What is the Derivative of the Infinite Series of Tangent Powers?

cj5892
Messages
2
Reaction score
0

Homework Statement


Consider the infinite series s(x)=1-tan^2(x)+tan^4(x)-tan^6(x)+... , where 0<x<pi/4
s'(x)=
A.sin2x
B.cos2x
C.-tan2x
D.-sin2x
E.-cos2x


The Attempt at a Solution



Attempting to derive the series, i get 1-2tanxsec^2x+4tan^3(x)sec^2(x)-6tan^5(x)sec^2(x)+...

am i missing something obvious here? i don't see any trig identities in that that would give any of the answers
 
Physics news on Phys.org
cj5892 said:

Homework Statement


Consider the infinite series s(x)=1-tan^2(x)+tan^4(x)-tan^6(x)+... , where 0<x<pi/4
s'(x)=
A.sin2x
B.cos2x
C.-tan2x
D.-sin2x
E.-cos2x

The Attempt at a Solution



Attempting to derive the series, i get 1-2tanxsec^2x+4tan^3(x)sec^2(x)-6tan^5(x)sec^2(x)+...

am i missing something obvious here? i don't see any trig identities in that that would give any of the answers

Hint: Sum that geometric series for s(x) first.
 
Last edited:
i still can't get it :(
 
cj5892 said:
i still can't get it :(

What's the series expansion of 1/(1+r)?
 
Or what is the sum of a geometric series with first term a and ratio r?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top