Author: zozio [ 31 Jul 2011, 16:31 ] Post subject: error on page 77 chapter 4.3 ? Sn= 1+x^2+x^4+ .....+x^2nsn*(1-x^2)= 1 + x^2 + x^4 + ..... + x^2(n-1) + x^2n -x^2 - x^4 - x^6 - ..... - x^2(n) - x^(2(n+1)) sn*(1-x^2)= 1 - x^(2(n+1))

 Author: vasco [ 31 Jul 2011, 16:51 ] Post subject: Re: error on page 77 chapter 4.3 ? HiWhat exactly do you think the error is here? Can you explain what you mean in more detail please?ThanksVasco

 Author: zozio [ 31 Jul 2011, 18:25 ] Post subject: Re: error on page 77 chapter 4.3 ? HelloI suposed there was a mistake, because it is written:Sn=(1-x^2)^-1.maybe there is something I do not understand ?.

 Author: vasco [ 01 Aug 2011, 06:21 ] Post subject: Re: error on page 77 chapter 4.3 ? What is written in the book is not S_n. It is the sum as n->infinity.The series given is a geometric series:a+ar+ar^2+ar^3+ar^4+....where a=1 and r=x^2S_n for a geometric series is:S_n=a(1-r^n)/(1-r) equation (1)which in this case where a=1 and r=x^2 is:S_n=(1-x^(2n))/(1-x^2) equation (2)However the limit of S_n as n->infinity is:1/(1-r) provided that r<1In this case 1/(1-x^2) provided that x^2<1Quote:Sn= 1+x^2+x^4+ .....+x^2nsn*(1-x^2)= 1 + x^2 + x^4 + ..... + x^2(n-1) + x^2n-x^2 - x^4 - x^6 - ..... - x^2(n) - x^(2(n+1)) sn*(1-x^2)= 1 - x^(2(n+1))What you did above is correct except that in the first line the last term is x^2(n-1)and then the last line becomes :Sn*(1-x^2)=1-x^2nwhich is the same as my equation 2.The main thing that you have misunderstood is that in the book Penrose is talking about the infinite series and about the sum when n->infinity.I hope this is helpful to you.Vasco

 Author: zozio [ 01 Aug 2011, 09:06 ] Post subject: Re: error on page 77 chapter 4.3 ? thank youbutS_0=x^(2*0)=1S_1=S_0+x^(2*1)S_n=S_0+ S_1+ .....+x^(2*n)and I'm not sure Penrose has the rigth to use the operator=it is not =, it's near and nearest as n --> infinite, but there is always an epsilon of difference, as litle as it can be.he should had written: lim S_n as n --> infinite=(1-x^2)^-1but not 1+x^2+x^4+ ..... =(1-x^2)^-1

 Author: vasco [ 01 Aug 2011, 12:15 ] Post subject: Re: error on page 77 chapter 4.3 ? HiWhen mathematicians write something like:c_0+c_1x^1+c_2x^2+c_3x^3+ .....it is a short way of writing: sum j=0 to infinity of c_jx^j. You will find this in many textbooks on mathematics.The "+...." at the end means "go on for ever".So it is OK to use = and write:sum j=0 to infinity of x ^2j = 1+x^2+x^4+ .....=1/(1-x^2)What IS missing on page 77 is the fact that in this particular example this is only true when x^2<1 or |x|<1. It is on page 78 under Fig 4.1VascoPS Notice that Penrose does not talk about S_n, which is the sum of the first n terms, he is always talking about the sum of the infinite series.

 Author: zozio [ 01 Aug 2011, 15:13 ] Post subject: Re: error on page 77 chapter 4.3 ? Sn= 1+x^2+x^4+ .....+x^2nsn*(1-x^2)= 1 + x^2 + x^4 + ..... + x^2(n-1) + x^2n -x^2 - x^4 - x^6 - ..... - x^2(n) - x^(2(n+1)) sn*(1-x^2)= 1 - x^(2(n+1))because of the inapropriate use of equal operator, we get the absurdity of page 78 1+2^2+2^4+2^6+2^8=(1-2^2)^-1=-1/3 !!! and a following non sens discussion !!just below the correct equation1+2^2+2^4+2^6+2^8=(1-x^2)^-1 *(1 - x^(2(n+1))) (with x=2 and n=4)we get -1/3*(1-2^(2*(4+1)))=-1/3*-1023=3411+4+16+64+256=341

 Author: zozio [ 01 Aug 2011, 15:22 ] Post subject: Re: error on page 77 chapter 4.3 ? same way on page 79----------------------------------------page 79sn=1+y^2+y^4+ .....+y^2n = 1 - y^(2(n+1))/(1-y^2)let i*x=y with i=square of -1 , so we getsn=1-x^2+x^4-x6 .....(+x^2n if n even or -x^2n if n odd) = ( 1 - (i*x)^(2(n+1)))/(1-(i*x)^2)=( 1 - (i*x)^2(n+1))/(1+x^2)lim when x<0 and n --> infinity of Sn= (1+x^2)^-1is it correct ?

 Author: vasco [ 02 Aug 2011, 05:55 ] Post subject: Re: error on page 77 chapter 4.3 ? zozio wrote:Sn= 1+x^2+x^4+ .....+x^2nsn*(1-x^2)= 1 + x^2 + x^4 + ..... + x^2(n-1) + x^2n -x^2 - x^4 - x^6 - ..... - x^2(n) - x^(2(n+1)) sn*(1-x^2)= 1 - x^(2(n+1))because of the inapropriate use of equal operator, we get the absurdity of page 78 1+2^2+2^4+2^6+2^8=(1-2^2)^-1=-1/3 !!! and a following non sens discussion !!just below the correct equation1+2^2+2^4+2^6+2^8=(1-x^2)^-1 *(1 - x^(2(n+1))) (with x=2 and n=4)we get -1/3*(1-2^(2*(4+1)))=-1/3*-1023=3411+4+16+64+256=341You have misquoted Penrose. On page 78 he writes:1+2^2+2^4+2^6+2^8+....=(1-2^2)^-1=-1/3NOT1+2^2+2^4+2^6+2^8=(1-2^2)^-1=-1/3In the last paragraph on page 78 he then writes:"...the above equation would be officially classified as 'nonsense'".He then discusses the possibility that the equation may make sense in some way. See here for the mathematics involved:http://en.wikipedia.org/wiki/Divergent_seriesAlso have a look here which talks about Euler's formula 1+2+3+...=-1/12 in string theory!!!!!!http://math.ucr.edu/home/baez/twf_ascii/week126To sum up (so to speak!) I do not think you have found an error in the book!

 Author: vasco [ 02 Aug 2011, 06:37 ] Post subject: Re: error on page 77 chapter 4.3 ? zozio wrote:same way on page 79----------------------------------------page 79sn=1+y^2+y^4+ .....+y^2n = 1 - y^(2(n+1))/(1-y^2)let i*x=y with i=square of -1 , so we getsn=1-x^2+x^4-x6 .....(+x^2n if n even or -x^2n if n odd) = ( 1 - (i*x)^(2(n+1)))/(1-(i*x)^2)=( 1 - (i*x)^2(n+1))/(1+x^2)lim when x<0 and n --> infinity of Sn= (1+x^2)^-1is it correct ?Yes it is correct, since (i*x)^2(n+1) = i^2(n+1).x^2(n+1) = -1^(n+1).x^2(n+1) then(1 - (i*x)^2(n+1))/(1+x^2)=(1-(-1)^(n+1).x^2(n+1))/1+x^2)If x^2<1 then x^2(n+1) -->0 as n-->infinity and so the limit for the sum is:1/(1+x^2)This is much easier to see if you substitute x-->ix on the right hand side as well as the left, so using1+x^2+x^4+x^6+x^8+... = (1-x^2)^-1and substituting ix for x gives:1-x^2+x^4-x^6+x^8-... = (1+x^2)^-1since 1-(ix)^2 =1- i^2.x^2=1-(-1).x^2=1+x^2Again I do not think you have found an error in the book!

 Author: zozio [ 02 Aug 2011, 13:53 ] Post subject: Re: error on page 77 chapter 4.3 ? ok , I agree now !, thanks for your explanations, and the time you spent.

 Author: vasco [ 02 Aug 2011, 16:01 ] Post subject: Re: error on page 77 chapter 4.3 ? Hi zozioYou're welcome!!Vasco

 Page 1 of 1 Archived: 07 Aug 2014 Powered by phpBB © 2000, 2002, 2005, 2007 phpBB Grouphttp://www.phpbb.com/