Question

Chapter, Sequence and series - Approximatiom by using binomial series. Explain it well plz

Answer 1

To show that if  x is very small  that is  x << 1, like  x = 0.01, 0.04 , 0.008 ...

square root of [(1+x)/(1-x) ]  = 1 - x + x²/2   (approx)

$LHS= \sqrt{\frac{1-x}{1+x}}\\\\=\sqrt{1-x}*\frac{1}{\sqrt{1+x}}\\\\=(1-x)^{\frac{1}{2}}*(1+x)^{- \frac{1}{2}}\\\\=(1-\frac{1}{2}x-\frac{1}{8}x^2+....)*(1+\frac{-1}{2}x+\frac{3}{8}x^2-....)\\\\=(1-\frac{x}{2}+\frac{3x^2}{8}-\frac{x}{2}+\frac{x^2}{4}-\frac{3x^3}{16}-\frac{x^2}{8}+\frac{x^3}{16}-\frac{3x^4}{64}....)\\\\=1-x+\frac{x^2}{2}+....$

We ignore the terms of power 3 and above to get the RHS.

We use the binomial expansion to get the above results:

$(1+x)^n=1+n*x+\frac{n(n-1)}{2}x^2+\frac{n(n-1)(n-2)}{3!}x^3+...\\\\Use\ this\ even\ if\ n\ is\ a\ fraction\ or\ negative.$

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We can also prove the above LHS = RHS in this following manner. It looks simple too.

$\sqrt{(1-x)*\frac{1}{(1+x)}}=\sqrt{(1-x)*(1-x)} \ approx..\\\\=\sqrt{1-2x+x^2}=\sqrt{1 -(2x-x^2)}\\\\=(1-(2x-x^2))^\frac{1}{2}=1 - \frac{(2x-x^2)}{2}\\\\=1-x+\frac{x^2}{2}$
Done.