Do there exist several positive real numbers such that their sum is $1$ and sum of their squares is less than $0.01$?

My Attempt: Let there are $n$ real numbers and we call them $x_{1},x_{2},..,x_{n}$. Since they are positive real so WLOG we can assume $x_{1}\geq x_{2}..\geq x_{n}$. The condition $x_{1}+x_{2}+..+x_{n}=n \implies x_{n} \leq \frac{1}{n}$. Also $x_{1}\geq x_{2}..\geq x_{n} \implies x_{1}^{2}\geq x_{2}^{2}..\geq x_{n}^{2}$
After this point I am stuck.

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    $\begingroup$ How can the sum of several positive integers be equal to $1$? $\endgroup$ – Jimmy R. Mar 26 '16 at 12:35
  • $\begingroup$ @JimmyR Thanks for the hint. $\endgroup$ – rugi Mar 26 '16 at 12:50
  • $\begingroup$ If $a < 1$ $a^2 <a$. If $a < .1$ $a^2 < .1a$. If $a < .01$ a^2< .01a $. So if a,b,c,d..... < .01 then $a^2+b^2+... < .01 (a+b+c....) $ $\endgroup$ – fleablood Mar 27 '16 at 1:23
  • $\begingroup$ Ten times $1/10$ would be enough... $\endgroup$ – CiaPan Mar 27 '16 at 6:44
  • $\begingroup$ @CiaPan No, because $(\frac1{10})^2+(\frac1{10})^2+\ldots+(\frac1{10})^2=\frac1{10}>0.01$. $\endgroup$ – Théophile Mar 27 '16 at 12:33

Just take positive reals $x_1,x_2,\dots ,x_n$ so that their sum is $1$ and all the numbers are smaller than $\frac{1}{100}$.

Then we have: $x_1^2+x_2^2+\dots x_n^2<\frac{1}{100}(x_1+x_2+\dots + x_n)=\frac{1}{100}\cdot1=0.01$


Yes. Take $x_n=\frac{\epsilon}{n}$, where $\epsilon=\left(\sum_{n=1}^N\frac{1}{n}\right)^{-1}$. We have $$\sum_{n=1}^Nx_n=\epsilon\sum_{n=1}^N\frac{1}{n}=1$$ by definition. On the other hand, $$\sum_{n=1}^Nx_n^2=\frac{\sum_{n=1}^N\frac{1}{n^2}}{\left(\sum_{n=1}^N\frac{1}{n}\right)^2}.$$ By taking $N$ large enough, this can be arbitrarily small since the numerator converges, while the denominator diverges to $\infty$.

  • $\begingroup$ Nice answer. But I have a minor doubt. This question is asked in a eighth standard , where concept of convergence is not taught. I suspect that there is an answer avoiding the use of convergence. $\endgroup$ – rugi Mar 26 '16 at 12:52
  • 1
    $\begingroup$ @rugi Rather than using the word convergence, it can be shown in an elementary way that $\sum_{n=1}^N\frac{1}{n^2}$ is always less $2$, while $\sum_{n=1}^N\frac{1}{n}$ can be arbitrarily large. This is all we need here. $\endgroup$ – Spenser Mar 26 '16 at 12:57
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    $\begingroup$ Why not just take $\frac{1}{101}$ one hundred and one times? Clearly the sum of the squares is $\frac{1}{101}$. $\endgroup$ – Yorch Mar 26 '16 at 17:32
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    $\begingroup$ Realize Spenser is showing a much stronger condition. Not only can you make it less than .01, you can make it as small as you want. $\endgroup$ – Nate 8 Mar 26 '16 at 23:48

Suppose all the numbers will be equal. Then we have:

$$n \times i^2 < 0.01$$ $$n \times i = 1$$

where $i$ is the number we're looking for and $n$ is the quantity. Substitute and get:

$$i < 0.01$$


The intuition:

When you square a number $x$, you multiply $x$ by $x$. That means that $x^2$ is always "$x$ times bigger" than $x$.

Suppose $x$ is large, like $100$. $100^2$ is much bigger than $100$. To be precise, it's $100$ times bigger. Examples like this give us the intuition "squaring makes a number bigger... and the larger the number, the larger the effect of squaring".

But if $x$ is smaller than $1$, "$x$ times bigger" is a confusing thing to say – multiplying a number by $x$ actually makes it smaller. So $x^2$ will be smaller than $x$ when $x$ is less than one.

For instance, suppose $x$ is $0.01$. $0.01^2$ is much smaller than $0.01$. To be precise, it's $1/0.01=100$ times smaller. By taking $x$ closer and closer to zero, you can make $x^2$ smaller than $x$ is, by bigger and bigger factors.

The problem at hand:

"Do there exist several positive real numbers such that their sum is 1 and sum of their squares is less than $0.01$?"

This problem asks us to find numbers that sum to $1$, even though their squares sum to less than $0.01$. That is, it is asking us to find numbers whose squares are really small, even though the numbers themselves aren't that small.

The intuition above tells us how to find these numbers: just take numbers much smaller than $1$!. We'll need a lot of them, to make them add up to $1$, but their squares will shrink by so much that this won't be a problem.

To start with, take ten copies of $0.1$. They add up to $1$. Each squared is $0.01$, so the ten squares added up is $0.1$. If the problem said "sum of their squares is less than $0.1$", we'd be done. But it wants us to be below $0.01$. So let's keep going – the smaller the numbers we pick, the more the squares will be less than the numbers.

Take one hundred copies of $0.01$. They add up to $1$. Each squared is $0.0001$, so the one hundred squares added up is $0.0001 \times 100 = 0.01$. Woah – right on the edge. Strictly speaking, the question wants us the sum of the squares to be less than $0.01$, so we need to go one more step.

Take one thousand copies of $0.001$. They add up to $1$. Each squared is $0.00001$, so the one thousand squares added up is $0.00001 \times 1,000 = 0.001$. We're done!

(Looking at the trend of the sum of squares, it's clear that if we kept going, we could get it as small as we wanted. That's what Spenser was saying, when they mentioned convergence. But it's really just a way of describing a trend.)

  • $\begingroup$ Nice explanation. Just right for a kid who is trying to grasp the non standard problem. thanks. $\endgroup$ – rugi Mar 27 '16 at 10:35

To combine Carry's and Spenser's answers, consider $a_i=\frac{1}{n}$ for $i=1\ldots n$. Then, $$\sum_{i=1}^{n}a_i = \sum_{i=1}^{n}\frac{1}{n}=1 \quad \quad \quad \quad \sum_{i=1}^{n}a_i^2 = \sum_{i=1}^{n}\frac{1}{n^2}=\frac{n}{n^2}=\frac{1}{n}$$

So by making n as large as you like (for example, at least $\frac{1}{\epsilon}$ for $\epsilon>0$), you can make the sum of squares as small as you like. In your case, n=100, 101, ... will work.


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