Using Taylor Polynomial for Laplace Transforms

Click For Summary
SUMMARY

This discussion focuses on the application of Taylor Polynomials in conjunction with Laplace Transforms, specifically for the function f(t)u(t-4). The correct approach involves expressing f(t) in powers of (t-4) and ensuring that the Taylor polynomial is derived up to the 4th derivative. After obtaining the Taylor polynomial, the Laplace Transform is applied, multiplying the result by e^{-4s} to account for the shift. The user confirmed that their arithmetic was correct but needed to include the last term for the 4th degree polynomial.

PREREQUISITES
  • Understanding of Taylor Series and Polynomials
  • Familiarity with Laplace Transforms
  • Knowledge of the Heaviside step function, u(t-a)
  • Ability to compute derivatives up to the 4th order
NEXT STEPS
  • Study the derivation and application of Taylor Series in calculus
  • Learn the properties and applications of Laplace Transforms
  • Explore the Heaviside step function and its role in piecewise functions
  • Practice computing higher-order derivatives for polynomial functions
USEFUL FOR

Students and professionals in mathematics, engineering, or physics who are working with differential equations and require a solid understanding of Taylor Polynomials and Laplace Transforms.

jofree87
Messages
37
Reaction score
0
Ive attached the problem and my work in the pic.

Questions:

Am I even applying the taylor polynomial the correct way? (I never learned taylor series, but I was supposed to be taught in the pre-requisite class)

Am I suppose to plug in c=4? I am not so sure about how the U4(t) works.

After I have found the taylor polynomial, do I just take the laplace transform of it, and then that is my answer?

thanks
 

Attachments

  • IMG_20110920_131227.jpg
    IMG_20110920_131227.jpg
    23.1 KB · Views: 473
Last edited:
Physics news on Phys.org
In this problem you want to make use of the forumula

\mathcal L(f(t-a)u(t-a) = e^{-as}\mathcal Lf(t)

but you need to take the transform of f(t)u4(t) = f(t)u(t-4), which isn't in that form. So to use the above formula, you need to express f(t) in powers of (t-4).

Once you have the Taylor polynomial, you would take its transform as though all the t-a terms were t and multiply the result by e-as.

I didn't check your arithmetic but be sure you go up through the 4th derivative since you have a 4th degree polynomial and the 5th derivative and above would be 0.

[Edit] Your numbers are OK but you need that last term to get the 4th degree.
 
Last edited:
ok, I see now. Thanks for the help!
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
9
Views
2K
Replies
5
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K