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\(\sqrt{-1}\) ?

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.

Note by Akash Shukla
4 months ago

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The 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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@Pi Han Goh Thank you very much. Now I got this. But the above series is divergence , then can we apply this? Akash Shukla · 3 months, 4 weeks ago

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

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@Svatejas Shivakumar Then what we use for expansion \(\sqrt{x+1}\) or such others quantities. Akash Shukla · 4 months ago

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Comment deleted 4 months ago

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@Svatejas Shivakumar Sorry, factorize- expansion Akash Shukla · 4 months ago

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@Akash Shukla You cannot expand those quantities. Svatejas Shivakumar · 4 months ago

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@Svatejas Shivakumar Why? Akash Shukla · 4 months ago

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@Akash Shukla It is a rule. Read this. Svatejas Shivakumar · 4 months ago

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@Svatejas Shivakumar Yes,thanks. I have edited the question. But the method is not wrong. Akash Shukla · 4 months ago

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