1. Sep 11, 2009

### Somefantastik

$$\int^{1}_{-1}f(x)dx = \sum^{n}_{j=-n}a_{j}f(x_{j})$$

Why does $$\sum_{j}a_{j} = 2$$ ?

I know that the aj's are weights, and in the case of [-1,1], they are calculated using the roots of the Legendre polynomial, but I don't understand why they all add up to 2.

2. Sep 11, 2009

### CFDFEAGURU

I believe that they add up to 2 because they are symmetric about the midpoint of the integration range of [-1,1]. For instance if you used a sampling of 10 points (on the same interval of integration) 5 would be positive and 5 would be negative. If you summed up the 5 positive points you would you arrive at 2.0

Thanks
Matt

Last edited: Sep 11, 2009
3. Sep 11, 2009

### Somefantastik

yeah when graphed out weights vs. abscissa, it looks like the attached photo, which is symmetric.

#### Attached Files:

• ###### Gaussian Weights 300.jpg
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4. Sep 12, 2009