Find the Taylor Polynomial:

Question

legomyego180 · Answer

Find$$P _{4}(x)$$
Centered at x=0 for the function:
$$f(x)=x-4\cos(x)$$

legomyego180 · Answer

$$f(x)=x-4\cos(x)$$
$$f'(x)=1+4\sin(x)$$
$$f''(x)=4\cos(x)$$
$$f ^{3}(x)=-4\sin(x)$$
$$f^4(x)=-4\cos(x)$$

Evaluated @ x=0
f(0)=-4
f'(0)=1
f''(0)=4
f'''(0)=0
f''''(0)=-4

$$\frac{ f^n(0) }{ n! }$$
f(0)=-4
f'(0)=1/2
f''(0)=4/6
f'''(0)=0
f''''(0)=-4/24

legomyego180 · Answer

I need to multiply all of these by (x-a)^n, and I think a=0 but my answer choices arent matching what Im getting

legomyego180 · Answer

My answer I get is:

-4+(x/2)+(2/3)x^2+0-(1/6)x^4

math&ing001 · Answer

Taylor for cosin at 0 is cos(x) = 1 - x2/2! + x4/4! - x6/6!

legomyego180 · Answer

Is that just something to have memorized? How do I work out this problem with that info?

math&ing001 · Answer

I memorized it from using it so much, and yes you can use it, no need to work it from the beginning.

legomyego180 · Answer

$$\cos(x)=1-\frac{ x^2 }{ 2! }+\frac{ x^4 }{ 4! }...$$

legomyego180 · Answer

just distribute the 4?

math&ing001 · Answer

Yep !

math&ing001 · Answer

and add the x...

legomyego180 · Answer

add the x?

math&ing001 · Answer

f(x)=x−4cos(x)
multiply by -4 and add x

legomyego180 · Answer

So it would look like

$$-4[1-\frac{ x^2 }{ 2! }+\frac{ x^4 }{ 4! }]+x$$

Which gave me the answer -4 +x+2x^2-(1/6)x^4

legomyego180 · Answer

Ty for your help @math&ing001

math&ing001 · Answer

You're welcome @legomyego180 =)
I like you username btw :P