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Series expansion help

  • #1
173
2

Homework Statement


Show that the first non-zero coefficient in the expansion of
##e^{-x}-\frac{1-x}{{\left(1-x^2\right)}^{\frac{1}{2}}{\left(1-x^3\right)}^{\frac{1}{3}}}##
in ascending powers of x is that of x^5

Homework Equations


Series expansion, logarithmic series

The Attempt at a Solution


Let ##y=e^{-x}-\frac{1-x}{{\left(1-x^2\right)}^{\frac{1}{2}}{\left(1-x^3\right)}^{\frac{1}{3}}}##

##{\ln y={\ln \left[\frac{1-x}{{\left(1-x^2\right)}^{\frac{1}{2}}{\left(1-x^3\right)}^{\frac{1}{3}}}\right]\ }\ }##

##{\ln y\ }={\ln \left(1-x\right)\ }-\frac{1}{2}{\ln \left(1-x^2\right)-\frac{1}{3}{\ln \left(1-x^3\right)\ }\ }##

##{\ln y\ }=\left(-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\frac{x^5}{5}\right)-\frac{1}{2}\left(-x^2-\frac{x^4}{2}\right)-\frac{1}{3}\left(-x^3\right)##

##{\ln y\ }=-x-\frac{x^5}{5}##

##y=e^{-x}\cdot e^{-\frac{x^5}{5}}##

##ye^x=e^{-\frac{x^5}{5}}##

##\left[\frac{1-x}{{\left(1-x^2\right)}^{\frac{1}{2}}{\left(1-x^3\right)}^{\frac{1}{3}}}\right]e^x=e^{-\frac{x^5}{5}}##



I know I'm incredibly close to the answer, however it seems like I can't solve it.
 

Answers and Replies

  • #2
RUber
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You seem to have left off the e^(-x) term in your first step with the log.
 
  • #3
173
2
You seem to have left off the e^(-x) term in your first step with the log.
I'm aware of this, however the reason why I do this is to find the series expansion of the (1-x)/((1-x^2)^(1/2) (1-x^3)^(1/3)) first, then use it to subtract from the series expansion of e^-x.
 
  • #4
RUber
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Then perhaps you simply mislabeled your y.
If ## y = \frac{ 1-x }{ (1-x^2)^{1/2}(1-x^3)^{1/3}} ## then I agree that you might write
## y = e^{-x - x^5/5}##
Your expression then is ##e^{-x} - e^{-x - x^5/5} ##
Take a look at the series representations of those two exponentials. The result should be clear.
 
  • #5
173
2
Then perhaps you simply mislabeled your y.
If ## y = \frac{ 1-x }{ (1-x^2)^{1/2}(1-x^3)^{1/3}} ## then I agree that you might write
## y = e^{-x - x^5/5}##
Your expression then is ##e^{-x} - e^{-x - x^5/5} ##
Take a look at the series representations of those two exponentials. The result should be clear.
How do you get e^-x -e^(-x -x^5 /5) ?
 
  • #6
RUber
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The way you expanded ln(y) was for the fraction part without the ##e^{-x}##. So ##e^{-x} - e^{ln y}## should be your expression.
As you said in post #3, you wanted to subtract the fractional part from the exponential part.
 
  • #7
173
2
The way you expanded ln(y) was for the fraction part without the ##e^{-x}##. So ##e^{-x} - e^{ln y}## should be your expression.
As you said in post #3, you wanted to subtract the fractional part from the exponential part.
Wasn't if I use e^(ln y), the expression would change to e^(-x -x^5 /5) ?
 
  • #8
RUber
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Back to my original point that in your first post you left ##e^{-x}## off when you took the log of y. So using the definition of y which I gave in post 4, you have the expression ##e^{-x}-y##. You did all the work to express ## y = e^{-x -x^5/5}##. Now you just have to subtract that from ##e^{-x}##.
 
  • #9
173
2
e
Now I have e^-x -e^(-x-x^5/5) Should I expand both as a series?
 
  • #10
RUber
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If you do, your answer will be evident.
 

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