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j2kun avatar j2kun commented on June 6, 2024 2

It is exactly the Lagrange interpolating polynomial, though I didn't hear that term when I first saw this. To the best of my memory, I originally discovered this way of writing the polynomial as a homework exercise in an undergraduate linear algebra class (the book was Axler's "Linear Algebra Done Right," the exercise was Chapter 4, Exercise 2).

Though I'm being somewhat post hoc since it was many years ago, my thought process was: can I break the problem down into more independent parts? The independence is across addition, and I made those parts "independent" because all but one of the terms is zero when plugging in the relevant x values.

Perhaps more practically, this is the kind of exercise for which mathematicians in my life have often said, "you just do it," meaning you pull out some scratch paper and experiment on small examples until you find an approach that generalizes.

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j2kun avatar j2kun commented on June 6, 2024 2

@abjrcode Those two are the same, and you can see this if you multiply -1 on the top and bottom of each term.

Also, if you suspect whether they're the same expression hidden in disguise, you can test it by evaluating the two functions at different values. We know from the polynomial interpolation theorem that if they agree on two or more inputs they have to be the same (even if they look different algebraically).

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perkss avatar perkss commented on June 6, 2024

Too true "you just do it," exactly why I was so happy to see your book come out! Thanks for the explanation of your thought process and the reference will check it out! Loving the book so far.

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abjrcode avatar abjrcode commented on June 6, 2024

I must be doing something stupid but I am arriving at a different representation:
WIN_20191229_17_15_52_Pro

So you are arriving at:
CodeCogsEqn (1)

and I am arriving at:
CodeCogsEqn

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abjrcode avatar abjrcode commented on June 6, 2024

Don't know why I didn't think of that 🤦‍♂
Rusty math mind
Thank you for your answer

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