Activity for Carefree Explorerâ€
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Edit | Post #287063 | Initial revision | — | about 2 years ago |
| Question | — |
Show that $\forall n \in \mathbb{Z}^{+}$, $25^n \equiv 25 \bmod{100}$. > Show that $\forall n \in \mathbb{Z}^{+}$, $25^n \equiv 25 \bmod{100}$. This was a simple observation I made when playing around and I came up with the following proof: > It follows from $(10a + 25)^2 = 100a^2 + 500a + 625 = 100(a^2 + 5a) + 625$ that any number ending in $25$ raised to a p... (more) |
— | about 2 years ago |
| Comment | Post #286908 |
Some insightful comments; thank you. You may want to consider moving some of it into your answer. (more) |
— | over 2 years ago |
| Comment | Post #286908 |
Thank you for the comments on proof writing; I will certainly keep them in mind. For inducting over the reals, I tried to show that $P(0)$ holds and that $\forall x \\; P(x) \implies P(x \pm \lim_{\epsilon \to 0}\epsilon)$ where $P(x): e^x \ge x + 1$ and $\lim_{\epsilon \to 0}\epsilon$ represents som... (more) |
— | over 2 years ago |
| Edit | Post #286907 |
Post edited: |
— | over 2 years ago |
| Edit | Post #286907 | Initial revision | — | over 2 years ago |
| Question | — |
Prove $e^x \ge x+1 \\\; \forall x \in \mathbb{R}$ using induction > (How) can we prove $e^x \ge x+1 \\; \forall x \in \mathbb{R}$ using induction (without using the derivative of $e^x$ at any stage)? Comments on my attempt are appreciated. I stumbled across a very nice proof of $\frac{\mathrm{d}}{\mathrm{d}x}e^x = e^x$ that uses the identity $e^x \ge x+1$. Br... (more) |
— | over 2 years ago |
| Comment | Post #286848 |
Thank you for the feedback. I have edited my post with clarifications. (more) |
— | over 2 years ago |
| Edit | Post #286848 |
Post edited: |
— | over 2 years ago |
| Edit | Post #286848 | Initial revision | — | over 2 years ago |
| Question | — |
Is there a way to encode a unique arrangement of vertices of a graph with a unique short word? I call graphs $G1$ and $G2$ distinct iff (i) $G1$ has a different arrangement1 of vertices than $G2$ and (ii) $G1$ and $G2$ have the same number of vertices. All other properties of $G1$ and $G2$ have no effect on distinguishing them. For example, $G1$ and $G2$ may be the same even if the number of e... (more) |
— | over 2 years ago |
| Comment | Post #286832 |
Thank you, that’s insightful. (more) |
— | over 2 years ago |
| Comment | Post #286833 |
I wish I could upvote this a 100 times :) (more) |
— | over 2 years ago |
| Edit | Post #286829 | Initial revision | — | over 2 years ago |
| Question | — |
Show that $f(x) = \arctan\left(\frac{x}{x+1}\right) + \arctan\left(\frac{x+1}{x}\right) = \frac{\pi}{2}$ > Show that $$f(x) = \arctan\left(\frac{x}{x+1}\right) + \arctan\left(\frac{x+1}{x}\right) = \frac{\pi}{2} \quad \forall x \in (-\infty, -1)\large\cup (0, \infty)$$ I observed this feature graphically, but am looking for a way to prove it. My attempt is as follows: > We can show that $$\arc... (more) |
— | over 2 years ago |