Re: the exponential function

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Sets, Maps and Symmetry Groups Course Discussion List

Re: the exponential function

Posted by Daniel Goroff on January 02, 19100 at 09:31:57:

In Reply to: the exponential function posted by Kevin Hall on December 30, 1999 at 19:20:51:

Yes, z^0=1 and 0!=1 are the n=0 cases that make the following recursive definitions work:

z^(n+1)=z*z^n and (n+1)!=(n+1)*n!

As for the other limit, just write out

(exp(h)-1)/h = 1 + h/2! + h^2/3! + ...

This equals 1 if you just plug in h=0, of course. But it is an infinite sum and so it may not be a continuous function of h. To be careful, you should show that you can bound h/2! + h^2/3! + ... by something that goes to zero as h does. What you know about the geometric series may help with that checking, for example.

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>>
>>Yes, z^0=1 and 0!=1 are the n=0 cases that make the following recursive definitions work:

>>z^(n+1)=z*z^n and (n+1)!=(n+1)*n!

>>As for the other limit, just write out 

>>(exp(h)-1)/h = 1 + h/2! + h^2/3! + ...

>>This equals 1 if you just plug in h=0, of course.  But it is an infinite sum and so it may not be a continuous function of h.  To be careful, you should show that you can bound h/2! + h^2/3! + ... by something that goes to zero as h does.  What you know about the geometric series may help with that checking, for example.

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