# Why is gradient the direction of steepest ascent?

$$f(x_1,x_2,...x_n):R^n \rightarrow R$$ The definition of the gradient is $$\frac{\partial f}{\partial x_1}e_1 +\ ... +\frac{\partial f}{\partial x_n}e_n$$

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 $e_i$.

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

Why do I get maximal value gain 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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Care to explain the downvote? –  AsinglePANCAKE Nov 13 '13 at 19:53
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? –  Pinocchio Feb 25 '14 at 1:51

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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+1 for “assuming differentiability”. –  Carsten Schultz Dec 26 '13 at 1:01

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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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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