In this section, we illustrate the fundamental link between derivative and optimization.

The arguably simplest functions are affine function so let's start with the minimization of an affine function. What is the minimum of $x + 1$ ? Let's plot this function:

```
using Plots
x = range(-8, stop=8, length=50)
p(x) = x + 1
plot(x, p.(x), label="")
ReLU(x) = (x < 0) ? 0.0 : x
plot(x, p.(x), label="")
```

In order to plot, we generate `100`

equally spaced numbers between `-8`

and `8`

:

`x`

We can see the elements in this range by converting into a vector with `collect`

:

```
collect(x)
p.(x)
```

What is the minimum of $x^2 - 2x + 1$ ?

```
x = range(0, stop=2, length=50)
p(x) = x^2 - 2x + 1 # (x - 1)^2
plot(x, p.(x), label="")
scatter!([1], [p(1)], label="")
plotly()
ε = 1e-1
c = 0.1
f = ReLU
x0 = range(c-ε, stop=c+ε, length=50)
plot(x0, f.(x0), label="", linewidth=4, ratio=:equal)
```

```
dp(x) = 2*(x - 1)
plot(x, dp.(x), label="")
scatter!([1], [dp(1)], label="")
plot(x, p.(x), label="")
tangent(x, y) = p(x) + (y - x) * dp(x)
@show dp(2)
y2 = range(1, stop=3, length=100)
plot!(y2, tangent.(2, y2))
@show dp(4)
y4 = range(3, stop=5, length=100)
plot!(y4, tangent.(4, y4))
@show dp(1)
y1 = range(0, stop=2, length=100)
plot!(y1, tangent.(1, y1))
y_1 = range(-2, stop=0, length=100)
plot!(y_1, tangent.(-1, y_1))
```

How to minimize $x^4 - 8x^3 + 4x^2 - 6x + 1$ ?

```
using Plots
x = range(-3, stop=8, length=1000)
p(x) = x^4 - 8x^3 + 4x^2 - 6*x + 1
plot(x, p.(x))
```

Is that related to its derivative ?

```
plot(x, x)
plot(x, x.^2)
plot(x, x.^5)
plot(x, x.^4)
plot(x, -x.^4)
dp(x) = 4x^3 - 24x^2 + 8x - 6
plot(x, dp.(x))
ddp(x) = 12x^2 - 48x + 8
plot(x, ddp.(x))
```

Is the zero-derivative condition sufficient ?

```
x = range(-3, stop=64, length=1000)
plot(x, x -> x * sin(x))
plot(x, x .* cos.(x) + sin.(x))
plot(x, cos.(x) - x .* sin.(x) + cos.(x))
```

Zero derivative is necessary for $x$ to be a global minimizer but there are 3 cases of zero derivative:

local minimum: when it's decreasing before $x$ and then increasing

local maximum: when it's increasing before $x$ and then decreasing

saddle point: when it's increasing (resp. decreasing) before $x$ and then increasing (resp. decreasing) after $x$

Say the first derivative of `f(x)`

that is nonzero is the `k`

-th derivative `f^{(k)}(x)`

then

if

`k`

is odd then it is not a local extremumif

`k`

is even and the value is positive then it is a local minimumif

`k`

is even and the value is negative then it is a local maximum

```
x = range(-1, stop=1, length=100)
plot(x, -x.^5)
```

```
using Plots
x = range(-4, stop=4, length=100)
y = range(-4, stop=4, length=100)
p(x, y) = 2x^2 + x*y + 2y^2 - 2x + 3y + 1
plotly()
surface(x, y, p)
```

$z = p(x, y) = 2x^2 + xy + 2y^2 - 2x + 3y + 1$
∂z/∂x ∂z/∂y ∂z/∂(x + y) = ∂z/∂x + ∂z/∂y ∂z/∂(x + 2y) = ∂z/∂x + 2∂z/∂y

$\frac{\partial p}{\partial x} = 4x + y - 2$ $\frac{\partial p}{\partial y} = x + 4y + 3$```
dx -> +1 -> 0 * 1 -> 0
dy -> +2 -> 0 * 2 -> 0
0 * 1 + 0 * 2 -> 0
```

The notation $\partial$ is defined such that $\frac{\partial y}{\partial x} = 0$ and $\frac{\partial x}{\partial y} = 0$

```
dpdx(x, y) = 4x + y - 2
dpdy(x, y) = x + 4y + 3
```

From the plot, it seems $(1, -1)$ is close to be a local minimizer, is it ?

```
p(1, -1)
dpdx(1, -1)
dpdy(1, -1)
```

$\frac{\partial p}{\partial x}$ is positive so it's increasing in $x$. So, should we increase or decrease $x$ to get closer to a local minimizer ?

```
p(1, -1)
p(1.1, -1)
p(0.75, -1)
p(0.74, -1)
p(0.77, -1)
p(0.78, -1)
dpdx(0.75, -1)
dpdy(0.75, -1)
```

Now it's constant in $x$ but decreasing in $y$

```
p(0.75, -0.94)
p(0.75, -0.93)
p(0.75, -0.938)
p(0.75, -0.939)
p(0.75, -0.937)
```

The *gradient* is defined as the concatenation of all the partial derivatives. Having a zero gradient is **necessary** to be a local minimizer. Again, it's not **sufficient** as it could also be a local maximizer or a saddle point.

```
∇p(x, y) = [dpdx(x, y), dpdy(x, y)]
-∇p(0.75, -0.938)
x = 0.75
y = -0.938
xnew = x - 0.2 * dpdx(x, y)
ynew = y - 0.2 * dpdy(x, y)
x, y = x - 0.2 * dpdx(x, y), y - 0.2 * dpdy(x, y)
v = [x, y]
∇p(v...)
function gradient_descent(h, num_iters = 100, v = [1, -1])
for _ in 1:num_iters
v = v - h * ∇p(v...)
end
return v
end
```

With a step size of `1`

, we diverge

`v = gradient_descent(1)`

With a step size of `0.1`

, we converge

```
v = gradient_descent(0.1)
v = @time gradient_descent(0.001, 1000000)
```

Looking at the solution, it looks rational

`rationalize.(v, tol=1e-14)`

It seems that with a step size above `0.4`

we do not converge and below `0.4`

you converge. What happens with a step size of exactly `0.4`

? With the step 0.4: you converge to an orbit of period 2:

```
v = gradient_descent(0.4, 100)
v1 = rationalize.(v, tol=1e-10)
```

At the next step, we get another one

```
v = gradient_descent(0.4, 101)
v2 = rationalize.(v, tol=1e-10)
```

And then we cycle

```
v = gradient_descent(0.4, 102)
rationalize.(v, tol=1e-10)
```

Indeed, here is the difference:

`v1 - v2`

With a step from `v1`

, we go form `v1`

to `v2`

`4 // 10 * ∇p(v1...)`

With a step from `v2`

, we go back to `v1`

`4 // 10 * ∇p(v2...)`

Now what happens with a step too small ?

`gradient_descent(0.001, 1000)`

With a bit more steps ?

`gradient_descent(0.001, 1000)`

And with even more steps ?

`gradient_descent(0.001, 10000)`

So, we observed that

With a step size that is

*too large*(in this case larger than 0.4), the gradient method may diverge.With a step size that is

*too small*, the method will converge but very slowly.

A local minimizer of $f(x)$ can be found numerically as follows:

Compute gradient $\nabla f(x)$

If approximately zero -> done

Follow direction $-\nabla f(x)$, find how long you follow it: line search and go to 1.

For our example, $\nabla p(x, y) = 0$ is a linear system so we can find the global minimizer by solving this system:

$4x + y = 2$ $x + 4y = -3$```
A = [
4 1
1 4
]
b = [2, -3]
x, y = A \ b
dpdx(x, y)
dpdy(x, y)
∇p(x, y)
rationalize(x)
rationalize(y)
```

```
using Pkg
pkg"add NLopt"
using JuMP
import NLopt
model = Model(NLopt.Optimizer)
set_optimizer_attribute(model, "algorithm", :LD_MMA)
set_optimizer_attribute(model, "ftol_rel", 1e-16)
set_optimizer_attribute(model, "xtol_rel", 1e-16)
@variable(model, x)
@variable(model, y)
p(x, y)
@objective(model, Min, p(x, y))
model
println(model)
JuMP.optimize!(model)
```

NLopt found a point with gradient zero and for which the function is locally increasing in every direction around the point. It therefore knows that it is a local minimizer but does not know whether it is a global minimizer. For this reason, its termination status is `LOCALLY_SOLVED`

.

```
solution_summary(model)
value(x)
value(y)
```

We can try as well with Ipopt without having to change the model

```
pkg"add Ipopt"
using Ipopt
set_optimizer(model, Ipopt.Optimizer)
JuMP.optimize!(model)
solution_summary(model)
value(x)
value(y)
```

It's important to check for the `solution_summary`

to check that it's `LOCALLY_SOLVED`

.

```
using JuMP
model = Model(Ipopt.Optimizer)
@variable(model, x)
@variable(model, y)
@objective(model, Max, p(x, y))
JuMP.optimize!(model)
value(x)
value(y)
solution_summary(model)
```

Here, it is `NORM_LIMIT`

because it is unbounded

*This page was generated using Literate.jl.*

© Benoît Legat. Last modified: September 10, 2024. Website built with Franklin.jl.