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asi123
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Homework Statement
Hey guys.
I need to develop this function into Laurent series.
I used the Sin Taylor series and got what I got.
Now, is there a trick or something to get the z-2 inside series or is this enough?
Thanks.
HallsofIvy said:Expand sin(z) about z= 2, not 0! And that can be done relatively easily by writing sin(z) as sin(u+ 2) where u= z- 2. sin(u+2)= cos(u)sin(2)+ sin(u)cos(2) so
[tex]sin(z)= \sum_{n=0}^\infty \frac{(-1)^n sin(2)}{(2n)!}(z-2)^{2n}+ \frac{(-1)^n cos(2)}{(2n+1)!}(z-2)^{2n+1}[/tex]
Where I have expanded sin(u) and cos(u) in the usual series around u= 0.
A Laurent series is a representation of a complex function as an infinite sum of terms involving powers of the independent variable, where some of the terms may have negative powers. It is used to approximate complex functions and is especially useful for functions with singularities.
Developing a Laurent series can be a complex and time-consuming process, especially for functions with multiple singularities. Assistance from experts can help ensure accuracy and efficiency in the development of the series.
The coefficients in a Laurent series can be determined by using the Cauchy integral formula, which involves calculating integrals around the singularities of the function. Alternatively, computer software such as Mathematica can also be used to calculate the coefficients.
While Laurent series can be used to approximate many complex functions, they may not converge for all functions. It is important to analyze the singularities and behavior of the function before attempting to develop a Laurent series.
One common mistake is to forget to include all singularities in the series, which can lead to inaccuracies in the approximation. It is also important to check for convergence and accuracy of the series, as well as to properly handle any branch cuts or discontinuities in the function.