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#### conscipost

##### Member

- Jan 26, 2012

- 39

- Thread starter conscipost
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- #1

- Jan 26, 2012

- 39

- Jan 29, 2012

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Well, it isn't! Unless u is just a power of x, $\sum\frac{u^n}{n!}$ is not even a power series.

What is the Maclaurin series for $e^{ln x}$? It

Because $e^x= \sum \frac{x^n}{n!}$ simple substitution gives $e^{u(x)}= \sum \frac{u^n}{n!}$ but, once again, that is not a "Maclaurin series" nor even a power series.

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- Jan 26, 2012

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Is it fair to say the two are equivalent, that is $\sum\frac{[ln(x)]^n}{n!}$ and the Maclaurin series for $e^{ln x}$? If they yield the same values for all x, are they different for all said purposes? I do agree that that is not in the form of a Maclaurin series. I guess this is all an unneeded relationship when it comes down to it. I thought that the series derived for e^x had some restriction due to the fact I had derived it using the fact that the derivative of e^x is e^x. But there is nothing but a substitution at hand. It almost seemed like a leap to just substitute, but why not?Well, it isn't! Unless u is just a power of x, $\sum\frac{u^n}{n!}$ is not even a power series.

What is the Maclaurin series for $e^{ln x}$? Itisn't$\sum\frac{[ln(x)]^n}{n!}$. That isn't the Maclaurin series of anything- as I said, it is not even a power series.

Because $e^x= \sum \frac{x^n}{n!}$ simple substitution gives $e^{u(x)}= \sum \frac{u^n}{n!}$ but, once again, that is not a "Maclaurin series" nor even a power series.

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- Jan 29, 2012

- 1,151

I will agree with your use of the word "equivalent". By the way, the Maclaurin series for $e^{ln x}$ is just "x".Is it fair to say the two are equivalent, that is $\sum\frac{[ln(x)]^n}{n!}$ and the Maclaurin series for $e^{ln x}$? If they yield the same values for all x, are they different for all said purposes? I do agree that that is not in the form of a Maclaurin series. I guess this is all an unneeded relationship when it comes down to it. I thought that the series derived for e^x had some restriction due to the fact I had derived it using the fact that the derivative of e^x is e^x. But there is nothing but a substitution at hand. It almost seemed like a leap to just substitute, but why not?

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- Jan 26, 2012

- 39

That's understood.I will agree with your use of the word "equivalent". By the way, the Maclaurin series for $e^{ln x}$ is just "x".