If $f''(x)=-f (x) $ , $g (x)=f'(x) $ and $F (x)=(f (x/2))^2+(g (x/2))^2$ and value of it at $x=5$ is $5$ then value of $F (10) $. From given conditions its known that $g'(x)=-f (x) $ . Then how to use the condition of $F (5)=5$
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$F$ is probably a constant function. – StubbornAtom Sep 03 '16 at 11:44
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Yes it is one of the solutions I also thought of it but how can we say that its the only solution from given conditions – Archis Welankar Sep 03 '16 at 11:48
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$F(x)=5 \forall x$ is the only solution from the conditions precisely because $F'(x)=0 \forall x$. – StubbornAtom Sep 03 '16 at 11:54
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I thought I had seen this before: http://math.stackexchange.com/q/427493/321264. (posed wrongly though) – StubbornAtom Sep 03 '16 at 12:02
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Consider $F (x)=(f (x/2))^2+(g (x/2))^2$ then $F'(x)=f(x/2)f'(x/2)+g(x/2)g'(x/2)$ But $g(x)=f'(x)$ and $g'(x)=f''(x)=-f(x)$ combining these we get $F'(x)=0$ and F is constant function.
sn24
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Minor pedantic note: a $1/2$ should appear when you use the chain rule. Of course this ends up not mattering since the result is zero. – User8128 Sep 03 '16 at 12:17
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From $f''(x) = -f(x)$, we can tell that $f$ is of the form $C\sin(x) + D\cos(x)$, which makes $g(x) = C\cos(x) - D\sin(x)$.
Inserting into $F$, we get $$ F(x) = (f(x/2))^2 + (g(x/2))^2 \\ = (C\sin(x/2) + D\cos(x/2))^2 + (C\cos(x/2) - D\sin(x/2))^2\\ = (C^2 + D^2)(\sin^2(x/2) + \cos^2(x/2)) = C^2 + D^2 $$ so $F(x)$ is constant, and therefore $F(10) = 5$.
Arthur
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