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...  Reply With Quote

2. 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?  Reply With Quote

3. Originally Posted by Kharakov 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...

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.  Reply With Quote

4. Originally Posted by steve_bank Do you understand complex numbers?
No. Ok, I might be being disingenuous. That image is generated with a form of dual (or bi) complex number, rotations, reflections, polygonal (or polyhedral?? I forget) transforms that I wrote a while back.  Reply With Quote

5. 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)  Reply With Quote

6. 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  Reply With Quote

7. 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?  Reply With Quote

8. Originally Posted by Kharakov 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.
Anyone can have a brain fart without warning. It smells bad and us=is soon forgotten.  Reply With Quote

9. Originally Posted by fast 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?
Dimensional analysis.2 cans per dollar. 6 cans/3 dollars is 2 cans per dollar. Normalizing to a unit value is what I would call it.

Bushels per acre. Pounds pw per square inch.  Reply With Quote

10. Thank you  Reply With Quote

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