Activity for Snoopy
| Type | On... | Excerpt | Status | Date |
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| Edit | Post #293104 |
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— | 5 months ago |
| Edit | Post #293104 |
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— | 5 months ago |
| Edit | Post #293104 | Initial revision | — | 6 months ago |
| Answer | — |
A: How to validate if the horizontal and vertical tangent lines exist for implicit functions? There are quite a few imprecise uses of terminology that could lead to confusion. Addressing these issues may help clarify the various questions raised. The first imprecise use of terminology appears early in the post: > Given an implicit function $x^2 + y^3 - 15xy = 0$, ... While the equati... (more) |
— | 6 months ago |
| Edit | Post #292410 |
Post edited: formatting for readability |
— | 9 months ago |
| Suggested Edit | Post #292410 |
Suggested edit: formatting for readability (more) |
helpful | 9 months ago |
| Comment | Post #292268 |
The hint provided is very sketchy. It indeed needs further explanation about the reduction, which I addressed at the beginning of my answer below. (more) |
— | 9 months ago |
| Edit | Post #292269 |
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— | 10 months ago |
| Edit | Post #292269 |
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— | 10 months ago |
| Edit | Post #292269 |
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— | 10 months ago |
| Edit | Post #292268 |
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| Edit | Post #292268 |
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— | 10 months ago |
| Edit | Post #292269 |
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— | 10 months ago |
| Edit | Post #292268 |
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— | 10 months ago |
| Edit | Post #292269 |
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— | 10 months ago |
| Edit | Post #292269 |
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— | 10 months ago |
| Edit | Post #292269 |
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— | 10 months ago |
| Edit | Post #292269 | Initial revision | — | 10 months ago |
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A: Holomorphic function on a connected bounded open subset of the complex plane First, as noted in the problem statement, we should explicitly assume that $\Omega$ is connected. Now, let’s proceed with the reduction suggested in the hint, specifically assuming $z0 = 0$. If $z0\ne 0$, we can consider the holomorphic function $\psi: \tilde{\Omega} \to \tilde{\Omega}$ where ... (more) |
— | 10 months ago |
| Edit | Post #292268 | Initial revision | — | 10 months ago |
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Holomorphic function on a connected bounded open subset of the complex plane > Problem. Let $\Omega$ be a bounded open subset of $\mathbb{C}$, and $\varphi: \Omega \rightarrow \Omega$ a holomorphic function. Prove that if there exists a point $z0 \in \Omega$ such that > $$ \varphi\left(z0\right)=z0 \quad \text { and } \quad \varphi^{\prime}\left(z0\right)=1 $$ > then... (more) |
— | 10 months ago |
| Edit | Post #292233 |
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— | 10 months ago |
| Edit | Post #292234 |
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— | 10 months ago |
| Edit | Post #292232 |
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— | 10 months ago |
| Edit | Post #292234 | Initial revision | — | 10 months ago |
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A: Does the series $\sum_{n=1}^{\infty} \frac{n^n}{n!} e^{-n}$ converge? If unfamiliar with Stirling's formula, one can still get a rough approximation for $n!$ using elementary calculations. The basic idea is to consider $\log(n!)$, which converts products into a sum that can then be analyzed using integrals. Observe that: $$ \log(n!) = \sum{m=1}^n\log m $$ This s... (more) |
— | 10 months ago |
| Edit | Post #292233 |
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— | 10 months ago |
| Edit | Post #292233 |
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— | 10 months ago |
| Edit | Post #292233 | Initial revision | — | 10 months ago |
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A: Does the series $\sum_{n=1}^{\infty} \frac{n^n}{n!} e^{-n}$ converge? The basic approach to determining the convergence of a series involves comparing it to a simpler series using either the direct comparison test or the limit comparison test. The latter indicates when a complicated summand can be substituted with a simpler one for easier analysis. Stirling's formul... (more) |
— | 10 months ago |
| Edit | Post #292232 | Initial revision | — | 10 months ago |
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Does the series $\sum_{n=1}^{\infty} \frac{n^n}{n!} e^{-n}$ converge? Problem: Does the series $\sum{n=1}^{\infty} \frac{n^n}{n!} e^{-n}$ converge? Note: In many calculus textbook examples, series involving the factorial term $n!$ are typically analyzed using the ratio test. These exercises often skip a detailed examination of how rapidly $n!$ grows. However, the ... (more) |
— | 10 months ago |
| Edit | Post #292197 |
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— | 10 months ago |
| Edit | Post #292197 | Initial revision | — | 10 months ago |
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A: Find the value of $\sum_{k=1}^\infty\frac{k^2}{k!}$ Another way to solve this problem is to exploit the context of power series, particularly term-by-term differentiation. Observe that: $$ e^z=\sum{k=0}^{\infty} \frac{z^k}{k!}, \quad e^z=\left(e^z\right)^{\prime}=\sum{k=1}^{\infty} k \cdot \frac{z^{k-1}}{k!}, \quad e^z=\left(e^z\right)^{\prime \... (more) |
— | 10 months ago |
| Edit | Post #292196 | Initial revision | — | 10 months ago |
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A: Find the value of $\sum_{k=1}^\infty\frac{k^2}{k!}$ One way to do it is based on the fact that $\displaystyle \sum{k=0}^\infty \frac1{k!} = e$ and one can do ''change of variable'' in series. Notice that \begin{align} \sum{k=1}^\infty\frac{k^2}{k!} &= \sum{k=1}^\infty\frac{k}{(k-1)!}&&\text{(cancellation)}\\ &=\sum{n=0}^\infty\frac{n+1}{n!}&&\... (more) |
— | 10 months ago |
| Edit | Post #292195 |
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— | 10 months ago |
| Edit | Post #292195 | Initial revision | — | 10 months ago |
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Find the value of $\sum_{k=1}^\infty\frac{k^2}{k!}$ Problem. Find the value of $\displaystyle \sum{k=1}^\infty\frac{k^2}{k!}$. Note: For most convergent series, proving convergence significantly differs from calculating their values, if it is even possible. One has tools such as convergence tests to determine whether a given series converges. On... (more) |
— | 10 months ago |
| Edit | Post #290537 |
Post edited: improve wording |
— | 10 months ago |
| Edit | Post #290538 |
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— | 11 months ago |
| Edit | Post #290528 |
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— | 11 months ago |
| Edit | Post #290527 |
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— | 11 months ago |
| Edit | Post #291938 | Initial revision | — | 11 months ago |
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A: Why does the method of separating variables work? You may be interested in reading these excellent lecture notes on the method of separation of variables by Steven Johnson. Quick Overview of Key Takeaways The notes highlight two main points: - The method of separation of variables is applicable primarily in scenarios characterized by symme... (more) |
— | 11 months ago |
| Edit | Post #290571 |
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— | about 1 year ago |
| Edit | Post #290570 |
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— | about 1 year ago |
| Edit | Post #290642 |
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— | over 1 year ago |
| Edit | Post #290643 |
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— | over 1 year ago |