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I am working on finding a Maclaurin series for this function.

$$f(x) =x^6e^{x^7}$$

So I think I have to evaulate the above function based on a Maclaurin series for $e^x$ = $\sum_{n=0}^\infty {x^n\over n!}$

I am just confused on how to connect the two series, and substitute the above function into the Maclaurin series for $e^x$

Can someone please help with explaining the first few steps? Thank Alot

user140749
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2 Answers2

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$$e^u=\sum_{n=0}^{\infty}\frac{u^n}{n!} ,$$ then if $u=x^7$, we have $$e^{x^7}=\sum_{n=0}^{\infty}\frac{(x^7)^n}{n!}=\sum_{n=0}^{\infty}\frac{x^{7n}}{n!}. $$

Then, $$f(x) =x^6 e^{x^7}=x^6\cdot \sum_{n=0}^{\infty}\frac{x^{7n}}{n!}=\sum_{n=0}^{\infty}\frac{x^{7n+6}}{n!}.$$

ZHN
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If you were just asked to write the Maclaurin series for $ \ e^{x^7} \ , $ you would insert $ \ x^7 \ $ in place of $ \ x \ $ in the series you have for $ \ e^x \ . $

Now, what is the derivative of $ \ e^{x^7} \ $ ? How could you then use the series you have for $ \ e^{x^7} \ $ to write one for $ \ x^6 \ e^{x^7} \ $ ?

colormegone
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