Recent content by rteng

What is the relative permittivity of gold at radio-frequencies or specifically around 15-20MHz?

Warren could you explain what was concluded here? I am in the same position as the guy who asked this question...

dammit ha k thanks

how would be the best way to do this? I mean, I know how to find the derivative... and it kind of makes a pattern...but I can't quite correlate that pattern to the 100th derivative

yes that part sorry

y=((x+sqrt(x))/x^{2}-2^{x})^{20} lny=20(ln(x+sqrt(x))-ln(x^{2}-2^{x})) y'/y=(20/(x+sqrt(x))*(1+1/2(x)^{-1/2}))-(20/x^{2}-2^{x})*(2x-derivative of 2^x)

yeah but then you have a term that is: [1/(x^2-2^x)]*(2x-d/dy2^x) thats where I get stuck

using logarithmic differentiation I can get it to a point where I have... ln(y)=20[ln(x+sqrt(x))-ln(x^2-2^x)] I do not know what to do at that point

yes that makes more sense thanks didn't consider that

Alright, I know how to find the horizontal asymptotes but the vertical asymptotes? I tried dividing the polynomials but maybe I am not doing it correctly because I cannot get a normal answer... and I am unsure how to factor the denominator.

the I get f=86Hz

I would say that the only way for which there be no destructive interference is if x is 0 but that revelation is really vague for me...could it be correct?

ok I do not understand c can anybody help?

ah yes...of course I think that should be the proper way to solve this problem...at least for a and b now on to c...

ok I found my path difference to be sqrt(4+x^2)-x so if I equate this to (n)(lambda)/2 then: sqrt(4+x^2)-x=(n)(lambda)/2 4+x^2-x^2=(n^2)(lambda^2)/4 there is my problem...