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:

Picture 3a1                                                                                         Picture 3a2

 x = 1.5 .. 2:

Picture 3b1                                                                                                         Picture 3b2

 x = 2 .. 2.5:

Picture 3c1                                                                                                         Picture 3c2

 x = 2.5 .. 3:

Picture 3d1                                                                                                            Picture 3d2

 x = 3 .. 3.25:

Picture 3e1                                                                                     Picture 3e2

 x = 3.25 .. 3.5:

Picture 3f1                                                                                              Picture 3f2

 x = 3.5 .. 4:

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"])

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 .

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

As seen here.

Again let .

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:

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 .

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

For Im(f(x)), the third zero to the right of x=-1 seems to be 

.

Indication that

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 .

The fifth zero seems to be .

The sixth zero seems to be .

The seventh zero seems to be .

To check :

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

Next

Next

To check

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 ,

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

indication that 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 ,

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

Indication that -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)

For ,

according to the above graphs, and this graph from Wolfram|Alpha there are an infinite number of values, x, from (-1,0) were the values of f are real numbers, having no imaginary part.

I need to restate my theorem in one coherent statement.

Conjecture MRBK 11.0:
when f(n)=(-1)^n* n^(1/n) and q(x) = (-1)^x*x^(1/x)
 sum(f(n), n = 1 .. x) = sum(f(n), n = 1 .. x-1) + q'(x)/(I*Pi+(1-ln(x))/x^2)

where q is picewise continuous throughout its domain.

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