Inspired by Adrian Pen

Calculus Level 5

$2\sum_{k=1}^{\infty}k^2f^{(k)}(x)=3f(x),\hspace{10mm}f(0)=2016$

Find the function $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfies the differential equation above for all real $x$ . Return $f'(0)$ as your answer.

If you come to the conclusion that no such function exists (or there is more than one), enter 666.

Clarification : $f^{(k)}(x)$ denotes the $k^\text{th}$ derivative of $f(x)$ .

Inspiration .

The answer is 672.

2 solutions

Andreas Wendler
May 29, 2016

The well-known entry $f(x)=e^{\lambda x}$ leads to the equation $\sum_{k=1}^{\infty}k^{2}\lambda^{k}=\frac{3}{2}$ Solution is $\lambda=\frac{1}{3}$ and we get $f(x)=2016\cdot e^{\frac{x}{3}}$ as well as $f'(0)=\frac{2016}{3}=\boxed{672}$

Yes, $f(x)=e^{\lambda x}$ is a good Ansatz (+1). How do we know that there aren't any other solutions of the form $f(x)=x^k e^{\lambda x}$ ? Also, the equation $\sum_{k=1}^{\infty}k^2\lambda^k=\frac{3}{2}$ has three complex solutions $\lambda$ ; why are you rejecting the other two, besides $\lambda=\frac{1}{3}$ ?

Otto Bretscher - 5 years ago

The characteristic equation $2\lambda-\lambda(1-\lambda)-\frac{3}{2}(1-\lambda)^{3}=0$ which follows after analyzing the infinite rows only delivers a real to real function for $\lambda_{1}=\frac{1}{3}$ . Of course there are additionally the solutions $\lambda_{2,3}=1^{+}_{-}\sqrt{2}i$ But the range of the according functions contains complex numbers! So they must be rejected and the solution for f(x) is unique.

Andreas Wendler - 5 years ago

The fact that the roots $1\pm \sqrt{2}i$ are non-real is not a sufficient reason to reject them. Think about $f''(x)+f(x)=0$ : non-real solutions give rise to real solutions thanks to Euler's marvelous formula ;)

Otto Bretscher - 5 years ago

@Otto Bretscher – Okay, so let us try to find a second solution $f(x)=2016\cdot e^{x}\cdot cos(\sqrt{2}x)$ constructed as a linear combination regarding the two complex values for $\lambda$ with valid initial condition.

But here we realize that the infinite sum doesn't converge. That's why we must reject the complex solutions for $\lambda$ .

Andreas Wendler - 5 years ago

@Andreas Wendler – Genau! Now your solution is just perfect

Otto Bretscher - 5 years ago

@Otto Bretscher – We need also to show that no other form is possible.

Abdelhamid Saadi - 5 years ago

@Abdelhamid Saadi – We needn't because another solution doesn't exist. This follows from the roots of the characteristic polynomial for $\lambda$ considering the initial condition!

Andreas Wendler - 5 years ago

Julian Poon
Jun 11, 2016

Let $f(x)=y$

We want to solve the equation

$\frac{3}{2}y=\sum _{n=1}^{ \infty}n^2y^{(n)} --(1)$

Differentiating both sides we get

$\frac{3}{2}y'=\sum _{n=1}^{\infty }n^2y^{(n+1)} --(2)$

Taking (1)-(2) we get:

$\frac{3}{2}\left(y-y'\right)=\sum _{n=1}^{ \infty}\left(2n-1\right)y^{(n)} --(3)$

Now doing the same thing ((3) - (3)') we get:

$\frac{3}{2}\left(y-2y'+y''\right)=2\sum _{n=1}^{ \infty}y^{(n)} - y'=2S - y'$

Rearranging:

$S=\frac{3}{4}\left(y-\frac{4}{3}y'+y''\right) --(4)$

Notice that

$S=(S+y)'=S'+y'$

Substituting that into (4) gives the differential equation:

$3y-11y'+7y''-3y'''=0$

Solving gives:

$y=c_3e^{\frac{x}{3}}+c_2e^x\sin \left(\sqrt{2}x\right)+c_1e^x\cos \left(\sqrt{2}x\right)$

It is then easy to show that $c_2$ and $c_1$ has to be 0 or the sum $\displaystyle \sum _{n=1}^{ \infty}n^2y^{(n)}$ diverges.

The rest follows suit.

Inspired by Adrian Pen

The answer is 672.

2 solutions

0 pending reports