Method of differences

Main post link -> https://brilliant.org/assessment/techniques-trainer/method-of-differences/

Hello everyone ,

I've just learnt the "Method of differences" and I'm really fascinated by the method !

Can somebody help me out how can we find the original polynomial f(x) f(x) ,

[ For example f(x) f(x) of the form f(x)=axn+bxn1+ f(x)= ax^n + bx^{n-1} + \dots ]

after drawing the difference table (i.e. after finding the values of f(a),D1(a) f(a) , D_1(a) \dots ) ?

Do elaborate 'cause I don't know much about Binomials .

Thank you.

Note by Priyansh Sangule
5 years, 10 months ago

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Are you talking about the values of f(n)f(n), or the closed form?

Tim Vermeulen - 5 years, 10 months ago

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The original polynomial of the form f(x)=axn+bxn1+ f(x) = ax^n + bx^{n-1} + \dots

Priyansh Sangule - 5 years, 10 months ago

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Okay. Could you give an example of such a table? I'd be happy to help you find the original polynomial.

Tim Vermeulen - 5 years, 10 months ago

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@Tim Vermeulen For example -

A cubic polynomial f(x)f(x) has the following values -

f(1)=13;f(2)=32;f(3)=69;f(4)=130f(1)=13 ; f(2)=32 ; f(3)=69 ; f(4)=130

Find f(x) f(x) .

So for this , first of all , I draw a difference table

nf(n)D1(n)D2(n)D3(n)113191862323724369614130 \begin{matrix} n & f(n) & D_1(n) & D_2(n) & D_3(n) & \dots \\ 1 & 13 & 19 & 18 & 6 & \dots \\ 2 & 32 & 37 & 24 & \dots \\ 3 & 69 & 61 & \dots \\ 4 & 130 & \dots \\ \vdots \end{matrix}

Now , at this point I run into trouble - that how can I find the polynomial of the form f(x)=axn+bxn1+ f(x) = ax^n + bx^{n-1} + \dots

Please help and do elaborate !

Thank you

Priyansh Sangule - 5 years, 10 months ago

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I'm encourage you to read through Worked Example 1, and do the (*) exercise, which gives you the answer.

Calvin Lin Staff - 5 years, 10 months ago

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Yes sir, I did note that but couldn't it be written in a simpler form ? I saw some people using combinatorics for that!

Please help , Thank you !

Priyansh Sangule - 5 years, 10 months ago

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