A math problem on Glenn Beck – can you solve it?

As I was flipping through the channels, I was surprised to see a math expression written on one of the chalkboards of the Glenn Beck program.

And it wasn’t something simple: the expression involved an infinite series. Can you solve the problem? (pictured below)


Here is a screen capture that I took from the Fox News website.

I magnified the equation, but you can click on the picture to see a larger image:

The problem reads:

 

Can you figure it out?

I’ll give you a hint: as with many things on Glenn Beck, you should double check the assumptions.



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  • Scott

    Given the use of sin and cos, I suspect some periodicity regarding this series.

    For n=(pi/2), (3pi/4), pi, (5pi/4), (3pi/2), (7pi/4), 2pi, (9pi/4), and (5pi/2);

    2n = pi, (3pi/2), 2pi, (5pi/2), 3pi, (7pi/2), 4pi, (9pi/2), 5pi; and

    sin(2n) = 0, -1, 0, 1, 0, -1, 0, 1, 0

    Likewise:
    cos(n) = 0, (-sqrt(2)/2), -1, (-sqrt(2)/2), 0, (sqrt(2)/2), 1, (sqrt(2)/2), 0;

    cos^4(n) = 0, (1/4), 1, (1/4), 0, (1/4), 1 (1/4), 0; and

    1+cos^4(n) = 1, (5/4), 2, (5/4), 1, (5/4), 2, (1/4), 1;

    This gives sin(2n)/(1+cos^4(n)):

    0, -(4/5), 0, (4/5), 0, (-4/5), 0, (4/5), 0

    The sum of which is, of course, 0.

    However, the any partial sum will only be zero if n is a multiple of (pi/2). For any arbitrary value of n, the sum of that partial series will fluxtuate from (-4/5) to (4/5), which to me says it is not convergent.

  • Faraz

    Since the index spans natural numbers, I would imagine sin and cos are in degrees. Under this assumption:

    Write f(n) = sin(2n)/(1+cos^4(n))

    Put n = k (mod 360) (congruence)

    Clearly, f(n) = f(k).

    Note that for k = 0, 1,… 180, f(k) + f(180-k) =
    f(k) + sin(360-2k)/(1+cos^4(180-k))=

    f(k) + -sin(2k)/(1+cos^4(k)) since the fourth power of cos means the sign of cos k doesn’t matter.

    But this equals zero.

    Likewise, for k = 181, 182… 359,
    f(k) + f(360-k) =
    f(k) + sin(720 – 2k)/(1+cos^4(360-k))=
    f(k) + -sin(2k)/(1+cos^4(k)) = 0.

    Hence SUM f(n) for n = 0,1,… 359 (mod 360) = 0.

    This is kind of a telescoping sum, but it’s not convergent by the true definition (due to the periodicity of f(n)).

    In radians, however, the story would be different. Note that if m and n are natural numbers, it’s not possible that sin (n) = sin (m) if its measured in radians. I’m not sure what the answer would be in that case.

  • sdfdsv

    ahahaha it’s april 1st, check the limit test

  • http://www.youtube.com/user/JuusoAlasuutari Juuso Alasuutari

    Number goes up, another number goes down. I can’t explain that.

  • uri vasas

    It can be solved using the integral test for convergence (I am cheating, because I saw the program and the the kid started it that way, but I believe I could’ve figured it out anyway)

    Take the integral of the stated function from 1 to infinity.substitute u = (cosx)^2 then du = -2cosxsinxdx.
    The numerator of the integrand is equivalent to 2cosxsinx. So the entire integrand is equivalent -1/(1+u^2).

    I will not change the bounds of the integral as this is tricky to do with the upper bound going to infinity and u being a periodic function.

    once you integrate you have arctan((cosx)^2) going from 1 to infinity. This is equal to arctan((cos1)^2) – lim ( t->infinity)arctan((cos(t))^2) =:I

    I just realized that this next step is iffy (and I think this means the sum doesn’t converge) but I will do it anyway. Here it is: because arctan(y) >= -pi/2, (or -arctan(y) <= pi/2) we get that
    I <= arctan((cos1)^2) + pi/2 < infinity. Which means that the sum converges.

    The reason this is iffy is that although IF I exists then it is less than infinity, this does not mean that I exists. In fact as t goes to infinity, cos(t)^2 oscillates from zero to one making arctan(cos(t)^2) oscillate as well making the limit divergent. Which means the sum diverges.

    Way to go, Glenn Beck. You know calculus about as well as you know politics. I wonder if the kid would have spotted the error.

  • guy

    Uri, recall the restrictions must be taken into taken consideration before we can use the integral test for convergence. Our series must be continuous, positive, and monotonically decreasing on [1,inf.) A quick graph of the series shows that it fails two of these criteria, so we cannot use the integral test for convergence in this case.

  • http://www.mindyourdecisions.com/blog/ Presh Talwalkar

    Someone found out the context of this question was Glenn Beck was interviewing a 12 year old math genius Jacob Barnett.

    Here is the clip: http://www.youtube.com/watch?v=hBW4S9xcTOk&t=9m39s

    Jacob starts to solve by using the integral test to show it does converge.

    I believe this is a mistake. As the sequence does not converge, the sum does not converge either by the limit test.

    The answer can also be calculated in Wolfram Alpha: http://www.wolframalpha.com/input/?i=sum+from+n%3D1+to+n%3Dinf+%28sin%282n%29%2F%281%2Bcos^4%28n%29%29%29

    Special thanks to Reddit comments for the Youtube and Wolfram Alpha links.

  • Pingback: Are Pillsbury Toaster Strudels tastier than Kellogg’s Pop Tarts – ads I overanalyze - Mind Your Decisions

  • http://asad123.wordpress.com Asad

    I showed this to my friend Junaid who explained it very simply. Look at the ranges of the numerator and denominator. The numerator, being a sine, ranges from -1 to 1. The demoninator ranges from 1 to 2. The limit of the series doesn’t approach 0, and it’s not an alternating series where positives and negatives cancel, so it must diverge.

    I noticed on YouTube people are saying this kid is so amazing. I commented, under the handle ibnsina786, that he has his math wrong and directed them to come here for proof. I hope you get some traffic out of it.

  • http://www.mindyourdecisions.com/blog/ Presh Talwalkar

    Thank you Asad, that was really nice to post a link on Youtube.

  • Cccil Rousseau

    Previous solvers have had everthing in place except a rigorous proof that f(n) doesn’t approach zero. Let’s review. (i) The integral test doesn’t apply because the integrand is not monotone decreasing.
    (ii) If we prove that f(n) doesn’t approach 0 the game is over and the series diverges, (iii) We might as well change the nth term to sin(n) because the denominator remains beween 1 and 2; sin(n)/(1+ cos^4(n)) is small iff sin(n) is small. (iv) Now for the slightly technical part. The terms of the series are sin(1), sin(2), sin(3), … where 1,2,3,… are radians measure. The sine function is periodic with period 2 pi. Take the domain to be [0, 2 pi]; then the terms are sin(1), … sin(6), sin(7-2 pi), sin(8-2 pi),… Think of a frog jumping from point to point on the interval [0, 2 pi]. If he is at point x, then his next jump will be to x + 1 mod 2 pi. Note that |sin x| is small iff x is in one of the neighborhoods (lilypads)|x| < delta, |x – pi| < delta, or |x- 2 pi| < delta where delta is small. Use contradiction. Suppose sin(n) converges to 0. At some point our frog is on a lilypad and needs to jump to another one a distance of approximately pi or 2 pi away and his only option is a jump of +1. This is impossible. The series diverges.

    Note. Think of the process of going from f(x) to f(n) as sampling; informatin is lost. (In effect that's why you need the monotone condition in the integral test.) The example f(x) = sin(pi x) has f(n) = 0 for all n and yet f(x) does not approach 0. It may seem like overkill but we have to have some kind of argument – we just can't assume that f(n) does not converge to 0 because f(x) does not converge to 0.

  • Cccil Rousseau

    Sorry. I was doing this from memory and I forgot that it was sin(2n). Please withdraw this and I will modify the argument.

  • Cccil Rousseau

    My failure to remember the factor of 2 is easily fixed. With the correct term sin(2n), there are 5 lilypads instead of 3, and these are neighborhoods of 0, pi/2, pi, 3pi/2, and 2pi (the points in [0, 2pi] where sin(2n) = 0). Again the frog finds himself on one of the lilypads and needing to jump to another one with a jump of +1. If the size of the lilypad is small enough then this is impossible. This argument can be extended to sin(kn) and ultimately depends on the irratioanality of pi.

  • Adrian

    the series can be evauluated using the integral test which proves it is divergent simply because at the end you’ll have arctan(cos^2(n)) as n approaches infinity …this already leaves an indeterminate bound between 0 and pi over two since cosine squared oscillates from 0 to 1 hence this is a divergent infinite series…. although its a nice one :)

  • Stephen

    When I saw the problem, I assumed that it was a typo, and they meant sinh and cosh instead of sin and cos.





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