Problem in solving a functional equation












1












$begingroup$


I have the following functional equation with me:



$f(x-y)=f(x)f(y)+g(x)g(y)$



$f(x)=cos x $ and $g(x)=sin x$ is an obvious solution but I have some trouble proving it. I have obtained the following simple results by simple substitutions



Firstly interchanging $x$ and $y$ tells me that the function is an even function.



Furthermore $x=y$ gives me



$f(0) = f(x)^2 + g(x)^2$.



Setting $y=0$ in the original equation gives me



$f(x) = f(x)f(0) + g(x)g(0)$.



Now from here I can clearly note that if $f(0)=0$ then both $f(x)$ and $g(x)$ are equal to zero. However I am unable to proceed from here. How do I proceed from here? My book just mentions that since $f(x)[1-f(0)]=g(x)g(0)$, $f(0) = 1$ and $g(0)=0$. How do I conclude this?






functional-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    The equation also admits the constant solutions $f(x) = c$ and $g(x) = sqrt{c-c^2}$ with $0le cle 1$.
    $endgroup$
    – Joey Zou
    Dec 1 '18 at 19:16
















1












$begingroup$


I have the following functional equation with me:



$f(x-y)=f(x)f(y)+g(x)g(y)$



$f(x)=cos x $ and $g(x)=sin x$ is an obvious solution but I have some trouble proving it. I have obtained the following simple results by simple substitutions



Firstly interchanging $x$ and $y$ tells me that the function is an even function.



Furthermore $x=y$ gives me



$f(0) = f(x)^2 + g(x)^2$.



Setting $y=0$ in the original equation gives me



$f(x) = f(x)f(0) + g(x)g(0)$.



Now from here I can clearly note that if $f(0)=0$ then both $f(x)$ and $g(x)$ are equal to zero. However I am unable to proceed from here. How do I proceed from here? My book just mentions that since $f(x)[1-f(0)]=g(x)g(0)$, $f(0) = 1$ and $g(0)=0$. How do I conclude this?






functional-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    The equation also admits the constant solutions $f(x) = c$ and $g(x) = sqrt{c-c^2}$ with $0le cle 1$.
    $endgroup$
    – Joey Zou
    Dec 1 '18 at 19:16














1












1








1





$begingroup$


I have the following functional equation with me:



$f(x-y)=f(x)f(y)+g(x)g(y)$



$f(x)=cos x $ and $g(x)=sin x$ is an obvious solution but I have some trouble proving it. I have obtained the following simple results by simple substitutions



Firstly interchanging $x$ and $y$ tells me that the function is an even function.



Furthermore $x=y$ gives me



$f(0) = f(x)^2 + g(x)^2$.



Setting $y=0$ in the original equation gives me



$f(x) = f(x)f(0) + g(x)g(0)$.



Now from here I can clearly note that if $f(0)=0$ then both $f(x)$ and $g(x)$ are equal to zero. However I am unable to proceed from here. How do I proceed from here? My book just mentions that since $f(x)[1-f(0)]=g(x)g(0)$, $f(0) = 1$ and $g(0)=0$. How do I conclude this?






functional-equations







share|cite|improve this question











$endgroup$




I have the following functional equation with me:



$f(x-y)=f(x)f(y)+g(x)g(y)$



$f(x)=cos x $ and $g(x)=sin x$ is an obvious solution but I have some trouble proving it. I have obtained the following simple results by simple substitutions



Firstly interchanging $x$ and $y$ tells me that the function is an even function.



Furthermore $x=y$ gives me



$f(0) = f(x)^2 + g(x)^2$.



Setting $y=0$ in the original equation gives me



$f(x) = f(x)f(0) + g(x)g(0)$.



Now from here I can clearly note that if $f(0)=0$ then both $f(x)$ and $g(x)$ are equal to zero. However I am unable to proceed from here. How do I proceed from here? My book just mentions that since $f(x)[1-f(0)]=g(x)g(0)$, $f(0) = 1$ and $g(0)=0$. How do I conclude this?






functional-equations




functional-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 21:36









the_fox

2,57711432




2,57711432










asked Dec 1 '18 at 6:30









saisanjeevsaisanjeev

947212




947212












  • $begingroup$
    The equation also admits the constant solutions $f(x) = c$ and $g(x) = sqrt{c-c^2}$ with $0le cle 1$.
    $endgroup$
    – Joey Zou
    Dec 1 '18 at 19:16


















  • $begingroup$
    The equation also admits the constant solutions $f(x) = c$ and $g(x) = sqrt{c-c^2}$ with $0le cle 1$.
    $endgroup$
    – Joey Zou
    Dec 1 '18 at 19:16
















$begingroup$
The equation also admits the constant solutions $f(x) = c$ and $g(x) = sqrt{c-c^2}$ with $0le cle 1$.
$endgroup$
– Joey Zou
Dec 1 '18 at 19:16




$begingroup$
The equation also admits the constant solutions $f(x) = c$ and $g(x) = sqrt{c-c^2}$ with $0le cle 1$.
$endgroup$
– Joey Zou
Dec 1 '18 at 19:16










1 Answer
1






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oldest

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2












$begingroup$

Assume that $g(0)$ is not zero. Then $g(x) = dfrac{1 - f(0)}{g(0)}f(x) = Cf(x)$, so $f(x-y) = (1 + C) f(x)f(y)$.



Setting $x = 2y$ gives $$f(y) = (1 + C) f(2y) f(y)$$ that is $f(2y)$ would be a constant (as long as $f(y)$ is not $0$).



So to get an interesting solution we must assume that $g(0) = 0$. $f(0) = 1$ follows.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the help. Could you please help me in solving the equation further to prove that cosx and sinx are the only possibilities for f(x) and g(x)?
    $endgroup$
    – saisanjeev
    Dec 2 '18 at 7:04











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1 Answer
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1 Answer
1






active

oldest

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active

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active

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2












$begingroup$

Assume that $g(0)$ is not zero. Then $g(x) = dfrac{1 - f(0)}{g(0)}f(x) = Cf(x)$, so $f(x-y) = (1 + C) f(x)f(y)$.



Setting $x = 2y$ gives $$f(y) = (1 + C) f(2y) f(y)$$ that is $f(2y)$ would be a constant (as long as $f(y)$ is not $0$).



So to get an interesting solution we must assume that $g(0) = 0$. $f(0) = 1$ follows.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the help. Could you please help me in solving the equation further to prove that cosx and sinx are the only possibilities for f(x) and g(x)?
    $endgroup$
    – saisanjeev
    Dec 2 '18 at 7:04
















2












$begingroup$

Assume that $g(0)$ is not zero. Then $g(x) = dfrac{1 - f(0)}{g(0)}f(x) = Cf(x)$, so $f(x-y) = (1 + C) f(x)f(y)$.



Setting $x = 2y$ gives $$f(y) = (1 + C) f(2y) f(y)$$ that is $f(2y)$ would be a constant (as long as $f(y)$ is not $0$).



So to get an interesting solution we must assume that $g(0) = 0$. $f(0) = 1$ follows.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the help. Could you please help me in solving the equation further to prove that cosx and sinx are the only possibilities for f(x) and g(x)?
    $endgroup$
    – saisanjeev
    Dec 2 '18 at 7:04














2












2








2





$begingroup$

Assume that $g(0)$ is not zero. Then $g(x) = dfrac{1 - f(0)}{g(0)}f(x) = Cf(x)$, so $f(x-y) = (1 + C) f(x)f(y)$.



Setting $x = 2y$ gives $$f(y) = (1 + C) f(2y) f(y)$$ that is $f(2y)$ would be a constant (as long as $f(y)$ is not $0$).



So to get an interesting solution we must assume that $g(0) = 0$. $f(0) = 1$ follows.






share|cite|improve this answer









$endgroup$



Assume that $g(0)$ is not zero. Then $g(x) = dfrac{1 - f(0)}{g(0)}f(x) = Cf(x)$, so $f(x-y) = (1 + C) f(x)f(y)$.



Setting $x = 2y$ gives $$f(y) = (1 + C) f(2y) f(y)$$ that is $f(2y)$ would be a constant (as long as $f(y)$ is not $0$).



So to get an interesting solution we must assume that $g(0) = 0$. $f(0) = 1$ follows.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 1 '18 at 20:34









user58697user58697

1,829512




1,829512












  • $begingroup$
    Thanks for the help. Could you please help me in solving the equation further to prove that cosx and sinx are the only possibilities for f(x) and g(x)?
    $endgroup$
    – saisanjeev
    Dec 2 '18 at 7:04


















  • $begingroup$
    Thanks for the help. Could you please help me in solving the equation further to prove that cosx and sinx are the only possibilities for f(x) and g(x)?
    $endgroup$
    – saisanjeev
    Dec 2 '18 at 7:04
















$begingroup$
Thanks for the help. Could you please help me in solving the equation further to prove that cosx and sinx are the only possibilities for f(x) and g(x)?
$endgroup$
– saisanjeev
Dec 2 '18 at 7:04




$begingroup$
Thanks for the help. Could you please help me in solving the equation further to prove that cosx and sinx are the only possibilities for f(x) and g(x)?
$endgroup$
– saisanjeev
Dec 2 '18 at 7:04


















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