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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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 \(-1\leq x<1 \). – Pi Han Goh · 3 months, 4 weeks ago

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– Akash Shukla · 3 months, 4 weeks ago

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

No. The binomial theorem is valid only when the power is an integer. – Svatejas Shivakumar · 4 months ago

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– Akash Shukla · 4 months ago

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

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– Akash Shukla · 4 months ago

Sorry, factorize- expansionLog in to reply

– Svatejas Shivakumar · 4 months ago

You cannot expand those quantities.Log in to reply

– Akash Shukla · 4 months ago

Why?Log in to reply

this. – Svatejas Shivakumar · 4 months ago

It is a rule. ReadLog in to reply

– Akash Shukla · 4 months ago

Yes,thanks. I have edited the question. But the method is not wrong.Log in to reply