1. ## Taylor Polynomials

Question given:

(I think you can click to enlarge it..)

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

2. ## Re: Taylor Polynomials

Isn't 5! = 120?

3. ## Re: Taylor Polynomials

yeah - I'm not understanding why they would put 120/5!

4. ## 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 $P_5(1)=(x-1)+(x-1)^5$