Discovering the Taylor Series for cos(x) at PI: Finding the Right Pattern

[bobber205]

Trying to find the Taylor Series for cos(x) where x0 is PI.
I've gotten

cos(x) -1
-sin(x) 0
-cos(x) 1
sin(x) 0
cos(x) -1

It's clearly 0 every other term so I need 2k or 2k-1. But the -1 term switches between -1 and 1
How in world do I deal with this? xD

Thanks for any suggestions. I am assuming I've made a mistake somewhere.
I've asked my fellow students and they're cheating with k+1 which does work.

Thanks! :)

 

[Mark44]

Trying to find the Taylor Series for cos(x) where x0 is PI.
I've gotten

cos(x) -1
-sin(x) 0
-cos(x) 1
sin(x) 0
cos(x) -1

It's clearly 0 every other term so I need 2k or 2k-1. But the -1 term switches between -1 and 1
How in world do I deal with this? xD

Thanks for any suggestions. I am assuming I've made a mistake somewhere.
I've asked my fellow students and they're cheating with k+1 which does work.
Thanks! :)

Why would you think k + 1 is cheating? I'm assuming you're talking about the exponent on -1 to give sign alternation.

 

[bobber205]

Is it correct?

 

[diazona]

Is what correct?

 

[HallsofIvy]

So far, you have, for the Taylor's series of cosine(x), about \pi,
-1 + (1/2)x²- (1/4!)x⁴+ ...

First, it should be clear that you will only have even powers of x and even numbers can be written "2n". That is, (1/(2(2))!)x⁴= (1/(2!) x²⁽²⁾ and (1/6!)x⁶ = 1/(3(2))! x²⁽³⁾. The only problem, then, is the power of -1.

The first term is n= 1 and that is positive. (-1)ⁿ would give (-1)¹= -1. To fix that you could just multiply the entire sum by -1:
(-1)(-1+ (1/2!)x²- (1/6!)x⁶+ ...+ (-1)^n/(2n)! x²ⁿ+ ...)

That is the same as multiplying the -1 in each term:
1- (1/2!)x²- (1/6!)x⁶+ ...+ (-1)(-1)ⁿ/(2n)! x²ⁿ+...

Which can also be written with (-1)ⁿ⁺¹:
1- (1/2!) x²- (1/6!)x<sup>6+ ...+ (-1)n+1/(2n)! x2n+ ...

Nothing "cheating" about that.</sup>

 

[Mark44]

So far, you have, for the Taylor's series of cosine(x), about \pi,
-1 + (1/2)x²- (1/4!)x⁴+ ...

Since this is a Taylor's series about \pi, all of the terms should be powers of x - \pi, not powers of x.

First, it should be clear that you will only have even powers of x and even numbers can be written "2n". That is, (1/(2(2))!)x⁴= (1/(2!) x²⁽²⁾ and (1/6!)x⁶ = 1/(3(2))! x²⁽³⁾. The only problem, then, is the power of -1.

The first term is n= 1 and that is positive. (-1)ⁿ would give (-1)¹= -1. To fix that you could just multiply the entire sum by -1:
(-1)(-1+ (1/2!)x²- (1/6!)x⁶+ ...+ (-1)^n/(2n)! x²ⁿ+ ...)

That is the same as multiplying the -1 in each term:
1- (1/2!)x²- (1/6!)x⁶+ ...+ (-1)(-1)ⁿ/(2n)! x²ⁿ+...

Which can also be written with (-1)ⁿ⁺¹:
1- (1/2!) x²- (1/6!)x<sup>6+ ...+ (-1)n+1/(2n)! x2n+ ...

Nothing "cheating" about that.</sup>