If \(\sqrt{-1}\) can be written as,

\(\sqrt{-1} = \sqrt{1-2} = (1-2)^\dfrac{1}{2}\).

Now ,

\((1-2)^\dfrac{1}{2} = 1-\dfrac{1}{2}*(2)- \dfrac{1*1}{2*2*2}*(2^2)-\dfrac{1*1*3}{2*2*2*3!}*(2^3)-\cdots\)

The above series will go upto infinite. So is this true? Is the value of \(\sqrt{-1}\) is possible.

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## Comments

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TopNewestThe convergence of the binomial expansion

\[ (1+x)^n = 1+ nx + \dfrac{n(n-1)}{2!} x^2 + \dfrac{n(n-1)(n-2)}{3!} x^3 + \cdots \]

is only true for \(-1\leq x<1 \).

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Thank you very much. Now I got this. But the above series is divergence , then can we apply this?

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No. The binomial theorem is valid only when the power is an integer.

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Then what we use for expansion \(\sqrt{x+1}\) or such others quantities.

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Comment deleted Jun 01, 2016

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this.

It is a rule. ReadLog in to reply

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