# Use taylor's THeorem to determine the accuracy

1. Apr 29, 2009

### razored

Use taylor's THeorem to determine the accuracy of the approximation: cos(.3) ~=1 - (.3)^2 / 2! + (.3)^4 / 4! when i use taylors theorem, i use (.3)^4 / 4! which gets me 2.03e-10 but the asnwer is R<=2.03e-5

2. Apr 29, 2009

### Dick

Re: Series

(.3)^4/4! is not equal to 2.03e-10. What form of the remainder term are you using?

3. Apr 29, 2009

### razored

Re: Series

Okay, it should be (.3)^5 / 5!

Last edited: Apr 29, 2009
4. Apr 29, 2009

### razored

Re: Series

But i still dont get it, why did i get hte right answer and ignore the f^(n+1)(c) of the formula? (Lagrange error bound) The derivative should have been cos , so i must have picked cos 0. Why?

5. Apr 30, 2009

### HallsofIvy

Staff Emeritus
Re: Series

The error does NOT involve f^(n+1)(c). It involves f^(n+1)(x) for some number between 0 and c (here .3). Since you don't know that number, you use the largest possible value of f^(n+1)(x) between 0 and c to get an upper bound on the error. Since cosine is decreasing, its largest value is cos(0)= 1.