Golden section inserted
What is the fundamental period of the continuous non-zero function f satisfying f ( x + 1 ) + f ( x − 1 ) = 2 5 + 1 ⋅ f ( x )
The answer is 10.
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2 solutions
Note: All that you have shown is that the period is 10. Why must the fundamental period be 10, as opposed to say 1 0 1 1 0 ?
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I think that 1 0 will be the fundamental period here. I am proving it :
From the note , we try to solve it by taking f ( x ) = λ x and In this case λ comes out to be e i . n 2 . π . From here I can say that λ is one of the n t h roots of unity. Hence it lies on the circle ∣ λ ∣ = 1 . So it will repeat its value once, when it covers the circle one time. So that will be the minimum value of the Period, that is called the Fundamental Period.
For eg. : let f ( x ) = i x where i = − 1 which is one of the fourth roots of unity. So it's fundamental period will be 4 . There exists no positive real number < 4 for which f ( x ) will keep repeating its value after that interval. On the basis of this, 4 will be the fundamental period here.
Similarly, λ is one of the n t h roots of unity. Hence its fundamental period will be n . @Calvin Lin @Sanjeet Raria
Yeah you're right. Can you tell us what's the fundamental period then? And how to show that that is indeed fundamental?
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I added further comments / details to the note that Sandeep linked to. I makes more sense for the discussion to occur there, because he already provided further details of how to proceed.
Nice solution Professor.!
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Your question deserved it!!
Any other way to deal with these kinds of question Sir? @Sandeep Bhardwaj
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If you want to solve it another way, observing the given equation carefully you will find that it is a homogenous linear difference equation of order 2. From there, first you need to find what exactly the function is. And then find the period of f ( x ) . But the solution posted by @Sanjeet Raria is elegant and very nice. Have a look at that.
Dont you think all multiples of 10 would be the period? Question must ask 'Fundamental period'
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Yeah right...sorry for that. I will edit. Thank you for pointing it out.!
The golden number φ = 2 1 + 5 is the larger root of x 2 − x − 1 = 0 . Therefore φ 2 − φ − 1 = 0 ⇒ φ 2 − φ = 1 ⇒ φ 2 − 1 = φ .
We have f ( x + 1 ) = φ f ( x ) − f ( x − 1 ) , therefore:
f ( 2 ) = φ f ( 1 ) − f ( 0 )
f ( 3 ) = φ f ( 2 ) − f ( 1 ) = φ 2 f ( 1 ) − φ f ( 0 ) − f ( 1 ) = φ f ( 1 ) − φ f ( 0 )
f ( 4 ) = φ 2 f ( 1 ) − φ 2 f ( 0 ) − φ f ( 1 ) + f ( 0 ) = f ( 1 ) − φ f ( 0 )
f ( 5 ) = φ f ( 1 ) − φ 2 f ( 0 ) − φ f ( 1 ) + φ f ( 0 ) = − f ( 0 )
f ( 6 ) = − φ f ( 0 ) − f ( 1 ) + φ f ( 0 ) = − f ( 1 )
f ( 7 ) = − φ f ( 1 ) + f ( 0 ) = − f ( 2 )
f ( 8 ) = − φ f ( 1 ) + φ f ( 0 ) = − f ( 3 )
f ( 9 ) = − f ( 1 ) + φ f ( 0 ) = − f ( 4 )
f ( 1 0 ) = − φ f ( 1 ) + φ 2 f ( 0 ) + φ f ( 1 ) − φ f ( 0 ) = f ( 0 )
f ( 1 1 ) = f ( 1 ) , f ( 1 2 ) = f ( 2 ) . . . ⇒ f ( 1 0 n + x ) = f ( x )
Therefore the period of f ( x ) is 1 0 .
Let f ( x ) = a , f ( x − 1 ) = b Now f ( x + 1 ) = 2 √ 5 + 1 a − b Replacing x with ( x + 1 ) & solving, ⇒ f ( x + 2 ) = 2 √ 5 + 1 ( a − b ) ⇒ f ( x + 3 ) = a − 2 √ 5 + 1 b ⇒ f ( x + 4 ) = − b ⇒ f ( x + 5 ) = − a ⇒ f ( x + 1 0 ) = − f ( x + 5 ) = − ( − a ) = a = f ( x ) Hence we get, f ( x + 1 0 ) = f ( x ) ⇒ P e r i o d i s 1 0
You can read a note regarding such difference equations by clicking here