1\sqrt{-1} ?

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

1=12=(12)12\sqrt{-1} = \sqrt{1-2} = (1-2)^\dfrac{1}{2}.

Now ,

(12)12=112(2)11222(22)1132223!(23)(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 1\sqrt{-1} is possible.

Note by Akash Shukla
3 years, 5 months ago

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The convergence of the binomial expansion

(1+x)n=1+nx+n(n1)2!x2+n(n1)(n2)3!x3+ (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 1x<1-1\leq x<1 .

Pi Han Goh - 3 years, 5 months ago

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

Akash Shukla - 3 years, 5 months ago

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

A Former Brilliant Member - 3 years, 5 months ago

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

Akash Shukla - 3 years, 5 months ago

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