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Googology.

Author: localhost433

posts/entries/googology.md

@@ -0,0 +1,150 @@
+---
+title: A Glimpse into Googology
+date: 2024-02-14
+tags: [math]
+author: R
+location: St. Catharines, ON, Canada
+---
+
+This is a problem really gets me wonder 'What is Googology?' during the summer, and I ended up reading about it. When I wrote this I was in my last days of high school, coming across a instagram reel gives the question of comparing $2^{100!}$ and $(2^{100})!$ (the reel was just making fun about math), we ended up arguing about this and come up with different methods to proof our ideas. 
+
+## Numerical Method
+
+### Stirling's Approximation
+
+According to Stirling's Approximation[^1]:
+
+$$
+n! \sim \sqrt{2\pi n} \left(\frac{n}{e}\right)^n
+$$
+
+[^1]: "\(\sim\)" here means that the two quantities are asymptotic, i.e., the ratio between these two terms tends to 1 as \(n \to \infty\).
+
+So:
+
+### For \(2^{100!}\)
+
+$$
+100! \sim \sqrt{2\pi \cdot 100} \left(\frac{100}{e}\right)^{100}
+\approx 9.3 \times 10^{57}
+$$
+
+$$
+2^{100!} \approx 2^{9.3 \times 10^{57}}
+$$
+
+### For \(\left(2^{100}\right)!\)
+
+$$
+2^{100}! \sim \sqrt{\pi \cdot 2^{101}} \left(\frac{2^{100}}{e}\right)^{2^{100}} \approx 2.82 \times 10^{15} \cdot \left(4.66 \times 10^{29}\right)^{1.26 \times 10^{30}}
+$$
+
+Both numbers are in the form of tetration (operation based on iterated exponentiation), so it's difficult to compare them directly by hand.
+
+## Logarithm Approach
+
+### \(\log_2(2^{100!})\)
+$$
+\log_2(2^{100!}) = 100!
+$$
+
+Using Stirling again:
+
+$$
+100! \sim \sqrt{2\pi \cdot 100} \left(\frac{100}{e}\right)^{100} > 25 \cdot \left(\frac{100}{e}\right)^{100} = 25 \cdot \frac{100^{100}}{e^{100}}
+$$
+
+Which approximates to:
+
+\[ > 9.3 \times 10^{57} \]
+
+### \(\log_2((2^{100})!)\) (code attached in [Appendix](#appendix))
+Using Stirling again:
+
+$$
+(2^{100})! \sim \sqrt{2\pi \cdot 2^{100}} \left(\frac{2^{100}}{e}\right)^{2^{100}}
+$$
+
+Take log base 2:
+
+$$
+\log_2((2^{100})!) \sim \log_2 \sqrt{2\pi \cdot 2^{100}} + 2^{100} \log_2 \left(\frac{2^{100}}{e}\right)
+$$
+
+Breaking this down:
+
+$$
+= \frac{1}{2}(\log_2 2\pi + 100) + 2^{100}(100 - \log_2 e)
+\approx 1.25 \times 10^{32}
+$$
+
+### Conclusion
+
+Since:
+
+$$
+9.3 \times 10^{57} > 1.25 \times 10^{32}
+$$
+
+We conclude:
+
+$$
+2^{100!} > (2^{100})!
+$$
+
+## Michael’s Method (Xiao 2024)
+
+Ngl, I think Michael make the best argument out of us all...
+
+For \(a \in \mathbb{Z}^+\), \(a > 6\), we know:
+
+\[
+\begin{align*}
+    a! &> a \cdot 2^a, \\
+    2^{a!} &> 2^{a \cdot 2^a}, \\
+           &> (2^a)^{2^a}, \\
+           &= \underbrace{2^a \cdot 2^a \cdot \dots \cdot 2^a}_{2^a \text{ terms}}.
+\end{align*}
+\]
+
+Then there is:
+
+$$
+(2^a)! = \underbrace{2^a \cdot (2^a - 1) \cdot \dots \cdot 1}_{2^a \text{ terms}}
+$$
+
+Clearly:
+
+$$
+2^{a!} > 2^{a \cdot 2^a} > (2^a)!
+$$
+
+## Alex’s Method (Li and Cheung 2024)
+
+$$
+100! \sim \sqrt{2\pi \cdot 100} \left(\frac{100}{e}\right)^{100} \sim 100^{100}
+$$
+
+And:
+
+$$
+(2^{100})! \sim \sqrt{2\pi \cdot 2^{100}} \left(\frac{2^{100}}{e}\right)^{2^{100}}
+$$
+
+Comparing growth rates, it's evident:
+
+$$
+2^{100!} > (2^{100})!
+$$
+
+
+
+## Appendix
+
+```python
+from sympy import pi, log, E
+
+expression = (1/2) * (log(2*pi, 2) + 100) + 2**100 * (100 - log(E, 2))
+evaluation = expression.evalf()
+print(evaluation)
+```

posts/entries/random.md

@@ -10,15 +10,15 @@ While reading the Wikipedia article about metric spaces just now, it was talking
 
 These are not really on the topic, what I want to say today is about the essential part of a linear transformation, and I figured out the proof and purpose just after looking over it in terms of domains and codomains. What I understand here is that
 
-$$
-A x = A(\proj{\ran A^*}x + ( x - \proj{\ran A^*}x ) ) = A \proj{\ran A^*} x
-$$
+\[
+Ax = A\left(\proj{\ran A^* } x + \left(x - \proj{\ran A^* } x \right)\right) = A\left(\proj{\ran A^* } x\right)
+\]
 
-just like doing an orthogonal decomposition on $x$, that 
+This uses orthogonal decomposition of $x$, where:
 
-$$
-x - \proj{\ran A^*}x \in (\ran A^*)^{\perp} = \ker A
-$$
+\[
+x - \proj{\ran A^* } x \in \left(\ran A^* \right)^\perp = \ker A
+\]
 
 so that 
 

posts/metadata/entries.json

@@ -3,7 +3,7 @@
         "title": "First Post",
         "date": "2025-03-30",
         "author": "R",
-        "location": "New York",
+        "location": "New York, NY",
         "tags": ["cs", "life"],
         "slug": "first"
     }
@@ -12,7 +12,7 @@
         "title": "Random stuff in linear algebra",
         "date": "2025-04-02",
         "author": "R",
-        "location": "New York",
+        "location": "New York, NY",
         "tags": ["math"],
         "slug": "random"
     }
@@ -21,7 +21,7 @@
         "title": "Seminar - Opensource",
         "date": "2025-04-04",
         "author": "R",
-        "location": "New York",
+        "location": "New York, NY",
         "tags": ["notes", "cs"],
         "slug": "seminar-opensource"
     }
@@ -30,8 +30,17 @@
         "title": "Seminar - Insights from the Financial Industry",
         "date": "2025-04-30",
         "author": "R",
-        "location": "New York",
+        "location": "New York, NY",
         "tags": ["notes", "finance"],
         "slug": "seminar-finance"
     }
+    ,
+    {
+        "title": "A Glimpse into Googology",
+        "date": "2024-02-14",
+        "author": "R",
+        "location": "St. Catharines, ON",
+        "tags": ["math"],
+        "slug": "googology"
+    }
 ]