math.txt

INTRO Math review

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Notation ($e$ is cool)

	In this course log or $\log_e$ means the {\em natural} logarithm (written {\em ln} in
	the textbooks)

		If I mean $\log_{10}$, I will say so

	exp() is a synonym for ``$e$ to the power of''

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Logarithms

	Multiplying on the original scale is equivalent to adding on the log
	scale

		$\log(ab) = \log(a) + \log(b)$

	Division is the opposite of multiplication and subtraction is the
	opposite of addition:

		$\bigeq \log(a/b) \equiv \log\left(\frac ab\right)
		= \log(a) - \log(b)$

	Exponentiation is to multiplication like multiplication is to
	addition:

		$\log(\lambda^k) = k\log(\lambda)$

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Complementarity

	$\log_e$ and exp are complementary functions

		$\exp(\log_e(x)) \equiv e^{\log_e(x)} = x$

		$\log_e(\exp(x)) \equiv \log_e(e^x) = x$

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Exponentiation

	The rules for exp are exactly complementary to those for log

		$\exp(a+b) = \exp(a) \cdot \exp(b)$

		$\bigeq \exp(a-b) = \exp(a) / \exp(b)$

		$\exp(k\log_e(\lambda)) = \lambda^k$

	Remember: you can't take the log or exponent of something with units

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Problems

	You should be able to use these rules to solve simple problems,
	algebraically, or on your calculator

		If $\lambda^{20} = 0.1$, what is $\lambda$?

	Also, word problems

		If a population declines by 90% in 20 generations, by 
		what proportion does it decline in one generation?

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Algebra

	If $\lambda^{20} = 0.1$, what is $\lambda$?

		ANS $20 \log_e (\lambda) = \log_e(0.1)$

		ANS $\log_e(\lambda) = \log_e(0.1)/20$

		ANS $\lambda = \exp(\log_e(0.1)/20)$

		ANS $\lambda=0.89$

	If $\exp(rt) = 5$, and $r=0.1/\yr$, what is $t$?

		ANS $rt = \log_e(5)$

		ANS $t = \frac{\log_e(5)}{0.1/\yr}$

		ANS $t = 16\yr$

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Units are our friends

	Keep track of units at all times

	Use units to confirm that your answers make sense

		Or to find quick ways of getting the answer

	Get used to manipulating and cancelling units

	36 mpg = ?? L/100km

		ANS
		$\bigeq \frac{\uname{gal}}{36 \uname{mi}} $ 
		$\bigeq  \cdot \frac{\uname{mi}}{1.6\uname{km}} $ 
		$\bigeq \cdot \frac{3.8\uname{L}}{\uname{gal}}$ 
		$\bigeq \cdot \frac{100\uname{km}}{100\uname{km}}$

		ANS 6.6 L/100km