Aso
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Holaaa,

Ok so I was just doing a bit of C4 today, and I dont understand how to get the range of a binomial expansion that doesnt converge, i.e. an expansion that has a power n which is -ve or a fraction (not a positive integer).

I understand that if i have (1+3x)^-2

then the range will be -1<3x<1

so that gives -1/3 < x < 1/3

which gives |x|<1/3 right?

HOWEVER.

the part I dont understand is why does the range of an expansion that doesn't converge start with -1<x<1 ? Why does the x value in a binomial expansion have to be between -1 and 1? I dont understand where that came from in the book :confused:

If someone could explain it to me I'd be very grateful

A
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Farhan.Hanif93
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Assuming that I've understood your question correctly, it's to do with the binomial series (1+x)^n (where n is not a positive integer) having a radius of convergence of 1. Which is something you don't need to worry about for C4 - it's something you'd come across in a first Analysis course at University.
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Aso
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Thank you!
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ztibor
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(Original post by Aso)
Holaaa,

Ok so I was just doing a bit of C4 today, and I dont understand how to get the range of a binomial expansion that doesnt converge, i.e. an expansion that has a power n which is -ve or a fraction (not a positive integer).

I understand that if i have (1+3x)^-2

then the range will be -1<3x<1

so that gives -1/3 < x < 1/3

which gives |x|<1/3 right?

HOWEVER.

the part I dont understand is why does the range of an expansion that doesn't converge start with -1<x<1 ? Why does the x value in a binomial expansion have to be between -1 and 1? I dont understand where that came from in the book :confused:

If someone could explain it to me I'd be very grateful

A
The general binomial expansion is an infinite funtion series
(1+x)^{\alpha}=\sum_{k=1}^{\inft  y} \binom{\alpha}{k}\cdot x^k
where \alpha is a complex number.
When \alpha is an positive integer f.e n, then the (n+1)th and following terms are 0 in the series, so the sum is finite for any real x
For other alpha the series will be convergent or divergent depending on x value
f.e x=-1 the series converges if and only if \alpha \ge 0
x>1 the series diverges ubless \alpha is non-negative integer
The question may be that for wich x will be the series convergent for any alpha
To answer this we have to calculate the radius of convergence of this function series around zero, because the terms a_k\cdot x^k=a_k \cdot (x-0)^k
the radius can be calculated with the ratio
\left | \frac{1}{\rho}\right |=\left |\lim_{k \rightarrow \infty}\frac{a_{k+1}}{a_k}\right |=\left | lim_{k \rightarrow \infty} \frac{\binom{\alpha}{k+1}}{ \binom{ \alpha }{ k } }\right|
Using that
\binom{\alpha}{k+1} =\binom{\alpha}{k}\cdot \frac{\alpha -k}{k+1}
So the radius
\left | \frac{1}{\rho}\right | =\left | \lim_{k \to \infty}\frac{\alpha -k}{k+1}\right |=|-1|=1
So \rho=1
That is the series convergent for any alpha when -1<x<1
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