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I'll make my question short, I am encountering a error when doing expansion. I am expanding $f(x)=2x^3+4x+1$ and after the expansion, things don't match.

Here's what I'm doing.

Let $a=5$

$$f(x)=\frac{f(5)}{1}+\frac{f'(5)}{1}(x-5)+\frac{f''(5)}{2}(x-5)^2+\frac{f'''(5)}{6}(x-5)^3$$

$$f(x)=\frac{271}{1}+\frac{154}{1}\left(x-5\right)+\frac{60}{2}\left(x^2-10x+25\right)+\frac{60}{6}\left(x^3-15x^2+75x-125\right)$$

And I immediately realize that the coefficient for $x^3$ is wrong.

Am I misunderstanding how to use Taylor series or my calculation just sucks?

Thanks in advance!

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6  
$f'''(x)\equiv 12$, so $f'''(5)=12$, not $60$. –  njguliyev Sep 29 '13 at 18:11
    
@njguliyev I'll go kill myself, I'm sorry for posting stupid question here. You wanna post it as an answer or delete it? –  Shane Hsu Sep 29 '13 at 18:13
1  
@njguliyev : Post it as an answer. –  Patrick Da Silva Sep 29 '13 at 18:14
1  
@ShaneHsu We all make mistakes sometimes, don't be dramatic. –  Pedro Tamaroff Sep 29 '13 at 19:08
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