the question is shown on the website http://home.attbi.com/~josephsieh/cp4.htm

Hi, wjs-ga:

Restrict the even function x^4 to the interval [-pi,pi].  The Fourier
expansion will thus consist only of even components:

x^4 = (a_0)/2 + (1/pi) SUM a_n cos(n x) [FOR n = 1,2,3,...]

where a_n = (1/pi) INTEGRAL x^4 cos(n x) dx [OVER x = -pi to pi]

When n = 0, we have simply:

a_0 = (1/pi) INTEGRAL x^4 dx [OVER x = -pi to pi] 

 = DIFF x^5/5 BETWEEN -pi and pi = 2(pi^4)/5

The other integrals, with n > 0, are only slightly more difficult with
the substitution u = nx:

a_n = (1/pi) INTEGRAL x^4 cos(n x) dx [OVER x = -pi to pi]

 = (1/pi) (1/n)^5 INTEGRAL (nx)^4 cos(n x) d(nx) [OVER x = -pi to pi]

 = (1/pi) (1/n)^5 INTEGRAL u^4 cos(u) du [OVER u = -pi*n to pi*n]

 = (1/pi) (1/n)^5 DIFF [ 4*u*(-6 + u^2)*cos(u) + (24 - 12*u^2 +
u^4)*sin(u) ]
BETWEEN -pi*n and pi*n

where I have taken advantage of the symbolic integration provided on
your link:

http://zen.uta.edu/cgi-bin/math/math_def_int.cgi?FUNCTION=x%5E4+Cos%5Bx%5D&VAR=x&LOW=0&HIGH=a

Notice that sin(pi*n) = 0 for all integers n, while cos(pi*n) =
(-1)^n.  So:

a_n = (1/pi) (1/n)^5 [ 8*(pi*n)*(-6 + (pi*n)^2)*((-1)^n) ]

 = (1/pi) (-1)^n (1/n)^5 [ 8(pi*n)^3 - 48(pi*n) ]

 = (1/pi) (-1)^n [ 8(pi^3)(1/n)^2 - 48pi(1/n)^4 ]

Putting these results together we have the Fourier series expansion of
x^4 on the interval [-pi,pi]:

x^4 = (a_0)/2 + SUM a_n cos(n x) [FOR n = 1,2,3,...]

 = (pi^4)/5 + SUM (-1)^n [ 8(pi^2)(1/n)^2 - 48(1/n)^4 ] cos(n x) [FOR
n = 1,2,3,...]

Your first hint suggests a substitution at x = pi, and we recall that:

cos(n x) = (-1)^n

Hence:

pi^4 = (pi^4)/5 + SUM 8(pi^2)(1/n)^2 - 48(1/n)^4 [FOR n = 1,2,3,...]

For simplicity we will understand SUM's to be indexed as above, over
the positive integers for n:

48 SUM (1/n)^4 = (pi^4)/5 - (pi^4) + 8(pi^2) SUM (1/n)^2

Now the last term is well known from:

SUM (1/n)^2 = (pi^2)/6

8(pi^2) SUM (1/n)^2 = 4(pi^4)/3

Thus:

48 SUM (1/n)^4 = (pi^4) * [ (1/5) - 1 + 4/3 ]

 = 8(pi^4)/15

and finally:

SUM (1/n)^4 = (pi^4)/90

This result may be confirmed by consulting any number of sources,
e.g.:

http://mathworld.wolfram.com/RiemannZetaFunction.html

where about 2/3's of the way down the page are listed values of the
Riemann zeta function for small integer arguments, i.e. zeta(2) and
zeta(4) for our purposes.

regards, mathtalk-ga

---

Clarification of Answer by mathtalk-ga on 17 Nov 2002 07:16 PST

Oops... in my very first equation I have a redundant (1/pi).  That
multiplier on the summation is actually absorbed into the individual
coefficients, as shown on the next line.

It should read:

x^4 = (a_0)/2 + SUM a_n cos(n x) [FOR n = 1,2,3,...] 
 
where a_n = (1/pi) INTEGRAL x^4 cos(n x) dx [OVER x = -pi to pi] 

regards, mathtalk-ga

---

Clarification of Answer by mathtalk-ga on 17 Nov 2002 07:17 PST

Another pesky factor of (1/pi), this time gone missing in the
computation of a_0, which should have read:

a_0 = (1/pi) INTEGRAL x^4 dx [OVER x = -pi to pi]  
 
 = (1/pi) DIFF x^5/5 BETWEEN -pi and pi = 2(pi^4)/5 

regards, mathtalk-ga

sorry, 

  but it does not make sense as stated.

   Can you elaborate?

The question and the coment are shown on the website
http://home.attbi.com/~josephsieh/cp.htm

Please check it out.

Hi wjs-ga,

I am interested in answering your question, but I believe that, to answer 
it well, your question will require more time and effort than the average 
amount of time and effort associated with this price. Here is a link to 
guidelines about pricing your question, 

https://answers.google.com/answers/pricing.html 

If you both raise your price and also post a clarification here, the 
system will notify me and I will take another look at your question.

I basically agree with rbnn-ga that the price offered is somewhat less
than the guidelines given by Google.  However since the service is
still in beta, and since the clarification included a symbolic
integration link, I felt it came closer to being a reasonable price
for the sake of customer good will.

BTW, rbnn-ga, congratulations on being selected Google's Researcher of
the Week last week!  Your work is very high quality.

regards, mathtalk

Thank you for your kind words, mathboy. Let me respond to your comment
though: the price at which you answered, at which price I also would
have endeavored to answer,  had been changed from an earlier price of
five; it was the first price, which is not viewable any more, to which
my earlier comment was directed.

I enjoyed reading your answer by the way; it's interesting and
complete.

Also, thank you wjs-ga for a very interesting and elegant question! I
enjoyed thinking about it and reading the answer, and I learned some
pretty mathematics as well.