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.
I'm not sure. It starts around 7:30 here: The Exponential Function | Highlights of Calculus (5 videos) | Highlights of Calculus | MIT OpenCourseWare
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!
It looks like it is a derivation of the Maclaurin series for e^x.
If, (second line) then can't simply equal 1, (first line), it must be plus some constant, which by some reasoning is taken to be 1.
But that means that also which in turn means that must contain an term, etc.. Ugh !
Yes, he's generating a Maclaurin series for e^x. Let me walk you through a few steps.
1) Definition: .
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
3) But note that if we have y = 1 then , 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 . If we integrate this equation we find that y = 1 + x (the arbitrary constant is 1).
5) But again we must have that
Recap: We now have that y = 1 + x and
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?