Foxtrot

MHB Math Helper

MarkFL

Staff member
Hey Balarka,

When I saw the topic title, I thought you were going to teach us to dance!

Even though it has been many years since I was in school, I still sometimes have the "anxiety dream," i.e., that it is the day of the final, I have done no preparation, brought no supplies, and I have shown up to school in underwear only...

mathbalarka

Well-known member
MHB Math Helper
MarkFL said:
When I saw the topic title, I thought you were going to teach us to dance!
Okay, let's Dance then : Cha-La, Head-Cha-La...

I never had any type of nervousness before or after my math tests. Maybe because math is my favourite subject?

MarkFL, wanna see how to make a funny topic serious? Write a closed form expression for the sum given in the comic in the question #2 of the test paper*.

PS* : This is called Foxtrot series and that was the thing I was looking for and accedentally found this page of the original comic.

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ZaidAlyafey

Well-known member
MHB Math Helper
I sometimes have a dream that the time of exam finishes before answering most of the questions . Also, an exam on a certain subject turns out to contain questions from other subjects .

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PS* : This is called Foxtrot series and that was the thing I was looking for and accedentally found this page of the original comic.
I was thinking whether this is solvable .

MarkFL

Staff member
I never had the dream before an actual final, just at seemingly random times.

That "Foxtrot series" looks like too much fun at 4:16 am...

mathbalarka

Well-known member
MHB Math Helper
ZaidAlyafey said:
I was thinking whether this is solvable
The sum reduces to $$\displaystyle \frac{1}{3} \left (1 - \ln(2) + \pi \sec(\sqrt{3} \pi/2) \right )$$

MarkFL

Staff member
The sum reduces to $$\displaystyle \frac{1}{3} \left (1 - \ln(2) + \pi \sec(\sqrt{3} \pi/2) \right )$$
Did you use the digamma function?

mathbalarka

Well-known member
MHB Math Helper
Did you use the digamma function?
I didn't do anything, but yes, there is an application of digamma in the derivation.

ZaidAlyafey

Well-known member
MHB Math Helper
The sum reduces to $$\displaystyle \frac{1}{3} \left (1 - \ln(2) + \pi \sec(\sqrt{3} \pi/2) \right )$$
That must be the hyperbolic function , I thought of solving the sum using partial fraction decomposition then solving by residues...

mathbalarka

Well-known member
MHB Math Helper
Yes, sorry that sec should be sech.

ZaidAlyafey" said:
I thought of solving the sum using partial fraction decomposition then solving by residues
I don't think residues are best method to do this; however, have you been able to use residues?

ZaidAlyafey

Well-known member
MHB Math Helper
Yes, sorry that sec should be sech.

I don't think residues are best method to do this; however, have you been able to use residues?
Ok, I will post the solution if I can figure that out ..