1. ## lagragian interpolation

Let Q(t) be a quadratic polynomial such that Q(0) = 0,
Q(1) = 2, and Q(3) = 6. Use Lagrangian interpolation to
obtain Q(2).

im not understanding these things at all.

any help?

2. Originally Posted by p00ndawg
Let Q(t) be a quadratic polynomial such that Q(0) = 0,
Q(1) = 2, and Q(3) = 6. Use Lagrangian interpolation to
obtain Q(2).

im not understanding these things at all.

any help?
Basically, they are giving you three points on the polynomial:

$Q({\color{red}0}) = 0$
$Q({\color{blue}1}) = 2$
$Q({\color{magenta}3}) = 6$

You need to determine the quadratic (i.e. degree 2) polynomial that will go through all three of these points. The formula looks like this:

$Q(t) = 0 \cdot \frac{t-{\color{blue}1}}{{\color{red}0}-{\color{blue}1}} \cdot \frac{t-{\color{magenta}3}}{{\color{red}0}-{\color{magenta}3}} + 2 \cdot \frac{t-{\color{red}0}}{{\color{blue}1}-{\color{red}0}} \cdot \frac{t-{\color{magenta}3}}{{\color{blue}1}-{\color{magenta}3}} + 6 \cdot \frac{t-{\color{red}0}}{{\color{magenta}3}-{\color{red}0}} \cdot \frac{t-{\color{blue}1}}{{\color{magenta}3}-{\color{blue}1}}$

I've color-coded the numbers to make it easier to see where each number comes from. For example, look at just the second term:

$2 \cdot \frac{t-{\color{red}0}}{{\color{blue}1}-{\color{red}0}} \cdot \frac{t-{\color{magenta}3}}{{\color{blue}1}-{\color{magenta}3}}$

In the front we multiply by the function value at t=1, which is 2. Then there is one fraction corresponding to each point given OTHER than t=1. Hopefully you can see the general format by studying what I am doing here. It's a bit hard to explain this online so perhaps you will get it just by studying the pattern in my work.

Once you have written out the terms like I did above, you can simplify it to get to a simple polynomial form, although it's not completely necessary. In the problem, it asks you to solve for $Q(2)$ which just means you take the polynomial $Q(t)$ we just created and then plug in $t=2$.

3. ahh i get it now, thank you for that.