Let f be a polynomial function such that _2(f(x))=( _2(2+(2)/(3)+(2)/(9)+ )) _3(1+(f(x))/(…

Question

Let $f$ be a polynomial function such that ${\log }_{2}(f(x))=({\log }_{2}(2+\frac{2}{3}+\frac{2}{9}+\ldots \ldots \infty))\cdot {\log }_{3}(1+\frac{f(x)}{f(1/x)}),x>0$ and $f(6)=37$. Then $\sum _{n=1}^{10}f(n)$ is equal to $____$ .

Answer: 395

Accepted Answer

Statement of Work :

1. $f(x)$ is a polynomial function

2. A logarithmic equation involving function $f:{\log }_{2}[f(x)]=[{\log }_{2}(2+\frac{2}{3}+\frac{2}{9}+\ldots ..\infty)]\cdot {\log }_{3}{1+\frac{f(x)}{f(\frac{1}{x})}}];x>0$ and $f(6)=37$

3. To evaluate the sum : $\sum _{n=1}^{10}f(n)$

$\begin{gathered} {\log }_{2}[f(x)] =[{\log }_{2}(2+\frac{2}{3}+\frac{2}{9}+\ldots .\infty)]\cdot [{\log }_{3}{1+\frac{f(x)}{f(\frac{1}{x})}}] \ =[{\log }_{2}2{1+\frac{1}{3}+{(\frac{1}{3})}^{2}+\ldots ..\infty}]\cdot |\, {\log }_{3}{1+\frac{f(x)}{f(\frac{1}{x})}}]. \ =[{\log }_{2}2(\frac{1}{1-\frac{1}{3}})]\cdot [{\log }_{3}{1+\frac{f(x)}{f(\frac{1}{x})}}] \ =[{\log }_{2}3]\cdot |\, {\log }_{3}{1+\frac{f(x)}{f(\frac{1}{x})}}] \end{gathered}$

$\frac{{\log }_{2}[f(x)]}{{\log }_{2}3}=[{\log }_{3}1+\frac{f(x)}{f(\frac{1}{x})})]$

$\Rightarrow {\log }_{3}[f(x)]={\log }_{3}{1+\frac{f(x)}{f(\frac{1}{x})}}$

$\Rightarrow f(x)=1+\frac{f(x)}{f(\frac{1}{x})}$

$\Rightarrow$ $f(x)\cdot f(\frac{1}{x})=f(x)+f(\frac{1}{x})$ .....(i)

The function $f(x)$ is a polynomial function stated in SOW where $f(6)=37=(6{)}^{2}+1\ldots ..(ii)$

Equation (ii) on logical analysis reduces the polynomial $f(x)$ to the possible form below :

$f(x)={x}^{2}+1\ldots \ldots (iii)\leftarrow $ Cubic powers of $x$ is not possible where $f(6)>216$

Now let us analyze the polynomial function $f(x)$ from preconditions set in SOW :

$f(x)\cdot f(\frac{1}{x})=(1+{x}^{2})\cdot (1+\frac{1}{{x}^{2}})=1+\frac{1}{{x}^{2}}+{x}^{2}+1=(1+{x}^{2})+(1+\frac{1}{{x}^{2}})=f(x)+f(\frac{1}{x})\ldots .(iv)$

Equation $(ii),(iii)$ and $(iv)$ mathematically certify the polynomial $f(x)=1+{x}^{2}$ :

$f(x)=1+{x}^{2}$

$\Rightarrow {s}_{10}=\sum _{n=1}^{10}f(n)=\sum _{n=1}^{10}(1+{n}^{2})=\sum _{n=1}^{10}(1)+\sum _{n=1}^{10}({n}^{2})$

$=10+\frac{(10)\cdot (10+1)\cdot {(2)\cdot (10)+1}}{6}=10+385=395$

Solutions · 1