User:StradivariusTV/Temp

A geometric series $$\sum^{\infty} ar^{n-1}$$ converges to finite S iff $$\{ar^{n-1}\} \,$$ converges to 0. Then $$S = \frac{1}{1-r}$$

Use integration to find arc length:

$$\int_a^b \sqrt{(\mathrm{d}x)^2 + (\mathrm{d}y)^2}$$

$$\int_a^b \sqrt{(\mathrm{d}x)^2 \cdot [1 + (\mathrm{d}y)^2/(\mathrm{d}x)^2]}$$

$$\int_a^b \sqrt{1 + f'(x)^2} \ \mathrm{d}x$$

Derive some trigonometric identities from Euler's formula:

$$\mathrm{e}^{\mathrm{i}2\theta} = \cos2\theta + \mathrm{i}\sin2\theta \,$$

$$(\mathrm{e}^{\mathrm{i}\theta})^2 = (\cos\theta + \mathrm{i}\sin\theta)^2 \,$$

$$(\mathrm{e}^{\mathrm{i}\theta})^2 = \cos^2\theta + 2\mathrm{i}\sin\theta\cos\theta - \sin^2\theta \,$$

Now associate terms:

$$\cos2\theta = \cos^2\theta - \sin^2\theta \,$$

$$\sin2\theta = 2\sin\theta\cos\theta \,$$

$$0.999\ldots = \lim_{n\rightarrow\infty} \sum^n \left(\frac{9}{10}\right)\left(\frac{1}{10}\right)^{n-1} = \frac{\frac{9}{10}}{1-\frac{1}{10}} = \frac{\frac{9}{10}}{\frac{9}{10}} = 1$$

$$\mathrm{e}^{\mathrm{i}x} = \cos x + \mathrm{i}\, \sin x$$

$$\frac{\mathrm{d}}{\mathrm{d}x}(\mathrm{e}^{\mathrm{i}x}) = \mathrm{i}\, \mathrm{e}^{\mathrm{i}x} = \mathrm{i}\, \cos x + \mathrm{i}^2\, \sin x = -\sin x + \mathrm{i}\, \cos x$$

$$\frac{\mathrm{d}}{\mathrm{d}x}(\cos x) = -\sin x$$

$$\frac{\mathrm{d}}{\mathrm{d}x}(\sin x) = \cos x$$

$$u = 3\sqrt{x} + 4$$

$$\frac{\mathrm{d}^n y}{\mathrm{d}x^n} = (-1)^{n-1} \frac{n!}{(1-2x)^{n+1}}$$