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Math Help - Someone please explain this wonderful thing I found.

  1. #1
    Member integral's Avatar
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    Someone please explain this wonderful thing I found.

    I was trying to explain taylor polynomials to someone and I came accross this:

    \displaystyle e^x=1+\int dx+\int \left [ \int dx \right ] dx+\int \left [ \int \left [ \int dx \right ] dx \right ] dx+...
    =1+x+\frac{x^2}{2}+\frac{x^3}{3!}+...

    This is a wonderful explanation for why the derivative of e^x=e^x (fundamental theorem of calculus) ... But what is the reason for this?

    the sum of the progressive derivatives of a function f(x) at f(0) times the progressive integrals of 1 yeilds f(x)!
    Last edited by integral; November 26th 2010 at 09:26 PM.
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  2. #2
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    Think back to the days before calculators. The only operations that you can do on numbers are addition, subtraction, multiplication, division and exponentiation. So that means that any function you can evaluate must be from some combination of these operations.

    The most general form of a combination of these numbers is a polynomial. So assume that the function \displaystyle e^x can be written as a polynomial.

    \displaystyle e^x = c_0 + c_1x + c_2x^2 + c_3x^3 + \dots. Now you need to try to evaluate these constants.


    You already know some information about the exponential function - that \displaystyle \frac{d}{dx}(e^x) = e^x and that \displaystyle e^0 = 1. You can use this information to evaluate the constants.

    \displaystyle e^0 = c_0 + c_1(0) + c_2(0)^2 + c_3(0)^3 + \dots

    \displaystyle 1 = c_0.

    Now differentiate both sides...


    \displaystyle e^x = c_1 + 2c_2x + 3c_3x^2 + 4c_4x^3 + \dots, and set \displaystyle x = 0 again...

    \displaystyle e^0 = c_1 + 2c_2(0) + 3c_3(0)^2 + 4c_4(0)^3 + \dots

    \displaystyle 1 = c_1.


    Differentiate both sides...

    \displaystyle e^x = 2c_2 + 3\cdot 2c_3x + 4\cdot 3c_4x^2 + 5\cdot 4c_5x^3 + \dots. Let \displaystyle x = 0...

    \displaystyle e^0 = 2c_2 + 3\cdot 2c_3(0) + 4\cdot 3c_4(0)^2 + 5\cdot 4c_5(0)^3 + \dots

    \displaystyle 1 = 2c_2

    \displaystyle \frac{1}{2} = c_2.


    Differentiate both sides...

    \displaystyle e^x = 3\cdot 2c_3 + 4\cdot 3 \cdot 2c_4x + 5\cdot 4\cdot 3 c_5x^2 + 6\cdot 5\cdot 4c_6x^3 + \dots. Let \displaystyle x=0...

    \displaystyle e^0 = 3\cdot 2c_3 + 4\cdot 3\cdot 2c_4(0) + 5\cdot 4\cdot 3c_5(0)^2 + 6\cdot 5 \cdot 4 c_6(0)^3 + \dots

    \displaystyle 1 = 3\cdot 2c_3

    \displaystyle \frac{1}{3!} = c_3.


    Differentiate both sides...

    \displaystyle e^x = 4\cdot 3\cdot 2c_4 + 5\cdot 4\cdot 3\cdot 2c_5 x+ 6\cdot 5\cdot 4\cdot 3c_6x^2 + 7\cdot 6\cdot 5\cdot 4c_7 x^3 + \dots. Let \displaystyle x = 0...

    \displaystyle e^0 = 4\cdot 3\cdot 2c_4 + 5\cdot 4\cdot 3\cdot 2c_5(0) + 6\cdot 5\cdot 4\cdot 3c_6(0)^2 + 7\cdot 6\cdot 5\cdot 4c_7(0)^3 + \dots

    \displaystyle 1 = 4\cdot 3\cdot 2c_4

    \displaystyle \frac{1}{4!} = c_4.


    Are you starting to see a pattern?

    \displaystyle e^x = 1 + x + \frac{1}{2}x^2 + \frac{1}{3!}x^3 + \frac{1}{4!}x^4 + \dots = \sum_{n = 0}^{\infty}\frac{1}{n!}x^n.
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