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its written that if you multiply the sum above by -9 and use the distributive law, all terms except "1" will cancel. I can't see that. I know that this is a divergent series. (the article I was reading was a layman's introduction to zeta regularization)

Similarly, how is $S=1+2+3+4... = -1/12$

Even if the series were to terminate somewhere, I dont see these value. This is completely baffling me.

Using $\frac{1}{1-x}$ I can put x=10 but that is not fair,isn't it?

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    $\begingroup$ These formulas hold for p-adic numbers. They are not your usual (real) numbers! See en.wikipedia.org/wiki/P-adic_number for more info. $\endgroup$
    – JavaMan
    Mar 17, 2011 at 5:20
  • $\begingroup$ that article is too advanced for my level. do you know of a more gentle introduction $\endgroup$
    – kuch nahi
    Mar 17, 2011 at 5:23
  • $\begingroup$ Unfortunately, I do not know of any less-technical introduction to p-adic numbers as it is not an introductory topic. One needs to have a good idea about some abstract algebra and some analysis to access the p-adic numbers. What is your background? $\endgroup$
    – JavaMan
    Mar 17, 2011 at 5:31
  • $\begingroup$ Linear algebra, calculus at the level of apostol. I also know a bit about differential equations, and the special functions that appear there. Applied fourier analysis... that's about all. Do any of these topic help? $\endgroup$
    – kuch nahi
    Mar 17, 2011 at 5:38
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    $\begingroup$ This is an interesting introduction to p-adic numbers from Edward Burger web.williams.edu/Mathematics/eburger/BurgerMathHorizons.pdf I saw him give a talk once where he used a p-adic "ruler" to always be able to win at limbo. $\endgroup$
    – Brian
    Mar 17, 2011 at 12:23

5 Answers 5

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This does not hold for real numbers, the numbers you are probably used to working with. Rather it works with a different type of number, called a p-adic number. I will get to what these are in a moment.

But first, what is a number? One might be tempted to give examples, such as $1$ or $-3/4$. But these are just symbols, which have no meaning on their own. Rather, mathematicians have defined them in a way that they all agree upon, which allows them to be added, multiplied, etc. It is by defining them that we give them meaning.

But why do we define them the way we do? We could just as easily define them in other ways; it just so happens that the typical definition is often the most useful. But this is not always the case. The p-adic numbers are a different way of defining "number".

To explain, I'll take the special case of 10-adic integers, which is where the equation you are asking about applies. The normal integers are what we are used to using daily, and if we take two integers $a$ and $b$ we can talk about how "close" $a$ is to $b$. Specifically, we say that $a$ is "close" to $b$ if they differ by a small power of ten, i.e. $40$ is sort of close to $30$ because $40-30=10=10^1$ while $4$ is very close to $3$ because $4-3=1=10^0$ (the same process works for numbers which do not differ by an integer power of ten if we allow non-integer powers). But why do we define "closeness" this way? What if we said $a$ is close to $b$ if they differ by a large power of ten? For example, $40,000$ would be sort of close to $30,000$ because $40,000-30,000=10,000=10^4$, while $4,000,000$ would be very close to $3,000,000$ because $4,000,000-3,000,000=1,000,000=10^6$. If we define "closeness" in this way, we get the 10-adic integers.

Now for the series you posted. Multiplying on both sides by $9$ gives $9+90+900+9000+90000+… = -1$. Well, we can see that this is giving us $\cdots999999 = -1$, with an infinite number of $9$'s to the left. But $9-(-1)=10=10^1$, $99-(-1)=100=10^2$, etc. so as we add more $9$'s onto the left we are getting closer and closer to $-1$, using the way we defined "closeness" for the 10-adic integers. In fact, we get infinitely close to $-1$, so this series must in fact be equal to $-1$ in the 10-adic integers. While this is not a rigorous argument, I hope it provides you with a rough sense of what's going.

Edit: Tying in what Ross said in his answer, the reason the formula $1/(1-10) = 1 + 10 + \cdots$ works in the 10-adic integers is that each successive term is getting closer to zero using the 10-adic notion of "closeness", so we can in fact ignore sufficiently late terms.

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  • $\begingroup$ Much more elegant than I could say on the subject! :) $\endgroup$
    – JavaMan
    Mar 17, 2011 at 5:52
  • $\begingroup$ @DJC: Thanks, I hope I struck the right balance between being intuitive and being formal. $\endgroup$ Mar 17, 2011 at 5:58
  • $\begingroup$ Alex: I am far from an expert on the topic, but 10 is not a prime number. I guess you can define 10-adic numbers but those will not form a field. Also, I might be completely wrong. $\endgroup$
    – Asaf Karagila
    Mar 17, 2011 at 7:26
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    $\begingroup$ @Asaf: Yes, you are correct that 10-adic numbers cannot form a field because zero divisors are present. However, this is not relevant to the value of the series. If needed, because $9$ is coprime to $10$ and $9 \in (3)$, we can localize at the prime ideal $(3)$ to get $9$ as a unit. $\endgroup$ Mar 17, 2011 at 7:34
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    $\begingroup$ Oops, I should have said localize at $\{1,3,3^2,\ldots\}$. $\endgroup$ Mar 17, 2011 at 14:04
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There are many formulas that fall in this category. The sum is not convergent, but you can do a formal manipulation that seems to sum it. Maybe the simplest is 1-1+1-1+1-1 ... If you represent it as $\sum (-1)^i$ you can sum it as $\frac {1}{1-(-1)}=\frac{1}{2}$. This is exactly in the spirit of your original comment "Using $\frac{1}{1−x}$ I can put $x=10$ but that is not fair,isn't it?" No, it's not fair, as the series is not convergent in the usual real number system, and as the p-adic numbers are not "normal" you should alert your readers when you use them. The formal power series manipulation that leads to the sum of the geometric series cancels all the intermediate terms just fine, but the last term doesn't go to zero so you can't ignore it.

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I recommend this post in Terry Tao's blog. It is not elementary, but can be understood by someone with knowledge of the calculus of one real varable.

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This formula is justified by analytic continuation (so you can, in fact plug in x=10 in $\frac{1}{1-x}$). See

http://en.wikipedia.org/wiki/1_%2B_2_%2B_4_%2B_8_%2B_%C2%B7_%C2%B7_%C2%B7

http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%C2%B7%C2%B7%C2%B7

The relation to p-adics is simple enough: $\frac{1}{1-x} - (1+x+x^2+\ldots +x^n)= x^{n+1}(1+x+\ldots)$, so to equate the two you want the difference to get small, which is true either for real numbers when $x<1$, or for p-apics when $x=p$.

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for your $1+2+3+4+...=-1/12$ this is the value of $\zeta(-1)$, the riemann zeta function at $-1$. so while it is not an equality as written, the definition is usually given as $\zeta(s)=\sum_{n=1}^{\infty}n^{-s}$ which only converges for real $s$ greater than 1. however it can be continued to the rest of the plane (with a pole at 1).

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