# Thread: Check my work on Taylor series?

1. ## 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?

2. 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

3. 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?

4. 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?
Opps, you are right, not reading the question closly enough.

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

CB

5. 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?
just add those?

6. 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?
just add those?
$\displaystyle f(x)=\sin(x)$

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

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

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

so:

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

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

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

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

so:

$\displaystyle \sin(x)=\sqrt{2}[1+ (x-\pi/4) - \frac{(x-\pi/4)^2}{2!} - \frac{(x-\pi/4)^3}{3!} + ...$

7. i though sin (pi/4) was (root(2))/2 ..?

8. Indeed it is.

9. 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