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Remedial Power Series Work

conscipost

Member
Jan 26, 2012
39
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

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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.
 
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conscipost

Member
Jan 26, 2012
39
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?
 
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HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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".
 

conscipost

Member
Jan 26, 2012
39
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.