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$$f(x_1,x_2,...x_n):\mathbb{R}^n \rightarrow \mathbb{R}$$ The definition of the gradient is $$ \frac{\partial f}{\partial x_1}\hat{e}_1 +\ ... +\frac{\partial f}{\partial x_n}\hat{e}_n$$

which is a vector.

Reading this definition makes me consider that each component of the gradient corresponds to the rate of change with respect to my objective function if I go along with the direction $\hat{e}_i$.

But I can't see why this vector (defined by the definition of the gradient) has anything to do with the steepest descent.

Why do I get maximal value again if I move along with the direction of gradient?

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Each component of the gradient tells you how fast your function is changing with respect to the standard basis. It's not too far-fetched then to wonder, how fast the function might be changing with respect to some arbitrary direction? Letting $\vec v$ denote a unit vector, we can project along this direction in the natural way, namely via the dot product $\text{grad}( f(a))\cdot \vec v$. This is a fairly common definition of the directional derivative.

We can then ask in what direction is this quantity maximal? You'll recall that $$\text{grad}( f(a))\cdot \vec v = |\text{grad}( f(a))|| \vec v|\text{cos}(\theta)$$

Since $\vec v$ is unit, we have $|\text{grad}( f)|\text{cos}(\theta)$, which is maximal when $\cos(\theta)=1$, in particular when $\vec v$ points in the same direction as $\text{grad}(f(a))$.

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    $\begingroup$ I was wondering, but how to you that grad(f(a)) gives the steepest change? How do you know there is not other vector that moving in its direction might lead to a steeper change? $\endgroup$ – Pinocchio Feb 25 '14 at 1:51
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    $\begingroup$ so we multiply the gradient with an arbitrary vector and then say that the product is maximum when the vector points in the same direction as the gradient? How does that answer the question? $\endgroup$ – novice Sep 3 '16 at 0:30
  • $\begingroup$ @novice It answers the question since the scalar product EQUALS the rate of change of $f$ along the direction of the unit vector. So, in maximizing the product, the rate of increase in $f$ is likewise maximized. $\endgroup$ – Mark Viola Sep 3 '16 at 1:16
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    $\begingroup$ @Dr.MV - sorry for coming back, but isnt the quantity being maximized rate of change of f ALONG the unit vector and not the rate of increase of f? $\endgroup$ – novice Sep 3 '16 at 4:26
  • $\begingroup$ is this proof only true in 3d-dimension? $\endgroup$ – hqt Oct 10 '17 at 9:49
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Other answers are correct in using the directional derivative to show that the gradient is the direction of steepest ascent/descent. However, I think it is instructive to look at the definition of the directional derivative from first principles to understand why this is so (it is not arbitrarily defined to be the dot product of the gradient and the directional vector).

Let $f(\mathbf{x}):\mathbb{R}^n \rightarrow \mathbb{R}$. The partial derivatives of $f$ are the rates of change along the basis vectors of $\mathbf{x}$:

$\textrm{rate of change along }\mathbf{e}_i = \lim_{h\rightarrow 0} \frac{f(\mathbf{x} + h\mathbf{e}_i)- f(\mathbf{x})}{h} = \frac{\partial f}{\partial x_i}$

Each partial derivative is a scalar. It is simply a rate of change.

The gradient of $f$ is then defined as the vector:

$\nabla f = \sum_{i} \frac{\partial f}{\partial x_i} \mathbf{e}_i$

We can naturally extend the concept of the rate of change along a basis vector to a (unit) vector pointing in an arbitrary direction. Let $\mathbf{v}$ be such a vector, i.e., $\mathbf{v} = \sum_{i} \alpha_i \mathbf{e}_i$ where $\sum_{i} \alpha_i^2 = 1$. Then:

$\textrm{rate of change along }\mathbf{v} = \lim_{h\rightarrow 0} \frac{f(\mathbf{x} + h\mathbf{v}) - f(\mathbf{x})}{h}$

Again, this quantity is a scalar.

Now, it can be proven that if $f$ is differentiable at $\mathbf{x}$, the limit above evaluates to: $(\nabla f) \cdot \mathbf{v}$. This is a dot product of two vectors, which returns a scalar.

We know from linear algebra that the dot product is maximized when the two vectors point in the same direction. This means that the rate of change along an arbitrary vector $\mathbf{v}$ is maximized when $\mathbf{v}$ points in the same direction as the gradient. In other words, the gradient corresponds to the rate of steepest ascent/descent.

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  • $\begingroup$ Are you missing a square root around $\sum_i \alpha_i^2$? The way I understand it, what you're trying to say is that the magnitude of $v$ should be 1, which only happens if the square root of the sum of alphas squared is 1 (I could easily be completely wrong, my apologies if that's the case). I like your explanation overall regardless. $\endgroup$ – jeremy radcliff Sep 24 '16 at 23:08
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    $\begingroup$ @jeremyradcliff Yes exactly, I'm saying the magnitude should be 1. I left out the square root precisely because $1^2 =1$. $\endgroup$ – MGA Sep 26 '16 at 16:23
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Consider a Taylor expansion of this function, $$f({\bf r}+{\bf\delta r})=f({\bf r})+(\nabla f)\cdot{\bf\delta r}+\ldots$$ The linear correction term $(\nabla f)\cdot{\bf\delta r}$ is maximized when ${\bf\delta r}$ is in the direction of $\nabla f$.

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The question you're asking can be rephrased as "In which direction is the directional derivative $\nabla_{\hat{u}}f$ a maximum?".

Assuming differentiability, $\nabla_{\hat{u}}f$ can be written as:

$$\nabla_{\hat{u}}f = \nabla f(\textbf{x}) \cdot \hat{u} =|\nabla f(\textbf{x})||\hat{u}|\cos \theta = |\nabla f(\textbf{x})|\cos \theta$$

which is a maximum when $\theta =0$: when $\nabla f(\textbf{x})$ and $\hat{u}$ are parallel.

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Each component of the derivative $$ \frac{\partial f}{\partial x_1}\ ... \frac{\partial f}{\partial x_n}$$ tells you how fast your function is changing with respect to the standard basis.
It's now possible to make a basetransformation to an orthogonal base with $ n-1 $ base Directions with $0$ ascent and the gradient direction. In such a base the gradient direction must be the steepest since any adding of other base directions adds length but no ascent.

For a 3 dimensional Vector space the base could look like this $$ \left( \left( \begin{matrix} \partial x_2 \\ -\partial x_1 \\ 0 \end{matrix} \right) \left( \begin{matrix} \partial x_1 \\ \partial x_2 \\ -\dfrac{(\partial x_1)²+(\partial x_2)²}{\partial x_3} \end{matrix} \right) \left( \begin{matrix} \partial x_1 \\ \partial x_2 \\ \partial x_3 \end{matrix} \right) \right) $$ By complete induction it can now be shown that such a base is constructable for an n-Dimensional Vector space. $$ \left( \left( \begin{matrix} \partial x_2 \\ -\partial x_1 \\ 0 \\ 0 \end{matrix} \right) \left( \begin{matrix} \color{blue}{\partial x_1 \\ \partial x_2} \\ -\dfrac{(\partial x_1)²+(\partial x_2)²}{\partial x_3} \\ 0 \end{matrix} \right) \left( \begin{matrix} \color{blue}{\partial x_1 \\ \partial x_2} \\ \color{green}{\partial x_3} \\ -\dfrac{(\partial x_1)²+(\partial x_2)²+(\partial x_3)²}{\partial x_4} \end{matrix} \right) \left(\begin{matrix} \color{blue}{\partial x_1 \\ \partial x_2} \\ \color{green}{\partial x_3} \\ \color{orange}{\partial x_4} \end{matrix} \right) \right) $$ One can see here that the first Basevector demands the first 2 Elements of the following Basevectors to be $\partial x_1$ & $\partial x_2$ because of the orthogonal condition,
similarly the 2nd vector demands all the 3rd elements of the following vectors to be $\partial x_3$
as does the 3rd vector for the 4th element them being $\partial x_4$.

If another dimension is added the n+1 Element of the n$th$ Vector needs to be $$-\dfrac{(\partial x_1)²+...+(\partial x_n)²}{\partial x_{n+1}}$$ to meet the $0$ ascension condition which in turn forces the new n+1$th$ Vector to be of the form $$\left(\begin{matrix}\partial x_1 \\ ... \\ \partial x_{n+1}\end{matrix}\right)$$ for it to be orthogonal to the rest.

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Let $\vec v$ be an arbitrary unit vector. Then the change of $f$ by moving in the direction of $v$, starting in point $a$, is given by $grad( f(a)) \cdot \vec v$. We want to find a $\vec v$ for which this inner product is maximal. For the inner product we have the Cauchy–Schwarz inequality $\vec a \cdot \vec b \leq |\vec a||\vec b|$. Now the equality holds when $\vec v = \lambda \; grad(f(a))$, for some $\lambda \in \mathbb{R}$.

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Let $v=\frac{s}{|s|}$ be a unit vector and assume that $v$ is a descent direction, i.e. $v^T\nabla f(x) <0$. Then $f(x+\lambda v)$ as a function of $\lambda$, describes how this function changes along the direction $v$.

The rate of descent at $x$ along $v$ is given by: $$ \frac{d}{d \lambda}f(x+\lambda v)|_{\lambda=0} = v^T \nabla f(x) =\frac{s^T}{|s|}\nabla f(x) \equiv \frac{s^T}{|s|}g$$ So we want to find the maximum of this quantity as a function of $s$. Differentiating the above wrt $s$ and setting it equal to zero, we get (noting that $\nabla_s|s| =\frac{s}{|s|}$): $g=(g^T v)v\equiv av$.

Taking the Euclidean norm: $|g|=|a||v|=|a| \Rightarrow a=\pm|g|$.

We choose the minus sign to satisfy that $v$ is descent. Hence the direction of the steepest descent is $$ v= \dfrac{1}{a}g = -\dfrac{g}{|g|}$$

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Just want to further clarify why the gradient provides the steepest ascent (instead of descent) here. Any differentiable $f$ can be approximated by the linear tangent plane, i.e., $$f(\mathbf{x} + h \mathbf{v}) = f(\mathbf{x}) + h \, \nabla f(\mathbf{x})^T \mathbf{v} $$ as $h \rightarrow 0$ for any unit-length direction $\mathbf{v}$ with $\parallel \mathbf{v} \parallel =1.$ As $h \downarrow 0$, consider the amount of change $$ f(\mathbf{x} + h \mathbf{v}) - f(\mathbf{x}) = h \, \left\{ \, \nabla f(\mathbf{x})^T \mathbf{v} \right\} ~~\in~~ \left[ - h \, \parallel \nabla f(\mathbf{x}) \parallel, ~ h \, \parallel \nabla f(\mathbf{x}) \parallel \right] $$ by Cauchy-Swcharz inequality, which reaches its maximum (increase) $(h \, \parallel \nabla f(\mathbf{x}) \parallel)$ when $\mathbf{v} = \nabla f(\mathbf{x}) / \parallel \nabla f(\mathbf{x}) \parallel$ and its minimum (i.e., maximum decrease) $ (-h \, \parallel \nabla f(\mathbf{x}) \parallel) $ if $ \mathbf{v}= - \nabla f(\mathbf{x})/\parallel \nabla f(\mathbf{x}) \parallel$ (the negative gradient direction).

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I think the most intuitive explanation is the following: Draw one vector in the X direction. Draw another vector in the Y direction. Make each vector any length you want. Make X longer than Y or Y longer than X or make them the same length. Doesn't matter. Now draw the vector that represents the sum of those two vectors, using the rectangle method. This is key to the intuition. You have to draw a rectangle. The resulting vector (representing the sum of the X-direction vector and the Y-direction vector) goes from one corner of the rectangle to the opposite corner. Remember that the length of the resulting vector represents the magnitude, which in this context represents slope. Now look at the drawing and ask yourself: is there any vector within this rectangle, starting at the origin, that is longer than the diagonal one? No. Any vector you draw that starts at the origin and goes to another side of the rectangle is shorter than the diagonal one. Of course, you can draw a vector at the origin that is longer than the diagonal one, but only if that vector leaves the rectangle. But if it leaves the rectangle, then it's no longer operating within the constraint of the given X vector and Y vector; it's working with some different X vector and Y vector. I hope that helps. It took me a while to understand this intuitively and the image of a diagonal within a rectangle finally switched the lamp on.

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  • $\begingroup$ It might help to mention what this has to do with the gradient, other than it being a vector. $\endgroup$ – snulty May 7 at 23:31

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