The 10,000 Post Thread v2

[airman15 4,111]

4381

[Frdm920 50]

4382

What nothing to add?

[astronelson 339]

4383.

You have to add something. It's against the rules of the thread to not do so.

[Duke87 1,070]

(4384)

How about if I multiply something?

[NMUSpidey 3,956]

4385

You can't multiply nothing. Aren't you like an engineer? You're supposed to know these things!

[astronelson 339]

4386.

Sure you can multiply nothing. You can multiply nothing by anything. You'll just get nothing as a result. That's the additive identity for you.

[Timmystwin 201]

4387

You can only not divide by nothing or divide by something and get nothing. I think. Theres probably some bizarre other example but meh.

[Larks2242 251]

4388.

[insert something intresting here]

[astronelson 339]

4389.

Division by zero is not defined.

Let's see if I can prove this using only axioms of the real numbers...

Part 1: multiplication by 0 yields 0.

a+0=a (definition of the additive identity)

(a+b)c=ab+ac (associativity)

Therefore, (a+0)b=ab and (a+0)b=ab+0b, so 0b=0.

Part 2: division by zero leads to a contradiction.

ab=ba (commutativity)

a^-1a=1 (multiplicative inverse)

1a=a (multiplicative identity)

Let us define division as multiplication by the multiplicative inverse, and let 0^-1a=b for some nonzero a. Then 00^-1a=0b.

So a=0b. But 0b=0 as shown in part 1. So a=0, which contradicts our statement that a is nonzero.

We can also derive a contradiction by letting a=0. 0^-1a=0^-10. Multiplication by the multiplicative inverse is 1, but multiplication by 0 is 0, a clear contradiction.

So the multiplicative inverse of 0 is not defined, and thus division by zero is not defined.

[NMUSpidey 3,956]

4390

I'm not sure I wanted to read all that at 7am, Astronelson...

[Duke87 1,070]

(4391)

A proof by contradiction just isn't complete without "Q.E.A."

  Edited June 28, 2011 by Duke87  

[RipRap 4,400]

4392

Hey Guys, I'm back from a terrible day!

@astronelson I have no idea what are you talking about...

[jacksunny 741]

4393.

I just attempted to fly a kite. Failed because of lack of wind.

[Larks2242 251]

4394.

Get a fan

[astronelson 339]

4395.

A proof by contradiction just isn't complete without "Q.E.A."

Q.E.A. (Quod Est Absurdum, "which is absurd") is rarely used. Q.E.D. (Quod Erat Demonstrandum, "what was to be demonstrated") is more popular.

But you are correct, I did neglect to add it, so I will now.

Q.E.D. (or A., as you prefer).

[Frdm920 50]

4396

Maths hmm?

Well I suppose I could just add to the discussion by saying that (theoritically) 0/0 is possible.

For example

Lim x -> 0 of sin(x)/x = sin(0)/0 = 0/0

At which point we say uh oh because obviously you can't divide by 0.

But in theory we can by using L'Hopitals rule.

Lim x -> 0 of f(x)/g(x) = Lim x -> 0 of f'(x)/g'(x) where f' and g' are the respective derivatives of f and g

f(x) = sin(x), f'(x) = cos(x)

g(x) = x, g'(x) = 1

So

lim x -> 0 of cos(x)/1 = cos(0)/1 = 1/1 = 1

Hence

Lim x -> 0 of sin(x)/x is 1. So 0/0 = 1

Thus in theory it is possible to divide by 0. Of course, this isn't entirely legal mathematics, as we're not ACTUALLY dividing by 0, rather we're taking the value of the function at a point infinately close to 0.

There, probably confused quite a few people but heh

[Duke87 1,070]

(4397)

That is not a valid application of L'Hopital's rule. The limit you've specified doesn't exist. Note that from the right the function goes to positive infinity and from the left it goes to negative infinity. In order for a limit to exist, the right and left hand limits have to both exist and be the same.

[NMUSpidey 3,956]

4398

[astronelson 339]

4399.

The limit looks fine to me. f(x)=Sin[x] and g(x)=x, and both go to zero as x goes to 0. L'Hopital's rule works.

Remember: L'Hopital's rule states that if the limit as x->c f(x) = lim x->c g(x) = 0 or +/- infinity, lim x->c f(x)/g(x) = lim x->c f'(x)/g'(x).

Lim x->0 Sin[x]/x =1 is a very well-known limit.

[jacksunny 741]

4400

Tell me about it

  Edited June 29, 2011 by jacksunny  

[Frdm920 50]

4401

Indeed my usage of L'Hopital's rule was correct. Duke perhaps you were confusing it with the Lim x -> 0 of cos(x)/x which meets the criteria you have described? In this case L'Hopital's rule is indeed invalid as negative infinity = lim x -> 0^- =/= lim x -> 0⁺ = infinity and hence the original limit does not exist.

Edit:

Also it can not be applied because lim x -> 0 of cos(x) = 1 but lim x -> 0 of x = 0 so you have 1/0 and hence not permitted as both limits have to tend to the same thing

  Edited June 29, 2011 by Frdm920  

[astronelson 339]

4402.

You wouldn't be able to apply it to lim x->0 Sin[x]/(1/x), but that fails to meet the conditions under which you would use L'Hopital's rule and, of course, you wouldn't need to anyway. It's the same as lim x->0 x*Sin[x], and that's easily calculable.

Somewhere L'Hopital's rule fails is for lim u->infinity u/(Sqrt[u²+1]). Applying it won't yield a useful result. You can, however, clearly see that as u increases, the "+1" term in the denominator becomes irrelevant and the function approaches u/u=1, which is the limit.

[RipRap 4,400]

4403

Well, y=mx+b

Wow, There's no need for a book-- I'll just read it off here. Thanks!

[Duke87 1,070]

(4404)

I'm just trying to picture here what the graph of this function looks like. Anything divided by x should have a vertical asymptote at x=0, and thus... wait, I see where my mistake is. to the left of 0, x is negative, but so is sin(x). So on both sides the lines go off to positive infinity, it isn't split with one positive and the other negative like 1/x. In which case, yes, the limit being 1 does sound familiar.

It now occurs to me that it's been been four years since I was last in a calculus class. Yikes!

[astronelson 339]

4405

I'm just trying to picture here what the graph of this function looks like. Anything divided by x should have a vertical asymptote at x=0, and thus... wait, I see where my mistake is. to the left of 0, x is negative, but so is sin(x). So on both sides the lines go off to positive infinity, it isn't split with one positive and the other negative like 1/x.

Vertical asymptote? What vertical asymptote? I see no vertical asymptote here.

It goes to 1. That is what a limit means.

Both Sin[x] and x are zero at x=0, so you can't say there's a vertical asymptote.

[RipRap 4,400]

4406

I have no idea what you guys are talking about.

[Frdm920 50]

4407

@ Astro, I think Duke was referring to the fact that most functions divided by x have a vertical asymptote at x = 0. Of course as per usual in Calc there's about a ballijion exceptions to the rule.

Also, I have no idea why it only just occured to me then but lim x->0 of sin(x)/x = 1 is the basis for the assumption that when x is very small sinx ~= x. Which is something we use an awful lot in my engineering classes.

@ 111222333444

University level Calculus

[astronelson 339]

4408.

You can write Sin[x] as a Taylor series, which makes that more obvious.

Sin[x] is the sum from n=0 to n=infinity of (-1)ⁿx⁽²ⁿ⁺¹⁾/((2n+1)!). For small x, the only term that really contributes to the function is when n=0, that is, x. So Sin[x] is approximately equal to x for small x.

See here for more information: http://en.wikipedia....i/Taylor_series

[Duke87 1,070]

(4409)

Vertical asymptote? What vertical asymptote? I see no vertical asymptote here.

!

Okay, I did not call that. Now all of a sudden it makes sense!

Now if only my TI-89 still worked...

[Frdm920 50]

You can write Sin[x] as a Taylor series, which makes that more obvious.

Sin[x] is the sum from n=0 to n=infinity of (-1)ⁿx⁽²ⁿ⁺¹⁾/((2n+1)!). For small x, the only term that really contributes to the function is when n=0, that is, x. So Sin[x] is approximately equal to x for small x.

4410

I assume the same can be done with cos and tan to get cos(x) ~ x and tan(x) ~ x for small x?