Taylor Series Expansion for f(z) = −1/z^2 about z = i + 1

In summary, the Taylor series expansion for f(z) = −1/z^2 about z = i + 1 is Ʃ\frac{1}{n!}f^{(n)}(1+i) * (z-i-1)^n, where n starts at 0 and goes to infinity. The second form with the powers of (z-i-1)^n is the more accurate representation.
  • #1
Applejacks
33
0

Homework Statement



Find the Taylor series expansions for f(z) = −1/z^2 about z = i + 1.

Homework Equations


The Attempt at a Solution



I'm just not sure what format I'm supposed to leave it in.

Is it meant too look like this:
f(z)=f(i+1)+f'(i+1)(x-i-1)...

or this
Ʃ[itex]\frac{1}{n!}[/itex][itex]f^{(n)}[/itex](1+i) * (z-i-1)^n (also is this correct?)
 
Last edited:
Physics news on Phys.org
  • #2
They are exactly the same thing. The first expression is just the first two terms of the second.
 
  • #3
Yeah I'm aware of that. I guess I should keep it in the second form though. Another question: what does the square do to the function. What I wrote can't be correct because -1/z would give the same thing.

Ʃ[itex]\frac{1}{n!}[/itex][itex]f^{(n)}[/itex](1+i) * (z-i-1)^2n
 
  • #4
Applejacks said:
Yeah I'm aware of that. I guess I should keep it in the second form though. Another question: what does the square do to the function. What I wrote can't be correct because -1/z would give the same thing.

Ʃ[itex]\frac{1}{n!}[/itex][itex]f^{(n)}[/itex](1+i) * (z-i-1)^2n

What?? Changing f changes f^(n)(i+1). That changes the series doesn't it? The power part (z-i-1)^n doesn't change. Those are the powers in the expansion of any function around i+1.
 
  • #5
ah right. Thanks for pointing that out.
 

What is a Taylor Series Expansion?

A Taylor Series Expansion is a mathematical concept used to approximate a function as a polynomial at a certain point. It is named after the mathematician Brook Taylor who first introduced it in the 18th century.

Why is a Taylor Series Expansion important?

A Taylor Series Expansion allows us to approximate complex functions with simpler polynomial equations, making calculations and analysis easier. It is also used in various fields such as physics, engineering, and economics to model real-world situations.

How is a Taylor Series Expansion calculated?

A Taylor Series Expansion is calculated by taking the derivatives of a function at a specific point and plugging them into a formula. The more derivatives included, the more accurate the approximation becomes.

What is the difference between a Taylor Series and a Maclaurin Series?

A Taylor Series is used to approximate a function at any point, while a Maclaurin Series is specifically used to approximate a function at the point 0. In other words, a Maclaurin Series is a special case of a Taylor Series when the point of approximation is 0.

What are the limitations of a Taylor Series Expansion?

A Taylor Series Expansion can only approximate a function within a certain radius of convergence. If the function is not analytic or has singularities, the Taylor Series will not converge to the original function. Additionally, including too many derivatives in the series can lead to large errors in the approximation.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
354
  • Calculus and Beyond Homework Help
Replies
1
Views
790
Replies
12
Views
827
  • Calculus and Beyond Homework Help
Replies
27
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
2K
  • Calculus and Beyond Homework Help
Replies
17
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
695
  • Calculus and Beyond Homework Help
Replies
9
Views
875
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Back
Top