Showing a root exists formalism

Formalising

Basically, when doing numerical methods proof of a root in an equation; for example:

"$f(x) = 2\sin (x^2) + x -2$

Show that $f(x) = 0$ has a root $\alpha$, between $x = 0.75$ and $x = 0.85$"

There are problems I have with the method that aren't really mentioned. For an interval $[a, b]$ the fact that there is a change in sign of a function from one bound to the other, in no way implies that a root exists. I brought this up to my maths teacher with the example of $f(x) = \frac{1}{x}$ and he said, well the function has to be continuous -- which made more sense. However, there are still cases in which a function can be discontinuous across an interval and yet a change of sign still correctly implies that a root exists. As well, shouldn't we state an assumption that the function is continuous within the interval when answering the question, if this is the case?

For example, the graph of $f(x) = \frac{x+1}{(x-1)(x-2)}$
From -5 to 0, the function is continuous and the change in sign implies a root of f(x) = 0; which turns out to be true. From 0 to 1.5, the change in sign does not imply a root because it is not continuous -- fits in with my teacher's statement. From -5 to 5, however, the change in sign correctly implies a root, yet the function is not continuous... so what ARE the requirements to correctly imply a root?

What I typically say is:There is a change in sign across the interval, therefore a root must exist in the interval.But despite getting the marks, it annoys me because it isn't true. So, how could I formally answer the question, using complicated maths notation [because it looks cool], that is rigorously true?

And I know the markers probably won't care; It's for my own peace of mind

Neatening it up

It's especially painful when they ask to show, for example, that $\alpha = 0.23512$ [not relating to the question above] is correct to 5 decimal places... then I have to write out 7 digit numbers over and over; it's a petty pet peeve but I'm all about efficiency. So, is there another way I can refer to the interval's lower\upper bound without having to keep using the numerical value over and over again, or would it be fine denoting my own notation at the start?

e.g. $\mathrm{Let\ }s=0.75,t=0.85\mathrm{\ and\ }I=[s,t]$, then I can atleast recite $s$ and $t$ instead of lengthy values

Finally, is there a way to indicate that there is a change in sign, that is concise and mathematical, instead of writing, "There is a change in sign across the interval therefore a root must exist"

I could say that $sgn(f(0.75)) \neq sgn(f(0.85))$ or I could say $f(0.75)\cdot f(0.85) < 0$, but technically speaking, couldn't there exist a root at x = 0.85 or x = 0.75, since the question only asks to show that a root lies 'between' these boundaries; in which case, the product would equal 0, and the sign would be undefined.

Thanks for any help if you actually managed to trawl through all of this and you're still alive.

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(sign change $\land$ function continuous on interval) $\Rightarrow$ root in interval.
Continuity is usually trivial so is often omitted as you say. You can use sgn notation if you want, or you can just write sign change.

continuous within the interval.... so obvious god dammit.

Am I right in reading that as:
If there's a change in sign and the function is continuous on the interval then there is a root in the interval? I've never seen the $\land$ notation before, is all

Yes.

...so what ARE the requirements to correctly imply a root?...

Ok, so I've written this, it hopefully conveys what I'm trying to say beneath it, does it make sense? *crosses fingers*

$\mathrm{for\ an\ interval}, I = [a, b] : f(a), f(b) \neq 0[br][br]( \lvert\left[ \mathrm{sgn}(f(x)) \right]_b^a\rvert = 2\ \land\ f:I \mapsto \mathbb{R}) \Rightarrow ( \alpha\in I: f(\alpha) = 0)$

For an interval I, from [and including] a to b, given f(a) and f(b) don't equal 0:
If the magnitude of the sign of f(a) minus the sign of f(b) is equal to 2 AND every value along the interval I maps to a real number, then there exists an $\alpha$ within the interval I such that f($\alpha$) is equal to 0.

$|\left[\mathrm{sgn}(f(x))\right]_b^a| = 2$ is right, but isn't very clear. You want whoever's reading your work to be able to understand it as easily as possible. $\mathrm{sgn}(f(a)) \neq \mathrm{sgn}(f(b))$ is much clearer.

$f:I \mapsto \mathbb{R}$ is incorrect use of $\mapsto$. It should be $\rightarrow$. Also, this statement is not equivalent to continuity on the interval, for example the sgn function satisfies this on [-1,1] and is not continuous on that interval.

There should also be a statement of existence on the RHS.

eg.
$(\mathrm{sgn}(f(a)) \neq \mathrm{sgn}(f(b)) \land f \text{ continuous on } [a, b]) \Rightarrow \exists \alpha \in [a, b] : f(\alpha) = 0$

I was reading from the Wikipedia page that a function is continuous if every value of x maps to a real number, and it used that/similar notation, as well as the fact that the intermediate value theorem is only true if the function is continuous. But I recently realised that a piecewise function could still easily be continuous, by this definition, and yet the IMT couldn't possibly be true -- so I kind of got the gist it wasn't right. I guess I'll read more up on continuity and differentiability next time :P

I have no idea why I didn't think of that way of representing the change in sign xD but thanks

What's the difference between the $\mapsto$ and the other thin arrow notations? I haven't been taught any of this so I just assumed they all meant the same thing!? And am I right in thinking that $\{ A:B\} \equiv \{ A | B \}$ (the : and | notation mean the same, or are they slightly different?)

~snip~
Yes, this is over the top, but I loooooove it.

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Why make life more complicated than it needs to be? People have enough trouble with Analysis without trying to make it look like someone spilled a bunch of arcane typography over the page.

...I probably look retarded from your perspective, most likely with multiple errors I've made, limping around trying to look fancy with notation and crap...

I'd sit here and beg people to teach me formal logic, but somehow I think people do not want to do that; instead I'm just using this stuff and trying to get the gist of how it works from the graduates/professors on here.

The fact that you've called it 'horrible' makes me think I'm better off watching lectures.

That definition isn't right.
The (limit) definition of continuity of a function of a point is that f(x) is continuous at x=a iff $\displaystyle\lim_{x\rightarrow a} f(x) = f(a)$.
A function f is continuous on an interval I iff f is continuous at all $a \in I$.
You're right, the IVT doesn't work for the definition you've found.

Piecewise functions can be continuous, just few of them will be.

$\rightarrow$ is used to specify the domain and codomain of a function, eg. $f: \mathbb{R} \rightarrow \mathbb{C}$(it's also used for limits in an unrelated way). $\mapsto$ is used to give the rule to get from a specific value in the domain to a value in the codomain, eg. $x\mapsto \sqrt{x}$. Putting these two together, we have a full function definition, $f: \mathbb{R} \rightarrow \mathbb{C},x\mapsto \sqrt{x}$.
| is the same as :. Sometimes one is preferred for example when you have a lot of modulus signs.

That makes much more sense! it seems I initially got the gist of what continuity was, then taught myself that it was something else and rammed the initial idea out of my own head but thanks for clearing that up

Edit: Just to clarify, is it equivalent/does that notation inherently carry that $\displaystyle\lim_{x \rightarrow a} = f(a) \Rightarrow \displaystyle\lim_{x \rightarrow a^+} = \displaystyle\lim_{x \rightarrow a^-}$?
(is the lim(x -> a) equivalent to saying, the limit from both sides)

That implication is literally true, yes, but note that continuity is an even stronger condition because it tells us that our function is defined at a point 'a' as well as the limits of f(x) approaching a from both sides being equal to each other.

(I'm pretty sure you've got this, but I'm posting in case (a) you haven't (!) and (b) for the benefit of anyone trying to decipher some of the 'horrible' formalism in this thread who isn't clear on the definition themselves )

The best way to get to grips with a lot of these ideas is to make up your own examples and see how far you can get with them by adjusting the definitions slightly - making sure you start with simple examples that aren't obscured by notation!

Then the limit of g(x) as x approaches 0 from either side is 0, but g(x) isn't continuous there because we've defined g(0) to have some random value not equal to the (two-sided) limit.