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crimeas [40]
4 years ago
8

A.If a new series is found by differentiating a known series with included endpoints, those endpoints might not be included in t

he new series. 
B.If a new series is found by differentiating a known series with included endpoints, those endpoints will never be included in the new series.
which statements are false?
Mathematics
1 answer:
Aleonysh [2.5K]4 years ago
8 0
Consider the power series

\displaystyle\sum_{n\ge1}\dfrac{x^n}n

By the ratio test, this series converges for

\displaystyle\lim_{n\to\infty}\left|\frac{x^{n+1}}{n+1}\cdot\frac n{x^n}\right|=|x|\lim_{n\to\infty}\frac n{n+1}=|x|

though we know by the alternating series test that the series converges for x=-1.

So this series converges for -1\le x.

Differentiating the series yields

\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n\ge1}\frac{x^n}n=\sum_{n\ge1}x^{n-1}=\sum_{n\ge0}x^n

which is the geometric series. We know this series converges for |x|, and this time the endpoints are not included.

This example shows that (A) is certainly possible; that is, x=-1 is valid in the first series, but not in the differentiated one.

- - -

Now consider the series

\displaystyle\sum_{n\ge0}\dfrac{x^n}{n!}

which we know to converge to e^x.

Differentiating, we get

\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n\ge1}\frac{x^{n-1}}{(n-1)!}=\sum_{n\ge0}\frac{x^n}{n!}=e^x

as expected. But both series converge everywhere, so this serves as a counter-example to the claim of B. So B is false.
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3 years ago
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scoundrel [369]

Answer:

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Step-by-step explanation:

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Answer:

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Answer:

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