Thanks very much for the explanation. But there are some doubts which persist for me . You say that I should use the definition of Riemann function on the positive real axis , and extentd the definition for complex numbers . But the thing is that I was only able to find the definition for...
Well , I think I have understood something wrong , because I can think of many examples of functions that are completely differential in a domain , but there exist more than 1 continuation of that function over a larger domain, which is still differentiable.
eg. f (x) = 1/ x^3 and f(x) =...
I would like to ask from where did the axioms of ZFC come from in the first place ?
Why is it that you consider the axioms of ZFC relevant , but adding a few more axioms to ZFC , just an exercise for set theorists ?
I am not a mathematician , and have very limited knowledge of mathematics...
Well , after looking at a couple of places , I came to know that the analytic continuation is a function that has the same value as the given function within the given functions domain , but is defined at points in a larger superset of the original domain too . Also that the analytic function is...
Could anyone tell me what is the Riemann zeta function. On Wikipedia , the definition has been given for values with real part > 1 , as :
Sum ( 1 / ( n^-s) ) as n varies from 1 to infinity.
but what is the definition for other values of s ? It is mentioned that the zeta function is the...
In the proof , it is not that p1 , p2 , p3 have been replaced by powers of 2 . All that it is saying is that :
p1.p2.p3.p4.....pn < 2 . (2^2) . (2^4) .... ( 2 ^ (2^n - 1 ) ).
This is because it is assuming the theorem to be true for p1 , p2 .. upto pn .
why at all should mn divide f(x) ?????
consider f(x) = 18 , thus possible values for m and n are 6 and 9 , but clearly 6 x 9 = 54 does not divide 18 , but had m and n been coprime then we would have a completely different answer.
It looks to me like your proof that legendre implies andrica , is implicitly assuming that there are 2 primes between n^2 and (n+1)^2 . But Legendre guarantees only the existence of a minimum of one. If there is only 1 prime bw n^2 and (n+1)^2 , then I am unable to follow how you conclude that...
I don't get it , why should the answer be r+p-1Cp-1 , if the coefficients are different .
Consider :
a + b = 5
There are 5 nonnegative integer solutions
Now consider :
a + 2b = 5
the only solutions that a can take are 1,3,5 - only 3 solutions .
Ok .. I got how to find the sum :
\sum_{r= 1}^\infty \frac{1}{2r(2r+1)}
\sum_{r= 1}^\infty \frac{1}{2r} - \sum_{r=1}^\infty \frac{1}{(2r+1)}
Now consider the series expansion of log(1 + x )
and in this expansion...
Thanks again .. what a fool I was .. Rather than considering the area under the curve from i to i+1 , I was considering the area under the curve from i-1/2 to i + 1/2 , thats why I had to take the difference and check for monotonicity and all ... it was pretty obvious .
By the way Ubibic ...
Thanks Robert for the explanation . Yes I was able to show that
\sum_{i=1}^r 1/i > \int_1^r\frac{1}{r} =ln(r)
But what bothers me is that I had formed some sort of argument in my mind to convince myself that if the integral is bounded for some function f(x) then the sum shall also...
Well I just came to know that \sum_{r=1}^\infty 1/r is itself divergent , so I am unsure of your question , it looks like some - \infty + \infty problem . I am at loss to see whats goin on ... does it mean the sum is undefined (because it doesnt seem to be right to say that the sum is...
Considering d(p,N) :
take remainder of N divided by p . Let it be r.
Now if r is a perfect square , then d(p.N) will have infinite solutions of the form r + kp
else there is no solution
I think you mean that N is fixed , but
If the question is :
Find all N < p such that n^2 = N ( mod p) for some n ( By some n , I mean that corresponding to each N there will be one n) , then :
d(p) = greatest integer less than square root of p
I was able to see that Andrica's Conjecture does indeed lead to Legendre's Conjecture .
Regarding the converse - i.e. given Legendre's Conjecture , then Andrica's Conjecture also holds . -- I can see that if Legendre,s Conjecture is stated for real values of n > 1 , and not just integer...
Ok ..... I was able to show that eq 2 ..i.e :
log ( [x] ! ) = \Psi(x) + \Psi(x/2) + \Psi(x/3) + ....
is indeed valid.
But still this was not at all obvious to me , and only after reading the lemma 2 part in the wikipedia proof of Bertrands postulate (...
Regarding Ramanujan's proof of the Bertrand postulate , I am unable to understand the 2nd equation that he uses ......
He starts the proof like this , let v(x) be the sum of logarithms of all primes less than or equal to x, Now consider :
\Psi(x) = v(x) + v (x^[1/2]) + v(x^[1/3]) + ...
What I am looking for is basically an algorithm ... Suppose I find one such base by exhaustive testing , now is it possible to find another base , or do I have to do exhaustive testing to find the second base too .
Or is there some way to rule out a certain sequence of digits from being a...
Well I really donno if ur heading in the right way .. but very nice thinking ....
If I can be of any help , You can express N(i+1) in terms of N(i) .. ( that is if my calculation is correct ) as follows :
N(i+k) = { (6^(3k)) * ( [ { N(i) + (2/3) } ^ 2 ] ^ {2k} ) } -...
Well .. I think u got it absolutely right here ..... I was saying the same thing basically .. that x < \Pi\pi(n) +1 < x! .... but nice to see your proof of it .....
by the way thanx for ur example (for instance 2*3*5*7*11*13+1=30031= 59*509) .... coz i was searching for such an example but...
I didnt understand the proof completely , but this is what I think u meant :
you have proved that ∏p¡(n) + 1 + 1 > n and < n!, and then concluded that ∏p¡(n) + 1 is a prime number due to some theorem by euclid.
Well if thats the case , then what I can say is ∏p¡(n) + 1 , need not be a prime...
OwlHoot,
A few questions here , how is it that if an integer == 1 mod each prime < n then that integer == 1 mod (n-1)! ??
eg. consider n = 6 ,
then the required integer = 2*3*5 + 1 ( becoz it gives rem 1 with all primes < 6)
but 2*3*5 + 1 mod (6-1)! , ie mod 120 is certainly not == 1 .
it...
Does anyone know if its possible to check whether a number is a perfect square in 2 different bases ? ( I dont mean the representation of a number in 2 different bases - coz that would be a ridiculous question , what I mean is consider a no say xyz - is it possible to prove / disprove that it is...