Taylor Polynomial Approximations

In summary, we used the Taylor series formula to approximate values for a given function, f, and its derivatives, using the values of the derivatives at a given point, c. By plugging in the appropriate values for c and the derivatives into the Taylor series formula, we were able to calculate the second and third degree Taylor polynomials for f and use them to approximate values for f(0.7) and f(1.2), respectively. Additionally, we were able to use the results from part a to calculate the second derivative, f', and approximate its value at x = 1.2 in part c.
  • #1
rjs123
90
0

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.
 
Physics news on Phys.org
  • #2
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
 
Last edited:
  • #3
ok, for c...I found this as the derivate for the taylor series...but i was wondering where the denominator went? n!


eq0009MP.gif
 
Last edited:
  • #4
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.
 
  • #5
Maru said:
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
 

1. What is a Taylor polynomial?

A Taylor polynomial is a mathematical function used to approximate a more complex function. It is constructed by using a series of derivatives of the original function at a specific point, called the center of the polynomial.

2. How is a Taylor polynomial different from a regular polynomial?

A Taylor polynomial is different from a regular polynomial because it is constructed using derivatives of the original function, making it a more accurate approximation. Regular polynomials are typically constructed using only a few terms and do not take into account derivatives.

3. Why are Taylor polynomials useful?

Taylor polynomials are useful because they allow us to approximate complex functions with simpler ones. This can make it easier to analyze and understand the behavior of a function. It also allows us to make predictions and estimates without having to use the original, more complex function.

4. What is the process of finding a Taylor polynomial?

The process of finding a Taylor polynomial involves using the derivatives of a function at a specific point to construct a polynomial. The more derivatives used, the more accurate the approximation will be.

5. Can a Taylor polynomial be used to find the exact value of a function?

No, a Taylor polynomial is an approximation of a function and will never give an exact value. However, as more terms are added to the polynomial, the approximation will become more accurate.

Similar threads

  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
770
  • Calculus and Beyond Homework Help
Replies
27
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
426
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
302
  • Calculus and Beyond Homework Help
Replies
6
Views
489
  • Calculus and Beyond Homework Help
Replies
4
Views
613
Back
Top