Do you know its property? -11
If the fundamental period of a continuous non-zero function f ( x ) satisfying
f ( x + 1 ) + f ( x − 1 ) = π . f ( x )
is a 1 a 2 a 3 a 4 a 5 a 6 a 7 . b 1 b 2 b 3 b 4 b 5 b 6 b 7 , find the value of i = 1 ∑ 7 ( a i + b i ) .
Assumptions:
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Round off your answer up to 7 decimal places.
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0 ≤ a 1 , a 2 , a 3 , a 4 , a 5 , a 6 , a 7 , b 1 , b 2 , b 3 , b 4 , b 5 , b 6 , b 7 ≤ 9
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{ a 1 , a 2 , a 3 , a 4 , a 5 , a 6 , a 7 , b 1 , b 2 , b 3 , b 4 , b 5 , b 6 , b 7 } ∈ Z
The answer is 35.
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1 solution
In this question, you will need the function to be continuous. There are a lot of discontinuous functions which satisfy the functional equation. Can you think of several?
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Now I've mentioned the function f ( x to be continuous. Earlier, I din't because I was by default continuous in my thoughts, because earlier I was working with sequences where "n" is supposed to be discrete i.e. non-negative integers, and after that I worked on "x" a continuous variable. So no doubt, the function f ( x ) in this case must be continuous. @Calvin Lin
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Sandeep sir, please clarify me what are a and b??
The f u n d a m e n t a l period of the function f ( x ) (non-zero) satisfying f ( x + 1 ) + f ( x − 1 ) = t . f ( x ) is c o s − 1 ( 2 t ) 2 . π
So taking t = π , we get Fundamental Period = 1 3 . 0 4 4 8 4 2 9 (after rounding off up to 7 decimal places) .
So { a 1 = a 2 = a 3 = a 4 = a 5 = 0 , a 6 = 1 , a 7 = 3 b 1 = 0 , b 2 = b 3 = 4 , b 4 = 8 , b 5 = 4 , b 6 = 2 , b 7 = 9 .
Hence i = 1 ∑ 7 ( a i + b i ) = 3 5
For the proof of the fundamental period, I'm writing a note. You can see it by clicking here .
enjoy!