Putnam Problem of the Day

[http://math.harvard.edu/putnam(...) No.10466046]

[math]
\text{For any square matrix }A\text{, we can define }\sin A\text{ by the usual power series:}
\\
\qquad \qquad \qquad \qquad \quad \quad ~ ~
\sin A = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} A^{2n+1}.
\\
\text{Prove or disprove: there exists a }2 \times 2\text{ matrix }A\text{ with real entries such that}
\\
\qquad \qquad \qquad \qquad \qquad \qquad
\sin A = \left( \begin{array}{cc} 1 & 1996 \\ 0 & 1 \end{array} \right).
[/math]

[http://math.harvard.edu/putnam(...) No.10466047]

Previous Thread >>10462198

[Anonymous No.10466108]

Idk. Let's say this matrix is [(1, c1) (0, 1)] so that its nth power is [(1, n * c1), (0, 1)].
Now, [math]\sum{\frac{(-1)^n}{(2n+1)!} c_1 (2 n + 1)} = 1996[/math].

[math]c_1 \sum{\frac{(-1)^n}{(2n+1)!} x^{2n + 1} (2n + 1)} = c_1 x cos(x)[/math] (differentiating and mplying by x). At point x = 1 this is exactly the power series above. So, c1 = 1996 divided by cos(1).

Does any of this make sense?
t. software ((E)) major subhuman

[Anonymous No.10466155]

>>10466046
No. To see this let's assume such a matrix exists and show it cannot
[eqn] A(\sin A) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} A*A^{2n+1}= \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} A^{2n+1}*A=(\sin A)A[/eqn]
This implies that the matrices commute. What does this imply? Some routine calculations shows that
[eqn] A = \left( \begin{array}{cc} a & b \\ 0 & a \end{array} \right)[/eqn]
From some [math]a,b \in \mathbb{R}[/math]. To see this just compute the commutator for some 2 by 2 matrix and use the fact that they commute to constrain the matrix elements. Now, another routine calculation shows that
[eqn] A^n = \left( \begin{array}{cc} a^n & na^{n-1}b \\ 0 & a^n \end{array} \right)[/eqn]
A simple induction proof suffices to prove it.
We now have all that we need
[eqn] \sin A = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} A^{2n+1} = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \left( \begin{array}{cc} a & b \\ 0 & a \end{array} \right)^{2n+1}= \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \left( \begin{array}{cc} a^{2n+1} & (2n+1)a^{2n}b \\ 0 & a^{2n+1} \end{array} \right)= \left( \begin{array}{cc} \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}a^{2n+1} & \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}(2n+1)a^{2n}b \\ 0 & \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}a^{2n+1} \end{array} \right)= \left( \begin{array}{cc} sin(a) & cos(a)b \\ 0 & sin(a) \end{array} \right)[/eqn]
Now, using this we see that
[eqn]\left( \begin{array}{cc} sin(a) & cos(a)b \\ 0 & sin(a) \end{array} \right)=\left( \begin{array}{cc} 1 & 1996 \\ 0 & 1 \end{array} \right)[/eqn]
Modulo the periodicity of the trig formulas we require that [math]a=\frac{\pi}{2}[/math] so that the diagonal elements match. But this means all off diagonal elements must be zero since [math]cos(\frac{\pi}{2})=0[/math]
There are slight gaps in the proof but I would not dare deprive the reader the pleasure of filling them in them self.

[Anonymous No.10466413]

I can't fucking do it

[Anonymous No.10466532]

>>10466046
sin A can't be 1996 because sin returns a value in range [-1, 1]
Disproved.

[Anonymous No.10466670]

>>10466532
A is a matrix and nobody said A is real matrix

[Anonymous No.10466766]

>>10466670
>there exists a 2x2 matrix A with real entries
>no one said A is a real matrix

[Anonymous No.10466915]

>>10466155
extraordinarily based and very nicely formatted

[Anonymous No.10467103]

>>10466155
Yeah the first step was the least obvious to me. I figured out the rest quite fast. Showing what form A is in when it commutes is also a little tricky.

[Anonymous No.10467199]

>>10467103
The more problems you do the more you get a feel for these things. I never really had a great showing at the putnam (didn't train at all) but surprisingly enough I got way more into problem solving as a recreational activity than as "training" for the actual exam.
>>10466915
Thanks anon. You may want to check out the "official" solution posted online. I checked it after the fact and it's pretty clever in its own way.