Recent content by saint_n

ok I am starting to get it,,but what's happens if its a double pole.Do you work out the residues for both functions and multiply the 2 residues together or does it depend eg Gamma[s]Zeta[1-s] so the pole would be at 0. Since the Gamma fn and the Zeta are multiplied together you multiply the...

Hey Is there a method in calculating the residues. Getting the poles is easy but i really don't know how my lecturer gets the residues eg. 1/2(Zeta[s]*Gamma[s/2] where at s = 1 it is Sqrt[Pi]/2 etc how does he do it?

for p = 1 |\frac{1}{2}(a+b)|\leq(\frac{1}{2}|(a+b)|)-ab isnt this false because you subtracting a ab on the RHS?

so you saying that |\frac{1}{2}(a+b)|^{p}\leq(\sqrt[p]{\frac{1}{2}}|(a+b)|)^{p}-pab \leq \frac{1}{2}(|a|^p+|b|^p)

Need help with a complex inequality?? hey! i been trying to do this inequality for a 2 hrs now and can't seem to prove it |\frac{1}{2}(a+b)|^p \leq \frac{1}{2}(|a|^p+|b|^p) where a,b are complex numbers Can anyone suggest a way?? thanks

ok,,thanks i also found the Fundamental Theorem of Algebra which I am reading up now. Thank you everyone! Cheers

I figured that we have to use complex numbers but we weren't told a method how to factorise complex polynomials with say n degree.Can you tell me a method of how to factorize a complex polynomial?? You mentioned that the quote above is the theorema egregium of Algebra which I am having...

the highest we ever gone up to is probably degree 5 that i know of..i'll try look up the theorem and see what you people are talking about

Do you know the name of it because of never heard of it.I wouldn't know where to start.If i have the name i can probably do some research on it and continue from there maybe

I don't see where you going.The roots you can get from the product..do want me to replace z with \cos\theta+i\sin\theta hmmm...with the LHS = (\cos\theta+i\sin\theta)^{n+1}-1=(\cos(n+1)\theta+i\sin(n+1)\theta)-1 and RHS =...

After learning a little latex maybe this looks better z^{n+1} - 1 =\Pi (z - e^\frac{2 \pi i(j+1)}{n+1}) where the product goes from 0<= j <= n

sorry!yes i was missing something in the original equation.So it suppose to look like this (z^(n+1))-1 = Product[(z-exp[2*Pi*i*(j+1)/(n+1)]) , 0<=j<=n] where Product[(z-exp[2*Pi*i*(j+1)/(n+1)]) , 0<=j<=n] = (z-exp[2*Pi*i*(0+1)/(n+1)]*(z-exp[2*Pi*i*(1+1)/(n+1)]*...*(z-exp[2*Pi*i*(n+1)/(n+1)]...

Hey!I have a tut question and I am having problems proving (z^(n+1))-1 = (z-exp[2*Pi*i*(j+1)/(n+1)]) where 0<=j<=n I tried doing it by induction which is easy for the 1st case with n = 1,I assume the case n-1 but then i get stuck with the last case. How will i know that all the z's with...

ok,,so what i have now is correct?Just so i can get started...

btw can you recommend a combinatorics book because he's notes help only a lil plus sometimes he doesn't explain himself properly(by the way we don't have a prescribed textbook) so anything we would be appreciated...