# Weighted Moving Average of Cubic

1. Feb 2, 2013

### Scootertaj

1. Show that applying a second-order weighted moving average to a cubic polynomial will not change anything.
$$X_t = a_0 + a_1t + a_2t^2 + a_3t^3$$ is our polynomial
Second-order weighted moving average: $$\sum_{i=-L}^{i=L} B_iX_{t+i}$$
where $$B_i=(1-i^2I_2/I_4)/(2L+1-I_2^{2}/I_4)$$
where $$I_2=\sum_{i=-L}^{i=L} i^2$$

2. I did a similar problem where applying a linear moving average to a linear equation returns the same thing, but I'm not sure how to proceed here.

2. Feb 2, 2013

### Ray Vickson

What is L supposed to be in your case?

3. Feb 2, 2013

### Scootertaj

L is any arbitrary number.
For any L, this should be true.

4. Feb 2, 2013

### SammyS

Staff Emeritus

What is I4 ?

5. Feb 2, 2013

### Scootertaj

Sorry, it just follows the same pattern as I2:

$$I_4=\sum_{i=-L}^{i=L} i^{4}$$

6. Feb 2, 2013

### Ray Vickson

So you have weighting factors of the form
$$B_i = A - B i^2, i=-L, \ldots, L.$$ What happens when you compute
$$\sum_{i=-L}^{L} (A - B i^2) [a_0 + a_1 (t+i) + a_2 (t+i)^2 + a_3 (t+i)^3]?$$

7. Feb 2, 2013

### Scootertaj

I'm unsure as to how you rearranged the weights to get $$B_i = A - B i^2$$, would you mind clarifying what A and Bi2 are?

Last edited: Feb 2, 2013
8. Feb 2, 2013

### Scootertaj

I mean, one way to write what we have is
$$\frac{1}{2L+1-I_2^{2}/I_4} \sum_{i=-L}^{i=L} (1-i^2\frac{I_2}{I_4}) [a_0 + a_1 (t+i) + a_2 (t+i)^2 + a_3 (t+i)^3]$$

9. Feb 2, 2013

### Scootertaj

I THINK I may have got it. I basically looked at what we had, and what we need.

In order for us to get back a0, for example, we need:

$$\frac{1}{2L+1-\frac{I_2^{2}}{I_4}}\sum_{i=-L}^{i=L}a_0(1-i^2\frac{I_2}{I_4}) = a_0$$
Well, let's multiply through:

$$\frac{a_0}{2L+1-\frac{I_2^{2}}{I_4}}\sum_{i=-L}^{i=L} 1 - a_0\frac{\sum_{i=-L}^{i=L} i^2\frac{I_2}{I_4}}{2L+1-\frac{I_2^{2}}{I_4}}$$
$$\frac{a_0(2L+1) - a_0\frac{I_2^{2}}{I_4}}{2L+1-\frac{I_2^{2}}{I_4}} = a_0$$

Then, do the same for the rest.