The exercise asks me to prove:
$$|e^{x+5y}-P_1(x,y)|< \frac{3}{2}(x+5y)^2$$
when $x+5y<1$
I don't understand what's the exercise suggesting but I tried this:
$e^{x+5y} - P_1(x,y)$ is just the error for this polynomial, which is:
$$\frac{1}{2}[\frac{∂^2 }{∂x^2}f(x^*,y^*)(x-0)^2+\frac{∂^2 }{∂xy}f(x^*,y^*)(x-0)(y-0)+\frac{∂^2 }{∂y^2}f(x^*,y^*)(y-0)^2] = \\ \frac{1}{2}[e^{x^*+5y^*}x^2+5e^{x^*+5y^*}xy+25e^{x^*+5y^*}y^2]$$
but how to bound it? why the contraint $x+5y<1$??
UPDATE:
I feel there's a way of ending with the LHS being less than $\frac{3}{2}(x^2+5xy+25y^2)$. I kinda can do it if I use $|a+b+c|\le |a|+|b|+|c|$ and if I assume somehow $e^{x^*+5y^*}\le 1$ but it's not exactly what's asked.
.....