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Hardest Math Contest EVER big prize





Dark-Tech
Given the equation as follows
f(x) = x^3-4*x+1

First I will award frih10$ to the first person who can tell me the number of places where the equation crosses the x-axis

Second
I will award frih30$ to the first person who can tell me the points to 10 decimal places where the equation crosses the X-axis

Finally, I will award frih$100 or 1/4 my total points, whichever is higher to the first person who can tell me all the equation that represent the exact points at which the equation crosses the X-axis

after 5 pages without the answer i will post a link to the answer
imagefree
i cant understand your third requirement, but the solution of the first two requirements is below.


First I will award frih10$ to the first person who can tell me the number of places where the equation crosses the x-axis


It croses X-axis 3 times.




Second I will award frih30$ to the first person who can tell me the points to 10 decimal places where the equation crosses the X-axis



Point 1 (when y=0) : X= 0.254102 , or, x= 1.860806 , or, x= -2.114908

Please clear the third requirement, so that i may be able to answer it too.
Dark-Tech
well you did indeed get the first two which i expected to happen pretty quickly
so i will send the prize after this post.

as for clarification of the third section, i am asking for 3 mathematical formulas each one being the exact formula that represents the positions that you pointed out in part 2

where \() represents the square root, cubic root \3()

EG:

1.4142135624 => (the square root of 2) => \(2)


For the points where the equation crosses the X-axis
these numbers are abbreviations for the equation answers, the full answer keeps going, so an equation is necessary to represent the exact posititions
(0.254102)
(1.860806 )
(-2.114908 )
MYP415
Is this what you are looking for on the third one:
0 = x^3-4*x+1
rvec
yes he needs that and there should be 3 answers like
x=0.254102
x=1.860806
x=-2.114908
but he needs the exact values
imagefree
MYP415 wrote:
Is this what you are looking for on the third one:
0 = x^3-4*x+1


no. But i still cant understand what is the requirement.


I think you are asking to write those frectional answers in p/q form or in the form of some mathematical expression. If yes, please confirm.
Blaster
haha algrabra 2

That is sooo easy with a graphing calculator.
Dark-Tech
imagefree wrote:
MYP415 wrote:
Is this what you are looking for on the third one:
0 = x^3-4*x+1


no. But i still cant understand what is the requirement.


I think you are asking to write those frectional answers in p/q form or in the form of some mathematical expression. If yes, please confirm.


yes, i want the answers as some sort of mathematical expression that is the exact answer for the positions,
x=0.254102
x=1.860806
x=-2.114908
are abbreviations for infinite values

Example: .333 => 1/3

the exact value is 0.3333333....
point 3 repeating

i need the mathematical expression for the points on the axis.

when i see them i will post a web page with the answers on it here

The answers are extremely complex and that is why the prize is so high
Dark-Tech
Blaster wrote:
haha algrabra 2

That is sooo easy with a graphing calculator.


The first 2 were meant to be easy to draw people in

the third one is a lot harder than it sounds

If there are 5 more posts of people unable to figure out what im asking for i will post 1 of the three and the overall prize will be reduced by 1/3
so

the prize would become
66 instead of 100 or 1/6 of my total amount
Dark-Tech
wow, this must really be a hard one if no one is even gonna try let alone get anywhere near the answer
I guess i was right that it was the hardest contest ever
inventor
I'll give it a try. It looks like you just want the three intercept values in fraction form, which may include pi, roots, or exponents. Is this correct?
imagefree
Dark-Tech wrote:
wow, this must really be a hard one if no one is even gonna try let alone get anywhere near the answer
I guess i was right that it was the hardest contest ever



Very Happy The hardest thing is that we cant understand the requirement. Once we get the requirement, may be most of all can figure out the solution.
Dark-Tech
inventor wrote:
I'll give it a try. It looks like you just want the three intercept values in fraction form, which may include pi, roots, or exponents. Is this correct?


That is exactly correct Smile


imagefree wrote:
Dark-Tech wrote:
wow, this must really be a hard one if no one is even gonna try let alone get anywhere near the answer
I guess i was right that it was the hardest contest ever



Very Happy The hardest thing is that we cant understand the requirement. Once we get the requirement, may be most of all can figure out the solution.


does that help any ?
Dark-Tech
Here is the first solution



There are still 2 more, but the prize has been dropped to only 100, the 1/4 no longer is available
Arnie
Quote:
the equation that represent the exact points at which the equation crosses the X-axis
No need for i... click here. Approximate values for these equations are:
x1 = 0.2541016884
x2 = -2.114907541
x3 = 1.860805853
Obviously, for all these points y = 0
Dark-Tech
yes those are the approximations but they are not the exact answers, i asked for the EXACT answers
and y does not = 0 it is close though

but close isn't good enough kinda like saying 300/900 = .3
which is wrong but close, .3333333 is close but still wrong, the best answer is 1/3
Arnie
Arnie wrote:
No need for i... click here.
Should I make the link larger? Click here
Now tell me what's inexact about those equations. I only stated their approximate values to show that they match what came out of the graphing calculators... punch in the equations from the link to verify.
Dark-Tech
Arnie wrote:
Quote:
the equation that represent the exact points at which the equation crosses the X-axis
No need for i... click here. Approximate values for these equations are:
x1 = 0.2541016884
x2 = -2.114907541
x3 = 1.860805853
Obviously, for all these points y = 0


i didn't check them all but for
x3 = sqr(16/3)*cos(arcsin(sqr(27/256)/3)+(pi/6))
i kept coming up with 1.863246715
Arnie
That's because you've got your parentheses messed up. The expression you're giving in your post does not equal the equation for x3 I supplied. You're first dividing sqrt(27/256) by 3 and then taking the arcsin, while you're supposed to first take the arcsin of sqrt(27/256) and then divide by 3. Compare my equation for x3 to your expression:
Code:
sqr(16/3)*cos(
              arcsin(
                     sqr(27/256)/3
                    )
              +(pi/6)
             )
Move the /3 to the outside of the ) that closes arcsin, and you have a correct expression for x3, yielding 1.860805853...

In my Console Calculator output below you can find all three valid expressions calculated to 50 digits. The used program is a freeware scientific calculator with 100-digit precision. Note that it uses asin instead of arcsin and sqrt instead of sqr.

I used the Options menu to set significant figures to 50. Here (click) is my output in a screenshot. Below it's in plain text, feel free to copy the commands and paste them into the calculator yourself.
Code:
> sqrt(16/3)*sin(asin(sqrt(27/256))/3)
ans = 0.25410168836505241212977787984307247090591526081114

> sqrt(16/3)*cos((asin(sqrt(27/256))/3)+((5/6)*pi))
ans = -2.1149075414767557985156140607085439681237562633872

> sqrt(16/3)*cos((asin(sqrt(27/256))/3)+((1/6)*pi))
ans = 1.860805853111703386385836180865471497217841002576

> ans^3-(4*ans)+1
ans = -2.6004811855643251224256043391428812219097448696602e-101
Note the e-101 in the final answer where I calculated f(x3) using the 50-digit value for x3 1.8680... as shown above. e-101 means 101 zeroes (as in 0.00000000... etc) so that's a very small number. If the calculator were able to handle infinite decimals, it would be exactly 0. But it's not because we're using 50-figure approximate values of the exact equations I supplied earlier. It's hard to deny now that those equations are the exact solutions...

Bottom line, you're awfully careless for an 'exact' guy. And I @#$%^%$ wasted my time.
Arnie
So DarkTech, where've you been lately? There's a promise in your starting post you know...
Dark-Tech
well i can't find anything wrong so i will give you th prize, and i just haven't been around
Arnie
All right, thanks Wink
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