Recent content by Wizlem

I believe all meromorphic functions can be written as the ratio of two holomorphic functions. A holomorphic function can be written as a (possibly infinite) product of monomials which are of the form (x-a) where a is a zero of the function. So if f(x) is meromorphic we can write it as g(x)/h(x)...

it appears as though r does not depend on t if Walter Lewin at MIT differentiates it correctly. The chain rule is required because \frac{d((f(t))^{2})}{dt} = 2 f \frac{d(f(t))}{dt} Thus, your derivative of 1/2mr^2 is wrong because you forgot the factor of dr/dt but that isn't really relavent...

I can't comment to specifically on whether or not you should take the class but when I wanted to skip over the first semester of physics I just asked the department chair if I could and he said ok.

if a is nxn and x is nx1 try Q = sum(sum(a.*repmat(x,[1 n]).*repmat(x',[n 1])))

The summation formula for the zeta function is only valid for real part of s greater than 1. Where the non-trivial zeros lye is not in this region hence your problem. Showing that the second term diverges wouldn't be so helpful either since it would probably diverge for any a<1 and not just a =...

-2\pi\sqrt {2n} - 2\pi < -2\pi\sqrt {2n} + 2\pi and so -2\pi(\sqrt {2n} + 1) < -2\pi(\sqrt {2n} - 1) and you still know -2\pi(\sqrt {2n} + 1) < N - \pi n

There are grammatical rules to follow but proof writing is just like any other writing and doesn't necessarily follow any format rules. It just needs to consist of logical steps. Most just consist of "We have this. Therefore, that is true. Thus, something else is true." and so on.

If you write this as x*(3x-1)-1 and do the chain rule you get (3x-1)-1-x*(3x-1)-2*3 which if you put (3x-1)-2 in both denominators gives you (3x-1)/(3x-1)2-(3x)/(3x-1)2 and the 3x cancels out which gives what you got so you got the right answer.

I guess it is odd to end up with the right answer.

|a| is defined as a if a \geq 0 and -a if a<0

By what you wrote above did you mean \sum^{\infty}_{n=-\infty} e^{2 \pi i n s} This sum doesn't converge because lim_{n\rightarrow\infty}e^{2 \pi i n s}\neq0

Isn't it true that A^TA is only not invertible if there are actually multiple solutions to Ax=b?

f(x)=x^(1/x)>1 for all x>e i believe so you have to restrict yourself to x such that f(x)>1.

Are you restricting x and y to positive values? This problem would appear to become quite tricky otherwise. For instance -2 and -4 are solutions also. If you can assume x>0 consider the function f(z) = z^(1/z). It's not hard to show this function is continuous and increases to a point and...

The taylor series is unique for a given function so if you find a series which fits the taylor series form, then it is the taylor series of that function. The problem with x5*ln(x) is that ln(x) doesn't have a taylor series expansion about 0.