- #1

nhrock3

- 415

- 0

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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- Thread starter nhrock3
- Start date

- #1

nhrock3

- 415

- 0

why the solution develops [tex]e^{-x}[/tex] and puts 0.5

and not [tex]e^{+x}[/tex] and putting -0.5

?

- #2

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 970

I suspect that both them method your book gives and your method would give the same answer. Have you tried it?

- #3

nhrock3

- 415

- 0

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

Science Advisor

Homework Helper

- 43,021

- 970

[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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