S.N.H.S. Mantis Presentation

[fleurdejoo]

It's a good thing you can imagine it!

Because you can't actually see it. :-(

 

[aNisip]

The picture isn't showing? Let me try again...

 

[fleurdejoo]

It's showing but it's tiny.

I am giving you grief because I want you to post some pictures.

It's no big deal though, even though I did send you a beautiful valida for it.

Just sayin'.

 

[aNisip]

k, here they are...my best fiednd was taking the pictures....he zoooooomed in on alot of things and didnt really understand the concept of "focus" ....and thats also why there are three pics of his gf holding the carolina, he was hiding behind the camera...lol

Enjoy!

 

[aNisip]

 

[aNisip]

 

[aNisip]

 

[aNisip]

Thanks for looking!

 

[ScienceGirl]

Very cool!

I like the one with you opening the net cage. Looks like an insect tamer coaxing out a foreign, dangerous species!

 

[fleurdejoo]

Awwwww yeah!!

That's what I'm talkin' about!!!

These are awesome Andrew!

Thanks for sharing!!!!

 

[aNisip]

basically at first sudents filled the room pretty evenly..but when i said id be bringing them out, they all moved to the back (incase y'all were wondering abt the sparsley occupied seats ) after the slide show, students were a little exhausted from all the info i just blessed them with...but when i took the variety of mantodea out they all perked up and came to life...and were all captivated and the last photo, i did tell her to smile, honest! but the best she gave me was a big stare and a litte smirk.

 

[Digger]

Great shots of your presentation Andrew. So that's the trick you use to pick up pretty blondes? Hand them a roach-munching mantis? Really unique !

 

[Digger]

Also love the equations in the background.......

Suppose that K, L and M form a tower of fields as in the degree formula above, and that both d = [L:K] and e = [M:L] are finite. This means that we may select a basis {u1, ..., ud} for L over K, and a basis {w1, ..., we} for M over L. We will show that the elements umwn, for m ranging through 1, 2, ..., d and n ranging through 1, 2, ..., e, form a basis for M/K; since there are precisely de of them, this proves that the dimension of M/K is de, which is the desired result.

First we check that they span M/K. If x is any element of M, then since the wn form a basis for M over L, we can find elements an in L such that

Then, since the um form a basis for L over K, we can find elements bm,n in K such that for each n,

Then using the distributive law and associativity of multiplication in M we have

which shows that x is a linear combination of the umwn with coefficients from K; in other words they span M over K.

Secondly we must check that they are linearly independent over K. So assume that

for some coefficients bm,n in K. Using distributivity and associativity again, we can group the terms as

and we see that the terms in parentheses must be zero, because they are elements of L, and the wn are linearly independent over L. That is,

for each n. Then, since the bm,n coefficients are in K, and the um are linearly independent over K, we must have that bm,n = 0 for all m and all n. This shows that the elements umwn are linearly independent over K. This concludes the proof.

 

[aNisip]

Also love the equations in the background.......

Suppose that K, L and M form a tower of fields as in the degree formula above, and that both d = [L:K] and e = [M:L] are finite. This means that we may select a basis {u1, ..., ud} for L over K, and a basis {w1, ..., we} for M over L. We will show that the elements umwn, for m ranging through 1, 2, ..., d and n ranging through 1, 2, ..., e, form a basis for M/K; since there are precisely de of them, this proves that the dimension of M/K is de, which is the desired result.

First we check that they span M/K. If x is any element of M, then since the wn form a basis for M over L, we can find elements an in L such that

Then, since the um form a basis for L over K, we can find elements bm,n in K such that for each n,

Then using the distributive law and associativity of multiplication in M we have

which shows that x is a linear combination of the umwn with coefficients from K; in other words they span M over K.

Secondly we must check that they are linearly independent over K. So assume that

for some coefficients bm,n in K. Using distributivity and associativity again, we can group the terms as

and we see that the terms in parentheses must be zero, because they are elements of L, and the wn are linearly independent over L. That is,

for each n. Then, since the bm,n coefficients are in K, and the um are linearly independent over K, we must have that bm,n = 0 for all m and all n. This shows that the elements umwn are linearly independent over K. This concludes the proof.

OR F2 Solve on my TI-89

but yes, since the bm,n coefficients are in K...we must have that bm,n = 0 for all m and all n...thusly elements umwn are linearly independent over K.

 

[Digger]

The answer is 42, Andrew.

42.

 

[ScienceGirl]

Haha, nice math, guys.

I also like how you let people hold them afterwards. Really adds an extra interest.

Way to represent the insect lovers and keepers!

 

[brancsikia339]

The answer is 42, Andrew.

42.

Nice pics Andrew! Love the expression on the girl holding the orchid :lol: