feat: some updates

15e640-twitter_post_sqrt_root_limit/main.typ

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+Notes on this post:
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+$
+  1=lim_(n->0)sqrt(n+sqrt(n+sqrt(n+ ...)))
+$
+
+Some argues it's $0$ or $1$, that's simply because the form is ill-defined and the post is misleading intentionally, here's another way to look at this:
+
+define $f_m$ as a series of functions:
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+$
+  f_m (x) = underbrace(sqrt(x+sqrt(x+sqrt(x+...))), "m times")
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+
+In this form we have two ways to write the original limit:
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+$
+  lim_(n->0) lim_(m->oo) f_m (n)
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+
+or
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+  lim_(m->oo) lim_(n->0) f_m (n)
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+
+Limits are not interchangeable by default, and orders matter, when $display(lim_(n->0) f_m (n))$ comes first it's undoubtedly $0$, when $display(lim_(m->oo) f_m (n))$ comes first it's $1$.
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+Here's how $f_m (x)$ looks like with $m=1,2,3,4,5$: (Credit: WolframAlpha)
+
+#image("image.png")
+
+I hope this clears things up.

2bc0c8-2024_spring_TA/green.typ

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+integral.double_(D) v laplace u - u laplace v dif x dif y = integral.cont_(diff D) v (diff u)/(diff bold(n)) - u (diff v)/(diff bold(n)) dif s
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+
+$
+integral.double laplace u dif x dif y = integral.cont (diff u)/(diff bold(n)) dif s
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+
+$
+nabla = bold(e)_r (diff)/(diff r) + bold(e)_theta 1/r (diff)/(diff theta) + 1/(r sin theta) bold(e)_phi (diff)/(diff phi)
+$