Author Topic: problem (Read 4808 times) Tweet Share

brendan · « **on:** March 27, 2008, 04:44:57 pm »

When $x = 0, y = 0.0499$
When $x = 3/12, y = 0.0533$
When $x = 6/12, y = 0.0575$
When $x = 9/12, y = 0.0653$

$y$ is a third degree polynomial. find $y$ in terms of $x$

Collin Li · « **Reply #1 on:** March 27, 2008, 05:47:18 pm »

$y = ax^3 + bx^2 + cx + d$

You have 4 pieces of information, you have 4 unknown variables, hence you can write 4 simultaneous equations and solve for unique solutions to $a,\, b,\, c,\, d$ .

From equation 1: $d=0.0499$

Use whatever method of solving linear equations you wish from here.

Ahmad · « **Reply #2 on:** March 27, 2008, 05:55:58 pm »

Possible alternative: Lagrange Interpolation

Mao · « **Reply #3 on:** March 27, 2008, 06:24:18 pm »

you can also find out the first coefficient with "finite difference" then solve a 2-var simultaneous system (which i find a lot easier):

it appeared briefly in the Mathworld MM 1/2 book, but I cant exactly explain why it works

x	u=4x	y	$\Delta y / \Delta u$	$\Delta^2 y / \Delta^2 u$	$\Delta^3 y / \Delta^3 u$
0	0	0.0499	0.0034	0.0008	0.0028
3/12	1	0.0533	0.0042	0.0036
6/12	2	0.0575	0.0078
9/12	3	0.0653

$\frac{d^3y}{du^3}=0.0028$

after a few antidifferentiation:

$y=\frac{0.0028}{6}u^3+\alpha u^2+ \beta u + 0.0499$

$y=\frac{0.0028}{6}\cdot (4x)^3++\alpha (4x)^2+ \beta (4x) + 0.0499$

so the leading coefficient will be:

$a=\frac{0.0028 \times 4^3}{6} = \frac{56}{1875}=0.029866666666..........$

(and if someone could tell me how to turn on borders in a table, that'll be great

)

jess3254 · « **Reply #4 on:** March 27, 2008, 07:02:58 pm »

Here goes nothing… I don’t know whether this will be correct but meh… good procrastination.

The general form for a 3rd degree polynomial is ax3+ bx2+cx+d=0
substitute the above x values into the equation.

X=0à 0a+0b+0c+d=0.0499
X=3/12à(1/64)a+(1/16)b+(1/4)c+d=0.0533
x=6/12à(8/64)a+(1/4)b+(1/2)c+d=0.0575
x=9/12à(27/64)a+(9/16)b+(3/4)c+d=0.0653

subtract the equations to eliminate d
(1/64)a+(1/16)b+(1/4)c=0.0034
(7/64)a+(3/16)b+(1/4)c=0.0042
(19/64)a+(5/16)b+(1/4)c=0.0078

subtract equations to eliminate c
(3/32)a+(1/8)b=0.0008
(3/16)a+(1/8)b=0.0036

subtract equations to eliminate b
(3/32)a=0.0028
a=0.0028´(32/3)
=(56/1875)

(3/32)a+(1/8)b=0.0008
(3/32)(59/1875)+(1/8)b=0.0008
0.0028+(1/8)b=0.0008
(1/8)b= -0.002
b= -2/125

(1/64)a+(1/16)b+(1/4)c=0.0034
(1/64)(56/1875)+(1/16)(-2/125)+(1/4)c=0.0034
(7/15000)+(-1/1000)+(1/4)c=0.0034
(1/4)c=(59/15000
c=59/3750

\equation is (56/1875)x3-(2/125)x2+(59/3750)x+0.0499
(I have used fractions for accuracy)

yay although I don't think I read the numbers correctly.
crap it doesn't copy from microsoft well...

jess3254 · « **Reply #5 on:** March 27, 2008, 07:04:23 pm »

I really don't think I did that correctly... I'm confused.

Can anyone explain it to me?

ed_saifa · « **Reply #6 on:** March 27, 2008, 07:11:32 pm »

subtracting the equations would take ages. Matrices would be awesome for this. Also, a beloved calculator

brendan · « **Reply #7 on:** March 27, 2008, 07:22:00 pm »

Quote from: ed_saifa on March 27, 2008, 07:11:32 pm

subtracting the equations would take ages. Matrices would be awesome for this. Also, a beloved calculator

yeah i sold my graphics calculator...and my linear algebra text D:

jess3254 · « **Reply #8 on:** March 27, 2008, 07:28:37 pm »

Quote from: ed_saifa on March 27, 2008, 07:11:32 pm

subtracting the equations would take ages. Matrices would be awesome for this. Also, a beloved calculator

haha yeah it did take ages.
Love procrastination.

Also I did this in IB Maths, and we weren’t allowed to use calculators! I’m used to doing it the long way

Ahmad · « **Reply #9 on:** March 27, 2008, 10:00:55 pm »

I do not think this is the fastest or most elegant way, however upon request: unsimplified using Lagrange Interpolation.

$f(x) = 0.0499\frac{(x-\frac{3}{12})(x-\frac{6}{12})(x-\frac{9}{12})}{(0 - \frac{3}{12})(0-\frac{6}{12})(0-\frac{9}{12})}+ 0.0533 \frac{(x-0)(x-\frac{6}{12})(x-\frac{9}{12})}{(\frac{3}{12}-0)(\frac{3}{12}-\frac{6}{12})(\frac{3}{12}-\frac{9}{12})}+ 0.0575 \frac{(x-0)(x-\frac{3}{12})(x-\frac{9}{12})}{(\frac{6}{12}-0)(\frac{6}{12}-\frac{3}{12})(\frac{6}{12}-\frac{9}{12})}+ 0.0653 \frac{(x-0)(x-\frac{3}{12})(x-\frac{6}{12})}{(\frac{9}{12}-0)(\frac{9}{12}-\frac{3}{12})(\frac{9}{12}-\frac{6}{12})}$

Ahmad · « **Reply #10 on:** March 27, 2008, 10:07:10 pm »

The basic idea behind the method above:

You'll see that we're given 4 data points, and our function f(x) is composed of the addition of 4 parts. You can think of each part separately. With the first part we concentrate on the first point (0,0.0499), we're looking for a polynomial that vanishes (is 0) at the other 3 points, that's why the numerator has $(x - \frac{3}{12})$ etc, so when I plug in $x = \frac{3}{12}$ this part is zero. Now, we also require that at x = 0, f(x) = 0.0499, this is why there are fractions in the denominator, so when I plug in x = 0, the numerator and denominators all cancel out, leaving the coefficient 0.0499.

The other parts are constructed similarly, and it is easy to see why this works.

ed_saifa · « **Reply #11 on:** March 27, 2008, 10:31:44 pm »

wow that method is crazy! i like it. good one Ahmad!

jess3254 · « **Reply #12 on:** March 28, 2008, 12:58:41 am »

Quote from: Ahmad on March 27, 2008, 10:07:10 pm

The basic idea behind the method above:

You'll see that we're given 4 data points, and our function f(x) is composed of the addition of 4 parts. You can think of each part separately. With the first part we concentrate on the first point (0,0.0499), we're looking for a polynomial that vanishes (is 0) at the other 3 points, that's why the numerator has $(x - \frac{3}{12})$ etc, so when I plug in $x = \frac{3}{12}$ this part is zero. Now, we also require that at x = 0, f(x) = 0.0499, this is why there are fractions in the denominator, so when I plug in x = 0, the numerator and denominators all cancel out, leaving the coefficient 0.0499.

The other parts are constructed similarly, and it is easy to see why this works.

Ah yes I think this makes sense

Thanks!

The problem in itself isn't hard at all, just extremely time consuming if you use my crappy way by subtracting

brendan · « **Reply #13 on:** March 28, 2008, 03:04:49 pm »

Find $78000 \times \exp\left(\displaystyle\int_0^1 \left(\displaystyle\frac{56}{1875}t^3+ \displaystyle\frac{-2}{125}t^2 +\displaystyle\frac{59}{2750}t +0.0499\right)\,dt\right)$ to 2 decimal places.

dcc · « **Reply #14 on:** March 28, 2008, 03:16:46 pm »

83051.21

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