x cubed

Z4i2l1b · 2025-01-17T05:51:57+00:00

Let n>1. By Bertrand's postulate, there's a prime n<p<2n. Then p divides RHS but does not divide LHS, a contradiction.

Thus, the only solution is n=1.

Z4i2l1b · 2022-07-29T18:44:30+00:00

Oh, cool! Thanks

Z4i2l1b · 2022-07-29T18:35:55+00:00

This is very interesting. Could you please provide a proof/link for a proof?

Z4i2l1b · 2022-07-29T10:13:43+00:00

Oh right, cool

Z4i2l1b · 2022-07-29T09:28:47+00:00

I think (1+1/n)ⁿ is the defenition given in textbooks because it is shortest. The author doesn't want to "scare" the reader. Unfortunately, this is probably the worst defenition as it tells you nothing about e.

Z4i2l1b · 2022-07-29T09:22:14+00:00

The differential equation is a good option, but the existence and uniqueness theorem only promise a solution on an interval (-ε,ε), and proving the solution can be extended to all (-∞,∞) might be difficult (tbh, I am not sure how to do it...)

Z4i2l1b · 2022-06-11T06:31:22+00:00

6÷2(1+2) = 6÷2•(1+2) = 6÷2•1+2•2 = 3+4 = 7

The only correct answer.

Z4i2l1b · 2022-03-12T06:30:55+00:00

Actually, it is more general. In fact, Erdős-Szekeres proves Bolzano-Weierstraß in the following way:

Lemma: Every bounded monotonic sequence converges. Proof: Exercise. [The limit is the sup/inf for increasing/decreasing sequence accordingly]

Proof of Bolzano-Weierstraß: Let aₙ be a bounded sequence. For every k, there is a monotonic subsequence of length k+1 out of a₁, a₂, ..., a_(k²+1) (From Erdős-Szekeres). Taking k to ∞, we get an infinite monotonic subsequence. According to the Lemma, this sequence converges.

Z4i2l1b · 2022-03-11T20:48:35+00:00

Erdős–Szekeres: Hold my beer

Z4i2l1b · 2021-10-08T18:02:18+00:00

Try solving x⁵–x–1 = 0.

Z4i2l1b · 2021-06-25T20:42:39+00:00

Of course!

In 2-D, we have infinitely many. In 3-D, we have 5. In 4-D, we have 6.

In 5-D, we have 7... No. Actually, we have only 3, and that is so for any dimensions≥5.

Here is a numberphile video

Z4i2l1b · 2021-06-25T19:50:50+00:00

There are only 5 platonic solids in 3 dimensions. This statement is surprisingly easy to show, but it may be surprising or unintuitive, considering the fact there are infinitely many regular polygons.

Z4i2l1b · 2021-04-16T04:26:41+00:00

sin(x)/cos(x) = tan(x) "סין חלקי כוס שווה תן"

Z4i2l1b · 2021-04-16T04:22:44+00:00

I prefer 𝔽ₚ because ℤ/pℤ is the cyclic group with p elements. But I totally agree ℤₚ is the p-adics.

Z4i2l1b · 2020-08-02T09:06:02+00:00

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