1. Your calculator will find 69!. Calculate 73! to 4s.f.

2. Estimate the first n! to exceed 10^200 and 10^300.

3. Multiplication is compound addition; index is compound multiplication. Use this to unpack 2^2^3 and 2^3^2 and so show that they have different values.

4. Consider the number series 2, 2^2, 2^2^2, 2^2^2^2. Evalauate these and attempt to evaluate the next element in the set. If you think this is ambiguous, work from the right end, 2^(2^2), 2^(2^(2^2)). I think you need to use logs. A delicious term for writing these is *power tower*.

5. If I were to write that number series as 2[1]2, 2[2]2, 2[3]2, 2[4]2 so that the middle number indicates how many twos are in the expression, then Q4 gives an value to 2[5]2 of about 2 x 10^(5 digit number), really large on one's usual idea of 'large'. Can you manage to approximate 2[6]2, perhaps as 10^(very large number)?

The reason for this sort of notation is that the numbers are quite difficult to write down and become very large very quickly.

6. Find 3[3]3 = 3^(3^3) and attempt 3[4]3.

3[3]3 might be called the second tower, 3 exponentiated three times. In which case the third tower would be 3^3....^3 where the number of threes is 3^3^3, which we know from Q6 to be already alarmingly large. This leads to the notation from Ronald Graham, where g₁ is the nth tower, made up of the number of 3s in the (n-1)th tower. You need a whole new idea of big for this.

Another way to write very large numbers succinctly is Steinhaus-Moser notation. Steinhaus wrote a number inside a triangle, so that triangle(n) = nⁿ but then moved to a number inside a square (rectangle, perhaps) which is n^n^n..... with the number n written n+1 times (I think it is n+1 not n).

7. Evaluate triangle(2) and square(2), Show that square(2)=2[4]2. Steinhaus went one further, the number inside a pentagon, though he later used just a number inside a circle, to indicate that the square process was to be nested, which makes the numbers explode.

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The 2 inside a circle is called a *mega,*which* * is already a very large number, since ② = square(square(2)) = square(triangle(triangle(2))) = square(triangle(2^{2})) = square(triangle(4)) = square(4^{4}) = square(256) .... (wikipedia) ..and square(256) is triangle (triangle...256 times) of 256. Remember that triangle 256 is 256^256, so keep on raising that to the power 256 until you've done it 256 times.

My instinct is that circle(n) would be infinitely large, but there is a range of numbers such that circle(n) has a value. Which points out that just *some* of these ridiculously large numbers are actually infinitely large. I suspect that circle(16) is infinitely large.

If you like collecting words, try *megiston,* which is a circle(10). [ The word is not *megistron* as you may find; read Hugo Steinhaus in the original or translation]. When we allow numbers to behave in such fashions as the circle function we hit extremely large numbers very very quickly, such that even counting the digits becomes very difficult, being only one operation fewer than evaluating the number itself.

You have already come across, I expect, googol, 10^100 and googolplex, 10^googol. These were named by a small boy, Milton Sirotta, asked by his uncle to give a name to a very large number. A googol is the practical limit to your calculator (probably), simply because only two character spaces have been left for index in scientific notation. Smaller than this is the Eddington number, 10^80, which is the (accepted) number of protons in the observable universe.

8. Given the density of space, that works out as about one particle per cubic metre. So suppose we tried to count how many neutrons would fill the universe? That would be 10^80 divided by the size of a neutron. Go on, then.... look stuff up and comment on the validity of 10^80 as a number to conjure with.

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You may like (Donald) Knuth's up arrow notation.

2↑4 = 2[3]4 = 2⁴; 2↑↑4 =2[4]4 = 2^16 = 65536

2↑↑↑4 = 2[5]4 = 2↑↑( 2↑↑4) = 2^(2^(2^(..... 2^16 times

using ↑ⁿ to indicate n arrows, clearly a↑ⁿb = a[n+2]b

9. Write a googol in arrow notation with one arrow and try to write it with two arrows. Find the nⁿ = n↑n = n↑↑2 closest to a googol, for integer n.

10. By taking logs twice, find the integer N for which N^N^N = N↑↑3 most closely matches a googolplex. For your answer, find the multiplicative error, larger/smaller, expressed in standard form.

Be clear: 10^100 might be written 10^10^2 but is not 10^2^10, which is 10^1024.

11. There is a large n such that nⁿ = n↑n = n↑↑2 = a googolplex. Approximate this n with (1+∂)x10^98 and write an expansion for the log of the log of nⁿ. Find an expression that includes log (1+∂) and (1+∂)⁻¹ and show that ∂ is close to 0.0202. I think this is first A-level content, written at STEP standard.

I say, having used a spreadsheet to do calculation of log(nlogn), that this n is 1.020 317 217 444 x 10^98, but this may be too large. My successive approximations gave a result of 1.020 199 973 137 Either I have made errors, or the calculation function is having issues. I'd like to receive correction or elucidation.

DJS 20220121

A1. 73! = 73x72x71x70x.69! = 3.9136x10^8 x 1.711x10^98 = 6.697x10^106

A2. Add 2 to the index from 70! (which is the first at 10^100, so add 100/2=50 to 70 and 120! ought to be close. But the many figures below unity, 0.70 to 0.99 will have a reduction effect, which is true: 120! is 0.0669x10^200. 121! is 8.09^10^200. Similarly, 170! will exceed 10^300 and 167! = 1.504x10^300

A3.Unpack from the right; 2^2^3= 2^*(2^3) 2^8 = 256; 2^3^2 = 2^9 = 512

A4. 2, 4, 16, 2[4]2=65536=2^16, 2[5]2 = 2^65536 = 2.00353 x10^19728

A5 This is difficult: 65536log2=19728.301796 and we want 10^19728.301796log2, so separate out the 10^integer and 0.302796x2=0.60359, so 2[6]2 = 10^(0.6...x10^19728) = 10^(6.0359x10^19727).

A6. 3[3]3 = 3^27= 10^27log3=10^12.88227 = 7.62559748497x10^12 = 7,625,597,484,987 (rounding issues corrected).

3[4]3 = 3^(3^27) = 10^((7.62...x10^13)log3) = 10^(3.638x10^13)

A7. triangle(2) =2² = 4; square(2)=2^2^2^2 = 2[4]2= 2^16=65536

A8. Size of a neutron is about 0.8x10^-15, so this number, filling all known space with neutrons,is still only about 10^95 and still smaller than a googol.

If we think the number of galaxies is at least 200 billion and we think our galaxy has around 2.4 x 10^67 atoms (75% hydrogen, 25% helium the rest forgettable), we're still around 10^80 protons in total. Would you want to argue with that statement?

A9. 10^100 = 10↑100 . 10↑↑2 = 10^10, 100↑100 = 10↑200=10^200; 10↑↑3=10^10^10 = 10 ^(ten billion). A two arrow 'near miss' of the form x↑↑3, yields 3↑↑3=7x10^12 or 4↑↑3=1.3x10^154, not at all close. A solution of the form x↑↑2 is of the form n^n, so 57^57= 1.216x10^100 is actually quite close.

A10. 10^10^100 is closer to 51^51^51 than 52^52^52. Take log(log), so that 100 log(10log10) =100 = Nlog(NlogN). A googolplex divided by 51↑↑3 is 10^11.5608... so the multiplicative error is 3.555x10^11. This answer is 355 billion times too small. The 52↑↑3 value is 5.461x10^26 times too big.

A11. 10^100 = nlogn and let n=(1+∂)*10^98.

n log n = (1+∂)*10^98 log ((1+∂)*10^98) = (1+∂)*10^98 (98+log(1+∂)) = 10^100

Now divide through by 10^98; (1+∂) (98+log(1+∂)) = 100

Divide by (1+∂) and then expand both functions, the reciprocal and the log;

98+log(1+∂) = 98 + ∂ -∂²/2 + ∂³/3 -... = 100/ (1+∂) = 100 (1-∂+∂²-∂³ +...)

Expanding only as far as quadratic terms, 98 + ∂ -∂²/2 = 100 - 100∂ + 100∂²

0= 2 -101∂ +201∂²/2 = 4-202∂ +201∂² => ∂ = 0.020208335 (and ignore larger value)

So n is close to 1.0202x10^98. The cubic approximation will be a bit smaller than the quadratic one.