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- Why is $1^{\\infty}$ considered to be an indeterminate form
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
- Proof that $(AA^{-1}=I) \\Rightarrow (AA^{-1} = A^{-1}A)$
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- Why is $1$ not a prime number? - Mathematics Stack Exchange
actually 1 was considered a prime number until the beginning of 20th century Unique factorization was a driving force beneath its changing of status, since it's formulation is quickier if 1 is not considered a prime; but I think that group theory was the other force
- Series expansion: $\frac{1}{(1-x)^n}$ - Mathematics Stack Exchange
What is the expansion for $(1-x)^{-n}$? Could find only the expansion upto the power of $-3$ Is there some general formula?
- I have learned that 1 0 is infinity, why isnt it minus infinity?
Thus the idea of $\frac{1}{0}$ can be interpreted as saying that if $\epsilon$ is infinitesimal then $\frac{1}{\epsilon}$ is infinite This resolves your problem because it shows that $\frac{1}{\epsilon}$ will be positive infinity or infinite infinity depending on the sign of the original infinitesimal, while division by zero is still undefined
- Double induction example: $ 1 + q + q^2 - Mathematics Stack Exchange
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- If $A A^{-1} = I$, does that automatically imply $A^{-1} A = I$?
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 basis vectors but after apply A-1 we again reach to the plane have basis i ^ (0,1) and j ^ (1,0) It means that after applying A-1 we reach to the transformation which does nothing
- 有问题，就会有答案 - 知乎
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