Question

expand log (1+mn) by maclaurin's theorem​

Answer 1

Answer:

Answer with explanation

Maclaurin's Theorem States that

  $f(x)=f(0)+x f'(0)+\frac{x^2}{2!}f"(0)+\frac{x^3}{3!}f^{3}(0)+\frac{x^4}{4!}f^{4}(0)+\frac{x^5}{5!}f^{5}(0)+.....$

                                          --------------------------------------(1)

y= log (1+m n), where, m n =x

So, the above equation can be written as

f(x)= y = log (1+x)

f(0)=log(1+0)=log 1 =0

$f'(x)=\frac{1}{1+x}\\\\f'(0)=\frac{1}{1+0}\\\\f'(0)=1\\\\f"(x)=\frac{-1}{(1+x)^2}\\\\f"(0)=-1\\\\f^{3}(x)=\frac{2}{(1+x)^3}\\\\f^{3}(0)=2\\\\f^{4}(x)=\frac{-6}{(1+x)^4}\\\\f^{4}(0)=-6\\\\f^{5}(x)=\frac{24}{(1+x)^5}\\\\f^{5}(0)=24\\\\.....$

So, y= log (1+x)

Putting these values in expression 1        

$f(x)=0 +x \times 1 + \frac{x^2}{2!} \times (-1)+\frac{x^3}{3!} \times 2+\frac{x^4}{4!} \times (-6)+\frac{x^5}{5!} \times 24+.......\\\\f(x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-..........\\\\ \text{Replacing x by , mn}\\\\f(x)= mn -\frac{m^2n^2}{2}+\frac{m^3n^3}{3}-\frac{m^4 n^4}{4}+\frac{m^5 n^5}{5}-.....$

This is required Expansion.