MaplePrimes - comments on Post, MRB constant K part 2

MaplePrimes - comments on Post, MRB constant K part 2 http://www.mapleprimes.com/posts/89030-MRB-Constant-K-Part-2 en-us 2026 Maplesoft, A Division of Waterloo Maple Inc. Maplesoft Document System Wed, 10 Jun 2026 17:51:43 GMT Wed, 10 Jun 2026 17:51:43 GMT The latest comments added to the Post, MRB constant K part 2 http://www.mapleprimes.com/images/mapleprimeswhite.jpg MaplePrimes - comments on Post, MRB constant K part 2 http://www.mapleprimes.com/posts/89030-MRB-Constant-K-Part-2 Theorem MRBK 9.0 http://www.mapleprimes.com/posts/89030-MRB-Constant-K-Part-2?ref=Feed:MaplePrimes:MRB constant K part 2:Comments#comment89134 g(n)=n^(1/n) f(n)=(-1)^n* g(n) h(n)=(-1)^n* (n^(1/n)-1) Theorem MRBK 9.0 sum(h(n), n = 1 .. x)-(sum(h(n), n = 1 .. x-1)) = h(x)   Proof:  sum(h(n), n = 1 .. x)-(sum(h(n), n = 1 .. x-1)) =(h(1)+h(2)+...+h(x-1)+h(x))-(h(1)+h(2)+...+h(x-1)) =h(x)  COROLLARY 9.05: sum(f(n), n = 1 .. x)-(sum(f(n), n = 1 .. x-1)) = f(x) The latest comments added to the Post, MRB constant K part 2 89134 Sun, 06 Jun 2010 03:54:48 Z Marvin Ray Burns Marvin Ray Burns Example MRBK 9.1 http://www.mapleprimes.com/posts/89030-MRB-Constant-K-Part-2?ref=Feed:MaplePrimes:MRB constant K part 2:Comments#comment89160 Example MRBK 9.1 A) restart; h := n-> (-1)^n*(n^(1/n)-1): sum(h(n), n = 1 .. 10)-(sum(h(n), n = 1 .. 9)); = 10^(1/10)-1 h(10) = 10^(1/10)-1   B) restart; h := n-> (-1)^n*(n^(1/n)-1): sum(h(n), n = 1 .. 101)-(sum(h(n), n = 1 .. 100)) = -101^(1/101)+1 h(101) = -101^(1/101)+1   Example MRBK 9.2 A) restart; f := n-> (-1)^n*(n^(1/n)): sum(f(n), n = 1 .. 10)-(sum(f(n), n = 1 .. 9)); = 10^(1/10) f(10) = 10^(1/10)-1   B) restart; f := n-> (-1)^n*n^(1/n): sum(f(n), n = 1 .. 101)-(sum(f(n), n = 1 .. 100)) = -101^(1/101) f(101)  = -101^(1/101)<a href="http://marvinrayburns.com"> </a> The latest comments added to the Post, MRB constant K part 2 89160 Sun, 06 Jun 2010 22:43:43 Z Marvin Ray Burns Marvin Ray Burns Conjecture MRBK9.91 http://www.mapleprimes.com/posts/89030-MRB-Constant-K-Part-2?ref=Feed:MaplePrimes:MRB constant K part 2:Comments#comment89168 A reminder: If a function is differentiable at some point c of its domain, then it is also continuous at c. However here we extend the notion of differentiability to be valid for individual points on the real number line, specifically positive integers.   f(n)=(-1)^n*n^(1/n) According to Theorem MRBK 4.0, When n is a positve integer, the derivative of f(n) is I*Pi*f+(1-ln(n))*f/n^2 = f*(I*Pi+(1-ln(n))/n^2) or f'(x) = f(x)*(I*Pi+(1-ln(x))/x^2). But by following Example MRBK 9.2,  we understand that sum(f(n), n = 1 .. x)-sum(f(n), n = 1 .. x-1)  = (1)^x*x^(1/x) =f(x) At x∈{1,2,3,...}, the instantaneous rate of change of f(x) is f(x)*(I*Pi+(1-ln(x))/x^2) and the chnge in the summing of all the positive integers from x-1 to x is f(x). Thus by Corollary MRBK 9.05 for f(n)=(-1)^n*n^(1/n), I have Hypothesis 9.90: The instantaneous rate of change of f(x) equals the chnge in the summing of all the positive integers, from x-1 to x, all times (I*Pi+(1-ln(x))/x^2). At n=x: f'(n)=(I*Pi+(1-ln(x))/x^2)*(sum(f(n), n = 1 .. x)-sum(f(n), n = 1 .. x-1)) f'(n)/(I*Pi+(1-ln(x))/x^2) = (sum(f(n), n = 1 .. x)-sum(f(n), n = 1 .. x-1)) Conjecture 9.91: sum(f(n), n = 1 .. x) = sum(f(n), n = 1 .. x-1) +  f'(n)/(I*Pi+(1-ln(x))/x^2)  The latest comments added to the Post, MRB constant K part 2 89168 Mon, 07 Jun 2010 02:01:49 Z Marvin Ray Burns Marvin Ray Burns Theorem MRBK 10.0 http://www.mapleprimes.com/posts/89030-MRB-Constant-K-Part-2?ref=Feed:MaplePrimes:MRB constant K part 2:Comments#comment89323  f(n)=(-1)^n* n^(1/n) Theorem MRBK 10.0: For f over the set of positive intgers: sum(f(n), n = 1 .. x) = sum(f(n), n = 1 .. x-1) + f'(x)/(I*Pi+(1-ln(x))/x^2) Proof: By THEOREM MRBK 8.0  f'(x)=I*Pi*f+(1-ln(x))*f/x^2. So sum(f(n), n = 1 .. x) = sum(f(n), n = 1 .. x-1) + f'(x) / (I*Pi+(1-ln(x))/x^2), concluding that sum(f(n), n = 1 .. x) = sum(f(n), n = 1 .. x-1) +f. While COROLLARY 9.05 says sum(f(n), n = 1 .. x)-(sum(f(n), n = 1 .. x-1)) = f(x) Which is the same as sum(f(n), n = 1 .. x) = sum(f(n), n = 1 .. x-1) +f. The latest comments added to the Post, MRB constant K part 2 89323 Wed, 09 Jun 2010 05:12:00 Z Marvin Ray Burns Marvin Ray Burns Example MRBK10.5 http://www.mapleprimes.com/posts/89030-MRB-Constant-K-Part-2?ref=Feed:MaplePrimes:MRB constant K part 2:Comments#comment89567 Example MRBK10.5 Given sum(f(n), n = 1 .. 999)=-.815607688 and f'(1000)=-0.5948705883e-5+3.163369134*I Using theorem 10.0, find sum(f(n), n = 1 .. 1000) By theorem 10.0, sum(f(n), n = 1 .. x) = sum(f(n), n = 1 .. x-1) + f'(x)/(I*Pi+(1-ln(x))/x^2) So sum(f(n), n = 1 .. 1000) =   .815607688 +(-0.5948705883e-5+3.163369134*I)/(I*Pi+(1-ln(1000))/1000^2)=.191323979+2.051982038*10^(-16)*I Verify: sum(f(n), n = 1 .. 1000)=0.191323979 which is the same as shown above, to machine precision in the complex plane.   Before I move on to using Theorem MRBk10.0 in MRB constant L, I wonder if there are any questions about or objections to how I wrote MRBk10.0      <a href="http://marvinrayburns.com">marvinrayburns.com </a> The latest comments added to the Post, MRB constant K part 2 89567 Sat, 12 Jun 2010 03:51:36 Z Marvin Ray Burns Marvin Ray Burns I should have clarified http://www.mapleprimes.com/posts/89030-MRB-Constant-K-Part-2?ref=Feed:MaplePrimes:MRB constant K part 2:Comments#comment89596 In <a href="89030-MRB-Constant-K-Part-2#comment89323">Theorem MRBK 10.0</a> I proved that for f(n)=(-1)^n* n^(1/n) sum(f(n), n = 1 .. x) = sum(f(n), n = 1 .. x-1) + f'(x)/(I*Pi+(1-ln(x))/x^2) However, I should have clarified that f' is derivative of the function, (-1)^x*x^(1.x) defined over real numbers,x. I really would like some feedback, (even if it is criticism,) before I move on to MRB constant L. I know a lot of people have read this blog but I have seen little opinion about it (only one thumbs down, which I don't mind). Let me know what you think, and thanks! The latest comments added to the Post, MRB constant K part 2 89596 Sun, 13 Jun 2010 06:28:58 Z Marvin Ray Burns Marvin Ray Burns