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Question Number 132935 by liberty last updated on 17/Feb/21

Answered by EDWIN88 last updated on 17/Feb/21

$$\mathrm{let}\mathrm{x}=2\Rightarrow \mathrm{f}\left(2\right).\mathrm{f}\left(\frac{1}{2}\right)=\mathrm{f}\left(2\right)+\mathrm{f}\left(\frac{1}{2}\right)$$$$\Rightarrow 65.\mathrm{f}\left(\frac{1}{2}\right)=65+\mathrm{f}\left(\frac{1}{2}\right)$$$$\Rightarrow \mathrm{f}\left(\frac{1}{2}\right)=\frac{65}{64}\Rightarrow \mathrm{f}\left(\frac{1}{2}\right)=1+\frac{1}{64}$$$$\Rightarrow \mathrm{f}\left(\frac{1}{2}\right)=1+\frac{1}{2^6}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=1+{\mathrm{x}}^6$$$$\mathrm{then}\mathrm{f}(-3)=1+{(-3)}^6=730$$

Answered by Ar Brandon last updated on 17/Feb/21

$$\mathrm{f}\left(2\right)=65\mathrm{therefore}\mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{a}\mathrm{non}\mathrm{zero}\mathrm{polynomial}$$$$\mathrm{Let}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_2{\mathrm{x}}^2+\cdot \cdot \cdot +{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}},{\mathrm{a}}_{\mathrm{n}}\ne 0$$$$\Rightarrow \left(\underset{\mathrm{r}=0}{\sum ^{\mathrm{n}}}{\mathrm{a}}_{\mathrm{r}}{\mathrm{x}}^{\mathrm{r}}\right)\left(\underset{\mathrm{r}=0}{\sum ^{\mathrm{n}}}\frac{{\mathrm{a}}_{\mathrm{r}}}{{\mathrm{x}}^{\mathrm{r}}}\right)=\underset{\mathrm{r}=0}{\sum ^{\mathrm{n}}}{\mathrm{a}}_{\mathrm{r}}{\mathrm{x}}^{\mathrm{r}}+\underset{\mathrm{r}=0}{\sum ^{\mathrm{n}}}\frac{{\mathrm{a}}_{\mathrm{r}}}{{\mathrm{x}}^{\mathrm{r}}}$$$$\Rightarrow \left(\underset{\mathrm{r}=0}{\sum ^{\mathrm{n}}}{\mathrm{a}}_{\mathrm{r}}{\mathrm{x}}^{\mathrm{r}}\right)\left(\underset{\mathrm{r}=0}{\sum ^{\mathrm{n}}}{\mathrm{a}}_{\mathrm{r}}{\mathrm{x}}^{\mathrm{n}-\mathrm{r}}\right)=\underset{\mathrm{r}=0}{\sum ^{\mathrm{n}}}{\mathrm{a}}_{\mathrm{r}}{\mathrm{x}}^{\mathrm{n}+\mathrm{r}}+\underset{\mathrm{r}=0}{\sum ^{\mathrm{n}}}{\mathrm{a}}_{\mathrm{r}}{\mathrm{x}}^{\mathrm{n}-\mathrm{r}}$$$$\Rightarrow ({\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_2{\mathrm{x}}^2+\cdot \cdot \cdot +{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}})({\mathrm{a}}_0{\mathrm{x}}^{\mathrm{n}}+{\mathrm{a}}_1{\mathrm{x}}^{\mathrm{n}-1}+{\mathrm{a}}_2{\mathrm{x}}^{\mathrm{n}-2}+\cdot \cdot \cdot {\mathrm{a}}_{\mathrm{n}})$$$$=({\mathrm{a}}_0{\mathrm{x}}^{\mathrm{n}}+{\mathrm{a}}_1{\mathrm{x}}^{\mathrm{n}+1}+{\mathrm{a}}_2{\mathrm{x}}^{\mathrm{n}+2}+\cdot \cdot \cdot +{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}})+({\mathrm{a}}_0{\mathrm{x}}^{\mathrm{n}}+{\mathrm{a}}_1{\mathrm{x}}^{\mathrm{n}-1}+{\mathrm{a}}_2{\mathrm{x}}^{\mathrm{n}-2}+\cdot \cdot \cdot +{\mathrm{a}}_{\mathrm{n}})$$$$\mathrm{Comparing}\mathrm{coefs}\mathrm{we}\mathrm{get}:$$$${\mathrm{a}}_0{\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_{\mathrm{n}},{\mathrm{a}}_{\mathrm{n}}\ne 0\Rightarrow {\mathrm{a}}_0=1$$$${\mathrm{a}}_0{\mathrm{a}}_{\mathrm{n}-1}+{\mathrm{a}}_1{\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_{\mathrm{n}-1}\Rightarrow {\mathrm{a}}_1=0$$$${\mathrm{a}}_0{\mathrm{a}}_{\mathrm{n}-2}+{\mathrm{a}}_1{\mathrm{a}}_{\mathrm{n}-1}+{\mathrm{a}}_2{\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_{\mathrm{n}-2}\Rightarrow {\mathrm{a}}_2=0$$$$\mathrm{Continuing}\mathrm{this}\mathrm{process}\mathrm{we}\mathrm{get}{\mathrm{a}}_1={\mathrm{a}}_2={\mathrm{a}}_3=\cdot \cdot \cdot ={\mathrm{a}}_{\mathrm{n}-1}=0$$$$\mathrm{For}\mathrm{coefs}\mathrm{of}{\mathrm{x}}^{\mathrm{n}}\mathrm{we}\mathrm{have};$$$${\mathrm{a}}_0^2+{\mathrm{a}}_1^2+{\mathrm{a}}_2^2+\cdot \cdot \cdot +{\mathrm{a}}_{\mathrm{n}-1}^2+{\mathrm{a}}_{\mathrm{n}}^2={2\mathrm{a}}_0$$$$\Rightarrow {\mathrm{a}}_0^2+{\mathrm{a}}_{\mathrm{n}}^2={2\mathrm{a}}_0\Rightarrow {\mathrm{a}}_{\mathrm{n}}^2=1\Rightarrow {\mathrm{a}}_{\mathrm{n}}=\pm 1$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=1\pm {\mathrm{x}}^{\mathrm{n}},\mathrm{f}\left(2\right)=65\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=1+{\mathrm{x}}^{\mathrm{n}}$$$$\Rightarrow 1+2^{\mathrm{n}}=65\Rightarrow \mathrm{n}=6\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=1+{\mathrm{x}}^6$$$$\Rightarrow \mathrm{f}(-3)=1+3^6=730$$

Commented by Ar Brandon last updated on 17/Feb/21

What level ?

Commented by Dwaipayan Shikari last updated on 17/Feb/21

$$Whenyouhaveadded?$$

Commented by Dwaipayan Shikari last updated on 17/Feb/21

$$Ayearinyourlife?$$

Commented by Ar Brandon last updated on 17/Feb/21

Haha ! Let's get inbox, bro. You'll tell me what you're talking about

Commented by Dwaipayan Shikari last updated on 17/Feb/21

$$Whatisyourbirthday?Iknowithasgone$$

Commented by Ar Brandon last updated on 18/Feb/21

13/01

Commented by Ar Brandon last updated on 18/Feb/21

When is that ? ��

Commented by Dwaipayan Shikari last updated on 20/Feb/21

$$(\frac{5^2-1}{4}+\frac{1}{8}{\left(\frac{\sqrt{5^2-1}}{2.1!}\right)}^2+\frac{1}{12}{\left(\frac{1.3\sqrt{5^2-1}}{2^2.2!}\right)}^2+\frac{1}{16}{\left(\frac{1.3.5\sqrt{5^2-1}}{2^3.3!}\right)}^2+....){\left({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{sinx}{x}dx\right)}^2$$

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