So here is the last of this series explaining the expressions on Newton’s clock:

We are now up to 8. Let’s look at

\[\mathop{\prod}\limits_{{k}{=}{0}}\limits^{1}{{(}{2}{k}{+}{2}{)}}

\]

This is another excellent example of how concise the words in maths can be. The symbol “????” is the capital version of ???? which corresponds to the english “P”. The “P” here stands for “product” which is the result of multiplying two or more numbers. The expression on the clock means : “Take the expression 2*k* + 2 and successively replace the *k *with the number at the bottom of the ???? symbol (in this case “0”), evaluate the expression to get a number, and increment *k* by 1 and repeat until you reach the number at the top of the ???? symbol (in this case “1”). Then multiply all these numbers together.”

You can see why maths expressions are much more concise than English. So to evaluate this expression, we first replace the *k* with 0, then work out 2(0) + 2. This equals 2. Now increment the *k* by 1 to get 1, then work out 2(1) + 2. This equals 4. Since *k* is now at the number at the top of ????, we are done increasing *k*. Now multiply these numbers together. 2 × 4 = 8 which is the correct number at this position on the clock.

Most of you now know what

\[\sqrt{81}

\]

means. It means “what number multiplied by itself equals 81?” The answer, of course is 9 as 9 × 9 = 81.

The next hour is

\[{\log}_{2}1024

\]

The basics of this expression have already been explained for position “2” on the clock. This expression is asking the question “what does the exponent of 2 have to be so that 2* ^{x}* = 1024?”. Hopefully, the answer is “10” and it is because 2

^{10}= 1024.

Now let’s look at *B*_{16}. Remember when I explained position “7” on the clock: 0111_{2}? That was a number in the base 2 system of counting. Another common base used with computers is the base 16 counting system. We are familiar with the base 10 counting system that has 10 symbols used to count with: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. The base 16 system needs 16 symbols. So what is used after 9 is reached? Well, we resort to the letters of the alphabet. The numbers up to 9 in base 16 correspond to the same numbers in base 10. The next number in base 16 is “A” which corresponds to 10 in base 10 and the next number is “B” which is 11 in base 10. So *B*_{16} = 11.

Finally, the last expression:

\[\mathop{\sum}\limits_{{i}{=}{1}}\limits^{3}{{(}{3}{i}{-}{2}{)}}

\]

The Σ symbol is the Greek capital “sigma” and corresponds to the english “S”. This letter stands for “Sum” which is the addition of two or more numbers. This expression is just like the one in position “8” on the clock except that you add the resulting numbers together instead of multiplying them. So starting at *i* = 1, 3(1) – 2 = 1, 3(2) -2 = 4, 3(3) – 2 = 7, and we are done as *i* now equals the numbers on top of the Σ. So now add these numbers together: 1 + 4 + 7 = 12. It is now high noon and that completes Newton’s clock.