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For every function which is linear with respect to x and y like f(x,y)=xy, the derivative is Ta(h1,h2)=f(a1,h2)+f(h1,a2) which satisfies $\lim_{h \to 0}\frac{\parallel f(x)-f(a)-T_a(h)\parallel}{\parallel h \parallel}=0$. this uses the fact that there exists some positive constant C such that...

Can you calculate the height from point C ?

When $f(x,y)=xy$ then if $a=({a}_{1},{a}_{2})$ and $h=({h}_{1},{h}_{2})$, $f(a+h)=f({a}_{1}+{h}_{1},{a}_{2}+{h}_{2})=({a}_{1}+{h}_{1})({a}_{2}+{h}_{2})$ $={a}_{1}{a}_{2}+{a}_{1}{h}_{2}+{a}_{2}{h}_{1}+{h}_{1}{h}_{2}$ $=f({a}_{1},{a}_{2})+{T}_{a}({h}_{1},{h}_{2})+{h}_{1}{h}_{2}$ where...

The Fourier transform of f is defined by $F(s)=\int_{-\infty}^{+\infty}f(t)e^{-i2\pi st}dt$. if f(t)=1 let $F_{1}$ be it's Fourier transform for $s\neq0$ you get $F_{1}(s)=\int_{-\infty}^{+\infty}e^{-i2\pi st}dt=0$ ( odd function ). And for s=0 $F_{1}(0)=\int_{-\infty}^{+\infty}1dt=+\infty$...

f(x)=(2x,3x+1) so f(a+h)=(2a+2h,3a+1+3h)=(2a,3a+1)+(2h,3h)=f(a)+Ta(h). Where Ta is a linear transformation which satisfies the limit zero. If you consider the example f(x)=(x2,3x+1) then f(a+h)=(a2+2ah+h2,3a+1+3h) so f(a+h)=(a2,3a+1)+(2ah,3h)+(h2,0) f(a+h)=f(a)+Ta(h)+S(h). Where Ta(h)=(2ah,3h)...