
What does $QAQ^{-1}$ actually mean? - Mathematics Stack …
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 …
What is the value of $1^i$? - Mathematics Stack Exchange
2010年8月30日 · There are infinitely many possible values for $1^i$, corresponding to different branches of the complex logarithm. The confusing point here is that the formula $1^x = 1$ is …
Formal proof for $(-1) \\times (-1) = 1$ - Mathematics Stack …
The Law of Signs $\rm\: (-x)(-y) = xy\:$ isn't normally assumed as an axiom. Rather, it is derived as a consequence of more fundamental Ring axioms $ $ [esp. the distributive law $\rm\,x(y+z) …
If $A A^{-1} = I$, does that automatically imply $A^{-1} A = I$?
2020年3月30日 · A-1 A means that first we apply A transformation then we apply A-1 transformation. When we apply A transformation we reach some plane having some different …
power series - How can we know the answer to 1-1+1-1+1 ...
Summing to infinity cannot truly be evaluated as a sum as it is a process of addition that never ends, rather than a final result. What we can say fairly categorically is that the only acceptable …
Why is $1/i$ equal to $-i$? - Mathematics Stack Exchange
2015年5月11日 · There are multiple ways of writing out a given complex number, or a number in general. Usually we reduce things to the "simplest" terms for display -- saying $0$ is a lot …
Formula for $1^2+2^2+3^2+...+n^2$ - Mathematics Stack …
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Proof that $(AA^{-1}=I) \\Rightarrow (AA^{-1} = A^{-1}A)$
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 …
abstract algebra - Prove that 1+1=2 - Mathematics Stack Exchange
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. …
General term formula of series 1/1 + 1/2 + 1/3 ... +1/n
$$\ln(n+1)\le\sum_{i=1}^n\frac1i\le\ln(n)+1$$ This is a rather tight upper limit and lower limit you can use to approximate your answer. One could also note that …