Calculus QuestionsDate: 27 Jan 1995 15:53:28 -0500 From: Anonymous Subject: Ask Dr. Math Why can't 4*sqr(1-x^2) be integrated from 0 to 1 to get an exact value of pi? If it can be integrated, how? What significance does the exact value of the integral of 2^x from 0 to .5 have? What is the formula for the integral of 2^x? Is there a formula or recursive sequence for estimating or finding the exact value of a non-decimal factorial? What is the formula for solving cubic equations for zero? I know it's long, but I'd like to know it. Date: 28 Jan 1995 14:22:32 -0500 From: Dr. Ken Subject: Re: Ask Dr. Math Hello there! I'll handle these questions one by one. 1. Why can't 4*sqr(1-x^2) be integrated from 0 to 1 to get an exact value of pi? If it can be integrated, how? Sure, we can integrate that. Try a substitution, and let x = Cos[u]. Then we can take the integral, and sure enough, we end up with exactly pi. But when you say "an exact value," you probably mean an exact decimal representation, right? Well, the answer is that we could get really close to the exact decimal representation, which is infinite. In fact, we could get as close as we want to it by taking approximations to this integral using Riemann sums and the like, but there is no method of determining how the decimal expansion for pi looks everywhere, and I doubt there ever will be. There are a number of other ways of plowing into the digits of pi. We can use the fact that Pi 2 x 2 x 4 x 4 x 6 x 6 x 8 x 8 x 10 x 10 x ... ---- = --------------------------------------------- 2 3 x 3 x 5 x 5 x 7 x 7 x 9 x 9 x 11 x 11 x ... which was stated by John Wallis in 1655. Or we could use the series Pi 1 1 1 1 ---- = 1 - --- + --- - --- + --- - ... 4 3 5 7 9 Which was discovered by Leibniz in 1674. Any of these methods can be used to find a good many digits of pi by just taking part of the big series or part of the fraction product. 2. What significance does the exact value of the integral of 2^x from 0 to .5 have? What is the formula for the integral of 2^x? Sure, we can do this integral. We usually write exponential things in terms of e (instead of 2), so 2^x becomes e^(Log[2^x]) = e^(x * Log[2]). Oh, by the way, when I say "Log," I mean the natural logarithm. So what happens if we make a guess of e^(x * Log[2]) for our integral? Well, its derivative is [e^(x * Log[2])] * Log[2]. And Log[2] is just a constant, so we get [e^(x * Log[2])]/Log[2] = (2^x)/Log[2] for our integral. So if we plug in your limits of integration, we get (2^(.5) - 2^0)/Log[2], which equals (Sqrt{2} - 1)/Log[2]. This gives an exact value for the area between the curve 2^x and the x-axis from 0 to 1/2. 3. Is there a formula or recursive sequence for estimating or finding the exact value of a non-decimal factorial? Sure, this is called the Riemann Zeta function. I'll denote it in this message as Z(x). When x is a positive integer, Z(x) matches up with x!. The formula for the Zeta function is 1 1 1 1 1 1 1 --- + --- + --- + --- + --- + --- + --- + .... 1^x 2^x 3^x 4^x 5^x 6^x 7^x Extension question: for what values of x is this function well-defined, meaning for what values of x will this function give you a real number? 4. What is the formula for solving cubic equations for zero? I know it's long, but I'd like to know it. Okay, here's how you do it. Let's say you have the equation ax^3 + bx^2 + cx + d = 0. The first thing you do is get rid of the a out in front by dividing the whole equation by it. Then we get something in the form x^3 + ex^2 + fx + g = 0. The next thing we do is get rid of the x^2 term by replacing x with (x - e/3). That will give us something of the form x^3 + px + q = 0. This is much easier to solve, although it's still hard. Now introduce two new variables, t and u, defined by u - t = q and tu = (p/3)^3. Then x = CubeRoot{t} - CubeRoot{u} will be a solution of x^3 + px + q = 0. Verify this result now, and make sure you see why it works. To find the other two solutions (if there are any) we could divide x^3 + px + q by its known factor (x - CubeRoot{t} + CubeRoot{u}), getting a quadratic equation that we could solve by the quadratic formula. So that's the basic idea behind the cubic. If you wanted to find the actual expression for t and u in terms of p and q, you could solve those two equations defining p and q (substitution would probably be easiest). Then you could obtain a real formula for x in terms of p and q. I hope this entertains you! Great questions! -Ken "Dr." Math Date: 30 Jan 1995 21:47:00 -0500 From: Anonymous Subject: Re: Ask Dr. Math Thank you for taking your time to reply. I am having trouble with the answers to my 2nd and 3rd questions. On the second question, I don't know if I'm doing something wrong or not, but zeta isn't coming close to factorial at all. For example, zeta(0) = infinity because x^0 =1 , making zeta(1) = 1+1+1.... You said it works for positive numbers, so 0 could be a counterexample, but zeta(1) goes beyond 1 very quickly. On the third question, you say to replace x with x-e/3. I don't understand this step or how it helps to simplify the equation. Is it some kind of system? I also assume that e = b/a, f = c/a, and g = d/a, and that e is not necessarily equal to about 2.71. Thank you once again for your time. Date: 31 Jan 1995 17:16:14 GMT From: Dr. Math Subject: Re: Ask Dr. Math Oh, goodness. You're right. Yup, it's not the Riemann-Zeta function at all. I apologize unreservingly. Rather, the function you're looking for is the Gamma function. It's defined as Gamma[x] = Integral from 0 to Infinity of (t^(x - 1) * e^-t dt). Now this should work a whole lot better. When you plug in 1 for x, what do you get? You get the Integral from 0 to Infinity of (e^-t dt), which is (-e^-t) from 0 to Infinity, which is 0 - (-1) = 1. Bingo. Just for kicks, try Gamma[2]. Do the integral by parts. Then try it for x+1, where x is general, evaluating the integral by parts: u = t^x du = x*t^(x-1) v = -e^-t dv = e^-t dt So Gamma[x + 1] = (-t^x * e^-t) from 0 to Infinity + Integral(x*t^(x-1)*e^-t dt) = 0 - 0 + x * Integral(t^(x-1) * e^-t dt) = x * Gamma[x]. That's why Gamma[n+1] matches up with n factorial. About the cubic stuff: yes, I just mean e as a new variable, not the base of the natural logorithm. So let's say you had the equation x^3 - 3x^2 + 4x - 6 = 0. Then replace x by y+1, and we get y^3 + 3y^2 + 3y + 1 - 3(y^2 + 2y + 1) + 4(y + 1) - 6 = 0, or more simply, y^3 + y - 8 = 0. Since this equation will be easier to solve (i.e., you and I know the method), we'll solve it, and then when we find y we'll add 1 to it to get x. By the way, the above equation wasn't one I specifically designed to come out well when I did the substitution, it was just one I thought of off the top of my head, making sure that the coefficient on x^2 was divisible by 3 so I wouldn't have to type in a lot of fractions. So the replacement x = y - e/3 was something designed to simplify the equations. And as far as whether e = b/a, f = c/a, and g = d/a, sure, that's what they are. I just divided through by a. Basically, the people who solved the cubic demonstrated that if you could solve equations of the form x^3 + px + q = 0, you could solve _any_ cubic equations, because you could always get them in that form. Anyway, I'm sorry about that screw-up in the Zeta/Gamma function area. -Ken "Dr." Math |
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