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Paul
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Re: Your rants here

February 20th, 2020, 8:43 pm

[$]\frac{d \,\mbox{Number}}{d \,\mbox{Anything}}=0[$]
 
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Paul
Posts: 10775
Joined: July 20th, 2001, 3:28 pm

Re: Your rants here

February 20th, 2020, 8:45 pm

I often explain vega via

[$]\frac{\partial \, \mbox{Area of a circle}}{\partial \pi}=r^2[$]
 
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Cuchulainn
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Re: Your rants here

February 20th, 2020, 9:12 pm

[$]\frac{\partial n!}{ \partial n} = (n-1)![$]

thus

[$]\frac{\partial n!}{ \partial (n-1)!} = n[$]
Last edited by Cuchulainn on February 20th, 2020, 9:24 pm, edited 3 times in total.
 
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katastrofa
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Joined: August 16th, 2007, 5:36 am
Location: Alpha Centauri

Re: Your rants here

February 20th, 2020, 9:14 pm

Haben Sie Zahl Schmerzen?
 
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Cuchulainn
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Re: Your rants here

February 20th, 2020, 9:21 pm

Haben Sie Zahl Schmerzen?
Yes, carnval.
I initially learned Dutch from such songs + ads.
www.youtube.com/watch?v=EfOgwRV8sGg

www.youtube.com/watch?v=ChU4gAS3NcE
 
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Cuchulainn
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Re: Your rants here

February 20th, 2020, 9:45 pm

[$]e^5 = 148.413[$]
[$]e^5 = 148.413[$]
[$]e^5 = 148.413[$]
[$]e^5 = 148.413[$]
[$]e^5 = 148.413[$]
[$]e^5 = 148.413[$]
[$]e^5 = 148.413[$]
[$]e^5 = 148.413[$]
[$]e^5 = 148.413[$]
[$]e^5 = 148.413[$]
[$]e^5 = 148.413[$]
[$]e^5 = 148.413[$]
[$]e^5 = 148.413[$]
[$]e^5 = 148.413[$]
[$]e^5 = 148.413[$]
[$]e^5 = 148.413[$]
 
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Paul
Posts: 10775
Joined: July 20th, 2001, 3:28 pm

Re: Your rants here

February 20th, 2020, 9:55 pm

That was in Stephen King's original draft. But his editor thought it too geeky.
 
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katastrofa
Posts: 9213
Joined: August 16th, 2007, 5:36 am
Location: Alpha Centauri

Re: Your rants here

February 20th, 2020, 10:05 pm

Haben Sie Zahl Schmerzen?
Yes, carnval.
I initially learned Dutch from such songs + ads.
www.youtube.com/watch?v=EfOgwRV8sGg

www.youtube.com/watch?v=ChU4gAS3NcE
Mein Deutch ich von Geothe gelernt habe. Oder Meister Yoda...
A small glass of local wine today, and some Martini tomorrow (30m+) :-D
Attachments
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Cuchulainn
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Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
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Re: Your rants here

February 20th, 2020, 10:20 pm

That was in Stephen King's original draft. But his editor thought it too geeky.
Stoopid question: who is SK?
 
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Paul
Posts: 10775
Joined: July 20th, 2001, 3:28 pm

Re: Your rants here

February 20th, 2020, 10:39 pm

Idea for a TV game show: "What's My Bubble?"
 
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katastrofa
Posts: 9213
Joined: August 16th, 2007, 5:36 am
Location: Alpha Centauri

Re: Your rants here

February 20th, 2020, 10:40 pm

Nitrogen
 
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bearish
Posts: 5535
Joined: February 3rd, 2011, 2:19 pm

Re: Your rants here

February 20th, 2020, 11:42 pm

Haben Sie Zahl Schmerzen?
Yes, carnval.
I initially learned Dutch from such songs + ads.
www.youtube.com/watch?v=EfOgwRV8sGg

www.youtube.com/watch?v=ChU4gAS3NcE
Mein Deutch ich von Geothe gelernt habe. Oder Meister Yoda...
A small glass of local wine today, and some Martini tomorrow (30m+) :-D
Wait - you are in the Canary Islands?  Not to be confused with Canary Wharf...
 
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katastrofa
Posts: 9213
Joined: August 16th, 2007, 5:36 am
Location: Alpha Centauri

Re: Your rants here

February 21st, 2020, 9:08 am

Sí, Señor :-) Visitando al diablo.
 
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Cuchulainn
Posts: 62126
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
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Re: Your rants here

February 21st, 2020, 9:24 am

That was in Stephen King's original draft. But his editor thought it too geeky.
Got it! Penny drops.
 
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Collector
Posts: 4613
Joined: August 21st, 2001, 12:37 pm

Re: Your rants here

February 21st, 2020, 11:52 am

I often explain vega via

[$]\frac{\partial \, \mbox{Area of a circle}}{\partial \pi}=r^2[$]
[$]\frac{\partial^2 \, \mbox{Area of a circle}}{\partial \pi \partial x}=ab[$]

this because a circle moved along the x axis is observed to be an ellipse due to length contraction, so we have the well unknown circle equation

[$]\frac{\partial^2 \, \mbox{Area of a circle}}{\partial \pi \partial t}-\frac{\partial^2 \, \mbox{Area of a circle}}{\partial \pi \partial x}=r^2-ab[$]

or more interesting

[$]\frac{\partial \, \mbox{Area of a circle}}{\partial t}-\frac{\partial \, \mbox{Area of a circle}}{ \partial x}=\pi r^2-\pi ab[$]

where a and b functions of v

and do we also get:?

[$]\frac{\partial \, \mbox{Area of a circle}}{\partial t}-\nabla \mbox{Area of a circle} =\pi r^2-\pi ab[$]

more complex under acceleration.
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