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November 01, 2025, 08:02:12 pm

Author Topic: hyperbolic trig xD  (Read 1757 times)  Share 

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pHysiX

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hyperbolic trig xD
« on: May 03, 2009, 08:08:47 pm »
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This is probably the wrong spot but since there are a lot of smart cookies in this part, I don't see any harm in posting this question =]

Verify that sinh(x)<0.5

What I did was:

since sinh(x)=,
i transposed it to

Is that sufficient as a verification? Please show other ways =]
« Last Edit: May 03, 2009, 08:10:31 pm by pHysiX »
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TrueTears

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Re: hyperbolic trig xD
« Reply #1 on: May 03, 2009, 09:22:55 pm »
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Yeah, your way is perfectly sufficient.

we have

But

So then we get - (some positive quantity)

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dcc

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Re: hyperbolic trig xD
« Reply #2 on: May 08, 2009, 09:59:27 pm »
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Technically, just because you can manipulate a relationship in some way into a relationship you know is true, doesn't mean it necessarily is.  This sort of proof is very prevalent with supposed '' proofs.

For ultra-super-rigor, you should probably frame this as a proof by contradiction.

Assume that .

Given that .

However, this is a contradiction, because we know that for all , we have , hence

kamil9876

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Re: hyperbolic trig xD
« Reply #3 on: May 08, 2009, 11:38:40 pm »
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This is probably the wrong spot but since there are a lot of smart cookies in this part, I don't see any harm in posting this question =]

Verify that sinh(x)<0.5

What I did was:

since sinh(x)=,
i transposed it to

Is that sufficient as a verification? Please show other ways =]

All you have proven is that "if then " which is different from "if then " (actually u might want to provide me with the details of your proof to be sure).

If A implies B and B is true, A doesn't neccesarily have to be true, an example:

"If a man is in Paris, he is in France" is different to "If a man is in France, he is in Paris".

In the first case, it may be true that the man is in France (B) but it may be false that he is in Paris (A).

Hence I would do you're exact proof but backwards.
for all x
Hence for all x:




In proofs, you have to be careful about converses. Converse of "A implies B" is "B imples A" and as you saw with the paris-france example, the converse is not neccesarily always true if the original statement is always true. Here is another good example:

If x=a satisfies , then satisfies . The converse is clearly not true here since satisfies the second but not the first equation.

There are heaps of other examples, but the main idea is that if you want to prove "If A then B" you must assume that A is true, do not assume that B is true and then try to prove A unless you can also prove that the steps are reversible.
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humph

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Re: hyperbolic trig xD
« Reply #4 on: May 09, 2009, 01:54:38 am »
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Technically, just because you can manipulate a relationship in some way into a relationship you know is true, doesn't mean it necessarily is.  This sort of proof is very prevalent with supposed '' proofs.

For ultra-super-rigor, you should probably frame this as a proof by contradiction.

Assume that .

Given that .

However, this is a contradiction, because we know that for all , we have , hence
People (in undergrad uni maths, anyway) always overuse proof by contradiction, when it's really not necessary most of the time.
The direct proof is (mildly) easier and just as rigorous:
for all , so dividing both sides by gives for all . Thus for all ,
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enwiabe

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Re: hyperbolic trig xD
« Reply #5 on: May 09, 2009, 02:25:04 am »
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i think robbo just got smashed

pHysiX

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Re: hyperbolic trig xD
« Reply #6 on: May 09, 2009, 07:41:08 pm »
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haha thanks guys.

my proof was that since e^-x is greater than zero for all real values of x, there is always a constant being taken away from 0.5e^x. Hence, sinh(x) must be less than 0.5e^x

thanks again guys...god i hate assignments >.<
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kamil9876

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Re: hyperbolic trig xD
« Reply #7 on: May 10, 2009, 12:08:06 am »
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yep that's the main idea, it's good that you started off with the fact -0.5e^-x <0 for all x.
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."