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Question-199620

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Question Number 199620 by Mingma last updated on 06/Nov/23
Answered by AST last updated on 06/Nov/23
f(x)=(x^2 /2)+cx;f(4)=10⇒c=(1/2)⇒f(x)=((x(x+1))/2)
$${f}\left({x}\right)=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{cx};{f}\left(\mathrm{4}\right)=\mathrm{10}\Rightarrow{c}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{f}\left({x}\right)=\frac{{x}\left({x}+\mathrm{1}\right)}{\mathrm{2}} \\ $$
Commented by Mingma last updated on 07/Nov/23
Nice one, sir!
Answered by witcher3 last updated on 06/Nov/23
f(4)=f(2+2)=2f(2)+4=10⇒f(2)=3 f(2)=2f(1)+1=3⇒f(1)=1,f(0)=0 f(x+1)−f(x)=1+x Σ_(k=0) ^(n−1) (f(k+1)−f(k))=((n(n+1))/2) f(x)=((x(1+x))/2) solution if f is continus we can show that is unique f∈C_0 solution suppose ∃g∈C_0 also solution g(4)=f(4)=10 h=g−f ⇒h(x+y)=h(x)+h(y) h(0)=h(1)=h(2)=h(4)=0,∀n∈N h(n)=0 because “f(n)=g(n) ” h(x+y)=h(x)+h(y) ⇒h(n)=nh(1)=0 h((p/q))=(p/q)h(1)=0⇒∀x∈IQ h(x)=0 by densite Q over IR and contiuity of h ∀x∈R ∃x_n ∈IQ x_n →x h(x_n )=0⇒lim_(n→∞) h(x_n )=lim_(n→∞) 0=0 ⇒h(x)=0,∀x∈R ⇒g(x)=f(x) ⇒ f is unique f(x)=((x(1+x))/2)
$$\mathrm{f}\left(\mathrm{4}\right)=\mathrm{f}\left(\mathrm{2}+\mathrm{2}\right)=\mathrm{2f}\left(\mathrm{2}\right)+\mathrm{4}=\mathrm{10}\Rightarrow\mathrm{f}\left(\mathrm{2}\right)=\mathrm{3} \\ $$$$\mathrm{f}\left(\mathrm{2}\right)=\mathrm{2f}\left(\mathrm{1}\right)+\mathrm{1}=\mathrm{3}\Rightarrow\mathrm{f}\left(\mathrm{1}\right)=\mathrm{1},\mathrm{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\mathrm{f}\left(\mathrm{x}+\mathrm{1}\right)−\mathrm{f}\left(\mathrm{x}\right)=\mathrm{1}+\mathrm{x} \\ $$$$\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\left(\mathrm{f}\left(\mathrm{k}+\mathrm{1}\right)−\mathrm{f}\left(\mathrm{k}\right)\right)=\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{2}}\:\:\mathrm{solution}\:\mathrm{if}\:\mathrm{f}\:\mathrm{is}\:\mathrm{continus}\:\mathrm{we}\:\mathrm{can}\:\mathrm{show}\:\mathrm{that}\:\mathrm{is} \\ $$$$\mathrm{unique} \\ $$$$\mathrm{f}\in\mathrm{C}_{\mathrm{0}} \mathrm{solution}\:\mathrm{suppose}\:\exists\mathrm{g}\in\mathrm{C}_{\mathrm{0}} \:\mathrm{also}\:\mathrm{solution} \\ $$$$\mathrm{g}\left(\mathrm{4}\right)=\mathrm{f}\left(\mathrm{4}\right)=\mathrm{10} \\ $$$$\:\mathrm{h}=\mathrm{g}−\mathrm{f}\:\Rightarrow\mathrm{h}\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{h}\left(\mathrm{x}\right)+\mathrm{h}\left(\mathrm{y}\right) \\ $$$$\mathrm{h}\left(\mathrm{0}\right)=\mathrm{h}\left(\mathrm{1}\right)=\mathrm{h}\left(\mathrm{2}\right)=\mathrm{h}\left(\mathrm{4}\right)=\mathrm{0},\forall\mathrm{n}\in\mathbb{N}\:\mathrm{h}\left(\mathrm{n}\right)=\mathrm{0}\:\mathrm{because} \\ $$$$“\mathrm{f}\left(\mathrm{n}\right)=\mathrm{g}\left(\mathrm{n}\right)\:'' \\ $$$$\mathrm{h}\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{h}\left(\mathrm{x}\right)+\mathrm{h}\left(\mathrm{y}\right) \\ $$$$\Rightarrow\mathrm{h}\left(\mathrm{n}\right)=\mathrm{nh}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{h}\left(\frac{\mathrm{p}}{\mathrm{q}}\right)=\frac{\mathrm{p}}{\mathrm{q}}\mathrm{h}\left(\mathrm{1}\right)=\mathrm{0}\Rightarrow\forall\mathrm{x}\in\mathrm{IQ}\:\mathrm{h}\left(\mathrm{x}\right)=\mathrm{0} \\ $$$$\mathrm{by}\:\mathrm{densite}\:\mathrm{Q}\:\mathrm{over}\:\mathrm{IR}\:\mathrm{and}\:\mathrm{contiuity}\:\mathrm{of}\:\mathrm{h} \\ $$$$\forall\mathrm{x}\in\mathbb{R}\:\exists\mathrm{x}_{\mathrm{n}} \in\mathrm{IQ}\:\mathrm{x}_{\mathrm{n}} \rightarrow\mathrm{x} \\ $$$$\mathrm{h}\left(\mathrm{x}_{\mathrm{n}} \right)=\mathrm{0}\Rightarrow\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}h}\left(\mathrm{x}_{\mathrm{n}} \right)=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}0}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{h}\left(\mathrm{x}\right)=\mathrm{0},\forall\mathrm{x}\in\mathbb{R} \\ $$$$\Rightarrow\mathrm{g}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)\:\Rightarrow\:\mathrm{f}\:\mathrm{is}\:\mathrm{unique} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{2}} \\ $$$$ \\ $$
Commented by Mingma last updated on 06/Nov/23
Nice one!
Commented by witcher3 last updated on 06/Nov/23
withe Pleasur
$$\mathrm{withe}\:\mathrm{Pleasur} \\ $$

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