Find the general formula for f(n) in terms of n.

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Aphelion_ts
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#1
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An investigation question from UK maths changes 2019 Q22, can someone please help=3

A function f satisfies the equation (n-2019)f(n) - f(2019-n) = 2019 for every integer n. Find the general formula for f(n) in terms of n.
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mqb2766
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Any ideas, have you worked out any of the simple values for instance to get some intuition.

Edit - looks like there is some symmetry that you could exploit for a general expression for f(n)?
Last edited by mqb2766; 1 month ago
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Aphelion_ts
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(Original post by mqb2766)
Any ideas, have you worked out any of the simple values for instance to get some intuition.

Edit - looks like there is some symmetry that you could exploit for a general expression for f(n)?
There are two questions before this, asking for f(1) and f(2019)
I was thinking of using fibonacci… dunno

Edit - f(2019)=2018*2019[official solution] and my answer to f(1) is zero (or complex if possible)
Last edited by Aphelion_ts; 1 month ago
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mqb2766
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(Original post by Aphelion_ts)
There are two questions before this, asking for f(1) and f(2019)
I was thinking of using fibonacci… dunno
Have you done the original question?
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Aphelion_ts
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(Original post by mqb2766)
Have you done the original question?
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mqb2766
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So any ideas based on the first two questions (the original smc one and 22.1)
Last edited by mqb2766; 1 month ago
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superharrydude09
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Consider the arguements in f(n), since making a certain substitution will reverse the argurments of f(n), and you get two equations with which you can cancel out f(n-2019), leaving you with a single equation with just f(n).
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Aphelion_ts
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(Original post by superharrydude09)
Consider the arguements in f(n), since making a certain substitution will reverse the argurments of f(n), and you get two equations with which you can cancel out f(n-2019), leaving you with a single equation with just f(n).
Indeed we can do that for each of the specific n values but what about general rules?
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mqb2766
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(Original post by Aphelion_ts)
Indeed we can do that for each of the specific n values but what about general rules?
When you sub n=2019-n you get
-n*f(2019-n) - f(n) = 2019
as well as the original equation
(n-2019)f(n) - f(2019-n) = 2019
You have two simultaneous equatins in two unknowns f(n) and f(2019-n), so solve them and get f(n) just as you did for f(2019) and f(1) in the first two questions.
Last edited by mqb2766; 1 month ago
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