Marvin Ray Burns

 I've been using Maple since 1997 or so.

MaplePrimes Activity

These are answers submitted by Marvin Ray Burns

Simply enter solve(1-sqrt(3)-sqrt(2)-2*x^2, x);


It would help if we could determine if, in the bottom right face of the cuboid frame, are those spirals or concentric circles inside of what looks like tires. Also I would like confirmation to my assumption that this is more than one equation graphed together[like z=Arccsh(x*I+y) and arccos(z*y+tan(z-y))=sinch(y*i+x^z)], or does all of this come from one equation[like z/(x+y)^(x/(Pi*y)=arccsh(x*I+log[z](y))+1/(z+x-y)*arccossinc(y*i+x)]?



From the right side, looking left, it looks like it is a 3D contour plot.

Notice the flat top of the surface; that could be from the z range being to small to include the top of the figure. I dislike saying it; but it looks like the top surface has been smoothed, as a visual effect.

For one thing we can compare the lower right with the upper left. They are not exactly symmetric but they are close to being so. Perhaps one is for positive domain and the other for negative domain of a function that treats positive numbers only slightly different than negative ones. That would be unlike the abs operation which treats positive and negative input diametrically. However the equation responsible for the graph could be taking the abs of an expression that gives one value for some positive inputs and a different value for negative inputs, like how abs(x-7) is nonmonotonic for some range of positive input and monotonic+ for other positive input and monotonic- for all negative input.


I sure have made the formula for the MRB constant appear quite different! It is interesting to note that except for convergence, the final form of the MRB constant has the same long-term properties of the original form, Limit(sum((-1)^n*n^(1/n),n=1..2*N),N=infinity).



A triple sum for the MRB constant

Below we extend notion of sum so that a series which does not converge is still subject to our analysis..


Using the theorem that says

Let t=(Pi*n^2-I*ln(n))/n), and then MRB constant = sum(sum((I*t)^k/k!,k = 0 .. infinity) ,n=1..infinity) +1/2

Not only will we formulate the MRB constant so there are no fractional powers, but we will write it so that there are no logarithmic functions either.



Previously we had

t=(Pi*n^2-I*ln(n))/n and


m=sum(exp(I*t), n = 1 .. infinity)+1/2.




We expanded exp(I*t) into a series  to get sum(I^k*t^k/factorial(k), k = 0 .. infinity).

Thus m=sum(Sum(I^k*t^k/factorial(k), k = 0 .. infinity), n = 1 .. infinity)+1/2


Now we have t=(Pi*n^2-I*ln(n))/n and

we can expand ln(n) into the series sum((-1)^(k+1)/k*(n-1)^k,k=1..infinity).

Thus t=(Pi*n^2-I*sum((-1)^(j+1)/j*(n-1)^j,j=1..infinity))/n.



Putting the expanded version of t into the formula for m gives


Distributing the I we get


Since I*(-1)^(j+1) = -I*(-1)^j, we have


Distributing the 1/n we get




Since I and the innermost series are both raised to the kth power, we can combine like powers and get


Since -(I^2) =1 we have


Moving the innermost series to the left gives us


Note: if you switch the sign of I*Pi*n you still get the same numeric result.




When computing the MRB constant,(sum((-1)^n*(n^(1/n)-1),n-1..infinity), the hardest part is the computation of all of the fractional powers i.e. (n^(1/n)), or even the exp(ln(n)/n).

The following form is by no means optimal, but by using series expansion, it does work around the need to compute the (n^(1/n))'s.

Let t=(Pi*n^2-I*ln(n))/n), and then MRB constant = sum(sum((I*t)^k/k!,k = 0 .. infinity) ,n=1..infinity) +1/2
OnFri, January 14, 2011 3:56:34 PM I showed that MRB constant = sum(exp(I*t),n=1..infinity) +1/2, where t=(Pi*n^2-I*ln(n))/n).
Now consider that exp(I*t) =sum(I^k/k!*t^k,k = 0 .. infinity)=sum(sum((I*t)^k/k!,k = 0 .. infinity) ,n=1..infinity), and so
 MRB constant = sum(Sum((I*t)^k/k!,k = 0 .. infinity) ,n=1..infinity) +1/2, where t=(Pi*n^2-I*ln(n))/n).

I thought I would mention another form of the MRB constant that I found. Here is a copy of my first mention of it.


From: Marvin To: Steven; Roland ; Carl ; zshen; Eric Cc: simon; Richard; psebah; support; Wolfram|Alpha Sent: Thu, January 13, 2011 6:41:45 PM Subject: Another form for the MRB constant


New Message ----

More esteemed colleagues, here are some second generation ponderings of Marvin Ray Burns.


I reshaped this new form for the MRB constant a little further in the right direction, I hope.

Remember, i am extending the notion of sum so that a series which does not converge may still have a well-defined Cesàro sum.


The MRB constant = sum((-1)^n*(n^(1/n)),n=1..infinity)+1/2.

Proof: The MRB constant

= sum((-1)^n*(n^(1/n)-1),n=1..infinity)

= limit(sum((-1)^n*n^(1/n),n=1..2*N),N=infinity)

= sum((-1)^n*(n^(1/n)),n=1..infinity)+1/2.



Now I will state a therom; the MRB constant = sum(exp(I*t),n=1..infinity) +1/2,

where t=(Pi*n^2-I*ln(n))/n).


We know that

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


n^(1/n) = exp(ln(n)/n).





Now also notice that

(I*Pi*n^2+ln(n))/n = I*(Pi*n^2-I*ln(n))/n.


(-1)^n*n^(1/n) = exp(I*(Pi*n^2-I*ln(n))/n)

= exp(I*t) where t=(Pi*n^2-I*ln(n))/n,

and so

sum((-1)^n*n^(1/n),n=1..infinity) = sum(exp(I*t),n=1..infinity), where t=(Pi*n^2-I*ln(n))/n.

Therefore, by the lemma, the MRB constant = sum(exp(I*t),n=1..infinity) +1/2, where t=(Pi*n^2-I*ln(n))/n).


With the arrangement sum(e^(i*t)), I hope to come closer to finding a closed form for the MRB constant.

Continued best wishes to all!

Marvin Ray Burns

Original investigator of the MRB constant


Since we have the series

1/2 sum((exp(w+I*m)+exp(w-I*m)),n=1 ..infinity) where m=Pi*n and w = ln(n)/n.

We can see from WolframAlpha that exp(w+I*m)+exp(w-I*m) =2 e^w cos(m)


Thus now our series looks like

 (1/2)*sum(2*exp(w)*cos(m),n=1..infinity)                 where m=Pi*n and w = ln(n)/n.

= sum(exp(w)*cos(m),n=1..infinity) .  

And of course exp(w) =exp(ln(n)/n) = n^(1/n).

Then we have

 sum(n^(1/n)*cos(m),n=1..infinity), and

cos(m) = cos(Pi*n) where

cos(Pi*n) = (-1)^n.

That gives us




The limsup of the function of partial sums of even n can be represented by


Also we know limit(sum((-1)^n*n^(1/n),n=1..2*N),N=infinity)

=sum((-1)^n*(n^(1/n)-1),n=1..infinity) which equals

0.1878596425...., which number we know.







In  fact since we have

1/2 sum((exp(w+I*m)+exp(w-I*m)),n=1 ..infinity) where m=Pi*n and w = ln(n)/n,

we can simplify this series to

1/2 sum((exp(w+x)+exp(w-x)),n=1 ..infinity) where x=I*Pi*n and w = ln(n)/n.

According to Wolfram Alpha I*Pi*n = 2 e^w cosh(x), so if Alpha is correct this series looks like

1/2 sum((exp(w+x)+exp(w-x)),n=1 ..infinity) where x=2*e^w cosh(x) and w = ln(n)/n, but I think with the 2*e^w cosh(x), it is too comlicated.


For integer n, since sin(n*Pi)=0, Isin(n*Pi)=0; so we can take that out of the formula for the series.

Let m=Pi*n and w = ln(n)/n. Then

sum((cos(m)+I*sin(m))*(cosh(w)+sinh(w)),n=1 ..infinity)

sum(cos(m)*(cosh(w)+sinh(w)),n=1 ..infinity); and

sum(cos(m)*(cosh(w)+sinh(w)),n=1 ..infinity)


sum(cos(m)*cosh(w)+cos(m)*sinh(w),n=1 ..infinity).

Then we can enter convert(cos(m)*cosh(w)+cos(m)*sinh(w), exp) into Maple and get

(1/2)*exp(w+I*m)+(1/2)*exp(w-I*m), which equals


So now the formula for our series looks like

1/2 sum((exp(w+I*m)+exp(w-I*m)),n=1 ..infinity) where m=Pi*n and w = ln(n)/n.




I havent tried any of these ideas, but the following might help.

I just found this by using the search box at the top of the page.

Say your sequence is {2*n+7}.

The easy way is to use the ?sum command.

restart; sum(2*n+7, n = 100 .. 100);

restart; sum(2*n+7, n = 1 .. 100);

Dear Rob,

M := Matrix([[1/2, 1], [1, 1/2]]); '1/2'*evalm(M)

will do it. The apostrophes keep the 1/2 from being evaluated.

Best wishes,



I found an example in Maple Primes' history, it is at

Here is what it looks like.






I hope this helps!

The technical answer is found in Those requirements, I assume, include using the more memory intensive Document mode, so I figure that the classic interface would, more so, run OK.

1 2 3 Page 1 of 3