# Represent the function $f(x)=x^{0.3}$ as a Taylor series centered at $5$

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: $$c_0$$, $$c_1$$, $$c_2$$, $$c_3$$

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

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

• I formatted the formulas in your question. See math notation guide.
– user147263
Dec 12 '14 at 4:39

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 $2!$.

$f(x) = x^{0.3}$, 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 $c_3 = \dfrac{0.357\cdot 5^{-2.7}}{6}$.

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