Can Integrating (x+1) Infinitely Yield e^x?

[Slayer37]

Hey I'm new here. I got a quick question about the e^x function.

I was noodling with my graphing calculator and noticed that every time i took the integral of (x+1) I kept getting a graph that was closer to the graph of e^x. So after this I took the integral of (x+1) about 10 times and got an equation almost exact to e^x.

If taking the integral of (x+1) infinitely gives you e^x, then shouldn't taking the derivative infinitely of e^x give you (x+1)? But this can't be right because the derivative of e^x is e^x... is there something I'm doing wrong?

 

[Defennder]

Look at the Taylor series expansion for e^x.

 

[HallsofIvy]

Indeed, that is precisely what you get using Poisson's iterative method for solving the differential equation x'= x, with x(0)= 1.

Take the x, on the right, to be the constant value 1: x'= 1=> x= t+ C and, since x(0)= 1, C= 1. x= t+ 1.
Now take the x, on the right, to be x+ 1: x'= t+ 1=> x= (1/2)t²+ t+ C and, since x(0)= 1, C= 1: x= (1/2)t²+ t+ 1.
Now take the x, on the right, to be (1/2)t²+ t+ 1: x'= (1/2)t²+ t+ 1=> x= (1/6)t³+ (1/2)t²+ t+ C and, since x(0)= 1, C= 1: x= (1/6)t³+ (1/2)t²+ t+ 1.

Continuing like that, x approaches, in the limit, the series
\sum_{n=0}^{\infty}\frac{1}{n!}x^n[/itex]<br /> <br /> And, of course, the function satisfying x&#039;= x, x(0)= 1 is e^x.