$$\sqrt{-1}$$ ?

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
2 years, 5 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 $$-1\leq x<1$$.

- 2 years, 5 months ago

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

- 2 years, 5 months ago

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

- 2 years, 5 months ago

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

- 2 years, 5 months ago

Comment deleted Jun 01, 2016

Sorry, factorize- expansion

- 2 years, 5 months ago

You cannot expand those quantities.

- 2 years, 5 months ago

Why?

- 2 years, 5 months ago

It is a rule. Read this.

- 2 years, 5 months ago

Yes,thanks. I have edited the question. But the method is not wrong.

- 2 years, 5 months ago