Math analysis course cheatsheet
Trigonometric functions
| Angle (°) | Radians | sin(θ) | cos(θ) | tan(θ) | csc(θ) | sec(θ) | cot(θ) |
|---|---|---|---|---|---|---|---|
| 0° | 0 | 0 | 1 | 0 | ∞ | 1 | ∞ |
| 30° | π/6 | 1/2 | √3/2 | 1/√3 | 2 | 2/√3 | √3 |
| 45° | π/4 | √2/2 | √2/2 | 1 | √2 | √2 | 1 |
| 60° | π/3 | √3/2 | 1/2 | √3 | 2/√3 | 2 | 1/√3 |
| 90° | π/2 | 1 | 0 | ∞ | 1 | ∞ | 0 |
| 120° | 2π/3 | √3/2 | -1/2 | -√3 | 2/√3 | -2 | -1/√3 |
| 135° | 3π/4 | √2/2 | -√2/2 | -1 | √2 | -√2 | -1 |
| 150° | 5π/6 | 1/2 | -√3/2 | -1/√3 | 2 | -2/√3 | -√3 |
| 180° | π | 0 | -1 | 0 | ∞ | -1 | ∞ |
Inverse Functions
A function has an inverse if it is bijective (one-to-one and onto). To verify that a linear function has an inverse, we ensure it is injective (one-to-one). For example,
\[f(x) = kx + b\] \[kx_1 + b = kx_2 + b\] \[kx_1 = kx_2\] \[x_1 = x_2\]This confirms that $f(x)$ has an inverse, as it passes the injectivity test.
To find the inverse function:
\[y = kx + b\] \[x = ky + b \implies ky = x - b\] \[y = \frac{x - b}{k}\]Thus, the inverse of $f(x)$ is:
\[f^{-1}(x) = \frac{x - b}{k}\]Even and Odd Functions
A function is even if:
\[f(x) = f(-x)\]Even functions are symmetric about the $y$-axis. Examples include $f(x) = x^2$ and $f(x) = \cos(x)$.
A function is odd if:
\[-f(x) = f(-x)\]Odd functions are symmetric about the origin $(0,0)$. Examples include $f(x) = x^3$ and $f(x) = \sin(x)$.
Limits
Some important limits include:
\[\lim_{x \to 0} \frac{\sin x}{x} = 1\] \[\lim_{x \to \infty} \left(1 + \frac{k}{x}\right)^x = e^k\] \[\lim_{x \to 0} \frac{\cos x - 1}{x^2} = -\frac{1}{2}\] \[\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\]Derivatives
\[(c)' = 0\] \[(x^n)' = nx^{n-1}\] \[(\sqrt{x})' = \frac{1}{2\sqrt{x}}\] \[\left(\frac{1}{x}\right)' = -\frac{1}{x^2}\] \[(e^x)' = e^x\] \[(a^x)' = a^x \ln a\] \[(\ln x)' = \frac{1}{x}\] \[(\log_a x)' = \frac{1}{x \ln a}\] \[(\sin x)' = \cos x\] \[(\cos x)' = -\sin x\] \[(\tan x)' = \frac{1}{\cos^2 x} = \sec^2 x\] \[(\cot x)' = -\frac{1}{\sin^2 x} = -\csc^2 x\] \[(\arcsin x)' = \frac{1}{\sqrt{1 - x^2}}\] \[(\arccos x)' = -\frac{1}{\sqrt{1 - x^2}}\] \[(\arctan x)' = \frac{1}{1 + x^2}\] \[(\text{arccot} x)' = -\frac{1}{1 + x^2}\] \[(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)\] \[\left( \frac{f(x)}{g(x)} \right)' = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}\] \[\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)\]The derivative at point $ x_0 $ is given by:
\[f'(x_0) = \lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}\]The equation of the tangent line at the point $ (x_0, y_0) $ is:
\[y - y_0 = f'(x_0)(x - x_0)\]The equation of the normal line (perpendicular to the tangent) is:
\[y - y_0 = -\frac{1}{f'(x_0)}(x - x_0)\]…this can be rearranged to express $f(x)$ as follows:
\[f(x) \approx f(x_0) + f'(x_0)(x - x_0)\]Taylor’s Theorem
Taylor’s theorem states that a function $ f(x) $ can be approximated by a polynomial $ M_n(f, x) $ near a point $ x_0 $.
Taylor Polynomial
The Taylor polynomial of degree $ n $ around the point $ x_0 $ is given by:
\[M_n(f, x) = f(x_0) + f'(x_0)(x - x_0) + \frac{f''(x_0)}{2!}(x - x_0)^2 + \cdots + \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n\]Remainder Term
The remainder term $ R_n $ quantifies the error between the actual function and the Taylor polynomial:
\[R_n(x) = \frac{f^{(n + 1)}(c)}{(n + 1)!}(x - x_0)^{n + 1}\]where $ c $ is some point between $ x_0 $ and $ x $.
Complete Taylor Expansion
The complete Taylor expansion of $ f(x) $ around $ x_0 $ can be expressed as:
\[f(x) = M_n(f, x) + R_n(x)\]This shows that the function can be represented as its polynomial approximation plus the error term.