Power series for complex function

In summary, the power series for the function f(z) = (1-z)^-m is Σa_n(n C m).(x^n-m), where the sum is from 0 to infinity. This can be obtained by solving the first order ODE provided and using the formula for differentiation of power series to determine the coefficients. However, there are still an infinite number of undetermined coefficients in the power series.
  • #1
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Homework Statement


Find the power series for the function
f(z) = (1-z)^-m

Hint: Differentiation gives:
f'(z) = m(1-z)^m-1
= m(1-z)^-1.f(z)

or:
zf'(z) + mf(z) = f'(z)

Use the formula for differentiation of power series to determine the coefficients of the power series for f.


Homework Equations





The Attempt at a Solution



Hi, here's my attempt so far:

From real analysis,
(1-x)^-1 = Σ(a_n).x^n, where the sum is from n=0 to infinity

Differentiating:f'(x) (1-x)^-2 = Σ(a_n).n.(x^n-1) , from n=0 to infinity

and so on until we get that the j-th derivative

i.e. f^j.(x) = Σa_n(n C j).(x^n-j), where (n C j) is "n choose j" (binomial coefficient.

Thus, substituting z for x and m for j,

(1-x)^-m = Σa_n(n C m).(x^n-m), sum from 0 to infinity.


---

I don't think this is correct; it seems too straightforward and I haven't used the hint.
Can someone please point me in the right direction?

Thanks for any help!
 
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  • #2
... not to mention that you still have an infinite number of undetermined coefficients.

They give you a first order ODE. Solve it using power series.
 

1. What is a power series for a complex function?

A power series for a complex function is an infinite series of the form f(z) = ∑n=0 an(z-z0)n, where z is a complex variable, z0 is a complex constant, and an are complex coefficients. It is a way of representing a complex function as a polynomial, and it can be used to approximate the values of the function near z0.

2. How is a power series for a complex function different from a Taylor series?

A power series for a complex function is a special case of a Taylor series, which is a way of representing a function as an infinite sum of terms involving the derivatives of the function evaluated at a specific point. The main difference is that a Taylor series is used for real-valued functions, while a power series is used for complex-valued functions.

3. What is the radius of convergence for a power series for a complex function?

The radius of convergence for a power series for a complex function is the distance from the center z0 to the nearest point where the series diverges. It can be calculated using the Cauchy-Hadamard formula, and it determines the interval of values for which the series will converge and accurately approximate the function.

4. How can a power series for a complex function be used to evaluate the function at a specific point?

A power series can be used to evaluate a complex function at a specific point by substituting the value of z into the series and summing the terms. However, this method only works if the value of z is within the radius of convergence. Otherwise, the series will diverge and will not accurately represent the function.

5. What is the relationship between the coefficients of a power series for a complex function and the derivatives of the function?

The coefficients of a power series for a complex function are related to the derivatives of the function by the formula an = f(n)(z0)/n!, where f(n) denotes the nth derivative of f(z) evaluated at z0. This relationship allows us to find the coefficients of a power series by taking derivatives of the function, and vice versa.

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