Difference between revisions of "Logarithm"

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<strong>Proposition:</strong> $\displaystyle\int \log(z) dz = z \log(z)-z$
 
<strong>Proposition:</strong> $\displaystyle\int \log(z) dz = z \log(z)-z$
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<strong>Proof:</strong> █
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<strong>Theorem:</strong> For $|z|<1$,
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$$\log(1+z) = -\displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^k z^k}{k}.$$
 
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<strong>Proof:</strong> █  
 
<strong>Proof:</strong> █  
 
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Revision as of 06:50, 11 February 2015

The logarithm is defined by the formula $$\log(x) = \displaystyle\int_1^x \dfrac{1}{t} dt.$$

Properties

Proposition: $\displaystyle\int \log(z) dz = z \log(z)-z$

Proof:

Theorem: For $|z|<1$, $$\log(1+z) = -\displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^k z^k}{k}.$$

Proof: