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Link between Pascal's triangle and integrals of trigonometric functions

  1. Nov 4, 2009 #1
    My lecturer keeps simplifying trigonometric integrals in one line such as
    [tex]\int^{2\pi}_{0}sin^{4}(t)dt=\frac{3\pi}{4}[/tex]
    and writes pascals triangle next to it. Just wondering what's the link between them? I'm sure it's obvious and easy, I'd just like to have an fast way of dealing with these
     
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  3. Nov 4, 2009 #2

    HallsofIvy

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    I have no idea what your teacher is doing! However you can express powers of sin(x) in terms of things like sin(4x). The "double angle identity", [itex]cos(2x)= cos^2(x)- sin^2(x)[/itex] can be rewritten as [itex]cos(2x)= cos^2(x)- (1- cos^2(x))= 2cos^2(x)- 1[/itex] or as [itex]cos(2x)= (1- sin^2(x))- sin^2(x)= 1- 2sin^2(x)[/itex].

    Those can be "solved" as [itex]cos^2(x)= (1/2)(1+ cos(2x))[/itex] and [itex]sin^2(x)= (1/2)(1- cos(2x))[/itex].

    So sin4(x)= (sin2(x))2= ((1/2)(1- cos(2x))2[/itex][itex]= (1/4)(1- 2cos(2x)+ cos^2(2x))[/itex]. And [itex]cos^2(2x)= (1/2)(1+ cos(2(2x)))= (1/2)(1+ cos(4x)[/itex] so
    [itex]sin^4(x)= 1/4- (1/2)cos(2x)+ (1/2)(1+ cos(4x))[/itex][itex]= 3/4- (1/2)cos(2x)+ (1/2)cos(4x)[/itex].

    Since the integrals of cos(2x) and cos(4x), from 0 to [itex]2\pi[/itex], are 0, the only thing remaining is that "1/4" and its integral from 0 to [itex]2\pi[/itex] is, of course, [itex](1/4)(2\pi)= \pi/2[/itex].
     
  4. Nov 4, 2009 #3

    arildno

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    Halls:
    That should be:
    [tex]\frac{1}{4}(1-2\cos(2x)+\frac{1}{2}(1+\cos(4x)))[/tex]
    yielding 3/8 to integrate, agreeing with the teacher's answer.
     
  5. Nov 4, 2009 #4
    Maybe your teacher is using the complex exponential version of trig functions...

    [tex]\sin^4 t = \left(\frac{e^{it}-e^{-it}}{2i}\right)^4[/tex]

    and Pascal's triangle is of obvious help in expanding the numerator of the power.
     
  6. Nov 4, 2009 #5

    arildno

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    The correct relation is fairly easy to gain by repeated use of partial integration:

    1. Let [tex]I_{n}=\int_{0}^{2\pi}\sin^{n}tdt=-\cos(t)\sin^{n-1}t\mid_{t=0}^{t=2\pi}+\int_{0}^{2\pi}(n-1)\cos^{2}t\sin^{n-2}t=(n-1)\int_{0}^{2\pi}(\sin^{n-2}t-\sin^{n}t)dt=(n-1)I_{n-2}-(n-1)I_{n}[/tex]

    2. Thus, we have gained the recurrence relation:
    [tex]I_{n}=(n-1)I_{n-2}-(n-1)I_{n}\to{I}_{n}=\frac{n-1}{n}I_{n-2}, n\geq{2}[/tex]

    (The case is identical if cosine constitutes our base, rather than sine, the exponent "n" remaining the same)
    3. We note that [tex]I_{0}=2\pi, I_{1}=0[/tex]
    Thus, we easily see that for "n" odd, the integral will be 0.
    Henceforth, we assume even n=2p for natural number p.

    4. We may write, for any particular "n", the value as a "sew-saw"-pattern starting with "n" as the value in the leftmost denominator:
    [tex]I_{n}=(\frac{n-1}{n}*\frac{n-3}{n-2}*.....*\frac{1}{2})*2\pi[/tex]

    5. This can again be twiddled into an explicit form as follows:
    With n=2p, the denominator [itex]n*(n-2)*(n-4)...2=2^{p}(p!)[/itex], where p! is the factorial of p.
    Multiplying both the numerator and denominator with this expression yields:
    [tex]I_{n}=\frac{n!}{2^{n}(p!)^{2}}*2\pi=\pi*2^{1-n}*\frac{n!}{(n-p)!p!}=\pi*2^{1-n}*\binom{n}{p}[/tex]

    The binomial coefficients are closely related to..Pascal's triangle. :smile:
     
    Last edited: Nov 4, 2009
  7. Nov 4, 2009 #6
    That looks like a good idea.

    I'd prefer a slight improvement
    [tex]\sin^4 t = \Im\left(e^{it}\left(\frac{e^{it}-e^{-it}}{2i}\right)^3\right)[/tex]
    Then it's easier to convert back to tri-functions.
     
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