Exponent error question

  • Thread starter nhrock3
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  • #1
nhrock3
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i need to calculate [tex]e^{-0.5}[/tex]

why the solution develops [tex]e^{-x}[/tex] and puts 0.5
and not [tex]e^{+x}[/tex] and putting -0.5
?
 

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  • #2
HallsofIvy
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I have no idea what you are talking about. What do you mean by "develops [itex]e^{-x}[/itex]? Writing as a Taylor's series? Approximating by the tangent line?

I suspect that both them method your book gives and your method would give the same answer. Have you tried it?
 
  • #3
nhrock3
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yes developing in taylor series
but in the first we have libnits series and on the other not
so its not the same

why they are not the same?
 
  • #4
HallsofIvy
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If [itex]f(x)= e^{-x}[/itex] then f(0)= 1, [itex]f'= -e^{-x}[/itex] so f'(0)= -1, [itex]f"(0)= e^{-x}[/itex] so f"(0)= 1, etc. The "nth" derivative, evaluated at x= 0, is 1 if n is even, -1 if n is odd. The Taylor's series, about x= 0, for [itex]e^{-x}[/itex] is
[tex]\sum_{n=0}^\infty \frac{(-1)^n}{n!}x^n[/itex].

In particular,
[tex]e^{-0.5}= \sum_{n=0}^\infty \frac{(-1)^n}{n!}(0.5)^n[/itex]

The usual Taylor's series for [itex]e^x[/itex] is, of course,
[tex]\sum_{n=0}^\infty \frac{1}{n!}x^n[/itex]

and now
[tex]e^{-0.5)= \sum_{n=0}^\infty \frac{1}{n!}(-0.5)^n= \sum_{n=0}^\infty \frac{1}{n!}(-1)^n(0.5)^n[/tex]
[tex]= \sum_{n=0}^\infty \frac{(-1)^n}{n!}(0.5)^n[/tex]

They are exactly the same.
 

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