For $-1\leq t_i\le 1$ and $1\le i\le n$, we have an $n\times n$ matrix $A$ such that $$A_{ij}=\exp(t_it_j)$$ Now, how can we use the Taylor expansion of $e^x $ to find a $\text{rank }10$ matrix $$B_{ij}=\sum_{k=1}^{10}a_{ki}b_{kj}$$ with $|A_{ij}-B_{ij}|\le 10^{-6}\:\forall i,j$.
My Thoughts:
First of all $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$
So this reminded me of the matrix exponential function for some reason. I don't know how much this will help for this problem though.
It's not really the matrix exponential because you're doing things entrywise rather than using matrix multiplication.
You want to try $$B_{ij} = \sum_{m=0}^9 (t_i t_j)^m/m! = \sum_{k=1}^{10} a_{ki} b_{kj}$$ where $a_{ki} = t_i^{k-1}/(k-1)!$ and $b_{kj} = t_j^{k-1}$. Note that for each $k$ the matrix with entries $a_{ki} b_{kj}$ has rank at most $1$, so $B$ has rank at most $10$. It's incorrect to say "rank $10$", since that's clearly impossible if $n < 10$.
For $-1 < t < 1$, by Taylor's theorem with Lagrange remainder $$\left|\exp(t) - \sum_{k=0}^9 \dfrac{t^k}{k!} \right| \le \dfrac{e}{10!} < 10^{-6}$$ so that $|A_{ij} - B_{ij}| < 10^{-6}$.