A question about Taylor Series

[Artusartos]

Find the Taylor series for cosx and indicate why it converges to cosx for all x in R.

The Taylor series for cosx can be found by differentiating $sum_{k=0}^{\infty} \frac{(-1)^k (x^{2k+1})}{(2k+1)!}$ on both sides...

But I'm not sure what the question means by "why it converges to cosx for all x in R". Isn't that just obvious, since the summation equals cosx...it obviously converges to it...

Thanks in advance

 

[tiny-tim]

Hi Artusartos!

Maybe they're asking you to prove the radius of convergence.

(you could try comparing it with the definition of e^ix)

 

[Artusartos]

Hi Artusartos!

Maybe they're asking you to prove the radius of convergence.

(you could try comparing it with the definition of e^ix)

Thanks. I'm not sure how I can compare it to e^(ix), but this is how I did it:

I just found $$lim |\frac{a_{n+1}}{a_n}|$$ and got infinity as the radius of convergence...

 

[tiny-tim]

yes, that should do!

(and look carefully, and you'll se that the cosx series is the real part of ∑ (ix)ⁿ/n! )

 

[HallsofIvy]

Be careful here! Showing that the radius of convergence is infinity shows that it converges for all x. That alone does not show that it converges to cos(x).

 

[Artusartos]

Be careful here! Showing that the radius of convergence is infinity shows that it converges for all x. That alone does not show that it converges to cos(x).

Thanks, but then what can I do to show that it converges to cos(x)?

 

[tiny-tim]

Thanks, but then what can I do to show that it converges to cos(x)?

Euler's formula? …

(and look carefully, and you'll se that the cosx series is the real part of ∑ (ix)ⁿ/n! )

 

[Artusartos]

Euler's formula? …

Thanks but I'm not sure what you mean. Doesn't the fact that cosx is equal to the Taylor series imply that they all converge to cosx?

 

[tiny-tim]

Doesn't the fact that cosx is equal to the Taylor series imply that they all converge to cosx?

if cosx is analytic, then yes

from http://en.wikipedia.org/wiki/Analytic_function …

A function is analytic if and only if it is equal to its Taylor series in some neighborhood of every point.​

but that's a rather circular answer …

there must be theorems that state when a function is analytic, but i can't offhand remember what they are

 

[Artusartos]

if cosx is analytic, then yes

from http://en.wikipedia.org/wiki/Analytic_function …

A function is analytic if and only if it is equal to its Taylor series in some neighborhood of every point.​

but that's a rather circular answer …

there must be theorems that state when a function is analytic, but i can't offhand remember what they are

Thanks, but we didn't learn about analytic functions yet...

 

[Ray Vickson]

Thanks but I'm not sure what you mean. Doesn't the fact that cosx is equal to the Taylor series imply that they all converge to cosx?

The classic example is f(x) = exp(-1/x^2) for x ≠ 0 and f(0) = 0. The function has derivatives of all orders, which all → 0 as x → 0, so if we define f and all its derivatives to = 0 at x = 0, then f is infinitely differentiable with all derivatives continuous. Its Taylor series around x = 0 is identically 0, with an infinite radius of convergence), but of course, f(x) is not the zero function. In this case the convergent Taylor series fails to represent the function. There are fancier examples, I believe.

One way to proceed is to note that
$$f(x) = T_n(x) + E_n(x),$$
where T_n(x)is the nth order Taylor polynomial and E_n(x) is the error. There are various formulas for E_n(x), and some of them may allow you to say whether or not you have E_n(x) → 0 as n → ∞ for all x, or at least for all x in an interval.