Maple and Calculus OperationsAlthough our initial Intro to Maple included discussion of a number of Maple functions related to calculus, you may not have noticed that fact if you didn't look carefully at what Maple can do. This handout is, in the first instance, just a summary of Maple's calculus-related functionality. As always, I'm only scratching the surface. The Maple help command (say ?diff; to get help on diff, for instance) offers a lot more information on any of the commands used in this worksheet.Basic CommandsLimitsMaple's command for limits is, logically enough, limit.limit(sin(x)/x, x=0);limit(1/x, x=0);limit(1/x^2, x=0);limit(sin(1/x), x=0);It is useful to plot all the functions above, and to make sure you understand what Maple is telling you with all these answers.Our investigation of derivatives could have been speeded up substantially had we been willing to use Maple uncritically as a black box for taking limits. Here, for instance, are calculations of the derivatives of the sine, exponential, and cube root functions. It would be useful review to try to remember how we obtained these results in class. Are you sure Maple doesn't have bugs that affect these calculations?limit((sin(x+h)-sin(x))/h, h=0);limit((exp(x+h)-exp(x))/h, h=0);g := x -> x^(1/3);(g(x)-g(a))/(x-a);limit(%, x=a);For limits, as for most calculus functions, Maple has an "inert" form of the limit command that writes the limit without evaluating it. This command is called Limit. It's often useful in writing down formulas in Maple worksheets.Limit(x^x, x=0);Limit(x^x, x=0) = limit(x^x, x=0);Thinks about why the value of this limit wasn't obvious. (What's NiMqJCUieEciIiE=? What's NiMpIiIhJSJ4Rw==? What should NiMqJCIiIUYk be, if anything?)DerivativesMaple has two functions for computing derivatives. If f is an expression (like NiMqJiUieEciIiMtJSRzaW5HNiNGJCIiIg==), then diff(f, x) is the derivative of f with respect to the variable x. If f is a function (like sin, or like the user-defined function g above), then D(f) is the derivative of f as a function. Using D often results in much less typing, and it is the easy way to get the derivative as a function.diff(sqrt(x+sqrt(x+sqrt(x))), x);f := x -> x^x;fprime := D(f);plot({f(x), fprime(x)}, x=0..2);D(sin)(x);D(D(sin))(x);D(D(D(sin)))(x);(D@@4)(sin)(x) = D(D(D(D(sin))))(x);These operators can also be used with undefined functions to get general differentiation rules. Right now, f and g have value, but we can remove thoseg(x);g := 'g';g(x);f := 'f'; f(x);Here are Maple versions of the product rule, the chain rule, and an extended chain rule for NiMtJSJmRzYjLSUiZ0c2Iy0lImhHNiMlInhH. Play with things like this, and make sure the notation makes sense. If you need help, try ?D or ?diff.D(f*g);D(f@g);diff(f(g(h(x))), x);IntegralsMaple does both indefinite and definite integrals with int. The inert form of int is, logically enough, Int. As with limits, the inert forms are useful in writing formulas.int(sin(x), x=0..Pi);int(x/(1+x^2), x);int(x/(1+x^4), x);int(x/(1+x^3), x);Int(x^3*exp(x),x);Int(x^3*exp(x),x) = int(x^3*exp(x),x);Int(x^n,x) = int(x^n,x);The values of inert forms can also be taken using the command value.Int(ln(x)^n/x, x) = value(Int(ln(x)^n/x, x));SumsSums are done with sum and Sum, just as integrals are done with int and Int.sum(k, k=1..10);sum(k, k=1..n);factor(%);sum(k^2, k=1..n);factor(%);Sum(k^3, k=1..n);Sum(k^3, k=1..n) = factor(sum(k^3, k=1..n));The range of sums Maple understands is pretty impressive. Here's an identity that should look beautiful and surprising:Sum(1/n^2, n=1..infinity) = sum(1/n^2, n=1..infinity);If Maple can do all this, then Maple ought to be able to handle the nasty work we encountered with Riemann sums. Here, for instance, is Maple's treatment of a computation ofNiMtJSRpbnRHNiQqJCUidEciIiQvRic7IiIhJSJ4Rw==using Riemann sums.f := x -> x^3;RSum := Sum(f(k*x/n)*(1/n), k=1..n);RSum = value(RSum);RSum = factor(value(RSum));This works out the Riemann sum for n slices. Now we want to take the limit of this sum as NiNmKjYjJSJuRzciNiQlKW9wZXJhdG9yRyUmYXJyb3dHNiIlKWluZmluaXR5R0YqRipGKg==.Int(t^3, t=0..x) = Limit(RSum, n=infinity);
Limit(RSum, n=infinity) = limit(value(RSum), n=infinity);The student calculus package.Maple also has a whole library of functions useful for the calculus student and contained in the package student. Let's load this package and see what functions it contains. You need to load the package using with(student) before you can use the functions.with(student);It's worth asking for help about most of these functions, just to see what they do. Here's a quick teaser:The midpoint approximation to the integral of NiMvJSJ5Ry0lJVpldGFHNiMlInhH between NiMvJSJ4RywkIiM/ISIi and NiMvJSJ4RywkIiM+ISIi is shown belowmiddlebox(Zeta(x), x=-20..-19, 10);The sum of the areas of the reactangles ismiddlesum(Zeta(x), x=-20..-19, 10);Numerically, this sum is approximately% = evalf(value(%));By comparison, a more exact value of the integral isInt(Zeta(x), x=-20..-19) = evalf(int(Zeta(x), x=-20..-19));The error in the midpoint sum is therefore onlymiddlesum(Zeta(x), x=-20..-19, 10) -
Int(Zeta(x), x=-20..-19) =
evalf(value(middlesum(Zeta(x), x=-20..-19, 10)))-
evalf(int(Zeta(x), x=-20..-19));Integration by SubstititionOne of the neatest uses of the student package is that it lets one explore possible choices of u when doing substitution. The command for doing this is changevar, which takes as arguments (1) the substitution you want to do, written either in the form NiMvJSJ1Ry0lImZHNiMlInhH or in the form NiMvJSJ4Ry0lImZHNiMlInVH, (2) the inert integral to be computed, and (3) the name of the new variable. For instance, here are a bunch of different substitutions applied to the integralpuzzle := Int(x^2/(1+x^6),x);changevar(u=x^2, puzzle, u);changevar(u=x^6, puzzle, u);changevar(u=1+x^6, puzzle, u);changevar(u=x^3, puzzle, u);The last of these has an obvious integral using the arctan function, so apparently that's the right substitution to have used. But isn't it nice to be able to prospect for a good choice so quickly?Integration by PartsThe analogous command letting you explore possible integrations by part is intparts, whose arguments are (1) the integral you are working on, and (2) your choice of u (the rest of the integral will automatically be taken to be v'). For instance, here are some possible ways to do integration by parts on the integralpuzzle2 := Int(x^3*exp(x^2),x);intparts(puzzle2, x^3);intparts(puzzle2, exp(x^2));intparts(puzzle2, x);intparts(puzzle2, x^2);Bingo! The last of these finally gave us an integral that doesn't involve weird functions like erf, and that we can do with the substitution NiMvJSJ1RyokJSJ4RyIiIw==. (Actually, I would have done this substitution first, and then used parts; but doing parts first shows off better the power of intparts as a prospecting tool.Problems for you.Here are a few problems you might want to do in order to reinforce the ideas in this handout, and to get comfortable using these commands. You need not turn these in, but I would be happy to read them for extra credit if you do want to turn them in.1. Use Maple to explore the sumsNiMtJSRTdW1HNiQqJiIiIkYnKSUibkclInNHISIiL0YpO0YnJSlpbmZpbml0eUc=for various values of s. What do you observe?2. As efficiently as possible, compute the 10th derivative of NiMtJSRzZWNHNiMlInhH.3. What happens when you use the substitution NiMvJSJ1Ry0lJ2FyY3Npbkc2IyUieEc= on the integralNiMtJSRJbnRHNiQtJSVzcXJ0RzYjLCYiIiJGKiokJSJ4RyIiIyEiIkYs?What if you use NiMvJSJ1Ry0lJ2FyY3Rhbkc2IyUieEc= on the integralNiMtJSRJbnRHNiQtJSVzcXJ0RzYjLCYiIiJGKiokJSJ4RyIiI0YqRiw=?Can you understand what's going on by doing these integrals by hand? Can you find other integral where these substitutions work? Be creative!4. Use Maple to explore sums likeNiMtJSRTdW1HNiQpJSJrRyUic0cvRic7IiIiJSJuRw==for various values of s, and see if you observe anything. Also, use Maple to look at sums likeNiMtJSRTdW1HNiQqJiUia0ciIiIsJkYnRihGKEYoRigvRic7RiglIm5H, NiMtJSRTdW1HNiQqKCUia0ciIiIsJkYnRihGKEYoRigsJkYnRigiIiNGKEYoL0YnO0YoJSJuRw==, NiMtJSRTdW1HNiQqKiUia0ciIiIsJkYnRihGKEYoRigsJkYnRigiIiNGKEYoLCZGJ0YoIiIkRihGKC9GJztGKCUibkc=, . . .See if you notice anything here. In both parts of this problem, it would be useful to factor the expressions you compute.5. Let f(x) be a parabolaf := x -> a*x^2 + b*x + c;What must a, b, and c be in order for f to have the same value, the same first derivative, and the same second derivative as NiMtJSRzaW5HNiMlInhH at NiMvJSJ4RyIiIQ==? (COmpute the derivatives of both functions at 0, set them equal to one another, and solve the resulting system of equations.) Plot sin(x) together with this parabola.Now take a generic cubic, NiMvLSUiZkc2IyUieEcsKiomJSJhRyIiIiokRiciIiRGK0YrKiYlImJHRisqJEYnIiIjRitGKyomJSJjR0YrRidGK0YrJSJkR0Yr, and try to make its value and its first, second, and third derivatives equal to the value and the first 3 derivatives of sin(x) at 0. Again, plot this cubic together with sin(x).Keep doing this with polynomials of increasing degree. You should be getting curves that approximate the sine function better and better.6. Ask Maple about the sumNiMtJSRTdW1HNiQqKCksJCIiIiEiIiUibkdGKSklInhHLCYqJiIiI0YpRitGKUYpRilGKUYpLSUqZmFjdG9yaWFsRzYjRi5GKi9GKzsiIiElKWluZmluaXR5Rw==.What has this got to do with Problem 5?