Taylor Polynomials

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June 11th 2012, 03:13 AM
cellae

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Taylor Polynomials

Question given:

Attachment 24062
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Both the question and solution are highlighter.. But I Don't unnderstand.. My answer would be:

P₅(x) = (x-1) + [(x²/2!)(x-1)²] + [(x³/3!)(X-1)³] + [(x⁴/4!)(X-1)⁴ + [(x⁵/5)(x-1)⁵

Where are all them zeros coming from, and where did the 120 come from D: !

thank you
June 11th 2012, 03:17 AM
Prove It

Re: Taylor Polynomials

Isn't 5! = 120?
June 11th 2012, 03:25 AM
cellae

Re: Taylor Polynomials

yeah - I'm not understanding why they would put 120/5!
June 11th 2012, 04:00 AM
simamura

Re: Taylor Polynomials

$f(1)=1^5-5*1^4+10*1^3-10*1^2+6*1-2=0$
Let's calculate first five derivatives:
$f'(x)=5x^4-20x^3+30x^2-20x+6$ , so $f'(1)=5*1^4-20*1^3+30*1^2-20*1+6=1$
$f''(x)=20x^3-60x^2+60x-20$ , so $f''(1)=20*1^3-60*1^2+60*1-20=0$
$f'''(x)=60x^2-120x+60$ , so $f'''(1)=60*1^2-120*1+60=0$
$f^{(4)}(x)=120x-120$ , so $f^{(4)}(1)=120*1-120=0$
$f^{(5)}(x)=120$ , so $f^{(5)}(1)=120$
Now, by definition
$P_5(1)=f(1)+\frac{f'(1)}{1!}(x-1)+\frac{f''(1)}{2!}(x-1)^2+\frac{f'''(1)}{3!}(x-1)^3+\frac{f^{(4)}(x)}{4!}(x-1)^4+\frac{f^{(5)}(x)}{5!}(x-1)^5$
Or
$P_5(1)=0+\frac{1}{1!}(x-1)+\frac{0}{2!}(x-1)^2+\frac{0}{3!}(x-1)^3+\frac{0}{4!}(x-1)^4+\frac{120}{5!}(x-1)^5=\frac{1}{1!}(x-1)+\frac{120}{5!}(x-1)^5$
Since 1!=1 and 5!=120 then $P_5(1)=(x-1)+(x-1)^5$

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