Calculate nth Degree Taylor Polynomial for f(x)=sqrtx | Taylor Polynomial Help

[mpgcbball]

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!

 

[Dick]

You would plug in .1 for x. You are writing the Taylor expansion of f(4+x).

 

[mpgcbball]

but is what i wrote for P(x) the nth degree polynomial?
Thanks

 

[Dick]

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).

 

[HallsofIvy]

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!

 

[mpgcbball]

is there a general method to finding the nth degree polynomial? or is it always just f^n(a)x^n/nfactorial ?? Thanks!

 

[mpgcbball]

for any function in general...thanks!