Inverse expressions

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 16 June 2011

Inverse expressions

$\sqrt{x}$ is the expression that undoes $x^2$. This means that

$\sqrt{x^2}=x\text{and}{\left(\sqrt{x}\right)}^2=x.$

$\ln x$ is the expression that undoes $e^x$. This means that

$\ln \left(e^x\right)=x\text{and}{e}^{\ln x}=x.$

${\sin }^{-1}x$ is the expression that undoes $\sin x$. This means that

${\sin }^{-1}\left(\sin x\right)=x\text{and}\sin \left({\sin }^{-1}x\right)=x.$

${\cos }^{-1}x$ is the expression that undoes $\cos x$. This means that

${\cos }^{-1}\left(\cos x\right)=x\text{and}\cos \left({\cos }^{-1}x\right)=x.$

${\tan }^{-1}x$ is the expression that undoes $\tan x$. This means that

${\tan }^{-1}\left(\tan x\right)=x\text{and}\tan \left({\tan }^{-1}x\right)=x.$

${\cot }^{-1}x$ is the expression that undoes $\cot x$. This means that

${\cot }^{-1}\left(\cot x\right)=x\text{and}\cot \left({\cot }^{-1}x\right)=x.$

${\sec }^{-1}x$ is the expression that undoes $\sec x$. This means that

${\sec }^{-1}\left(\sec x\right)=x\text{and}\sec \left({\sec }^{-1}x\right)=x.$

${\csc }^{-1}x$ is the expression that undoes $\csc x$. This means that

${\csc }^{-1}\left(\csc x\right)=x\text{and}\csc \left({\csc }^{-1}x\right)=x.$

${\log }_{a}x$ is the expression that undoes $a^x$. This means that

${\log }_{a}x\left(a^x\right)=x\text{and}{a}^{{\log }_{a}x}=x.$

WARNING: ${\sin }^{-1}x$ is VERY DIFFERENT from ${\left(\sin x\right)}^{-1}$. For example,

${\sin }^{-1}0={\sin }^{-1}\left(\sin 0\right)=0,\text{BUT}{\left(\sin x\right)}^{-1}={\displaystyle \frac{1}{\sin 0}}={\displaystyle \frac{1}{0}},$

which is undefined.

Example. Explain why $\ln 1=0$. $$\ln 1=\ln \left(e^0\right)=0.$$

Example. Explain why $\ln \left(ab\right)=\ln a+\ln b$. $$\ln \left(ab\right)=\ln (e^{\ln a}\cdot e^{\ln b})=\ln \left(e^{\ln a+\ln b}\right)=\ln a+\ln b.$$

Example. Explain why $\ln \left({\displaystyle \frac{1}{a}}\right)=-\ln a$. $$\ln \left({\displaystyle \frac{1}{a}}\right)=\ln \left({\displaystyle \frac{1}{e^{\ln a}}}\right)=\ln \left(e^{-\ln a}\right)=-\ln a.$$

Example. Explain why $\ln \left(a^b\right)=b\ln a$. $$\ln \left(a^b\right)=\ln \left({\left(e^{\ln a}\right)}^b\right)=\ln \left(e^{b\ln a}\right)=b\ln a.$$

Thus

$e^0=1$	turns into	$\ln 1=0$,
$e^x{e}^y=e^{x+y}$	turns into	$\ln ab=\ln a+\ln b$,
$e^{-x}={\displaystyle \frac{1}{e^x}}$	turns into	$\ln {\displaystyle \frac{1}{a}}=-\ln a$,
${\left(e^x\right)}^y=e^{yx}$	turns into	$\ln \left(a^b\right)=b\ln a$.

Notes and References

These important proofs should be learned at the same time that the notations for inverse expressions are learned. These expressions exist and make sense without real numbers (in the context of elements of $\mathbb{Q}\left(\right(x\left)\right)$ and other similar settings). Convergence of these expressions on evaluation in topological groups and fields is quite a different issue from the identities that these expressions satisfy. The answers to the convergence on evaluation questions depends very heavily on the topological ring where the evaluation map is being applied.

References

[Bou] N. Bourbaki, Algebra II, Chapters 4–7 Translated from the 1981 French edition by P. M. Cohn and J. Howie, Reprint of the 1990 English edition, Springer-Verlag, Berlin, 2003. viii+461 pp. ISBN: 3-540-00706-7. MR1994218

[Mac] I.G. Macdonald, Symmetric functions and Hall polynomials, Second edition, Oxford University Press, 1995. ISBN: 0-19-853489-2 MR1354144

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