Page 1 of 1 [ 12 posts ]
 Print view Previous topic | Next topic
error on page 77 chapter 4.3 ?
Author Message

Joined: 14 May 2011, 15:38
Posts: 6
error on page 77 chapter 4.3 ?
Sn= 1+x^2+x^4+ .....+x^2n

sn*(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))

31 Jul 2011, 16:31
Supporter

Joined: 07 Jun 2008, 08:21
Posts: 235
Re: error on page 77 chapter 4.3 ?
Hi
What exactly do you think the error is here? Can you explain what you mean in more detail please?
Thanks
Vasco

31 Jul 2011, 16:51

Joined: 14 May 2011, 15:38
Posts: 6
Re: error on page 77 chapter 4.3 ?
Hello
I suposed there was a mistake, because it is written:
Sn=(1-x^2)^-1.
maybe there is something I do not understand ?.

31 Jul 2011, 18:25
Supporter

Joined: 07 Jun 2008, 08:21
Posts: 235
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^2
S_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<1

In this case 1/(1-x^2) provided that x^2<1

Quote:
Sn= 1+x^2+x^4+ .....+x^2n

sn*(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^2n
which 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

01 Aug 2011, 06:21

Joined: 14 May 2011, 15:38
Posts: 6
Re: error on page 77 chapter 4.3 ?
thank you
but
S_0=x^(2*0)=1
S_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)^-1
but not 1+x^2+x^4+ ..... =(1-x^2)^-1

01 Aug 2011, 09:06
Supporter

Joined: 07 Jun 2008, 08:21
Posts: 235
Re: error on page 77 chapter 4.3 ?
Hi
When 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 4.1

Vasco

PS 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.

01 Aug 2011, 12:15

Joined: 14 May 2011, 15:38
Posts: 6
Re: error on page 77 chapter 4.3 ?
Sn= 1+x^2+x^4+ .....+x^2n

sn*(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 equation
1+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=341
1+4+16+64+256=341

01 Aug 2011, 15:13

Joined: 14 May 2011, 15:38
Posts: 6
Re: error on page 77 chapter 4.3 ?
same way on page 79
----------------------------------------page 79
sn=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 get

sn=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)^-1
is it correct ?

01 Aug 2011, 15:22
Supporter

Joined: 07 Jun 2008, 08:21
Posts: 235
Re: error on page 77 chapter 4.3 ?
zozio wrote:
Sn= 1+x^2+x^4+ .....+x^2n

sn*(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 equation
1+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=341
1+4+16+64+256=341

You have misquoted Penrose. On page 78 he writes:

1+2^2+2^4+2^6+2^8+....=(1-2^2)^-1=-1/3
NOT
1+2^2+2^4+2^6+2^8=(1-2^2)^-1=-1/3

In 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_series
Also 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/week126

To sum up (so to speak!) I do not think you have found an error in the book!

Last edited by vasco on 02 Aug 2011, 07:07, edited 1 time in total.

02 Aug 2011, 05:55
Supporter

Joined: 07 Jun 2008, 08:21
Posts: 235
Re: error on page 77 chapter 4.3 ?
zozio wrote:
same way on page 79
----------------------------------------page 79
sn=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 get

sn=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)^-1
is 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 using

1+x^2+x^4+x^6+x^8+... = (1-x^2)^-1

and substituting ix for x gives:

1-x^2+x^4-x^6+x^8-... = (1+x^2)^-1

since 1-(ix)^2 =1- i^2.x^2=1-(-1).x^2=1+x^2

Again I do not think you have found an error in the book!

02 Aug 2011, 06:37

Joined: 14 May 2011, 15:38
Posts: 6
Re: error on page 77 chapter 4.3 ?
ok , I agree now !, thanks for your explanations, and the time you spent.

02 Aug 2011, 13:53
Supporter

Joined: 07 Jun 2008, 08:21
Posts: 235
Re: error on page 77 chapter 4.3 ?
Hi zozio
You're welcome!!
Vasco

02 Aug 2011, 16:01
 Page 1 of 1 [ 12 posts ]