TT's Maths Thread

[kamil9876]

Yeah but Bernoulli's discovery of it is much less cheap than the application of elementary algebra.

[TrueTears]

yeah ok, but still fuck l'hopital's rule.

[kamil9876]

LOL
I see calculus isn't liked very much around here

Actually Generalized Binomial theorem is pretty calculusy, again more irony.

[/0]

LOL
I see calculus isn't liked very much around here

Actually Generalized Binomial theorem is pretty calculusy, again more irony.

oh my bad

still, nice non-standard solution TT

[TrueTears]

Use an proof to show that

Just started on , was wondering if someone can show me how to go through this proof.

I get the basic jist but I don't really know how to set it out etc and what exactly do we have to prove...

[kamil9876]

so we want to show that for any given we can find a such that:



We can naturally find this by showing that we seek to find the x values that satisfy:



But this already suggests we should set .

since if (1)

Then multiplying everything by 4 yields:

(2) as required. Thus for all x that satisfy (1), they also satisfy (2)

[/0]

We need to prove that IF , THEN

First we can analyse the second expression and work backwards



Then





Now if we set , we can sub into the original proposition:

IF THEN

Since and are exactly the same statement, the IF-THEN construction is satisfied.

[TrueTears]

Thanks for the reply guys but I still don't get quite what we need to prove exactly.

From Stewarts:

The precise definition of a limit: is

"iff for every there is a number such that if then "

So what EXACTLY out of that statement do we have to prove?


Actually I think I kinda get it...

So the first thing you have to prove is that there does EXIST a for every . In this case we worked backwards to find that

Then we must prove that if then

Which we showed by subbing back into the original inequality which does yield the result of

So we have Q.E.D

Is that interpretation right?


But the only dilemma is, how do we prove "for every "?

We just proved for one fixed and proved for every number LESS than it, but there are numbers bigger than , so the question is how do we prove for ALL ?

[kamil9876]

The trick is that was arbitrary. In other words we have answered this:

for , set and the condition is satisfied.

for set and the condition is satisfied.

for set and the condition is satisfied.

.
.
.

ie we have answered an uncountably infinite number of questions, with just one sentence.

[TrueTears]

The trick is that was arbitrary. In other words we have answered this:

for , set and the condition is satisfied.

for set and the condition is satisfied.

for set and the condition is satisfied.

.
.
.

ie we have answered an uncountably infinite number of questions, with just one sentence.

So basically whatever we sub in first acts like an "upper bound", and thus everything smaller than this upper bound satisfies the condition.

However there are infinitely many upper bounds we can sub in, thus all 0 is satisfied.

Is that right?

[TrueTears]

1. How to use proof to show ?

2. How to use proof to show ?

[/0]

I'll give these a go, hope they're right

1.

if the following statement is true:

"If then "

"If then "

Since is always true, the limit is c.

2.

"If then "

"If then "

So we can choose

Then the proof goes:

If

Pick

[TrueTears]

Thanks I think I'm getting the hang of these now, can check if the following is right?  8-)

Prove

We need to prove that for every there exists a such that if then

First we need to find that there does exist a for every





If we can find some constant such that

Then if we can find that to satisfy

We can have

Now assume is fairly close to so we can have

Thus

Thus

Which means a suitable is

So we have and

Thus

So for every if then

Case 1: If

We have

Case 2: If

We have

[kamil9876]

case1 is unnecesary (why?)

I once gave birth to a nicer proof for the case that I like more It's basically proving continuity of over and also over !!

[TrueTears]

Prove that

This one is a bit tricky because I can't seem to get the term to contain a term :idiot2:

kamil, shed some ingenuity pls