update

some typos.

quarto/alternatives/SciML.qmd

@@ -129,7 +129,7 @@ This problem can also be solved using a bracketing method. The package has both
 
 ```{julia}
 u0 = (1.0, 2.0)
-prob = NonlinearProblem(f, u0)
+prob = IntervalNonlinearProblem(f, u0)
 ```
 
 And
@@ -149,7 +149,7 @@ Incorporating parameters is readily done. For example to solve $f(x) = \cos(x) -
 f(x, p) = @. cos(x) - x/p
 u0 = (0, pi/2)
 p = 2
-prob = NonlinearProblem(f, u0, p)
+prob = IntervalNonlinearProblem(f, u0, p)
 solve(prob, Bisection())
 ```
 
@@ -159,7 +159,7 @@ The *insignificant* difference in stopping criteria used by `NonlinearSolve` and
 
 
 ```{julia}
-an = solve(NonlinearProblem{false}(f, u0, p), Bisection())
+an = solve(IntervalNonlinearProblem{false}(f, u0, p), Bisection())
 ar = solve(Roots.ZeroProblem(f, u0), Roots.Bisection(); p=p)
 nextfloat(an[]) == ar, f(an[], p), f(ar, p)
 ```
@@ -418,7 +418,7 @@ prob = OptimizationProblem(sys, u0, p; grad=true, hess=true)
 soln = solve(prob, LBFGS())
 ```
 
-We used an initial guess of $x=4$ above. The `LBFGS` method is a computationally efficient modification of the Broyden-Fletcher-Goldfarb-Shanno algorithm ... It is a quasi-Newton method that updates an approximation to the Hessian using past approximations as well as the gradient."  On this problem it performs similarly to `Newton`, though in general may be preferable for higher-dimensional problems.
+We used an initial guess of $x=4$ above. The `LBFGS` method is a computationally efficient modification of the Broyden-Fletcher-Goldfarb-Shanno algorithm. It is a quasi-Newton method that updates an approximation to the Hessian using past approximations as well as the gradient. On this problem it performs similarly to `Newton`, though in general may be preferable for higher-dimensional problems.
 
 
 ### Two dimensional
@@ -515,7 +515,7 @@ For a simple definite integral, such as $\int_0^\pi \sin(x)dx$, we have:
 
 ```{julia}
 f(x, p) = sin(x)
-prob = IntegralProblem(f, 0.0, pi)
+prob = IntegralProblem(f, 0.0, 1pi)
 soln = solve(prob, QuadGKJL())
 ```
 
@@ -546,7 +546,7 @@ The `Integrals` solution is a bit more verbose, but it is more flexible. For exa
 
 ```{julia}
 f(x, p) = sin.(x)
-prob = IntegralProblem(f, [0.0], [pi])
+prob = IntegralProblem(f, [0.0], [1pi])
 soln = solve(prob, HCubatureJL())
 ```
 
@@ -574,7 +574,7 @@ Consider $d/dp \int_0^\pi \sin(px) dx$. We can do this integral directly to get
 \frac{d}{dp} \int_0^\pi \sin(px)dx
 &= \frac{d}{dp}\left( \frac{-1}{p} \cos(px)\Big\rvert_0^\pi\right)\\
 &= \frac{d}{dp}\left( -\frac{\cos(p\cdot\pi)-1}{p}\right)\\
-&= \frac{\cos(p\cdot \pi) - 1)}{p^2} + \frac{\pi\cdot\sin(p\cdot\pi)}{p}
+&= \frac{\cos(p\cdot \pi) - 1}{p^2} + \frac{\pi\cdot\sin(p\cdot\pi)}{p}
 \end{align*}
 
 Using `Integrals` with `QuadGK` we have:
@@ -583,7 +583,7 @@ Using `Integrals` with `QuadGK` we have:
 ```{julia}
 f(x, p) = sin(p*x)
 function ∫sinpx(p)
-    prob = IntegralProblem(f, 0.0, pi, p)
+    prob = IntegralProblem(f, 0.0, 1pi, p)
     solve(prob, QuadGKJL())
 end
 ```
@@ -614,7 +614,7 @@ The power of a common interface is the ability to swap backends and the uniformi
 #### $f: R^n \rightarrow R$
 
 
-The area under a surface generated by $z=f(x,y)$ over a rectangular region $[a,b]\times[c,d]$ can be readily computed. The two coding implementations require $f$ to be expressed as a function of a vector–*and* a parameter–and the interval to be expressed using two vectors, one for the left endpoints (`[a,c]`) and on for the right endpoints (`[b,d]`).
+The area under a surface generated by $z=f(x,y)$ over a rectangular region $[a,b]\times[c,d]$ can be readily computed. The two coding implementations require $f$ to be expressed as a function of a vector–*and* a parameter–and the interval to be expressed using two vectors, one for the left endpoints (`[a,c]`) and one for the right endpoints (`[b,d]`).
 
 
 For example, the area under the function $f(x,y) = 1 + x^2 + 2y^2$ over $[-1/2, 1/2] \times [-1,1]$ is computed by:
@@ -664,7 +664,7 @@ To compute this transformed integral, we might have:
 ```{julia}
 function vol_sphere(ρ)
     f(rθ, p) = sqrt(ρ^2 - rθ[1]^2) * rθ[1]
-    ls = [0,0]
+    ls = [0.0,0.0]
     rs = [ρ, 2pi]
     prob = IntegralProblem(f, ls, rs)
     solve(prob, HCubatureJL())
@@ -677,7 +677,7 @@ If it is possible to express the region to integrate as $G(R)$ where $R$ is a re
 
 
 $$
-\iint_{G(R)} f(x) dA = \iint_R (f\circ G)(u) |det(J_G(u)| dU
+\iint_{G(R)} f(x) dA = \iint_R (f\circ G)(u) |det(J_G(u))| dU
 $$
 
 turns the integral into the non-rectangular domain $G(R)$ into one over the rectangular domain $R$. The key is to *identify* $G$ and to compute the Jacobian. The latter is simply accomplished with `ForwardDiff.jacobian`.