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This is simple math but only taught in A-level (with proof by induction). Hint: Watch free Khan Academy Math lecture to learn more …. Note: If is changed to , this is Euler’s “Basel Problem” which leads to the unsolved Riemann Hypothesis. Below is a 400-year-old walnut tree: ...
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Proving Statements About All Natural Numbers Induction comes in many flavors, but the goal never changes. We use induction when we want to prove something is true about all natural numbers. These statements will look something like this: For all natural numbers n, 1+2+⋯+n=n(n+1)/2. ...
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So we can start in the critical strip and remain bounded by induction. Thus [itex]h[/itex] is entire and bounded. By Liouville's theorem, [itex]h[/itex] is constant. Since [itex]h(1) = G(1)g(0) = 0[/itex] we must have [itex]g=0[/itex], hence [itex]F(z) \equi...
If the top coefficient of is rational, say , then by partitioning the natural numbers into residue classes modulo , we see that the claim follows from the induction hypothesis; so we may assume that the top coefficient is irrational. In order to use the van der Corput inequality as stated...
Cantor’s theory of well-orderings furnished him with transfinite induction; with this, you can show that the answer is “yes” for any well-ordered (or well-orderable) set. Conversely, if M̿ is an aleph, then M is well-orderable. Cantor firmly believed that every set can be well-...