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I'm struggling with this former Putnam Exam problem:

Suppose $f$ and $g$ are nonconstant, differentiable, real-valued functions on $R$. Furthermore, suppose that for each pair of real numbers $x$ and $y$, $f(x + y) = f(x)f(y) - g(x)g(y)$ and $g(x + y) = f(x)g(y) + g(x)f(y)$. If $f'(0) = 0$, prove that $(f(x))^2 + (g(x))^2 = 1$ for all $x$.

Right. So obviously, $f(x) = \cos x$ and $g(x) = \sin x$ satisfy the conditions and also the conclusion of the problem. But are these the unique such functions, and if so, how to prove it? And if not, then how to prove the conclusion otherwise?

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Actually, $\cos(ax)$ and $\sin(ax)$, for arbitrary $a$, would be the general solutions. –  J. M. Aug 26 '10 at 11:59
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One key observation would be to show that $f(2x)^2+g(2x)^2=\left(f(x)^2+g(x)^2\right)^2$ –  J. M. Aug 26 '10 at 12:02
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Did you try to introduce f+ig? –  Pierre-Yves Gaillard Aug 26 '10 at 12:19
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Putnam solutions are available online; can you post the exam year and problem number? Following Pierre-Yves Galliard's suggestion: the formulas say that E=f+ig satisfies E(x+y)=E(x)E(y). Hence E(0)=0 or 1. Nonconstant implies E(0)=1. Differentiability should imply E(t)=exp((A+Bi)t) for real A,B, differentiation at 0 gives E'(0)=A+Bi so that A=0. So from this point of view the problem is to show that differentiable solutions are exponential functions. –  T.. Aug 26 '10 at 17:53
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Several published solutions are at: books.google.com/books?id=AA-lOA1nPDcC&pg=PA142 –  T.. Aug 26 '10 at 19:59
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3 Answers

up vote 10 down vote accepted

Let $u(x) = f(x)^2 + g(x)^2$. It should be easy to show that $u(x+y) = u(x)u(y)$, thus $u$ must be an exponential function $u(x) = e^{Bx}$.

Now, as $f(2x) = f(x)^2 - g(x)^2$, we should get $0 = g(0) g'(0)$. Comparing this with $u'(0)$ we should get $B = 0$, thus $f(x)^2 + g(x)^2 = 1$.

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It's important to remark here that the second step only works under the differentiability hypothesis. –  Qiaochu Yuan Aug 26 '10 at 15:47
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And the first step also? (At least, it uses some degree of regularity of $u$.) But that's okay, of course: the differentiability is given! –  Matt E Aug 27 '10 at 4:43
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How to fill in necessary steps and conditions to make this proof rigorous is left as an exercise for readers. –  KennyTM Aug 27 '10 at 6:14
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Here is a different way.

We have

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

Differentiate wrt $y$

$f'(x+y) = f(x)f'(y) - g(x) g'(y)$ , put $y = 0$.

$f'(x) = -g(x) g'(0)$

Similarly we get

$g'(x) = f(x) g'(0)$

Thus $f(x)f'(x) + g(x)g'(x) = 0$

Thus the function $f^{2}(x) + g^{2}(x)$ is constant, as it's derivative is zero.

Now

$f(0) = f^2(0) - g^{2}(0)$

and

$g(0) = 2f(0)g(0)$

Squaring and adding both we get

$f^{2}(0) + g^{2}(0) = (f^{2}(0) + g^{2}(0))^2$

Now if $f^{2}(0) + g^{2}(0) = 0$ then because $f^{2}(x) + g^{2}(x)$ is a constant, we get $f(x) = 0$ which implies $f$ is constant.

Thus $f^{2}(0) + g^{2}(0) = 1$ and hence $f^{2}(x) + g^{2}(x) = 1$

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Since KennyTM posted a very elegant proof I will post an uncreative one that requires no thought to produce.

First write out all the obvious identities that come from the information told:

  • $f(x + y) = f(x)f(y) - g(x)g(y)$

  • $f(x) = f(x)f(0) - g(x)g(0)$

  • $f(0) = f(0)f(0) - g(0)g(0)$

  • $g(x + y) = f(x)g(y) + g(x)f(y)$

  • $g(x) = f(x)g(0) + g(x)f(0)$

  • $g(0) = f(0)g(0) + g(0)f(0)$

  • $Df(x + c) = Df(x)f(c) - Dg(x)g(c)$

  • $Dg(x + c) = Df(x)g(c) + Dg(x)f(c)$

  • $Df(0) = 0 = Df(0)f(0) - Dg(0)g(0) = -Dg(0)g(0)$

  • $Dg(0) = Df(0)g(0) + Dg(0)f(0) = Dg(0)f(0)$

The $Df(0)$ equation tells us that $Dg(0) = 0 \lor g(0) = 0$ Since we don't want $Dg(0) = 0$ lets discharge that automatically now by proving it contradictory: It would imply that $Dg(0 + c) = Df(0)g(c) + Dg(0)f(c) = 0$ so $g(x)$ would be a constant but that is not allowed.

We have seen that $g(0) = 0$ and now we update the identifies with this information, and that tells us is that $f(0) = 1$. We cannot determine the value of $Dg(0)$ so just call it some constant $k$. Now it is easy to recognize that we have a simple ODE problem that could be solved symbolically by a calculator or by taking eigenvalues/vectors of the matrix that defines the system. Giving the solutions $\sin(kx)$ & $\cos(kx)$.

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