maths at uni (1 Viewer)

[mojako]

This is out of topic, but, if you do maths at uni.. or something with physics.. or anything that uses maths, do you deal with very loong equations like this:
http://mathworld.wolfram.com/HypergeometricFunction.html
(I mean, the complexity is comparable to that)

I've heard that they use computers, but do the computers take care of all complex things or do u stil have to work out some of them?
how exactly do they use computers??

 

[McLake]

We use computers to preform functions. So say you need to intergrate an expression, f(x). You can type in intergrate(f(x)) and it will integrate it for you. Or you can get it to substitute numbers in. Or calculate factors. Just about any maths you can think of it can do.

 

[Affinity]

Pure maths -> paperwork
applied/stats -> computer does the job for you

 

[xiao1985]

pure maths ---> fun
applied maths --->boring, aprt from the bit which is associated with pure maths...

 

[mojako]

hmm.. I thought applied maths would be more interesting
coz it's "applied"

 

[xiao1985]

nah pure maths way more interesting... u'd see when u do calculus...

any way, just for fun, here's a q in my integral calc... (usyd, pplz plz don't ruin the fun =p)

/ 1
| 1/x^2 dx
/ -1

use fundamental theorem of calc:

1
[- 1/x]
-1

evaluate u get -2...

but the graph of 1/x^2 are all above x axis, so everything should be positve... ??@@=p

 

[withoutaface]

nah pure maths way more interesting... u'd see when u do calculus...

any way, just for fun, here's a q in my integral calc... (usyd, pplz plz don't ruin the fun =p)

/ 1
| 1/x^2 dx
/ -1

use fundamental theorem of calc:

1
[- 1/x]
-1

evaluate u get -2...

but the graph of 1/x^2 are all above x axis, so everything should be positve... ??@@=p

Is probability within the domain of applied maths? Cos if so I've decided I like both.
EDIT: Am I meant to be able to do this?

Is the point that you can't integrate in a domain which includes an undefined value (ie x=0)?

 

[xiao1985]

yea probabilty is in the domain of applied maths...

as a matter of fact, i was just learn prob this morn in my stat lecture... @@ oh the boredom... because of i dun like applied and also the lecturer too...

regardin the q, u should be able to get it...

hint: at most of the time, when u are dealing with calculus, the fn needs to be continous...

 

[Archman]

well its easy to see that
1
[-1/x]
n
diverges as u pick really small positive n
same thing will happen when applied to the negative side
so the area doesnt have a limit.
i think you get the -2 pretty much the same way as how we "prove"
1+2+4+8+... = -1
as your summing a divergent sequence dodgily.

 

[jm1234567890]

yeah, just split it into 2 parts and integrate it.

so you don't integrate over the discontinunity

 

[withoutaface]

But then you end up with an undefined value as part of the integral (ie -1/0), so you could only hope to take a value such as .00001 or something and get an estimate.

 

[jm1234567890]

But then you end up with an undefined value as part of the integral (ie -1/0), so you could only hope to take a value such as .00001 or something and get an estimate.

no, you take the limit to it

 

[withoutaface]

ahh i see, my bad.

 

[xiao1985]

as u take limit tho, each integral approaches infinity... addition of infinity is not defined...

well, strictly speakin =)

 

[jm1234567890]

Well then the integral doesn't exist?

I can't see how you can 'make' that integral have a value.

 

[sammeh]

the integral exists but i would think only as a limit, that is the value of 2 is the limiting value of the area function -1/x ?? meh i dunno

 

[xiao1985]

the integral exists but i would think only as a limit, that is the value of 2 is the limiting value of the area function -1/x ?? meh i dunno

if you define the integral as the area under the curve, then the integral does not exist...

 

[mojako]

if you define the integral as the area under the curve, then the integral does not exist...

so what is the other definition of integral??

 

[jm1234567890]

if you define the integral as the area under the curve, then the integral does not exist...

What do you define the integral as then?

the anti-differential? that can't be done for a definate integral though.

 

[sammeh]

the anti-differential? that can't be done for a definate integral though.

mmm i was under the impression that you could take the anti-differential of a definite integral under certain conditions, like if the function was continous over the interval between the limits but discontinous at one of them? in this case you could do this from 0->1 which would give you a result that was a limited value with f(x) convergent?

mmm its late and this is all half rememberd crap from reading here and there, pls correct me if im wrong ;x