# Tangent plane approximation: $f_1(x_0,y_0)h+f_2(x_0,y_0)k + \epsilon(h,k)\sqrt{h^2 + k^2}$ can be rewritten with $|h|+|k|$ instead of $\sqrt{h^2+k^2}$

As a condition for a function $$f:\mathbb{R}^2 \to \mathbb{R}$$ to be differentiable at $$(x_0, y_0)$$, the author asks that it be possible to write $$f(x_0 +h, y_0+k) - f(x_0, y_0)=f_1(x_0,y_0)h+f_2(x_0,y_0)k + \epsilon(h,k)\sqrt{h^2 + k^2}$$ where $$\epsilon(h,k) \to 0$$ as $$\sqrt{h^2 + k^2} \to 0$$.

Notice that this is comes from the tangent plane function $$T(x,y) = f(x_0,y_0) + f_1(x_0,y_0)(x-x_0) + f_2(x_0,y_0)(y-y_0)$$ which approximates $$f(x)$$ at $$(x_0, y_0)$$; and also from the corresponding relative error $$\frac{|f(x,y) - T(x,y)|}{\sqrt{(x-x_0)^2 + (y-y_0)^2}}$$ which goes to $$0$$ as $$(x,y)$$ approaches $$(x_0, y_0)$$.

The author offers the alternative condition $$f(x_0 + h, y_0 +k) - f(x_0, y_0) = f_1(x_0, y_0)h + f_2(x_0,y_0)k + \epsilon(h,k)(|h| + |k|)$$ by observing that $$\sqrt{h^2 + k^2} \leq |h| + |k| \leq \sqrt{2}\sqrt{h^2 +k^2}$$.

I'm not sure, but I think that the substitution is justified because the initial error function: $$\epsilon(h,k)= \frac{f(x_0 +h, y_0+k) - f(x_0, y_0) - f_1(x_0,y_0)h-f_2(x_0,y_0)k}{\sqrt{h^2 + k^2}}$$ is bigger than the resulting error function: $$\epsilon(h,k)= \frac{f(x_0 +h, y_0+k) - f(x_0, y_0) - f_1(x_0,y_0)h-f_2(x_0,y_0)k}{|h| + |k|}$$

My question is: why does $$|h| + |k|$$ have to have the upper bound $$\sqrt{2}\sqrt{h^2 + k^2}$$?

This question comes from reading page 711 of Elementary Real Analysis by Thomson, Bruckner and Bruckner.

• Observe $\sqrt{h^2k^2}\le \frac{h^2+k^2}2$ so $(|h|+|k|)^2\le2(|h|^2+|k|^2)$ Commented May 16, 2021 at 12:19
• @PNDas thank you! Commented May 16, 2021 at 12:27
• All norms in a finite-dimensional Cartesian space $S$ are uniformly equivalent: If $N$ and $N'$ are norms for $S$ then there are positive $k,k'$ such that $kN(p)\le N'(p)\le k'N(p)$ for every $p\in S$. Commented May 16, 2021 at 23:36

$$(|h|+|k|)^2 = h^2+k^2+2|hk|$$
For any 2 real numbers $$a,b,~~(a-b)^2 \ge0 \Rightarrow \boxed{a^2+b^2\ge2ab}$$.
$$(|h|+|k|)^2 = h^2+k^2+2|hk| \le h^2+k^2+ h^2+k^2 = 2(h^2+k^2)\Rightarrow \boxed{|h|+|k|\le\sqrt2\sqrt{h^2+k^2}}$$
The inequality is derived using $$2ab \leq a^2 + b^2$$ for $$a,b \in \Bbb R$$ $$(|h| + |k|)^2 = |h|^2 + |k|^2 + 2|h||k| \leq 2(|h|^2 + |k|^2)$$ Taking the square root we get $$|h| + |k| \leq \sqrt{2}\sqrt{|h|^2 + |k|^2}$$