Mathematics - 2012-12-02 (page 4 of 7)

True wisdom comes from deeply observing and reflecting upon things oneself, not how many years you have lived, not learning all the stupid cliches from people that have been used a million times.

Jordan

I am almost done with math anyways, just need to take calculus 2 again, calc 3 and like 3 more math classes and I am done with math

Peter Sheldrick

good that true wisdom isn't tested on his next exam

doing lots of problems is what gets him good scores

Argon

@Jordan If you don't like math, why are you taking it?

Jordan

@Argon College requirement

I like math, I am just really bad at it

Argon

Oh. What are you doing in university?

Peter Sheldrick

8:21 PM

there are books full of questions with solutions but a CAS is more flexible since you look for your own problems at your level (as many as desired) - a bit like the CAS has also superseeded tables of integrals

Jordan

Computer science

Argon

Compsci, nice

Jordan

I am not very good at programming either but I like it

user19161

If I learn programming I would start with Python!

Jordan

I have to learn python now actually

I learned a tiny bit of c++, scheme and now I need to learn python

Peter Sheldrick

8:23 PM

Java?

Jordan

I need to learn java next semester

user19161

@PeterSheldrick I see you changed your picture again, now there are two Peters in this chat!

Argon

@PeterSheldrick Cash money, ain't nothin' funny!

Peter Sheldrick

@Will, yup - i suddenly had panic about using graphics from proprietary math software :/

Jordan

man I suck

Peter Sheldrick

8:26 PM

big companies scare me

Jordan

how do I do $(3x^{3/2})^2$?

user19161

@jordan If you are really bad at algebra you can think of many operations geometrically, for example, ab is just the area of a rectangle with sides a and b, so from that you get the intuitive idea why ab=ba.

Jordan

I am even worse at algebra

err geometry

Argon

@Jordan $(a^b)^c = a^{bc}$

Peter Sheldrick

@WillHunting, dude, you probably mean well, but stuff like that does not help on tests

Argon

8:27 PM

@WillHunting I think that complicates stuff...

user19161

@PeterSheldrick Dude, one can do well for the tests but so what? After that, everything will fall apart. The rain will come and wash the house away...

Peter Sheldrick

what is 'area' anyway? it just goes down the philosophy spiral

Jordan

I doubt I will need to know much math beyond basic algebra really

user19161

@Argon It's a better way of remembering ab=ba anyway.

user19161

@PeterSheldrick Well, we aren't discussing that now.

Argon

8:28 PM

@WillHunting I guess.....

Jordan

man I am shit at algebra, still can't get this basic problem

Peter Sheldrick

it's fine to write oh tests are meaningless, but then please be consistent and don't look down on people with poor results on horrible mechanical test scores

Jordan

this is math for 10 year olds and I cant do it

Argon

@Jordan If I were 10, I couldn't do it

Jordan

If I was 25 I couldn't do it :P

Nimza

8:33 PM

good evening!

Jordan

is the exponent in this positive or negative $(y^{-1/2})^2$

I can't remember that algebra rule

Argon

Hi @Nimza

user19161

The thing is if you do the algebra without understanding, sooner or later you will forget the rules, or you may mess up because you will apply the rules wrongly.

Nimza

Help me please, if $\mu$ is a signed measure such that $|\mu|$ is finite is it true that $\left| \int f(x) \mu(dx) \right| \leqslant \int |f(x)| \, |\mu|(dx)$?

@Argon hi!

Jordan

what is to understand about the rules? they are pretty basic and they exist because they work

there is no understanding

Argon

8:34 PM

@Jordan $(a^b)^c=a^{bc}$

Jordan

I can't conceptualize why a negative fraction exponent does a specific thing to a number, I just memorize that it does

Argon

@Jordan $2^1 2^1 = 2^{1+1}$

right?

user19161

@Jordan Well, that is not true. For example, why is a to the b times a to the c equal to a to the b plus c?

Argon

$2^{-1} 2^1 = 2^{-1+1}= 2^0 = 1$

Jordan

I don't know why, it just works

Argon

8:36 PM

Then $\frac{1}{2}=2^{-1}$

Does that make sense?

Jordan

no

user19161

@Jordan That's the problem, that's why you are still having problems with algebra. If you learn by understanding first, it will be slower at first but faster later on.

Argon

@Jordan $2^{a} = 2\cdot 2 \cdots 2$, $a$ times

Jordan

I understand that is works, but I don't see any bigger picture to it, it works because it does

Argon

Right?

Jordan

8:38 PM

yes

but $2^{1/3} = 2 * 1/3$?

Argon

So, $2^a 2^b = (2\cdot 2 \cdots 2 ) (2\cdot 2 \cdots 2 )$

And now there are $a+b$ $2$s

$2^a 2^b = 2^{a+b}$

Makes sense?

Jordan

no

Argon

$2^a 2^b = \overbrace{(2 \cdot 2 \cdots 2)}^n \overbrace{(2 \cdot 2 \cdots 2)}^m = \overbrace{2 \cdot 2 \cdots 2}^{n+m}$

Jordan

isn't it n * m?

Argon

Which is $2^{n+m}$

Peter Sheldrick

8:41 PM

it's not even rendering the latex properly on my browser

Argon

@Jordan Say I have $2^2 = 2\cdot 2$ and $2^3 = 2\cdot 2\cdot 2$

Peter Sheldrick

it does on the actual site, just not in the chat

Jordan

this is why I can't learn math in school, I learn way too slowly and I never learn anything in class and I spend an immense amount of time on homework only to learn nothing and fall behind

Argon

Then $2^2 \cdot 2^3 = \overbrace{2\cdot 2\cdot 2\cdot 2\cdot 2}^5$

Makes sense?

Old John

@Kalima - why did you delete the Diophantine question? - you should have left it, as other visitors might have found it interesting.

Jordan

8:43 PM

yes that makes sense

Argon

Does this make sense now:

3 mins ago, by Argon

$2^a 2^b = \overbrace{(2 \cdot 2 \cdots 2)}^n \overbrace{(2 \cdot 2 \cdots 2)}^m = \overbrace{2 \cdot 2 \cdots 2}^{n+m}$

Jordan

I get it now, I was thinking something weird

yes

Argon

Ok. Now, of course, this makes sense only for nonzero, natural numbers, right?

Assume it holds true for all rational numbers too.

So now we may say

$2^{1/2}2^{1/2} = 2^1$

Peter Sheldrick

writing the equations in plain text does have advantages to

Argon

@Jordan Makes sense so far?

Jordan

8:46 PM

yes

Limitless

@all, ciao! I'm out to the hospital to check on my ears.

Argon

So, $2^{1/2}2^{1/2} = (2^{1/2})^2 = 2$

@Limitless Bye!

Peter Sheldrick

@Limitless, bye get well soon

Argon

Or in other words, $2^{1/2}$ is a number that, when squared, equals $2$, i.e. $\sqrt 2$

Now, we can use the same logic to show

$2^0 2^1 = 2^1$

Jordan

I have too much to learn to ask questions here, I will just make poeple angry

Argon

8:49 PM

So

$2^0 = 1$

Similarly,

$2^{-1}2^{1}= 2^0 = 1$

thus

$2^{-1} = \frac{1}{2}$

etc.

Makes sense?

Jordan

why is something to the zero power always 1? I know that it works, and I know that I can work backwards to it but that is all i know

Peter Sheldrick

definition

Jordan

like 2*1 = 2 I get that it needs to start at 1

Argon

@Jordan $a^0 a^1 = a^{1+0} = a$

Peter Sheldrick

like 0! = 1

Argon

8:52 PM

@PeterSheldrick Let's not get too confusing!

@Jordan What do you mean?

Peter Sheldrick

it helps you write stuff like exp(x) = x^0/0!+x^1/1!+x^2/2!+...

Jordan

$2 ^ 1 = 2$ $2^2 = 2* 2$

Argon

@Jordan $2 \neq 2^2$

Jordan

I don't know if this makes sense to anyone but me but if you keep dividing down from higher powers by the base number you get the previous power

Argon

$2^1 = 2^{-1} 2^2$

@Jordan Sure, if you mean $a^b/a=a^{b-1}$

Jordan

8:54 PM

yes

Argon

This is true, and can be generalized to $\frac{a^b}{a^c} = a^{b-c}$

because $\frac{a^b}{a^c} = a^b \frac{1}{a^c} = a^{b}a^{-c}=a^{b-c}$

Jordan

that makes sense

I think I need to quit school for a year to learn all of this

Peter Sheldrick

school is boring enough if you just keep going

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$Mathematics$ Mathematics