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Question Number 218609 by MrGaster last updated on 13/Apr/25
Can y=x be expressed as them su of two periodic functions?
$$\mathrm{Can}\:{y}={x}\:\mathrm{be}\:\mathrm{expressed}\:\mathrm{as}\:\mathrm{them} \\ $$$$\mathrm{su}\:\mathrm{of}\:\mathrm{two}\:\mathrm{periodic}\:\mathrm{functions}? \\ $$
Answered by MrGaster last updated on 13/Apr/25
R∈x=f(x)+g(x) ∃T_(1 ) ,T2>0:f(x+T_1 )−f(x)∧g(x+T_(2 ) )=g(x) ∀x∈R ∀n∈Z^+ :x+nT_1 T_2 =f(x+nT_1 T_2 )+g(x+nT_1 T_2 )=f(x)+g(x)=x ⇒lim_(n→∞) ((x+nT_1 T_2 )/n)=lim_(n→∞) (x/n)⇒T_1 T_2 =0 ⊥(T_1 ,T_2 >0) ∴∄f,g periodic:f(x)+g(x)=x determinant (((┐(∃f,g∈C(R,R):∀x∈R,∃T_1 T_2 >0,f(x+T_1 )=f(x)∧g(x+T_2 )=g(x)∧f(x)+g(x)=x))))
$$\mathbb{R}\in{x}={f}\left({x}\right)+{g}\left({x}\right)\:\exists{T}_{\mathrm{1}\:} ,{T}\mathrm{2}>\mathrm{0}:{f}\left({x}+{T}_{\mathrm{1}} \right)−{f}\left({x}\right)\wedge{g}\left({x}+{T}_{\mathrm{2}\:} \right)={g}\left({x}\right)\:\forall{x}\in\mathbb{R} \\ $$$$\forall{n}\in\mathbb{Z}^{+} :{x}+{nT}_{\mathrm{1}} {T}_{\mathrm{2}} ={f}\left({x}+{nT}_{\mathrm{1}} {T}_{\mathrm{2}} \right)+{g}\left({x}+{nT}_{\mathrm{1}} {T}_{\mathrm{2}} \right)={f}\left({x}\right)+{g}\left({x}\right)={x} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{x}+{nT}_{\mathrm{1}} {T}_{\mathrm{2}} }{{n}}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{x}}{{n}}\Rightarrow{T}_{\mathrm{1}} {T}_{\mathrm{2}} =\mathrm{0}\:\bot\left({T}_{\mathrm{1}} ,{T}_{\mathrm{2}} >\mathrm{0}\right) \\ $$$$\therefore\nexists{f},{g}\:\mathrm{periodic}:{f}\left({x}\right)+{g}\left({x}\right)={x} \\ $$$$\begin{array}{|c|}{\urcorner\left(\exists{f},{g}\in{C}\left(\mathbb{R},\mathbb{R}\right):\forall{x}\in\mathbb{R},\exists{T}_{\mathrm{1}} {T}_{\mathrm{2}} >\mathrm{0},{f}\left({x}+{T}_{\mathrm{1}} \right)={f}\left({x}\right)\wedge{g}\left({x}+{T}_{\mathrm{2}} \right)={g}\left({x}\right)\wedge{f}\left({x}\right)+{g}\left({x}\right)={x}\right)}\\\hline\end{array} \\ $$
Commented by MrGaster last updated on 13/Apr/25
The axiom of choice isa equivlent to the existence of aHamel basis for any vectorp sace. If we assume the axiomof choice we can choosea Hamel basis A,then for any x∈R,there exists x=Σ_(a∈A) (x,a)a when (x,a)satisfies (x+b,a)=(x,a)+(b,a) when b∈A,b≠a,then(b,a)=0,and b is the period of f(x)=(x,a)Therefore we can express x as x(x,a)a+Σ_(b∈A−{a}) (x,b)b Thus any nthdegreei polynomal can be expresseds a the sum of n+1 periodicc funtions.
$$\mathrm{The}\:\mathrm{axiom}\:\mathrm{of}\:\mathrm{choice}\:\:\mathrm{isa} \\ $$$$\mathrm{equivlent}\:\mathrm{to}\:\mathrm{the}\:\mathrm{existence}\:\mathrm{of}\: \\ $$$$\mathrm{aHamel}\:\mathrm{basis}\:\mathrm{for}\:\mathrm{any}\:\mathrm{vectorp} \\ $$$$\mathrm{sace}.\:\mathrm{If}\:\mathrm{we}\:\mathrm{assume}\:\mathrm{the}\: \\ $$$$\mathrm{axiomof}\:\mathrm{choice}\:\mathrm{we}\:\mathrm{can}\: \\ $$$$\mathrm{choosea}\:\mathrm{Hamel}\:\mathrm{basis}\:{A},\mathrm{then}\:\mathrm{for}\:\mathrm{any} \\ $$$${x}\in\mathbb{R},\mathrm{there}\:\mathrm{exists} \\ $$$${x}=\underset{{a}\in{A}} {\sum}\left({x},{a}\right){a} \\ $$$$\mathrm{when}\:\left({x},{a}\right)\mathrm{satisfies}\:\left({x}+{b},{a}\right)=\left({x},{a}\right)+\left({b},{a}\right) \\ $$$$\mathrm{when}\:{b}\in{A},{b}\neq{a},\mathrm{then}\left({b},{a}\right)=\mathrm{0},\mathrm{and}\:{b}\:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{period}\:\mathrm{of}\:{f}\left({x}\right)=\left({x},{a}\right)\mathrm{Therefore}\:\mathrm{we}\:\mathrm{can}\:\mathrm{express}\:{x}\:\mathrm{as} \\ $$$${x}\left({x},{a}\right){a}+\underset{{b}\in{A}−\left\{{a}\right\}} {\sum}\left({x},{b}\right){b} \\ $$$$\mathrm{Thus}\:\mathrm{any}\:\mathrm{nthdegreei} \\ $$$$\mathrm{polynomal}\:\mathrm{can}\:\mathrm{be}\:\mathrm{expresseds} \\ $$$$\mathrm{a}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:{n}+\mathrm{1}\:\mathrm{periodicc} \\ $$$$\mathrm{funtions}. \\ $$
Commented by MrGaster last updated on 13/Apr/25
If axiom of choice is admitted, y=x can indeed be expressed as the sum of the periods of two discontinuous functions.

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