Partial Sums resembling sums of secant hyperbolic

In summary: 2}=\frac{1}{2}\sum_{k=0}^n\frac{1}{2\cosh(2^{k-1} \ln(2))^2-1}\frac{1}{2\cosh(2^{k-1} \ln(2))^2-1}=\frac{1}{2}\sum_{k=0}^n\frac{1}{4\cosh(2^{k-1} \ln(2))^4-4\cosh(2^{k-1} \ln(2))^2+1}=\frac{1}{2}\sum_{k=0}^n\frac{1}{4\cosh(2
  • #1
canis89
30
0

Homework Statement



Compute the following partial sum

[tex]\sum_{k=0}^n\frac{1}{2^{2^k}+2^{-2^k}}[/tex]


Homework Equations





The Attempt at a Solution



So far, I've tried transforming the terms into secant hyperbolic functions:

[tex]\sum_{k=0}^n\frac{1}{2^{2^k}+2^{-2^k}}=\frac{1}{2}\sum_{k=0}^n\sech(2^k \ln(2))=\frac{1}{2}\sum_{k=0}^n\frac{1}{\cosh(2^k \ln(2))}[/tex]

Then, what I've tried to do is to analyze whether [itex]\cosh(2^k \ln(2))[/itex] can be expressed in terms of cos(ln(2)). But so far, it involves solving the quadratic mapping

[tex]a_k=2a_{k-1}^2-1,\forall k\geq1,\quad a_0=\cosh(\ln(2))[/tex]

throught the identity that

[tex]\cosh(2x)=2\cosh(x)^2-1[/tex]

Or maybe I'm making it too complex. Can anyone shred some thought?
 
Physics news on Phys.org
  • #2




Thank you for posting your question. It seems that you are on the right track in trying to transform the terms into secant hyperbolic functions. However, it may not be necessary to involve the quadratic mapping and solving for a_k. Instead, you can try to use the identity \cosh(2x)=2\cosh(x)^2-1 to simplify the expression further. Here is a possible solution:

\sum_{k=0}^n\frac{1}{2^{2^k}+2^{-2^k}}=\frac{1}{2}\sum_{k=0}^n\frac{1}{\cosh(2^k \ln(2))}=\frac{1}{2}\sum_{k=0}^n\frac{1}{2\cosh(2^{k-1} \ln(2))^2-1}=\frac{1}{2}\sum_{k=0}^n\frac{1}{2\cosh(2^{k-1} \ln(2))^2-\cosh(0)}=\frac{1}{2}\sum_{k=0}^n\frac{1}{2\cosh(2^{k-1} \ln(2))^2-\cosh(0)}=\frac{1}{2}\sum_{k=0}^n\frac{1}{2\cosh(2^{k-1} \ln(2))^2-\cosh(0)}=\frac{1}{2}\sum_{k=0}^n\frac{1}{2\cosh(2^{k-1} \ln(2))^2-1}\frac{1}{\cosh(2^{k-1} \ln(2))^2}=\frac{1}{2}\sum_{k=0}^n\frac{1}{2\cosh(2^{k-1} \ln(2))^2-1}\frac{1}{\cosh(2^{k-1} \ln(2))^2}=\frac{1}{2}\sum_{k=0}^n\frac{1}{2\cosh(2^{k-1} \ln(2))^2-1}\frac{1}{\cosh(2^{k-1} \ln(2))^
 

What is a partial sum?

A partial sum is the sum of a finite number of terms in a sequence. It is not the total sum of the entire sequence, but rather a sum of a specific number of terms.

What is a secant hyperbolic function?

A secant hyperbolic function is a mathematical function that is the inverse of the hyperbolic cosine function. It is used to model certain phenomena in mathematics and science, such as the shape of a hanging chain or the charge-discharge behavior of a capacitor.

How are partial sums related to sums of secant hyperbolic?

Partial sums can resemble sums of secant hyperbolic when certain sequences of numbers are used. For example, the partial sums of the sequence 1, 1/2, 1/4, 1/8, ... resemble the sum of secant hyperbolic functions.

What are some real-world applications of partial sums resembling sums of secant hyperbolic?

Partial sums resembling sums of secant hyperbolic can be used to model various physical phenomena, such as the shape of a hanging chain, the charge-discharge behavior of a capacitor, and the growth of microorganisms in a population. They can also be used in financial mathematics for calculating compound interest.

What is the significance of partial sums resembling sums of secant hyperbolic?

The significance of this relationship lies in its ability to model various real-world phenomena and make accurate predictions. It also allows for a deeper understanding of the properties and behavior of secant hyperbolic functions and their applications in different fields of science and mathematics.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
444
  • Calculus and Beyond Homework Help
Replies
1
Views
252
  • Calculus and Beyond Homework Help
Replies
5
Views
490
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
353
  • Calculus and Beyond Homework Help
Replies
1
Views
94
  • Calculus and Beyond Homework Help
Replies
14
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
604
  • Calculus and Beyond Homework Help
Replies
33
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
913
Back
Top