Marvin Ray Burns

 I've been using Maple since 1997 or so.

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These are replies submitted by Marvin Ray Burns



The MRB constant

=

The MKB constant

=

 

 



Download mrb_equals.mw

 

marvinrayburns.com

This constant, MKB, has made it into Steven Finch's Mathematical Constants book addenda page. It is located at the end of 6.11.

The link is http://algo.inria.fr/csolve/erradd.pdf .

Well I guess thats it; no big mystery!

The function f is a spiral and ploting with the floor operator simply plots points for the floor of f(x). The points are simply in an order that, when all put together, gives the shape of picture 1.

marvinrayburns.com

It might be easier to see what is going on if we break picture 1 into parts. We will also plot the expressions without the floor command.

 

 x = 1 .. 1.5:

plots[complexplot](evalf(floor(cos(Pi*x)*x^(1/x)+I*sin(Pi*x)*x^(1/x))), x = 1 .. 1.5, labels = ["Re", "Im"])plots[complexplot](evalf((cos(Pi*x)*x^(1/x)+I*sin(Pi*x)*x^(1/x))), x = 1 .. 1.5, labels = ["Re", "Im"])

Picture 3a1                                                                                         Picture 3a2

 x = 1.5 .. 2:

plots[complexplot](evalf(floor(cos(Pi*x)*x^(1/x)+I*sin(Pi*x)*x^(1/x))), x = 1.5 .. 2, labels = ["Re", "Im"])plots[complexplot](evalf((cos(Pi*x)*x^(1/x)+I*sin(Pi*x)*x^(1/x))), x = 1.5 .. 2, labels = ["Re", "Im"])

Picture 3b1                                                                                                         Picture 3b2

 x = 2 .. 2.5:

plots[complexplot](evalf(floor(cos(Pi*x)*x^(1/x)+I*sin(Pi*x)*x^(1/x))), x = 2 .. 2.5, labels = ["Re", "Im"])plots[complexplot](evalf((cos(Pi*x)*x^(1/x)+I*sin(Pi*x)*x^(1/x))), x = 2 .. 2.5, labels = ["Re", "Im"])

Picture 3c1                                                                                                         Picture 3c2

 

 x = 2.5 .. 3:

plots[complexplot](evalf(floor(cos(Pi*x)*x^(1/x)+I*sin(Pi*x)*x^(1/x))), x = 2.5 .. 3, labels = ["Re", "Im"])plots[complexplot](evalf((cos(Pi*x)*x^(1/x)+I*sin(Pi*x)*x^(1/x))), x = 2.5 .. 3, labels = ["Re", "Im"])

Picture 3d1                                                                                                            Picture 3d2

 

 

 x = 3 .. 3.25:

plots[complexplot](evalf(floor(cos(Pi*x)*x^(1/x)+I*sin(Pi*x)*x^(1/x))), x = 3 .. 3.25, labels = ["Re", "Im"])plots[complexplot](evalf((cos(Pi*x)*x^(1/x)+I*sin(Pi*x)*x^(1/x))), x = 3 .. 3.25, labels = ["Re", "Im"])

Picture 3e1                                                                                     Picture 3e2

 

 x = 3.25 .. 3.5:

 plots[complexplot](evalf(floor(cos(Pi*x)*x^(1/x)+I*sin(Pi*x)*x^(1/x))), x = 3.25 .. 3.5, labels = ["Re", "Im"])plots[complexplot](evalf((cos(Pi*x)*x^(1/x)+I*sin(Pi*x)*x^(1/x))), x = 3.25 .. 3.5, labels = ["Re", "Im"])

Picture 3f1                                                                                              Picture 3f2

 

 

 x = 3.5 .. 4:

plots[complexplot](evalf(floor(cos(Pi*x)*x^(1/x)+I*sin(Pi*x)*x^(1/x))), x = 3.5 .. 4, labels = ["Re", "Im"])plots[complexplot](evalf((cos(Pi*x)*x^(1/x)+I*sin(Pi*x)*x^(1/x))), x = 3.5 .. 4, labels = ["Re", "Im"])

Picture 3g1                                                                                                                      Picture 3g2

 


This should have the same shape as Picture 1.

 

plots[complexplot](evalf(floor(cos(Pi*x)*x^(1/x)+I*sin(Pi*x)*x^(1/x))), x = 1 .. 4, labels = ["Re", "Im"])

plots[complexplot](evalf(floor(cos(Pi*x)*x^(1/x)+I*sin(Pi*x)*x^(1/x))), x = 1 .. 4, labels = ["Re", "Im"])

However, the third quadrant is different.

Using maple 12, the first graph looks differnt in the first and fourth quadrants.

Picture 2

 

Simply find a formula for your sequence. Hint, it looks a lot like the formula I gave you. It is something times n plus something else.

Figure out what x and y satisfies

x*0 +y=1

x*1+y=5

etc.

Then notice you are summing the 0th term to the 100th term instead of summing the 1st through the 100th, so you will have to make one small change in the second line following:

restart; sum(x*n+y, n = 100 .. 100);

restart; sum(x*n+y, n = 1 .. 100);

You will also have to fill in the x and y with the answers you found above.

Now you probably ought to tell your teacher that I helped you some.

marvinrayburns.com

 

Simply find a formula for your sequence. Hint, it looks a lot like the formula I gave you. It is something times n plus something else.

Figure out what x and y satisfies

x*0 +y=1

x*1+y=5

etc.

Then notice you are summing the 0th term to the 100th term instead of summing the 1st through the 100th, so you will have to make one small change in the second line following:

restart; sum(x*n+y, n = 100 .. 100);

restart; sum(x*n+y, n = 1 .. 100);

You will also have to fill in the x and y with the answers you found above.

Now you probably ought to tell your teacher that I helped you some.

marvinrayburns.com

Again let f(x)=(-1)^x*x^(1/x).

The cumlative sum of the zeros ∈ <ℝ,[-1,0)> of Im(f(x)) diverge like -log(1.2*x)

As seen here.

plot(-log(1.2*x),x=2..1000)

plot(-log(1.2*x),x=2..10000)

plot(-log(1.2*x),x=2..100000)

Again let f(x)=(-1)^x*x^(1/x).

I had hoped to find some combination of known constants to equal the sum of the zeros ∈ <ℝ,[-1,0)> of Im(f(x), but, the sum of the zeros ∈ <ℝ,[-1,0)> of Im(f(x)) is (very slowly) divergent, as in the following:

 



 

 

 

(1)

 

 

(2)

 

 

(3)

 

 

 

x=1/2 (-W + sqrt[-4 + W^2])

limit(1/2 *(-W + sqrt(-4 + W^2)),W= infinity)=limit(1/2*(-W+sqrt(-4+W^2)),W= infinity)

and sum(1/2 *(-W + sqrt(-4 + W^2)),W=2 ..infinity) = -infinity

 

 



Download june20a,10.mw


Dear Tom Paris, :)

Will you be a little more explicit in what you did "by hand;" what was the derivatives of your sums?

I would like to help.

 

Again let f(x)=(-1)^x*x^(1/x).

All of the zeros ∈ <,[-1,0)> of Im(f(x)) are the roots ∈ <,[-1,0)> of x^2+W*x=-1, were W ∈< ,[2,∞)>.

I will try to put it this way:

{ zeros ∈ <ℝ,[-1,0)> of Im(f(x)) } = { roots ∈ <ℝ,[-1,0)> of x^2+W*x=-1 }, were W ∈<ℕ ,[2,∞)>.

The two sets are equal!

ℕ stands for natural numbers.

s ∈ <ℝ,[-1,0)> means s is in the set of all real numbers >=-1 and <0 .

W ∈<ℕ ,(1,∞)> means W is in the set of all natural numbers greater than 1.

Again let f(x)=(-1)^x*x^(1/x),

For Im(f(x)), the third zero to the right of x=-1 seems to be1/2*(sqrt(21)-5) 

.

Indication that 1/2*(sqrt(21)-5)

is a most likely a zero:

Digits := 10; evalf(eval(Im(f(x)), x = 1/2*(sqrt(21)-5)))=0.00006274148290
Digits := 30; evalf(eval(Im(f(x)), x = 1/2*(sqrt(21)-5)))=-5.15516567057872417416934365318*10^(-25)
...

 

The fouth zero seems to be srt(2)-1.

The fifth zero seems to be 1/2*(sqrt(357)-19).

The sixth zero seems to be 12*sqrt(2)-17.

The seventh zero seems to be 6*sqrt(10)-19.

To check 6*sqrt(10)-19:

Digits := 80; evalf(eval(Im(f(x)), x = 6*sqrt(10)-19))=1.4149889409...*10^(-15)
Digits := 160; evalf(eval(Im(f(x)), x = 6*sqrt(10)-19))=-2.034233693511...10^(-96)

...

 

The next zero seems to be 4*sqrt(33)-23

Next 5*sqrt(23)-24

Next 4*sqrt(39)-25

To check 4*sqrt(39)-25

Digits := 160; evalf(eval(Im(f(x)), x = 4*sqrt(39)-25))=-2.5063583022939...*10^(-70)
Digits := 200; evalf(eval(Im(f(x)), x = 4*sqrt(39)-25))=-4.02129897436313282..*10^(-111)

...

marvinrayburns.com

Again let f(x)=(-1)^x*x^(1/x),

For Im(f(x)), the second zero to the right of x=-1 seems to be sqrt(3)-2.

indication that sqrt(3)-2 is a most likely a zero:

Digits := 10; evalf(eval(Im(f(x)), x = sqrt(3)-2))=-0.000002580544291
 Digits := 30; evalf(eval(Im(f(x)), x = sqrt(3)-2))= -2.12846188527463303702449027759 10^(-26)
...


Again let f(x)=(-1)^x*x^(1/x),

For Im(f(x)), the first zero to the right of x=-1 seems to be phi-1 where phi is the silver ratio.

Indication that phi-1 is most likely a zero:

 evalf(eval(Im(f(x)), x = (sqrt(5)-1)*(1/2)-1))=-1.560570058*10^(-7)

and

Digits := 30; evalf(eval(Im(f(x)), x = (sqrt(5)-1)*(1/2)-1))=-1.15017467791296590399352342471*10^(-27)


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