# e^x

• July 23rd 2013, 05:09 AM
Paze
e^x
I'm watching some MIT OCW and the professor says in relation with the function y=e^x :

We start the function at y=1

This has to be equal to its derivative so dy/dx=1

But I can't stop here...I must add an x to the function so that its slope will be one...So y=1+x...But now they need to be equal so I must add that x to dy/dx so dy/dx=1+x...

What I don't understand is...Why do we add that x? I understand why we start the differential equation at 1 but I do not understand why he insists that this x has to be added.

Thank you.
• July 23rd 2013, 05:40 AM
topsquark
Re: e^x
Quote:

Originally Posted by Paze
I'm watching some MIT OCW and the professor says in relation with the function y=e^x :

We start the function at y=1

This has to be equal to its derivative so dy/dx=1

But I can't stop here...I must add an x to the function so that its slope will be one...So y=1+x...But now they need to be equal so I must add that x to dy/dx so dy/dx=1+x...

What I don't understand is...Why do we add that x? I understand why we start the differential equation at 1 but I do not understand why he insists that this x has to be added.

Thank you.

I'm not quite sure why he's doing that. Is this supposed to be a derivation of a power series for e^x?

-Dan
• July 23rd 2013, 05:41 AM
Paze
Re: e^x
• July 23rd 2013, 05:45 AM
Paze
Re: e^x
So far I've gathered that he is attempting to show how e^x relates to compound interest. That's what he's showing with the calculations...I don't understand it quite fully though.

Oh, hold up. It's because it's simply the derivative..He just has a weird way of wording it. I think I get it!
• July 23rd 2013, 05:53 AM
BobP
Re: e^x
It looks like it is a derivation of the Maclaurin series for e^x.

If, (second line) $\frac{dy}{dx}}=1,$ then $y$ can't simply equal 1, (first line), it must be $x$ plus some constant, which by some reasoning is taken to be 1.

But that means that $\frac{dy}{dx}}=1+x$ also which in turn means that $y$ must contain an $x^{2}/2$ term, etc.. Ugh !
• July 23rd 2013, 06:01 AM
Paze
Re: e^x
Quote:

Originally Posted by BobP
It looks like it is a derivation of the Maclaurin series for e^x.

If, (second line) $\frac{dy}{dx}}=1,$ then $y$ can't simply equal 1, (first line), it must be $x$ plus some constant, which by some reasoning is taken to be 1.

But that means that $\frac{dy}{dx}}=1+x$ also which in turn means that $y$ must contain an $x^{2}/2$ term, etc.. Ugh !

Yup, I got it! Thanks!
• July 23rd 2013, 06:09 AM
topsquark
Re: e^x
Yes, he's generating a Maclaurin series for e^x. Let me walk you through a few steps.
1) Definition: $\frac{dy}{dx} = y$.

So we can start this equation by posting y(0) = 1, which is the starting point for the compound interest. (I'm assuming any.)

2) Using 1) we know that $\frac{dy}{dx} = y = 1$

3) But note that if we have y = 1 then $\frac{dy}{dx} = 0$, which is not true. We need it to be 1.

4) So now we are going to construct a new function y(x) such that $\frac{dy}{dx} = 1$. If we integrate this equation we find that y = 1 + x (the arbitrary constant is 1).

5) But again we must have that $\frac{dy}{dx} = y = 1 + x$

Recap: We now have that y = 1 + x and $\frac{dy}{dx} = 1 + x$

But again we need another term to add to y that will make the above dy/dx correct. Repeat steps 3) to 5). Rinse and repeat.

He's creating the power series one term at a time. Does this help?

-Dan