What does convex in this figure mean?

What does the convex(Single global minimum) mean in this picture?
Is it indicated that the whole Loss fuction equation could be plotted as convex(single global minimum)?

Convex means that the 2nd partial derivative is always positive or zero.
In practice, it means there is only one minimum.

Hi. TMosh,

Thanks for your reply.

what is the 2nd partial derivative as you mentioned above?

It’s the partial derivative of the first partial derivative. :nerd_face: But seriously this is a term from multivariate calculus. If you don’t already know what that means, then it is beyond the scope of this course. The idea of derivatives in calculus is that they express the rate of change of a function at each point in its domain. So the derivative is another function related to the original function which expresses the rate of change of the first function, which is the slope of the curve of the first function. The second derivative is the derivative of the first derivative, so it expresses the rate of change of the slope of the function. For the more concrete example that the original function gives the position of an object in space, the first derivative is the speed of the object and the second derivative is the acceleration (rate of change of the speed).

Example:
Consider the derivatives of the equation of a parabola (y = x^2).
The first derivative is y’ = 2*x.
The second derivative is y’’ = 2.
Since ‘2’ is always positive, we know that a parabola is convex.

Hi, Paul,

Thanks for your reply.

I know the concepts of first and second partial derivative. I think I know the second partial derivative in the loss function equtition.

Oh, sorry, the way you asked the question I thought you were asking about the concept, not the actual derivative. But if you know multivariate calculus already, then presumably you know the Chain Rule and then you just need to know the derivatives of the individual functions there.

The only one that is not just an application of the Chain Rule is:

h(z) = log(z)
h'(z) = \displaystyle \frac {1}{z}
h''(z) = -\displaystyle \frac {1}{z^2}

It’s just algebra from there.

Hi,@Tmosh,

Thank very much for your example.

I think I am cleared that the convex indicates only 1 minimum in the loss function as shown in the screenshot.

Hi. Paul,

Well Noted.

Thank you very much.