Show that $p'(x_0) = {{4y_1 - 3y_0 - y_2} \over 2h}.$

Question

It was part of an bigger excercise to calculate the Newton polynomial

$$p(x) =y_0 + {{y_1 - y_0} \over h} (x-x_0) + (-2y_0 + 4y_1 - 2y_2)(x-x_0)(x-x_1).$$

Now, I need to show that

$$p'(x_0) = {{4y_1 - 3y_0 - y_2} \over 2h}.$$

But solving this by simplifying $p(x)$ seems way over the top, given the fact that this was an old exam question and that it would take too long. Does anyone see a trick to do this easier?

score 0 · Answer 1 · answered Jun 28 '17 at 10:29

0

Just compute of $$ p(x)=y_0+\frac{y_1-y_0}h(x-x_0)+\frac{y_2-2y_1+y_0}{2h^2}(x-x_0)(x-x_1) $$ the term-wise derivative $$ p'(x)=\frac{y_1-y_0}h+\frac{y_2-2y_1+y_0}{2h^2}(2x-x_1-x_0) $$ and evaluate at $x=x_0$.

Of course it is helpful to start with an interpolation formula that actually has $p(x_2)=y_2$.

answered Jun 28 '17 at 10:29

Lutz Lehmann

126,666

Thanks! Is it possible that my Interpolation formula is wrong then? – Julian Jun 28 '17 at 11:20
Why don't you check yourself what you get for $p(x_2)$? – Lutz Lehmann Jun 28 '17 at 11:58
Well, it's not $y_2$ at least. I should check the formula again then. – Julian Jun 28 '17 at 14:45
The way I see it, the formula doesn't even hold for $p(x_1) = y_1.$ – Julian Jun 28 '17 at 15:03
No, that part is correct as $x_1-x_0=h$ and the third term has a zero factor. – Lutz Lehmann Jun 28 '17 at 16:05

Show that $p'(x_0) = {{4y_1 - 3y_0 - y_2} \over 2h}.$

1 Answers1