Precalc Summary

Publish Date

This isn't completely related to computer science so I didn't include the tag :P. Functions in this article only have single variables.

Limit

limxa+f(x)=b\lim_{x \rightarrow a^{+}}{f(x)} = b

represents the fact that when xx approaches aa from right, ff approaches bb, which is called a "right limit".

limxaf(x)=b\lim_{x \rightarrow a^{-}}{f(x)} = b

represents the fact that when xx approaches aa from left, ff approaches bb, which is called a "left limit". When both of them are the same, that is

limxa+f(x)=limxaf(x)=b\lim_{x \rightarrow a^{+}}{f(x)} = \lim_{x \rightarrow a^{-}}{f(x)} = b

, we say that when xx approaches aa, ff approaches bb, which is

limxaf(x)=b\lim_{x \rightarrow a}{f(x)} = b

, this also shows that ff has a limit at x=ax=a.

Continuity

For a function to be continuous in a certain interval, it must first have limit in that interval. If f:RR,f(x)f: R \rightarrow R, f(x) is continuous given x[a,b]x \in [a,b], then it must satisfy the following facts:

  1. For all cc in [a,b][a,b], ff at x=cx=c must have a limit that is defined.
  2. For all cc in [a,b][a,b], f(c)=limxcf(x)f(c) = \lim_{x \rightarrow c}{f(x)}

Derivative

A function's derivative with respect to a specific variable is the function's change regarding the change of the specified variable. For example, the derivative of f:RR,f(x)f: R \rightarrow R, f(x) regarding xx is dfdx\frac{df}{dx}. It can be defined like this:

dfdx=limh0f(x+h)f(x)h\frac{df}{dx} = \lim_{h \rightarrow 0}{\frac{f(x+h) - f(x)}{h}}

Differentiable

When a function is differentiable in a certain interval, it must first be continuous in that interval. If f:RR,f(x)f: R \rightarrow R, f(x) is differentiable, then its derivative must be continuous, so finally we can say that

Differentiable    Continuous    Limit Existing\text{Differentiable} \implies \text{Continuous} \implies \text{Limit Existing}

IVT Intermediate Value Theorem

For a continuous function ff in interval [a,b][a,b], it must satisfy the following fact: c[min(f(a),f(b)),max(f(a),f(b))]\forall c \in [\min(f(a), f(b)), \max(f(a), f(b)) ], there exists at least one rr that fulfills f(r)=c{r[a,b]}f(r) = c \{r \in [a,b]\}.

Boundedness Theorem

For a function ff in interval [a,b][a,b], UU denotes the set of upper bounds (scalar) and LL denotes the set of lower bounds (scalar). Define v,v[a,b]v, \forall v \in [a, b],

lf(v)u,lL,uUl \leq f(v) \leq u, l \in L, u \in U

EVT Extreme Value Theorem

When we limit the upper bound set and the lower bound set to only include values that exist as ff's output, we get EVT, where LL and UU are each reduced respectively to a single value, f(c)f(c) and f(d)f(d).

Exponential Differentiation

Let ee be a number, bb be a constant scalar, that results in

f(x)=ebxdfdx=bebx=bf(x)f(x) = e^{bx} \\ \frac{df}{dx} = be^{bx} = b \cdot f(x)

, we can say that for all aR,a>0a \in R, a > 0,

f(x)=ax=eln(a)xdfdx=ln(a)eln(a)x=ln(a)f(x)f(x) = a^x = e^{\ln(a)x} \\ \frac{df}{dx} = \ln(a) \cdot e^{\ln(a)x} = \ln(a) \cdot f(x)

Logarithmic Differentiation

We know that

d(lnx)dx=1x\frac{d(\ln x)}{dx} = \frac{1}{x}

, so for function ff,

d(lnf(x))dx=d(lnf(x))dfdfdx=1f(x)dfdx\frac{d(\ln f(x))}{dx} = \frac{d(\ln f(x))}{df} \cdot \frac{df}{dx} = \frac{1}{f(x)} \cdot \frac{df}{dx}

. The above technique is actually chain rule if you look closely!

Linear Approximation

Given function f(x)f(x), the linear approximation (tangent lines) at (r,f(r))(r, f(r)) is y=f(r)(xr)+f(r)=g(x)y = f'(r)(x-r)+f(r) = g(x). (ff' is the derivative of ff)

Cheatsheet

(1+x)r1+rx where x0(1+x)^r \approx 1 + rx \text{ where } x \rightarrow 0 sin(x)x where x0\sin(x) \approx x \text{ where } x \rightarrow 0 cos(x)1 where x0\cos(x) \approx 1 \text{ where } x \rightarrow 0 ex1+x where x0e^x \approx 1 + x \text{ where } x \rightarrow 0 ln(1+x)x where x0\ln(1+x) \approx x \text{ where } x \rightarrow 0

Deduction

(1+x)r(1+0)r+r(1+0)r1x=1+rx(1+x)^r \approx (1+0)^r + r(1+0)^{r-1}x=1+rx sin(x)sin(0)+cos(0)x=x\sin(x)\approx \sin(0) + \cos(0)x = x cos(x)cos(0)+sin(0)x=1\cos(x)\approx \cos(0) + -\sin(0)x = 1 exe0+e0x=1+xe^x \approx e^0 + e^0 x = 1 + x ln(1+x)ln(1)+x1+0=x\ln(1+x) \approx \ln(1) + \frac{x}{1+0}=x

Quadratic Approximation

Think of a function f(x)=ax2+bx+cf(x)=ax^2+bx+c, we want to see its derivatives at x=0x=0:

f(0)=c,f(0)=b,f(0)=2af(0)=c, f'(0)=b, f''(0)=2a

which can represent f(x)f(x) as f(0)+f(0)x+f(0)x2(2!)1f(0) + f'(0)x + f''(0)x^2 \cdot (2!)^{-1}. Generalizing this observation, we can get that the quadratic approximation at x=rx=r is

f(x)f(r)+f(r)(xr)11!+f(r)(xr)22!f(x) \approx f(r) + \frac{f'(r)(x-r)^1}{1!} + \frac{f''(r)(x-r)^2}{2!}

The notation for quadratic approximation of function ff is Q(f)Q(f).

Approximation Laws

Q(f(x)g(x))=Q(Q(f(x))Q(g(x)))Q(f(x) \cdot g(x)) = Q(Q(f(x))\cdot Q(g(x)))

Taylor Series

It's not particularly difficult to see the pattern from above approximations. For function ff, its approximation at x=rx=r would be

f(x)=f(r)+n=1f(n)(x)(xr)nn!f(x) = f(r) + \sum^{\infty }_{n=1}{\frac{f^{(n)}(x) \cdot (x-r)^n}{n!}}

, which is called Taylor series.