The Math Help Topic
Forum rules
Please keep the forum rules and guidelines in mind when creating or replying to a topic.
Please keep the forum rules and guidelines in mind when creating or replying to a topic.
-
stan423321

- Posts: 7312
- Joined: Sat Jan 12, 2008 3:22 pm
- Location: Poland, of Union of Socialistic European Republics
- Tings: 63250
Re: The Math Help Topic
I've got a question!
What's faster way of counting that?
(1*n) + (2*(n-1)) + ... + ((n-1)*2) + (n*1)
What's faster way of counting that?
(1*n) + (2*(n-1)) + ... + ((n-1)*2) + (n*1)
-
Hunchman801

- Posts: 87649
- Joined: Thu Aug 07, 2003 6:50 pm
- Location: Paris, France
- Contact:
- Tings: 640357
Re: The Math Help Topic
n(n+1)²/2 - n(n+1)(2n+1)/6 = n(n+1)(n+2)/6.
-
stan423321

- Posts: 7312
- Joined: Sat Jan 12, 2008 3:22 pm
- Location: Poland, of Union of Socialistic European Republics
- Tings: 63250
Re: The Math Help Topic
Are you sure? Thanks.
-
Hunchman801

- Posts: 87649
- Joined: Thu Aug 07, 2003 6:50 pm
- Location: Paris, France
- Contact:
- Tings: 640357
Re: The Math Help Topic
Well yeah, you just have to write it as the sum of i(n+1-i) and to develop the whole thing.
Re: The Math Help Topic
I had my math finals today. It went well but there's one thing I couldn't figure out, and it's been bugging me all day. Could someone please solve this differential equation:
y''+2xy'+2y=2
y(0)=0 and y'(0)=0
This is probably ridicolously easy by international standards, but don't blaim me; blaim the crappy level of science and math education in the Norwegian high school.
y''+2xy'+2y=2
y(0)=0 and y'(0)=0
This is probably ridicolously easy by international standards, but don't blaim me; blaim the crappy level of science and math education in the Norwegian high school.
-
Hunchman801

- Posts: 87649
- Joined: Thu Aug 07, 2003 6:50 pm
- Location: Paris, France
- Contact:
- Tings: 640357
Re: The Math Help Topic
There is no general expression for the solutions of a second-order linear differential equation with non-constant coefficients. There's probably a simple way but I studied this three or four years ago so I don't really remember anything 
WolframAlpha found it anyway: 1-e-x².
Edit: after a few minutes I found it: y''+2xy'+2y = (y'+2xy)'. The first-order linear equation can be solved easily, and then I'm pretty sure you can do the rest with the constants and everything.
Good to know I can still do math after so many years
WolframAlpha found it anyway: 1-e-x².
Edit: after a few minutes I found it: y''+2xy'+2y = (y'+2xy)'. The first-order linear equation can be solved easily, and then I'm pretty sure you can do the rest with the constants and everything.
Good to know I can still do math after so many years
Re: The Math Help Topic
guys can somebody help me with this? it's probably not too hard but I keep getting bullshit in the end
it's the exam period and I'm pretty desperate
it's the exam period and I'm pretty desperate
- Attachments
-
- onoez
- integral.JPG (11.13 KiB) Viewed 679 times
Re: The Math Help Topic
I assume you've tried to integrate it by parts ?
-
Hunchman801

- Posts: 87649
- Joined: Thu Aug 07, 2003 6:50 pm
- Location: Paris, France
- Contact:
- Tings: 640357
Re: The Math Help Topic
Well I don't really have time to give it a try tonight but I quickly found out that ∫xdx/√((x-a)(b-x)) = ((a+b)/2)×∫dx/√((x-a)(b-x)) (you just have to integrate (a+b-2x)dx/√((x-a)(b-x))). I'm going to bed now but I'll see what can be done tomorrow when I get back from work.
Re: The Math Help Topic
Yeah that's what I was doing too, and after some work with the demoninator and integrating it I got a couple of -infinities which can't be right... I'll have to recheck it
edit
yes apparently I'd fucked up before, now I get arcsin((2x-(a+b))/(a-b)) which seems better ;O
wolframalpha doesn't agree though
edit
yes apparently I'd fucked up before, now I get arcsin((2x-(a+b))/(a-b)) which seems better ;O
wolframalpha doesn't agree though
Re: The Math Help Topic
I got (a+b)*Pi/2 : (a+b)/2 * Arcsin (2u/(b-a)) from (a-b)/2 to (b-a)/2. (With u = x - (a+b)/2)
I'm sure this is good. If you derivate this Arcsin, you get (a+b-2x)dx/√((x-a)(b-x))), which is, as Hunch said, equal to your integral.
I'm sure this is good. If you derivate this Arcsin, you get (a+b-2x)dx/√((x-a)(b-x))), which is, as Hunch said, equal to your integral.
-
Hunchman801

- Posts: 87649
- Joined: Thu Aug 07, 2003 6:50 pm
- Location: Paris, France
- Contact:
- Tings: 640357
Re: The Math Help Topic
I have to say guys, we are definitely the most awesome Rayman forum ever 
Re: The Math Help Topic
I need a strike tag !Hunchman801 wrote:I have to say guys, we are definitely the most awesome forum ever
Do you have some softwares you can use to write maths (with all the symbols needed, sums, integrals, limits, fractions...) ?
Edit : nevermind, even that piece of shit known as Word provides this.
Re: The Math Help Topic
OK thank you guys =)
P.S.: strike tag = do want
P.S.: strike tag = do want
-
Hunchman801

- Posts: 87649
- Joined: Thu Aug 07, 2003 6:50 pm
- Location: Paris, France
- Contact:
- Tings: 640357
Re: The Math Help Topic
No way.Matyuv wrote:P.S.: strike tag = do want
Re: The Math Help Topic
Cheers.Hunchman801 wrote:No way.Matyuv wrote:P.S.: strike tag = do want
Re: The Math Help Topic
Yaay ! Thanks HunchMatyuv wrote:Cheers.Hunchman801 wrote:No way.Matyuv wrote:P.S.: strike tag = do want
-
PluMGMK

- Posts: 40514
- Joined: Fri Jul 31, 2009 9:00 pm
- Location: https://www.youtube.com/watch?v=cErgMJSgpv0
- Contact:
- Tings: 136636
Re: The Math Help Topic
Thank you!
Re: The Math Help Topic
For those of you who are into programming and maths here's a link - http://projecteuler.net/index.php?section=about
It's got some nice problems to solve... I've solved 4 so far lel.Project Euler is a series of challenging mathematical/computer programming problems that will require more than just mathematical insights to solve. Although mathematics will help you arrive at elegant and efficient methods, the use of a computer and programming skills will be required to solve most problems.
Re: The Math Help Topic
Thanks Matyuv, that seems funny ! Solved the first one, took me some time cause I added the multiples of both 3 and 5 twice, forgot to substract them.
I hope the next ones are harder
I hope the next ones are harder


