# Determining which of the following is a Maclaurin Series

1. Apr 26, 2014

### MathewsMD

1. The problem statement, all variables and given/known data

In attached image.

2. The attempt at a solution

Now, after looking at the solution, the only real conclusion I can come up with is that a Maclaurin series must have x's with non-negative integer value as the exponents, correct? This is because for the the general representation of the Maclaurin series, $c_n = \frac {f^{(n)}(0)}{n!}$ and if n is negative, then you're taking the antiderivative and the factorial is also negative (undefined). Is this not allowed? Is this not allowed for Maclaurin series only or any Taylor series in general, since in this case a = 0? This is the only conclusion I've been able to come up with I believe it's flawed.

I was also looking at option D, and isn't $\frac {1-e^x}{x} = \frac {1}{x} - ∑^∞_{n=0} \frac {x^{n-1}}{n!}$ and wouldn't this too have a negative exponent on x for the first term at n = 0?

Is there anything I am blatantly missing here?
Any explanations for the question and its solution would be much appreciated!

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Last edited: Apr 26, 2014
2. Apr 26, 2014

### Dick

Yes, the series must have all nonnegative integer exponents. And for option D the series expansion of e^x is 1+x+x^2/2!+... That 1 cancels the other 1.

3. Apr 26, 2014

### MathewsMD

I missed that cancellation. Thank you!