Power Expansion (Complex variables)

In summary, using the power series for e^z and the definition of sin(z), it can be shown that the sum of (-1)^k z^(2k+1)/(2k+1)! is equal to sin(z). In the given attempt at a solution, it is shown that when n is even, i^n-(-i)^n is equal to 0, and when n is odd, it is equal to 2i. This leads to the final solution for sin(z) as shown in the conversation.
  • #1
tmlfan_17
11
0

Homework Statement



Use the power series for e^z and the def. of sin(z) to check that
sum ((-1)^k z^(2 k+1))/((2 k+1)!)

Homework Equations





The Attempt at a Solution



I apologize, but I am not particularly good with latex. Therefore, I attached a picture of my solution thus far. I've tried many methods, but this is where I get stuck and I can't seem to get sin(z) to equal its power expansion. Any help would be very much appreciated.
 

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  • #2
So you have:
$$\frac{1}{2i}\sum_{n=0}^{\infty} \frac{z^n}{n!}\left(i^n-(-i)^n\right)$$
Clearly, if ##n## is even, ##i^n-(-i)^n=0##. Can you figure out what happens if ##n## is odd i.e ##n## is of the form ##2k+1##?
 
  • #3
Yes. Thank you sir!
 
  • #4
tmlfan_17 said:
Yes. Thank you sir!

Glad to help but please don't call me sir, I am a student myself. :smile:
 

1. What is a power expansion in complex variables?

A power expansion in complex variables is a mathematical technique used to express a complex function as a series of simpler functions. It involves breaking down a complex function into simpler terms using the properties of complex numbers and evaluating the function at different points.

2. Why is power expansion important in complex analysis?

Power expansion is important in complex analysis because it allows us to simplify complex functions and express them in terms of simpler functions. This makes it easier to study and analyze complex functions, and can also help us approximate the value of a function at a given point.

3. What is the general form of a power expansion in complex variables?

The general form of a power expansion in complex variables is given by the Taylor series, which is a representation of a function as an infinite sum of terms that are calculated from the values of the function's derivatives at a single point.

4. What is the difference between a Taylor series and a Laurent series?

A Taylor series is a power expansion that is used to approximate a function around a single point, while a Laurent series is used to approximate a function around an annulus (a region between two concentric circles). A Taylor series has only positive powers of the variable, while a Laurent series can have both positive and negative powers.

5. How is a power expansion used to evaluate complex integrals?

A power expansion can be used to evaluate complex integrals by first expressing the integrand as a power series, then integrating each term of the series separately. This allows us to break down a complex integral into simpler integrals that can be easily evaluated. In some cases, the power series may converge to the exact value of the integral, while in others it may provide an approximation.

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