How many terms t of e^ia are needed for cosine(a) to be accurate to n decimal places?
f(n,a):= # of terms t of e^ia needed for n decimal accuracy for cosine(a)
disregarding the initial 1...
How many terms t of e^ia are needed for cosine(a) to be accurate to n decimal places?
f(n,a):= # of terms t of e^ia needed for n decimal accuracy for cosine(a)
disregarding the initial 1...
2) Are there an infinite amount of (non trivial) alternating infinite series in the following form, with r being a variable ratio, and a being a set sequence of numbers (like 2!,4!,6!... or 1!,3!,5!... or 1,20,120, 1200... etc.)
.. that have an infinite amount of zeros as r is increased, similar to sine and cosine?
Do you understand complex numbers?
You are invoking Euler's Formula. e^ix represents a single complex frequency or sin/cos wave. They are real numbers as such infinite. In engineering we convert to phasors to make math easier.
Scroll down to trigonometry
https://en.wikipedia.org/wiki/Euler%27s_formula
https://en.wikipedia.org/wiki/Phasor
I used the Taylor Series to calculate trig functions in embedded computing. I dervived the number of terms needed by calculation.
To answer more simply the relationship between sin-cos and an a complex exponential is an identity. The transformation to complex exponentials is analogous to transforming to logarithms to perform multiplication and division.
a*b = log(a) + log(b)
Steve, are you fucking with me because I derail threads, or do you really think you answered any question that I asked?
I didn't want to write a formula myself. Damnit. Fuck you guys. I'll do it.
termspart:1-a^2/2, tp1:2,
while floor [(cosine(a) - termspart)*10^n] >0 do [
termspart:termspart+ (-1)^tp1*a^(2*tp1)/((2*tp1)!),
tp1:tp1+1,
disp ("fuck you guys")
],
disp (sconcat("fuck you guys: ", tp1-1))
easier method good enough for my purposes: calculate when 10^(n+1) *[a^(2(t+2))/(((t+2)2)!)] <5
Last edited by Kharakov; 01-12-2019 at 09:10 PM.
I paid three dollars for a six pack of Pepsi.
If I divide by putting $3 in the numerator and 6 in the denominator, I can interpret the result of the calculation as having paid on average fifty cents per can of Pepsi.
If I divide by putting 6 in the numerator and $3 in the denominator, I can calculate the resulting value, but I have no idea how to articulate an interpretation for my answer. Any idea?
Thank you