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I am having a hard time coming up with a solution to this problem.

Suppose that $f$ is twice differentiable and that $f'' + f = 0$. If $f(0) = f'(0) = 0$, use Taylor's theorem to show that $f = 0$.

The definition of Taylor's theorem we were given uses the Lagrange remainder $f^{(n+1)}(c) \frac{(x-a)^{n+1}}{(n+1)!}$. Technically we don't have to use Taylor's theorem, but my professor said he doesn't know of a way to do it without it.

The obvious use of Taylor's theorem is that for all $x$ we have $f(x) = f''(c) \cdot x^2/2$ for some $|c| < |x|$. And then this means $f(x) = -f(c) \cdot x^2/2$. Then I don't know how to proceed. I can't think of a way except to show that $f(x) \ne 0$ leads to a contradiction. But what shall I contradict? Here are some of the promising ideas I've had that don't seem to lead anywhere: (suppose WLOG that $x>0$ in the following)

  • I can construct a sequence $(c_n)$ where $f(x) = (-1)^n f(c_n)\cdot x^{2n}/(2n)!$ since it is easy to show that $f = -f'' = f^{(4)} = \cdots$. So I have all these nonzero $f(c_n)$. But where will that take me?

  • How about another sequence? There is a $c_1$ such that $0 < c_1 < x$ and $f(c_1) = -f(x) \cdot 2/x^2 \ne 0$. Then there's a $c_2$ with $0<c_2<c_1$ and $f(c_2) = -f(c_1) \cdot 2/{c_1}^2 \ne 0$, and so on. This leads to another sequence $(c_n)$ with nonzero image. It converges (since it's strictly decreasing and bounded below by 0), but does anything useful happen there?

  • Since each of the $c_i$ are distinct, and $f(c_i)$ has opposite sign to $f(c_{i-1})$ for all $i$, with the IVT I can show that $f$ has infinitely many zeros in the interval $(0, x)$. I don't see how this can be manipulated to a contradiction though.

How can I proceed?

louiswins
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  • Wait, how can you show that $f=-f''$? I can see a way to show that $f=-f'+C$ and having $f(0)=f'(0)=0$ resolves $C=0$ in that case, but this means (to me) that $f=f''$... – abiessu Nov 05 '13 at 23:41
  • If $f''+f=0$, then $f=-f''$. – Martin Argerami Nov 05 '13 at 23:43
  • It looks like I made a dumb mistake in typing the question. The correct condition is that $f'' + f = 0$. – louiswins Nov 05 '13 at 23:43
  • Okay, that makes more sense, thank you! – abiessu Nov 05 '13 at 23:44
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    As far as doing it without Taylor's theorem, multiply by $f'$ and integrate, getting $f^2 + f'^2 = \text{const}$. The initial condition implies that the constant is $0$. But since squares are nonnegative, we must have $f\equiv 0$. – Sean Eberhard Nov 06 '13 at 17:18

4 Answers4

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From $f''=-f $ we get that $f $ is infinitely differentiable.

Let us first work on an interval $[-m,m]$. As $f$ is continuous, there exists $M>0$ with $|f(t)|\leq M$ for all $t\in[-m,m]$.

Now we write the degree-$n$ Taylor expansion: $$ f(x)=f(0)+f'(0)x+f''(0)\frac{x^2}2+\cdots+f^{(n-1)}(0)\frac{x^{n-1}}{(n-1)!}+f^{(n)}(c)\,\frac{x^n}{n!}, $$ where $c$ is between $0$ and $x$. The condition $f''=-f$, together with $f(0)=f'(0)=0$, guarantees that $f^{(k)}(0)=0$ for all $k$. Then $$ f(x)=\frac{f^{(n)}(c)}{n!}, $$ and so for $n $ even we have $|f^{(n)}|=|f| $, and then $$|f(x)|\leq\frac{M}{n!}.$$ As we can get this estimate for all even $n$, we conclude that $f(x)=0$ for all $x\in [-m,m]$. As the choice of $m$ was arbitrary, $f(x)=0$ for all $x$.

Martin Argerami
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Hint: So I think you've deduced from the given that the Taylor polynomial of degree $n$ at $0$ is $0$ for every $n$. Now deduce that on any interval $[0,a]$ we have a constant $M$ so that $|f^{(n)}(x)|\le M$ for all $x$ and all $n$. Now consider the remainder formula for the error, and let $n\to\infty$.

Ted Shifrin
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If $f$ is twice differentiable and $f+f''=0$, it follows that $f\in C ^\infty$ and $f^{(k)}(0)=0$. Choose $x\in \mathbb R$. For all $n\in \mathbb N$, expanding around zero, Taylor's theorem gives $$f(x)=\dfrac{f^{(n+1)}(tx)}{(n+1)!}x^{n+1}, \qquad t\in (0,1).$$ Let $M_i=\max _{u^2 \leq x^2} |f^{(i)}(u)|$ for $0\leq i \leq 3$, and $M=\max \{M_i\}$. Then:$$|f(x)|\leq M \frac{x^{n+1}}{(n+1)!}.$$ Letting $n\rightarrow\infty$, you get $|f(x)|\leq 0$.

pppqqq
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  • Thank you, I didn't even think to use that $f^{(n+1)}$ is bounded. Since the continuous image of a compact set is compact. – louiswins Nov 05 '13 at 23:57
  • I've made a minor modification in the way the bound $M$ is defined; previously it was $M=\max f^{(n)}$ and so we had a (a priori) $n$-dependent bound. – pppqqq Nov 06 '13 at 17:11
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Unless I am grossly mistaken, we don't need Taylor's theorem. This is a second order differential equation with constant coefficients. The solution is $c_1 \cos(x) + c_2 \sin(x)$. The initial conditions give that $c_1 = c_2 = 0$. Your Professor doesn't know of a way to solve this without Taylor's theorem?

Euler....IS_ALIVE
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    How do you prove that every solution is of this form? – Ted Shifrin Nov 05 '13 at 23:47
  • @TedShifrin Existence and Uniqueness theorem for ODE with constant coefficients. – Euler....IS_ALIVE Nov 05 '13 at 23:49
  • This is for a first-semester undergraduate Real Analysis course, so we're mostly just using the definitions of derivative and Taylor's theorem. So nothing from differential equations unless I can quickly prove it as a lemma. – louiswins Nov 05 '13 at 23:50
  • Ok well I was just going off the fact that you said you don't have to use Taylor's theorem. Usually, one takes Differential Equations before Real Analysis. – Euler....IS_ALIVE Nov 05 '13 at 23:51
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    How can one handle differential equations without real analysis at all? Differentiation and integration would be absolutely required for that, wouldn't they? – Daniel Fischer Nov 06 '13 at 00:01
  • @DanielFischer Not sure? The usual sequence is Calc I,II,III, Ode, Linear Algebra, Analysis. – Euler....IS_ALIVE Nov 06 '13 at 00:08
  • What would "Analysis" refer to in that listing? Where I come from, one starts (or at least used to) with calculus (which here is called Analysis) and linear algebra, and ODEs are usually treated in the third semester (Ana III). – Daniel Fischer Nov 06 '13 at 00:16
  • @DanielFischer Something on the level of baby Rudin. – Euler....IS_ALIVE Nov 06 '13 at 00:20
  • Okay, then what is calculus? I thought PMA was about calculus; continuity, differentiation, integration, that sort of stuff. – Daniel Fischer Nov 06 '13 at 00:23
  • @DanielFischer I'm not sure what PMA means. Yes, the good students will have the sequence integrated, but the normal students (ie, engineering, science etc) will go Calc 123, diff eq, linear algebra (sometimes concurrently), then two sequences of "Advanced Calculus" (also called analysis) if necessary for their program. – Euler....IS_ALIVE Nov 06 '13 at 00:40
  • PMA = Principles of Mathematical Analysis. If I interpret things correctly, affectionately known as "Baby Rudin". – Daniel Fischer Nov 06 '13 at 00:41
  • -1: the question is basically asking for a proof of the uniqueness of the solution of the second order constant coefficient initial value problem. Answering "because of the uniqueness of the solution of the second order constant coefficient initial value problem" does not contribute a lot. – Martin Argerami Mar 08 '17 at 05:17