My book mentioned that
"If -1<x<1, then (1+x)^n = 1 + nx + n(n-1)(x^2)/2! + n(n-1)(n-2)(x^3)/3! + ... + n(n-1)(n-2)...(n-r+1)(x^r)/r! + ..."
Why -1<x<1? Why is this restriction on x necessary?
Thanks in advance.
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My book mentioned that
"If -1<x<1, then (1+x)^n = 1 + nx + n(n-1)(x^2)/2! + n(n-1)(n-2)(x^3)/3! + ... + n(n-1)(n-2)...(n-r+1)(x^r)/r! + ..."
Why -1<x<1? Why is this restriction on x necessary?
Thanks in advance.
Because it's possible that infinite series might diverge (go to infinity). You need to check the interval for which the series is convergent.Quote:
Originally Posted by startanewww
In that case, shouldn't the restriction be "x =/= +infinity or -infinity (when n is just a small number)"?
If -1<x<1, will the value of (1+x)^n be very small?
I really don't understand..
The series depends on x. But the series itself is only convergent if it follows specific conditions, which means that it will only be convergent for particular values of x (the ones which will satisfy the conditions). I suggest you research convergence tests for infinite series.
I think that you are missing the point.Quote:
Originally Posted by startanewww
for any
.
So it would be true for
The OP was actually referring to the GENERALISED Binomial Theorem, which is an infinite series.Quote:
Originally Posted by PlatoI think that you are missing the point.
So it would be true for