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snyy

Member

Registered: 2005-12-29

Posts: 8

Equation Problem.. t=??

a = x0 * (1-t)³ + x1 * 3 * t * (1-t)² + x2 * 3 * t² * (1-t) +x3 * t³

if x0 = 138, x1=116, x3= 171, x4=198 while t=0.01
then a =137.934235

but for now, i want to find t, i have the all the value of x0,x1,x2,x3 and a. so how to reverse the formula so that i can find t??
thanks..

krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

Re: Equation Problem.. t=??

It would be something very VERY ugly.
And probably it's complexity will make it useless.

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IPBLE:  Increasing Performance By Lowering Expectations.

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Equation Problem.. t=??

a = x0 * (1-t)³ + x1 * 3 * t * (1-t)² + x2 * 3 * t² * (1-t) +x3 * t³

    1-2t+t^2(1-t)              1-2t+t^2        3x2(t^2-t^3)       x3t^3
                          3x1 (t-2t^2+t^3)         |                |
     1-2t+t^2                   |                  |                |
      -t +2t^2 -t^3             |                  |                |
    --------------              v                  v                v
  x0(1-3t+3t^2-t^3)

     (-x0+3x1-3x2+x3)t^3 + (3x0-6x1+3x2)t^2 + (-3x0+3x1)t + x0

I don't know what to do next to this cubic equation.

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igloo myrtilles fourmis

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Equation Problem.. t=??

 I found a method to possibly get close to an answer.

a =  (-x0+3x1-3x2+x3)t^3 + (3x0-6x1+3x2)t^2 + (-3x0+3x1)t + x0

Get t on left side, and put a on right, leave t^3 and t^2 on right side.

 -(-3x0+3x1)t = (-x0+3x1-3x2+x3)t^3 + (3x0-6x1+3x2)t^2 + x0  - a

Divide by -(-3x0+3x1)

   t =  (-1/(-3x0+3x1)) ((-x0+3x1-3x2+x3)t^3 + (3x0-6x1+3x2)t^2 + x0  - a )

 Make a guess at t and plug it into the t^3 and t^2 and solve right side of equation.

 Take answer and put it back in again.

 Repeat until you appear to be getting closer and closer to last result.

 I don't know if this will work, but I just read this on a web page about solving a cubic.

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igloo myrtilles fourmis

krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

Re: Equation Problem.. t=??

ax^3+bx^2+cx+d=0
Here's the real solution:

Last edited by krassi_holmz (2005-12-29 14:10:46)

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IPBLE:  Increasing Performance By Lowering Expectations.

krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

Re: Equation Problem.. t=??

I told you it will be ugly.

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IPBLE:  Increasing Performance By Lowering Expectations.

snyy

Member

Registered: 2005-12-29

Posts: 8

Re: Equation Problem.. t=??

sorry, don't really understand how you do this.
can show me how you get this formula (-x0+3x1-3x2+x3)t^3 + (3x0-6x1+3x2)t^2 + (-3x0+3x1)t + x0 ??
from x0(1-3t+3t^2-t^3)  + 3x1 (t-2t^2+t^3) + 3x2(t^2-t^3) + x3t^3

krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

Re: Equation Problem.. t=??

OK,Here is the answer direct from the original equation:

Last edited by krassi_holmz (2005-12-29 20:20:30)

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IPBLE:  Increasing Performance By Lowering Expectations.

MathsIsFun

Administrator

Registered: 2005-01-21

Posts: 7,713

Re: Equation Problem.. t=??

Ye Gods!

You did better than this guy:

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"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Equation Problem.. t=??

sorry, don't really understand how you do this.
can show me how you get this formula (-x0+3x1-3x2+x3)t^3 + (3x0-6x1+3x2)t^2 + (-3x0+3x1)t + x0 ??
from x0(1-3t+3t^2-t^3)  + 3x1 (t-2t^2+t^3) + 3x2(t^2-t^3) + x3t^3

Method used is similar to this example:
  (A+B)C + (A+B)D
  See how I am skipping a step and not writing it down.
  This is the equivalent.
  (C+D)A + (C+D)B
  The step I didn't ever write down is because you just organize by
  jumping around and choosing the A's first, and the B's second.
  But it would be:
  AC + BC   +  AD + BD
  Then get the A's on the left, and B's on the right.

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igloo myrtilles fourmis