Mathematics - 2012-04-21 (page 2 of 2)

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6:03 PM

@HenningMakholm I keep thinking about Lebesgue integrals. I will have to revisit that answer and see if it needs to be changed.

http://i.stack.imgur.com/KSSY4.png

by Paulo Cereda, TeXSE

x4d33746153706c306974

Hi, I need help with probability : 200 coins are tossed , what is the probability of getting 30 heads?

Henning Makholm

@x4d33746153706c306974 Google "binomial distribution"

@robjohn I think it becomes easier. My intuition is that there would be an entire interval where $f$ is strictly positive, and the integral of $P^2f$ over that interval would necessarily contribute positively unless $P$ is identically zero.

x4d33746153706c306974

@HenningMakholm Hi, Thanks , Yes I'm already aware of that , but is there way to do it without binomial distribution ?

Henning Makholm

@x4d33746153706c306974 Why? The binomial distribution is not just a particular solution, it is the name of your problem -- except for the particular constants 200 and 30.

x4d33746153706c306974

6:17 PM

@HenningMakholm just curious to know whether it's possible or not

Henning Makholm

That's like asking whether there's a way to compute 33+29893 without using addition.

user19161

@x4d33746153706c306974 Whichever way you do it boils down to using the binomial distribution because that is what it is.

x4d33746153706c306974

@HenningMakholm arhhhh that's not fair, well there must be some way to do it without using binomial distribution

user19161

@x4d33746153706c306974 You misspelled "fair" which is not the same as "fare". Also life is not fair.

x4d33746153706c306974

for example getting 2 heads out 3 coins tossed can be easily done without using binomial coefficients

@JasperLoy Thanks :)

user19161

6:21 PM

@x4d33746153706c306974 Are you having difficulty with a homework problem?

x4d33746153706c306974

@JasperLoy well I was practicing probability (not homework )but I just am curious to know the other way

user19161

@x4d33746153706c306974 Whichever other way there is is just a variation and not really another method. There is nothing mysterious about the binomial distribution.

user19161

@x4d33746153706c306974 One can always list all possible cases and count manually. It is still a binomial distribution in the end.

why does my calculator tell me that $-2^2$ is a negative number?

Heya

x4d33746153706c306974

6:24 PM

@JasperLoy aah , Thanks :)

@Jordan it's (-1) 2^2 I guess :P

@Jordan -2^2 is not the same as (-2)^2

I know, but why does it assume the calculation that no one would ever do?

x4d33746153706c306974

lol

Easy

It computes left to right

user19161

@Jordan Because it does the exponentiation first.

6:26 PM

if I am doing $-2^2$ I will always want $(-2)^2$ otherwise I would do -1(2^2)

user19161

@Jordan That's not how it should be written. If I write $-a^2$ I mean the negative of the square.

Henning Makholm

@Jordan You're welcome to that, but if you want to communicate with people (or machines) that use the usual convention, you'll have to write is as $(-2)^2$.

I dont understand

so in my book when I am suppose to find $x^2$ with x = -2 it actually wants me to do $-1(2^2)$?

@HenningMakholm How about this approach?

user19161

@Jordan That is asking for the square of the negative, the way it is written in the question.

6:30 PM

but why is that assumed?

x^2 = (x)^2

user19161

@Jordan Because $x^2$ means the square of $x$.

or when I have $f(x) = 3x - 2x$ that is actually 1(3x) + (-1)(2x)?

user19161

@Jordan When substituting values into variables, you don't just literally replace the variable with the value into the expression physically. You have to ask yourself what the expression means.

so if that is true then when I have $f(x) = 3 - x^2$ it is actually $3 + (-1)(x^2)$

user19161

6:32 PM

@Jordan That is one way to look at it.

user19161

@Jordan Yes, you can say that.

I dont know how I have been doing math and geting correct answers at all then if this is true

Magic

user19161

@Jordan Sometimes one can get lucky.

so if I want the derivative of -x^2 it is actually the product rule with (-1) * x^2

6:34 PM

Well if one only looks at positive numbers and positive operations, it does not really matter.

It just happens to work out that if you do it either way the answer is the same regardless

user19161

@Jordan You can see it as the product rule used with $-1$ and $x^2$ if you are talking about $-x^2$.

@Jordan -1 is a constant and can be put outside of the derivation. If you really want to use the product rule you can write it as $-1 \cdot x^{0}$, but the derivative of a constant is always zero. So it all works out...

well I have to relearn 20 years of math

user19161

@Jordan You just have to rethink certain fundamentals.

6:37 PM

You must unlearn what you have learned -Yoda

user19161

Get the foundation right and everything will be right. Get the foundation wrong and everything will be wrong.

user19161

@robjohn Thanks, I am only a Padawan.

I am still weak in the fork.

user19161

@N3buchadnezzar You may use a spoon instead.

Henning Makholm

@robjohn I think that will work -- it took some time to notice that $P$ being a polynomial necessarily makes the $U_k$s nice enough that it makes sense to Riemann integrate over them. (Somehow I had gotten into my head that you were integrating over preimages under $f$ rather than $P$).

6:38 PM

@N3buchadnezzar when eating in polite company, remember to use the forks.

so $-x^2 where x = 2$ and $x^2 where x = -2$ are not the same

@robjohn So no drinking from the bowls, and singing on the top of the tables?

@HenningMakholm Yes, the continuity of P makes them open sets.

@Jordan Correct.

Henning Makholm

@robjohn Not only open, but the union of finitely many open intervals.

6:40 PM

@HenningMakholm yes.

how have I even done math up to this point without knowing that?

Maybe I did know it but I forgot overnight, that seems to happen a lot

@HenningMakholm and that works for Riemann or Lebesgue integrals.

Magic

Of course it all depends on the OP confirming that $f$ is non-negative.

Henning Makholm

@Jordan It seems that you have mentally kept track of the boundaries of the subexpressions you inserted instead of $x$, and then evaluated the result as if there had been the necessary parentheses.

6:43 PM

Are there other forms of integrals than Riemann , (Stjelts?) and Lebesgue. And what are the differences?

Henning Makholm

@Jordan Then it only bit you when you expected the calculator to know about your mental bookkeeping.

user19161

@N3buchadnezzar There is the Henstock integral.

well I still can't get this simple problem right anyways

@N3buchadnezzar There are many. See Wikipedia.

I need to make 3.2 -1 + 2 to equal 4.2

user19161

6:44 PM

Denjoy, Perron, Henstock and Kurzweil had different formulations of the Henstock integral but they are equivalent in particular contexts. There is a book on these 4 formulations.

I am so fucking bad at math, not sure why I could add that

user19161

@Jordan You could choose to keep working at it or do something else.

user19161

@Jordan Er what do you mean?

I have been doing math homework for 3 hours now and I am not even a quarter of the way done lol

Oh, stop cursing is also an idea. It keeps you from being banned from chat =)

6:46 PM

I spent like 20 minutes trying to make 3.2 -1 + 2 = 4.2 and I couldnt do it for some reason

user19161

@Jordan I don't understand the question.

that is why I fail all my tests I think, I just mess up simple stuff for some reason and waste unreasonable amounts of time

what is $3.2 -1 + 2 equal to$

user19161

@Jordan That is why one needs to get the fundamentals right, not just keep doing problems blindly. That is what is happening in the world today. Students practise blindly without understanding.

user19161

@Jordan Depends on whether the dot means decimal or multiplication of course.

3 and 2 tenths

user19161

6:49 PM

@Jordan So isn't it obvious that it becomes 4.2 after subtracting 1 and then adding 2?

I can get it now, but I just spent like 20 minutes on it for some reason

Sorry, my connection is very poor now.

I want to teach my students the formula for solving third degree equations, that will stop them from wanting formulas for everything.

Henning Makholm

@JasperLoy Apparently, some unfortunate middle schoolers are taught mnemonics that imply that addition must be done before subtraction

yes, every school in the US teaches that

Henning Makholm

@Jordan You've been to every one of them to check?

6:53 PM

No but I do know that all teachers are paid very little and most do not care or are qualified to be good teachers

so they teach math as something you memorize

I know nearly every school is held to standardized testing

so they teach that

Then get your ass over to MIT or something =)

I will be lucky to get accepted to a for profit school :P

@robjohn indeed.

@leo indeed? are you referring to the comments?

I was thinking that one can prove that then $P^2f\equiv 0$ and then since $\int f\gt 0$, $P$ must be 0

@robjohn yes

I was wrong

6:59 PM

brb gotta restart my browser. I am getting no pings.

I know there is a general feeling that math isn't about memorization but then why is so much of calculus memorization based?

At least where I go to school we are expected to memorize so many derivatives and anti derivatives and then all the derivative and integral rules we need to memorize but we only go through the proof once really fast and then we are just expected to memorize it without ever really learning why the proof works

Memorization is the beginning of intelligence.

@Jordan There are some formulas that are useful to know by heart since they are used frequently, but you first should understand them.

there is too much material in calculus to understand it all

@Jordan There are basic rules, and if you memorize a few, you should be able to combine them to handle almost any derivative.

2

7:08 PM

I just barely have enough time to get the homework done between classes, where would I find the time to dig through proofs and other in depth things?

integrals are a different story.

still no ping. :-(

$8x^3 + 2x - 12x^2 - 3$ x = 2 not sure what I am doing wrong

maybe I am thinking of exponents wrong or something

68-51?

@robjohn ping test...how about now? I remember you had this problem before, what did you do then?

@Rob as I said, no ping :-(

nevermind I was doing this ocmpletely wrong

7:11 PM

Need to undo stars gah!

@N3buchadnezzar which ones?

@Rob I restarted my browser. I have done that. Let me turn if off and back on.

Reboot @robjohn

@robjohn ALL OF THEM

GWhen computing the probability that P(X>Y) (or P(X-Y>0)) how should I set my bound for my double integral? I haven't taken multivariable calculus yet, so I haven't formally learned to do this. $\int_a^b \int_c^d f(x)dy dx$ Should I integrate $\int_a^b \int_(-x^1 f(x)dy dx$

I wonder if this has anything to do with having disabled Java (not Javascript) in my browser to avoid the virus problems.

7:16 PM

If you go to the tab where you starred them
you should see an option to "unstar." :-)

@robjohn This site has viruses?

Given that $f_Y(y) and f_X(x) $ are both less than one.

@Rob No, but there are some vulnerabilities in the Java plugin for my OS

I need to upgrade my OS to 10.6, I guess.

user19161

@robjohn The ping uses Adobe Flash I believe.

@robjohn My shockwave program crashes sometimes here?

user19161

@robjohn Sometimes my ping is delayed by half a minute.

user19161

7:21 PM

I am on Debian Squeeze.

Henning Makholm

@Jordan By your own account you're wasting a lot of time attempting to do homework without first understanding the principles the homework is supposed to train. That time would be a lot better invested in understanding first.

In mathematics you don't understand things. You just get used to them

How do I understand the concepts? I spend a lot of time reading the book but it doesnt really make sense

user19161

@Rob Not true.

user19161

@Jordan You need a good book and/or a good teacher.

7:29 PM

@robjohn ping test

user19161

Unfortunately most books these days are just full of colourful pictures.

@JasperLoy Are you saying that memorization plays no role in math?

user19161

@Rob No, I am not saying that. How do you know that 2 times 3 is 6? Memory.

Success! I had to install a new Flash plugin. The old one was blocked due to security issues.

user19161

@robjohn So I was right! Yay! Plus one for me!

7:30 PM

+1 @JasperLoy

@JasperLoy I Lay my marbles in front of me and count them.

user19161

@N3buchadnezzar Then remember to bring enough marbles to the exam hall!

and don't forget to bring your infinitely divisible marbles

You need many marbles for Calculus 3

@JasperLoy Indeed

ie the marble with size 0.99...

7:34 PM

@Rob I do not need those, I just need infinite amounts of ordinary marbles

An infinite amount of ordinary marbles must weigh a lot :-)

Pictures are really helpful in calculus I think

A picture is worth a thousand words.

user19161

@Jordan Indeed. One can solve multidimensional problems too using 2D pictures as a special case. Pictures help motivate more formal proofs.

user19161

@Rob Sometimes a word is worth a thousand pictures.

7:48 PM

Stewart is just really good at picking the worst pictures

@JasperLoy A word has to be written in a language, a picture does not.

My question was closed as "too localized" in MathOverflow. May I re-state it in math.StackExchange?

user19161

@porton Er, why post a research level question here?

user19161

Anyway you may always try, no harm.

@JasperLoy: It seems nevertheless simple enough (despite I have not yet solved it myself)

@JasperLoy: I may try. But won't this be considered as kinda spamming?

user19161

7:55 PM

@porton Well, if it is well-written I think it is alright. And mathoverflow is not part of SE unless I am mistaken.

@robjohn ping test ;-)

user19161

@Rob He seems to have vanished after the ping was fixed.

@JasperLoy Probably afk...

like I should be.

In fact need to be.

8:40 PM

@Rob I was afk so I didn't hear

@robjohn self-ping

still works

is $x(x^\frac{1}{3}) = x^\frac{4}{3}$?

Henning Makholm

Hah! I'm going to get a Disciplined badge.

@Jordan Yes

should that be intuitive for me to imagine? more specifically should I be able to picture what is going on with $\frac{x^2}{x^\frac{1}{2}}$

I guess the way I need to look at it is in terms of mulitples and roots

like (2 * 2 ) / (2/2) is that correct?

Henning Makholm

@Jordan I don't think pictures are a good way to understand that. It's just an application of exponent rules.

I know I have memorized that rule

I was jsut wondering if there was a more natural way to look at it

Henning Makholm

8:59 PM

Well, you could picture an row of x·x·x···x, where a negative exponent (or a denominator) means to take x's away from an already existing row. The trouble then is to imagine "partial" x's in the row when there are fractional exponents. You wouldn't happen to be familiar with slide rules, would you?

no idea what a slide rule is

user19161

9:18 PM

@Jordan One way to look at this is try to cube both sides.

@HenningMakholm We already got one. :-)

of course it is hard to find the answer you deleted later :-)

user19161

@robjohn We?

@JasperLoy it's from Monty Python and the Holy Grail

playing on the royal "we"

user19161

@robjohn Ah never watched.

@JasperLoy That's too bad. Unless you don't like British humor, it is worth a watch.

user19161

9:23 PM

@robjohn I definitely like Mr Bean!

@Jordan that is a sad sign of the times (at least it is to me).

@JasperLoy then you will most likely enjoy the Holy Grail

user19161

@robjohn Wait are you talking about the slide rule? Actually I am not too sure what that is either! A ruler that slides?

Henning Makholm

@robjohn Ah, but later I can just search the chat transcript and find the link here :-)

user19161

@HenningMakholm Oh well, that is just misapplication of a mnemonic!

Henning Makholm

@JasperLoy A slide rule is a mechanical device for doing multiplications, used before electronic calculators became common. In its basic form it consists of two rulers with identical logarithmic scales engraved on them, which can slide along each other. To multiply $a$ with $b$, slide the upper scale such that $1$ on the upper scale coincides with $a$ on the lower scale. Then $b$ on the upper scale will coincide with $ab$ on the lower scale.

user19161

9:30 PM

@HenningMakholm Er, I tell you what I use to do multiplications before I used calculators. I learnt to multiply by hand!

user19161

@HenningMakholm I hate the graphing calculators they use in schools these days. Students don't understand why a graph looks the way it does because they just use the graphing calculator.

user19161

Technology can aid in learning, but sometimes it actually hinders learning.

Henning Makholm

@JasperLoy Yes, but the slide rule was quicker if you just needed a few digits of precision.

user19161

Computers are used not as a means to the end of making students understand, but as an end in itself sometimes I feel.

Henning Makholm

Wikipedia article

9:34 PM

@HenningMakholm You can. I didn't make a note of mine in chat, so I am not sure where it is.

Henning Makholm

And the reason I brought up slide rules is that if one happens to have familiarity with them, it gives an immediate feel for how the exponent rules work and what to do about negative exponents (they move the same about on the scale, but to the left instead of to the right) and how factional exponents work.

user19161

@HenningMakholm Thanks, I have actually used a ruler that slides too. It rolls actually and there is a scale on the roller so you know the distance rolled.

@HenningMakholm I'm sure that someone who's used a slide rule better understands Benford's Law

I wrote a page trying to explain Benford's Law

user19161

@robjohn Wow, now you have made the internet a better place!

@JasperLoy do you think that Benford's Law is that important? or is that sarcasm?

user19161

9:41 PM

@robjohn No, I mean literally what I said. I don't like sarcasm!

@JasperLoy Thanks. I hope that someone finds it useful.

user19161

@robjohn By the way, I never heard of this law. I was just expressing my appreciation for someone who contributes articles to the internet.

@JasperLoy There are many pages on that same site. A lot are for reference from sci.math posts.

@Henning: pinging you to make sure you saw Qiaochu's edit to your deleted answer.

Sorry gotta go again, have a nice week-end y'all!

user19161

@tb Enjoy your weekend too!

Henning Makholm

9:46 PM

@tb Yes, I did. Presently editing and undeleting.

Argh. Preview is acting up again. Wants to re-render all TeX for each key I press.

Well, I must leave as well. Going for a night with the Lyrid meteor shower and a bunch of telescopes. See everybody tomorrow!

@HenningMakholm do you have the bookmarks that turn on and off the updating?

user19161

@robjohn Have fun!

@JasperLoy I will :-)

@robjohn Have fun and see you tomorrow : )

@HenningMakholm Look here for "rendering off" and "rendering on"

9:49 PM

@HenningMakholm Congratulations! And welcome to the club : )

Closing down.

Henning Makholm

Now undeleted. Wonder whether I get to keep the badge.

user19161

@HenningMakholm Badges are never taken away once awarded I think.

Henning Makholm

@robjohn Why, it works! :-)

Though I first had to figure out whether I could get my aggressively trimmed Firefox to display bookmarks, a feature I don't normally use.

user19161

10:05 PM

@HenningMakholm First thing I do when I install FF is to delete all the bookmarks!

10:22 PM

Can someone explain something to me about separable polynomials? In Dummit & Foote there is a Corollary and a proposition that together seem to imply than every irreducible polynomial over a field is separable.

very basic dumb question about fourier analysis. if you want to do an integral $\int_\mathbb{R} f(x) e^{-2\pi i \xi y}$ or something, in order to avoid complex analysis should i be breaking this up the exponential into $i\cdot sin$ and $\cos$?

The Corollary states that "every irreducible polynomial over a field of characteristic 0 is separable". And the proposition states that "every polynomial over a finite field $\mathbb{F}$ is separable.

folland never seems to mention that one uses the trigonometric functions to compute fourier series, so i wonder if he's doing the exponential using some other technique

But every field either has characteristic 0 or characteristic $p$ for some prime $p$. So isn't every field either characteristic 0 or finite with prime characteristic?

Henning Makholm

@DavidK That seems to leave open the possibility of a non-irreducible polynomial over a characteristic-0 field being non-separable.

Also, there are infinite fields of prime characteristic -- for example the field of rational functions over $\mathbb F_p$.

10:26 PM

@all anyone know basic fourier analysis? i think my question is fairly simple if you've read folland or beyond

@HenningMakholm Hmm. Yes I hadn't thought of that case.

@EricGregor Why not?

Ok well riddle me this: Suppose $f(x)\in F[x]$ is an irreducible polynomial of degree $n$ over an arbitrary field $F$, let $L$ be the splitting field of $f(x)$ over $F$ and let $\alpha$ be a root of $f(x)$ in $L$.

Does it follow from these hypotheses that $f(x)$ is separable?

@MattN my real question is about an alternative. do i have to use sin and cos? folland doesn't seem to ever suggest this approach

Henning Makholm

@EricGregor Well, should you avoid complex analysis in the first place? In any case, since the variable stays real, the usual complications and niceties of complex analysis do not really apply here anyway. For a function R->C things work out mostly excactly like R->R^2.

10:31 PM

@EricGregor What exactly do you want to do?

i'm trying to do a problem in folland, 8.45

I want to show (for $n=1$) that $$u(x,t)=\frac12[f(x+t)+f(x-t)]+\frac12\int_{x-t}^{x+t}g(s)\ ds,$$where $$(\partial_t^2-\Delta)u=0,\ \ u(x,0)=f(x), \ \ \partial_t u(x,0)=g(x).$$

$\hat u (x,t)=f*\partial_t W_t(x)+g*W_t(x),$for $W_t=[\frac{\sin (2\pi t|\xi|)}{2\pi|\xi|}]^\vee.$

Oh my. @JonasTeuwen to the rescue!

Henning Makholm

@DavidK Beats me. I'm making this up as I go along. Not sure what a non-separable (but irreducible) polynomial anywhere would look like.

ducks away from tumeni symbols

i can put my work up here and ask my question

i've written some stuff up and i'm getting stuck. i was avoiding the use of sins and cosines. i may be making some elementary blunders

10:33 PM

Jonas is the man for you.

Or ask it on main.

well i'll write what i have and if anyone comments that would be great

i was going to, but then i thought my question might be way more basic

actually my problem isn't really with that problem in all its glory

i'm trying to compute $W_t$ really

We have a hint: use the identity $ \chi_{[-a,a]}^\vee(x) = \frac{\sin(2\pi a x)}{\pi x },$ where the superscript $\vee$ denotes the inverse Fourier transform.

**Work toward a solution**: In order to understand $u(x,t)$ I need to first understand $W_t(x)$ in order to take $\partial_t$ or the convolution with $g$. For some reason I can't get a convergent expression.

We have

$\begin{eqnarray*}
W_t&=&\int_\mathbb{R} \frac{\sin (2\pi t|\xi|)}{2\pi|\xi|}\cdot e^{-2\pi i |\xi|x}\ dx \\
&=&\frac12 \int_\mathbb{R} \chi_{[-t,t]}^\vee(|\xi|)\cdot e^{-2\pi i |\xi|x}\ dx \\

@HenningMakholm I'm pretty sure something like $(x-\sqrt{2})^{2}$ would be irreducible and non-separable over $\mathbb{Q}(\sqrt{2})$.

that's what i was going to post on main

i must be doing something stupid because the final thing doesn't converge

Oh um I'm an idiot

@JonasTeuwen, i call upon your vast understanding

10:36 PM

@MattN Hi.

@EricGregor Can't you just fill in?

@JonasTeuwen Hi there : ) Did you hear Fourier transform? : )

Henning Makholm

@DavidK Doesn't look very irreducible to me.

@jonas hi. what do you mean fill in?

Yes, but I'm very tired and Ilya has my Folland.

@EricGregor $\hat u \to u$.

@HenningMakholm I know. That's what happens when you come up with exemplary counter-examples

10:38 PM

@jonas this is my first day of looking at fourier analysis. i don't know what you mean.

Take the inverse Fourier transform.

i like to start with the problems and work backwards

am i not doing that?

my mistake could be very elementary

@EricGregor You can fill in your $u$ into the PDE?

trig identities are so hard, so much memorization needed

@jonas i'm not sure what you mean. i would love to know why i can't get $W_t$ to converge.

10:41 PM

GUYS I'M SO PISSED THAT EVERYONE IS DOWNVOTING

(meta)

what can I do with $\frac{1+cos^2 x}{cos^2 x}$

the numerator is -sin^2 x isn't it?

and the bottom is cosxcosx

@Jordan No.

I would split it into two fractions: \dfrac{1}{cos^2} + \dfrac{cos^2}{cos^2}

But then again, I'm not sure where you're heading with this!

integration

is that then sec^2 x or is that wrong?

sec^2 PLUS ONE!

And that is something :)

also @JonasTeuwen, i'm embarassed to admit that i don't really know PDEs at all. i can only blindly follow the road folland has laid out for me without learning more background

10:44 PM

or maybe it's sec^2 -1 ...

it is -1

What is the original problem? @Jordan

so this isn't any good

Evaluate the integral from 0 to pi/4 $\frac{1+cos^2 x}{cos^2 x}$

Then we're doing fine.

sec^2 x + 1 = tan^2 x + 2. Integrate term by term!

how are those equal?

10:48 PM

You don't have to have the entire integrand undergo metamorphosis by a slick trig identity...

subtract 1

(or subtract 2, depending on which way you're used to seeing it)

I dont know what you are talking about

@EricGregor I'm sorry, can I help you tomorrow. I'm too tired now :-).

@Jordan What did you mean when you said "it is -1"?

(I literally want you to think back to what we were talking about, if you don't mind)

oh sec^2 -1 = tan^2

sec^2 x + 1 = tan^2 x + 2

subtract 2 from both sides

10:51 PM

I dont know how to do that

ok, sure. do you have a recommendation for something online i could read that might help me out? i literally don't know the basics

So you do know what I'm talking about :)

folland is good conceptually, but right now i need to just calculate, i think

@Jordan ???? You don't know how to subtract 2? I'm not sure what you mean.

I would never think of doing that on a test

or on my own, ever

10:52 PM

Ah. Well you don't have to have a genius insight. Like I was saying earlier, you just need to use some trig identity (or substitution) to get the integrand into something that you do recognize

So you have this (1 + cos^2)/(cos^2). This is sec^2 + 1. You can take the 1 out. You know what sec^2 is...

Hi Kann

Kannappan Sampath

Hi @TheChaz

@EricGregor Read Folland!

(Is "Kann" an acceptable abbreviation of your name?)

not sure what to do from here

Kannappan Sampath

@TheChaz Sure, but if it is not more work, you may considering adding an a just that it makes a complete name in local languages here, but I am fine with Kan too. :)

10:56 PM

@KannappanSampath Henceforth, your name shall have 3, 5 or 9 letters when I type it!

@Jordan Do you agree that the integrand equals (2 + tan^2 x)dx?

no

Do you agree that the integrand equals (1 + sec^2 x)dx?

over cos^2?

Kannappan Sampath

@TheChaz :)

ok I see

10:59 PM

So we have (1 + sec^2 x)dx

Do you agree that sec^2 x = 1 + tan^2 x?

yes

So we have (1 + [sec^2 x])dx = (1 + [1 + tan^2 x])dx after substituting.

But that's just 2 + tan^2 x

@JonasTeuwen, math.stackexchange.com/questions/135002/…

You with me, @Jordan?

yes

2x+ sec^4 x?

11:05 PM

Actually.

Let's go back to (1 + sec^2)dx

ok

What is the integral of sec^2 x ??

tanx

Yup.

So 1x + tan(x) is an antiderivative

No crazy trig identities needed. Sorry, lost track of which direction we were headed for a minute!

I still wouldnt have been able to do it by myself

I spent like 20 minutes trying to do some crazy stuff

11:07 PM

If you asked yourself "do I know the integral of each term on the inside" (after the trig ID), you would have got it.

I spent like 2 hours the other night on a problem about a stupid equilateral triangle, and still haven't solved it!

I just need to spend more time memorizing this page of identities

Don't worry. Your confidence is possibly the worst I have ever seen in a math student, so you have a lot to overcome in that department. You can do a lot more than you think you can.

(At the same time, you're venturing into pretty deep water... not sure what to tell ya!)

About the identities, you could try to learn how to derive them

Do you know how to derive tan^2 x + 1 = sec^2 x from the identity sin^2 x + cos^x = 1? That's what I mean... how to get from one to the next.

I am not sure how to derive integral identities

yes

(I meant trig identities... guess you meant integral tables!)

both really

I really do not understand trig much

11:11 PM

See you tomorrow!

I took it in a 4 week summer class

You should fire your guidance counselor.

It is acommunity college :P

Didn't really have one back then

I am going to do the same thing with calc 2

Then to Calc 3?

Oh crap. I have to get to the gym before it closes or I won't be able to keep my herculean physique. ttyl

I won't take calc 3 until(if) I transfer

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