• Jul 24th 2006, 11:32 AM
Use taylor polynomials about 0 to evaluate cos(0.52) to 4 deciamal places, showing all your working.

What I did;

Since the taylor polynomial about 0 for cosine function is

p n (x) = 1-1/2!x^2 + 1/4!x^4- 1/6!x^6+.....1/n!x^n

Hence (did all working)

p2(0.52) = 0.8646

P4(0.52) = 0.867846

p6(0.52) = 0.867819

p8(0.52) = 0.867819

So cos(0.52) = 0.8678
• Jul 24th 2006, 11:44 AM
ThePerfectHacker
Quote:

Use taylor polynomials about 0 to evaluate cos(0.52) to 4 deciamal places, showing all your working.

What I did;

Since the taylor polynomial about 0 for cosine function is

p n (x) = 1-1/2!x^2 + 1/4!x^4- 1/6!x^6+.....1/n!x^n

Hence (did all working)

p2(0.52) = 0.8646

P4(0.52) = 0.867846

p6(0.52) = 0.867819

p8(0.52) = 0.867819

So cos(0.52) = 0.8678

But you have to prove that is accurate to 4 decimal places ;)

By Taylor's theorem,
$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}+R_4$
Where, (the remainder term)
$R_4=\frac{f^{(5)}(z_n)(x-c)^{n+1}}{(n+1)!}$ for some, $z_n$ between zero and $x$
But $|f^{(5)}(z_n)=-\sin (z_n)|\leq 1$
Thus, what you have is that the remainder term is,
$|R_n|\leq \frac{(.52)^5}{5!}=.00041$
This is correct up to three decimal places.
• Jul 24th 2006, 12:41 PM
ThePerfectHacker
Thanx very much.

At this point proof of approximations found by evaluating Taylor Polynomials is beyond the scope of my course. I will definitely keep the workings for later ref.

Thanx once more.
• Jul 24th 2006, 01:55 PM
ThePerfectHacker
Quote: