View Poll Results: What is F(x) of f(x)=(3-e^-x)²? (Note: "^" is supposed to be a power, i.e. x^3=x^³)
F(x)=(9x-e^-x)³ 0 0%
F(x)=(1/3x-e^-x)³ 1 12.50%
F(x)=(x-e^-x)³ 0 0%
None of the above (please explain) 7 87.50%
Voters: 8. You may not vote on this poll

 
 
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Old October 26, 2003, 19:38   #1
Stefan Härtel
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What do you think of this?
Here's the first couple of graphics from my coming scenario. No details are yet to be revealed, but guesses are premitted
What do you think of this?
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Old October 26, 2003, 19:45   #2
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That math stuff makes my head hurt

The graphics look great though I'd guess that it's some kind of exploration scenario.
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Old October 26, 2003, 19:47   #3
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I don't do math.

Nice graphic.
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Old October 26, 2003, 19:51   #4
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really nice waterfall!
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Old October 26, 2003, 19:55   #5
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That's supposed to be [i]easy](i] stuff I've got to know for my final exams in May (Meaning: I won't ever make it).

I just noticed you can't properly see the unit in the lower left corner after I've cut off the shield from the screenshot graphic.

The waterfall, BTW, will be only a small one of many elements in this scenario.
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Old October 26, 2003, 20:47   #6
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Is he a stone-age hunter?
A Crusoe-type stranded on some island perhaps?
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Old October 26, 2003, 20:57   #7
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Looks a bit Aztec!

Maths?

That's all Von Braun talk to me!
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Old October 26, 2003, 21:52   #8
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You forgot about the 3 * (-e^-x) terms in the binomial square
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Old October 26, 2003, 22:16   #9
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(3-(e^-x))(3-(e^-x))=9-3(e^-x)-3(e^-x)e^x Owwwww, a blood vessel just burst....
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Old October 27, 2003, 06:59   #10
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Well, the basic point is that the result must be:
f'(x)=(3-e^-x)² if f(x) is what would be F(x).
I personally think it must be F(x)=(x-e^-x)³; but I'd love to be proved otherwise (basically this is just a maths problem that's been keeping me occupied the other night, and I wanted to hear a second opinion).

Quote:
Is he a stone-age hunter?
A Crusoe-type stranded on some island perhaps?
Nope. No island.

Quote:
Looks a bit Aztec!
Closer, but still far away.
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Old October 27, 2003, 13:36   #11
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The integral will be much more complicated than that... You also have the chain rule to worry about.

Try typing in the formula on this site:
http://integrals.wolfram.com/index.en.cgi

(Be sure to use uppercase E for e)
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Old October 27, 2003, 13:55   #12
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The result doesn't show (it won't load the image)

If F(x) would be (x-e^-x)³, then f(x) would be, according to the chain rule: f(x)=(3-3^-x)²; or would it? According to
f'(x)=2*(3-e^-x)*e^-x if f(x)=(3-e-^x)²

In another calculation, however, f'(x) would be f'(x)=3*(x-e^-x)*(3-e^-x)

This is sick
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Old October 27, 2003, 14:47   #13
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OK, Stefan!

Spill the beans!

What is the scenario about?
Not mathematics surely?
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Old October 27, 2003, 15:11   #14
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Quote:
Originally posted by Stefan Härtel
The result doesn't show (it won't load the image)
I attached the image I got... It looks rather more complicated than the possible outcomes you gave. I'm not sure if that's because the program tries to write out everything in full... I.e. it would say x2 - 2x + 1 rather than (x - 1)2.

Quote:
If F(x) would be (x-e^-x)³, then f(x) would be, according to the chain rule: f(x)=(3-3^-x)²; or would it?
No, it would be:

3 * (x - e-x)2 * (1 + e-x)

So you made two mistakes:
- The multiplier of 3 is outside the square. I.e. the derivative of x3 is 3(x2), not (3x)2.
- You have to take the derivative of the part in parentheses as well, and use the chain rule... that's the (1 + e-x) at the end.

Quote:
This is sick
I'll say... I'm glad I got that behind me. Not that the statistical maths I have to deal with is any easier, though.

EDIT: I forgot the minus in (1 + e-x)
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Last edited by Mercator; October 27, 2003 at 15:19.
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Old October 27, 2003, 15:32   #15
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I have much to learn (I've got one point out of 15 at best, that's a 5- mark, but it's enough since I'm good at English).

Quote:
OK, Stefan!

Spill the beans!

What is the scenario about?
Not mathematics surely?
No, I just posted the poll because I went nuts about the problem

I can't tell you exactly what the scenario will be about because that'd be telling too much. But I can tell you that it's an adventure scenario set in pre-Colombian Ecuador. I've got some very cool ideas so far, but I don't want to tell because that'd spoil too many surprises.
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Old October 27, 2003, 15:36   #16
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Yikes... You'd better ignore that attached image. The site doesn't seem to be too good at resolving those fractions (?).

If you try to get the integral of 3 * (x - e-x)2 * (1 + e-x) the result looks really ugly...

Try this site, instead:
http://www.calc101.com/

You have to pay if you want a full step-by-step explanation for the integration. Without a password, it still shows the end result.
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Old October 27, 2003, 15:40   #17
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calc101.com makes the solution to your original question look a lot better (although still not quite as simple as you hoped):
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Old November 3, 2003, 17:44   #18
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Quote:
Originally posted by Stefan Härtel I can't tell you exactly what the scenario will be about because that'd be telling too much. But I can tell you that it's an adventure scenario set in pre-Colombian Ecuador. I've got some very cool ideas so far, but I don't want to tell because that'd spoil too many surprises.
I'm unable to guess what your scen will be about, but what I can guess for sure is that I'll have MUCH pleausre playing it!

The graphical extract looks great, BTW!
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Old November 4, 2003, 06:55   #19
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:)
the pic looks wonderful Stefan! good work
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Old November 4, 2003, 13:26   #20
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Ummmmm . . . . .
An official "no comment" on the mathematics from hell, Stefan.

Sounds like an exploration/conquest scenario about the rise of the Inca empire? The Chimu culture?

Alternatively, if it's about the European exploration, then it could be either Pizzarro's or Orelana's expeditions into the region. The term "pre-Columbian" seems to deny this last guess, however.
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Old November 4, 2003, 13:44   #21
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I know!

The scen is about conquistadors using advanced mathematics to defeat the Incas!
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Old November 5, 2003, 11:58   #22
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Quote:
The scen is about conquistadors using advanced mathematics to defeat the Incas!
You'd play the Incas in that one, no doubt!

But... no. It's neither about the Incas or the Europeans (not even the Chimu).
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