Interesting plot of the solution set!

n=(1/((sqrt(2))+2)),n1=((((n*2)+2)^2)-2)/4,n1/n = 4

On Sat, 27 Aug 2022 10:13:58 -0700 (PDT), Dan joyce
<danj4084@gmail.com> wrote:

>Interesting plot of the solution set!
>
>n=(1/((sqrt(2))+2)),n1=((((n*2)+2)^2)-2)/4,n1/n = 4

Solution set of what? A solution set implies there is something to
solve.

You have two constants that happen to have a ratio of 4? What is the
significance?

--
Remove del for email

On Saturday, August 27, 2022 at 1:50:34 PM UTC-4, Barry Schwarz wrote:
> On Sat, 27 Aug 2022 10:13:58 -0700 (PDT), Dan joyce
> <danj...@gmail.com> wrote:
>
> >Interesting plot of the solution set!
> >
> >n=(1/((sqrt(2))+2)),n1=((((n*2)+2)^2)-2)/4,n1/n = 4
> Solution set of what? A solution set implies there is something to
> solve.
>
> You have two constants that happen to have a ratio of 4? What is the
> significance?
>
> --
> Remove del for email
The plot!

On 8/27/2022 10:13 AM, Dan joyce wrote:
> Interesting plot of the solution set!
>
> n=(1/((sqrt(2))+2)),n1=((((n*2)+2)^2)-2)/4,n1/n = 4

Show us a plot so we have something to test against!

;^)

On Sat, 27 Aug 2022 11:43:33 -0700 (PDT), Dan joyce
<danj4084@gmail.com> wrote:

>On Saturday, August 27, 2022 at 1:50:34 PM UTC-4, Barry Schwarz wrote:
>> On Sat, 27 Aug 2022 10:13:58 -0700 (PDT), Dan joyce
>> <danj...@gmail.com> wrote:
>>
>> >Interesting plot of the solution set!
>> >
>> >n=(1/((sqrt(2))+2)),n1=((((n*2)+2)^2)-2)/4,n1/n = 4
>> Solution set of what? A solution set implies there is something to
>> solve.
>>
>> You have two constants that happen to have a ratio of 4? What is the
>> significance?
>>
>> --
>> Remove del for email
>The plot!

What plot? On what coordinate system? What is your independent
variable?

All you have is three constants:
0.292893218813
1.17157287525
4.0

--
Remove del for email

On Saturday, August 27, 2022 at 2:50:30 PM UTC-4, Chris M. Thomasson wrote:
> On 8/27/2022 10:13 AM, Dan joyce wrote:
> > Interesting plot of the solution set!
> >
> > n=(1/((sqrt(2))+2)),n1=((((n*2)+2)^2)-2)/4,n1/n = 4
> Show us a plot so we have something to test against!
>
> ;^)

Enter in it Wolfram and you will see the plot!

On 8/28/2022 7:26 AM, Dan joyce wrote:
> On Saturday, August 27, 2022 at 2:50:30 PM UTC-4, Chris M. Thomasson wrote:
>> On 8/27/2022 10:13 AM, Dan joyce wrote:
>>> Interesting plot of the solution set!
>>>
>>> n=(1/((sqrt(2))+2)),n1=((((n*2)+2)^2)-2)/4,n1/n = 4
>> Show us a plot so we have something to test against!
>>
>> ;^)
>
> Enter in it Wolfram and you will see the plot!

This one:

https://i.ibb.co/Z6jRfVS/image.png

?

On Sunday, August 28, 2022 at 6:09:13 PM UTC-4, Chris M. Thomasson wrote:
> On 8/28/2022 7:26 AM, Dan joyce wrote:
> > On Saturday, August 27, 2022 at 2:50:30 PM UTC-4, Chris M. Thomasson wrote:
> >> On 8/27/2022 10:13 AM, Dan joyce wrote:
> >>> Interesting plot of the solution set!
> >>>
> >>> n=(1/((sqrt(2))+2)),n1=((((n*2)+2)^2)-2)/4,n1/n = 4
> >> Show us a plot so we have something to test against!
> >>
> >> ;^)
> >
> > Enter in it Wolfram and you will see the plot!
> This one:
>
> https://i.ibb.co/Z6jRfVS/image.png
>
> ?
Yes, exactly!

On Sunday, August 28, 2022 at 6:09:13 PM UTC-4, Chris M. Thomasson wrote:
> On 8/28/2022 7:26 AM, Dan joyce wrote:
> > On Saturday, August 27, 2022 at 2:50:30 PM UTC-4, Chris M. Thomasson wrote:
> >> On 8/27/2022 10:13 AM, Dan joyce wrote:
> >>> Interesting plot of the solution set!
> >>>
> >>> n=(1/((sqrt(2))+2)),n1=((((n*2)+2)^2)-2)/4,n1/n = 4
> >> Show us a plot so we have something to test against!
> >>
> >> ;^)
> >
> > Enter in it Wolfram and you will see the plot!
> This one:
>
> https://i.ibb.co/Z6jRfVS/image.png
>
> ?
Chris,
Try this in Wolfram that uses this formula in ABS for negative squares and roots and also to find -y value
of -x^2- (-x)= -y
sqrt(n)=(((sqrt((n*4)+2))-2)/2), n^2=((((((sqrt(n))*2)+2)^2)-2)/4), n--->oo

-x+-y= -x^2 in the third negative quadrant.
-x^2 -(-x) =-y the two points needed to plot in the third negative quadrant starting @ x=0 and -y=(-0.5)
x goes negative with the first iteration of (sqrt(0.6+0.5))-1 = 0.0488088481...
-x = - 0.0488088481... Then incrementing +0.1 for the next plot point of (sqrt(0.7+0.5))-1
All in ABS of coerce.
Dan

On 8/29/2022 4:40 PM, Dan joyce wrote:
> On Sunday, August 28, 2022 at 6:09:13 PM UTC-4, Chris M. Thomasson wrote:
>> On 8/28/2022 7:26 AM, Dan joyce wrote:
>>> On Saturday, August 27, 2022 at 2:50:30 PM UTC-4, Chris M. Thomasson wrote:
>>>> On 8/27/2022 10:13 AM, Dan joyce wrote:
>>>>> Interesting plot of the solution set!
>>>>>
>>>>> n=(1/((sqrt(2))+2)),n1=((((n*2)+2)^2)-2)/4,n1/n = 4
>>>> Show us a plot so we have something to test against!
>>>>
>>>> ;^)
>>>
>>> Enter in it Wolfram and you will see the plot!
>> This one:
>>
>> https://i.ibb.co/Z6jRfVS/image.png
>>
>> ?
> Chris,
> Try this in Wolfram that uses this formula in ABS for negative squares and roots and also to find -y value
> of -x^2- (-x)= -y
> sqrt(n)=(((sqrt((n*4)+2))-2)/2), n^2=((((((sqrt(n))*2)+2)^2)-2)/4), n--->oo
>
>
> -x+-y= -x^2 in the third negative quadrant.
> -x^2 -(-x) =-y the two points needed to plot in the third negative quadrant starting @ x=0 and -y=(-0.5)
> x goes negative with the first iteration of (sqrt(0.6+0.5))-1 = 0.0488088481...
> -x = - 0.0488088481... Then incrementing +0.1 for the next plot point of (sqrt(0.7+0.5))-1
> All in ABS of coerce.

I might have some time tonight. Fwiw, check this plot out:

___________________________________
x(t) = cos(t) * abs(cos(t))
y(t) = sin(t) * abs(sin(t))
___________________________________

t from 0 to pi2

https://www.wolframalpha.com/input?i=x+%3D+cos%28t%29+*+abs%28cos%28t%29%29%2C+y+%3D+sin%28t%29+*+abs%28sin%28t%29%29%2C+t+from+0+to+pi2

;^)

On Monday, August 29, 2022 at 7:46:41 PM UTC-4, Chris M. Thomasson wrote:
> On 8/29/2022 4:40 PM, Dan joyce wrote:
> > On Sunday, August 28, 2022 at 6:09:13 PM UTC-4, Chris M. Thomasson wrote:
> >> On 8/28/2022 7:26 AM, Dan joyce wrote:
> >>> On Saturday, August 27, 2022 at 2:50:30 PM UTC-4, Chris M. Thomasson wrote:
> >>>> On 8/27/2022 10:13 AM, Dan joyce wrote:
> >>>>> Interesting plot of the solution set!
> >>>>>
> >>>>> n=(1/((sqrt(2))+2)),n1=((((n*2)+2)^2)-2)/4,n1/n = 4
> >>>> Show us a plot so we have something to test against!
> >>>>
> >>>> ;^)
> >>>
> >>> Enter in it Wolfram and you will see the plot!
> >> This one:
> >>
> >> https://i.ibb.co/Z6jRfVS/image.png
> >>
> >> ?
> > Chris,
> > Try this in Wolfram that uses this formula in ABS for negative squares and roots and also to find -y value
> > of -x^2- (-x)= -y
> > sqrt(n)=(((sqrt((n*4)+2))-2)/2), n^2=((((((sqrt(n))*2)+2)^2)-2)/4), n--->oo
> >
> >
> > -x+-y= -x^2 in the third negative quadrant.
> > -x^2 -(-x) =-y the two points needed to plot in the third negative quadrant starting @ x=0 and -y=(-0.5)
> > x goes negative with the first iteration of (sqrt(0.6+0.5))-1 = 0.0488088481...
> > -x = - 0.0488088481... Then incrementing +0.1 for the next plot point of (sqrt(0.7+0.5))-1
> > All in ABS of coerce.
> I might have some time tonight. Fwiw, check this plot out:
>
> ___________________________________
> x(t) = cos(t) * abs(cos(t))
> y(t) = sin(t) * abs(sin(t))
> ___________________________________
>
> t from 0 to pi2
>
> https://www.wolframalpha.com/input?i=x+%3D+cos%28t%29+*+abs%28cos%28t%29%29%2C+y+%3D+sin%28t%29+*+abs%28sin%28t%29%29%2C+t+from+0+to+pi2
>
> ;^)
Nice, but it ends @pi

This one I plotted years ago in the third quadrant starts @ n= (sqrt(.6 +.0.5))-1 then incrementing by 0.1 after each
iteration (0.5+0.1) etc. Starting the curve @ -x=-0.048.8088481..and -x^2 = -1.2 all in ABS of course.
-x^2-(-x)=-y for the two points -x and -y on the curve. This means -x=-0.0488088481... -x =-0.048.8088481... --->-x ---->oo

On Monday, August 29, 2022 at 8:08:24 PM UTC-4, Dan joyce wrote:
> On Monday, August 29, 2022 at 7:46:41 PM UTC-4, Chris M. Thomasson wrote:
> > On 8/29/2022 4:40 PM, Dan joyce wrote:
> > > On Sunday, August 28, 2022 at 6:09:13 PM UTC-4, Chris M. Thomasson wrote:
> > >> On 8/28/2022 7:26 AM, Dan joyce wrote:
> > >>> On Saturday, August 27, 2022 at 2:50:30 PM UTC-4, Chris M. Thomasson wrote:
> > >>>> On 8/27/2022 10:13 AM, Dan joyce wrote:
> > >>>>> Interesting plot of the solution set!
> > >>>>>
> > >>>>> n=(1/((sqrt(2))+2)),n1=((((n*2)+2)^2)-2)/4,n1/n = 4
> > >>>> Show us a plot so we have something to test against!
> > >>>>
> > >>>> ;^)
> > >>>
> > >>> Enter in it Wolfram and you will see the plot!
> > >> This one:
> > >>
> > >> https://i.ibb.co/Z6jRfVS/image.png
> > >>
> > >> ?
> > > Chris,
> > > Try this in Wolfram that uses this formula in ABS for negative squares and roots and also to find -y value
> > > of -x^2- (-x)= -y
> > > sqrt(n)=(((sqrt((n*4)+2))-2)/2), n^2=((((((sqrt(n))*2)+2)^2)-2)/4), n--->oo
> > >
> > >
> > > -x+-y= -x^2 in the third negative quadrant.
> > > -x^2 -(-x) =-y the two points needed to plot in the third negative quadrant starting @ x=0 and -y=(-0.5)
> > > x goes negative with the first iteration of (sqrt(0.6+0.5))-1 = 0.0488088481...
> > > -x = - 0.0488088481... Then incrementing +0.1 for the next plot point of (sqrt(0.7+0.5))-1
> > > All in ABS of coerce.
> > I might have some time tonight. Fwiw, check this plot out:
> >
> > ___________________________________
> > x(t) = cos(t) * abs(cos(t))
> > y(t) = sin(t) * abs(sin(t))
> > ___________________________________
> >
> > t from 0 to pi2
> >
> > https://www.wolframalpha.com/input?i=x+%3D+cos%28t%29+*+abs%28cos%28t%29%29%2C+y+%3D+sin%28t%29+*+abs%28sin%28t%29%29%2C+t+from+0+to+pi2
> >
> > ;^)
> Nice, but it ends @pi
>
> This one I plotted years ago in the third quadrant starts @ n= (sqrt(.6 +.0.5))-1 then incrementing by 0.1 after each
> iteration (0.5+0.1) etc. Starting the curve @ -x=-0.048.8088481..and -x^2 = -1.2 all in ABS of course.
> -x^2-(-x)=-y for the two points -x and -y on the curve. This means -x=-0.0488088481... -x =-0.048.8088481... --->-x ---->oo

Also another one ---
This is the only way I can express these negative squares and roots in a polar plot.

Enter this in Wolfram alpha as a polar plot---

plot r^2 = e^(-0.224744871391589049098642037352..theta) - (.908295745839672839274080366999...

Where -0.2247448713915... is the square root of -1 and .9082957458396... is the square root of -pi. (is the ABS value)

On Monday, August 29, 2022 at 8:38:53 PM UTC-4, Dan joyce wrote:
> On Monday, August 29, 2022 at 8:08:24 PM UTC-4, Dan joyce wrote:
> > On Monday, August 29, 2022 at 7:46:41 PM UTC-4, Chris M. Thomasson wrote:
> > > On 8/29/2022 4:40 PM, Dan joyce wrote:
> > > > On Sunday, August 28, 2022 at 6:09:13 PM UTC-4, Chris M. Thomasson wrote:
> > > >> On 8/28/2022 7:26 AM, Dan joyce wrote:
> > > >>> On Saturday, August 27, 2022 at 2:50:30 PM UTC-4, Chris M. Thomasson wrote:
> > > >>>> On 8/27/2022 10:13 AM, Dan joyce wrote:
> > > >>>>> Interesting plot of the solution set!
> > > >>>>>
> > > >>>>> n=(1/((sqrt(2))+2)),n1=((((n*2)+2)^2)-2)/4,n1/n = 4
> > > >>>> Show us a plot so we have something to test against!
> > > >>>>
> > > >>>> ;^)
> > > >>>
> > > >>> Enter in it Wolfram and you will see the plot!
> > > >> This one:
> > > >>
> > > >> https://i.ibb.co/Z6jRfVS/image.png
> > > >>
> > > >> ?
> > > > Chris,
> > > > Try this in Wolfram that uses this formula in ABS for negative squares and roots and also to find -y value
> > > > of -x^2- (-x)= -y
> > > > sqrt(n)=(((sqrt((n*4)+2))-2)/2), n^2=((((((sqrt(n))*2)+2)^2)-2)/4), n--->oo
> > > >
> > > >
> > > > -x+-y= -x^2 in the third negative quadrant.
> > > > -x^2 -(-x) =-y the two points needed to plot in the third negative quadrant starting @ x=0 and -y=(-0.5)
> > > > x goes negative with the first iteration of (sqrt(0.6+0.5))-1 = 0.0488088481...
> > > > -x = - 0.0488088481... Then incrementing +0.1 for the next plot point of (sqrt(0.7+0.5))-1
> > > > All in ABS of coerce.
> > > I might have some time tonight. Fwiw, check this plot out:
> > >
> > > ___________________________________
> > > x(t) = cos(t) * abs(cos(t))
> > > y(t) = sin(t) * abs(sin(t))
> > > ___________________________________
> > >
> > > t from 0 to pi2
> > >
> > > https://www.wolframalpha.com/input?i=x+%3D+cos%28t%29+*+abs%28cos%28t%29%29%2C+y+%3D+sin%28t%29+*+abs%28sin%28t%29%29%2C+t+from+0+to+pi2
> > >
> > > ;^)
> > Nice, but it ends @pi
> >
> > This one I plotted years ago in the third quadrant starts @ n= (sqrt(.6 +.0.5))-1 then incrementing by 0.1 after each
> > iteration (0.5+0.1) etc. Starting the curve @ -x=-0.048.8088481..and -x^2 = -1.2 all in ABS of course.
> > -x^2-(-x)=-y for the two points -x and -y on the curve. This means -x=-0.0488088481... -x =-0.048.8088481... --->-x ---->oo
> Also another one ---
> This is the only way I can express these negative squares and roots in a polar plot.
>
> Enter this in Wolfram alpha as a polar plot---
>
> plot r^2 = e^(-0.224744871391589049098642037352..theta) - (.908295745839672839274080366999...
>
>
> Where -0.2247448713915... is the square root of -1 and .9082957458396... is the square root of -pi. (is the ABS value)

Notice the parabolic curve at the bottom of the screen that gives the best depiction of this curve beginning @ -y=-0.5
and -x=0

On Tuesday, August 30, 2022 at 12:20:57 AM UTC-4, Dan joyce wrote:
> On Monday, August 29, 2022 at 8:38:53 PM UTC-4, Dan joyce wrote:
> > On Monday, August 29, 2022 at 8:08:24 PM UTC-4, Dan joyce wrote:
> > > On Monday, August 29, 2022 at 7:46:41 PM UTC-4, Chris M. Thomasson wrote:
> > > > On 8/29/2022 4:40 PM, Dan joyce wrote:
> > > > > On Sunday, August 28, 2022 at 6:09:13 PM UTC-4, Chris M. Thomasson wrote:
> > > > >> On 8/28/2022 7:26 AM, Dan joyce wrote:
> > > > >>> On Saturday, August 27, 2022 at 2:50:30 PM UTC-4, Chris M. Thomasson wrote:
> > > > >>>> On 8/27/2022 10:13 AM, Dan joyce wrote:
> > > > >>>>> Interesting plot of the solution set!
> > > > >>>>>
> > > > >>>>> n=(1/((sqrt(2))+2)),n1=((((n*2)+2)^2)-2)/4,n1/n = 4
> > > > >>>> Show us a plot so we have something to test against!
> > > > >>>>
> > > > >>>> ;^)
> > > > >>>
> > > > >>> Enter in it Wolfram and you will see the plot!
> > > > >> This one:
> > > > >>
> > > > >> https://i.ibb.co/Z6jRfVS/image.png
> > > > >>
> > > > >> ?
> > > > > Chris,
> > > > > Try this in Wolfram that uses this formula in ABS for negative squares and roots and also to find -y value
> > > > > of -x^2- (-x)= -y
> > > > > sqrt(n)=(((sqrt((n*4)+2))-2)/2), n^2=((((((sqrt(n))*2)+2)^2)-2)/4), n--->oo
> > > > >
> > > > >
> > > > > -x+-y= -x^2 in the third negative quadrant.
> > > > > -x^2 -(-x) =-y the two points needed to plot in the third negative quadrant starting @ x=0 and -y=(-0.5)
> > > > > x goes negative with the first iteration of (sqrt(0.6+0.5))-1 = 0.0488088481...
> > > > > -x = - 0.0488088481... Then incrementing +0.1 for the next plot point of (sqrt(0.7+0.5))-1
> > > > > All in ABS of coerce.
> > > > I might have some time tonight. Fwiw, check this plot out:
> > > >
> > > > ___________________________________
> > > > x(t) = cos(t) * abs(cos(t))
> > > > y(t) = sin(t) * abs(sin(t))
> > > > ___________________________________
> > > >
> > > > t from 0 to pi2
> > > >
> > > > https://www.wolframalpha.com/input?i=x+%3D+cos%28t%29+*+abs%28cos%28t%29%29%2C+y+%3D+sin%28t%29+*+abs%28sin%28t%29%29%2C+t+from+0+to+pi2
> > > >
> > > > ;^)
> > > Nice, but it ends @pi
> > >
> > > This one I plotted years ago in the third quadrant starts @ n= (sqrt(.6 +.0.5))-1 then incrementing by 0.1 after each
> > > iteration (0.5+0.1) etc. Starting the curve @ -x=-0.048.8088481..and -x^2 = -1.2 all in ABS of course.
> > > -x^2-(-x)=-y for the two points -x and -y on the curve. This means -x=-0.0488088481... -x =-0.048.8088481... --->-x ---->oo
> > Also another one ---
> > This is the only way I can express these negative squares and roots in a polar plot.
> >
> > Enter this in Wolfram alpha as a polar plot---
> >
> > plot r^2 = e^(-0.224744871391589049098642037352..theta) - (.908295745839672839274080366999...
> >
> >
> > Where -0.2247448713915... is the square root of -1 and .9082957458396... is the square root of -pi. (is the ABS value)
> Notice the parabolic curve at the bottom of the screen that gives the best depiction of this curve beginning @ -y=-0.5
> and -x=0
Correction --- and x=0