- #1

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I suppose Newton's theorem is not a way since it requires factorials.

Thanks!

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- Thread starter Juan Pablo
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- #1

- 40

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I suppose Newton's theorem is not a way since it requires factorials.

Thanks!

- #2

rock.freak667

Homework Helper

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nC1=n!/(n-1)!1!

and n!=n(n-1)!

so nC1 simplifies to n

Similarly

nC2=n!/(n-2)!2!

[tex]=\frac{n(n-1)(n-2)!}{(n-2)!2!}=\frac{n(n-1)}{2!}[/tex]

and so forth for nC3,nC4,etc

- #3

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One use, or was so before calculators, is to approximate certain square roots. Take this case,

[tex]\sqrt{1+a^2} = a +\frac{1}{2a}-\frac{1}{8a^3} +-+[/tex]

In the case of [tex]\sqrt{101} = 10 + 1/20-1/8000 + -[/tex]

This is just a little less that 10.05 and can be easily carried out.

Last edited:

- #4

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(1+x)^(1/2) = Polynomial_function(x) ??

(1+x)^(n) easily works out for n>0, n<0,n=0...........but what about for n = 1/m form ??

m is an Integer.

- #5

uart

Science Advisor

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(1+x)^(1/2) = Polynomial_function(x) ??

(1+x)^(n) easily works out for n>0, n<0,n=0...........but what about for n = 1/m form ??

m is an Integer.

[tex](1+x)^{a}=1+ax+a(a-1)\,\frac{x^{2}}{2!}+a(a-1)(a-2)\,\frac{x^{3}}{3!}+... [/tex]

This works for integer and non integer "a". So a=1/m is no problems.

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