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Represent the function $f(x)=x^{0.3}$ as a power series $\sum_{n=0}^\infty c_n(x-5)^n$

Find the following coefficients: $c_0$, $c_1$, $c_2$, $c_3$

Here are my answers:

  • $c_0= 5^{0.3} $
  • $c_1= 0.3 \cdot 5^{-0.7} $
  • $c_2= -0.2 \cdot 5^{-1.7} $
  • $c_3 = 0.35 \cdot 5^{-2.7}$

What am I doing wrong? I know $c_0$ and $c_1$ are correct, but what is wrong with $c_2$ and $c_3$?

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  • $\begingroup$ I formatted the formulas in your question. See math notation guide. $\endgroup$
    – user147263
    Dec 12 '14 at 4:39
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We find the coefficient $c_2$ of $x^2$.

We have $f''(x)=(0.3)(-0.7)x^{-1.7}$. Evaluate at $x=5$, divide by $2!$. You have two little errors, replacing $-0.21$ by $-0.2$, and forgetting to divide by $2!$.

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$f(x) = x^{0.3}$, and $a = 5$. We have: $c_n = \dfrac{f^{(n)}(5)}{n!}$. Thus:

$f'(x) = 0.3x^{-0.7}$, $f''(x) = -0.21x^{-1.7}$, $f'''(x) = 0.357x^{-2.7}$. Thus:

$c_2 = -\dfrac{0.21\cdot 5^{-1.7}}{2}$, and $c_3 = \dfrac{0.357\cdot 5^{-2.7}}{6}$.

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  • $\begingroup$ Ok, I see it, I forgot the denominator. Thanks! $\endgroup$
    – Kris
    Dec 12 '14 at 4:18

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