Recent content by thomas430

Perfect, thanks I like Serena. It's working very nicely! :-D

Oh, I see! So supposing x represents 2 parameters q and w, I should end up with a 1x2 Jacobian matrix like: [f1(a1,x)2/dq + f2(a1,x)2/dq + f3(a1,x)2/dq + f1(a2,x)2/dq + ... f3(an,x)2/dq | | f1(a1,x)2/dw + f2(a1,x)2/dw + f3(a1,x)2/dw + f1(a2,x)2/dw + ... f3(an,x)2/dw]

Awesome, now I can get coding and test it out. So should I use (f1)^2+(f2)^2+f(3)^2 as the function for the Jacobian matrix, or just f1+f2+f3? I think the latter because the derivatives in the Jacobian are how the optimisation works, right? Thanks so much :-D

Thanks, I like Serena! Your reply helped a lot, but I'm still trying to get my head around how to relate it back to my problem. My problem looks like this... three equations (I haven't put the actual ones because they're very long): f1(a,b) = 0 f2(a,b) = 0 f3(a,b) = 0 where a are...

Hi there, I have been testing out the Levenberg-Marquardt algorithm. I've successfully coded a method in MATLAB for the example I saw on wikipedia: f(x) = a*cos(bx)+b*sin(ax) and this has worked well. The application I'm using the algorithm for is a system of 3 equations, however. Does...

Hi there, I've just read the following: The expression that is given is: \int_{A \infty} e_j \times h_k* \cdot \widehat{z} dA = 0 where * denotes the complex conjugate, and z^ is the unit vector in the direction of propagation (along the axis of the fibre). Can anyone explain...

So... g = h^k, \:for\: (k,p-1)=1 substituting: g^s \equiv \left( g^k \right)^t \: mod \: p or g^s \equiv \left( g^t \right)^k \: mod \: p We know that if p|ab, then p|a or p|b.. so: g^s \equiv g^t \: mod \: p Is that correct? What would come next?

If k is the index of h, then h \equiv g^k \: mod \: p and: g^s \equiv (\left g^k )\right ^t \: mod \: p Is that the right idea?

Hi all, I've been staring at this question on and off for about a month: Suppose that p is an odd prime, and g and h are primitive roots modulo p. If a is an integer, then there are positive integers s and t such that a \equiv g^s \equiv h^t mod p. Show that s \equiv t mod 2. I feel as...

Oh! So d /should/ divide c, because d=(a,b)... but this contradicts d=mc, which suggests d cannot divide c, as d > c. (as m>1 because m=(x,y) and we said that (x,y) != 1) Thanks, HallsofIvy :-)

Hi everyone, I saw that for the linear diophantine equation d=ax+by, where d=(a,b), that x and y must be coprime. Why is this? I feel like there are properties of coprime numbers that I am not aware of, because there are a few things like this that I have encountered. Any help...

Thanks to all of you for your discussion, you've helped a great deal! Bingk and Petek, your proofs next to one another gave me great insight :-D

Hi all, so I was looking at Legendre symbols, and I saw that \left(\frac{2}{p}\right)=(-1)^{\frac{p^2-1}{8}}. How does one show that \frac{p^2-1}{8} is always an integer? That is, how can we show that 8 | p^2-1? Can a similar method be applied to show that 24 | p^3-p? Thanks :-)...

Thanks Robert, that makes good sense.

I see how that works.. and I expanded for n=3 to get a taste. Thanks, ramsey!