Taylor Polynomial Approximations

[rjs123]

Homework Statement

Let f be a function that has derivatives of all orders for all real numbers.
Assume f(1) = 3, f'(1) = -2, f''(1) = 2, and f'''(1) = 4

a. Write the second-degree Taylor polynomial for f about x = 1 and use it to approximate f(0.7)

b. Write the third-degree Taylor polynomial for f about x = 1 and use it to approximate f(1.2)

a. Write the second-degree Taylor polynomial for f', derivate of f about x = 1 and use it to approximate f'(1.2)

The Attempt at a Solution

I want to use the taylor series formula:

Pn(x) = f(c) + f'(c)(x-c) + f''(c)(x-c)^2/(2!) + f'''(c)(x-c)^3/(3!) + ...

It seems just a matter of plugging in for a and b...but I'm not sure how to start this.

 

[Maru]

There is a reason why you are given the values f(1) = 3, f'(1) = -2, f''(1) = 2, and f'''(1) = 4. This is because you are expected to simply plug them in, along with each constant c, which is just the point you are "expanding about," in this case, 1. Seems simple enough huh? Thus, you plug the various derivatives and c into the Taylor Series:

Pn(x) = f(c) + f'(c)(x-c) + f''(c)(x-c)^2/(2!) + f'''(c)(x-c)^3/(3!)
Pn(x) = 3 - 2(x-1) + 2(x-1)^2/(2!) + 4(x-1)^3/(3!)

For part b, now we have a third-degree Taylor Series expansion polynomial to approximate f(1.2)!
Now we just plug in 1.2 for X into the equation we've developed and solve:
Pn(1.2) = 3 - 2(1.2-1) + 2(1.2-1)^2/(2!) + 4(1.2-1)^3/(3!) = some number

Do the same thing for part a except use Pn(x) = f(c) + f'(c)(x-c) + f''(c)(x-c)^2/(2!)
since they only want a second degree expansion. The degree simply means up to which derivative will we carry out the expansion.

For part C, we can actually use the result from part a and just take it's derivative to get f', then evaluate that at x = 1.2

 

[rjs123]

ok, for c...I found this as the derivate for the taylor series...but i was wondering where the denominator went? n!

 

[eczeno]

the n!'s are in the c's. you are mixing notation. in your first post, c represents the center; in your second post, a represents the center. my suggestion is to read Maru's post again; everything you need to know is in there.

 

[rjs123]

There is a reason why you are given the values f(1) = 3, f'(1) = -2, f''(1) = 2, and f'''(1) = 4. This is because you are expected to simply plug them in, along with each constant c, which is just the point you are "expanding about," in this case, 1. Seems simple enough huh? Thus, you plug the various derivatives and c into the Taylor Series:

Pn(x) = f(c) + f'(c)(x-c) + f''(c)(x-c)^2/(2!) + f'''(c)(x-c)^3/(3!)
Pn(x) = 3 - 2(x-1) + 2(x-1)^2/(2!) + 4(x-1)^3/(3!)

For part b, now we have a third-degree Taylor Series expansion polynomial to approximate f(1.2)!
Now we just plug in 1.2 for X into the equation we've developed and solve:
Pn(1.2) = 3 - 2(1.2-1) + 2(1.2-1)^2/(2!) + 4(1.2-1)^3/(3!) = some number

Do the same thing for part a except use Pn(x) = f(c) + f'(c)(x-c) + f''(c)(x-c)^2/(2!)
since they only want a second degree expansion. The degree simply means up to which derivative will we carry out the expansion.

For part C, we can actually use the result from part a and just take it's derivative to get f', then evaluate that at x = 1.2

ok, I'm pretty sure i got c now. I didnt bring up the denominator

f = 3 - 2(x-1) + 1(x-1)^2 + 2/3(x-1)^3

f' = 0 - 2 + 2(x - 1) + 2(x-2)^2

f'(1.2) = -2 + 2(1.2 - 1) + 2(1.2 - 2)^2
f'(1.2) = -1.52

for a I got 3.69
for b I got 2.6453