Recently I have been thinking a lot about variations of the Basel Problem, and methods to solve them. Here I found the following solution to the Basel Problem by Alfredo Z. (I include the entire answer due to its brevity)
Define the following series for x>0
sinxx=1−x23!+x45!−x67!+⋯.
Now substitute x=√y to arrive at
sin√y √y =1−y3!+y25!−y37!+⋯.
if we find the roots of sin√y √y =0 we
find that
y=n2π2 for n≠0 and n in the integers
With all of this in mind, recall that for a polynomial
P(x)=anxn+an−1xn−1+⋯+a1x+a0 with roots
r1,r2,⋯,rn
1r1+1r2+⋯+1rn=−a1a0
Treating the above series for sin√y √y
as polynomial we see that
112π2+122π2+132π2+⋯=−−13!1
then multiplying both sides by π2 gives the desired series.
112+122+132+⋯=π26
This solution fascinated me. Although it takes a few things for granted (such as that the Fundamental Theorem of Algebra applies to infinite polynomials) I nevertheless thought the proof was one of the most beautiful I've seen. In attempting to generalize this I stumbled across the following function and its associated power series:
cos(x√2)cosh(x√2)=∞∑k=0x4k(4k)!(−1)k
By a simple substitution we find that
cos(x1/4√2)cosh(x1/4√2)=∞∑k=0xk(4k)!(−1)k
Which is a polynomial in x; noting that cosh is nowhere zero (on the real line) we solve for the roots as such:
cos(x1/4√2)=0⟹x1/4√2=nπ+π2⟹x=π4(2n+1)44
Using the argument for Viete's Formula in the solution above, we find that
4π4∞∑k=01(2k+1)4=−−14!1
Upon manipulation we find that
∞∑k=01(2k+1)4=π496
From here we can bring this into a form more like that of the Basel Problem by noting that
∞∑k=01(2k+2)4=116∞∑k=01(k+1)4=116∞∑k=11k4
Noting that this computes the even terms of ∑∞k=1k−4 we find that the odd terms must equal 1516∑∞k=1k−4 which is our series calculated above
Applying this, we find that, as desired,
∞∑k=11k4=π490
However, I have been unable to generalize this approach further. Intuition tells me that the functions desired would have power series similar to those used, and would be a combination of trigonometric functions. Nevertheless, I have only been able to solve this for the case above (which required me to switch the power series from sin(x) to cos(x) from Alfredo's proof, as neglecting cosh(x) was desirable).
My question is thus: Can this proof format be applied to solve series of the form ∑∞k=1k−n for any n other than 2 and 4?
Note: I apologize for the length of this post, but I felt that a full presentation of the proof might assist in generalizing it. If anyone has any suggestions concerning this please let me know!
Note 2: In case anyone is troubled by this seemingly naive application of Viete's Formula to infinite polynomials, know that this is perfectly valid, and is known as the Root Linear Coefficient Theorem.
Answer
To start, the "factoring" step is known as the Weierstrass factorization theorem, which asserts that you can express some functions as products of their factors.
From here, take note that you had
sin(x)x=…(1+x222π2)(1+x212π2)(1−x212π2)(1−x222π2)…
which is basically what you noted.
From here, I multiply similar terms to get
f(x):=sin(x)x=(1−x212π2)(1−x222π2)(1−x232π2)…
and define this as f(x).
using (1−ue2πin)(1−ue4πin)…(1−ue2πi)=1−un, we can then get
f(xe2πin)f(xe4πin)…f(xe2πi)=(1−x2n12nπ2n)(1−x2n22nπ2n)(1−x2n32nπ2n)…
And since this is not generalizable to n∉N, we can only solve for
∞∑k=11k2n
By using this method. As for the odd values, no closed form yet exists.
No comments:
Post a Comment