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Taylor polynomial

  • Thread starter meganlz09
  • Start date
  • #1
1
0
I need to find the Taylor polynomial of degree 4 expanded about a=4 for the function f(x)=squareroot of (x)=x^(1/2)

This is what I've started with but I'm not sure how to proceed and if I even started correctly:
f'(x)(-1/2)x^(-1/2)=1/2sqrt(x)
f"(x)=(-1/4)x^(-3/2)=-1/4x^3/2
f"'(x)=(3/8)x^(-3)=3/8sqrt(x)
f""(x)=(-9/8)x^(-4)
and then i just plug 4 in for x

any explanation toward the correct answer would be great,thanks
 

Answers and Replies

  • #2
378
2
it's

P(x) = f(x) + (x-a)*f'(a) + 1/2! * (x-a)^2 * f''(a) + .... +1/4! * (x-a)^4 * f''''(x)

so you need to plug in a and f(a),.. values

and
if f(x) = (x)^1/2
then f'(x) = 1/2(x)^-0.5 .. (your differentiation seems wrong)
 
  • #3
11
0
Well now just plug into the taylor series...

f(x) = f(4) + f'(4)(x-4)+f''(4)(x-4)^2/2!+f'''(4)(x-4)^3/3!+...

Hope this helps...
 
  • #4
However, you may want to check those derivatives again before proceeding, especially the last two.
 
  • #5
29
0
Well now just plug into the taylor series...

f(x) = f(4) + f'(4)(x-4)+f''(4)(x-4)^2/2!+f'''(4)(x-4)^3/3!+...

Hope this helps...
dont forget to divide by 0!, 1!, 2!, 3!,... accordingly
 

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