Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**snyy****Member**- Registered: 2005-12-29
- Posts: 8

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..

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

It would be something very VERY ugly.

And probably it's complexity will make it useless.

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

```
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.
```

**igloo** **myrtilles** **fourmis**

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

```
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.
```

**igloo** **myrtilles** **fourmis**

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

ax^3+bx^2+cx+d=0

Here's the real solution:

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

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

I told you it will be ugly.

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**snyy****Member**- Registered: 2005-12-29
- Posts: 8

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

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

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

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

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,619

Ye Gods!

You did better than this guy:

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

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

snyy wrote:

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.

**igloo** **myrtilles** **fourmis**

Offline

Pages: **1**