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Exponential Derivative Problem

  1. Sep 2, 2008 #1
    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?
     
  2. jcsd
  3. Sep 2, 2008 #2

    Defennder

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    Look at the Taylor series expansion for e^x.
     
  4. Sep 2, 2008 #3

    HallsofIvy

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    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
    [tex]\sum_{n=0}^{\infty}\frac{1}{n!}x^n[/itex]

    And, of course, the function satisfying x'= x, x(0)= 1 is ex.
     
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