# Check my work on Taylor series?

• May 10th 2009, 06:15 PM
pinkprincess08
Check my work on Taylor series?
Find the first 4 terms (k=0,1,2,3) of the Taylor polynomial for f(x) = sin(x) centered at c=(pi/4)

sin(x) = Sum(k-->0 to inf)[((-1)^k * x^(2k+1))/((2k+1)!)]

[f^(k) * (c) * (x^(2k+1))] / (k!)
^^^^ f^(k) = kth derivative of f
I get
x - ((x^3)/3!) - ((x^5)/5!) - ((x^7)/7!)

Is that correct?
• May 10th 2009, 07:10 PM
CaptainBlack
Quote:

Originally Posted by pinkprincess08
Find the first 4 terms (k=0,1,2,3) of the Taylor polynomial for f(x) = sin(x) centered at c=(pi/4)

sin(x) = Sum(k-->0 to inf)[((-1)^k * x^(2k+1))/((2k+1)!)]

[f^(k) * (c) * (x^(2k+1))] / (k!)
^^^^ f^(k) = kth derivative of f
I get
x - ((x^3)/3!) - ((x^5)/5!) - ((x^7)/7!)

Is that correct?

No the signs should alternate.

CB
• May 10th 2009, 07:24 PM
pinkprincess08
so then it's
x - ((x^3)/3!) + ((x^5)/5!) - ((x^7/7!) ?

I thought that was the Maclaurin series for sin(x)
What is the difference between Taylor series of sin(x) centered at (pi/4) and Maclaurin series?(Thinking)
• May 10th 2009, 07:31 PM
CaptainBlack
Quote:

Originally Posted by pinkprincess08
so then it's
x - ((x^3)/3!) + ((x^5)/5!) - ((x^7/7!) ?

I thought that was the Maclaurin series for sin(x)
What is the difference between Taylor series of sin(x) centered at (pi/4) and Maclaurin series?(Thinking)

Opps, you are right, not reading the question closly enough.

Then no, since $\sin(\pi/4)=\sqrt{2}$ there should be a constant term, and square roots of 2 all over the place.

CB
• May 10th 2009, 08:10 PM
pinkprincess08
so when k = 0, i get (pi*root(2))/8
k=1, {(pi*root(2))/8} * {x - (pi/4)}
k=2, [-pi*root(2) * (16x^2 - 8 pi x + pi^2)] / 128
k=3, [pi*root(2) * (64x^3 - 48 pi x^2 + 12 pi^2 x - pi^3)] / 3072

What do I do now?
• May 10th 2009, 08:26 PM
CaptainBlack
Quote:

Originally Posted by pinkprincess08
so when k = 0, i get (pi*root(2))/8
k=1, {(pi*root(2))/8} * {x - (pi/4)}
k=2, [-pi*root(2) * (16x^2 - 8 pi x + pi^2)] / 128
k=3, [pi*root(2) * (64x^3 - 48 pi x^2 + 12 pi^2 x - pi^3)] / 3072

What do I do now?

$f(x)=\sin(x)$

$f'(x)=\cos(x)$

$f''(x)=-\sin(x)$

$f'''(x)=-\cos(x)$

so:

$f(\pi/4)=\sqrt{2}$

$f'(\pi/4)=\sqrt{2}$

$f''(\pi/4)=-\sqrt{2}$

$f'''(\pi/4)=-\sqrt{2}$

so:

$
\sin(x)=\sqrt{2}[1+ (x-\pi/4) - \frac{(x-\pi/4)^2}{2!} - \frac{(x-\pi/4)^3}{3!} + ...
$
• May 10th 2009, 08:38 PM
pinkprincess08
i though sin (pi/4) was (root(2))/2 ..?
• May 11th 2009, 03:59 PM
TiRune
Indeed it is.
• May 11th 2009, 07:37 PM
CaptainBlack
Quote:

Originally Posted by pinkprincess08
i though sin (pi/4) was (root(2))/2 ..?

Good, then you can use that to get the correct answer.

CB