The Student Room Group

Official TSR Mathematical Society

Scroll to see replies

Reply 3560
Original post by Oh I Really Don't Care

You do not require 'anything additional', this is more than approachable with the material you appear to have net.


I see. I was referring to the "last step" in finding suitable way to use the tests, which didn't seem obvious.
The differentiation of the power series is what I want to justify. In that case, I would have another go with it, but its not going to be tonight. Thanks. :wink:
(edited 12 years ago)
Anyone know of notation for integrals over all space?
I'm getting bored of writing the infinity symbol up to 6 times per line so I've been using this notation in my notes:


But does anyone know of a better notation, if there is a standard notation for this?
Original post by Manitude
Anyone know of notation for integrals over all space?
I'm getting bored of writing the infinity symbol up to 6 times per line so I've been using this notation in my notes:


But does anyone know of a better notation, if there is a standard notation for this?


If you're using known spatial co-ordinates (say x=(x,y,z)\mathbf{x} = (x,y,z)) then we've been told we can write:

xdS\iiint_{\mathbf{x}} dS

Don't know whether that's a just-one-lecturer thing though, as he's taken most of my Applied lectures.
Original post by marcusmerehay
If you're using known spatial co-ordinates (say x=(x,y,z)\mathbf{x} = (x,y,z)) then we've been told we can write:

xdS\iiint_{\mathbf{x}} dS

Don't know whether that's a just-one-lecturer thing though, as he's taken most of my Applied lectures.


Would that not require an additional (x,y,z)\forall (x,y,z), so x=(x,y,z)(x,y,z)\mathbf{x} = (x,y,z) \forall (x,y,z) for additional clarity?
Though I guess that is quite concise. Cheers.
Original post by Manitude
Would that not require an additional (x,y,z)\forall (x,y,z), so x=(x,y,z)(x,y,z)\mathbf{x} = (x,y,z) \forall (x,y,z) for additional clarity?
Though I guess that is quite concise. Cheers.


I guess if you state it once at the start it wouldn't hurt - then you can let it follow on from there.
Original post by marcusmerehay
I guess if you state it once at the start it wouldn't hurt - then you can let it follow on from there.


Sounds like a plan. I'll see how my lecturer reacts next time I hand in a worksheet where we actually do integrals over all space. From the looks of the current worksheet there are none.
Reply 3566
Interesting way to denote all triples of real numbers: R3\mathbb{R}^3.

I think it is possible to abbreviate it as follows.

f(x,y,z) dxdydz\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x,y,z)\ dxdydz can be the same as R3f(x,y,z) dV\int_{\mathbb{R}^3} f(x,y,z)\ dV.

In general, it can be Rnf(x) dx\int_{\mathbb{R}^n} f(\mathbf{x})\ d\mathbf{x}, where xRn\mathbf{x} \in \mathbb{R}^n and dxd\mathbf{x} is the corresponding n-dimensional differential.

At least, that's what the literature says. :tongue:

EDIT: I like the infinity symbol in between them though. :tongue:
(edited 12 years ago)
Original post by gff
Interesting way to denote all triples of real numbers: R3\mathbb{R}^3.

I think it is possible to abbreviate it as follows.

f(x,y,z) dxdydz\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x,y,z)\ dxdydz can be the same as R3f(x,y,z) dV\int_{\mathbb{R}^3} f(x,y,z)\ dV.

In general, it can be Rnf(x) dx\int_{\mathbb{R}^n} f(\mathbf{x})\ d\mathbf{x}, where xRn\mathbf{x} \in \mathbb{R}^n and dxd\mathbf{x} is the corresponding n-dimensional differential.

At least, that's what the literature says. :tongue:

EDIT: I like the infinity symbol in between them though. :tongue:

This sounds good too, ta :smile:
anyone doing/done a maths degree?,
I love maths; but the job prospects are anti-me finance, banking accounting :no: is this true or just the majority of people go in to this field; how can I help people with a maths degree, I wan't a challenging job, maybe work in NASA I mean maths was involved for 'man landing on moon'

HELP.
Can somebody help me with this quesiton please:

cos3x = 4cos^2x

Meant to solve for 0 < x < 2pi

I know that cos3x = 4cos^3x-3cosx, so at first I rearranged and got:

4cos^3x - 4cos^2x -3cox = 0, then I divided everything by cosx ( which I'm not sure I'm actually meant to do), then factorised and got x=3/2 and x=-1/2

I'm pretty sure I've used the wrong method though, so could somebody point me in the right direction please?

Sorry about the untidiness aswell, I don't know how to actually write out the equation like I see you lot do in this thread

Thanks in advance
Original post by Entrepreneur123
Can somebody help me with this quesiton please:

cos3x = 4cos^2x

Meant to solve for 0 < x < 2pi

I know that cos3x = 4cos^3x-3cosx, so at first I rearranged and got:

4cos^3x - 4cos^2x -3cox = 0, then I divided everything by cosx ( which I'm not sure I'm actually meant to do), then factorised and got x=3/2 and x=-1/2

I'm pretty sure I've used the wrong method though, so could somebody point me in the right direction please?

Sorry about the untidiness aswell, I don't know how to actually write out the equation like I see you lot do in this thread

Thanks in advance


Create a thread here and post your problem for a quicker response. As a hint though, since it's a cubic you should expect to get 3 points for x, not 2 unless it's a repeated root (cosx = 0 is another one you seem to have missed if your working is correct)
Original post by translucent
anyone doing/done a maths degree?,
I love maths; but the job prospects are anti-me finance, banking accounting :no: is this true or just the majority of people go in to this field; how can I help people with a maths degree, I wan't a challenging job, maybe work in NASA I mean maths was involved for 'man landing on moon'

HELP.


Generally speaking NASA would recruit physicists for that kind of thing. The maths involved in a physics degree is more appropriate for space exploration than the maths in a physics degree. I guess the difference is that maths is more about proving statements, whereas physics simple accepts they are true and uses them to solve problems. At least that's my impression from speaking with mathematics students and from studying physics myself.
Original post by Manitude
Generally speaking NASA would recruit physicists for that kind of thing. The maths involved in a physics degree is more appropriate for space exploration than the maths in a physics degree. I guess the difference is that maths is more about proving statements, whereas physics simple accepts they are true and uses them to solve problems. At least that's my impression from speaking with mathematics students and from studying physics myself.

okay you've addressed the NASA issue, but what about job prospects that is the BIG issue here?
Original post by translucent
okay you've addressed the NASA issue, but what about job prospects that is the BIG issue here?


A maths degree probably closes the fewest number of doors compared to pretty much any other degree.
That's just about all that needs to be said.
Reply 3574
Original post by translucent
okay you've addressed the NASA issue, but what about job prospects that is the BIG issue here?


If you'd want to work in NASA or anything, I suspect you'd have to be quite good(and work on stuff like fluid dynamics or w/e).
Original post by Slumpy
If you'd want to work in NASA or anything, I suspect you'd have to be quite good(and work on stuff like fluid dynamics or w/e).

apart from finance, banking and accounting what else could I do, anything related to computers?
Reply 3576
Original post by translucent
apart from finance, banking and accounting what else could I do, anything related to computers?


There's a fair market for software engineers I think, and maths seems ok for that.
Also, this isn't really the place for this question.
Reply 3577
The following is another nice question.


[*] For a positive integer nn, compute the integral

xn1+x+x22!++xnn! dx\displaystyle \int \frac{x^n}{1 + x + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!}}\ dx
Original post by gff
The following is another nice question.


[*] For a positive integer nn, compute the integral

xn1+x+x22!++xnn! dx\displaystyle \int \frac{x^n}{1 + x + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!}}\ dx

Spoiler

(edited 12 years ago)
Reply 3579
Original post by Blutooth
A nice approachable question :smile:


Somebody who didn't manage to do it has negged you. :tongue:
This solution will look nicer in a spoiler, in my opinion. Give strangers the chance to have a go with it. :biggrin:

Quick Reply

Latest