Gradient Ascent: Finding Maximum of a Scalar Function

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SUMMARY

The discussion centers on the mathematical proof that for a scalar function f(x) = f(x1, x2, ..., xn), the maximum value occurs when the change Δx is aligned with the gradient ∇f(x). The user initially struggled with the problem but ultimately demonstrated that the direction of maximal growth is determined by the gradient, confirming that ∇f must be parallel to the direction vector n for maximization. This conclusion is supported by applying the chain rule and analyzing the function g_n(t) to find the maximum of the directional derivative.

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  • Understanding of scalar functions and their gradients
  • Familiarity with partial derivatives and the notation ∇f
  • Knowledge of the chain rule in calculus
  • Concept of directional derivatives and unit vectors
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gaganaut
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hi all,

Couple of months ago I had an entrance exam wherein this problem appeared. (I hope this is what it was).

For a scalar function [tex]f\left(x\right)=f\left(x_{1},x_{2},...,x_{n}\right)[/tex] the gradient is given as
[tex]\nabla f=\left(\frac {\partial f \left(x\right)} {\partial x_1},\frac {\partial f \left(x\right)} {\partial x_2},...,\frac {\partial f \left(x\right)} {\partial x_n}\right)[/tex]
Then show that for any small change [tex]\Delta x[/tex], [tex]f\left(x+\Delta x\right)[/tex] is maximum if [tex]\Delta x[/tex] lies along [tex]\nabla f\left(x\right)[/tex].

Frankly, I did not get the question then. So I did few preliminary steps as follows.
[tex]f\left(x+\Delta x\right)=f\left(x\right)+\nabla f\left(x\right)\cdot\Delta x + O\left(\left(\Delta x\right)^2\right)[/tex]

Hence,
[tex]\nabla f\left(x+\Delta x\right)=\nabla f\left(x\right)+\nabla \left(\nabla f\left(x\right)\cdot\Delta x\right)[/tex]

For [tex]f\left(x+\Delta x\right)[/tex] to be maximum, [tex]\nabla f\left(x+\Delta x\right)=0[/tex] is maximum

Hence,
[tex]\nabla f\left(x\right)=-\nabla \left(\nabla f\left(x\right)\cdot\Delta x\right)[/tex]

This is where I gave up then. Bu intuition I could see that the dot product would give a maximum answer if [tex]\Delta x[/tex] lies along [tex]\nabla f\left(x\right)[/tex]. But I could not prove it.

So can somebody help me with this as this question is haunting me for last two months now.

Thanks in advance.
 
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Let [tex]g_{n}(t)=f(x_{1}+tn_{1},x_{2}+tn_{2},...,x_{n}+tn_{n})[/tex]
where [tex]\vec{n}=(n_{1},n_{2},...,n_{n})[/tex] is a unit vector in direction n.

Thus, g_{n}(t) measures the increase of f along some direction.

At t=0, we are at the point [tex]\vec{x}_{0}=(x_{1},x_{2}...,x_{n})[/tex]

Now, the direction for maximal growth of f at [itex]\vec{x}_{0}[/itex] will be finding the maximum of [tex]\frac{dg_{n}}{dt}|_{t=0}[/tex], considered as a function of [itex]\vec{n}[/tex]<br /> <br /> By aid of the chain rule, this transforms to finding the maximum of:<br /> [tex]\nabla{f}_{\vec{x}=\vec{x}_{0}}\cdot\vec{n}[/tex]<br /> <br /> Clearly, [itex]\vec{n}[/itex] must be parallell to the gradient in order to maximize the expression, which proves the result.[/itex]
 
Thanks for the reply

Thank you so much for the reply. I appreciate that. We can now say that this question has been resolved.
 

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