real analysis – Mathemafrica

Convex functions

What I learnt in class today:

A convex function $f:\mathbb R\to\mathbb R$ is defined as satisfying

$f(\lambda x + (1-\lambda )y)\leq \lambda f(x)+(1-\lambda )f(y) \quad \forall x,y\in \mathbb R,\ \forall \lambda \in [0, 1].$

Thus, the shape of a convex function is like $\smallsmile$ . An example of a convex function is f(μ)=μ²:

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By Aidan Horn| August 15th, 2016|Level: Simple, Undergraduate|0 Comments

The confusion about discontinuity

Whilst reading Mícheál Ó Searcóid’s book Metric Spaces, I found out about a nuance in the definition of continuity that I was not previously aware of, and something which may be taught incorrectly at high schools. M. Searcóid states that a function such as tan(x) is continuous (read the page here). The definition of continuity at a point is based on the fact that the function has a value at that point (if a function is continuous at x = a, then f(a) has a value in the expression |f(x)-f(a)|<ϵ). However, following M. Searcóid’s line of thought about continuous functions, it does not make sense to consider points at which a function is not defined. If we were asked to prove that the function is ‘discontinuous’ at a point, we would need to show that the condition for continuity at that point is false. And the negation* of the condition for continuity would not make sense at a point where the function value is not defined.…

By Aidan Horn| July 21st, 2016|Courses, Level: intermediate|2 Comments

The domain of a composite function

In this article I outline a systematic way of finding the domain of a composite function. A definition that can be used for this purpose follows:

$D(f \circ g) = \{x|x\in D(g) \wedge g(x) \in D(f)\}$

(Vaught, 1995:18)

Where $D(\lambda) =$ $\text{the domain of }\lambda$

The explanatory method which follows is to show how to use this definition in different examples.

Example 1

Solve $D(\ln(\ln(\ln x)))$ .

Solution:

Let $\ln(\ln(\ln x)) = f(g(h(x)))$

$\begin{minipage}{3in} \begin{align*} \text{Firstly, }& D(g\circ h) = \{x|x\in (0,\infty)\wedge \ln x\in (0,\infty)\} \\ & \ln x\in (0,\infty) \Leftrightarrow x\in (1,\infty) \\ \therefore \; \; & D(g\circ h) = x \in (1, \infty) \\ ~\\ \text{Now, }& D(f \circ (g\circ h)) = \{x|x\in (1,\infty)\wedge \ln(\ln x)\in (0,\infty)\} \\ & \ln(\ln x)\in (0,\infty) \Leftrightarrow \ln x\in (1,\infty) \Leftrightarrow x \in (e,\infty) \\ \therefore \; \; & D(f \circ g\circ h) = x \in (e, \infty) \end{align*} \end{minipage}$

$\square$

Example 2.1

Let $f(x)=x+1$ and $g(x)=x^2$ where $D(g)=[-2,2]$ .

Find $D(f\circ g)$

Solution:

$\begin{minipage}{2in} \begin{align*} f\circ g(x) &= x^2+1 \\ D(f\circ g) &= \{x|x \in [-2,2] \wedge x^2 \in \mathbb{R} \} \\ & x^2 \in \mathbb{R} \Leftrightarrow x \in \mathbb{R} \\ \therefore \; \; & x \in [-2,2] \end{align*} \end{minipage}$

$\square$

Example 2.2

Consider the same constraints as in Example 2.1, but with $D(f)=[-2,1]$

Solution:

$\begin{minipage}{3in} \begin{align*} D(f\circ g) &= \{x|x \in [-2,2] \wedge x^2 \in [-2,1] \} \\ & x^2 \in [-2,1] \Leftrightarrow x^2 \in [0,1] \Leftrightarrow x \in [-1,1] \\ \therefore \;\; & x \in [-1,1] \end{align*} \end{minipage}$

$\square$

References

Vaught, RL. 1995. Set theory: An introduction. 2nd edition. Boston: Birkhäuser.

LaTeX and PDF format here

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By Aidan Horn| March 4th, 2015|English, Level: intermediate|0 Comments