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Question Number 199620 by Mingma last updated on 06/Nov/23

Answered by witcher3 last updated on 06/Nov/23

$$\mathrm{f}\left(4\right)=\mathrm{f}(2+2)=2\mathrm{f}\left(2\right)+4=10\Rightarrow \mathrm{f}\left(2\right)=3$$$$\mathrm{f}\left(2\right)=2\mathrm{f}\left(1\right)+1=3\Rightarrow \mathrm{f}\left(1\right)=1,\mathrm{f}\left(0\right)=0$$$$\mathrm{f}(\mathrm{x}+1)-\mathrm{f}\left(\mathrm{x}\right)=1+\mathrm{x}$$$$\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}-1}}(\mathrm{f}(\mathrm{k}+1)-\mathrm{f}\left(\mathrm{k}\right))=\frac{\mathrm{n}(\mathrm{n}+1)}{2}$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}(1+\mathrm{x})}{2}\mathrm{solution}\mathrm{if}\mathrm{f}\mathrm{is}\mathrm{continus}\mathrm{we}\mathrm{can}\mathrm{show}\mathrm{that}\mathrm{is}$$$$\mathrm{unique}$$$$\mathrm{f}\in {\mathrm{C}}_0\mathrm{solution}\mathrm{suppose}\mathrm{\exists }\mathrm{g}\in {\mathrm{C}}_0\mathrm{also}\mathrm{solution}$$$$\mathrm{g}\left(4\right)=\mathrm{f}\left(4\right)=10$$$$\mathrm{h}=\mathrm{g}-\mathrm{f}\Rightarrow \mathrm{h}(\mathrm{x}+\mathrm{y})=\mathrm{h}\left(\mathrm{x}\right)+\mathrm{h}\left(\mathrm{y}\right)$$$$\mathrm{h}\left(0\right)=\mathrm{h}\left(1\right)=\mathrm{h}\left(2\right)=\mathrm{h}\left(4\right)=0,\mathrm{\forall }\mathrm{n}\in \mathbb{N}\mathrm{h}\left(\mathrm{n}\right)=0\mathrm{because}$$$$‘‘\mathrm{f}\left(\mathrm{n}\right)=\mathrm{g}\left(\mathrm{n}\right){}^{″}$$$$\mathrm{h}(\mathrm{x}+\mathrm{y})=\mathrm{h}\left(\mathrm{x}\right)+\mathrm{h}\left(\mathrm{y}\right)$$$$\Rightarrow \mathrm{h}\left(\mathrm{n}\right)=\mathrm{nh}\left(1\right)=0$$$$\mathrm{h}\left(\frac{\mathrm{p}}{\mathrm{q}}\right)=\frac{\mathrm{p}}{\mathrm{q}}\mathrm{h}\left(1\right)=0\Rightarrow \mathrm{\forall }\mathrm{x}\in \mathrm{IQ}\mathrm{h}\left(\mathrm{x}\right)=0$$$$\mathrm{by}\mathrm{densite}\mathrm{Q}\mathrm{over}\mathrm{IR}\mathrm{and}\mathrm{contiuity}\mathrm{of}\mathrm{h}$$$$\mathrm{\forall }\mathrm{x}\in \mathbb{R}\mathrm{\exists }{\mathrm{x}}_{\mathrm{n}}\in \mathrm{IQ}{\mathrm{x}}_{\mathrm{n}}\to \mathrm{x}$$$$\mathrm{h}\left({\mathrm{x}}_{\mathrm{n}}\right)=0\Rightarrow \underset{\mathrm{n}\to \mathrm{\infty }}{\lim h}\left({\mathrm{x}}_{\mathrm{n}}\right)=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim 0}=0$$$$\Rightarrow \mathrm{h}\left(\mathrm{x}\right)=0,\mathrm{\forall }\mathrm{x}\in \mathbb{R}$$$$\Rightarrow \mathrm{g}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)\Rightarrow \mathrm{f}\mathrm{is}\mathrm{unique}$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}(1+\mathrm{x})}{2}$$$$$$

Commented by Mingma last updated on 06/Nov/23

Nice one!

Commented by witcher3 last updated on 06/Nov/23

$$\mathrm{withe}\mathrm{Pleasur}$$

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