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Who belives SOMETHING = NOTHING ? - HSN forum

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Who belives SOMETHING = NOTHING ?


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#1 cK_hoangan

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Posted 29 November 2007 - 02:04 PM

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 unsure.gif . But on the left hand side, so funny, it's definetly 0. blink.gif ( what else can be if subtract 1/2 S by itself ? dry.gif )


So, can u now get the point ? Clearly, NOTHING = SOMETHING , MATHS is miracle !!!! ohmy.gif


#2 Marcus

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Posted 06 January 2008 - 01:07 AM

That puzzled me for quite a while, the problem is that your S changes its value, you have got an infinite series...but the number of terms you specify it to changes, therefore changing the value...

keeping it to 3 terms

if S = 1/2 + 1/3 +1/4
then 1/2S = 1/4 + 1/6 + 1/8 = 13/24

S-1/2S = 1/2 + 1/3 +1/4 - 1/4 - 1/6 - 1/8
1/2S= 1/2 + 1/3 - 1/6 - 1/8 = 13/24

therefore if you subtracted the first 1/2S from the second 1/2S you would get 0 = 0

QED
=-=-=Marcus=-=-=

#3 cK_hoangan

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Posted 17 January 2008 - 03:47 PM

but ... then, wat is the error ? unsure.gif

If you agree that it is an infinite serie, how could you know the number of its terms is changed ? smile.gif



#4 Marcus

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Posted 20 January 2008 - 01:57 AM

QUOTE(cK_hoangan @ Jan 17 2008, 03:47 PM) View Post
but ... then, wat is the error ? unsure.gif

If you agree that it is an infinite serie, how could you know the number of its terms is changed ? smile.gif


the problem is: you have solved an infinite series using a set number of terms, your S , and both \frac{1}{2}Ss are all to different numbers of terms, it is the differences in these numbers of terms that the error arises...

To do this properly you would have to specify S in such a way that it can be infinite, ie S equals the recurrence relation:
U_{n+1} = U_{n} + \frac{1}{n+1}
which you half:
\frac{1}{2}U_{n+1} = \frac{1}{2}U_{n} + \frac{1}{2(n+1)}
then you subtract them:


\begin{align*}
U_{n+1} - \frac{1}{2} U_{n+1} &= U_{n} + \frac{1}{n+1} -(\frac{1}{2}U_{n} + \frac{1}{2(n+1)}) \\
\frac{1}{2} U_{n+1} &= \frac{1}{2}U_{n} + \frac{1}{n+1}(1-\frac{1}{2}) \\
\frac{1}{2} U_{n+1} &= \frac{1}{2}U_{n} + \frac{1}{2(n+1)} \\
\end{align*}

now you should be able to see that if you subtract the two \frac{1}{2}U_{n}s now you would get 0=0
=-=-=Marcus=-=-=





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