Main result and example with a nonlinear function
- A nonlinear function is a function where the relationship between the input and output is not a straight line. The graph of a nonlinear function looks more like an S-shape or an L-shape than a straight line.
- The graph of a linear function looks like a straight line. The graph of any other type of function can be classified as either linear or nonlinear.
So that was a warm-up for understanding non-linear functions. The ideas you learned can be applied to a non-linear function because it is well approximated by linear function. For example, let us retell the previous story to understand this nonlinear function. We can go back to the function fat the beginning that we approximated with a linear model. a The function f of (%, y) is y squared minus cubed minus x. The approximation for f of (1 plus delta x, plus delta y) was approximately negative minus 4 delta x plus delta y. f(x,y)=y^2-x^3-x f(1+Δx_1,1+Δy)~~-1-4Δx+2Δy Here is a question that is to similar to this one but concerns this more complex function. Let us begin at (1, 1), the point at which we did our linear approximation. We can move in any direction by a small distance, 0.1. How can we maximize the function f? And why does it say 0.1 instead of 1? The function f looks a lot like a linear function as long as you are in a small box-for example, if you only move by about 0.1. And it may not appear to be a linear function of your move by 1. And the effect of your move be 10 may not appear to be linear. If we are only talking about moving a little bit, then under certain circumstances we can apply this linear approximation and transform our problem into one that is easier to solve. Therefore, the vector that we have moved by is (delta x, delta y). F(1+Δx,1+Δy) Its lenght is 0.1. 1(Δx,Δy)1=0.1 The value of f equals negative 1 minus 4 times delta x plus 2 times delta y. F(1+Δx,1+Δy)~~-1-4Δx+2Δy We would like the change in fto be as big as possible. It appears to be a dot product once again. Thus. the change in f is negative 4 delta x plus 2 delta y. This can be written as (negative 4, comma 2) dotted with (delta X, delta y). Change in F is -4Δx+2Δy=(-4,2)*(Δx,Δy) Actually, the derivatives of fat that point with respect to x and y are the gradient of fat (1, 1). We can then use the dot product formula to find the norm of the gradient of f times the norm of (delta x, delta y) times cos(theta), where (theta) is the angle between f and (delta X. delta y). And we’re moving by 0.1, so we know that this is 0.1. Therefore, it’s the norm of the gradient off times 0.1 times cosine of theta. -4Δx+2Δy=(-4,2)*(Δx,Δy)=|∇f(1,1)||(Δx,Δy)|cosθ=0.1|∇f|cosθ This is most pronounced when the cosine of theta equals 0. When the vector (delta x, delta y) is in the same direction as the gradient, the function is largest. Now we can make a summary that is similar to the one we made for linear functions. Where shall we put it? In summary, the gradient points straight uphill in the direction of the steepest increase. The magnitude of a gradient is the slope. A steep hill has a large magnitude, and a shallow hill has a small magnitude. Another way to say this is that if we move a distanced in the direction of the gradient, then f increases by the slope times d–the norm of this gradient times d. We can see this by looking at our formula for f over here. To get the biggest increase, we want to choose theta equals 0. This means that they will go in the direction of the gradient. The cosine of theta equals 1. The change in f is the size of the gradient times how far we moved You may ask. why is the slope the magnitude of the gradient and not the product of the magnitude of the gradient and distance? The question is why the slope of a line is equal to the magnitude of its gradient, and not equal to the magnitude of its gradient times 0.1. Let’s vizualize a slope. What is it? Slope is here. Consider measuring it. If this is 1, this one probably is 1.5. So if this is 2, that appears to be 3. [Graph] The slope of a line is rise over run, which in this case is 1.5 slope=rise/run=1,5 The slope formula is versatile because it can be used with any combination of run and rise. We would have different vertical displacements, but we would always have the same slope. This slope is important because it describes how the line is angled. When we look back at the expression cosine and theta being 1, this is the rise. |∇f|(0,1)cosθ The expression is the magnitude of delta x delta y. This is the run. So. this is the graph of f, and this is height. And then we changed x and y, and moved a distance 0.1 there. And then we saw the change in f. What if we have problems with realizing what the rise and run are from this 2D picture? How does a flat, two-dimensional picture translate into a three-dimensional representation of x, y, and z? So the question is, how does a 2D picture compare to a 3D object? The truth is that x and y are down here. z is coming this way, and what do we have. We begin with a point in the xy-plane, and then another. The resulting graph is displayed below. At these points, the function has a value a and the value clearly is bigger. The difference between the values of each point is the rise, while the distance between the two points themselves is called run. Is the rise just cosine theta equals 1? Entire blue thing is the rise. This IS the graph of f. To compute the rise, we determine how much larger f is at this point compared to f at that point. Rephrasing, it’s the change in f. We computed the change in f with linear approximation. This iS the old value of f. This is the change in f and we simplify this and we got that.
