Talk Nerdy To Me ;)
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08-01-2017, 03:37 PM
RE: Talk Nerdy To Me ;)
[Image: nerds-home-funny-demotivational-poster-1272618780.jpg]
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08-01-2017, 03:41 PM
RE: Talk Nerdy To Me ;)
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08-01-2017, 03:59 PM
RE: Talk Nerdy To Me ;)
The Israeli government once officially invited Albert Einstein to be its president, a figurehead position that doesn't really have any executive power. He politely declined this generous offer, citing his lack of qualifications for the role.

If we came from dust, then why is there still dust?
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08-01-2017, 04:23 PM
RE: Talk Nerdy To Me ;)
(08-01-2017 02:35 PM)Kernel Sohcahtoa Wrote:  Let G be a group with the general operation *; thus, we can denote this as <G,*>. By the definition of a group, the elements in <G,*> will be associative [a*(b*c)=(a*b)*c.] and will possess an identity element e (for example, a*e=a and e*a=a) along with an inverse (for example, a*a'=e and a'*a=e).

Now, IMO, here's a pretty cool fact: if G is a group and a,b are elements in G, then (ab)^-1 (or the inverse of ab) is equal to b^-1a^-1 (or the inverse of b times the inverse of a). Now, let the operation in G be multiplicative. To prove this, multiply the left side of b^-1a^-1 by ab. Thus, ab(b^-1a^-1)
=a(bb^-1)a^-1 (via the associative law of algebra)
=aea^-1 (note that bb^-1= e)
=aa^-1 (note ae=a)
=e.

Since the product of ab and b^-1a^-1 resulted in the identity element e, then this means that ab and b^-1a^-1 are inverses of each other. Consequently, the inverse of ab is b^-1a^-1
or (ab)^-1=b^-1a^-1.

pffftt .... that's not nerdy. Tongue

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#sigh
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08-01-2017, 04:49 PM
RE: Talk Nerdy To Me ;)
(08-01-2017 04:23 PM)GirlyMan Wrote:  
(08-01-2017 02:35 PM)Kernel Sohcahtoa Wrote:  Let G be a group with the general operation *; thus, we can denote this as <G,*>. By the definition of a group, the elements in <G,*> will be associative [a*(b*c)=(a*b)*c.] and will possess an identity element e (for example, a*e=a and e*a=a) along with an inverse (for example, a*a'=e and a'*a=e).

Now, IMO, here's a pretty cool fact: if G is a group and a,b are elements in G, then (ab)^-1 (or the inverse of ab) is equal to b^-1a^-1 (or the inverse of b times the inverse of a). Now, let the operation in G be multiplicative. To prove this, multiply the left side of b^-1a^-1 by ab. Thus, ab(b^-1a^-1)
=a(bb^-1)a^-1 (via the associative law of algebra)
=aea^-1 (note that bb^-1= e)
=aa^-1 (note ae=a)
=e.

Since the product of ab and b^-1a^-1 resulted in the identity element e, then this means that ab and b^-1a^-1 are inverses of each other. Consequently, the inverse of ab is b^-1a^-1
or (ab)^-1=b^-1a^-1.

pffftt .... that's not nerdy. Tongue

[Image: math_zpsfckpkjbj.png]

[Image: tumblr_ohuanvSHzC1vw7r6zo1_400.gif]
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08-01-2017, 05:31 PM
RE: Talk Nerdy To Me ;)
(08-01-2017 03:41 PM)jennybee Wrote:  [Image: 82348ae914c8259c7de3aa834250c66c.jpg]

I would shower you with joules. Yes

Skepticism is not a position; it is an approach to claims.
Science is not a subject, but a method.
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08-01-2017, 07:44 PM
RE: Talk Nerdy To Me ;)
(08-01-2017 05:31 PM)Chas Wrote:  
(08-01-2017 03:41 PM)jennybee Wrote:  [Image: 82348ae914c8259c7de3aa834250c66c.jpg]

I would shower you with joules. Yes

Watt a Btu-ful thing to say Tongue
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08-01-2017, 07:48 PM
RE: Talk Nerdy To Me ;)
I am truly drawn to mathematical beauty, and this here graph is about as sexy as it gets. It's weird, but I get immense satisfaction from solving mathematical problems. I love to kill the Sudoku in the paper in the morning. I do them in ink. Ohmy When I taught math in high school, I didn't try to solve the problems ahead of time. I'd just work them out on the chalkboard in front of the students, unrehearsed. Tremendous satisfaction in that. But then, I had to do some crazy derivations in university, both in math and physics, which made those fairly trivial (but didn't remove the beauty or satisfaction).

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08-01-2017, 07:50 PM (This post was last modified: 08-01-2017 07:53 PM by cactus.)
RE: Talk Nerdy To Me ;)
(08-01-2017 07:48 PM)Fireball Wrote:  [Image: zAZmleB.png]
The fact that e is in that equation still blows my damn mind. lol

I was introduced to all of these concepts separately with no context, and then one day the teacher was like, "Here's an equation that relates imaginary numbers, trigonometry, and natural logs" and I was like wuuuuhhhhhhh Gasp

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08-01-2017, 10:14 PM
RE: Talk Nerdy To Me ;)
(08-01-2017 07:48 PM)Fireball Wrote:  I am truly drawn to mathematical beauty, and this here graph is about as sexy as it gets. It's weird, but I get immense satisfaction from solving mathematical problems. I love to kill the Sudoku in the paper in the morning. I do them in ink. Ohmy When I taught math in high school, I didn't try to solve the problems ahead of time. I'd just work them out on the chalkboard in front of the students, unrehearsed. Tremendous satisfaction in that. But then, I had to do some crazy derivations in university, both in math and physics, which made those fairly trivial (but didn't remove the beauty or satisfaction).

[Image: zAZmleB.png]

This is awesome. Thank you for posting Euler's formula. I absolutely enjoyed using Euler's formula to find complex roots to second order differential equations. I also thought that the following variant of Euler's formula, which was retrieved via Paul's online notes, was very cool: e^-iθ = cos(-θ) + isin(-θ)=cos(θ) - isin(θ). Live long and prosper, Fireball.

"I'm fearful when I see people substituting fear for reason." Klaatu, from The Day The Earth Stood Still (1951)
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