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April 27, 2024, 11:05:53 pm

Author Topic: Bazza's 3/4 Question Thread  (Read 30022 times)  Share 

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Planck's constant

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Re: Bazza's 3/4 Question Thread
« Reply #30 on: January 15, 2012, 06:51:52 pm »
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I will not attempt a response to your question because I noticed that kamil9876 is lurking in this thread and he is certain to provide the definitive answer :)

kamil9876

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Re: Bazza's 3/4 Question Thread
« Reply #31 on: January 15, 2012, 07:02:20 pm »
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Actually feel free to provide an answer. I was just going to explain about how I don't really know the precise definition of assymptote for graphs ( I guess for functions it's easy to define). TT's link doesn't really provide a rigorous definition.

I was also going to ask about the silly function:



does it have an assymptote at ?
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Planck's constant

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Re: Bazza's 3/4 Question Thread
« Reply #32 on: January 15, 2012, 07:17:13 pm »
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Based on my (non-expert) interpretation of TT link, I would suggest that x=1 is not an asymptote in this case because it fails the "crossing the line infinitely often" test

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Re: Bazza's 3/4 Question Thread
« Reply #33 on: January 15, 2012, 07:18:10 pm »
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Actually feel free to provide an answer. I was just going to explain about how I don't really know the precise definition of assymptote for graphs ( I guess for functions it's easy to define). TT's link doesn't really provide a rigorous definition.

I was also going to ask about the silly function:



does it have an assymptote at ?

Odd question! My money would be on a hollow dot at x=1 and obviously 0 elsewhere. An asymptote wouldn't make sense because it can't 'approach' anything given it is zero for all values.


Quote
For something like

y=\frac {a} {f(x)} + \frac {b} {g(x)} + c


Do f(x) and g(x) become asymptotes as well ? (i think they do but the book doesn't mention it anywhere)

I'm pretty sure they do also. Just by common sense - even if c=0, the first term can never be 0 (it's a fraction) so the function can never equal the second term, and vice versa.
For the record though, it's almost certainly not mentioned in the book because (somebody correct me if I'm wrong) you never have to sketch or mark in a non-linear asymptote. Though it will still be useful for getting the correct shape, I guess.
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kamil9876

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Re: Bazza's 3/4 Question Thread
« Reply #34 on: January 15, 2012, 07:32:59 pm »
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^ I know what the graph looks like, just asking for what the definition of asymptote is. You're contradicting yourself there by saying no to my example but yes to:

Quote
For something like



Do f(x) and g(x) become asymptotes as well ? (i think they do but the book doesn't mention it anywhere)

Voltaire: "There is an astonishing imagination even in the science of mathematics ... We repeat, there is far more imagination in the head of Archimedes than in that of Homer."

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Re: Bazza's 3/4 Question Thread
« Reply #35 on: January 15, 2012, 07:40:45 pm »
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Actually feel free to provide an answer. I was just going to explain about how I don't really know the precise definition of assymptote for graphs ( I guess for functions it's easy to define). TT's link doesn't really provide a rigorous definition.

I was also going to ask about the silly function:



does it have an assymptote at ?

This is a 'removable singularity' if my memory serves me correctly.

edit: also known as a 'pole of order 0'. Point is, things don't get this complicated in spesh so I wouldn't worry lol
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Re: Bazza's 3/4 Question Thread
« Reply #36 on: January 15, 2012, 07:46:51 pm »
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Yeah it is. Though I dunno if that relates to asymptotes. I guess usually whenever confusion arises on atarnotes, we always turn to some 'math gun' to clarify and then he/she pulls out their rusty copy of Walter Rudin etc. to check the formal definition, then water it down for vcelevel and get respect (or karma as it was known in the good ole days). Unfortunately no one in higher mathematics cares about asymptotes(we usually care more about 'asymptotic analysis' if anything) in the same way that vce students do, so it doesn't work as well here.
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Re: Bazza's 3/4 Question Thread
« Reply #37 on: January 15, 2012, 07:49:51 pm »
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Yeah that was basically my point about overcomplicating things. Maybe if bazza just gives us an actual example and then someone can explain why so and so function 'crosses the asymptote'.

edit: btw what is Walter Rudin lol
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Re: Bazza's 3/4 Question Thread
« Reply #38 on: January 15, 2012, 07:58:30 pm »
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Quote
btw what is Walter Rudin lol

He is a guy who wrote what many people (not so much me) treat as the bible of undergraduate Analysis.
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Planck's constant

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Re: Bazza's 3/4 Question Thread
« Reply #39 on: January 15, 2012, 08:14:52 pm »
+1

For the record though, it's almost certainly not mentioned in the book because (somebody correct me if I'm wrong) you never have to sketch or mark in a non-linear asymptote. Though it will still be useful for getting the correct shape, I guess.


Unless I am totally missing your point, you are quite often required to sketch non-linear asymptotes in spesh, eg

f(x) = x^2 + 1/x


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Re: Bazza's 3/4 Question Thread
« Reply #40 on: January 15, 2012, 08:35:15 pm »
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For something like




Do f(x) and g(x) become asymptotes as well ? (i think they do but the book doesn't mention it anywhere)

thanks again :)


So lol... uh... what's the answer xD ?
yea i was thinking along incomm's line i think... but i'm not sure;does anyone know?; can pi explain your answer a bit more?

(and i figured out the "crossing the asymptote" one i think, the books like sketch the graphs of and
so c is the asymptote for the second one but not the overall one... i think...

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« Last Edit: January 03, 2017, 12:45:22 am by pi »

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Re: Bazza's 3/4 Question Thread
« Reply #41 on: January 15, 2012, 08:39:34 pm »
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Unless I am totally missing your point, you are quite often required to sketch non-linear asymptotes in spesh, eg

f(x) = x^2 + 1/x

Right, but questions like that (at least on the exam, SACs can be nastier) won't require you to draw in the asymptote for any mark, and our teacher told us we'd never have to actually mark that y=x^2 asymptote on the graph. Like I said, useful for shape of that graph.
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Re: Bazza's 3/4 Question Thread
« Reply #42 on: January 15, 2012, 08:49:22 pm »
+1
^ I know what the graph looks like, just asking for what the definition of asymptote is. You're contradicting yourself there by saying no to my example but yes to:

Quote
For something like



Do f(x) and g(x) become asymptotes as well ? (i think they do but the book doesn't mention it anywhere)

Okay well answering that question I kind of implicitly assumed that f(x) and g(x) weren't the same and that b wasn't -a...
Asymptotes are defined as a curve which the function approaches, so knowing what the graph looks like is basically enough to tell if something's an asymptote - if there's no approaching, then there's no asymptote.

Of course I'm not actually super-pro at maths and if anybody does want to pull out their Rudin and prove me wrong then I'll wear it :)
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Re: Bazza's 3/4 Question Thread
« Reply #43 on: January 15, 2012, 08:50:31 pm »
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i have absolutely no idea what is going on :(

For something like




Do f(x) and g(x) become asymptotes as well ? (i think they do but the book doesn't mention it anywhere)

thanks again :)


So lol... uh... what's the answer xD ?
yea i was thinking along incomm's line i think... but i'm not sure;does anyone know?; can pi explain your answer a bit more?

(and i figured out the "crossing the asymptote" one i think, the books like sketch the graphs of and
so c is the asymptote for the second one but not the overall one... i think...

anyone xD

Moderator action: removed real name, sorry for the inconvenience
« Last Edit: January 03, 2017, 12:45:37 am by pi »

dc302

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Re: Bazza's 3/4 Question Thread
« Reply #44 on: January 15, 2012, 09:06:38 pm »
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You mean, does f(x)=0 and g(x) = 0 become asymptotes? The answer is it depends. I personally think there is no reason to try and see a set rule, and rather just sketch the functions using known methods wherever required. Although usually, yes they do become asymptotes too.


edit: Maybe I should provide more reasoning. When f(x) = 0, g(x) = 0, you get vertical asymptotes (because diving by zero gets you infinity). These asymptotes are almost always preserved, and the only time it doesn't is when they cancel each other out (think about infinity minus infinity). This can result in the the 'asymptote' going to some constant number, or even to infinity again (think about if you get infinity(1) minus infinity(2) where infinity(1) is 'bigger', so the overal effect goes to 1).

example: 1/x^2 - 1/x
« Last Edit: January 15, 2012, 09:12:31 pm by dc302 »
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