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Let,$f(x, y, z) =x^3+y^3+z^3$

$L$ be a linear map from $\mathbb R^3$ to $\mathbb R$ Satisfying $$\displaystyle\lim_{(x, y, z) \to (0, 0,0)} \frac{f(1+x, 1+y, 1+z) -f(1, 1,1) -L(x, y, z) }{\sqrt{x^2+y^2+z^2}}=0$$

Then find the value of $L(1, 2,4)$

I am unable to understand how to approach. Pls help

IIT JAM 2023

Nope
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3 Answers3

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Your problem describes the Frechet Derivative of the function $f(x,y,z) = x^3 + y^3 + z^3$ at the point $(1,1,1)$ evaluated in the direction $(1,2,4)$.

Here is how to do this from first principles. If such a map $L$ exists, then we have

$$ \begin{align*} 0 &= \lim_{(h,0,0) \to 0} \frac{f(1+h,1,1) - f(1,1,1) - L(h,0,0)}{h}\\ &= \lim_{h \to 0} \frac{f(1+h,1,1) - f(1,1,1)}{h} - \frac{L(h,0,0)}{h}\\ &= \lim_{h \to 0} \frac{(1+h)^3 + 1 + 1 - (1+1+1)}{h} - \frac{hL(1,0,0)}{h}\\ &= \lim_{h \to 0} \frac{3h+3h^2+h^3}{h} - L(1,0,0)\\ &= 3 - L(1,0,0) \end{align*}$$

So $L(1,0,0) = 3$. Note that this is exactly $\frac{\partial f}{\partial x} \big|_{(1,1,1)}$.

Similarly, by symmetry, we have $L(0,1,0) = L(0,0,1) = 3$.

Thus $L$ must be the linear map with the matrix $\begin{bmatrix} 3 & 3 & 3\end{bmatrix}$ if it exists. To show that $L$ actually does satisfy the definition you need to go back to the original limit and verify that it actually does equal $0$.

Once you very this, $L(1,2,4) = \begin{bmatrix} 3 & 3 & 3\end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 4\end{bmatrix} = 3(1) + 2(3) + 4(3) = 21$.

Steven Gubkin
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Hint:

For the 1st part the gradient $$L(1,1,1)= \left[\frac{\partial f}{\partial x}|_{(1,1,1)}\qquad \frac{\partial f}{\partial y}|_{(1,1,1)}\qquad \frac{\partial f}{\partial z}|_{(1,1,1)} \right],$$

will do the task.

janmarqz
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    Yes Sir, L(1, 1,1) it is. How can I find L(1, 2,4)? Is it L(1, 2,4) = 3(1) +3(2) +3(4) = 21 ? – Nope Feb 18 '23 at 15:35
  • Not exactly, because $$L(1,2,4)= \left[3x^2|{(1,2,4)}\qquad 3y^2|{(1,2,4)}\qquad 3z^2|_{(1,2,4)} \right],$$ – janmarqz Feb 18 '23 at 15:55
  • Actually Sir, In question it is asked about value. But is L(1, 2,4) a vector? or just a value? – Nope Feb 18 '23 at 16:10
  • This is the gradient at $(1,2,4)$ but if you used to have directional derivative, see the @Steph 's post. – janmarqz Feb 18 '23 at 16:13
  • This is also incorrect. $L(1,1,1)$ should be a number, not a vector. You have written the matrix for $L$. We do have that $L = Df\big|_{(1,1,1)}$. We then need to apply this matrix to the vector $(1,2,4)$. – Steven Gubkin Feb 18 '23 at 23:36
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The definition of the linear map is rather strange ? I will go this way: Define the vector $\mathbf{x}_0 = (1,1,1)$ and $\mathbf{v} = (x,y,z)$. Let $t=\| \mathbf{v} \|$, and define the unit vector $\mathbf{w}= \frac{\mathbf{v}}{\|\mathbf{v}\|}$

Thus the directional derivative writes $$ Df(\mathbf{x}_0)[\mathbf{w}] = \lim_{t\to 0} \frac{f(\mathbf{x}_0+t\mathbf{w})-f(\mathbf{x}_0)}{t} = \nabla f(\mathbf{x}_0):\mathbf{w} $$ where the colon operator stands for the inner product between two vectors. I think you are asked to compute the directional derivative in the direction $\mathbf{w}= \frac{\mathbf{v}}{\|\mathbf{v}\|}$ with $\mathbf{v}=(1,2,4)$ To do so, you need to compute the gradient $\nabla f(\mathbf{x}_0)=3(x_0^2,y_0^2,z_0^2) =3(1,1,1)$. Finally $$ 3(1,1,1):\frac{1}{\sqrt{261}}(1,2,4) = \frac{21}{\sqrt{261}} $$

UPDATE :

Define the vector $\mathbf{x}_0 = (1,1,1)$ and $\mathbf{v} = (x,y,z)$.

Define the unit vector $\mathbf{w}= \frac{\mathbf{v}}{t}$ with $t=\| \mathbf{v} \|$.

The directional derivative at point $\mathbf{x}_0$ in the direction $\mathbf{w}$ writes $$ Df(\mathbf{x}_0)[\mathbf{w}] = \lim_{t\to 0} \frac{f(\mathbf{x}_0+t\mathbf{w})-f(\mathbf{x}_0)}{t} $$

We can write $$ \lim_{t\to 0} \frac{f(\mathbf{x}_0+t\mathbf{w})-f(\mathbf{x}_0)- t Df(\mathbf{x}_0)[\mathbf{w}]}{t} =0 $$

Because $t Df(\mathbf{x}_0)[\mathbf{w}] = Df(\mathbf{x}_0)[t\mathbf{w}]$, it holds $$ \lim_{t\to 0} \frac{f(\mathbf{x}_0+t\mathbf{w})-f(\mathbf{x}_0)- Df(\mathbf{x}_0)[t\mathbf{w}]}{t} =0 $$ or equivalently $$ \lim_{\| \mathbf{v} \|\to 0} \frac{f(\mathbf{x}_0+\mathbf{v})-f(\mathbf{x}_0)- Df(\mathbf{x}_0)[\mathbf{v}]}{\| \mathbf{v} \|} =0 $$

We deduce that $L(\mathbf{v}) =Df(\mathbf{x}_0)[\mathbf{v}] =\nabla f(\mathbf{x}_0):\mathbf{v} $ and thus the correct result is 21.

Steph
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  • Sir, it was the actual question. I also don't understand what it actually means. What is the correct answer according to you? Actually ans is scaler value or a vector? – Nope Feb 18 '23 at 16:13
  • The formulation is quite awkward but I think the answer is a scalar in any case (presumably the one I gave to you). – Steph Feb 18 '23 at 16:31
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    The definition of the linear map in question is exactly the correct definition of the multivariate derivative. We can evaluate the derivative on vectors which are not unit vectors. See https://en.wikipedia.org/wiki/Fr%C3%A9chet_derivative – Steven Gubkin Feb 18 '23 at 23:20
  • @StevenGubkin, thank you for pointing me the Frechet derivative. I have updated my answer accordingly. – Steph Feb 20 '23 at 19:25