Back in the day it was really simple. God whispered secrets in the prophets' ears, and everyone know what to do, right? Wrong!

Happy nine days to you all.

## Wednesday, July 22, 2009

## Wednesday, July 15, 2009

### I wanted to find this for a while.

## Tuesday, July 14, 2009

### Chassidish!!

I just got an email from my nutiritionist (the one who got me started on this gluten-free diet). She says I can't have beer or whiskey, but I can have vodka. Oh darn! I forgot to ask about Everclear.

## Tuesday, July 7, 2009

### You are not safe!

I was floating around Google Reader today, and noticed that Reader had some posts from a friend's blog which I had thought said friend had deleted. I scrolled down further and found many more extremely personal posts, which I'm certain said friend had deleted

This reminded me of when a dear fellow blogger posted something nasty about me and immediately took it down. Later that night, I was chatting with TRS about a post which TRS had removed from his own blog:

In those days of youth and naivete I thought that TRS meant that those who had subscribed already would always have the post in their RSS readers. But seeing those supposedly deleted posts made me think that maybe RSS readers really do keep things forever. I subscribed to the blog in which I had been maligned way back in the day, scrolled down to September 12, and viola! The following was seen:

Sep 12, 2008 2:57 PM

I have this visceral belief that e is pure evil. Does anybody else feel that way some time?

The point of this post is not to laugh at people whose secrets are buried in Google Reader or to badmouth the guy who thought I'm viscerally evil. The point is that you gotta be careful. In the past I toyed with the idea of posting stuff that might be a tad too personal for public consumption but would be interesting to the friendly regular readers of this here humble blog. I figured I could delete it after everyone had read it, and my confidentiality would be safe. Say I to you, au contraire! Anybody can read anything anybody has ever written on any blog, merely by subscribing in Google reader!

*before I ever dreamt of subscribing to his blog*.This reminded me of when a dear fellow blogger posted something nasty about me and immediately took it down. Later that night, I was chatting with TRS about a post which TRS had removed from his own blog:

TRS: It's online, and on the rss feeds...

1:03 AM me: listen, it'll last as long as it lasts, and that's that

TRS: forever

me: is [OTHER BLOGGER]'s post about me forever?

TRS: probably

1:04 AM me: take it down before it feeds into any more feeds and gets cached and all that nasty stuff

In those days of youth and naivete I thought that TRS meant that those who had subscribed already would always have the post in their RSS readers. But seeing those supposedly deleted posts made me think that maybe RSS readers really do keep things forever. I subscribed to the blog in which I had been maligned way back in the day, scrolled down to September 12, and viola! The following was seen:

Sep 12, 2008 2:57 PM

**E**from CENSORED by CENSOREDI have this visceral belief that e is pure evil. Does anybody else feel that way some time?

The point of this post is not to laugh at people whose secrets are buried in Google Reader or to badmouth the guy who thought I'm viscerally evil. The point is that you gotta be careful. In the past I toyed with the idea of posting stuff that might be a tad too personal for public consumption but would be interesting to the friendly regular readers of this here humble blog. I figured I could delete it after everyone had read it, and my confidentiality would be safe. Say I to you, au contraire! Anybody can read anything anybody has ever written on any blog, merely by subscribing in Google reader!

## Saturday, July 4, 2009

### Guys and Girls

To some of you this post will restate the obvious. To some of you, this post will make no sense. But some of you will find this post enlightening. So if you're in that third group, this post is for you.

Guys like girls. Girls like guys. But they can't just meet each other and start romantic relationships. There's a whole game they need to play first. I don't know the rules of the game, but the main idea is that you can't say what you mean.

Now, when you meet someone and there's no chance of any romance developing, you're excused from that whole game. You're allowed to be yourself. A girl can buy a guy a coke. A guy doesn't need to act like Mister Cool. This is what the pick-up artists warn against: the worst place to be, they say, is in a girl's "friend zone." You'll stay "just friends" forever and never get the real sechoirah ("merchandise"). But I don't care. The friend zone happens to coincide with my comfort zone.

VD"L

In unrelated news, I've had ads on the blog for the past two days, and I already made one cent! Soon I'll be rich!

Guys like girls. Girls like guys. But they can't just meet each other and start romantic relationships. There's a whole game they need to play first. I don't know the rules of the game, but the main idea is that you can't say what you mean.

Now, when you meet someone and there's no chance of any romance developing, you're excused from that whole game. You're allowed to be yourself. A girl can buy a guy a coke. A guy doesn't need to act like Mister Cool. This is what the pick-up artists warn against: the worst place to be, they say, is in a girl's "friend zone." You'll stay "just friends" forever and never get the real sechoirah ("merchandise"). But I don't care. The friend zone happens to coincide with my comfort zone.

VD"L

In unrelated news, I've had ads on the blog for the past two days, and I already made one cent! Soon I'll be rich!

## Friday, July 3, 2009

### Update about Anti-Sociality and Coco

Last semester I was taking algebra and sitting in on calc I. This semester I'm taking calc II and sitting in on pre-calc. There's this dude who was in calc I with me and in chemistry with me (although that barely counts, because there were over a hundred people in chem with me). Now he's taking calc I again and he's sitting in on pre-calc with me. So we've been seeing each other around the tutoring center and in pre-calc. Polite gentleman that I am, I always raised my eyebrows in greeting when I saw him, and he would nod. We may have even ventured a "hi, how are you?" a few times. Yesterday we actually spoke! A full fledged conversation! The professor in pre-calc was giving an exam, and we discussed whether we should take it, being that we're just sitting in. Then we discussed optimization problems and what a sneaky trick it was for Prof. Destina to ask the calc I students to graph a rational function which had no vertical asymptotes. To those who say it can't be done: I socialized with a classmate, without any preliminary small talk bullsh*t.

In barely related news, below is the paper on which I disproved the Coco theorem. My writing is on the left and on top. Her writing is on the bottom right.

In barely related news, below is the paper on which I disproved the Coco theorem. My writing is on the left and on top. Her writing is on the bottom right.

## Wednesday, July 1, 2009

### E vs. Coco (If you don't know math, just skip the mumbo-jumbo. You'll get the story regardless.)

In my school, there's a tutoring center, where you can get help with math and physics. They lady in charge has this nasty high-pitched voice, painted-on eye lashes, and wears the horriblest, loudest, most-mismatched clothes you've ever seen. When you come in she screams, "Sign in! Sign in!" Imagine what would happen if people didn't sign in at the tutoring center... or even worse, they might sign in but leave out the last four digits of their social security numbers... Oh! God save us from such horrors! Thank God we have Coco to prevent such tragedies.

Despite the b*tch at the door, I like studying there. The atmosphere is kind of like a zal, and it's harder to space out or get too involved with checking my email on my phone when I'm with other people.

Anyhow, Coco had just finished screaming at one of the tutors for not offering his help. (That exchange was actually kind of funny.

Coco: Why you don't ask students if they need help?

Tutor: I aksed 'em all, two minutes ago.

Girl on the other side of room: (raising her hand) I need help)

After that exchange, Coco felt like making herself useful before she resumed her post at her computer to look out for students trying to sneak in without signing in. So she moseyed over to me and asked if needed help. I accepted her offer, and we started working through a problem together. My trig is horrible. I learned it all myself, so I'm missing lots of basics.

Me: (hoping she'll tell me the formula) OK, now it's cos2t = 0, so we need the double angle formula...

Coco: (mutters something unintelligible with her foreign accent)

Me: Yeah, the double angle formula, which is...

Coco: (smirking) the double angle formula

Me: which is...

Coco: (smirks silently)

I turn to cheat sheet on the inside from cover, and shamefacedly copy the double angle formula. But I copied the wrong one. So I copy the second one, even more shamefacedly.

I wasn't going to let this slide. I figured I'd ask her a question that's been bothering me since yesterday.

∫ e^[ln(2x)] dx Looks like a scary integral. It's not.

∫ e^[ln(2x)] dx = ∫ [e^ln(x)]^(2) dx

= ∫ x^2 dx

= x^3/3 + c

But suppose you didn't simply it first, could you still solve the problem by taking the ln of both sides? Here's what I did yesterday:

y = ∫ e^[ln(2x)] dx

ln(y) = ln ∫ e^[ln(2x)] dx

= ∫ ln e^[ln(2x)] dx

= ∫ [ln(2x)] dx

u=2x; du=2

= 1/2 ∫ [ln(2x)] 2dx

= 1/2 ∫ ln(u) du

using integration by parts, you get

1/2 [x*ln(2x) - x] + c

remember, that all this is ln(y). To get y, we raise that whole mess to e, so the final answer is:

e^(1/2 [x*ln(2x) - x] + c)

This is obviously not the right answer. But what had I done wrong? Perhaps the mistake was taking the ln of both sides?

I showed Coco the original expression.

Coco: That simplifies to X^2.

Me: Yeah, but if I don't simplify it, could I get the integral by taking the ln of both sides, and then raising everything to e?

Coco: Don't do it.

Me: I understand it's harder, but is it allowed?

Coco: Don't do it.

Me: Yeah, but could I? Suppose you have an integral that you can't evaluate, are you allowed to take the ln and then raise it to e?

Coco: Yes, but-

Me: Great. So I want to see how you would do that here.

Coco: No

Me: But I want to see how it would work out. Could you show me?

Coco: No

There I was, thirsting for knowledge, and she was refusing and making me feel like a spoiled child who wants the parent to give it unnecessary stuff.

So I bided my time and thought evil thoughts about Coco. And thought about this rule some more, and realized that Coco was wrong. Here's a simple proof:

∫2^x dx

According to the Coco Theorem, you proceed as follows:

∫2^x dx = y

∫ln 2^x dx = ln(y)

= ∫ x*ln2 dx

ln2 is a constant, so you take it outside:

= ln2 ∫ x dx

= ln2*x2/2 + c. That is complete BS. The Coco theorem is wrong

But by now Coco was ensconced near her computer, heavily engrossed in Yahoo news. I showed her my work and asked her what I had done wrong? Each step had been legal, according to her, but the answer was wrong. Everyone knows that ∫2^x dx = 2^x/ln(2) + c. Where had we gone wrong? She got all confused. She started to scribble, pulled a bunch of u-substations but couldn't get anywhere. She asked for my book, and started to flip through chapter 7. I calmly reminded her that Logarithms and exponentials are discussed in chapter 6. She violently flipped to chapter six, where they were discussing the derivatives of exponentials. I knew that the formula for integrating exponentials was on the next page, but I kept quiet. Let the master of the tutoring center plumb the mysteries of mathematics in peace.

She scribbled some more, and decided that ∫2^x dx is in fact 2^x/ln(2) + c, as I had predicted. So what was wrong with what I had done? Is it possible that you're not allowed to take the ln of both sides and the raise the equation to e? Eh? Was I perhaps misinformed a few minutes ago? Oh yeah, I had suspected as such. Well, Coco, don't feel bad. I hear you got the double-angle formula down pat.

Despite the b*tch at the door, I like studying there. The atmosphere is kind of like a zal, and it's harder to space out or get too involved with checking my email on my phone when I'm with other people.

Anyhow, Coco had just finished screaming at one of the tutors for not offering his help. (That exchange was actually kind of funny.

Coco: Why you don't ask students if they need help?

Tutor: I aksed 'em all, two minutes ago.

Girl on the other side of room: (raising her hand) I need help)

After that exchange, Coco felt like making herself useful before she resumed her post at her computer to look out for students trying to sneak in without signing in. So she moseyed over to me and asked if needed help. I accepted her offer, and we started working through a problem together. My trig is horrible. I learned it all myself, so I'm missing lots of basics.

Me: (hoping she'll tell me the formula) OK, now it's cos2t = 0, so we need the double angle formula...

Coco: (mutters something unintelligible with her foreign accent)

Me: Yeah, the double angle formula, which is...

Coco: (smirking) the double angle formula

Me: which is...

Coco: (smirks silently)

I turn to cheat sheet on the inside from cover, and shamefacedly copy the double angle formula. But I copied the wrong one. So I copy the second one, even more shamefacedly.

I wasn't going to let this slide. I figured I'd ask her a question that's been bothering me since yesterday.

∫ e^[ln(2x)] dx Looks like a scary integral. It's not.

∫ e^[ln(2x)] dx = ∫ [e^ln(x)]^(2) dx

= ∫ x^2 dx

= x^3/3 + c

But suppose you didn't simply it first, could you still solve the problem by taking the ln of both sides? Here's what I did yesterday:

y = ∫ e^[ln(2x)] dx

ln(y) = ln ∫ e^[ln(2x)] dx

= ∫ ln e^[ln(2x)] dx

= ∫ [ln(2x)] dx

u=2x; du=2

= 1/2 ∫ [ln(2x)] 2dx

= 1/2 ∫ ln(u) du

using integration by parts, you get

1/2 [x*ln(2x) - x] + c

remember, that all this is ln(y). To get y, we raise that whole mess to e, so the final answer is:

e^(1/2 [x*ln(2x) - x] + c)

This is obviously not the right answer. But what had I done wrong? Perhaps the mistake was taking the ln of both sides?

I showed Coco the original expression.

Coco: That simplifies to X^2.

Me: Yeah, but if I don't simplify it, could I get the integral by taking the ln of both sides, and then raising everything to e?

Coco: Don't do it.

Me: I understand it's harder, but is it allowed?

Coco: Don't do it.

Me: Yeah, but could I? Suppose you have an integral that you can't evaluate, are you allowed to take the ln and then raise it to e?

Coco: Yes, but-

Me: Great. So I want to see how you would do that here.

Coco: No

Me: But I want to see how it would work out. Could you show me?

Coco: No

There I was, thirsting for knowledge, and she was refusing and making me feel like a spoiled child who wants the parent to give it unnecessary stuff.

So I bided my time and thought evil thoughts about Coco. And thought about this rule some more, and realized that Coco was wrong. Here's a simple proof:

∫2^x dx

According to the Coco Theorem, you proceed as follows:

∫2^x dx = y

∫ln 2^x dx = ln(y)

= ∫ x*ln2 dx

ln2 is a constant, so you take it outside:

= ln2 ∫ x dx

= ln2*x2/2 + c. That is complete BS. The Coco theorem is wrong

But by now Coco was ensconced near her computer, heavily engrossed in Yahoo news. I showed her my work and asked her what I had done wrong? Each step had been legal, according to her, but the answer was wrong. Everyone knows that ∫2^x dx = 2^x/ln(2) + c. Where had we gone wrong? She got all confused. She started to scribble, pulled a bunch of u-substations but couldn't get anywhere. She asked for my book, and started to flip through chapter 7. I calmly reminded her that Logarithms and exponentials are discussed in chapter 6. She violently flipped to chapter six, where they were discussing the derivatives of exponentials. I knew that the formula for integrating exponentials was on the next page, but I kept quiet. Let the master of the tutoring center plumb the mysteries of mathematics in peace.

She scribbled some more, and decided that ∫2^x dx is in fact 2^x/ln(2) + c, as I had predicted. So what was wrong with what I had done? Is it possible that you're not allowed to take the ln of both sides and the raise the equation to e? Eh? Was I perhaps misinformed a few minutes ago? Oh yeah, I had suspected as such. Well, Coco, don't feel bad. I hear you got the double-angle formula down pat.

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