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Taylor Series Remainder Theorem

1. Prove that the MacLaurin series for cosx converges to cosx for all x.

2. Relevant equations
Ʃ(n=0 to infinity) ((-1)^n)(x^2n)/((2n)!) is the MacLaurin series for cosx
|Rn(x)|[itex]\leq[/itex]M*(|x|^(n+1))/((n+1)!) if |f^(n+1)(x)|[itex]\leq[/itex]M
lim(n->infinity)Rn=0 then a function is equal to its Taylor series within the interval of convergence.

3. The attempt at a solution. I already know what the solution is but I can't figure out why we are using this solution.
First you get: |f^(n+1)(x)|[itex]\leq[/itex]1 because |cosx| and |sinx| functions are always bounded by 1
Then you write |Rn(x)|[itex]\leq[/itex](|x|^(n+1))/((n+1)!)
lim as n->infinity of (|x|^(n+1))/((n+1)!)=0, therefore lim as n-> infinity of |Rn|=0 so the MacLaurin series of cosx converges to cosx for all x.
My question
Why don't we just write Rn = [itex]\Sigma[/itex] from (i=n+1 to infinity) ((-1)^i)*(x^2i)/((2i)!) and we can see from this that the limit as n->infinity makes Rn go to zero, therefore cosx converges to cosx? Why did we have to go through all of that other stuff?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution


Staff Emeritus
Science Advisor
Gold Member
If you write that the remainder is equal to the terms that you didn't sum up yet, then you're assuming that the infinite series is equal to cos(x) in the first place.

There are functions where the Taylor series converges, but does not converge to the function, so you can't assume that the function is equal to the Taylor series and just calculate the interval of convergence

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