If #F(x) = 2F(6-x)# for all #x# then what is #F(1)# ?

2 Answers
George C.
Nov 9, 2017

#F(1) = 0#

Explanation:

Given:

#F(x) = 2F(6-x)#

we find:

#F(1) = 2F(6-1) = 2F(5) = 2(2F(6-5)) = 4F(1)#

Subtracting #F(1)# from both ends and transposing, we find:

#3F(1) = 0#

Dividing both sides by #3# we find:

#F(1) = 0#

In fact, for any #x# we have #6-(6-x) = x#

Hence #F(x) = 4F(x)# and #F(x) = 0#

Footnote

The above proof works not only for normal number systems, but for most rings, e.g. modulo #7# arithmetic, except that it relies on dividing by #3#. So the proof and the result fail in rings that have no multiplicative inverse for #3#, e.g. #ZZ_3#, #ZZ_9#, any field of characteristic #3#.

Cesareo R.
Nov 9, 2017

See below.

Explanation:

We have

#F(x) = 2F(6-x)# and making #y = 6-x#

#F(y-6)=2F(y)# or analogously #F(x-6) = 2F(x)# then

#F(x) = 4 F(x)# or

#3F(x) = 0# or

#F(x) = 0#

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