Discover the First Non-Zero Coefficient of e^-x Series Expansion | Homework Help

In summary: Otherwise, you can simplify it.In summary, 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.
  • #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.
 
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  • #2
You seem to have left off the e^(-x) term in your first step with the log.
 
  • #3
RUber said:
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
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
RUber said:
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
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
RUber said:
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
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
RUber said:
e
Now I have e^-x -e^(-x-x^5/5) Should I expand both as a series?
 
  • #10
If you do, your answer will be evident.
 

1. What is series expansion and why is it useful?

Series expansion is a mathematical technique used to represent a function as a sum of terms, typically involving powers of a variable. It is useful in simplifying complicated functions and in approximating values for functions that are difficult to evaluate directly.

2. How do you perform a series expansion?

To perform a series expansion, you typically start with a known function and use mathematical techniques such as Taylor series or Maclaurin series to represent the function as a sum of terms. This allows you to simplify the function and make approximations for values.

3. What is the difference between a Taylor series and a Maclaurin series?

A Taylor series is a type of series expansion that represents a function as a sum of terms, typically including all powers of a variable. A Maclaurin series is a special case of a Taylor series where the expansion is centered at 0, meaning that the constant term is 0. In other words, a Maclaurin series is a Taylor series with the center of expansion at 0.

4. How do you determine the accuracy of a series expansion?

The accuracy of a series expansion depends on the number of terms included in the series. The more terms included, the more accurate the approximation will be. Additionally, the accuracy can be improved by using higher-order terms or by using a different type of series, such as a Laurent series.

5. Can series expansion be used for all functions?

No, series expansion can only be used for functions that have a finite number of terms in their expansion. Functions that have an infinite number of terms, such as trigonometric functions, cannot be represented as a series. Additionally, some functions may have a series expansion that only converges within a certain interval.

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