The Maple input, solve(y1-y2, {omega1, omega2, theta1, theta2}), gave, basically, that result; at least I believe that is the correct interpretation.

In Theorem MRBK 10.0

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!



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

marvinrayburns.com

 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.

<math xmlns='http://www.w3.org/1998/Math/MathML'><semantics><mrow xref='id4'><mo>-</mo><mrow xref='id3'><mroot><mn xref='id2'>101</mn><mn xref='id1'>101</mn></mroot></mrow></mrow><annotation-xml encoding='MathML-Content'><apply id='id4'><minus/><apply id='id3'><root/><degree><cn id='id1' type='integer'>101</cn></degree><cn id='id2' type='integer'>101</cn></apply></apply></annotation-xml><annotation encoding='Maple'>-101^(1/101)</annotation></semantics></math>

Above I tried pasting straight into the editor. Are we the only two people with this problem? Come on!

Not in Firefox/3.6.3 or in IE 8.0.

P. S. Below I pasted the following, straight from Maple, without the quotes, into the HTML code and look what came out.

"<math xmlns='http://www.w3.org/1998/Math/MathML'><semantics><mrow xref='id4'><mo>-</mo><mrow xref='id3'><mroot><mn xref='id2'>101</mn><mn xref='id1'>101</mn></mroot></mrow></mrow><annotation-xml encoding='MathML-Content'><apply id='id4'><minus/><apply id='id3'><root/><degree><cn id='id1' type='integer'>101</cn></degree><cn id='id2' type='integer'>101</cn></apply></apply></annotation-xml><annotation encoding='Maple'>-101^(1/101)</annotation></semantics></math>"

-101101101101-101^(1/101)

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)

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)

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)

 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)

THEOREM MRBK 8.0

By THEOREM MRBK 4.0, When n is in the set of (positive) integers the derivative of f is exactly I*Pi*f+(1-ln(n))*f/n^2.

So f' = I*Pi*f+(1-ln(n))*f/n^2| n ∈ {1,2,3,...}

Solving for f, we have the following:

f' = I*Pi*f+(1-ln(n))*f/n^2

f' = f*(I*Pi+(1-ln(n))/n^2)

f=f' / (I*Pi+(1-ln(n))/n^2)

 THEOREM MRBK 7.0

Given f(n)=(-1)^n* n^(1/n)

and g(n)= n^(1/n),

by Theorems 5.0 and 6.0,

(Re(f(n)')/g(n)')'=(f(n)/g(n))'=I*exp(I*Pi*n)*Pi | n ∈ {1,2,3,...},

Where I*exp(I*Pi*n)*Pi=(-1)^n*Pi*I

Pi is just a Greek letter ;)

f:=n->(-1)^n*n^(1/n)

g:=n->n^(1/n)

THEROM MRB 6.0

(f/g)' = I*exp(I*Pi*n)*Pi

Proof:

> simplify(`assuming`([diff(f(n)/g(n), n)-I*exp(I*Pi*n)*Pi], [n::posint]));
0

 Note that for n>0, the cumulative distribution function formula appears similar to 1+ (f/g)':

1+ I*exp(I*Pi*n)*Pi where I is a variable and Pi is -1.

See http://en.wikipedia.org/wiki/Exponential_distribution .
 

f(n)=(-1)^n* n^(1/n)

g(n)= n^(1/n)

k(n)' =d/dn k(n)

Re(f(n)')/g(n)'=f(n)/g(n)=(-1)^n | n ∈ {1,2,3,...},
 

Proof:

> simplify(`assuming`([Re(diff(f(n), n))/(diff(g(n), n))-(-1)^n], [n::posint]));
0

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.

COROLLARY MRBK 1.1a doesn't seem to follow readily enough so I will make it a theorem.

 

DEFINITIONS

f(n)=(-1)^n* n^(1/n)

g(n)= n^(1/n)

k(n)' =d/dn k(n)

∈ means in.

n  In the set of integers is n ∈ {1,2,3,...}.

Im(f(n)) is the imaginary part of f(n)

f := proc (n) options operator, arrow ... end proc; is f:=n->

THEOREM MRBK 2.0  

When n is in the set of integers the derivative of f has an imaginary part of exactly Pi*f.

Proof:

> f := proc (n) options operator, arrow; (-1)^n*n^(1/n) end proc; simplify(`assuming`([Im(diff(f(n), n))-Pi*f(n)], [n::posint]));
0