MHB Why is the Maclaurin series of $e^u$ simply $\sum\frac{u^n}{n!}$?

[poissonspot]

It just occurred to me now that I never asked why the Maclaurin series of $e^u$, u being some function of x, is simply $\sum\frac{u^n}{n!}$. I should say I understand why the Maclaurin series of $e^x$ is $\sum\frac{x^n}{n!}$, due to the derivative of the exponential function being itself, but why this series expansion should apply to functions of x is less clear. I've played with this expansion a little and see what might be an inductive argument, but once again its not clear. Any suggestions?

 

[HallsofIvy]

It just occurred to me now that I never asked why the Maclaurin series of $e^u$, u being some function of x, is simply $\sum\frac{u^n}{n!}$. I should say I understand why the Maclaurin series of $e^x$ is $\sum\frac{x^n}{n!}$, due to the derivative of the exponential function being itself, but why this series expansion should apply to functions of x is less clear. I've played with this expansion a little and see what might be an inductive argument, but once again its not clear. Any suggestions?

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 isn'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.

 

[poissonspot]

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 isn'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.

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?

 

[HallsofIvy]

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?

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

 

[poissonspot]

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

That's understood.