Taylor expand (1+z)^n where |z | < 1 and n is any complex #

In summary, the problem is that the complex part of the Taylor series expansion is attached to a real coefficient and cannot be split.
  • #1
Vitani11
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Homework Statement


Same as title.

Homework Equations


Taylor expansion.

The Attempt at a Solution


Okay - what?! I don't even know where to begin. I taylor expanded the function and pretended like n was just some number and that doesn't help. I've never learned this. How? Can you point me in some direction?
 
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  • #2
Vitani11 said:
I taylor expanded the function and pretended like n was just some number and that doesn't help.
Why not? You should show the work you did.
 
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  • #3
Sorry if this is difficult to see (and sideways)
 

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  • #4
I assume that this is a problem from a complex analysis class.
I don't know what you used as the definition of a Taylor series (should have been shown in Relevant equations) or if you know how to take the derivatives of that analytic function.
If you take the derivatives of an analytic function and use the definition of the Taylor series, it should be valid within it's radius of convergence.
 
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  • #5
Vitani11 said:
Sorry if this is difficult to see (and sideways)

If you are serious about wanting help, you need to respect the helpers by typing out your solution.
 
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  • #6
Okay. Sorry.

1+nz/1!+(n(n-1)z2)/2!+(n(n-1)(n-2)z3)/3!+(n(n-1)(n-2)(n-3)z4)/4! is the expansion I get. I see the expansion for ez in there. What do I do about the complex parts (n?)
 
  • #7
the complex parts of the expansion are n+n(n-1)+n(n-1)(n-2)+n(n-1)(n-2)(n-3) but they are all attached to a real coefficient Z so I can't split them. Do I need to generalize this pattern? because that is what my intuition tells me.
 
  • #8
What bothers you about the complex parts? Are you in a complex analysis class?
 
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  • #9
Mathematical physics - nothing bothers me about them I just need to know how to deal with them in an expansion.
 
  • #10
When dealing with a function of complex variables, you should expect the Taylor series expansion to have complex coefficients. n-1 is a perfectly fine complex number if n is complex. But you should change the notation because everyone will assume that n is a natural number.
 
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  • #11
Yes I know. So I expanded the function, as the question asked, but the next question then is show ln(1+z) = z-z2/2+z3/3-z4/4 using the results from the expansion I just did (I am not aloud to Taylor expand). I can't see a relationship between the expansion I just did and that question because the expansion I just did involves complex numbers whereas the next question has none in there. This is what I mean by I need to find a way to deal with the complex part of the expansion. I'm just trying to find relationships.
 
  • #12
The best environment for Taylor series expansions are in complex numbers. Don't try to separate out the real and imaginary parts unless there is some unusual reason to.
n doesn't have to be complex, it just can be complex. You should be able to think of a value of n that is directly relates (1+z)n and ln(1+z).

I will not say more than that on a homework problem.
 
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  • #13
Vitani11 said:
the complex parts of the expansion are n+n(n-1)+n(n-1)(n-2)+n(n-1)(n-2)(n-3) but they are all attached to a real coefficient Z so I can't split them. Do I need to generalize this pattern? because that is what my intuition tells me.

The algebra is exactly the same as it would be if ##n## were a real number. The fact are that standard algebra holds for complex quantities, so that if some or all of the quantities are complex the basic laws ##a+b = b+a##, ##ab = ba##, ##a(b+c) = ab + ac##, ##a+0=a##, ##a \times 1 = a##, ##1/(1/a) = a##, etc., etc. all hold.
 
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  • #14
Got it. Thanks amigos
 

FAQ: Taylor expand (1+z)^n where |z | < 1 and n is any complex #

1) What is the purpose of Taylor expanding (1+z)^n?

The purpose of Taylor expanding (1+z)^n is to approximate the function at a specific point by using a series of derivatives evaluated at that point.

2) How do you determine the coefficients in the Taylor expansion of (1+z)^n?

The coefficients in the Taylor expansion of (1+z)^n are determined by taking the nth derivative of the function at the given point and dividing by n!. The coefficients can also be found using the binomial theorem.

3) Can the Taylor expansion of (1+z)^n be used for values of z outside the range of |z| < 1?

No, the Taylor expansion of (1+z)^n is only valid for values of z within the range of |z| < 1. This is because the function may behave differently outside this range and the Taylor expansion may not accurately approximate the function.

4) How does the accuracy of the Taylor expansion of (1+z)^n change as n increases?

The accuracy of the Taylor expansion of (1+z)^n increases as n increases. This is because as n increases, the series includes more terms and therefore provides a better approximation of the function at the given point.

5) Can the Taylor expansion of (1+z)^n be used for complex values of n?

Yes, the Taylor expansion of (1+z)^n can be used for complex values of n. The same principles apply, where the coefficients are determined by taking the nth derivative of the function at the given point and dividing by n!. However, the series may converge to a different function for complex values of n.

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