If you agree that it is an infinite serie, how could you know the number of its terms is changed ?
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There have been 16 items by cK_hoangan (Search limited from 16-May 21)
#100668 Who belives SOMETHING = NOTHING ?
Posted by
on 17 January 2008 - 03:47 PM
in
Mathematics
but ... then, wat is the error ?
If you agree that it is an infinite serie, how could you know the number of its terms is changed ?
If you agree that it is an infinite serie, how could you know the number of its terms is changed ?
#100621 Quark and Bohr model
Posted by
on 19 December 2007 - 11:10 PM
in
Physics
I got stuck at 2 things.
Pls, is there anyone explain briefly for me:
What is quark ?
What the meaning of Bohr model ?
Pls, is there anyone explain briefly for me:
What is quark ?
What the meaning of Bohr model ?
#100600 2001 Q21
Posted by
on 04 December 2007 - 07:05 PM
in
Physics
Definitely, you're right.
Acceleration always goes the same direction with its own force. Therefore, the horizontal acceleration has to work out from the horizontal force. There is no point to pick up the horizontal acceleration with the 4N force, they are all different direction
Generally, You are right, well-done!
: .
Acceleration always goes the same direction with its own force. Therefore, the horizontal acceleration has to work out from the horizontal force. There is no point to pick up the horizontal acceleration with the 4N force, they are all different direction
Generally, You are right, well-done!
: .
#100596 Pls HELP on differentiation
Posted by
on 30 November 2007 - 08:26 PM
in
Mathematics
1. y = cos (3x - pi/6)- cos2 ( tan (x) ) - sin (pi/3) , dy/dx = ?
2. y= (ln2 (x) + lnx + 1)^1/2 , dy/dx = ?
3. y= ( tan (3x) - cotg (3x) )^1/2, dy/dx = ?
2. y= (ln2 (x) + lnx + 1)^1/2 , dy/dx = ?
3. y= ( tan (3x) - cotg (3x) )^1/2, dy/dx = ?
#100592 Who belives SOMETHING = NOTHING ?
Posted by
on 29 November 2007 - 02:04 PM
in
Mathematics
SOMETHING = NOTHING, it's possible !!! Let see the magic of Maths now :
Consider :
S = 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 +.....
=> 1/2 S = 1/4 + 1/6 + 1/8 + .....
=>S - 1/2S = 1/2 + 1/3 + 1/5 + 1/7 + .... (canceled respectly 1/4; 1/6; 1/8; ... from S by 1/2 S)
=> 1/2 S = 1/2 + 1/3 + 1/5 + 1/7 + ...
Now we subtract the second 1/2 S for the first 1/2 S :
1/2 S - 1/2 S = (1/2 + 1/3 + 1/5 + 1/7 + ... ) - ( 1/4 + 1/6 + 1/8 + ... )
1/2 S - 1/2 S = 1/2 + 1/3 + 1/5 + 1/7 + ... - 1/4 - 1/6 - 1/8 - ...
0 = 1/2 + ( 1/3 - 1/4) + ( 1/5 - 1/6) + ( 1/7 - 1/8 ) + ....
0 = 1/2 + 1/12 + 1/30 + 1/56 + ....
From now. obviously seeing that the right hand side would end up with a positive number which (of course) is greater than 0 . But on the left hand side, so funny, it's definetly 0. 
( what else can be if subtract 1/2 S by itself ? 
)
So, can u now get the point ? Clearly, NOTHING = SOMETHING , MATHS is miracle !!!!
Consider :
S = 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 +.....
=> 1/2 S = 1/4 + 1/6 + 1/8 + .....
=>S - 1/2S = 1/2 + 1/3 + 1/5 + 1/7 + .... (canceled respectly 1/4; 1/6; 1/8; ... from S by 1/2 S)
=> 1/2 S = 1/2 + 1/3 + 1/5 + 1/7 + ...
Now we subtract the second 1/2 S for the first 1/2 S :
1/2 S - 1/2 S = (1/2 + 1/3 + 1/5 + 1/7 + ... ) - ( 1/4 + 1/6 + 1/8 + ... )
1/2 S - 1/2 S = 1/2 + 1/3 + 1/5 + 1/7 + ... - 1/4 - 1/6 - 1/8 - ...
0 = 1/2 + ( 1/3 - 1/4) + ( 1/5 - 1/6) + ( 1/7 - 1/8 ) + ....
0 = 1/2 + 1/12 + 1/30 + 1/56 + ....
From now. obviously seeing that the right hand side would end up with a positive number which (of course) is greater than 0
. But on the left hand side, so funny, it's definetly 0. ( what else can be if subtract 1/2 S by itself ? )So, can u now get the point ? Clearly, NOTHING = SOMETHING , MATHS is miracle !!!!
#100591 1 + 1 = 1 ? ? ?
Posted by
on 29 November 2007 - 01:45 PM
in
Mathematics
Well-done !!!! 
as I said, it's all for relaxing :D
You wanna try once more with 1 > 4 ? I will prove it soon
as I said, it's all for relaxing :
D You wanna try once more with 1 > 4 ? I will prove it soon
#100588 1 + 1 = 1 ? ? ?
Posted by
on 28 November 2007 - 10:29 PM
in
Mathematics
This is a "trick stuff" in Maths which quite interesting. Hope that anyone will enjoy it.
Assume : A = B
=> A^2 = AB (multiply both by A)
=> A^2 - B^2 = AB - B^2 (take away both to B^2)
=> (A-B)(A+B)= B (A-B)
=> A+B = B
=> 2B = B (hence A = B)
=> 2 = 1
=> 1 + 1 = 1 BINGO !!!
I think we shouldn't study Maths anymore cause It's all wrong now ...
Assume : A = B
=> A^2 = AB (multiply both by A)
=> A^2 - B^2 = AB - B^2 (take away both to B^2)
=> (A-B)(A+B)= B (A-B)
=> A+B = B
=> 2B = B (hence A = B)
=> 2 = 1
=> 1 + 1 = 1
BINGO !!! I think we shouldn't study Maths anymore cause It's all wrong now ...
#100587 differential questions ...
Posted by
on 28 November 2007 - 09:39 PM
in
Mathematics
To ad absurdum :
Thank you. I'm really grateful.
I'm in exciting mood and a wee bit apprehensive state. Somehow, I'm quite confident of my Maths knowledge, it's due to our 'crazy' curriculum about Maths in my country e.g. We have to remember over 100 rules and theorems of Maths such as A^3 + B^3 = ... ; cos2x=cos2 (x) - sin2 (x)=2cox2 (x)-1 ; ...etc and etc ...
All my trouble now is about language. I haven't got enough "maths terms" to use in the BMO, therefore I'm worry about my explanation when solving problems.
I did look up of some past paper of BMO. It's required a lot of knowledge in Arithmetic which is so ironically my weakness ...
Anyway, let see wat happen ...
Thank you once again.
Thank you. I'm really grateful.
I'm in exciting mood and a wee bit apprehensive state. Somehow, I'm quite confident of my Maths knowledge, it's due to our 'crazy' curriculum about Maths in my country e.g. We have to remember over 100 rules and theorems of Maths such as A^3 + B^3 = ... ; cos2x=cos2 (x) - sin2 (x)=2cox2 (x)-1 ; ...etc and etc ...
All my trouble now is about language. I haven't got enough "maths terms" to use in the BMO, therefore I'm worry about my explanation when solving problems.
I did look up of some past paper of BMO. It's required a lot of knowledge in Arithmetic which is so ironically my weakness ...
Anyway, let see wat happen ...
Thank you once again.
#100581 differential questions ...
Posted by
on 28 November 2007 - 07:25 PM
in
Mathematics
Wikipedia isn't an ideal one. There's only the definition of it. What I need further about its properties, how it works and applies to solve problems.
BMO stands for British Mathematics Olympiad. I suppose you heard about it.
I have plenty of question in Maths but they're quite beyond the AH level. I wonder is it OK if I post it here...
BMO stands for British Mathematics Olympiad. I suppose you heard about it.
I have plenty of question in Maths but they're quite beyond the AH level. I wonder is it OK if I post it here...
#100579 differential questions ...
Posted by
on 28 November 2007 - 06:00 PM
in
Mathematics
To Geogre:
I'm Vietnamese. I've just been in UK for a couple of months: exchanged student ...
My English's terrible. First few days coming here, I couldn't hear anythings, neither speaking ...
But I'm learning ... so far...
I'm 17. I attempt to go to uni in UK. With 5 subjects I'm taking now, 2 AH and 3 H I'm quite confident.
I'm really keen on Maths, so interested in me. This Friday I'm going to sit in BMO ... Do you know any website or resource I can search for Modulo. I need further about it but couldn't find out any books mention about it.
I'm Vietnamese. I've just been in UK for a couple of months: exchanged student ...
My English's terrible. First few days coming here, I couldn't hear anythings, neither speaking ...
But I'm learning ... so far...
I'm 17. I attempt to go to uni in UK. With 5 subjects I'm taking now, 2 AH and 3 H I'm quite confident.
I'm really keen on Maths, so interested in me. This Friday I'm going to sit in BMO ... Do you know any website or resource I can search for Modulo. I need further about it but couldn't find out any books mention about it.
#100578 differential questions ...
Posted by
on 28 November 2007 - 05:46 PM
in
Mathematics
For iain :
I got your point. But the question is really y=(lnx)^(x^0,5) which George's done very well.
but u's right, if the question's in the form y = ln (x^(x^0.5)), it's really similar to AH maths.
I got your point. But the question is really y=(lnx)^(x^0,5) which George's done very well.
but u's right, if the question's in the form y = ln (x^(x^0.5)), it's really similar to AH maths.
#100575 how do you integrate...
Posted by
on 28 November 2007 - 01:01 PM
in
Mathematics
Oh thanks, silly me, forgot the chain rule, so careless !!!
#100574 differential questions ...
Posted by
on 28 November 2007 - 01:00 PM
in
Mathematics
what u got for the first question? I mean the final result ... I got it but it's so messy, a loads ...
#100573 differential questions ...
Posted by
on 28 November 2007 - 12:55 PM
in
Mathematics
I'm really impressed. How old r u ?
By the way, actually, when u reach to sin3 (x) + cos3 (x) / ( cos (x) + sin (x)) we do have a shorter way to simplify :
Remember : sin3 (x) + cos3 (x) = (sin (x) + cos (x)) (sin2 (x) - sinxcosx + cos2 (x)) A^3 + B^3=(A+B)(A^2-AB+B^2)
= (sin (x) + cos (x)) (1 - sinxcosx) sin2 (x) + cos2 (x) = 1
= (sin (x) + cos (x)) (1 - 1/2 sin (2x)) sin (2x) = 2 sinx cosx
hence ((sin (x) + cos (x)) will be simplified to end up with the form y = 1-1/2 sin (2x). YEAH ! You're right !!!
P/s: how could u type power and fraction when posting ? I'm a newcomer ...
By the way, actually, when u reach to sin3 (x) + cos3 (x) / ( cos (x) + sin (x)) we do have a shorter way to simplify :
Remember : sin3 (x) + cos3 (x) = (sin (x) + cos (x)) (sin2 (x) - sinxcosx + cos2 (x)) A^3 + B^3=(A+B)(A^2-AB+B^2)
= (sin (x) + cos (x)) (1 - sinxcosx) sin2 (x) + cos2 (x) = 1
= (sin (x) + cos (x)) (1 - 1/2 sin (2x)) sin (2x) = 2 sinx cosx
hence ((sin (x) + cos (x)) will be simplified to end up with the form y = 1-1/2 sin (2x). YEAH ! You're right !!!
P/s: how could u type power and fraction when posting ? I'm a newcomer ...
#100569 differential questions ...
Posted by
on 27 November 2007 - 06:18 PM
in
Mathematics
Pls, help me these 2 questions
1. y = (lnx)^(√x) , dy/dx ?
2. y= sin2 (x) / (1+cotg x) + cos2 (x) / (1+tan x) , dy/dx = ?
1. y = (lnx)^(√x) , dy/dx ?
2. y= sin2 (x) / (1+cotg x) + cos2 (x) / (1+tan x) , dy/dx = ?
#100567 how do you integrate...
Posted by
on 27 November 2007 - 06:01 PM
in
Mathematics
It isn’t too bad.
Let rearrange it first
1/(1+cosx) = 1/( 1 + 2cos2(x/2)-1) cosx = 2cos2 (x/2)-1
= 1/ 2 cos2 (x/2)
=1/2 sec2 (x/2)
From now, if I weren’t wrong, integrate sec2 (x/2), we get tan (x/2) + C !!!
So the answer is 1/2 tan (x/2) + C
Let rearrange it first
1/(1+cosx) = 1/( 1 + 2cos2(x/2)-1) cosx = 2cos2 (x/2)-1
= 1/ 2 cos2 (x/2)
=1/2 sec2 (x/2)
From now, if I weren’t wrong, integrate sec2 (x/2), we get tan (x/2) + C !!!
So the answer is 1/2 tan (x/2) + C
