2013年1月15日 · The main reason that it takes so long to get to $1+1=2$ is that Principia Mathematica starts from almost nothing, and works its way up in very tiny, incremental steps. The work of G. Peano shows that it's not hard to produce a useful set of axioms that can prove 1+1=2 much more easily than Whitehead and Russell do.

$$\ln(n+1)\le\sum_{i=1}^n\frac1i\le\ln(n)+1$$ This is a rather tight upper limit and lower limit you can use to approximate your answer. One could also note that $$\sum_{i=1}^n\frac1i=\int_0^1\sum_{i=0}^{n-1}x^i\ dx=\int_0^1\frac{1-x^n}{1-x}\ dx$$ We also have the Euler-Maclaurin expansion:

Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

The indeterminate forms are often abbreviated with stuff like "$1^\infty$" but that's not what they mean. This "$1^\infty$" (in regards to indeterminate forms) actually means: when there is an expression that approaches 1 and then it is raised to the power of an expression that approaches infinity we can't determine what happens in that form.

2020年4月28日 · Let me provide some context. I was specifically looking at the application of linear algebra to the stress tensor $\sigma$ and how we get the transformed stress state in some rotated coordinate system by the following multiplication: $\sigma'=Q\sigma Q^{-1}(Q$ is a rotation matrix defined by the angle we want to rotate the axes).

Total result is (using inclusion-exclusion principle): $$ 6^{3n+1} - 3\cdot 3^{3n+1} + 3\cdot 1^{3n+1}.$$ Edit 1: I also assume that each person has to participate somewhere. As deinst pointed out, it may not be the case, and then the result is slightly different (see his comment ).

If we take the set of real numbers, and look at "raise to the power of", then $1^x$ is equal to 1 for any x, as x -> infinity. So in that case, you could have a convention of saying that $1^\infty = 1$. But $\frac{1}{1} = 1$, so $1^{-\infty}$ would also equal 1. However, when you go about defining these new conventions, you have to be extremely ...

2021年2月3日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.