### SICP: Broken Math

Here's a great little math trick from SICP:
`v = 1 + 2 + 4 + 8 + ...`
Let's play around with that equation a bit:
`v - 1 = 2 + 4 + 8 + ...(v - 1) / 2 = 1 + 2 + 4 + 8 + ...`
But that's the same as the first line. Hence, we see that:
`(v - 1) / 2 = v`
Solving for v, we get:
`v = -1-1 = 1 + 2 + 4 + 8 + ...`
Crazy, eh? Sussman said:
Arguments that are based on convergence are flaky, if you don't know the convergence beforehand. You can make wrong arguments. You can make deductions as if you know the answer and not be stopped somewhere by some obvious contradiction.
I think the bad assumption in this case is thinking that some v could exist and be a normal integer. Clearly, there is no such normal integer v that equals 1 + 2 + 4 + 8 + ...

kib said…
Funny post,

Sussman is right, we have to be very careful when handling limits.

But for this one, it seems rather natural that v will not be an integer.
stan said…
Although, if you write a C program to evaluate that sum using signed integers, it does converge to -1. :)
Anonymous said…
Why do you call it a trick? Seems more like an anti-trick to me, if anything. As in, no, really, don't try this at home or anywhere else.
kib said…
A nice one was found by Gauss (it is said to, be maybe that's a legend...) when he was a child and asked to find this (finite) sum :

Sn = 1 + 2 + 3 + ... + n

Gauss started by writing the numbers like this :

Sn = 1 2 3 4 5 ... n

then in reversed order :
Sn = n n-1 n-2 2 1

What's the point ? He could now add each column and see that they all give n+1:

2*Sn = n+1 n+1 n+1 ... n+1

As with have n equal terms, 2*Sn is easily computed :
2*Sn = = n*(n+1)

So, Sn = n*(n+1)/2

Nice no ?!
jjinux said…
> Although, if you write a C program to evaluate that sum using signed integers, it does converge to -1. :)

Haha, I thought of that too ;)

> Why do you call it a trick?

I didn't know what else to call it ;)
jjinux said…
> A nice one was found by Gauss

Ah, yes, I know that story. The funny is when I showed my wife this "trick", she came up with the same story ;) So let me see:

n = infinity
(infinity (infinity - 1)) / 2 = infinity

(But note, it's the same "kind" of infinity, since there's a 1-1 mapping between the two.)
Anonymous said…
There is quite a theory about divergent series. Have you found http://en.wikipedia.org/wiki/Divergent_series
where they give a bibliography, including a book by G. H. Hardy.

I think this particular sum is due to Euler.

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