Evaluating Hermite Identity with Integers: Help with Homework Equations

  • Thread starter Thread starter rattanjot14
  • Start date Start date
  • Tags Tags
    Identity
AI Thread Summary
The discussion focuses on evaluating a sum involving integers a, b, and m, specifically the expression related to the Hermite identity. The user attempted to apply the identity but faced difficulties due to the non-integer nature of a/m. References to "Concrete Mathematics" highlight the importance of the greatest common divisor (gcd) of a and m in determining the result. Additionally, integrals related to the problem are mentioned, suggesting a deeper mathematical connection. The conversation encourages further attempts and sharing of methods to solve the equation.
rattanjot14
Messages
16
Reaction score
0

Homework Statement



Let a and b be integers and m an integer >1 Evaluate

[b/m] + [(b+a)/m]+ [(b+2a)/m]+ [(b+3a)/m]+ [(b+4a)/m]+ [(b+5a)/m]+...+ [(b+(m-1)a)/m]

Homework Equations





The Attempt at a Solution


i tried to use hermite identity.

[x] + [x + 1/n] + [x + 2/n] +...+ [x + (n-1)/n] = [nx]

assuming x = b/m and 1/n = a/m.. but a/m is not an integer so i can't use it. I m stuck what to do?
 
Physics news on Phys.org
Funny you should mention that, just last week I read about it in pp 90-94 of Concrete Mathematics by Ronald L. Graham , Donald E. Knuth, and Oren Patashnik. Several interesting things are the dependence of the result on gcd(a,m) ,the fact that

$$\sum_{0 \le k < m} \left[ \frac{b+k \, a}{m}\right] = \sum_{0 \le k < a} \left[ \frac{b+k \, m}{a}\right], \, \, \, \, \, \, \, \, \, \mathop{integers \, \, \, a,m>0} $$

the closely related integrals

$$\frac{1}{m} \int_0^m (a \, x+b) \! \mathop{dx}=\frac{1}{a} \int_0^a (m \, x+b) \! \mathop{dx}=\frac{a \, m}{2}+b$$

The book uses several special cases to deduce the general one. Give it another try. If you have trouble describe the methods you tried and those you have know.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

Similar threads

Back
Top