Taylor polynomial help

  • #1
11
0

Homework Statement



I'm trying to make the nth degree taylor polynomial for f(x)=sqrtx centered at 4 and then approximate sqrt(4.1) using the 5th degree polynomial


I know that the polynomials are found using the form:
P(x)= f(x)+f'(x)x+f''(x)x^2/2factorial...f^n(x)x^n/nfactorial

so would P(4) just be:

f(4)+f'(4)x + f''(4)x^2/2factorial + f'''(4)x^3/3factorial...

and then would i just plug in 4.1 for x?

thanks for your help...i would also appreciate any general comments on taylor polynomials as I don't really understand them. Thanks!
 

Answers and Replies

  • #2
You would plug in .1 for x. You are writing the Taylor expansion of f(4+x).
 
  • #3
but is what i wrote for P(x) the nth degree polynomial?
Thanks
 
  • #4
What you wrote is a bit garbled. Here's a correction. Notice the different roles of x and a. a is the point you are expanding around and x is the displacement from a.

P(a,x)= f(a)+f'(a)x+f''(a)x^2/2factorial...f^n(a)x^n/nfactorial

is the nth degree approximation to f(a+x).
 
Last edited:
  • #5

Homework Statement



I'm trying to make the nth degree taylor polynomial for f(x)=sqrtx centered at 4 and then approximate sqrt(4.1) using the 5th degree polynomial


I know that the polynomials are found using the form:
P(x)= f(x)+f'(x)x+f''(x)x^2/2factorial...f^n(x)x^n/nfactorial

so would P(4) just be:

f(4)+f'(4)x + f''(4)x^2/2factorial + f'''(4)x^3/3factorial...
No. The taylor series "centered on 4" is f(4)+ f'(4)(x-4)+ f"(4)/2 (x-4)2+ ...+ f(n)(4)/n! (x- 4)^n
Now let x= 4.1.

and then would i just plug in 4.1 for x?
Or use your polynomial with x= 0.1

thanks for your help...i would also appreciate any general comments on taylor polynomials as I don't really understand them. Thanks!
 
  • #6
is there a general method to finding the nth degree polynomial? or is it always just f^n(a)x^n/nfactorial ?? Thanks!
 
  • #7
for any function in general...thanks!
 

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