Differentiable functions

[void sky]

When proving that a function is differentiable can we always apply p->q and use the trivial conditional proof to show that p->q is always true regardless of the output of p when q is true? p refers to 0<| x-c| < delta and q is |f(x)-L|< epsilon.

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[void sky]

Differentiable functions

[frail igloo]

You cannot just assume q is true. You assume p is true and show that forces q to be true. If you want to think about it in terms of the truth table, you are showing the scenario where p->q would be false cannot occur.

[fallow horizon]

are you assuming the function is continuous?

[fallow horizon]

p->q is true if and only if either p is false or q is true

unless you know that p is true, you can't say more than that

[void sky]

but using the |f(x)-L| definition q can be true right?

[void sky]

yes

[void sky]

if it is can we use the same approach for any type of these questions?

[obsidian mountain]

@void sky

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[obsidian mountain]

@fallow horizon The user still needs help with this help request.

[turbid topaz]

you mentioned differentiability but introduced a definition of continuity. Perhaps youre alluding to the theorem that all diffable functions are continuous?

Anyway, your question is unclear. However, regarding what aL said:

p->q is true if and only if either p is false or q is true
This means that you can interchange these two forms of a logical implication:

• p -> q
• ~p v q
  since "v" is commutative, you also have:
• ~q -> ~p
  and this is very often used. this last one is called the contrapositive of the original statement p->q
  that means the following are equivalent:
• a function, if differentiable, is continuous
• a function, if not continuous, is not differentiable

[void sky]

Right I get it, but can we always use trivial conditional proof to show that a function is differentiable if we know either q is true or p is false?

[turbid topaz]

dont know what a trivial conditional proof is, or p, or q

[void sky]

P is 0<|x-c|<delta while q is |f(x)-L|< epsilon

From epsilon delta proofs

Trivial conditional proof just refers to how u use p->q in proofs

[turbid topaz]

okay. i don't know why you're calling it that. can you give me an example of the type of situation youre asking about? your question is not clear

[void sky]

Sure can I send my working when I get home? It's at home 😭

[turbid topaz]

yes

[void sky]

@turbid topaz this is what I meant

[turbid topaz]

Oh, you're asking, is there always a delta that can be chosen that is CONSTANT wrt to epsilon

the answer is, definitely not.

Also, your proof technique assumes that df/dx is continuous. Though that happens to be true in this case, it assumes more than what you need to prove differentiability

the easiest way to prove differentiability in exercises like this is just examine:

$\dfrac{f(x)-f(c)}{x-c} \text{ as } x \to c}$
or
$\dfrac{f(x+h)-f(x)}{h} \text{ as } h \to 0$

[noble shoal]

gfauxpas
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[turbid topaz]

let me look at your proof again actually, im not sure what you did or that i understood what you did correctly

scratch what I said at first

okay, revised answer:

a) if you are going to use an epsilon-delta approach, you cannot assume delta is constant WRT epsilon
b) but also, you dont need to use epsilon-delta for this kind of exercise, though you can

[frail igloo]

q can be true, but you can't assume q is true. The only case where p->q is false is when q is F and p is T. So you assume p is T and show that forces q to be T as well. Then you know p->q is T, because the only case where it can be F can't happen.

[reef steppe]

red circle: no

by writing f'(1), you are already assuming that f is differentiable at 1, but that is exactly what you want to prove

red square: just go straight to the point, the limit of (f(x) - f(1))/(x-1) exists and is equal to 3 (not f'(1) at this point!)

Though, as mentioned by gfauxpas, delta-epsilon proofs of limits are extremely cumbersome, that is typically why you have composition theorems and some basic limits for stuff like that

So generally, avoid them unless there is no choice

[turbid topaz]

though here you did prove that f' is continuous at 1, once you established that it exists at 1

[reef steppe]

I don't think so, f' existing at a point does not entail it is continuous at that point

[turbid topaz]

"since |difference quotient-f'(1)|<epsilon"

[reef steppe]

e.g. f : x l-> x²sin(1/x) if x =/= 0, 0 otherwise

f is differentiable at 0, but f' is not continuous at 0

[turbid topaz]

in this particular case I mean

it's because f is affine

[void sky]

ohh

[turbid topaz]

that the epsilon-delta formulation has |difference quotient - f'(1)|=0

[void sky]

isn't that the formula though?

[turbid topaz]

no the formulas are what I said:

[turbid topaz]

.

[reef steppe]

$f'(1) = \lim_{x \to 1} \frac{f(x) - f(1)}{x - 1}$

[noble shoal]

Rion

[turbid topaz]

they are equivalent, by making the substitution x-c = h

[void sky]

this?

[reef steppe]

so first of all, before you even talk about f'(1), you need to show the limit exists

[turbid topaz]

no,

[reef steppe]

which is exactly what you are asked to prove

so you are not allowed to mention f'(1) until the bottom line

that is, "f is differentiable at 1 and f'(1) = 3"

[turbid topaz]

i'm saying that once he proves it exists, the work he did above proved f' is continuous at 1

which is not what th eassignment asked for

[void sky]

by just evaluating or by proving?

[void sky]

ohh

[reef steppe]

It is what you are effectively proving

you are showing here that (f(x) - f(1))/(x - 1) has a limit at x = 1

[void sky]

oh

[reef steppe]

once you have proven that this limit exists

[reef steppe]

you establish that f'(1) exists

[void sky]

So then what is wrong with this?

[reef steppe]

your first definition is trolling you

[turbid topaz]

also has math mistakes

(c-delta, c+delta) in Domain(f)

instead of subseteq

[void sky]

I prob shldn't trust my professor on everything he says frm now on 😭

[reef steppe]

$f$ is differentiable at $c$ \textit{when} there exists $L$ such that for all $\epsilon > 0$, there exists $\delta > 0$ such that for all $x \in \mathrm{Domn}(f)$, $0 < \lvert x - c \rvert < \delta$ implies $\left \lvert \frac{f(x) - f(c)}{x-c} - L \right\rvert < \epsilon$. Then you denote $f'(c) = L$

[void sky]

so by proving that the limit exists, what should I do next?

[noble shoal]

Rion

[reef steppe]

(note that here you implicitly invoke the limit uniqueness property)

[turbid topaz]

.... therefore, proving the limit exists IS proving f is differentiable at L

[void sky]

ohhh so f'(c) is irrelevant?

my sir said f'(c)= L but I didnt get that

[void sky]

I see

[turbid topaz]

it's not irrelevant, it's that it's misleading to introduce the symbol "f'(c)" at a stage of the exercise before you showed it exists

[reef steppe]

it means that f'(c) is the limit when it exists

which entails that you have to prove its existence

before even talking about it

[void sky]

ohh

[turbid topaz]

it's like using the notation $f^{-1}$ before you prove $f$ has an inverse

[noble shoal]

gfauxpas

[turbid topaz]

since not all functions have inverses

[reef steppe]

yep, big trap

[void sky]

so in my working which I was just supposed to replace the f'(1) with L?

[reef steppe]

yes, in our case 3

[void sky]

damnn

[reef steppe]

only at the bottom line

you conclude that f is differentiable at 1 (since you've shown the limit of ((f(x) - f(1))/(x-1) at 1 exists)

and that f'(1) = the limit you found

[void sky]

Ohhh I get it now

also can the trivial proof be used wherever possible?

[reef steppe]

What do you call "trivial proof"?

[void sky]

the red box here

[reef steppe]

well I don't know if it's relevant

but by that point, you are already done

[turbid topaz]

in general epsilon delta proofs have delta a function of epsilon, or a function of both delta and x

[reef steppe]

you've shown that there exists $\delta > 0$ such that for all $x$ verifying $0 < \lvert x - c \rvert < \delta$, $\left\lvert \frac{f(x) - f(1)}{x-1} - 3 \right\rvert < \epsilon$

[noble shoal]

Rion

[void sky]

I see

[reef steppe]

and by that point you're done

by definition

not by anything else

[void sky]

so by proving it u show that f'(1) exists and that f(x) is differetiable

not by implementing f'(1) early on

[turbid topaz]

so in general you'll see something like "so we let delta = max{1, epsilon/|x|}". In your case it happened to be delta was a constant

[reef steppe]

yes, you are not allowed to mention f'(1) before showing that the limit exists

so you show the limit exists, find its value

by definition, f will be differentiable at 1, and f'(1) = the limit you found

[void sky]

okayy I got it now, I learnt a lot more abt this from u guys from my professor, thanks a lot!

[turbid topaz]

do you know about khan academy?

[reef steppe]

I'm just nitpicking a lot

[turbid topaz]

check it out if you havent

[reef steppe]

but this is important

[void sky]

yeaa

[reef steppe]

so that you don't make wrong steps

faux pas, get it

hahaha

[void sky]

lol

[turbid topaz]

[void sky]

yea I got a lot more to learn lol

[reef steppe]

Also, something that might be helpful to you in the future

that I would actually encourage most math profs to do

Whenever you write a definition, do not use the word "if" or "iff"

use the word "when"

if and iff are words that represent logical entailment, but it isn't really there when you define something

[void sky]

thanks a lot I'll keep that in mind

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