I'm trying to show that:
$\sum_{a=0}^b\text{GCD}(a,b)s(a,b)=0$
where $s$ is the Dedekind sum. Any ideas?
I have already managed to show the above without the GCD factor, i.e.
$\sum_{a=0}^bs(a,b)=0$
I'm quite stuck on this. How would you prove the following by induction?
$d/dx(x e^{-x}) = (-1)^n (e^{-x})(x-n)$
.
I did:
$d/dx(x e^{-x})=(e^{-x}) - x(e^{-x})$
Now I have no idea how to proceed and prove this by induction. I would really appreciate if anyone could guide me. A clear answer wo...
I'm quite stuck on this. How would you prove the following by induction?
$\frac{d}{dx}(x e^{-x}) = (-1)^n (e^{-x})(x-n)$
.
I did:
$\frac{d}{dx}(x e^{-x})=(e^{-x}) - x(e^{-x})$
Now I have no idea how to proceed and prove this by induction. I would really appreciate if anyone could guide me. A...
I'm quite stuck on this. How would you prove that the $n^{th}$ derivative of $x e^{-x}$ if the $(-1)^n (e^{-x})(x-n)$ by induction?
I did:
$\frac{d}{dx}(x e^{-x})=(e^{-x}) - x(e^{-x})$
Now I have no idea how to proceed and prove this by induction. I would really appreciate if anyone could guid...
For induction first of all give a value to the formula to see whether it gives good feedback.Second write the formula for the $i$ and prove it for the $i+1$
@robjohn A couple of approaches? I have no approach there, it's over my head. There are some other things that I might try, but I'm not sure if they are useful.
@Sawarnik Clearly divisible by 6. Divisible by another 3 and another 2, thus by 36. Dividing out by that (took me a calculator) we get 1666850183335016835. Clearly, it's divisible by only one 5. Thus, not a perfect square.
@DanielFischer Also I know that Gauss said, if it was about conjectures, I could make lots of conjectures to bamboozle all of the mathematicians, he believed that making unsolvable conjectures in math is not quite hard.
@BalarkaSen Not all elementary number theory is "kid math", I'd say. Although, if you mean that the problem can be understood by kids (even if they don't know as much about maths as you do), then the Collatz conjecture is a fitting example.
The roots of the problem essentially goes very deep in NT and probably to something called Ergodic number theory, which even pros can't understand sometimes =P
@BalarkaSen There is no "best" algorithm for all uses. And it depends on what form you have the number given in. And the size of the number, of course.
@BalarkaSen It's mathematical problems where you create a program to solve it. Example calculate the sum of all integers from 1-100 inclusive(which can be done using progression, and in programming it's a loop)
@Sabಠ_ಠ Well, if you multiply by the factor $(-1)^n$ we left out above, you have $$\left(\frac{d}{dx}\right)^{n+1}\left(xe^{-x}\right) = (-1)^{n+1}(x-(n+1))e^{-x}.$$
@DanielFischer For finding inflection points of a function these conditions should be fulfilled: 1)$f''(x)=0$ or undefined 2)the function should have $f'(x)$ defined for it at the point $x$ 4)$f"(x)$ should change sign in the $x$ , is there anything left?
@MrWho I would have to read up on the definition of an inflection point. Instead of me reading wikipedia and relaying it here, it may be better if you read it yourself, less risk of forgetting something.
@DanielFischer No I just want to know whether I'm right about the conditions and if I am, we've got a special type of inflection point which is vertical and so $f'(x)=+ OR -\infty$ and it doesn't fulfill the conditions, how is it then?
Can it be that both comes from Mittag-Leffler expansion theorem?
Sorry, I typoed on the second one. It should have been $$\wp(z) = \frac1{z^2} + \sum_{(m, n) \in \Bbb Z^2 \setminus \{(0, 0)\}} \frac{1}{(z - m - n\tau)^2} + \frac1{(m + n\tau)^2}$$
Given a positive (you may assume that it isn't descending at any point) arithmetic function f such that for every function g=o(n) it follows that g(n)=o(f(n)).
@Sabಠ_ಠ You should differentiate $(-1)^n(x-n)e^{-x}$. The first is a constant factor, so $$\frac{d}{dx}\left((-1)^n(x-n)e^{-x}\right) = (-1)^n \left[\left(\frac{d}{dx}(x-n)\right)e^{-x} + (x-n)\frac{d}{dx}e^{-x}\right].$$
@DanielFischer Anyway any idea how to map $D = \{ z : -B < \Im(z) < B \}$ into the right half plane? I know I have to change the "angle" / but I do not quite see how. Just a hint is fine (as usual..) =)
Ask your favourite programming language what it thinks a/b*c means. And $\cos \sin x + 3$ is a parse error even for humans. While it's common to not parenthesize products in arguments of trigonometric (and similar) functions, e.g. $\sin (n+\frac{1}{2})\pi$ does not mean $\pi\sin (n+\frac{1}{2})$ usually, there's not much of such a convention for sums. Does $\cos^2 x + \sin^2 x$ mean $\cos^2 \left(x+\sin^2x\right)$?
And I actually preffer to write trigonometric functions as $(\cos x)^2 + (\sin x)^2$ to avoid all confusing. But extensive use of parenthesis is bad from a typographical point, even though it might be slightly better pedagogicaly.
@N3buchadnezzar The ultimate pedagogical challenge: Explain why it is a good idea to use $\sin^2 x$ for $(\sin x)^2$ and $\cos^{-1} x$ for $\arccos x$ in the same expression.
I am working on a problem, which would possibly relate the Fourier transform/series with the jump singularities of the function where the function itself or one of its derivatives jump. ((some kind of logarthimic blow ups too, possibly as a corollary).
Consider a BV function $f(t)$ in $L^2(\math...
@DanielFischer But then you don't get an opportunity to test students on their memory of an arbitrarily chosen standard. (Assuming you're going the full nine yards and making it set-valued)
@KarlKronenfeld The topology induced by the Hausdorff pseudo-metric? Would be my best guess without further information. (And I find it a terrible idea to denote the space $C(X)$.)
@KarlKronenfeld No, doesn't sound plausible to me. On the space of compact subsets of a metric space, you have the Hausdorff-metric, and the topology induced by that is sometimes called the Hausdorff topology. It must be something analogous here, but I'm not sure how the topology would be defined without a metric on $X$.
I'm trying to show that:
$\sum_{a=0}^b\text{GCD}(a,b)s(a,b)=0$
where $s$ is the Dedekind sum. Any ideas?
I have already managed to show the above without the GCD factor, i.e.
$\sum_{a=0}^bs(a,b)=0$
I'm quite stuck on this. How would you prove the following by induction?
$d/dx(x e^{-x}) = (-1)^n (e^{-x})(x-n)$
.
I did:
$d/dx(x e^{-x})=(e^{-x}) - x(e^{-x})$
Now I have no idea how to proceed and prove this by induction. I would really appreciate if anyone could guide me. A clear answer wo...
