## seth3 2 years ago In this video http://ocw.mit.edu/resources/res-18-005-highlights-of-calculus-spring-2010/highlights_of_calculus/the-exponential-function/ prof. strang multiplies two equations (series) e^x = 1+x+1/2x^2+... and e^X = 1+X+1/2X^2+... to prove that the result equals e^(x+X). But when I multiply the two equations (e^x)x(e^X) using the distributive rule, I get the following result: (1+x+1/2x^2)x(1+X+1/2X^2) =(1+X+1/2X^2)+(x+xX+1/2xX^2)+(1/2x^2+1/2Xx^2+1/4x^2 X^2) which does not equal e^(x+X) = 1+(x+X)+(1/2x^2+xX+1/2X^2). I would be very grateful if someone could point out where I went wrong! Thanks

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1. rosina

Assume f(x)=e^x, what you need to to know is that the exponential series is infinite, so multiplying several items of the f(x) and f(X) is not enough. The professor just checked the start point and several items. You're doing very well for the first 3 terms of f(x+X). The better result is to show the infinity of the series using "...", so the better result is (1+x+1/2x^2 +...)x(1+X+1/2X^2+...) =(1+X+1/2X^2)+(x+xX+1/2xX^2)+(1/2x^2+1/2Xx^2+1/4x^2 X^2)+... . If you adjust the order of the terms, it will be 1+(x+X)+(1/2x^2+xX+1/2X^2)+1/2xX^2+1/2Xx^2+1/4x^2X^2+...=1+(x+X) +(x+X)^2+1/2xX^2+1/2Xx^2+1/4x^2X^2+... For 1+(x+X) +(x+X)^2, they are equal to the first three items of f(x+X). For 1/2xX^2+1/2Xx^2+1/4x^2X^2+... , they would be the items of (x+X)^3 +...+(x+X)^n+... The professor went on to check the (x+X)^3, and it was proved. So he deduced that the result f(x)*f(X)=f(x+X). The deduction is not so strict as it doesn't show the case when x and X approach n. Here's another way to do it http://www.pa.msu.edu/~stump/champ/exp.pdf.