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Question Number 4102 by prakash jain last updated on 28/Dec/15

Find all function f:R→R  such that  f(x+1)=xf(x)  f (0)=1   f(x)≠Γ(x)

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{function}\:{f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$$\mathrm{such}\:\mathrm{that} \\ $$$${f}\left({x}+\mathrm{1}\right)={xf}\left({x}\right) \\ $$$${f}\:\left(\mathrm{0}\right)=\mathrm{1}\: \\ $$$${f}\left({x}\right)\neq\Gamma\left({x}\right) \\ $$

Commented by prakash jain last updated on 28/Dec/15

Γ(x) is gamma function.

$$\Gamma\left({x}\right)\:\mathrm{is}\:\mathrm{gamma}\:\mathrm{function}. \\ $$

Commented by 123456 last updated on 28/Dec/15

i think that if function need to be continuous  then Γ is the only function

$$\mathrm{i}\:\mathrm{think}\:\mathrm{that}\:\mathrm{if}\:\mathrm{function}\:\mathrm{need}\:\mathrm{to}\:\mathrm{be}\:\mathrm{continuous} \\ $$$$\mathrm{then}\:\Gamma\:\mathrm{is}\:\mathrm{the}\:\mathrm{only}\:\mathrm{function} \\ $$

Commented by Filup last updated on 29/Dec/15

x=0  f(0+1)=0f(0)  ∴f(1)=0  Γ(1)=1  ∴f(x)≠Γ(x)???

$${x}=\mathrm{0} \\ $$$${f}\left(\mathrm{0}+\mathrm{1}\right)=\mathrm{0}{f}\left(\mathrm{0}\right) \\ $$$$\therefore{f}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\Gamma\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$\therefore{f}\left({x}\right)\neq\Gamma\left({x}\right)??? \\ $$

Commented by 123456 last updated on 29/Dec/15

f(1)=1 for gamma function

$${f}\left(\mathrm{1}\right)=\mathrm{1}\:\mathrm{for}\:\mathrm{gamma}\:\mathrm{function} \\ $$

Commented by Filup last updated on 29/Dec/15

but if f(x)=Γ(x)  x=1  f(1)=0  Γ(1)=1  ∴f(x)≠Γ(x)  I am confused

$$\mathrm{but}\:\mathrm{if}\:{f}\left({x}\right)=\Gamma\left({x}\right) \\ $$$${x}=\mathrm{1} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\Gamma\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$\therefore{f}\left({x}\right)\neq\Gamma\left({x}\right) \\ $$$$\mathrm{I}\:{am}\:{confused} \\ $$

Commented by prakash jain last updated on 29/Dec/15

Find f meeting th conditions given above.  Functions other than Γ function.

$$\mathrm{Find}\:{f}\:\mathrm{meeting}\:\mathrm{th}\:\mathrm{conditions}\:\mathrm{given}\:\mathrm{above}. \\ $$$$\mathrm{Functions}\:\mathrm{other}\:\mathrm{than}\:\Gamma\:\mathrm{function}. \\ $$

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