User blog comment:Planterobloon/How do you compare huge numbers to each other?/@comment-11227630-20170812012304

Some notations are compatible with another. e.g. a^^…^b (with c up arrows) = a→b→c (in chained arrow notation) = {a,b,c} (in Bird's array notation) = E(a)1#1#…1#b (with c-1 1's). In this case, an expression can convert into another notation holding its value. e.g. 2^^^^4 = E(2)1#1#1#4.

But some notations are not compatible with another, and this is more common than the prior one. For example, in fast-growing hierarchy (FGH), $$f_2(n)=n\times2^n$$, and $$f_3(n)$$ will have no compact expression in up-arrow notation or hyper-E notation. In this case, don't try calculating the number fully out because it's usually impossible in the real world. Instead, find upper bounds and lower bounds of it in other notations.

Let's find bounds of $$f_{\omega+1}(3)$$ in extended-E notation. Next, let $$n=f_3(3)\approx6859\times10^{121210691}$$, and it's between E121210691 and E121210692, so we have E100##(E121210691)#2 < $$f_{\omega+1}(3)$$ < E100##(E121210692)#2.
 * $$E[2]n=2^n 1)
 * E[2]2#n#2 < $$f_3^2(n)$$ < E[3]n#(n+E[3]n#n) < E[3](n+1)#n#2 (n > 1)
 * E[2]2#2#n ≤ E[2]2#n#n < $$f_4(n)$$ < E[3](n+1)#n#n (n > 1)
 * E[2]2#2#…2#n (with k-2 #'s) < $$f_k(n)$$ < E[3](n+1)#(n+1)#…(n+1)#n#n (with k-2 #'s)
 * E[2]2##(n-1) < $$f_\omega(n)$$ < E[3](n+1)##(n-1)
 * E[2]2##(n-1)#2 < $$f_\omega^2(n)$$ < E[3](n+2)##(n-1)#2

The inequalities are a bit complex, and will be much more complex when the comparisons go higher. To get a quick approximately image, you can use the idea of "functional approximation": It can flatten all the +1, +2, -1 details and give you an approximation.
 * $$f\ge^*g$$ if there exists a and b such that $$f(an+b)>g(n)$$ for all n
 * $$f\approx g$$ if $$f\ge^*g$$ and $$g\ge^*f$$