Simple Riemann zeta function algebra help

In summary, the conversation discussed a mathematical problem involving series of reciprocals and how to find the sum by subtracting one series from another. The final solution was to subtract the series of reciprocals of even powers from the series of reciprocals of all integers, leaving only the reciprocals of odd powers.
  • #1
tim9000
867
17
Hi
1.PNG

It's just that last step I'm not getting, so you have:
[1 / Kz] - [1 / (2K)z]
= [ (2K)z - Kz ] / [(2K)z * Kz]
= [ (2)z - 1 ] / [(2K)z*]
Then what?
Thanks
 
Mathematics news on Phys.org
  • #2
tim9000 said:
Hi
View attachment 80246
It's just that last step I'm not getting, so you have:
[1 / Kz] - [1 / (2K)z]
= [ (2K)z - Kz ] / [(2K)z * Kz]
= [ (2)z - 1 ] / [(2K)z*]
Then what?
Thanks
Expand the first series, and you get ##\frac{1}{1^k} + \frac{1}{2^k} + \frac{1}{3^k} + ... + \frac{1}{n^k} ... ##
Expand the second series, and you get ##\frac{1}{2^k} + \frac{1}{4^k} + ... + \frac{1}{(2n)^k} ... ##
If you subtract the second series from the first, term by term, what do you get?
 
  • #3
Mark44 said:
Expand the first series, and you get ##\frac{1}{1^k} + \frac{1}{2^k} + \frac{1}{3^k} + ... + \frac{1}{n^k} ... ##
Expand the second series, and you get ##\frac{1}{2^k} + \frac{1}{4^k} + ... + \frac{1}{(2n)^k} ... ##
If you subtract the second series from the first, term by term, what do you get?
Sorry, I'm not sure, I'm too rusty.
 
  • #4
Can you see that every term in the second series is also present in the first series?
 
  • #5
Mark44 said:
Can you see that every term in the second series is also present in the first series?
Ah, I've been trying to use factorisation and cancellation.
So when you minus S1 - S2 it leaves only 1/1k
but how does that explain the final line of working they got for the sum?
 
  • #6
tim9000 said:
Ah, I've been trying to use factorisation and cancellation.
So when you minus S1 - S2 it leaves only 1/1k
No, that's not right. You still end up with a series with an infinite number of terms. Every term in the 2nd series is in the first series, but not every term in the 1st series is in the second. Expand both series like I did above, but with more terms, and then do the subtraction.
tim9000 said:
but how does that explain the final line of working they got for the sum?
 
  • #7
tim9000 said:
Sorry, I'm not sure, I'm too rusty.
The second series consists of the reciprocals of powers of even integers. The first series consists of the reciprocals of all integers. Subtract the second from the first leaves the reciprocals of the powers of all odd integers.
 
  • #8
Mark44 said:
No, that's not right. You still end up with a series with an infinite number of terms. Every term in the 2nd series is in the first series, but not every term in the 1st series is in the second. Expand both series like I did above, but with more terms, and then do the subtraction.
mathman said:
The second series consists of the reciprocals of powers of even integers. The first series consists of the reciprocals of all integers. Subtract the second from the first leaves the reciprocals of the powers of all odd integers.

Woops, yes I see what you were saying. Aaah ok, no wonder I couldn't rearrange it, I was contemplating partial fractions etc.
Thanks
 

1. What is the Riemann zeta function?

The Riemann zeta function is a mathematical function that was discovered by Bernhard Riemann in the 19th century. It is defined as the infinite sum of the reciprocals of all positive integers raised to a certain power. In simpler terms, it is a function that takes in a complex number and outputs a complex number.

2. What is the significance of the Riemann zeta function?

The Riemann zeta function is significant because it has connections to various areas of mathematics, including number theory, complex analysis, and physics. It also plays a crucial role in the study of prime numbers and the distribution of prime numbers.

3. How is the Riemann zeta function related to the Riemann hypothesis?

The Riemann hypothesis is a famous unsolved problem in mathematics that states that all non-trivial zeros of the Riemann zeta function lie on the critical line where the real part of the input is equal to 1/2. In other words, if the Riemann hypothesis is true, it would provide us with valuable insight into the behavior of the Riemann zeta function.

4. Can the Riemann zeta function be simplified?

Yes, the Riemann zeta function can be simplified for certain values of the input. For example, when the real part of the input is greater than 1, the function can be expressed as an infinite product of prime numbers. However, for other values of the input, the function cannot be simplified and must be evaluated numerically.

5. How is the Riemann zeta function used in real-world applications?

The Riemann zeta function is used in various real-world applications, such as cryptography, physics, and engineering. For example, it is used in the development of efficient algorithms for finding large prime numbers, which is essential in cryptography. It is also used in the study of quantum chaos and the behavior of electrons in a magnetic field.

Similar threads

B Does this help solve the Riemann Hypothesis?
    • General Math
Replies
4
Views
1K
Insights Computing the Riemann Zeta Function Using Fourier Series
    • General Math
Replies
5
Views
3K
I Hausdorff dimension of Riemann zeta function assuming RH
    • General Math
Replies
11
Views
2K
I Connection of zeta to primes: less Euler than error or L-functions?
    • General Math
Replies
2
Views
1K
I A thought about the Riemann hypothesis
    • General Math
Replies
7
Views
1K
I Riemann zeta: regularization and universality
    • General Math
Replies
1
Views
1K
A Paper About the Riemann Zeta Function
    • General Math
Replies
2
Views
1K
I Is there any way to simplify my proof by induction?
    • General Math
Replies
13
Views
1K
I Geometry of series terms of the Riemann Zeta Function
    • Topology and Analysis
Replies
3
Views
1K
I No Way to Solve π(x) from Riemann's Zeta Function?
    • General Math
Replies
9
Views
2K

Hot Threads

Recent Insights

Back
Top