Full disclosure: This was part of my homework that I couldn't solve completely. Also, I wasn't sure how to give a proper title to this question.
Before going to the statement to prove, I think some context is necessary, and hopefully through my explanation you would also get an idea on the gaps in my understanding of the problem.
The context:
To numerically solve an ODE given by $y'=f(x,y),\;\;y(x_0)=y_0$
The BDF method is constructed using the interpolating polynomial $p$ of degree $k$ from the following $k+1$ (not equidistant) data points:
$$(x_{i-k+1},y_{i-k+1}),...,(x_i,y_i),(x_i+1,y_i+1)$$
with the unknown value $y_{k+1}$. The unknown value is determined by the demand $(A)$:
$$p'(x_{i+1})=f(x_{i+1},p(x_{i+1}))$$
For adaptive step-size, we introduce another polynomial $q$ of degree $k$ from the following $k+1$ (not equidistant) data points: $$(x_{i-k},y_{i-k}),(x_{i-k},y_{i-k}),\dots,(x_i,y_i)$$
The polynomials are constructed using Newton's Interpolation.
The problem:
Prove the following: The two conditions, $(A)$ and $p(x_{i+1}) = y_{i+1}$ imply:
$$0 = q'(x_{i+1}) + \alpha_{i+1}(y_{i+1} - q(x_{i+1})) - f(x_{i+1},y_{i+1})$$
where, $$\alpha_{i+1} = \dfrac{1}{x_{i+1}-x_{i}} + \dfrac{1}{x_{i+1}-x_{i-1}} + \dots + \dfrac{1}{x_{i+1}-x_{i-k+1}} $$
My attempt:
I recognized how the two conditions can be substituted into the statement:
$$
0 = q'(x_{i+1}) + \alpha_{i+1}(y_{i+1} - q(x_{i+1})) - f(x_{i+1},y_{i+1})\\
0 = q'(x_{i+1}) + \alpha_{i+1}(p(x_{i+1}) - q(x_{i+1})) - p'(x_{i+1}) \\
p'(x_{i+1}) - q'(x_{i+1}) = \alpha_{i+1}(p(x_{i+1}) - q(x_{i+1}))\\
$$
finally forming:
$$ \alpha_{i+1} = \dfrac{p'(x_{i+1}) - q'(x_{i+1})}{(p(x_{i+1}) - q(x_{i+1}))} $$
In the next step I can construct the two polynomials $p$ and $q$ and subtract them to hopefully get the expression for $\alpha$.
Through Newton's Interpolation method, I constructed $q$ to be: $$q(x) = [y_{i-k}] + [y_{i-k},y_{i-k+1}](x-x_{i-k}) + [y_{i-k},y_{i-k+1},y_{i-k+2}](x-x_{i-k})(x-x_{i-k+1}) + ... + [y_{i-k},...,y_{i}](x-x_{i-k})\dots(x-x_{i-1})$$ where $[y_0,y_1,...y_k]$ are determined by divided difference
$p(x)$ is constructed similarly by just adding 1 in the subscripts to match its corresponding data points.
$$p(x) = [y_{i-k+1}] + [y_{i-k+1},y_{i-k+2}](x-x_{i-k+1}) + ... + [y_{i-k+1},...,y_{i+1}](x-x_{i-k+1})\dots(x-x_{i})$$
Now I am unsure how to proceed from here, since the subtraction of $p$ and $q$ doesn't seem to be the right idea. Also, I have to take derivative of the polynomials as well? I think there might be an easier way.
Your help would be appreciated, thank you! :)
The interpolation polynomial over all data points from $i-k$ to $i+1$ can be written in Newton form in different ways, depending on the order of the points. So you get by introducing $x_{n-k}$ last on the left side and $x_{i+1}$ last on the right side $$ p(x)+y[x_{i-k},...,x_{i+1}](x-x_{n-k+1})...(x-x_i)(x-x_{i+1}) \\=q(x)+y[x_{i-k},...,x_{i+1}](x-x_{n-k})(x-x_{n-k+1})...(x-x_{i}) $$ or $$ p(x)-q(x)=y[x_{i-k},...,x_{i+1}](x-x_{n-k+1})...(x-x_{i})\cdot(x_{i+1}-x_{n-k}). $$
Now take the logarithmic derivative $$ \frac{p'(x)-q'(x)}{p(x)-q(x)}=\frac1{x-x_{n-k+1}}+...+\frac1{x-x_i} $$ as constant factors reduce to zero under the logarithmic derivative.
Finally, evaluating at $x=x_{i+1}$ gives the claim.