# binomial series - why is there a range of validity?

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#1
in the form (1 + x)^n the expansion is only valid between -1<x<1 so that the terms get progressively smaller in the series

I dont get why do they need to get smaller?
0
1 month ago
#2
(Original post by m.s124)
in the form (1 + x)^n the expansion is only valid between -1<x<1 so that the terms get progressively smaller in the series

I dont get why do they need to get smaller?
Because you expand it as an infinite series (for suitable n) so the sum only exists if the terms decrease to zero. It should be in your textbook?
0
1 month ago
#3
(Original post by m.s124)
in the form (1 + x)^n the expansion is only valid between -1<x<1 so that the terms get progressively smaller in the series

I dont get why do they need to get smaller?
It is possible to express (1+x)^n as an infinite series of terms in increasing powers of x.

The problem with infinite series is that they only converge and give us a meaningful finite value under very particular conditions.

The infinite series for (1+x)^n is a geometric series with common ratio x.

If |x| > 1 it means that every successive term is bigger in magnitude than the term before it. Summing these up blows the sum up to infinity.

If |x| < 1 it means that successive terms get smaller, and it turns out to be sufficient for convergence.

When n is a positive integer, this sum is finite therefore it will always give a finite value with no restriction on x.
Last edited by RDKGames; 1 month ago
1
#4
(Original post by RDKGames)
It is possible to express (1+x)^n as an infinite series of terms in increasing powers of x.

The problem with infinite series is that they only converge and give us a meaningful finite value under very particular conditions.

The infinite series for (1+x)^n is a geometric series with common ratio x.

If |x| > 1 it means that every successive term is bigger in magnitude than the term before it. Summing these up blows the sum up to infinity.

If |x| < 1 it means that successive terms get smaller, and it turns out to be sufficient for convergence.

When n is a positive integer, this sum is finite therefore it will always give a finite value with no restriction on x.
so the range of validity are the values that it will make the series converge to a value?
but the series still works outside that range... when |x|>1... so why does there have to be a range if it works for alll values of x
0
1 month ago
#5
(Original post by m.s124)
so the range of validity are the values that it will make the series converge to a value?
but the series still works outside that range... when |x|>1... so why does there have to be a range if it works for alll values of x

The range of validity is when the series converges, yes.

When |x| > 1 the infinite series does not converge.
0
1 month ago
#6
(Original post by m.s124)
so the range of validity are the values that it will make the series converge to a value?
but the series still works outside that range... when |x|>1... so why does there have to be a range if it works for alll values of x
This isn't hard to see with a specific example.

(1+x)^(-1) = 1 - x + x^2 - x^3 + x^4 - ...

The RHS converges if |x|<1 and converges to 1/(1+x).

If you tried setting x = -2 then the LHS is -1 and the RHS is 1+2+4+8+16+... which diverges to infinity.
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