \frac{1}{3} \int_{0}^{1} \frac{dt}{t^3 + 1} \\
&= \frac{1}{6} - \frac{1}{3} \int_{0}^{1} \left( \frac{1}{3(t+1)} -\frac{2 t-1}{6(t^2-t+1)}+\frac{1}{2(t^2-t+1)} \right) \; dt \\
&= \frac{1}{6} - \frac{1}{3} \left( \frac{1}{3} \log 2 + 0 + \frac{\pi}{3\sqrt{3}} \right) \\
&= \frac{1}{6} - \frac{1}{9} \left( \log 2 + \frac{\pi}{\sqrt{3}} \right).
\end{align*}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Try to prove $\lim_{n \to \infty}n(\ln 2-A_n) = \frac{1}{4}$ $$A_n=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$$
Try to prove $$\lim_{n \to \infty}n(\ln 2-A_n) = \frac{1}{4}$$
I try to decompose $\ln 2$ as $$\ln(2n)-\ln(n)=\ln\left(1+\frac{1}{2n-1}\right)+\dots+\ln\left(1+\frac{1}{n}\right)\;,$$ but I can't continue, is that right?
|
Trying to follow the idea of the OP, write $$ \log 2 = \log (2n+2) - \log (n+1) = \log \left( 1+ \frac{1}{2n+1} \right) + \log \left( 1+ \frac{1}{2n} \right) + \cdots + \log \left( 1+ \frac{1}{n+1} \right) $$
so then $$ n( \log 2 - A_n) = n\log \left(1+ \frac{1}{2n+1} \right) + n\sum_{k=1}^n \left( \log \left( 1+ \frac{1}{n+k} \right) - \frac{1}{n+k} \right) .$$
Since near $x=0$ we have $\displaystyle \log(1+x) = x - \frac{x^2}{2} + \mathcal{O}(x^3) ,$ the first term tends to $1/2$ and the summand is $ \displaystyle \frac{-1}{2 (n+k)^2 } + \mathcal{O}(1/n^3) .$ Thus, $$n\sum_{k=1}^n \left( \log \left( 1+ \frac{1}{n+k} \right) - \frac{1}{n+k} \right) = \frac{-1}{2} \cdot \frac{1}{n} \left( \sum_{k=1}^n \frac{1}{\left(1+ \frac{k}{n} \right)^2}\right) + \mathcal{O}(1/n) $$
$$ \to \frac{-1}{2} \int^1_0 \frac{1}{(1+x)^2} dx= -\frac{1}{4}. $$
Thus, $$ n(\log 2 - A_n) \to \frac{1}{2} - \frac{1}{4} = \frac{1}{4}.$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
What is the least value of the function $y= (x-2) (x-4)^2 (x-6) + 6$? What is the least value of function:
$$y= (x-2) (x-4)^2 (x-6) + 6$$
For real values of $x$ ?
Does $\frac{dy}{dx} = 0$, give the value of $x$ which will give least value of $y$?
Thanks in advance.
|
$$\frac{dy}{dx} = (x-4)^2 (x-6)+(x-4)^2 (x-2)+2(x-2)(x-4)(x-6) = 0 $$
$$ \Rightarrow (x-4) \left[ (x-4)(2x-8)+2(x-2)(x-6) \right] = 0$$
$$ \Rightarrow (x-4) \left[ 2x^2-16x+32+2x^2-16x+24 \right] = 0$$
$$ \Rightarrow (x-4) \left[ 4x^2-32x+56 \right] = 0$$
$$ \Rightarrow 4(x-4) \left[ (x-4+\sqrt{2})(x-4-\sqrt{2}) \right] = 0$$
There is a maxima or minima at $x=4, \hspace{5pt}x=(4-\sqrt{2}),\hspace{5pt} x=(4+\sqrt{2})$
If you take the second derivative and find where the second derivative is positive, that is where the function value has a minima.
Second derivative is
$$8(x-4)^2+4(x-4+\sqrt{2})(x-4-\sqrt{2})$$
And at $x=4\pm\sqrt{2}$ for instance the second derivative is positive and therefore at it is minimum at those points.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
If $f(x)f(y)=f(\sqrt{x^2+y^2})$ how to find $f(x)$ As we know, for the $$f(x)f(y)=f(x+y)$$ $f(x)=\mathrm e^{\alpha x}$ is a solution.
What about
$f(x)f(y)=f(\sqrt{x^2+y^2})$?
Does anybody know about the solution of the function equation?
I tried to find $f(x)$.
See my attempts below to find $f(x)$.
$$f(x)=a_0+a_1x+\frac{a_2x^2}{2!}+\frac{a_3x^3}{3!}+\cdots$$
$$f(y)=a_0+a_1y+\frac{a_2y^2}{2!}+\frac{a_3y^3}{3!}+\cdots$$
$$f(x)f(y)=a_0f(y)+a_1f(y)x+\frac{a_2f(y)x^2}{2!}+\frac{a_3f(y)x^3}{3!}+\cdots$$
$$f(\sqrt{x^2+y^2})=a_0+a_1\sqrt{x^2+y^2}+\frac{a_2(x^2+y^2)}{2!}+\frac{a_3(x^2+y^2)^{3/2}}{3!}+\cdots=$$
$$f(\sqrt{x^2+y^2})=a_0+a_1y\sqrt{1+(x/y)^2}+\frac{a_2(x^2+y^2)}{2!}+\frac{a_3y^2(1+(x/y)^2)^{3/2}}{3!}+\cdots=f(x)f(y)=a_0f(y)+a_1f(y)x+\frac{a_2f(y)x^2}{2!}+\frac{a_3f(y)x^3}{3!}+\cdots$$
if we use binom expansion for $(1+(x/y)^2)^{m}$
$$(1+(x/y)^2)^{m}=1+\frac{mx^2}{y^2}+\frac{m(m-1)x^4}{2!y^4}+\frac{m(m-1)(m-2)x^6}{3!y^6}+\cdots$$
Let's put the expansion to the equation $f(\sqrt{x^2+y^2})$
$$
\begin{align}
& f(\sqrt{x^2+y^2}) =a_0 + a_1 y \left( 1 + \frac{(1/2)x^2}{y^2} + \frac{(1/2)((1/2)-1)x^4}{2!y^4} \right. \\ \\
& \left. {} + \frac{(1/2)((1/2)-1)((1/2)-2)x^6}{3!y^6} + \cdots\right) + \frac{ a_2 (x^2+y^2)}{2!} \\ \\
& + \frac{a_3y^2 \left(1+\frac{(3/2)x^2}{y^2}+\frac{(3/2)((3/2)-1)x^4}{2!y^4}+\frac{(3/2)((3/2)-1)((3/2)-2)x^6}{3!y^6}+\cdots\right)}{3!} +\cdots \\ \\
& = a_0f(y)+a_1f(y)x+\frac{a_2f(y)x^2}{2!}+\frac{a_3f(y)x^3}{3!}+\cdots
\end{align}
$$
If we equal for all $x^n$ terms in both sides
we can see $a_{2n-1}=0$, but to find $a_{2n}$ seems hard for me.
Any idea to find $a_{2n}$
Thanks in advice.
|
Let' assume you want solution to $f:\mathbb{R}\mapsto\mathbb{R}$ that satisfies this equation $$\forall x,y\in\mathbb{R},f(x)f(y)=f\left(\sqrt{x^2+y^2}\right)\tag{1}$$
Let's plug in $(x,y)=(0,0)$ in the equation $(1)$,
$$f(0)^2=f(0)\Rightarrow f(0)\in\{0,1\}$$
Case-1 : $f(0)=0$
For $x>0, f(x)=f\left(\sqrt{x^2+0^2}\right)=f(x)f(0)=0$.
For $x<0, f(x)^2=f(x)f(x)=f\left(\sqrt{x^2+x^2}\right)=f\left(\sqrt{2}|x|\right)=0\Rightarrow f(x)=0$.
So, if $f(0)=0$, then $\forall x\in\mathbb{R}, f(x)=0$
Case-2 : $f(0)=1$
In this case, we claim that this function must be even.
Proof :
$f(x)=f(x)f(0)=f\left(\sqrt{x^2+0^2}\right)=f\left(\sqrt{(-x)^2+0^2}\right)=f(-x)f(0)=f(-x)\tag{2}$
Now, we will consider two assumptions under this case.
Assumption-A :
$\exists a>0, f(a)=0$.
Then, $\forall x\geq a$, $$f(x)=f\left(\sqrt{a^2+\left(\sqrt{x^2-a^2}\right)^2}\right)=f(a)f\left(\sqrt{x^2-a^2}\right)=0$$
Assumption-B :
$\exists b>0, f(b)>0$
Then, $$f\left(\sqrt2b\right)=f\left(\sqrt{b^2+b^2}\right)=f(b)^2>0$$
Proceeding this way, we can find arbitrarily large real number $x$ for which $f(x)>0$. Which contradicts with Assumption-A that says after reaching a bound, we can't find and real number $x$ for which $f(x)>0$.
So, we conclude that $a$ and $b$ can't exist simultaneously. If we assume the existence of $a$, We get the solution : $$f(x) =\left\{\begin{array}{10}1 & \mbox{if } x=0\\0 & \mbox{otherwise}\end{array}\right.$$
We can check and indeed, this is a valid solution.
Now we assume the existence of $b$, that is, $\exists b>0, f(b)>0$. Since this assumption excludes the possibility of the existence of real number $a$ for which $f(a)=0$, we only have two options : either $f(x)>0$ or $f(x)<0$ for any real number $x$. We claim that $\forall x, f(x)>0$.
Proof :
$$f(x)=f\left(|x|\right)=f\left(\sqrt{\left(\frac{x}{\sqrt2}\right)^2+\left(\frac{x}{\sqrt2}\right)^2}\right)=\left(f\left(\frac{x}{\sqrt2}\right)\right)^2>0$$
So, $\forall x, f(x)>0$.
Now, we will plug in $(\sqrt{x},\sqrt{y})$ where $x,y\geq0$ in equation $(1)$,
$$f\left(\sqrt{x}\right)f\left(\sqrt{y}\right)=f\left(\sqrt{x+y}\right)\\\Leftrightarrow \ln\left(f\left(\sqrt{x}\right)\right)+\ln\left(f\left(\sqrt{y}\right)\right)=\ln\left(f\left(\sqrt{x+y}\right)\right)$$
Let's define $\phi:\mathbb{R}_{\geq0}\mapsto\mathbb{R}$ such that $\phi(x)=\ln\left(f\left(\sqrt{x}\right)\right)$. Note that $\phi(x)+\phi(y)=\phi(x+y)$ which satisfies the Cauchy's functional equation. So, $\phi$ is a solution to Cauchy's functional equation. The most trivial solution is $\forall x\geq0,\phi(x)=kx$ where $k\in\mathbb{R}$. There are non trivial solutions which are highly pathological functions.
So, analyzing all the cases we got this solutions :
*
*$\forall x, f(x)=0$
*$f(x) =\left\{\begin{array}{10}1 & \mbox{if } x=0\\0 & \mbox{otherwise}\end{array}\right.$
*$\forall x, f(x)=e^{\phi(x^2)}$ where $\phi:\mathbb{R}_{\geq0}\mapsto\mathbb{R}$ and $\phi$ satisfies Cauchy's functional equation.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Help to evaluate determinant I want to evaluate the determinant of the $n \times n$ matrix
$\left|\begin{array}{ccccc} 1 & 0 & \ldots & 0& 0 \\
0 & 0 & \ldots & 0 & -a\\
0 & 0 & \ldots & -a & 0\\
&&&\vdots \\
0 & -a & 0 &\dots & 0
\end{array}\right|.$
So I try to say that it is $(-1)^{ n + (n-1) + \ldots n-(n-2)}(-a)^{n-1}$. So power of -1 should be $\frac{(n-1)(n+2)}{2} + n-1$. However answer given is $n(n-1)/2$. Where is wrong?
|
Using cofactor expansion, we have:
$$
\left|\begin{array}{ccccc} 1 & 0 & \ldots & 0& 0 \\
0 & 0 & \ldots & 0 & -a\\
0 & 0 & \ldots & -a & 0\\
&&&\vdots \\
0 & -a & 0 &\dots & 0
\end{array}\right|
=
\left|\begin{array}{ccccc}
0 & \ldots & 0 & -a\\
0 & \ldots & -a & 0\\
&&&\vdots \\
-a & 0 &\dots & 0
\end{array}\right|
.
$$
The determinant of the anti-diagonal matrix in the RHS is give by (note that RHS is $n-1\times n-1$):
$$ \det = \dfrac{(-1)^{\frac{n(n-1)}{2}} (-a)^n }{(-a)} = (-1)^{\frac{n(n-1)}{2}}(-a)^{n-1}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\int \cos(\cos x)~dx$
Evaluate $\int \cos(\cos x)~dx$
I tried to use chain rule but failed. Can anyone help me please?
|
$\int\cos\cos x~dx=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\cos^{2n}x}{(2n)!}~dx=\int\left(1+\sum\limits_{n=1}^\infty\dfrac{(-1)^n\cos^{2n}x}{(2n)!}\right)~dx$
For $n$ is any natural number,
$\int\cos^{2n}x~dx=\dfrac{(2n)!x}{4^n(n!)^2}+\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin x\cos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+C$
This result can be done by successive integration by parts.
$\therefore\int\left(1+\sum\limits_{n=1}^\infty\dfrac{(-1)^n\cos^{2n}x}{(2n)!}\right)~dx$
$=x+\sum\limits_{n=1}^\infty\dfrac{x}{4^n(n!)^2}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n((k-1)!)^2\sin x\cos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+C$
$=\sum\limits_{n=0}^\infty\dfrac{x}{4^n(n!)^2}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n((k-1)!)^2\sin x\cos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+C$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$ In Spivak's Calculus 3rd Edition, there is an exercise to prove the following:
$$x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$$
I can't seem to get the answer. Either I've gone wrong somewhere, I'm overlooking something, or both. Here's my (non) proof:
$$\begin{align*}
x^n - y^n &= (x - y)(x^{n-1} + x^{n-2}y +\cdots+ xy^{n-2} + y^{n-1}) \\
&= x \cdot x^{n-1} + x \cdot x^{n-2} \cdot y + \cdots + x \cdot x \cdot y^{n-2} + x \cdot y^{n-1}\\
&\qquad + (-y) \cdot x^{n-1} + (-y) \cdot x^{n-2} \cdot y + \cdots + (-y) \cdot x \cdot y^{n-2} + (-y) \cdot y^{n-1}\\
&= x^n + x^{n-1} y + \cdots + x^2 y^{n-2} + x y^{n-1} - x^{n-1}y - y^2 x^{n-2} - \cdots- x y^{n-1} - y^n \\
&= x^n + x^2 y^{n-2} - x^{n-2} y^2 - y^n \\
&\neq x^n - y^n
\end{align*}$$
Is there something I can do with $x^n + x^2 y^{n-2} - x^{n-2} y^2 - y^n$ that I'm not seeing, or did I make a mistake early on?
EDIT:
I should have pointed out that this exercise is meant to be done using nine of the twelve basic properties of numbers that Spivak outlines in his book:
*
*Associate law for addition
*Existence of an additive identity
*Existence of additive inverses
*Commutative law for additions
*Associative law for multiplication
*Existence of a multiplicative identity
*Existence of multiplicative inverses
*Commutative law for multiplication
*Distibutive law
|
Here is the inductive step, presented more conceptually
$$\rm\frac{x^{n+1}-y^{n+1}}{x-y}\: =\ x^n\: +\ y\ \frac{x^n-y^n}{x-y}$$
So, intuitively, proceeding inductively yields
$$\rm\:x^n + y\: (x^{n-1} + y\: (x^{n-2} +\:\cdots\:))\ =\ x^n + y\: x^{n-1} + y^2\: x^{n-2} + \:\cdots $$
Use this intuition to compose a formal proof by induction.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Fitting a sine wave of known frequency through three points We have a computationally expensive function of a large set of data and an angle, that is known to result in a sine wave:
$$ f(\text{data}, \theta) \approx a \sin (\theta + b) + c$$
We want to find the constants $a$, $b$, and $c$, executing the function as few times as possible.
|
Yes, you can do it in 3 points, considering $f(0)$, $f(\pi)$, and $f(\frac{\pi}{2})$.
$$
\begin{aligned}
f(0) &= a\sin b + c & \\
f(\tfrac{\pi}{2}) &= a\sin(b + \tfrac{\pi}{2}) + c & &= a\cos b + c \\
f(\pi) &= a\sin(b + \pi) + c & &= - a\sin b + c
\end{aligned}
$$
To find $c$:
$$
\begin{aligned}
f(0) + f(\pi) &= 2c \\
\implies c &= \frac{f(0) + f(\pi)}{2}
\end{aligned}
$$
To find $a$:
$$
\begin{aligned}
(f(0) - c)^2 + (f(\tfrac{\pi}{2}) - c)^2 &= a^2\sin^2 b + a^2\cos^2 b= a^2 \\
\implies a &= \sqrt{(f(0) - c)^2 + (f(\tfrac{\pi}{2}) - c)^2}
\end{aligned}
$$
And finding $b$:
$$
\begin{aligned}
\frac{f(0) - c}{f(\tfrac{\pi}{2}) - c} &= \frac{a\sin b}{a\cos b} = \tan^{-1} b \\
\implies b &= \operatorname{atan2}\left(f(0) - c, f(\tfrac{\pi}{2}) - c\right)
\end{aligned}
$$
|
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|
Sums of Double series where $a_{m,n}=\frac {m-n} {2^{m+n}}\frac {\left( m+n-1\right) !} {m!n!}$ I am trying to show if $$a_{m,n}=\dfrac {m-n} {2^{m+n}}\dfrac {\left( m+n-1\right) !} {m!n!}$$ such that $$\left( m,n>0\right) $$ that $\sum _{m=0}^{\infty }\left( \sum _{n=0}^{\infty }a_{m,n}\right) =-1$, $\sum _{n=0}^{\infty }\left(\sum _{m=0}^{\infty } a_{m,n}\right) =1$.
Now we observe that $a_{m,0}=2^{-m}$,$a_{0,n}=-2^{-n}$ and $a_{0,0}=0$.
We can rewrite the first series as
that $\sum _{m=0}^{\infty }\left( \sum _{n=0}^{\infty }a_{m,n}\right)$.
$\Rightarrow \sum _{m=0}^{\infty }\left(\dfrac {1} {2^{m}}+\dfrac {\left( m-1\right) } {2^{m+1}}+\dfrac {\left( m-2\right) \left( m+1\right) } {2^{m+2}2!}+\dfrac {\left( m-3\right) \left( m+1\right) \left( m+2\right) } {2^{m+3}3!}+\ldots\right)$
$\Rightarrow \sum _{m=0}^{\infty }\dfrac {1} {2^{m}}\left(1+\dfrac {\left( m-1\right) } {2^{1}}+\dfrac {\left( m-2\right) \left( m+1\right) } {2^{2}2!}+\dfrac {\left( m-3\right) \left( m+1\right) \left( m+2\right) } {2^{3}3!}+\ldots\right)$
I can n't recognize the expression inside the brackets, any help of how to proceed forward would be much appreciated.
|
Here is a possible approach:
$$\frac{1}{(1-x)^{m+1}} = \sum_{n=0}^{\infty} \binom{m+n}{n} x^n$$
Write $\displaystyle a_{m,n}$ as $\displaystyle \frac{m}{(m+n)2^{m+n}} \binom{m+n}{n}$ - $\displaystyle \frac{n}{(m+n)2^{m+n}} \binom{m+n}{n}$
For the first term, multiply by $x^m$ and integrate between $0$ and $1/2$.
For the second terms, differentiate first, the multiply by $x^{m+1}$ and integrate.
When you sum over $m$, you can move the sum into the integral and can possibly simplify it (I haven't worked out the details...)
|
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|
How to evaluate $\displaystyle\int {1\over (1+kx^2)^{3/2}}dx$ What change of variable should I use to integrate $$\displaystyle\int {1\over (1+kx^2)^{3/2}}dx$$
I know the answer is $$\displaystyle x\over \sqrt{kx^2+1}.$$ Maybe a trig or hyperbolic function?
|
$\int \dfrac{1}{(1+kx^2)^{3/2}}dx$
Put $1+kx^2=t$, Then,
$$2kx\cdot dx = dt \text{. Also, } x = \sqrt{t-1\over k}$$
$$2kx\cdot dx = dt$$
$$dx = \frac{dt}{2kx} = \frac{\sqrt{k}dt}{2k\sqrt{t-1}}= \frac{dt}{2\sqrt{k}\sqrt{t-1}}$$
$\int \dfrac{1}{(1+kx^2)^{3/2}}dx=$ $\int \dfrac{1}{2t^{1.5}\sqrt{k}\sqrt{t-1}}dt$
$$\dfrac{1}{2\sqrt{k}}\int \dfrac{1}{t^{1.5}.\sqrt{t-1}}dt=\dfrac{1}{2\sqrt{k}}\dfrac{2\sqrt{t-1}}{\sqrt{t}}=\dfrac{1}{\sqrt{k}}\dfrac{\sqrt{t-1}}{\sqrt{t}}$$
$$=\dfrac{1}{\sqrt{k}}\dfrac{\sqrt{t-1}}{\sqrt{t}}$$
$$=\dfrac{1}{\sqrt{k}}\dfrac{\sqrt{kx^2}}{\sqrt{1+kx^2}}$$
$$=\dfrac{x}{\sqrt{1+kx^2}}$$
|
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|
Find a three digit number ($\overline{xyz}$)? (Excuse me for my english: I'm spanish speaker)
"Find a three digit number $\overline{xyz}$ such that $x^2 +y^2 + z^2$ is equal to the number (xyz). "
I have this equation:
$$x^2 -100x +(y^2-10y+z^2-z)=0$$
And the discriminant (in $x$):
$$\Delta_x = -4y^2 +40y-4z^2+4z+10000$$
Solutions for $x_i$ are:
$$x_i = \frac{100 \pm \sqrt{\Delta_x}}{2}$$
in which ${\Delta_x} \geq 0$. The obvious conditions are:
i) $x \neq 0$ and $x=1,..., 9$;
ii) $y, z = 0, ..., 9$.
I need an orientation because pick $(y,z) \in \{0,...9\} \times \{0,...9\}$ I believe it's arduous task.
Another "attemp" from myself is to find $(x, y, z) \in \{1,...9\} \times \{0,...9\} \times \{0,...9\} $ such that
$$(x-50)^2 + (y-5)^2 +\left(z- \frac{1}{2}\right)^2 = \frac{10101}{4}$$
but I'm not skilfull in multivariable calculus :(
Any advice is welcome.
|
It seems that solving it for $y$ is a better idea, as we are going to have $y=5\pm\sqrt{something}$.
Let us solve it for a number with any number of digits: $\overline{...vwxyz}$.
$y^2-10y+(z^2-z)+(x^2-100x)+(w^2-1000w)+...=0$
$y=5\pm\sqrt{25-(z^2-z)-(x^2-100x)-(w^2-1000w)-...}$.
Note, that $0\le y\le9$, therefore, $0\le(z^2-z)+(x^2-100x)+(w^2-1000w)+...\le25$.
But $z^2-z\le9^2-9=72$, while $x^2-100x\le1^2-100=-99$ for $1\le x\le9$, $w^2-1000w\le-999$ for $1\le w\le9$ etc.
Hence, $x=w=...=0$ and $25-z(z-1)$ is a square. For $z=0,1,2,3,4,5$ we get $25,25,23,19,13,5$ and only the first two are squares.
The only two numbers are $0$ and $1$.
|
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|
Show that $2^{3^k}+1$ is divisible by $3^k$ for all positive integers $k$ I attempted this by induction: Here is what I did
For $k=1, 2^{3^{1}}+1 = 2^{3}+1 = 9 = 3^{2}$ is divisible by $3^{1}$ , so the result is true for $k=1$
Now I assume the result to be true for $k=m$,
$2^{3^{m}}+1$ is divisible by $3^m$. To show the result to be true for $k=(m+1)$,
$2^{3^{m+1}}+1 = 2^{3^m} \times 2^3+1$ and I was stuck here.
|
The last step you did can be corrected as
$$2^{3^{m+1}}+1 = (2^{3^m})^3+1$$ because if you look at it this way
$$(2^{3^m})^3 = (2^{3^m}) \times (2^{3^m}) \times (2^{3^m}) = 2^{3 \times 3^m} = 2^{3^{m+1}}$$
Using the fact that the result is true for $k=m$
$$2^{3^{m}}+1 = x \times 3^{m} \implies 2^{3^{m}} = x3^{m}-1 \tag{B}$$
$$
\begin{align*}
2^{3^{m+1}} &= (2^{3^m})^3 \\
&= (x3^{m}-1)^3 \\
&= x^3 3^{3m}-x^2 3^{2m+1}+x3^{m+1}-1\\
\end{align*}
$$
Which means
$$2^{3^{m+1}}+1 = x^3 3^{3m}-x^2 3^{2m+1}+x3^{m+1} \tag{A}$$
And clearly $(A)$ shows the right side is divisible by $3^{m+1}$ because each term is divisible by $3^{m+1}$
which means $2^{3^{m + 1}}+1 = y3^{m+1}$ for some $y$, or the result is true for $k=(m+1)$.
|
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|
Calculate the integral of $1/(2x^2+2x+1)$ I need to calculate the integral of $1 / (2x^2+2x+1)$.
I used WolframAlpha and get this answer: $$\tan^{-1}(2x+1)$$ but I don't understand how to get there.
Can you help?
|
First note that $\displaystyle \int \frac{dy}{a^2+y^2} = \frac1a \tan^{-1}\left(\frac{y}{a} \right)$. This is obtained by substituting $y = a \tan( \theta)$ in the integrand and integrating.
For your problem, $I = \displaystyle \int \frac{dx}{2x^2 + 2x+1}$. First complete the square in the denominator i.e. rewrite $2x^2 + 2x + 1$ as $2 \left(x^2 + x + \frac12 \right) = 2 \left( \left( x + \frac12 \right)^2 + \left(\frac12 \right)^2 \right)$. Hence, we get $$I = \displaystyle \int \frac{dx}{2x^2 + 2x+1} = \int \frac{dx}{2 \left( \left( x + \frac12 \right)^2 + \left(\frac12 \right)^2 \right)} = \frac12 \int \frac{dx}{\left( \left( x + \frac12 \right)^2 + \left(\frac12 \right)^2 \right)}.$$ Setting $y = x + \frac12$, we get $$I = \frac12 \int \frac{dy}{y^2 + \left(\frac12 \right)^2} = \tan^{-1}(2y) = \tan^{-1}(2x+1).$$
|
{
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|
Please help me prove by induction that $n^n>1\cdot3\cdot5\cdot\ldots\cdot(2n-1)$ Please help me prove by induction that
$$
n^n>1\cdot3\cdot5\cdot\ldots\cdot(2n-1)
$$
|
We need to prove $1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)< n^n$
For $n=1$ we have $1 < 1$, what is wrong.
For $n=2$ we have $3 < 4$ what is true.
Note that $1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1) (2\cdot (n+1)-1)=n^n(2(n+1)-1)$
then prove $(n+1)^{n+1} - n^n\cdot (2\cdot (n+1)-1) > 0$
|
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|
Limit of sequence $x_n^n$ Let $x_1=2$, $x_{n+1}=\sqrt{x_n+\frac{1}{n}}$ for all $n\geq 1$. Prove that $\lim\limits_{n\to\infty}x_n=1$ and evaluate $\lim\limits_{n\to\infty}x_n^n$.
|
For $1$, first observe that if $x_n>1$ then $x_{n+1}=\sqrt{x_n+\frac{1}{n}}>\sqrt{1+\frac{1}{n}}>1$, so the sequence $(x_n)$ is bounded below by $1$. Note that
$$\begin{eqnarray}x_n>x_{n+1}&\iff&x_n>\sqrt{x_n+\frac{1}{n}}\\
&\iff&x_n^2>x_n+\frac{1}{n}\\
\end{eqnarray}$$
and that this is clearly true when $n=1$. If it holds for $n$ then
$$\begin{eqnarray}x_{n+1}^2&=&x_n+\frac{1}{n}\\
&=&\sqrt{x_n^2}+\frac{1}{n}\\
&>&\sqrt{x_n+\frac{1}{n}}+\frac{1}{n}\\
&>&x_{n+1}+\frac{1}{n}\\
&>&x_{n+1}+\frac{1}{n+1}\\
\end{eqnarray}$$
so it holds for $n+1$, thus by induction it holds for all $n$. Thus we have a decreasing sequence which is bounded below, so it must come to some limit. Since $x_{n+1}=\sqrt{x_n+\frac{1}{n}}$ is a continuous function of $x_n + \frac{1}{n}$, the limit must satisfy $L=\sqrt{L}$, so it must be either $1$ or $0$, and since the sequence is bounded below by $1$ the limit must be $1$.
For $2$, I suggest using the fact that
$$x_{n+1}^{n+1}=x_{n+1}\sqrt{\left(x_n+\frac{1}{n}\right)^{n}}$$
and applying the binomial theorem and looking at the resulting terms.
|
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|
Polynomial equation Is it possible to find polynomials with rational coefficients $P(x),Q(x)$ such that $Px^3+Px^2+Qx+2Q=1$? I have trying in vain to find one by inspection, but that might just be me.
|
Just in case the OP does not know the Euclidean algorithm, I'll compute it.
Divide $(x^3 + x^2)$ by $x+2$. This is just ordinary high school long division. So we find that
$$\begin{equation}(x^3 + x^2) = (x+2)q(x) + r(x)\end{equation}$$
where $q(x) = x^2 - x$ and $r(x) = 2x$. Now by the Euclidean algorithm we then divide $(x+ 2)$ by $2x$ to get
$(x+2) = 2x(\frac{1}{2}) + 2$ so that $2 = (x+2) - 2x(\frac{1}{2})$. Now from the first expression we found that $2x = (x^3 + x^2) - (x+2)(x^2 - x)$, so that substituting it in here we get
$$\begin{eqnarray*}2 &=& (x+2) - \frac{\bigg[(x^3 + x^2) - (x+2)(x^2 - x) \bigg]}{2}\\
&=& (x+2)\frac{(2+ (x^2 - x))}{4} - \frac{[x^3 + x^2]}{4} \end{eqnarray*} $$
Reading off this your $P$ is $-1/4$ while $$ Q= \frac{(2+ (x^2 - x))}{4} .$$
|
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|
Computation of a residue Consider the function $\cos\left(\frac{z}{z+1}\right)$, which has an essential singularity at the point $z=-1$.
How does one compute its residue at $z=-1$.
|
$$ A=\cos\left(\frac{z}{z+1}\right) = \cos\left(\frac{z\color{red}{+1-1}}{z+1}\right)=\cos \left( 1- \frac{1}{z+1} \right) $$
$$ \cos(a\pm b)=\cos(a)\cos(b) \mp \sin(a)\sin(b)$$
$$A=\cos(1)\cos\left(\frac{1}{z+1}\right)+\sin(1)\sin\left(\frac{1}{z+1}\right) $$
$$ \cos(z)=1-\frac{z^2}{2!}+\frac{z^4}{4!}-... \hspace{10mm} \sin(z)=z-\frac{z^3}{3!}+\frac{z^5}{5!}-... $$
$$ \cos\left(\frac{1}{z+1}\right)=1-\frac{1}{(z+1)^2 2!}+\frac{1}{(z+1)^4 4!}-... $$
$$ \sin\left(\frac{1}{z+1}\right)=\frac{1}{z+1}-\frac{1}{(z+1)^3 3!}+\frac{1}{(z+1)^5 5!}-... $$
$$ A=\cos(1) \left[ 1-\frac{1}{(z+1)^2 2!}+\frac{1}{(z+1)^4 4!}-...\right]+\sin(1)\left[ \frac{1}{z+1}-\frac{1}{(z+1)^3 3!}+\frac{1}{(z+1)^5 5!}-... \right]$$
$$ \text{ so residue is coefficient of term } \frac{1}{z+1} \rightarrow \sin(1)$$
|
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|
Sum of the reciprocal of sine squared I encountered an interesting identity when doing physics homework, that is,
$$ \sum_{n=1}^{N-1} \frac{1}{\sin^2 \dfrac{\pi n}{N} } = \frac{N^2-1}{3}. $$
How is this identity derived? Are there any more related identities?
|
Let's denote $ S_N=\sum_{n=1}^{N-1} \frac{1}{\sin^2 \dfrac{\pi n}{N} } $
Consider an equality:
$$\frac{\pi^2}{\sin^2(\pi s)}=\int_0^{\infty}\frac{x^{s-1}}{1-x}\ln\frac{1}{x}dx;0<s<1$$
Because of $0<\frac{n}{N}<1$, the integral form applies. Thus:
$$S_N=\frac{1}{{\pi}^2} \int_0^{\infty}\frac{\ln\frac{1}{x}}{1-x}\sum_{n=1}^{N-1}x^{\frac{n}{N}-1}dx=$$
$$= \frac{1}{{\pi}^2} \int_0^{\infty}\frac{\ln\frac{1}{x}}{1-x}\frac{x^{\frac{1}{N}-1}-1}{1-x^{\frac{1}{N}}}dx=$$
$$=\frac{N^2}{{\pi}^2} \int_0^{\infty}\frac{\ln\frac{1}{x}}{1-x}\frac{1-x^{N-1}}{1-x^{N}}dx=$$
$$=2\frac{N^2}{{\pi}^2} \int_0^{1}\frac{\ln\frac{1}{x}}{1-x}\frac{1-x^{N-1}}{1-x^{N}}dx= \frac{N^2-1}{3} $$
because of $$\int_0^{1}\frac{\ln\frac{1}{x}}{1-x}\frac{1-x^{N-1}}{1-x^{N}}dx= \frac{N^2-1}{N^2}\frac{{\pi}^2}{6}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How can I determinate the bases for the most simple representation of a linear transformation? Imagine a linear transformation $\Phi : \mathbb{R}^4 \rightarrow \mathbb{R}^3$ with the ordered standard basis:
$B = (\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix})$ and
$C = (\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix})$. The transformation is defined by $M^B_C (\Phi) = \begin{pmatrix} 0 & 0 & 0 & 0 \\1 & 2 & 3 & 4 \\5 & 6 & 7 & 8 \end{pmatrix}$
Lets define some other basis:
$D = (\begin{pmatrix} 1 \\ 2 \\ 3 \\ 4 \end{pmatrix}, \begin{pmatrix} 1 \\ 3 \\ 3 \\ 7 \end{pmatrix}, \begin{pmatrix} 3 \\ 1 \\ 4 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ 7 \\ 1 \\ 8 \end{pmatrix})$ and $E = (\begin{pmatrix} 2 \\ 3 \\ 5 \end{pmatrix}, \begin{pmatrix} 3 \\ 5 \\ 7 \end{pmatrix}, \begin{pmatrix} 5 \\ 7 \\ 11 \end{pmatrix})$
In this case you get this transformation matrix:
$M^D_E (\Phi) =
\begin{pmatrix}
110 & 156 & 93 & 195 \\
10 & 16 & 3 & 15 \\
-50 & -72 & -39 & -87
\end{pmatrix}$
If you have $M^D_E(\Phi)$ given and you want to find the two Basis B and C so that the linear transformation $M^B_C(\Phi)$ is as simple as possible. How would you do that?
(simple means as many zeros as possible)
I guess one step could be to determine the null space.
$\left \{ \begin{pmatrix}-39\\ 15\\ 0\\ 10\end{pmatrix}, \begin{pmatrix}-51\\ 30\\ 10\\ 0\end{pmatrix} \right \}$
|
Try the singular value decomposition, which gives orthonormal bases in the domain and in the codomain with respect to which the linear transformation is given by a diagonal matrix.
|
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Let $a,b$ be positive real numbers. Prove $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}} \geq \frac{2}{\sqrt{1+ab}}$ Let $a,b$ be positive real numbers. Prove $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}} \geq \frac{2}{\sqrt{1+ab}}$
if either
$(1) 0 \leq a,b \leq 1$
OR
$(2) ab \geq 3$
Since this question was under Trigonometry, I assumed the following.
Since $a,b$ are positive real numbers with $0 \leq a,b \leq 1$, I can assume that for some $x,y, a=\tan(x), b=\tan(y)$ and therefore it is to be shown that
$$\frac{1}{\sec x} + \frac{1}{\sec y} = \cos x+ \cos y \geq \frac{2\cos x \cos y}{\sqrt{cos(x-y)}}$$
(Originally posted without that $2$ on the right - Sorry!)
I do know that
$$\cos x + \cos y \geq 2 \sqrt{\cos x \cos y}$$
Now how to proceed? Just give me hints !
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Let $f(x)=\frac{1}{\sqrt{1+e^{2x}}}$.
Hence, $f''(x)=\frac{e^{2x}(e^{2x}-2)}{\sqrt{(e^{2x}+1)^5}}\geq0$ for all $x\geq\frac{1}{2}\ln2$.
Id est, for all $\{a,b\}\subset[\sqrt2,+\infty)$ by Jensen we obtain:
$$\sum_{cyc}\frac{1}{\sqrt{1+a^2}}=\sum_{cyc}f(\ln{a})\geq2f\left(\frac{\ln{a}+\ln{b}}{2}\right)=\frac{2}{\sqrt{1+ab}}$$
|
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|
Inter-causal reasoning: How to solve probability with two conditions? Below is the scheme of conditional dependence and the probabilities of events:
P(A=1) = 0.01
P(A=0) = 0.99
P(B=1) = 0.1
P(B=0) = 0.9
P(C=1|A=0,B=0) = 0.1
P(C=1|A=0,B=1) = 0.5
P(C=1|A=1,B=0) = 0.6
P(C=1|A=1,B=1) = 0.9
Given the probabilities above I wanted to calculate P(B=1|C=1) and P(B=1|C=1,A=1) but didn't get the correct result.
I wrote the probabilistic function the following way:
P(A, B, C) = P(A)P(B)P(C|A, B)
and then set the variables
P(B=1, C=1) = P(A=0, B=1, C=1) + P(A=1, B=1, C=1)=
=P(A=0)P(B=1)P(C=1|A=0, B=1) + P(A=1)P(B=1)P(C=1|A=1, B=1)=
=0.99*0.1*0.5 + 0.01*0.1*0.9 = 0.0495
The result however is not correct and don't know where is the error. I would be very thankful if anyone could correct/explain what's wrong.
|
The typical way I do inter-causal reasoning is to flip the conditional probabilities around --
$$
\begin{align}
P(B = 1 \vert C = 1) & = \frac{P(B = 1, C = 1)}{P(C = 1)} \\
& = \frac{P(C = 1 \vert B = 1)P(B = 1)}{P(C = 1)} \\ \\
P(B = 1 \vert C = 1, A = 1) & = \frac{P(B = 1, C = 1, A = 1)}{P(C = 1, A = 1)} \\
& = \frac{P(C = 1 \vert B = 1, A = 1)P(B = 1)P(A = 1)}{P(C = 1, A = 1)} \\
& = \frac{P(C = 1 \vert B = 1, A = 1) P(B = 1) P(A = 1)}{P(C = 1 \vert A = 1)P(A = 1)} \\
& = \frac{P(C = 1 \vert B = 1, A = 1) P(B = 1)}{P(C = 1 \vert A = 1)}
\end{align}
$$
Does that help?
|
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|
Plot $|z - i| + |z + i| = 16$ on the complex plane
Plot $|z - i| + |z + i| = 16$ on the complex plane
Conceptually I can see what is going on. I am going to be drawing the set of points who's combine distance between $i$ and $-i = 16$, which will form an ellipse. I was having trouble getting the equation of the ellipse algebraically.
I get to the point:
$x^2 + (y - 1) ^2 + 2\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} + x^2 + (y+ 1)^2 = 256$
$2x^2 + 2y^2 + 2\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} = 254$
It seems like I'm and doing something the hard way.
|
Maybe it's quicker way.
Equation $|z-i| + |z+i| = 16$ is equivalent to
$$
\sqrt{x^2 + (y-1)^2} = 16 - \sqrt{x^2 + (y+1)^2}.
$$
Squaring both sides you obtain
$$
x^2 + (y-1)^2 = 256 - 32\sqrt{x^2+(y+1)^2} + x^2 + (y+1)^2.
$$
Some terms cancel out hence you get
$$
8\sqrt{x^2 + (y+1)^2} = 64 + y,
$$
and
$$
64(x^2 + (y+1)^2) = 64^2 + 128y + y^2.
$$
Finally
$$
64x^2 + 63y^2 = 64 \cdot 63.
$$
Now it is easy to write ellipse equation
$$
\frac{x^2}{63} + \frac{y^2}{64} = 1.
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proof- central term of recursive pyramid
Moderator Note: At the time that this question was posted, it was from an ongoing contest. The relevant deadline has now passed.
How can we prove $4^{-n}(n+1)\left(2^{2n+1}-{2n+1\choose n+1}\right)$ produces the central term in the $2n+1$th row of the following pyramid?
The pyramid is produced in this way: atop the triangle is 1 and each number below it is determined by summing half of each of the numbers that it is supporting, and adding 1 to the sum.
1
3/2 3/2
7/4 10/4 7/4
15/8 25/8 25/8 15/8
31/16 56/16 66/16 56/16 31/16
I have attempted induction and it does not work well as I am sure you can see. I do not know how to prove the formula... even an intuitive proof would work.
Please see my comment below about constructing the pyramid as a whole using coefficients. Another question to all(ESP joriki) : what can be said in general about any term in the pyramid? Can we make a generalization of any term in the pyramid based on The row number and the $k+1$th or kth term in a row, perhaps, depending on how you define k? I have been looking at it and it seems that writing it in terms of k and z, where z=n-1 might make it simple, but I don't know.
|
It appears that joriki has answered the original question. I got bogged down on that when I got sick, but I did manage to establish that $4^{-n}(n+1)\left(2^{2n+1}-{2n+1\choose n+1}\right)$ gives the expected number of tosses of a fair coin required to get either $n+1$ heads or $n+1$ tails, as I mentioned in the comments. It’s too long for a comment, but it might be of some interest, so here it is.
Suppose that we toss a fair coin until we get either $n+1$ heads or $n+1$ tails. Say that this happens on the $k$-th toss; clearly $n+1\le k\le 2n+1$. There are $\binom{k-1}n$ sequences that terminate with a head on the $k$-th toss and another $\binom{k-1}n$ sequences that terminate with a tail on the $k$-th toss. Each of these occurs with probability $2^{-k}$, so the expected number of tosses is
$$2\sum_{k=n+1}^{2n+1}2^{-k}k\binom{k-1}n\;.$$
To get rid of the negative exponents I’m going to multiply through by $2^{2n}$ to get
$$\begin{align*}
2^{2n+1}\sum_{k=n+1}^{2n+1}2^{-k}k\binom{k-1}n&=\sum_{k=n+1}^{2n+1}2^{2n+1-k}(k-n)\binom{k}n\\
&=\sum_{k=1}^{n+1}2^{n+1-k}k\binom{n+k}n\\
&=\sum_{k=1}^{n+1}2^{n+1-k}(n+k)\binom{n+k-1}{k-1}\\
&=\sum_{k=0}^n2^{n-k}(n+1+k)\binom{n+k}k\\
&=\sum_{k=0}^n2^{n-k}(n+1)\binom{n+1+k}k\\
&=(n+1)\sum_{k=0}^n2^{n-k}\binom{n+1+k}k\;.
\end{align*}$$
Thus, we’d like to show that $$\sum_{k=0}^n2^{n-k}\binom{n+1+k}k=2^{2n+1}-\binom{2n+1}{n+1}\;.\tag{1}$$
Let $$s_n=\sum_{k=0}^n2^{n-k}\binom{n+1+k}k\;.$$ Then
$$\begin{align*}s_{n+1}&=\sum_{k=0}^{n+1}2^{n+1-k}\binom{n+2+k}k\\
&=\sum_{k=0}^{n+1}2^{n+1-k}\left(\binom{n+1+k}k+\binom{n+1+k}{k-1}\right)\\
&=2\sum_{k=0}^n2^{n-k}\binom{n+1+k}k+\binom{2n+2}{n+1}+\sum_{k=0}^n2^{n-k}\binom{n+2+k}k\\
&=2s_n+\binom{2n+2}{n+1}+\frac12\left(s_{n+1}-\binom{2n+3}{n+1}\right)\;,
\end{align*}$$
so $$s_{n+1}=4s_n+2\binom{2n+2}{n+1}-\binom{2n+3}{n+1}\;.$$ Taking $(1)$ as induction hypothesis, we have
$$\begin{align*}
s_{n+1}&=2^{2n+3}-4\binom{2n+1}{n+1}+2\binom{2n+2}{n+1}-\binom{2n+3}{n+1}\\
&=2^{2n+3}-\binom{2n+3}{n+2}+2\left(\binom{2n+1}{n+1}+\binom{2n+1}n\right)-4\binom{2n+1}{n+1}\\
&=2^{2n+3}-\binom{2n+3}{n+2}+2\left(\binom{2n+1}{n+1}+\binom{2n+1}{n+1}\right)-4\binom{2n+1}{n+1}\\
&=2^{2n+3}-\binom{2n+3}{n+2}\;,
&=
\end{align*}$$
as desired.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is this limit correct: $\lim_{x \to+\infty} \frac{\log_{2}(x-1)}{x} = 0$?
Find $\space\ \begin{align*} \lim_ {x \to+\infty} \left [ \frac{\log_{2}(x-1)}{x}\right]
\end{align*}$.
After some minutes around this limit I did it this way:
$\log_{2}(x-1)=y \Leftrightarrow 2^y=x-1$
So,$\space x=2^y+1$.
When $x \to +\infty$,$\space y \to +\infty$ also. By substitution:
$\begin{align*} \lim_ {y \to+\infty} \left [ \frac{\log_{2}(2^y+1-1)}{2^y+1}\right]=\lim_ {y \to+\infty} \left [ \frac{\log_{2}(2^y)}{2^y+1}\right]=\end{align*}$
$\begin{align*}\lim_ {y \to+\infty} \left [ \frac{y}{2^y+1}\right]=\lim_ {y \to+\infty} \left [ \frac{1}{\frac{2^y+1}{y}} \right]=\lim_ {y \to+\infty} \left [ \frac{1}{\frac{2^y}{y}+\frac{1}{y}}\right]= \frac{1}{+\infty+0}=0
\end{align*}$
Is this correct?Are there any other easy way to find this limit?Thanks
|
Your way is fine, as an alternative we have that
$$ \frac{\log_{2}(x-1)}{x}= \frac{\log_{2}(x-1)}{x-1}\frac{x-1}{x}\to 0\cdot 1 =0$$
indeed eventually $\log_{2}(x-1) \le \sqrt{x-1}$ and then
$$\frac{\log_{2}(x-1)}{x-1} \le \frac{\sqrt{x-1}}{x-1} \to 0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/127750",
"timestamp": "2023-03-29T00:00:00",
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|
Linear Algebra Working with Linear Transformations Let $v_1=[-3;-1]$ and $v_2= [-2;-1]$
Let $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ be the linear transformation satisfying:
$T(v_1)=[15;-6]$ and $T(v_2)=[11;-3]$
Find the image of an arbitrary vector $[x;y]$
|
Note sure if (homework) yet. So hint:
Let
$$
T =
\begin{pmatrix}
a & b \\ c & d
\end{pmatrix}
$$
We can re-interpret the given $T(v_1)$ and $T(v_2)$ as:
$$
\begin{pmatrix}
a & b \\ c & d
\end{pmatrix}
\begin{pmatrix}
-3 \\ -1
\end{pmatrix}
=
\begin{pmatrix}
15 \\ -6
\end{pmatrix} ,
\\
\begin{pmatrix}
a & b \\ c & d
\end{pmatrix}
\begin{pmatrix}
-2 \\ -1
\end{pmatrix}
=
\begin{pmatrix}
11 \\ -3
\end{pmatrix}
$$
Or more succinctly as,
$$
\begin{pmatrix}
a & b \\ c & d
\end{pmatrix}
\begin{pmatrix}
-3 & -2 \\ -1 & -1
\end{pmatrix}
=
\begin{pmatrix}
15 & 11 \\ -6 & -3
\end{pmatrix}
\tag{1}
$$
Can you take it from here?
|
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|
Differentiate respect to $x$ $(x^2+2x+1)^3$
let u=$x^2+2x+1$ $\frac{du}{dx} = 2x$ or $2x+2$ $\frac{dy}{dx}=3u^2 $
if $\frac{du}{dx} = 2x$ then
$3(x^2+2x+1)^2 (2x)$
answer is $6x(x^2+2x+1) $
Or if $\frac{du}{dx} = 2x+2$ then
$3(x^2+2x+1)^2 (2x+2)$
$6x(x^2+2x+1)+2$
however the right answer is $6(x+1)^5$
can please help me out?
thanks in advance!
|
by Chain rule :
$f'(x)=3(x^2+2x+1)^2 \cdot(x^2+2x+1)'=3\cdot(x+1)^4\cdot(2x+2)=$
$=6\cdot(x+1)^4\cdot(x+1)=6(x+1)^5$
|
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"timestamp": "2023-03-29T00:00:00",
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|
First Four Nonzero Terms of Taylor Series I'm not acing Calculus II and while this problem might be easy to some, it's not to me. Any help will do.
Problem: Find the first nonzero terms of the Taylor Series for $f(x) = \cos x$ where
$$a = {\pi\over 6}$$
My work:
\begin{array}{rlrll}
f(x) =& \cos x & f(a) =& \cos {\pi\over 6} &= {\sqrt 3\over 2} \\
f'(x) =& - \sin x & f'(a) =& - \sin {\pi\over 6} &= -{1\over 2} \\
f ''(x) =& - \cos x & f ''(a) =& - \cos {\pi\over 6} &= -{\sqrt 3\over 2} \\
f ''' (x) =& - (-\sin x) = \sin x & f ''' (a) =& \sin {\pi\over 6} &= {1\over 2} \\
f^4(x) =& \cos x & f^4(a) =& \cos {\pi\over 6} &= {\sqrt 3\over 2}
\end{array}
Taylor Series:
$$a_n={f^n(a)\over n!}(x-a)^n$$
$$\eqalign{
& \frac{{{f^0}(a)}}{{0!}}{(x - a)^0} + \frac{{f'(a)}}{{1!}}{(x - a)^1} + \frac{{f''(a)}}{{2!}}{(x - a)^2} + \frac{{f'''(a)}}{{3!}}{(x - a)^3} + \frac{{{f^4}(a)}}{{4!}}{(x - a)^4} \cr
& = \frac{1}{{0!}}\;\frac{{\sqrt 3 }}{2}(1) + \left( { - \frac{1}{2}} \right)\frac{1}{{1!}}\left( {x - \frac{\pi }{6}} \right) + \left( { - \frac{{\sqrt 3 }}{2}} \right)\frac{1}{{2!}}{\left( {x - \frac{\pi }{6}} \right)^2} + \frac{1}{{3!}}\frac{1}{2}{\left( {x - \frac{\pi }{6}} \right)^3} + \frac{1}{{4!}}\frac{{\sqrt 3 }}{2}{\left( {x - \frac{\pi }{6}} \right)^4} \cr
& = \;\frac{{\sqrt 3 }}{2} - \frac{1}{2}\left( {x - \frac{\pi }{6}} \right) - \frac{{\sqrt 3 }}{2}\frac{1}{2}{\left( {x - \frac{\pi }{6}} \right)^2} + \frac{1}{2}\frac{1}{{3 \cdot 2}}{\left( {x - \frac{\pi }{6}} \right)^3} + \frac{{\sqrt 3 }}{2}\frac{1}{{4 \cdot 3 \cdot 2}}{\left( {x - \frac{\pi }{6}} \right)^4} \cr
& = \;\frac{{\sqrt 3 }}{2} - \frac{1}{2}\left( {x - \frac{\pi }{6}} \right) - \frac{{\sqrt 3 }}{4}{\left( {x - \frac{\pi }{6}} \right)^2} + \frac{1}{{12}}{\left( {x - \frac{\pi }{6}} \right)^3} + \frac{{\sqrt 3 }}{{48}}{\left( {x - \frac{\pi }{6}} \right)^4} \cr} $$
Is this problem solved correctly? If not, where did I go wrong? If you have any tips for the Taylor Series I'll take it also. Thank you.
|
Recall that $f(h+a)=\cos(h+(\pi/6))=\cos(h)\cos(\pi/6)-\sin(h)\sin(\pi/6)$. Hence, $$2f(h+a)=\sqrt3\cos(h)-\sin(h).
$$
Now, $\cos(h)=1-\frac12h^2+O(h^4)$ and $\sin(h)=h-\frac16h^3+O(h^4)$, hence
$$
2f(h+a)=\sqrt3(1-\tfrac12h^2)-(h-\tfrac16h^3)+O(h^4).
$$
The first four terms of the expansion of $f(h+(\pi/6))$ are the $1$, $h$, $h^2$ and $h^3$ terms, namely,
$$
f\left(h+\frac\pi6\right)=\frac{\sqrt3}2-\frac12h-\frac{\sqrt3}4h^2+\frac1{12}h^3+O(h^4).
$$
More generally, the $h^{2n}$ term of the Taylor expansion is $$\frac{(-1)^n\sqrt3}{2(2n)!}$$ and the $h^{2n+1}$ term of the Taylor expansion is $$\frac{(-1)^{n+1}}{2(2n+1)!}.$$
|
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|
Finding $\int \frac{1}{x}\sqrt{1+\frac{1}{x^4}}\,dx$ Finding $$\int \frac{1}{x}\sqrt{1+\frac{1}{x^4}}\,dx$$
I suppose I need integration by parts and trigo substitution
Let $u=\frac{1}{x} \Rightarrow du = -\frac{1}{x^2} dx$
Let $dv = \sqrt{1+(\frac{1}{x^2})^2}$, $\frac{1}{x^2} = \tan{\theta}$. Is my substitution OK?
So $x=\frac{1}{\sqrt{\tan{\theta}}} \Rightarrow dx = -\frac{\sec^2{\theta}}{2\sqrt{\tan{\theta}}}\,d\theta$. But this will be very complicated to integrate later?
Am I supposed to be trying something else?
UPDATE: An attempt
$$\int \frac{1}{x} \sqrt{1+\frac{1}{x^4}} dx$$
Let $u = \frac{1}{x} \Rightarrow du = -\frac{1}{x^2} dx$
Let $dv = \sqrt{1+(\frac{1}{x^2})^2} dx$
Let $\alpha = \frac{1}{x^2} \Rightarrow d\alpha = -\frac{1}{2x^3}$
$dv=\sqrt{1+\alpha^2} d\alpha$
Let $\alpha = \tan{\theta} \Rightarrow d\alpha = \sec^2{\theta} d\theta$
$dv = \sqrt{1+\tan^2{\theta}} \sec^2{\theta} d\theta = \sec^3{\theta}$. Looks wrong here ?
|
Usually we want to get rid of the square root. One way to do so is to bring in into the form $\sqrt{1+\cosh^2(u)}$ so lets try this in two steps
$$\frac{1}{x^2}=v \Rightarrow dv = \frac{-1}{2x^3}dx \Leftrightarrow dx = -2x^3 \, dv$$
$$ \int \frac{1}{x}\sqrt{1+\frac{1}{x^4}}\,dx = \int \frac{-2x^3}{x}\sqrt{1+v^2}\,dv = -2\int \frac{1}{v}\sqrt{1+v^2}\,dv$$
With $v=\cosh(u) \Rightarrow dv = \sinh(u)\,du$ and $1+\cosh^2(u) = \sinh^2(u)$ follows
$$ ... = -2\int\frac{\sinh^2(u)}{\cosh(u)} \,du$$
This will become a bit messy from here on. Seeing how the final result is ...
$$ -\frac{1}{2} \sqrt{1+\frac{1}{x^4}}+\frac{1}{2}\sinh^{-1}\left(x^2\right) $$
|
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|
Prove via mathematical induction that $4n < 2^n$ for all $ n≥5$. I did the following base case $n = 5$
$$\begin{align*}
4(5) &\lt 2 ^5\\
20 &\lt 32
\end{align*}$$
So true.
$$\begin{align*}
4n &\lt 2^n\\
n &\lt 2^{n-2}\\
\log_2(n)+2 &\lt n
\end{align*}$$
But I don't think this is right. Where do I add in the $(n+1)$.
|
Given $5\leq{n}$
Assume $4n<2^n$
Prove $4(n+1)<2^{n+1}$:
*
*$4(n+1)=4n+4$
*$4n+4<2^n+4$ assumption used here
*$2^n+4<2^n+32$
*$2^n+32=2^n+2^5$
*$2^n+2^5\leq2^n+2^n$ given fact used here
*$2^n+2^n=2^{n+1}$
|
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|
Fallacy in Fermat's Little Theorem? Fermat's Little Theorem: If $p$ is prime, then for every $1 ≤ a < p$,
$a^{p-1} ≡ 1$ $(mod$ $p)$
Let $p$ be 9 (a composite number), and let $a$ be 2.
Let $S$ be the nonzero integers modulo $9$
$S = (1, 2, 3, 4, 5, 6, 7, 8)(mod$ $9)$
$2^8S$
$=2^8(1, 2, 3, 4, 5, 6, 7, 8)$$=
(1*2,2*2,3*2,4*2,5*2,6*2,7*2,8*2)$
$= (2, 4, 6,8,10,12,14,16)$$
= (1,2,3,4,5,6,7,8)(mod$ $9$)
∴ $8! = 2^8*8!$
Divide both sides by $8!$ and you arrive at:
$2^8 ≡ 1$ $(mod$ $9)$
Doesn't Fermat's Little Theorem only apply when $p$ is a prime number? The above calculations showed when $p = 9$ (composite), $a = 2$ still fulfills the equation.
|
i) No see Fermat pseudoprime.
ii) $2^8=256$ and $2+5+6=13$ so that $2^8= 1+3 \pmod{9}$ (the other one is right)
$2^{340} = 1\pmod{341}$ and $341=11\cdot 31$
|
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|
why does $\frac{1}{z\cdot \sin{z}} $ only have pole when clearly its undefined at $n\pi$ I am having trouble with a specific problem actually. I have a function $$f(z) = \frac{1}{z\cdot \sin{z}}$$
Now I want to find the residues of this. The Laurent series expanded about $0$ shows that $0$ is a pole of order $2$. The expansion looks something like this $$\frac{1}{z^2}+\frac{1}{6} + \frac{7z^2}{360} + \cdots $$
so since the first coefficient of $z$ is just zero, the residue of this function is $0$.
BUT I want to know why zero is the ONLY pole. Clearly $2\pi$ is a singularity point. Then when you expand about $2\pi$ you get the following expansion $$\frac{1}{2\pi (z - 2\pi)} - \frac{1}{4\pi^2} + \frac{(3+2\pi^2)(z-2\pi)}{24\pi^3} + \cdots $$
Again, it looks to me that the first negative power of $z$ has the coefficient $\frac{1}{2\pi}$.
So why is it that when I type in "poles of function 1/(z*sin(z))" wolfram only identifies 0 as the pole. If I type in "poles of function 1/(sin(z))" then it identifies the poles as $n\pi$. Furthermore if you type in "residues of 1/(z*sin(z))" it only identifies 0 as a residue when we just saw above that $\frac{1}{2\pi}$ is also a residue. Whats even more weird is that if you type in "residues of 1/(z*sin(z)) at 2pi" it does give the right residue. Weird.
|
Rather than expanding the function $f(z)$ around the point $z=2\pi$, let's rearrange $f(z)$ instead by looking at the Taylor series expansion of $\sin(z)$
$$ f(z) = \frac{1}{(z)(z-\frac{z^3}{3!}+\frac{z^5}{5}+\cdots)}$$
$$= \frac{1}{(z^2)(1-\frac{z^2}{3!}+\frac{z^4}{5}+\cdots)}$$
$$ f(z) = \frac{1}{(z^2)}\left(1+\left(\frac{z^2}{3!}-\frac{z^4}{5!}+\cdots\right)+\left(\frac{z^2}{3!}-\frac{z^4}{5!}+\cdots\right)^2+\cdots\right)$$
$$ f(z) = \left(\frac{1}{z^2} + \frac{1}{z}\right) \left(1+\frac{z^2}{3!}+\cdot\right) = \frac{1}{z^2}+\frac{1}{z}+\cdots
$$
Rather than expanding at (what seems) an undefined point, a quick series alteration gives us a different "picture" of this function.
|
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|
How do you integrate $\int \frac{1}{a + \cos x} dx$? How do you integrate $\int \frac{1}{a + \cos x} dx$? Is it solvable by elementary methods? I was trying to do it while incorrectly solving a homework problem but I couldn't find the answer.
Thanks!
|
This doesn't help you to evaluate the indefinite integral, but I though I would add that the definite integral $\int_0^{2 \pi
} \frac{1}{a + \cos x} \ dx$ can also be evaluated using methods from complex analysis. Let us make the simplifying assumption that $a > 1$ to avoid a blow up in the integral. Other values of $a$ (including complex ones!) can be made to work too.
We have
\begin{align*}
\int_0^{2 \pi
} \frac{dx}{a + \cos x} &= \int_0^{2 \pi} \frac{dx}{a + \frac{e^{ix} + e^{-ix}}{2}} \\
&= 2\int_0^{2 \pi} \frac{e^{ix} \ dx}{2ae^{ix} + e^{2ix} + 1} && \text{Let } z=e^{ix}, \text{ so } dz = ie^{ix} \ dx. \\
&= \frac{2}{i} \int_{|z|=1} \frac{dz}{z^2 + 2az + 1} \\
&= \frac{2}{i} \int_{|z|=1} \frac{dz}{(z-z_1)(z-z_2)}
\end{align*}
where the circle $|z|=1$ is parametrized counterclockwise and
\begin{align*} z_1 = -a - \sqrt{a^2-1} && z_2 = -a + \sqrt{a^2-1}.
\end{align*}
Clearly $z_1 < -a < -1$, so $z_1$ is outside of the unit circle. As for $z_2$, it is convenient to notice that
\begin{align*}
z_1 z_2 = 1.
\end{align*}
Thus, since $z_1$ is outside the unit circle, its inverse $z_2$ is inside the unit circle. So, applying the residue theorem, we have
\begin{align*}
\frac{2}{i} \int_{|z|=1} \frac{dz}{(z-z_1)(z-z_2)}
&= \frac{2}{i} \ 2 \pi i \ \mathrm{Res}\left( \frac{1}{(z-z_1)(z-z_2)}, z_2\right) \\
&= 4 \pi \frac{1}{z_2 - z_1} \\
&= 4 \pi \frac{1}{2 \sqrt{a^2 -1}} \\
&= \frac{2 \pi}{\sqrt{a^2-1}}.
\end{align*}
One can check this result against the one obtained from Arybatha's antiderivative, although a bit of care needs to be taken as an improper integral crops up while making the various substitutions.
\begin{align*}
\int_0^{2 \pi
} \frac{dx}{a + \cos x}
&= \int_0^\pi \frac{2 \ dy}{a+1 + (a-1) \tan^2(y)} && y= \frac{x}{2}\\
&= \int_0^\infty \frac{ 2 dt}{ a + 1 + (a-1)t^2} +\int_{-\infty}^0 \frac{ 2 dt}{ a + 1 + (a-1)t^2} && t = \tan(y) \\
&= \frac{2}{\sqrt{a^2-1}} \arctan\left( \sqrt{ \frac{a-1}{a+1}} t \right) \big|_0^\infty + \frac{2}{\sqrt{a^2-1}} \arctan\left( \sqrt{ \frac{a-1}{a+1}} t \right) \big|_{-\infty}^0 \\
&= \frac{2 \pi}{\sqrt{a^2-1}}
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/134577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 5,
"answer_id": 2
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|
Linear algebra - Dual, functionals Given three functionals
*
*$f_1(p) = \int_0^1 p(t)\,dt$
*$f_2(p) = \int_0^2 p(t)\, dt$
*$f_3(p) = \int_0^{-1} p(t)\, dt$
defined on $V = P_2$, the space of all polynomials over $\mathbb R$ of degree not greater than 2. Asking to prove that
$\{f_1, f_2, f_3\}$ is a basis for $V^*$ by finding the basis for $V$ of which it is the dual.
|
You want to find polynomials $p_1,p_2,p_3$ of degree at most $2$ with the property that $f_i(p_j) = 1$ if $i=j$, and $f_i(p_j) = 0$ if $i\neq j$.
For instance, you want to find $p_1(x) = a+bx+cx^2$ such that
$$\begin{align*}
1=f_1(p_1) &=\int_0^1(a+bx+cx^2)\,dx \\
&= \left.\left( ax + \frac{b}{2}x^2 + \frac{c}{3}x^3\right)\right|_{0}^1\\
&= a + \frac{b}{2}+\frac{c}{3}.\\
0 =f_2(p_1) &= \int_0^2(a+bx+cx^2)\,dx\\
&= \left.\left( ax + \frac{b}{2}x^2 + \frac{c}{3}x^3\right)\right|_{0}^2\\
&= 2a + 2b + \frac{8c}{3}.\\
0 = f_3(p_1) &= \int_0^{-1}(a+bx+cx^2)\,dx\\
&= \left.\left( ax + \frac{b}{2}x^2 + \frac{c}{3}x^3\right)\right|_{0}^{-1}\\
&= -a + \frac{b}{2} -\frac{c}{3}.
\end{align*}$$
So this gives us a system of three linear equations in three unknowns:
$$\begin{array}{rcccccl}
a & + & \frac{1}{2}b & + & \frac{1}{3}c & = & 1\\
2a & + & 2b & + & \frac{8}{3}c & = & 0\\
-a & + & \frac{1}{2}b & - & \frac{1}{3}c & = & 0.
\end{array}$$
Solving it will give the value of $a$, $b$, and $c$.
Replacing the solution vector with $(0,1,0)^T$ gives $p_2$; replacing the solution vector with $(0,0,1)^T$ gives $p_3$.
So, in essence, we are trying to find the inverse of a matrix; that matrix is related to the functions $f_1$, $f_2$, $f_3$. In fact, it is the inverse of the matrix corresponding to the linear transformtion $T\colon V\to V$ given by $T(p) = (f_1(p),f_2(p),f_3(p))$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/135372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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|
Solving the recursion $3a_{n+1}=2(n+1)a_n+5(n+1)!$ via generating functions I have been trying to solve the recurrence:
\begin{align*}
a_{n+1}=\frac{2(n+1)a_n+5((n+1)!)}{3},
\end{align*}
where $a_0=5$, via generating functions with little success. My progress until now is this:
Let $A(x)=\sum_{n=0} ^{\infty} a_nx^n$. By multiplying both sides of our recurrence relation by $x^n$ and summing over $n$ from $0$ to $\infty$, we see that
\begin{align}
\sum_{n=0} ^{\infty} a_{n+1} x^n = \frac{2}{3}\sum_{n=0} ^{\infty} (n+1)a_nx^n + \sum_{n=0} ^{\infty} (n+1)!x^n.
\end{align}
Using our definition of $A(x)$ we can rewrite the left hand side as
\begin{align*}
\sum_{n=0} ^{\infty} a_{n+1} x^n=\frac{A(x)-a_0}{x}.
\end{align*}
Such manipulations of the right hand side have been difficult because of the coefficients of the power series.
Is there anyway to proceed from here, or are generating functions not suited to solve such a recurrence?
|
Exponential generating function of sequence $\{a_n\}$ is $f(x) = \sum_{n=0}^\infty a_n \frac{x^n}{n!}$. Rewriting the recurrence equation as
$$
\frac{a_{n+1}}{n+1} = \frac{2}{3} a_n + \frac{5}{3} n!
$$
Now multiplying both sides by $\frac{x^n}{n!}$ and using recurrence relation for factorial:
$$
a_{n+1} \frac{x^n}{(n+1)!} = \frac{2}{3} a_n \frac{x^n}{n!} + \frac{5}{3} x^n
$$
Summing from $n=0$ to infinity:
$$
\frac{1}{x} \sum_{n=1}^\infty a_n \frac{x^n}{n!} = \frac{2}{3} \sum_{n=0}^\infty a_n \frac{x_n}{n!} + \frac{5}{3} \sum_{n=0}^\infty x^n
$$
or
$$
\frac{1}{x} \left( f(x) - a_0 \right) = \frac{2}{3} f(x) + \frac{5}{3} \frac{1}{1-x}
$$
Solving for $f(x)$ we readily get:
$$
f(x) = \frac{5}{1-x} = \sum_{n=0}^\infty (5 \cdot n!) \frac{x^n}{n!}
$$
Thus the solution is $a_n = 5 \cdot n!$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/135803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 1
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|
A 'should be simple' derivative Help! I can't seem to prove that $$\Bigl[\log(x+\sqrt{x^2 + 1})\Bigr]' = \frac{1}{\sqrt{x^2 + 1}}$$ I keep getting some horrible answer namely $$\frac{x + \sqrt{x^2 + 1}}{x\sqrt{x^2 + 1} + 1 + x^2}$$ does this cancel down at all?
|
The other answers are quite what's necessary, but note that working tidely you get
$$\frac{d}{{dx}}\log \left( {x + \sqrt {1 + {x^2}} } \right) = \frac{{\frac{d}{{dx}}\left( {x + \sqrt {1 + {x^2}} } \right)}}{{x + \sqrt {1 + {x^2}} }}=$$
$$\frac{{1 + \frac{x}{{\sqrt {1 + {x^2}} }}}}{{x + \sqrt {1 + {x^2}} }}$$
Now, before doing anything else, take ${\sqrt {1 + {x^2}} }$ as a factor in the denominator. You get
$$\frac{{1 + \frac{x}{{\sqrt {1 + {x^2}} }}}}{{x + \sqrt {1 + {x^2}} }} = \frac{1}{{\sqrt {1 + {x^2}} }}\frac{{1 + \frac{x}{{\sqrt {1 + {x^2}} }}}}{{1 + \frac{x}{{\sqrt {1 + {x^2}} }}}} = \frac{1}{{\sqrt {1 + {x^2}} }}$$
Sometimes it is useful not to "rush it" when solving problems or evaluating expressions.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/135920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Summation by parts of $\sum_{k=1} ^n \frac{2k+1}{k(k+1)}$ Let $\{f_k\}$ and $\{g_k\}$ be sequences of real numbers. The formula for summation by parts is given by:
$\sum_{k=m} ^n f_k \Delta g_k=(f_{n+1}g_{n+1}-f_mg_m)-\sum_{k=m} ^n g_{k+1}\Delta f_k$,
where $\Delta f_k=f_{k+1}-f_k$.
Letting $f_k=2k+1$ and $g_k=-\frac{1}{k}$. One then computes $\Delta f_k=2$ and $\Delta g_k=\frac{1}{k(k+1)}$. Therefore, using the partial summation formula, we have
\begin{align*}
\sum_{k=1} ^n f_k \Delta g_k=\sum_{k=1} ^n \frac{2k+1}{k(k+1)}&=-\frac{2n+3}{n+1}+3+\sum_{k=1} ^n \frac{2}{k+1} \\
&=\frac{n}{n+1}+2(H_n-1)\\
&=2H_n-\frac{n+2}{n+1},
\end{align*}
where $H_n$ denotes the $n^{th}$ harmonic number.
I have check my answer multiple times, but I am convinced it is incorrect. Could anyone point out a flaw in my reasoning?
Here is the full solution:
Let $f_k=2k+1$ and $g_k=-\frac{1}{k}$. One then computes $\Delta f_k=2$ and $\Delta g_k=\frac{1}{k(k+1)}$. Therefore,
\begin{align*}
\sum_{k=1} ^n f_k \Delta g_k=\sum_{k=1} ^n \frac{2k+1}{k(k+1)}&=-\frac{2n+3}{n+1}+3+\sum_{k=1} ^n \frac{2}{k+1} \\
&=\frac{n}{n+1}+2(H_{n+1}-1)\\
&=2H_n+\frac{n}{n+1}-2\frac{n}{n+1}\\
&=2H_n-\frac{n}{n+1}.
\end{align*}
|
It does look to be your final conversion to harmonic numbers is at fault. In particular,
$$\begin{align*}
\sum_{k=1}^n \frac1{k+1}&=\sum_{k-1=1}^n \frac1{k-1+1}\\
&=\sum_{k=2}^{n+1} \frac1{k}\\
&=H_{n+1}-1\\
&=H_n+\frac1{n+1}-1\\
&=H_n-\frac{n}{n+1}
\end{align*}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/138454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Absolute max for $f(x,y,z)=x^ay^bz^c$, with constraint $g(x,y,z)=x+y+z-1$ I need to show absolute max for $f(x,y,z)=x^ay^bz^c$, with constraint $g(x,y,z)=x+y+z-1$ is $$\frac{a^ab^bc^c}{(a+b+c)^{a+b+c}}$$
So I I do have $$ax^{a-1}y^bz^c = bx^ay^{b-1}z^c = cx^ay^bz^{c-1} =\lambda$$
then I went on to equating each of the 2 equations giving
$$y=\frac{b}{a}x, z=\frac{c}{a}x$$
So I have $x+\frac{b}{a}x+\frac{c}{a}x=1$ but I am not so sure how to continue
The answer instead had
$$\lambda x = ax^ay^bz^c, \lambda y = bx^ay^bz^c, \lambda z = cx^ay^bz^c$$, then making observation that $x:y:z=a:b:c$, which means
$$x=\frac{a}{a+b+c}, y=\frac{b}{a+b+c}, z=\frac{c}{a+b+c}$$
I don't quite get this ... how do I derive this?
The rest ...
$$(\frac{a}{a+b+c})^a(\frac{b}{a+b+c})^b(\frac{c}{a+b+c})^c=1$$
...
|
When you have
$$ax^{a-1}y^bz^c = bx^ay^{b-1}z^c = cx^ay^bz^{c-1} =\lambda$$
$ax^{a-1}y^bz^c =\lambda \hspace{4pt}$ will give you $\hspace{4pt} \lambda x = a x^a y^b z^c$
Similarly
$bx^ay^{b-1}z^c =\lambda \hspace{4pt} \Rightarrow \hspace{4pt} \lambda y = b x^a y^b z^c$
$cx^ay^bz^{c-1} =\lambda \hspace{4pt} \Rightarrow \hspace{4pt} \lambda z = c x^a y^b z^c$
$$ \frac{\lambda x}{\lambda y} = \frac{x}{y} = \frac{a x^a y^b z^c}{b x^a y^b z^c} = \frac{a}{b}$$
$$ \frac{\lambda y}{\lambda z} = \frac{y}{z} = \frac{b x^a y^b z^c}{cx^a y^b z^c} = \frac{b}{c}$$
$$ x : y : z = a : b : c$$
What happens to $$\lambda x + \lambda y + \lambda z \hspace{5pt} ?$$
Is it
$$ x^a y^b z^c (a+b+c)$$
Now can you relate to the answer you were supposed to get?
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/139251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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|
Evaluating $\int \frac{x+8}{\sqrt {x+12}}dx$ How should I go about integrating the function $$\frac{x+8}{\sqrt {x+12}}$$
I have tried substituting $u = \sqrt{ x + 12 }$, but that leads me nowhere...
Could somebody possibly just tell me which steps have to be followed in order to evaluate this integral?
|
Let $u = \sqrt{x+12}$ and $du = \frac{1}{2\sqrt{x+12}} dx$
$$
\begin{align*}
\int \frac{x+8}{\sqrt{x+12}} \hspace{3pt}dx &= \int {\hspace{3pt}\frac{u^2-12+8}{u}}\cdot2u\hspace{3pt} du\\
&= \int {\hspace{3pt}2u^2-8}\hspace{3pt} du\\
&= 2\int {\hspace{3pt}u^2-4}\hspace{3pt} du\\
&= 2\cdot \frac{u^3}{3} -8u {\hspace{3pt}}\\
&= \frac{2}{3}{(x+2)^{3/2}} -8\sqrt{x+12} {\hspace{3pt}}\\
&= \frac{2}{3}x\sqrt{x+12} +C\\
\end{align*}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/139598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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|
How to prove that $\int_0^1\left(\sum\limits_{k=n}^\infty {x^k\over k}\right)^2\,dx = \int_0^1 2x^{n-1}\log\left(1+{1\over\sqrt{x}}\right)\,dx$ American Mathematical Monthly problem 11611 essentially asks you to show that
$$\lim_n\ n \int_0^1\left(\sum_{k=n}^\infty {x^k\over k}\right)^2\,dx=2\log(2).\tag1$$
This would follow easily from (2) below, which is true for small values of $n$
according to Maple.
But I couldn't prove equation (2) in general, so I found a direct solution for (1) instead.
$$\int_0^1\left(\sum_{k=n}^\infty {x^k\over k}\right)^2\,dx = \int_0^1 2x^{n-1}\log\left(1+{1\over\sqrt{x}}\right)\,dx.\tag2$$
But I'm still curious about (2). What am I missing? How can equation (2) be proven?
|
Here's a way to do it by brute force. First write
$$
\begin{align}
\int_0^1\left(\sum_{k = n}^\infty \frac{x^k}{k}\right)^2\,dx & = \sum_{k,m\geq n}\frac{1}{km}\int_0^1x^{k+m}\,dx \\
& = \sum_{k,m \geq n} \frac{1}{km(k+m+1)}.
\end{align}
$$
Put $r = k+m$, so that $m = r-k$, and transform the sum:
$$
\begin{align}
\sum_{k,m\geq n} \frac{1}{km(k+m+1)} &= \sum_{r = 2n}^\infty \frac{1}{r+1}\sum_{k = n}^{r-n}\frac{1}{k(r-k)} \\
&= \sum_{r = 2n}^\infty \frac{1}{r(r+1)}\sum_{k=n}^{r-n}\left(\frac{1}{k} + \frac{1}{r-k}\right) \\
& = \sum_{r = 2n}^\infty \frac{2}{r(r+1)}\sum_{k = n}^{r-n}\frac{1}{k} \\
& = 2\sum_{r = 2n}^\infty\left(\frac{1}{r}\sum_{k = n}^{r-n}\frac{1}{k} - \frac{1}{r+1}\sum_{k = n}^{r-n}\frac{1}{k}\right) \\
& = 2\sum_{r = 2n}^\infty\left(\frac{1}{r}\sum_{k = n}^{r - n}\frac{1}{k} - \frac{1}{r+1}\sum_{k = n}^{r+1 - n} \frac{1}{k}\right) + 2 \sum_{r = 2n}^\infty \frac{1}{(r+1)(r+1 - n)}. \\
\end{align}
$$
The first sum in the last line telescopes, so it can be evaluated as
$$
2\sum_{r = 2n}^\infty \left(\frac{1}{r}\sum_{k = n}^{r - n}\frac{1}{k} - \frac{1}{r+1}\sum_{k = n}^{r+1 - n}\frac{1}{k}\right) = \lim_{r \to \infty} \left(\frac{1}{n^2} - \frac{2}{r+1}\sum_{k = n}^{r+1 - n}\frac{1}{k}\right) = \frac{1}{n^2}.
$$
Thus we need to prove that
$$
\frac{1}{n^2} + 2\sum_{r = 2n}^\infty \frac{1}{(r+1)(r+1 - n)} = 2\int_0^1 x^{n-1}\log(1+x^{-1/2})\,dx. $$
Since
$$
\begin{align}
2\int_0^1 x^{n-1}\log(1+x^{-1/2})\,dx &= 2\int_0^1 x^{n-1}\log(1 + x^{1/2})\,dx - \int_0^1 x^{n-1}\log{x}\,dx \\
& = 2\int_0^1 x^{n-1}\log(1+x^{1/2})\,dx + \frac{1}{n^2},
\end{align}
$$
we need only prove that
$$
\sum_{r = 2n}^\infty \frac{1}{(r+1)(r+1 - n)} = \int_0^1 x^{n-1}\log{(1+x^{1/2})}\,dx.
$$
This can be done by developing $\log{(1+x^{1/2})}$ in powers of $x^{1/2}$:
$$
\begin{align}
\int_0^1 x^{n-1}\log{(1+x^{1/2})}\,dx & = \sum_{m = 1}^\infty\frac{(-1)^{m+1}}{m} \int_0^1 x^{n- 1 + m/2}\,dx \\
& = 2\sum_{m = 1}^\infty \frac{(-1)^{m+1}}{m(2n+m)} \\
& = 2 \sum_{\text{$m$ odd}} - 2\sum_{\text{$m$ even}} \frac{1}{m(2n+m)} \\
& = 2 \sum - 4\sum_{\text{$m$ even}} \frac{1}{m(2n+m)} \\
& = 2 \sum_{m = 1}^\infty \frac{1}{m(2n+m)} - \sum_{m = 1}^\infty \frac{1}{m(n+m)} \\
& = 2 \sum_{m = 1}^\infty \frac{1}{m}\left(\frac{1}{2n+m} - \frac{1}{2n+2m}\right) \\
& = \sum_{m = 1}^\infty \frac{1}{(2n+m)(n+m)} \\
& = \sum_{r = 2n}^\infty \frac{1}{(r + 1)( r +1 - n)},
\end{align}
$$
which proves the identity. The other methods are obviously more concise, but after working through this I couldn't resist posting it.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "30",
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|
How Can One Prove $\cos(\pi/7) + \cos(3 \pi/7) + \cos(5 \pi/7) = 1/2$ Reference: http://xkcd.com/1047/
We tried various different trigonometric identities. Still no luck.
Geometric interpretation would be also welcome.
EDIT: Very good answers, I'm clearly impressed. I followed all the answers and they work! I can only accept one answer, the others got my upvote.
|
Hint: start with $e^{i\frac{\pi}{7}} = \cos(\pi/7) + i\sin(\pi/7)$ and the fact that the lhs is a 7th root of -1.
Let $u = e^{i\frac{\pi}{7}}$, then we want to find $\Re(u + u^3 + u^5)$.
Then we have $u^7 = -1$ so $u^6 - u^5 + u^4 - u^3 + u^2 -u + 1 = 0$.
Re-arranging this we get: $u^6 + u^4 + u^2 + 1 = u^5 + u^3 + u$.
If $a = u + u^3 + u^5$ then this becomes $u a + 1 = a$, and rearranging this gives $a(1 - u) = 1$, or $a = \dfrac{1}{1 - u}$.
So all we have to do is find $\Re\left(\dfrac{1}{1 - u}\right)$.
$\dfrac{1}{1 - u} = \dfrac{1}{1 - \cos(\pi/7) - i \sin(\pi/7)} = \dfrac{1 - \cos(\pi/7) + i \sin(\pi/7)}{2 - 2 \cos(\pi/7)}$
so
$\Re\left(\dfrac{1}{1 - u}\right) = \dfrac{1 - \cos(\pi/7)}{2 - 2\cos(\pi/7)} = \dfrac{1}{2} $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/140388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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|
Finding where the slope of tangent line is = 1 I am suppose to find the points on the curve $x^2+2y^2 = 1$ where the tangent line has a slope of 1.
I am not sure how to do this but I guessed that if I set it is
$y = \sqrt{\frac{1-x^2}{2}}$ then I can find that derivative and work from there.
I find the derivative as $y = (1-x^2)(\frac{1-x^2}{2})^\frac{3}{2}$
I am not sure how to work from here but just looking at the function it looks like x=1 would make it zero. I have absolutely no idea how to factor that, it just seems like a mess.
|
Don't solve for $y$ first; use implicit differentiation!
If $x^2+2y^2=1$, then
$$\begin{align*}
\frac{d}{dx}(x^2+2y^2) &= \frac{d}{dx} 1\\
2x + 4y\frac{dy}{dx} &= 0\\
4y\frac{dy}{dx} &= -2x\\
\frac{dy}{dx} &= -\frac{2x}{4y}\\
\frac{dy}{dx} &= -\frac{x}{2y}.
\end{align*}$$
If the tangent has slope $1$, then $\frac{dy}{dx}=1$, so
$$\begin{align*}
1 &= -\frac{x}{2y}\\
-2y &= x.
\end{align*}$$
Plugging into the original equation and solving for $y$, we get:
$$\begin{align*}
x^2 + 2y^2 &= 1\\
(-2y)^2 + 2y^2 &= 1\\
4y^2 + 2y^2 &= 1\\
6y^2 &= 1\\
y^2 &=\frac{1}{6}\\
|y| &=\sqrt{\frac{1}{6}}
\end{align*}$$
giving two values of $y$, and two corresponding values of $x$.
|
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"url": "https://math.stackexchange.com/questions/140985",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
A trigonometric identity: $(\sin x)^{-2}+(\cos x)^{-2}=(\tan x+\cot x)^2$ I've been trying to prove it for a while, but can't seem to get anywhere.
$$\frac{1}{\sin^2\theta} + \frac{1}{\cos^2\theta} = (\tan \theta + \cot \theta)^2$$
Could someone please provide a valid proof?
I am not allowed to work on both sides of the equation.
Work so far:
RS:
$$
\begin{align}
& \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} + 2 \\[10pt]
& = \frac{\sin^4\theta}{(\cos^2\theta)(\sin^2\theta)} + \frac{\cos^4\theta}{(\sin^2\theta) (\cos^2\theta)} + \frac{(\sin^4\theta)(\cos^2\theta)}{(\sin^2\theta)(\cos^2\theta)} + \frac{(\sin^2\theta)(\cos^4\theta)}{(\sin^2\theta)(\cos^2\theta)} \\[10pt]
& = \frac{\sin^4\theta + \cos^4\theta + (\sin^4\theta)(\cos^2\theta) + (\sin^2\theta)(\cos^4\theta)}{(\cos^2\theta)(\sin^2\theta)}
\end{align}
$$
I am completely lost after this.
|
Hint:
$$\tan(\theta) + \cot(\theta)= \frac{\sin(\theta)}{\cos(\theta)}+\frac{\cos(\theta)}{\sin(\theta)}=\frac{\sin^2(\theta)}{\sin(\theta)\cos(\theta)}+\frac{\cos^2(\theta)}{\sin(\theta)\cos(\theta)} $$
|
{
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"url": "https://math.stackexchange.com/questions/142252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Multiplication convention rules Student was asked to convert the following statement into multiplication format
$$7+7+7+7+7+7$$
She wrote the answer as $7\times 6=42$ and was marked wrong as the teacher expected $6\times 7=42$.
Is there any rule that can clarify the answer format?
The same with converting a multiplication sum into adding $6\times 3$
she wrote $3+3+3+3+3+3$ and once again was marked as wrong. Teacher expected $6+6+6=18$
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I suspect there is no commonly accepted convention on whether $2+2+2$ is $2\times 3$ or $3\times 2$. But note that $2\times 2\times 2$ is $2^3$, so I personally prefer to say that $2+2+2=2\times 3$.
|
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|
Polynomial inequality I found the following problem on a website and would be curious to find a solution.
Let $a_1\ge a_2\ge\cdots\ge a_n$ be real numbers such that for all integer $k>0$:
$$a_1^k+a_2^k+\cdots+a_n^k\ge 0$$
Let $p=\max\{|a_1|,\ldots,|a_n|\}$. Show that $p=a_1$, and that
$$(x-a_1)(x-a_2)\cdots(x-a_n)\le x^n-a_1^n$$
for all $x>a_1$.
The first part is rather simple since we can normalize by $p$ and consider the limit $k\to\infty$. However, I can't seem to get the second part right...
POSSIBLE SOLUTION FOR $n\ge 6$:
Without loss of generalization, assume that $a_1=1$. Since $x>|a_i|$ for all $i$, the following expansions converge:
$$\sum_i\ln\left(1-\frac{a_i}{x}\right)=-\sum_{i}\sum_{k=1}^\infty\frac{a_i^k}{kx^k}.$$
Since the sum over $i$ for a given $k$ is non-negative, and we have
$$\sum_i\ln\left(1-\frac{a_i}{x}\right)\le-\sum_{k=1}^\infty\sum_i\left(a_i^{2k}+a_i^{2k+1}\right)\frac{1}{(2k+1)x^{2k+1}}.$$
Notice that $a_i^{2k}+a_i^{2k+1}\ge0$ since $|a_i|\le1$. Then
$$\sum_i\left(a_i^{2k}+a_i^{2k+1}\right)\ge2.$$
We have
$$\sum_i\ln\left(1-\frac{a_i}{x}\right)\le-\sum_{k=1}^\infty\frac 2{(2k+1)x^{2k+1}}\le-\sum_{k=1}^\infty\frac{2}{(2k+2)x^{2k+2}}.$$
The last inequality then reduces to
$$\sum_i\ln\left(1-\frac{a_i}{x}\right)\le-\sum_{k=2}^\infty\frac{1}{kx^{2k}}=-\sum_{k=1}^\infty\frac{1}{kx^{2k}}+\frac{1}{x^2}.$$
Therefore,
$$\prod\left(x-a_i\right)\le x^n\left(1-\frac{1}{x^2}\right)\exp(1/x^2).$$
We can actually show that this is a sharper bound that the one required in the question, for $n\ge 6$. Consider the functions $f(y)=(1-y)\exp(y)$ and $g(y)=(1-y^3)$ on the interval $[0,1]$. We have $f(0)-g(0)=f(1)-g(1)=0$. Furthermore, $f'(y)-g'(y)=y(-\exp(y)+3y)$, $f'(0)-g'(0)<0$, and let $h(y)=-\exp(y)+3y$. Then, $h(0)<0$, $h(1)>0$, and $h'(y)=-\exp(y)+3>0$. All this implies that $f'(y)-g'(y)$ has a single root on the interval $(0,1)$, which in turn means that $f(y)-g(y)<0$ on the same interval. Therefore,
$$\prod\left(x-a_i\right)\le x^n\left(1-\frac{1}{x^6}\right)\le x^n-x^{n-6}\le x^n-1$$
for $n\ge 6$. As @bgins said earlier, it is easy to show that the above holds for $n=2$ and $3$. However, it seems very tedious to show this for $n=4$ and $5$.
POSSIBLE SOLUTION FOR $n=4$: Assume again $a_1=1$, and let $a_2=a$,$a_3=b$,$a_4=c$. The case where all of these are positive is easy to do by induction, so assume at least $c$ is negative. Then, $$(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ac)x-abc=\\x^3+\frac{a+b+c}{2}(a+b+c-2x)x-\frac{a^2+b^2+c^2}{2}x-abc.$$
Under the assumption that $c<0$, we have $-1\le a+b+c\le 2<2x$, and it follows that
$$\frac{a+b+c}{2}(a+b+c-2x)x\le \frac{-1}{2}(-1-2x)x=x^2+x/2.$$
Therefore,
$$(x-a)(x-b)(x-c)\le x^3+x^2+x+1$$
and this proves it.
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Part a) is indeed simple. Let me give a proof for part b). For $k=1$ our condition gives $a_1\ge -\sum_{k=2}^{n}a_k=S.$ Now we apply AM-GM to estimate $$\prod_{k=2}^{n}(x-a_k)\le \left(\frac{(n-1)x-S}{n-1}\right)^{n-1}\le \left(x+\frac{a_1}{n-1}\right)^{n-1}.$$
So we are left to show that $$(x-a_1)\left(x+\frac{a_1}{n-1}\right)^{n-1}\le x^n-a_1^n.$$
Binomial theorem now implies the result, since $$\left(x+\frac{a_1}{n-1}\right)^{n-1}=x^{n-1}+\sum_{k = 2}^{n-2}{\binom{n-1}{k}x^{n-1-k}\left(\frac{a_1}{n-1}\right)^k}+a_1^{n-1}\le \sum_{k=0}^{n-1}x^{n-1-k}a_1^k$$ and $$\binom{n-1}{k}\cdot\frac{1}{(n-1)^k}< 1.$$
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|
Logarithms - Get the second solution to a logarithm equation Here's the original equation:
$\ln(11x-10) + \Big(\ln(11x-10)\Big)^2$ = 6
I've managed to obtain one solution: $x = \frac{e^2 + 10}{11}$
through those steps:
*
*$\ln\Big((11x-10)(11x-10)^2\Big) = 6$
*$\ln\Big((11x-10)^3\Big) = 6$
*$(11x-10)^3 = e^6$
*$11x-10 = e^2$
*$11x = e^2+10$
*$x = \dfrac{e^2+10}{11}$
The textbook shows a second solution: $x = \dfrac{10e^3 + 1}{11e^3}$, but how do you get to that result?
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Your solution is incorrect because :
$\ln(11x-10) +(\ln(11x-10))^2$ is not equal to $\ln(11x-10 \times {(11x-10)}^2)$
To solve
$$\ln(11x-10) + (\ln(11x-10))^2 = 6$$
Let $t=\ln(11x-10)$
$$t^2 +t -6=0$$
this gives , $t=2$ and $t=-3$
when $t=2 \longrightarrow \ln(11x-10)=2 \longrightarrow x=\frac{e^2+10}{11}$
when $t=-3 \longrightarrow \ln(11x-10)=-3 \longrightarrow x=\frac{10e^3+1}{11e^3}$
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Using contour integration, or other means, is there a way to find a general form for $\int_{0}^{\infty}\frac{\sin^{n}(x)}{x^{n}} \, dx$? While studying some CA, I ran across methods of evaluating $$\int_0^\infty \frac{\sin x}{x} \, dx, \;\ \int_0^\infty \frac{\sin^2 x}{x^2} \, dx, \;\ \text{and} \ \int_0^\infty \frac{\sin^3 x}{x^{3}} \, dx.$$
Is there a way to find a closed form for $$\int_0^\infty \frac{\sin^n x}{x^n} \, dx, \ n \in \mathbb{N}_{>0} ?$$
Rather it be contour integration or some clever method using real analysis.
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$$
\int_0^\infty \frac{\sin^n(x)}{x^n} \mathrm{d} x = \frac{\pi}{2^{n+1} \cdot (n-1)!} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} (2k-n)^{n-1} \operatorname{sign}(2k-n)
$$
where $\operatorname{sign}(x) = \cases{ 1 & $x > 0$ \\ 0 & $x = 0$\\ -1 & $x < 0$}$.
As to the (probabilistic) proof, notice that $\frac{\sin(t)}{t}$ is the characteristic function of a uniform random variable on $(-1,1)$. The sum of $n$ independent identically distributed such uniform random variables is known as Irwin-Hall random variable $Y_n$, and the integral in question is a multiple of its PDF evaluated at $x=0$:
$$
\phi_{Y_n}(x) = \frac{1}{2 \pi} \int_{-\infty}^\infty \frac{\sin^n(t)}{t^n} \mathrm{e}^{-i t x} \mathrm{d} t = \frac{1}{\pi} \int_{0}^\infty \frac{\sin^n(t)}{t^n} \cos(t x) \mathrm{d} t
$$
The closed form for the PDF is given on the wikipedia with the reference.
As to more explicit derivation. We first integrate by parts, $n-1$ times, then use binomial theorem for $\sin^n(x)$:
$$ \begin{eqnarray}
\int_0^\infty \frac{\sin^n(x)}{x^n} \mathrm{d} x &=& \int_0^\infty \frac{\mathrm{d}^{n-1}}{\mathrm{d} x^{n-1}}\left( \sin^n(x) \right) \frac{1}{(n-1)!}\frac{\mathrm{d} x}{x} \\
&=& \frac{1}{(n-1)!} \int_0^\infty \frac{1}{2^n i^n} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} \frac{\mathrm{d}^{n-1}}{\mathrm{d} x^{n-1}}\left( \mathrm{e}^{i (2k-n)x} \right) \frac{\mathrm{d} x}{x} \\
&=& \frac{1}{(n-1)!} \int_0^\infty \frac{1}{2^n i^n} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} \left(i (2k-n)\right)^{n-1} \mathrm{e}^{i (2k-n)x} \frac{\mathrm{d} x}{x} \\
&=& \frac{1}{(n-1)!} \int_0^\infty \frac{1}{2^n} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} \left((2k-n)\right)^{n-1} \sin((2k-n)x) \frac{\mathrm{d} x}{x}
\end{eqnarray}
$$
In the last line, $\mathrm{e}^{i (2k-n) x}$ was expanded use Euler's formula, and since the sum is real, only real summands are retained. Then, integrating term-wise nails it:
$$
\begin{eqnarray}
\int_0^\infty \frac{\sin^n(x)}{x^n} \mathrm{d} x &=& \frac{1}{2^n} \frac{1}{(n-1)!} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} \left((2k-n)\right)^{n-1} \int_0^\infty \sin((2k-n)x) \frac{\mathrm{d} x}{x} \\
&=& \frac{1}{2^n} \frac{1}{(n-1)!} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} \left((2k-n)\right)^{n-1} \frac{\pi}{2} \operatorname{sign}(2k-n)
\end{eqnarray}
$$
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How to show that $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $ using induction? I am attempting a question, where I have to show $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $.
I have managed to solve the base case, which gives 9, which is a multiple of 3.
From here on,
I have $(n+1)((n+1)^2 + 8)$
$n^3 + 3n^2 + 11n + 9$
$n(n^2 + 8) + 3n^2 + 3n + 9$
How can I show that $3n^2 + 3n + 9$ is a multiple of 3?
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Without induction:
Since $8=-1\pmod{3}$, $n(n^2+8)=n(n^2-1)=(n-1)n(n+1)\pmod{3}$. Since $n-1$, $n$ and $n+1$ are three consecutive integers, at least (and in fact, exactly) one of them is a multiple of $3$, hence their product is a multiple of $3$.
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|
Evaluating $\int \sqrt{\frac{x^a}{(x+b)^c}} dx$ for $a,b,c\in\mathbb N$ and $a \lt c$ For $a,b,c\in\mathbb N$ are they any procedures how to calculate the following indefinite integral?
$$\int \sqrt{\frac{x^a}{(x+b)^c}} dx$$
it is also possible to assume that $a \lt c$ but it would be great if it works more general.
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It seems in any particular case, Maple knows how to do this.
$$
\int \frac{x^{\frac{3}{2}}}{(x + 5)^{\frac{5}{2}}} d x = \frac{-2\sqrt{5} x^{\frac{5}{2}}
\Biggl(4 x^{\frac{7}{2}} + 15 x^{\frac{5}{2}} - 15 \sqrt{5} \operatorname{arcsinh} \biggl(\frac{\sqrt{5} \sqrt{x}}{5}\biggr) x^{2} \biggl(\frac{x}{5} + 1\biggr)^{\frac{3}{2}}\Biggr)}{
75 x^{\frac{9}{2}} \biggl(\frac{x}{5} + 1\biggr)^{\frac{3}{2}}}
$$
So I assume you can get some reduction formulas through integration by parts, then do a few beginning cases, like $a=1,2$, $c=1,2$
added substitution $y=\mathrm{arcsinh}\sqrt{x/b}$ leads to
$$
\int \frac{x^{\bigl(\frac{a}{2}\bigr)}}{(x + b)^{\bigl(\frac{c}{2}\bigr)}} d x = 2 b^{\Bigl(1 + \frac{a}{2} - \frac{c}{2}\Bigr)} \int {\operatorname{sinh} (y)^{(1 + a)}}{\operatorname{cosh} (y)^{(1-c)}} d y
$$
and the method for this trig-type integral is known.
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Is $n = k \cdot p^2 + 1$ necessarily prime if $2^k \not\equiv 1 \pmod{n}$ and $2^{n-1} \equiv 1 \pmod{n}$? $p$ is an odd prime and $k$ is a positive integer.
Let $n=k \cdot p^2+1$.
If $2^k \not\equiv 1 \pmod n$ and $2^{n-1} \equiv 1 \pmod n$, is $n$ prime? If it is, why?
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I'm able to find a proof of the $k=2$ case, which is the first non-trivial case.
Result: Let $p$ be an odd prime and $n=2p^2+1$ such that $2^{n-1} \equiv 1 \pmod n$. Then $n$ is prime.
This proof splits into four steps:
*
*We will prove that if $p$ divides $\varphi(n)$, then $n$ is prime.
*We will prove that if $2^{2p} \equiv 1 \pmod n$, then $p$ divides $\varphi(n)$.
*We will prove that if $2^{p^2} \equiv 1 \pmod n$, then $p$ divides $\varphi(n)$.
*The Lucas primality test implies that if $2^{2p} \not\equiv 1 \pmod n$ and $2^{p^2} \not\equiv 1 \pmod n$, then $n$ is prime.
Lemma: Let $p$ be a prime and $n=2p^2+1$. If $p$ divides $\varphi(n)$, then $n$ is prime.
Proof: Since $\varphi(n)$ is multiplicative and $p$ is prime, $p$ divides $\varphi(q^t)$ for some prime power divisor $q^t$ of $n$. Since $\varphi(q^t)=q^{t-1}(q-1)$, and $q$ divides $n$ while $p$ does not divide $n$, we must have that $p$ divides $q-1$.
Since $q$ divides $n$, we know $n=bq$ for some $b \geq 1$. Since $p$ divides $q-1$, we know $q=cp+1$ for some $c \geq 1$.
Hence $$n=b(cp+1)=bcp+b$$ and from before $$n=2p^2+1.$$ By taking these equations modulo $p$, we must have that $b \equiv 1 \pmod p$.
Case I: $b \geq 2p+1$. Then $n=bcp+b>2p^2+1$, giving a contradiction.
Case II: $b=p+1$. Then $n=cp(p+1)+p+1=cp^2+2cp+1$. Since $n$ is odd, we must have that $c$ is even, and thus $n>2p^2+1$, giving a contradiction.
Thus $b=1$, and hence $c=2p$, and hence $q=2p^2+1=n$. Hence $n$ is prime (since $q$ is a prime divisor of $n$). End proof.
*
*Now assume $2^{2p} \equiv 1 \pmod n$. Then $4^p \equiv 1 \pmod n$. Since $\gcd(4,n)=1$, we know $4$ is a member of the group $(\mathbb{Z}_n)^{\times}$. Hence the order of $4$, which we will denote $o_4$, divides $|(\mathbb{Z}_n)^{\times}|=\varphi(n)$. However, since $4^p \equiv 1 \pmod n$, we know that $o_4$ divides $p$. But since $n \geq 19$, we know that $o_4>1$, so $o_4=p$. Hence $p$ divides $\varphi(n)$.
*Now assume $2^{p^2} \equiv 1 \pmod n$. Since $\gcd(2,n)=1$, we know $2 \in (\mathbb{Z}_n)^{\times}$. Hence the order $o_2$ of $2$ divides $|(\mathbb{Z}_n)^{\times}|=\varphi(n)$. However, since $2^{p^2} \equiv 1 \pmod n$, we know that $o_2$ divides $p^2$. But since $n \geq 19$, we know that $o_2>1$, so $p$ divides $o_2$ (more specifically, we know $o_2 \in \{p,p^2\}$). Hence $p$ divides $\varphi(n)$.
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Triangle related question My question is:
In $\Delta ABC$, let $AE$ be the angle bisector of $\angle A$. If $\frac{1}{AE} = \frac{1}{AC} + \frac{1}{AB}$ then prove that $\angle A = 120^\circ$.
What I tried: I extended side $AB$ and took a point $M$ on it such that $AC$ is congruent to $AM$. Then I proved that $AE$ is parallel to $MC$. I was trying to prove that $\triangle AMC$ is equilateral so that I get $\angle MAC=60^\circ$. But I am not able to prove it.
Any help to solve this question would be greatly appreciated!
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Mark point $D$ on side $AC$ such that $AE=AD$.
Rewrite the given relation as follows:
$$\frac{1}{AE}-\frac{1}{AC}=\frac{1}{AB}$$
$$\frac{AC-AE}{AE\cdot AC}=\frac{1}{AB}$$
$$\frac{AC-AD}{AD\cdot AC}=\frac{1}{AB}$$
$$\frac{DC}{AD}=\frac{AC}{AB}$$
Now by the angle bisector theorem:
$$\frac{BE}{EC}=\frac{AB}{AC}$$
Combining we obtain:
$$\frac{DC}{AD}=\frac{EC}{BE}$$
Therefore,
$$\frac{AC}{DC}=\frac{AD+DC}{DC}=1+\frac{AD}{DC}=1+\frac{BE}{EC}=\frac{BE+EC}{BE}=\frac{BC}{EC}$$
Hence $\Delta ABC$ is isometric to $\Delta CDE$. Hence $\angle CDE=\angle CAB$.
Now $AE=AD$ hence in $\Delta AED$ we have $\angle AED = \angle ADE=\beta$. $\angle DAE = \frac{1}{2}\angle CAB = \frac{1}{2} \angle CDE = \alpha$. We obtain:
$$\angle DAE + \angle ADE + \angle AED = \alpha + 2\beta = \pi$$
$$\angle CDE +\angle ADE = 2\alpha + \beta = \pi$$
Solving we find $\alpha = \beta$, hence $\angle DAE = \angle ADE = \angle AED =\frac{\pi}{3}$, hence $\angle CAB = \frac{2\pi}{3}$.
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How to find the closed form formula for this recurrence relation $ x_{0} = 5 $
$ x_{n} = 2x_{n-1} + 9(5^{n-1})$
I have computed: $x_{0} = 5, x_{1} = 19, x_{2} = 83, x_{3} = 391, x_{4} = 1907$, but cannot see any pattern for the general $n^{th}$ term.
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Note that Ayman’s technique of ‘unwinding’ the recurrence works even without the preliminary division by $2^n$:
$$\begin{align*}
x_n&=2x_{n-1}+9\cdot5^{n-1}\\
&=2\left(2x_{n-2}+9\cdot5^{n-2}\right)+9\cdot5^{n-1}\\
&=2^2x_{n-2}+2\cdot9\cdot5^{n-2}+9\cdot5^{n-1}\\
&=2^2\left(2x_{n-3}+9\cdot5^{n-3}\right)+2\cdot9\cdot5^{n-2}+9\cdot5^{n-1}\\
&=2^3x_{n-3}+2^2\cdot9\cdot5^{n-3}+2\cdot9\cdot5^{n-2}+9\cdot5^{n-1}\\
&\qquad\qquad\qquad\vdots\\
&=2^kx_{n-k}+2^{k-1}\cdot9\cdot5^{n-k}+2^{k-2}\cdot9\cdot5^{n-k+1}+\ldots+9\cdot5^{n-1}\\
&=2^kx_{n-k}+9\sum_{i=0}^{k-1}2^i5^{n-1-i}\\
&\qquad\qquad\qquad\vdots\\
&=2^nx_0+9\sum_{i=0}^{n-1}2^i5^{n-1-i}\\
&=5\cdot2^n+9\sum_{i=0}^{n-1}\left(\frac25\right)^i5^{n-1}\\
&=5\cdot2^n+9\cdot5^{n-1}\left(\frac{1-(2/5)^n}{1-2/5}\right)\\
&=5\cdot2^n+3\cdot5^n\left(1-\frac{2^n}{5^n}\right)\\
&=5\cdot2^n+3\cdot5^n-3\cdot2^n\\
&=2^{n+1}+3\cdot5^n\;.
\end{align*}$$
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Prove that the product of four consecutive positive integers plus one is a perfect square I need to prove the following, but I am not able to do it. This is not homework, nor something related to research, but rather something that came up in preparation for an exam.
If $n = 1 + m$, where $m$ is the product of four consecutive positive
integers, prove that $n$ is a perfect square.
Now since $m = p(p+1)(p+2)(p+3)$;
$p = 0, n = 1$ - Perfect Square
$p = 1, n = 25$ - Perfect Square
$p = 2, n = 121$ - Perfect Square
Is there any way to prove the above without induction? My approach was to expand $m = p(p+1)(p+2)(p+3)$ into a 4th degree equation, and then try proving that $n = m + 1$ is a perfect square, but I wasn't able to do it. Any idea if it is possible?
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Recursion on the square root, k(n)
n = lead integer
k(n) = square root [n(n+1)(n+2)(n+3) + 1]
k(1)^2 = 25
k(2)^2 = 121
k(3)^2 = 361
k(4)^2 = 841
k(5)^2 = 1681
k(1) = 5
k(2) = 11 = 5+ 6 = 5 + 2*(2+1)
k(3) = 19 = 11+ 8 = 11 + 2*(3+1) = 5 + 2 * ((2+1) + (3+1))
k(4) = 29 = 19+10 = 19 + 2*(4+1) = 5 + 2 * ((2+1) + (3+1) + (4+1))
k(5) = 41 = 29+12 = 29 + 2*(5+1) = 5 + 2 * ((2+1) + (3+1) + (4+1) + (5+1))
k(n) = 5 + 2 * ( (n+1)(n+2)/2 - (1+2) )
= (n+1)(n+2) - 1
induction step
k(n+1) = k(n) + 2(n+2)
= (n+1)(n+2) - 1 + 2(n+2)
= (n+2)(n+3) - 1
k(n)^2 - 1 = ((n+1)(n+2))- 1)^2 - 1
= ((n+1)(n+2))^2 - 2(n+1)(n+2)
= ((n+1)(n+2)) * ((n+1)(n+2) - 2)
= ((n+1)(n+2)) * (n^2 + 3n)
= n(n+1)(n+2)(n+3)
!!!
Thinking a little more about k(n).
A rough approximation of the root of ( n(n+1)(n+2)(n+3) + 1 ) would be n^2.
We might improve it a bit by averaging the products of the outer and inner pair of factors,
k(n) = (n(n+3) + (n+1)(n+2)) / 2
= (n+1)(n+2) -1
Dead on. Surprising, no?
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Evaluating $\int \frac{x^3}{(x^2 + 1)^\frac{1}{3}}dx$ $$\int \frac{x^3}{(x^2 + 1)^\frac{1}{3}}dx$$
I am suppose to make a $u$ substitution and to make this a rational integral and then evaluate it from there but I have no idea how to do that. There aren't any good examples of this in the book and I can not find any us that make this doable.
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Let $u = x^2 + 1$, $du = 2x\,dx$:
\begin{align*}
\int \frac{x^3\,dx}{(x^2 + 1)^\frac{1}{3}} &= \int \frac{x^2\,x\,dx}{(x^2 + 1)^\frac{1}{3}} = \frac{1}{2} \int \frac{u-1}{u^\frac{1}{3}}\,du = \frac{1}{2} \int \left(u^\frac{2}{3} - u^{-\frac{1}{3}}\right) \,du \\
&= \frac{3}{10} u^{\frac{5}{3}} - \frac{3}{4} u^{\frac{2}{3}} + C \\
&= \frac{3}{5} \left(x^2+1\right)^{\frac{5}{3}} - \frac{3}{4} \left(x^2+1\right)^{\frac{2}{3}} + C
\end{align*}
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"url": "https://math.stackexchange.com/questions/156553",
"timestamp": "2023-03-29T00:00:00",
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|
Simplifying a radical? How would I simplify the following two radicals.
$$\sqrt{\frac{X^3}{50}}$$
For my answer I got $\frac{X^2}{50X}$ but I am not sure if this is correct.
My second question is how would I simplify $\sqrt{\frac{1}{12}}$
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For your first one,
$$\sqrt\frac{x^3}{50} = \frac{\sqrt{x^2}\sqrt{x}}{\sqrt{25}\cdot\sqrt{2}} = \frac{x\sqrt{x}}{5\sqrt{2}}$$
At this point, you could multiply top and bottom through by $\sqrt{2}$ (to make the $\sqrt{2}$ part of the bottom "go away"):
$$\frac{x\sqrt{x}\cdot\sqrt{2}}{5\sqrt{2}\cdot\sqrt{2}} = \frac{x\sqrt{2x}}{10}$$
For the second one,
$$\sqrt\frac{1}{12} = \frac{1}{\sqrt{12}} = \frac{1}{\sqrt{4}\cdot\sqrt{3}} = \frac{1}{2\sqrt{3}}$$
From here, you could opt to multiply top and bottom through by $2\sqrt{3}$ as shown below (which is called rationalizing the denominator - to get rid of square root terms in the denominator.)
$$\frac{1\cdot{2\sqrt{3}}}{2\sqrt{3}\cdot{2\sqrt{3}}} = \frac{2\sqrt{3}}{4\cdot\sqrt{9}} = \frac{\sqrt{3}}{2\cdot{3}} = \frac{\sqrt{3}}{6}$$
When dealing with problems like these, it is helpful to split what is under the square root into two separate square root terms (assuming $x$ is nonnegative.) For example,
$$\sqrt\frac{x^3}{50} = \frac{\sqrt{x^3}}{\sqrt{50}}$$
Then, you want to try to rationalize the denominator OR try to manipulate the denominator as I did above to make it more friendly (i.e. $\sqrt{25}\cdot{\sqrt{2}} = \sqrt{50}$, and since you know that $\sqrt{25} = 5$, $\sqrt{50}$ must equal $5\sqrt{2}$.
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"language": "en",
"url": "https://math.stackexchange.com/questions/157948",
"timestamp": "2023-03-29T00:00:00",
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|
$1 +1$ is $0$ ?
Possible Duplicate:
-1 is not 1, so where is the mistake?
$i^2$ why is it $-1$ when you can show it is $1$?
So:
$$
\begin{align}
1+1 &= 1 + \sqrt{1} \\
&= 1 + \sqrt{1 \times 1} \\
&= 1 + \sqrt{-1 \times -1} \\
&= 1 + \sqrt{-1} \times \sqrt{-1} \\
&= 1 + i \times i \\
&= 1 + (-1) \\
&= 1 - 1\\
&= 0
\end{align}
$$
I can't see anything wrong there, and I can't see anything wrong in $1+1=2$ too. Clearly, $1+1$ is $2$, but I really want to know where is the incorrect part in the above.
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$$\sqrt{ab} = \sqrt{a} \times \sqrt{b}$$ is valid only for non-negative real numbers $a$ and $b$. Hence, the error is in the step $$\sqrt{(-1) \times (-1)} = \sqrt{-1} \times \sqrt{-1}$$
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"language": "en",
"url": "https://math.stackexchange.com/questions/158409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Positive definite quadratic form $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ Is $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ positive definite?
Approach:
The matrix of this quadratic form can be derived to be the following
$$M := \begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{2} & \cdots & \frac{1}{2} \\
\frac{1}{2} & 1 & \frac{1}{2} & \cdots & \frac{1}{2} \\
\cdots & \cdots & \cdots & \cdots & \cdots \\
\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \cdots & \frac{1}{2} \\
\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \cdots & 1 \\
\end{pmatrix}$$
It suffices to show that $\operatorname{det}M > 0$, then the claim follows.
Any hints how to show the positivity of this determinant?
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A direct approach would note that your form is
$${1\over2}\left[\left(\sum_{i=1}^n x_i\right)^2+\sum_{i=1}^n x_i^2\right].$$
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"language": "en",
"url": "https://math.stackexchange.com/questions/159506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Integration of $\int\frac{1}{x^{4}+1}\mathrm dx$ I don't know how to integrate $\displaystyle \int\frac{1}{x^{4}+1}\mathrm dx$. Do I have to use trigonometric substitution?
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HINT :
Use
$$x^4+1=(x^2+1)^2-(\sqrt 2x)^2=(x^2+\sqrt 2x+1)(x^2-\sqrt 2x+1)$$
to get
$$\begin{align}\int\frac{dx}{1+x^4}&=\int\frac{\frac{1}{2\sqrt 2}x+\frac 12}{x^2+\sqrt 2x+1}dx+\int\frac{-\frac{1}{2\sqrt 2}x+\frac 12}{x^2-\sqrt 2x+1}dx\\&=\frac{1}{2\sqrt 2}\int\frac{x+\sqrt 2}{x^2+\sqrt 2x+1}dx-\frac{1}{2\sqrt 2}\int\frac{x-\sqrt 2}{x^2-\sqrt 2x+1}dx\\&=\frac{1}{\sqrt 2}\int\frac{x+\sqrt 2}{(\sqrt 2x+1)^2+1}dx-\frac{1}{\sqrt 2}\int\frac{x-\sqrt 2}{(\sqrt 2x-1)^2+1}dx.\end{align}$$
Here, set $\sqrt 2x+1=\tan\alpha$ for the first and set $\sqrt 2x-1=\tan\beta$ for the second.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/160157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 4
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.