The following is a very real conversation between three very real human beings that took place this evening on tinychat.com/ssr. All I can tell you is that I used "waveocean" for my nickname, and I had the other two people use "californiawaves" and "creatureofthenight" as their nicknames to protect their identity in case anyone else showed up. For those of you who enjoy math, I hope you have fun reading this conversation. For those of you who do not, I hope you laugh at what you are about to read.

And to clarify, the California Waves and CreatureOfTheNight SBNation accounts do not belong to anyone else. They are alternative accounts that were created by me, WaveOcean. WaveOcean, California Waves, and CreatureOfTheNight are all the same person on SBNation. In the conversation on tinychat, they were just nicknames used by me and two other people.

[18:17] guest-8225 changed nickname to waveocean

[18:20] guest-8261 entered the room

[18:20] guest-8270 entered the room

[18:20] waveocean: hi

[18:21] waveocean: is this who i think it is?

[18:21] guest-8270 changed nickname to californiawaves

[18:21] waveocean: All right

[18:21] waveocean: ready for some Math 527?

[18:21] californiawaves: Okay

[18:21] waveocean: This is tiny chat, a site where you can chat with anyone without an account

[18:21] californiawaves: cool

[18:22] waveocean: i don't think anyone else will stumble in on our conversation

[18:22] waveocean: so how did you get the sixth root of e?

[18:22] californiawaves: they be really bored if they did!

[18:22] waveocean: maybe i could get creatureofthenight to come here

[18:22] californiawaves: cool

[18:23] californiawaves: did you start by using logs

[18:23] waveocean: er....

[18:24] californiawaves: you know, log L = blah blah

[18:24] waveocean: I started by writing e^[(1/x^2)ln(x/sin(x))]

[18:25] californiawaves: okay, I used log L = (1/x^2) log(x/sin(x))  

[18:25] waveocean: fair enough

[18:25] waveocean: continue

[18:25] californiawaves: I had to use L'hopital several times

[18:25] californiawaves: you?

[18:26] waveocean: i used it 3 times

[18:26] californiawaves: yeah, I used it once to show x/sinx = 1 

[18:26] waveocean: all righ

[18:26] waveocean: it becomes 1/cosx

[18:27] californiawaves: yep

[18:27] waveocean: now i have ln(1/cosx)/x^2

[18:28] waveocean: e to the power of all that

[18:28] californiawaves: okay, I have log(x/sinx) / x^2 which is of the form 0/0

[18:28] californiawaves: same thing

[18:29] waveocean: so use l'hopital's rule to get tanx/2x

[18:30] californiawaves: I get lim  (-xcosx + sinx)/2 x^2 sinx

[18:31] guest-8518 entered the room

[18:31] waveocean: what....

[18:31] guest-8518 changed nickname to creatureofthenight

[18:31] waveocean: hi creatureofthenight

[18:31] creatureofthenight: hi

[18:31] waveocean: we're discussing 4.4.5c

[18:31] californiawaves: hi

[18:32] creatureofthenight: ok...i got e^1/6

[18:32] waveocean: i got square root of e

[18:32] waveocean: DAMNIT

[18:32] waveocean: Then i must be wrong

[18:32] californiawaves: me too!

[18:32] creatureofthenight: I had to apply l'Hopital 3 times I think

[18:32] californiawaves: yeah

[18:32] waveocean: so you're saying the derivative of ln(1/cosx) is not tanx?

[18:33] creatureofthenight: i don't have any tangents involved in mine...just sin and cos

[18:33] californiawaves: me too

[18:33] californiawaves: I have two terms

[18:34] waveocean: i don't get it

[18:34] waveocean: it's cos times the derivative of sec

[18:34] waveocean: isn't it?

[18:34] creatureofthenight: did you ln both sides first, before you started applying l'hopital?

[18:35] waveocean: i didn't...but...i'm trying to take the derivative of ln(1/cosx). That's what i want to do

[18:35] waveocean: right?

[18:36] creatureofthenight: the first time you take derivatives you should be taking the derivaive of ln(x/sinx) on th

[18:36] creatureofthenight: e top

[18:36] californiawaves: I used the original log(x/sinx)

[18:36] creatureofthenight: and x^2 on the bottom

[18:36] waveocean: how?

[18:36] waveocean: you cannot take log(x/sinx)

[18:36] waveocean: it's negative infinity

[18:37] creatureofthenight: that's why you l'hopital

[18:37] californiawaves: differentiate the original

[18:37] waveocean: but then you have negative infinity on the top and 0 on the bottom

[18:37] waveocean: you can't use l'hopital's rule

[18:38] waveocean: they both have to be 0

[18:38] waveocean: or they both have to be infinity

[18:38] creatureofthenight: hmm...

[18:38] creatureofthenight: I dunno...I just did it anyway

[18:39] waveocean: shit, where's californiawaves

[18:39] creatureofthenight: ?

[18:40] waveocean: it kicked her off for no reason

[18:40] waveocean: well,

[18:41] waveocean: i think i'm going to live and die by the answer i put

[18:41] waveocean: but thanks for talking with me about it

[18:41] creatureofthenight: yeah...I'm leaving mine too...I wrote too much to rewrite it...lol

[18:41] waveocean: all right

[18:41] waveocean: shall we do 2b or 4 first?

[18:42] creatureofthenight: we can do 4

[18:42] creatureofthenight: I typed it so I could copy and paste it

[18:42] creatureofthenight: here's what i did:

[18:42] creatureofthenight: Let y=sin^(-1)(x)Then siny=xDifferentiate both sides (using implicit differentiation) of t

[18:43] creatureofthenight: the equation to get  Cosy(dy/dx)=1(dy/dx)=1/cosy

[18:43] creatureofthenight: after the 1, there was supposed to be space

[18:43] creatureofthenight: that dy/dx=1/cosy is the next step

[18:44] waveocean: all right

[18:44] waveocean: i follow

[18:45] creatureofthenight: ok..then  (dy/dx)=1/(cos(sin^(-1)(x)))     (as y=sin^(-1)(x))

[18:45] creatureofthenight: Use trig identity cos^2+sin^2=1 to get:    (dy/dx)=1/(√(1-sin(sin^(-1)(x))^2))

[18:45] creatureofthenight: So    (dy/dx)=(f^-1)’(x)=1/√(1-x^2)

[18:46] guest-8810 entered the room

[18:46] guest-8810 changed nickname to californiawaves

[18:46] creatureofthenight: it sucks to have to write it like that

[18:46] waveocean: Yeah...

[18:46] creatureofthenight: I hope it's understandable

[18:46] waveocean: I'm having trouble.

[18:46] creatureofthenight: where?

[18:47] californiawaves: damn comcast!

[18:47] waveocean: so we use the trig identity to get dy/dx equals what?

[18:47] waveocean: sorry californiawaves

[18:47] creatureofthenight: I've been having trouble with comcast too...

[18:47] californiawaves: no worries

[18:47] creatureofthenight: the denominator should become the square root of 1 minus sin^2(sin^-1(x))

[18:48] creatureofthenight: and those sines canceld to get x^2

[18:48] waveocean: oh....you replace cos

[18:48] waveocean: i get it

[18:48] creatureofthenight: yeah

[18:48] waveocean: wow

[18:48] waveocean: that's pretty badass

[18:48] creatureofthenight: I thought it was pretty cool

[18:49] waveocean: all right then. That's that.

[18:49] waveocean: Now for 2b

[18:49] californiawaves: mine's a mess! ha

[18:49] creatureofthenight: at least you've done something. I'm stumped

[18:49] waveocean: first of all, what is f'?

[18:50] waveocean: how would we differentiate f?

[18:50] californiawaves: just 1 + e g'  

[18:50] waveocean: exactly

[18:50] waveocean: now what do we know about g'?

[18:51] creatureofthenight: bounded by M

[18:51] waveocean: exactly

[18:51] californiawaves: and could use MVT on g

[18:51] waveocean: oh

[18:51] waveocean: that's not what i did

[18:51] creatureofthenight: can we go from the beginning?

[18:51] waveocean: all right

[18:51] creatureofthenight: or is differentiating f the beginning?

[18:52] waveocean: it's where i started

[18:52] creatureofthenight: ok

[18:52] californiawaves: ok

[18:52] waveocean: so anyway, how can we set epsilon so that 1 + episilon g' is positive?

[18:53] waveocean: knowing that -M <= g' <= M

[18:53] creatureofthenight: I'm guessing we use the positive M somehow....

[18:54] waveocean: how about 1/(2M)?

[18:54] creatureofthenight: ok...that makes sense

[18:54] waveocean: then f' is greater than 1/2

[18:54] waveocean: so it's always positive

[18:54] waveocean: and what did we learn from 2a when f' is positive?

[18:55] creatureofthenight: strictly increasing

[18:55] waveocean: in other words, monotone

[18:55] waveocean: and thus one-to-one

[18:55] creatureofthenight: which makes it one to one

[18:55] creatureofthenight: ok

[18:55] waveocean: there you go!

[18:55] creatureofthenight: oh that's not so bad!

[18:55] californiawaves: wow!

[18:55] waveocean: no it's not lol

[18:55] creatureofthenight: I was having the hardest time

[18:55] californiawaves: me too

[18:55] waveocean: well, that's what i'm here fore

[18:55] waveocean: for

[18:56] californiawaves: what thm # is that

[18:56] waveocean: Uh...

[18:56] waveocean: it's not really a theorem.

[18:56] waveocean: it's what the professor would call "second nature".

[18:56] waveocean: i mean, if something is strictly increasing,

[18:57] waveocean: how could f(x) = f(y) without x=y?

[18:57] californiawaves: right

[18:57] creatureofthenight: ok...so I'm a little confused on how we got f' > 1/2....f'=1+eg'<=1+eM?

[18:58] waveocean: here's the deal

[18:58] waveocean: -M <= g' <= M

[18:58] waveocean: so set e = 1/(2M)

[18:58] waveocean: then 1/2 <= f' <= 3/2

[18:59] waveocean: so f' is positive

[18:59] creatureofthenight: oh ok

[18:59] waveocean: all clear?

[18:59] californiawaves: neat

[18:59] creatureofthenight: i think so

[18:59] waveocean: i'm going to stay logged in while working, so let me know if you want to talk some more.

[19:00] creatureofthenight: if you could help me with 3.3.5 on the prev hw real quick that would be great

[19:00] waveocean: all right

[19:00] creatureofthenight: I know let e=(M+f(a))/2....

[19:00] creatureofthenight: but I'm kinda stuck

[19:00] waveocean: well, we know e > 0

[19:00] californiawaves: was it the last one?

[19:01] waveocean: and we know f is continuous at a

[19:01] creatureofthenight: right....and no californiawaves, it was in the middle somewhere

[19:01] waveocean: so there exists d > 0 such that |x - a| < d implies |f(x) - f(a)| < e

[19:02] creatureofthenight: yeah I got all that

[19:02] waveocean: well, then -e < f(x) - f(a) < e

[19:02] creatureofthenight: oh ok...I'm a tard...I think i can take it from there

[19:03] waveocean: oh can you?

[19:03] creatureofthenight: maybe not...

[19:03] waveocean: we can talk about this again tomorrow if you prefer

[19:03] waveocean: or shall i continue?

[19:04] creatureofthenight: you can continue....lol

[19:04] waveocean: so we have f(x) < e + f(a)

[19:05] waveocean: which is less than (M - f(a))/2 + f(a)

[19:05] waveocean: which is less than M

[19:05] creatureofthenight: oh ok

[19:06] waveocean: So we want to pick x such that |x - a| < d

[19:06] waveocean: In other words, x must lie in the open interval (a - d, a + d)

[19:06] waveocean: which clearly contains a

[19:06] creatureofthenight: yeah

[19:06] waveocean: i wrote that down and got 5/5

[19:06] californiawaves: sweet

[19:07] creatureofthenight: nice. 

[19:07] waveocean: all right. I need to get to work. I've only written down all the problems that weren't on

[19:07] waveocean: sheet 6

[19:07] waveocean: i'll stay here

[19:08] creatureofthenight: good luck!

[19:08] californiawaves: for sheet 6.4 can't get + or - sign to go away at the end

[19:08] waveocean: if you want to talk later

[19:08] creatureofthenight: for #4 I let y=sin^-1(x) and then implicitly differentiated both sides...is that how you

[19:08] creatureofthenight: did it californiawaves?

[19:09] californiawaves: wow, no I used inverse fcn thm

[19:10] creatureofthenight: oh...well if you do it like I did, then you end up with 1/sqrt(1-cos(sin^-1(x)) and you 

[19:10] creatureofthenight: can replace cos using cos=sqrt(1-sin^2) and then you get exactly what you want

[19:11] californiawaves: cool

[19:13] californiawaves: thanks guys! good night

[19:13] creatureofthenight: good night!

[19:13] waveocean: good night!