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Consider these two lists

A = {1,25,31,84,87,134,158,182,198}
B = {2,18,42,66,113,116,169,175,199}

Now for both lists, add 1,000,000 to each of the numbers then take the sum of their square roots. For example, list A gives us

$\sqrt{1000001}+\sqrt{1000025}+...\sqrt{1000198} = 9000.44998...$

Now what is interesting to note here is that the number obtained from list A and the number obtained from list B differ first in the 37th decimal place. A gives us the digit 2 and B gives us 5. Any explanation as to how/why does this happen? Or rather how were these numbers chosen? Is there some theory behind this, an algorithm perhaps to compile such lists or is it just some random number theory abomination?

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2 Answers 2

up vote 6 down vote accepted

A quick check shows that $$\sum_{a \in A} a^k = \sum_{b \in B} b^k$$ for $k \in \{1,2,\ldots,8\}$. So you can approximate it using binomial theorem $$\sqrt{a+x} = \sqrt{a} \sqrt{1+\dfrac{x}a} = \sqrt{a} \times \left(1 + \dfrac12 \dfrac{x}a - \dfrac12 \times \dfrac12 \left(\dfrac{x}{a} \right)^2 + \dfrac12 \times \dfrac12 \times \dfrac32 \left(\dfrac{x}{a} \right)^3 \mp \cdots \right)$$ to see that the difference between the two is of the order of $$1000 \times \dfrac12 \times \dfrac12 \times \dfrac32 \times \dfrac52 \times \dfrac72 \times \dfrac92 \times \dfrac{11}2 \times \dfrac{13}2 \times \dfrac{15}2 \times \left(\dfrac{\sum_{a \in A} a^9 - \sum_{b \in B} b^9}{1000000^9} \right) \approx 10^{-37}$$ (since the first nine terms in both the expansions are the same).

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It's worth noting that integer sets satisfying the property in the first line are studied under the name Prouhet-Tarry-Escott. –  Erick Wong Oct 11 '12 at 23:54
    
Could you please explain the rest of the line following 1000 and how it is related to 3rd line? Thanks. –  Emmad Kareem Oct 12 '12 at 0:06
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This is really a comment on Marvis' answer, but too large for one. It shows a bit more detail about how it works.

Let's take a simpler set: $A=\{1,6,8\}, B=\{2,4,9\}$. We can see that $1+6+8=15=2+4+9, 1^2+6^2+8^2=101=2^2+4^4+9^2$. Then if we form $\sqrt{1000001}+\sqrt{1000006}+\sqrt{1000008}-\sqrt{1000002}-\sqrt{1000004}-\sqrt{1000009}$ we get about $4.5 \cdot 10^{-15}$. The point is that we have canceled out the leading terms in the Taylor series. As Marvis says, $$\sqrt {1000006}= 1000\sqrt{1+\frac 6{1000000}}\approx 1000\left(1+\frac 12\frac 6{1000000}-\frac 12 \cdot \frac 12 \left(\frac 6{1000000}\right)^2+\frac 12 \cdot \frac 12 \cdot \frac 32 \left(\frac 6{1000000}\right)^3\right)$$ where we have kept the first four terms. Because the sum of each set and the sum of squares of each set are the same, the terms linear and quadratic in $\frac 1{1000000}$ will cancel and we will be left with the term in $1000\left(\frac 1{1000000}\right)^3$, which is $10^{-15}$

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+1, Thank you for your clarification. –  Emmad Kareem Oct 12 '12 at 3:05
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