Question Number 3789 by 123456 last updated on 21/Dec/15
$${f}\left({nx}\right)={f}\left({x}+{n}\right)−{f}\left({x}\right) \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}\left({x}\right)=? \\ $$
Commented by Rasheed Soomro last updated on 21/Dec/15
$${f}\left({nx}\right)={f}\left({x}+{n}\right)−{f}\left({x}\right) \\ $$$$−−−− \\ $$$${x}=\mathrm{0} \\ $$$${f}\left(\mathrm{0}\right)={f}\left({n}\right)−{f}\left(\mathrm{0}\right) \\ $$$${f}\left({n}\right)=\mathrm{0} \\ $$$${n}\:\:{is}\:\:{an}\:\:{arbitrary}\:{constant} \\ $$$${So} \\ $$$${f}\left({x}\right)=\mathrm{0}\:\:\:\:??? \\ $$
Commented by Rasheed Soomro last updated on 21/Dec/15
$${f}\left({nx}\right)={f}\left({x}+{n}\right)−{f}\left({x}\right) \\ $$$${n}=\mathrm{1} \\ $$$${f}\left({x}\right)={f}\left(\mathrm{1}+{n}\right)−{f}\left({x}\right) \\ $$$${f}\left({x}\right)=\frac{{f}\left(\mathrm{1}+{x}\right)}{\mathrm{2}} \\ $$$${x}=\mathrm{0}\:{in}\:{above} \\ $$$${f}\left(\mathrm{0}\right)=\frac{{f}\left(\mathrm{1}+\mathrm{0}\right)}{\mathrm{2}}=\mathrm{0} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$${f}\left(\mathrm{1}\right)=\frac{{f}\left(\mathrm{2}\right)}{\mathrm{2}}=\mathrm{0} \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{0} \\ $$$${f}\left(\mathrm{3}\right)=\mathrm{0} \\ $$$$\vdots \\ $$$${f}\left({x}\right)=\mathrm{0} \\ $$
Commented by 123456 last updated on 21/Dec/15
$${f}\left({x}\right)=\frac{{f}\left({x}+\mathrm{1}\right)}{\mathrm{2}} \\ $$
Commented by prakash jain last updated on 21/Dec/15
$${n}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{treated}\:\mathrm{as}\:\mathrm{variable}\:\mathrm{since} \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{for}\:\mathrm{all}\:{x}. \\ $$$${for}\:{example} \\ $$$${g}\left({x}\right)={x}−{n} \\ $$$${g}\left({n}\right)=\mathrm{0}\:{but}\:{g}\left({x}\right)\neq\mathrm{0} \\ $$
Commented by Rasheed Soomro last updated on 21/Dec/15
$$\mathcal{T}{hank}\mathcal{S}! \\ $$
Commented by Rasheed Soomro last updated on 21/Dec/15
$${for}\:{x}=\mathrm{0}\:\:,\:\:{f}\left({nx}\right)={f}\left({x}+{n}\right)−{f}\left({x}\right)\Rightarrow{f}\left({n}\right)=\mathrm{0}\:\: \\ $$$$\mathcal{W}{hat}\:{is}\:{meant}\:{by}\:\:{definition}\:\:{f}\left({n}\right)=\mathrm{0} \\ $$$${Couldn}'{t}\:{we}\:{use}\:{this}\:{definition}\:{for} \\ $$$${n}=\mathrm{1},{n}=\mathrm{4},{n}={anyvalue} \\ $$$${Couldn}'{t}\:{we}\:{say}\:\:{f}\left(\mathrm{6}\right)=\mathrm{0},\:{f}\left({anyvalue}\right)=\mathrm{0} \\ $$$${If}\:\:{yes}\:{then}\:\:{n}\:{is}\:{just}\:{like}\:{a}\:{variable}? \\ $$
Answered by Rasheed Soomro last updated on 21/Dec/15
![f(nx)=f(x+n)−f(x)...................(i) f(0)=0 f(x)=? −−−−−−−− n=1 in (i) f(x)=f(x+1)−f(x)⇒2f(x)=f(x+1) x→x−1 in above f(x)=2f(x−1) f(0)=0 [given] f(1)=2f(0)=2(0)=0 f(2)=2f(1)=0 f(3)=2f(2)=0 f(x)=0 This can be proved by mathematical induction.](https://www.tinkutara.com/question/Q3806.png)
$${f}\left({nx}\right)={f}\left({x}+{n}\right)−{f}\left({x}\right)……………….\left({i}\right) \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}\left({x}\right)=? \\ $$$$−−−−−−−− \\ $$$${n}=\mathrm{1}\:\:{in}\:\:\:\left({i}\right) \\ $$$${f}\left({x}\right)={f}\left({x}+\mathrm{1}\right)−{f}\left({x}\right)\Rightarrow\mathrm{2}{f}\left({x}\right)={f}\left({x}+\mathrm{1}\right) \\ $$$${x}\rightarrow{x}−\mathrm{1}\:{in}\:{above} \\ $$$${f}\left({x}\right)=\mathrm{2}{f}\left({x}−\mathrm{1}\right) \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0}\:\:\left[{given}\right] \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{2}{f}\left(\mathrm{0}\right)=\mathrm{2}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{2}{f}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$${f}\left(\mathrm{3}\right)=\mathrm{2}{f}\left(\mathrm{2}\right)=\mathrm{0} \\ $$$$ \\ $$$${f}\left({x}\right)=\mathrm{0} \\ $$$${This}\:{can}\:{be}\:{proved}\:{by}\:{mathematical}\:{induction}. \\ $$