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Tagged: Functional equations

- This topic has 2 replies, 2 voices, and was last updated 11 months, 2 weeks ago by Agamdeep Singh.

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- June 12, 2019 at 12:42 am #29158Agamdeep SinghParticipant
Is the following method correct?

I mean is it enough to imply that f(n)=N

Do we have to prove wether f is ever increasing or ever decreasing before doing this. Like could it be that f is sometimes greater than n and sometimes less, and we proved that is not greater when it is less.

Kind of like sin(x). It increases and decreases. Do we have to consider the that f might also be increasing and decreasing above and below n.

or is it correct and that is taken care of when we take the case f(n)>n.

Help would really be appreciated.

Thanks.

June 12, 2019 at 10:39 am #29183Nitin PrasadParticipantNo, just checking for the case where f is decreasing or increasing is not sufficient. As you pointed out, we need to consider the case (if possible) where f is just some random arrangement and doesn’t have monotonic property.

Hints

- If ‘f’ a one-one function? Observe that if f(m)=f(n) implies m=n then, the given function is one-one.
- Observe that $$f[ f(m+k)+f(n-k) ]=m+n=f[ f(m)+f(n) ]$$. What can we conclude?

Using further manipulations you can finally show that f(n)=n. The complete solution has been attached below

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You must be logged in to view attached files.June 16, 2019 at 12:50 pm #29266Agamdeep SinghParticipantThanks sir

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