Wikipedia:Reference desk/Archives/Mathematics/2014 December 26

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December 26

Is there any trick or shortcut for solving integration problems?

Following is an integration problem with four options, out of which only one option is correct. I know how to solve this problem by general method, but I want to know how this can be solved by using some shortcut method or any other trick. Is it possible to select the correct option even without solving the original question? One way is - differentiate all four options and check which one matches with the given question.
integral((5 x^8+7 x^6)/(2 x^7+x^2+1)^2, x) =
A. (-x^7)/(2 x^7+x^2+1)+c
B. ln(x^7/(2 x^7+x^2+1))+c
C. -ln(x^7/(2 x^7+x^2+1))+c
D. (x^7)/(2 x^7+x^2+1)+c
Answer: D. (x^7)/(2 x^7+x^2+1)+c — Preceding unsigned comment added by 106.215.150.121 (talk) 20:30, 26 December 2014 (UTC)

Since it is multiple choice, an easy first pass is to look at the asymptote and sign behavior. In this case the integrand is positive everywhere, thus the corresponding integral must be an increasing function of x. This rules out A and C which decrease from x = 0 to x = infinity. We can also observe that the integrand is well-defined and finite everywhere that $2x^{7}+x^{2}+1\neq 0$; however, answer B is ill-defined for small negative numbers, e.g. x = -0.1, $ln(x^{7}/(2x^{7}+x^{2}+1))\approx ln(-10^{-7})$ which is imaginary, so we can also rule out B. Hence D. I'm sure there are other ways to argue this as well with doing any integrals or derivatives. In general, it is often the case that multiple choice questions can be solved (or at least some options eliminated) without bothering to do significant math. Of course, taking that approach doesn't necessarily prove any mathematical skill (just savvy test taking ability), so personally, I would say multiple choice math tests often aren't very useful for measuring mathematical ability. Assuming you are a student, I would encourage you to also learn the direct way to do the computation as that is likely to be more useful in your future pursuits than simply figuring out how to beat the test. Dragons flight (talk) 00:00, 27 December 2014 (UTC)

@Dragons flight: When it appears in an antiderivative, at least for the real number system, it is assumed that natural logarithm's argument has its absolute value taken, $\int {\frac {dx}{x}}=\ln \vert x\vert +C$, so we cannot use this to rule out B.--Jasper Deng (talk) 03:48, 27 December 2014 (UTC)

Yes, absolute value would be typical for an antiderivative, but on a multiple choice test I would never assume a given answer option contains any operations aside from what was actually written, so I would stand by my reading that B is not a reasonable answer. (Unless, the original poster omitted the absolute value indicators by accident when writing here or the question explicitly limited the range of x in some other way.) Dragons flight (talk) 04:14, 27 December 2014 (UTC)

If you want to poke at things a different way, you could note that B at x=0 is +infinity, but the integrand is finite at x=0, which is a different way of noting the natural log is a nonsense answer. Dragons flight (talk) 04:47, 27 December 2014 (UTC)

I assume you want $\int {\frac {5x^{8}+7x^{6}}{(2x^{7}+x^{2}+1)^{2}}}dx$. I would recommend working it out using long division followed by partial fraction decomposition; one thing I dislike about multiple choice integration problems is that they encourage shortcuts instead of actually learning the general method. In real life you do not get antiderivatives by guessing or shortcutting.--Jasper Deng (talk) 01:49, 27 December 2014 (UTC)

How about if you graph it, between convenient limits, like 0 and 1, and look at the area under the curve ? That should match one of the answers (where C = 0). If it happens to match more than one, then repeat with different limits until it only matches one answer. This might be a quick method, if you are allowed a graphing calculator on the test. StuRat (talk) 06:29, 27 December 2014 (UTC)

It's rather easy to see that differentiation of B or C cannot give the integrand, so it's just down to determining the sign. That's easy since the integral over [0,1] (e.g.) must be positive. Sławomir Biały (talk) 14:26, 27 December 2014 (UTC)

This is the trick:

www.wolframalpha.com/input/?i=integral++%285x^8%2B7x^6%29%2F%282x^7%2Bx^2%2B1%29^2+dx

Answer:

integral (5 x^8+7 x^6)/(2 x^7+x^2+1)^2 dx = -(x^2+1)/(4 x^7+2 x^2+2)+constant

Is it equal to option D?

www.wolframalpha.com/input/?i=%28x^7%29%2F%282+x^7%2Bx^2%2B1%29%2B%28x^2%2B1%29%2F%284+x^7%2B2+x^2%2B2%29

Input: x^7/(2 x^7 + x^2 + 1) + (x^2 + 1)/(4 x^7 + 2 x^2 + 2)

Answer: 2^(-1)

Yes, the difference is constant. Bo Jacoby (talk) 14:29, 28 December 2014 (UTC).

If you're doing this for homework or an exam, WA isn't (typically) allowed. For teaching purposes I could only ever suggest it to check existing answers.--Jasper Deng (talk) 09:43, 30 December 2014 (UTC)

In response to

In real life you do not get antiderivatives by guessing or shortcutting

posted a little bit up: I partly disagree. As opposed to differentiation, integration has an element of guessing to it. An integration problem is really a differential equation, and there is certainly no catch-all algorithm. Of course, the "guessing" can be more or less ingenious. YohanN7 (talk) 11:31, 30 December 2014 (UTC)

Yes, integration requires constant guessing. :-) StuRat (talk) 00:55, 1 January 2015 (UTC)

See Risch algorithm for a catch-all algorithm (kinda). Staecker (talk) 01:03, 1 January 2015 (UTC)

Just to add to this, it was suggested that partial fractions be used to evaluate the integral. This seems to be very impractical: the denominator has degree 14, and the numerators that one must deal with get much worse after long division. In a best case scenario, one must factor a degree six polynomial (optionally over either the reals or complex numbers). I don't see any obvious simplifications. (Also, I don't really know much about the internals of the Risch algorithm, but it is surely dependent on some non-trivial facts about field theory that allow solutions of algebraic equations to be represented exactly.) Sławomir Biały (talk) 14:17, 1 January 2015 (UTC)

Bridge hand probability

In a particular bridge hand, the contract can be made, in one way, by playing a finesse which will work 50% of the time depending upon whether E or W has the missing higher card.

There is an alternative 'plan' that needs the splits between the E&W hands to be within certain limits: E&W have 7 spades, 8 hearts, 5 diamonds and 6 clubs. For this plan to work, the spades must be split 1:6 or more evenly; the hearts must be split 2:6 or more evenly; the diamonds must be split 3:2; and the clubs must be split 4:2 or more evenly. It doesn't matter which way round the split is, but obviously each hand must have 13 cards in it.

What is, and how do I work out, the probability of scheme 2 working?

Merry Christmas. -- SGBailey (talk) 22:53, 26 December 2014 (UTC)

Not a very formal answer for the math desk, but using Monte Carlo methods I estimate that the split required by the second method will occur about 56.7% of the time. Dragons flight (talk) 23:37, 26 December 2014 (UTC)

So better than the finesse, but not much better. Thanks -- SGBailey (talk) 19:04, 27 December 2014 (UTC)

For an exact answer, I don't think there's an easier way (sorry, no cite for a reference to say so!) than to enumerate all the possible distributions, compute the probability for each one, and add. It turns out that there are 28 possible distributions, and if East-West have S spades, H hearts, etc. (S+H+D+C = 26), then the probability that East has exactly s of the S spades, h of the H hearts, etc. (s+h+d+c = 13) is just C(S,s)C(H,h)C(D,d)C(C,c)/C(26,13). This Perl code does the job:

    use Math::Counting ("combination");
     
    my ($S, $H, $D, $C) = (7, 8, 5, 6);
    my ($tot, $cases) = (0, 0);
    for my $s (1 .. 6) {
      for my $h (2 .. 6) {
        for my $d (2 .. 3) {
          for my $c (2 .. 4) {
            next unless ($s + $h + $d + $c == 13);
             
            ++$cases;
            $tot += combination($S,$s) * combination($H,$h) * combination($D,$d) * combination($C,$c);
          }
        }
      }
    }
    printf "%d cases evaluated, probability is %.4f%%\n", $cases, 100*$tot/combination(26,13);

And the output confirms Dragons' simulation to 3 signficant digits:

    28 cases evaluated, probability is 56.7237%

--65.94.50.4 (talk) 06:38, 28 December 2014 (UTC)

This J code avoids the for-loops.

  s=.1 2 3 4 5 6
  h=.2 3 4 5 6
  d=.2 3
  c=.2 3 4
  (+/,(13=s+/h+/d+/c)*(s!7)*/(h!8)*/(d!5)*/(c!6))%13!26
0.567237

Bo Jacoby (talk) 21:43, 28 December 2014 (UTC).