Please edit post #2, this is probably a typo Cos(A-B) - Cos(A-B) = 2SinASinB

Error spotted by Thuhina.

]]>"13. Introduction of the product. If M is a set different from 0 and a is anyone of its elements, then according to No.5 it is definite whether M = {a} or not. It is therefore always definite whether a given set consists of a single element or not.

Now let T be a set whose elements, M, N, R, . . ., are various (mutually disjoint) sets, and let S1 be any subset of its union ST. Then it is definite for every element M of T whether the intersection [M, 8 1 ] consists of a single element or not. Thus all those elements of T that have exactly one element in common with 8 1 are the elements of a certain subset T 1 of T, and it is again definite whether T 1 = T or not. All subsets S1 of ST that have exactly one element in common with each element of T then are, according to Axiom III, the elements of a set P = T, which, according to

Axioms III and IV, is a subset of union T and will be called the connection set [Verbindungsmenge] associated with T or the product of the sets M, N, R, . . .. If T = {M, N}, or T = {M, N, R}, we write T = MN, or T = MNR, respectively, for

short. "

I just do not understand why it is called "product" and how {M,N} can become MN here. Not in general therefore, but in this text. Thank you.]]>

for every natural number *n*. We call this function the *Dirichlet convolution* of *f* and *g*.

Yes, I should have clarified. After you use the specified formula, you get the Rational # in between by: x+d, x+2d, x+3d...x+nd. To give an example, lets say we have to find 3 rational numbers between 2 and 3. Therefore: x=2 y=3 n=3. Thus, d= (3-2)/(3+1)=1/4. So the n numbers are: 2 1/4, 2 2/4 and 2 3/4. This is fairly obvious, I just wanted to give the generalized formula.

Yes I agree with you I also use this formula of rational number and I got the actual answer.

]]>Thank for sharing. Now i can also calculate only in 3 minutes.]]>