Exponential Function

Reillyman · 05-05-2011 03:08PM #1

Anyone have any ideas on this? I've been studying compounding interest and came across the problem;

Sketch a proof of the fact that;

[latex]\lim_{n\to \infty} (1+\frac{x}{n})^n = e^x [/latex]

hivizman · 05-05-2011 04:27PM

This is a standard result in finance theory, which forms the basis of what is often referred to as "continuous compounding".

There's an explanation and sketch of a proof here.

Fremen · 05-05-2011 05:08PM

when x=1, that's actually the definition of e.

Just take the log of both sides, then taylor expand the log function. The limit commutes with the log because log is continuous.

I think I missed this question on my finals. In retrospect, it's not hard.

Michael Collins · 05-05-2011 06:06PM

Or take the natural log of both sides then use L'Hôpital's rule.

Reillyman · 05-05-2011 06:35PM

hivizman wrote: »

This is a standard result in finance theory, which forms the basis of what is often referred to as "continuous compounding".

There's an explanation and sketch of a proof here.

Yeah I know that's the subject I'm studying just going over it all again for exams.

Thanks alot mate.

Reillyman · 05-05-2011 07:05PM

Here's what I have;

I expanded [latex](1+\frac{x}{n})^n[/latex]

applied the limit to all terms in the sequence and am left with;

[latex]\frac{1}{0!} + \frac{x}{1!} + \frac{x}{2!} + \frac{x}{3!} ....)[/latex]

Then I took out [latex]x[/latex] and am left with;

[latex] 1 + x(\frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} ....)[/latex]

I know as n approaches infinity this equals;

[latex] 1 + x(e^x -1) [/latex]

Any ideas?

This could be really simple to solve or maybe I made a mistake but I'm drawing a blank!

hivizman · 05-05-2011 11:10PM

Reillyman wrote: »

Here's what I have;

I expanded [latex](1+\frac{x}{n})^n[/latex]

applied the limit to all terms in the sequence and am left with;

[latex]\frac{1}{0!} + \frac{x}{1!} + \frac{x}{2!} + \frac{x}{3!} ....)[/latex]

Note that [latex](1+\frac{x}{2})^2[/latex] when expanded gives:

[latex]1+2\frac{x}{2}+\frac{x^2}{4}[/latex]

More generally, [latex](1+\frac{x}{n})^n[/latex] expands to:

[latex]1+n\frac{x}{n}+\frac{n(n-1)}{2}\frac{x^2}{n^2}+...+\frac{x^n}{n^n}[/latex]

It looks like you have omitted the exponents for the various powers of x. You should have got:

[latex]\frac{1}{0!} + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!}+ ....[/latex]

Reillyman · 06-05-2011 09:32AM

hivizman wrote: »

Note that [latex](1+\frac{x}{2})^2[/latex] when expanded gives:

[latex]1+2\frac{x}{2}+\frac{x^2}{4}[/latex]

More generally, [latex](1+\frac{x}{n})^n[/latex] expands to:

[latex]1+n\frac{x}{n}+\frac{n(n-1)}{2}\frac{x^2}{n^2}+...+\frac{x^n}{n^n}[/latex]

It looks like you have omitted the exponents for the various powers of x. You should have got:

[latex]\frac{1}{0!} + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!}+ ....[/latex]

Yeah **** your right! Thanks

Exponential Function

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