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John Stillwell - Naive Lie Theory.pdf - Index of

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7.1 Logarithm and exponential 141<br />

the same terms occur in the expansion <strong>of</strong> log(e x ),when|e x − 1| < 1, and<br />

their sum is zero because log(e x )=x under these conditions.<br />

Thus log(e X )=X as required.<br />

□<br />

The inverse property allows us to derive certain properties <strong>of</strong> the matrix<br />

logarithm from corresponding properties <strong>of</strong> the matrix exponential. For<br />

example:<br />

Multiplicative property <strong>of</strong> matrix logarithm. If AB = BA, and log(A),<br />

log(B), and log(AB) are all defined, then<br />

log(AB)=log(A)+log(B).<br />

Pro<strong>of</strong>. Suppose that log(A)=X and log(B)=Y ,soe X = A and e Y = B by<br />

the inverse property <strong>of</strong> log. Notice that XY = YX because<br />

(A − 1)2 (A − 1)3<br />

X = log(1 +(A − 1)) = (A − 1) − + −···,<br />

2 3<br />

(B − 1)2 (B − 1)3<br />

Y = log(1 +(B − 1)) = (B − 1) − + −···,<br />

2 3<br />

and the series commute because A and B do. Thus it follows from the<br />

addition formula for exp proved in Section 5.2 that<br />

AB = e X e Y = e X+Y .<br />

Taking log <strong>of</strong> both sides <strong>of</strong> this equation, we get<br />

log(AB)=X +Y = log(A)+log(B)<br />

by the inverse property <strong>of</strong> the matrix logarithm again.<br />

Exercises<br />

The log series<br />

log(1 + x)=x − x2<br />

2 + x3<br />

3 − x4<br />

4 + ···<br />

was first published by Nicholas Mercator in a book entitled Logarithmotechnia in<br />

1668. Mercator’s derivation <strong>of</strong> the series was essentially this:<br />

∫ x ∫<br />

dt x<br />

log(1 + x)=<br />

0 1 + t = (1 − t + t 2 − t 3 + ···)dt = x − x2<br />

0<br />

2 + x3<br />

3 − x4<br />

4 + ···.<br />

Isaac Newton discovered the log series at about the same time, but took the idea<br />

further, discovering the inverse relationship with the exponential series as well.<br />

He discovered the exponential series by solving the equation y = log(1 + x) as<br />

follows.<br />

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