Question

f(x)+f(x+2)=root3 f(x+1) f(5)=3 so find f(5+12r) where r =0,1,2.........99

Answer 1

sqrt is only on 3 right ? not f(x+1)

Answer 2

i think yes\[f(x)+f(x+2)=\sqrt{3} f(x+1)\]

Answer 3

Dear Sir, Do you require a one-off special case solution or a METHOD to SOLVE THIS and MANY OTHER SIMILAR PROBLEMS ?

Answer 4

The METHOD is well known its:

Answer 5

u want to say its periodic?\[f(5+12r)=f(5)=3 \ \ \ r=0,1,2,...\]

Answer 6

i've seen it before but i cant remember how to do it :(

Answer 7

FOURIER TRANSFORM ( or a related technique of Laplace transform). AS is well known shift OPERATOR IN TIME DOMAUN is equivalent to PHASE MULTIPLIER IN FREAUENCY DOMAIN

Answer 8

So let us begin the solution

Answer 9

Let us TRANSFORM BOTH SIDES OF THIS EQUATION INTO THE FREQUENCY DOMAIN

Answer 10

isn't this a linear recurrence relation?

Answer 11

where you use a characteristic equation to solve it?

Answer 12

Shifting by D like f(x+D) is equivalent to MULTIPLICATION by exp(-iDf)

Answer 13

Of COURSE IN THE FOURIER DOMAIN

Answer 14

SO your equation becomes:

Answer 15

yes asnaseer its similar but there is only one given condition

Answer 16

I thought you only needed one given condition to solve a linear recurrence relation?

Answer 17

F(f) + exp(-iDf)*F(f) = sqrt5 * exp(-if)F(f)

Answer 18

as one can see this is a trivial linear equation on F(f)

Answer 19

for what its worth, this technique is described here: http://hcmop.wordpress.com/2012/04/20/using-characteristic-equation-to-solve-general-linear-recurrence-relations/

Answer 20

by the way the generating function approach is equivalent - via Z-transform, which is a complex variant of Fourier transform

Answer 21

F(f) (1 + exp(-i5f)) = sqrt(5)*exp(-if)

Answer 22

F(f) =\[F(f) = \frac{ \sqrt{5}e^{-if}}{ 1 + e^{-i5f} }\]

Answer 23

Now @gauti1989 this is a simple fraction - ITS INVERSE FOURIER TRANSFORM IS WELL KNOWN, and you can find it in any table and possibly even Wolfram Alpha will compute the inverse transform of this. Then you require that you condition will also be true and obtain the final answer

Answer 24

If you want 'just the answer" without understanding the function - then notice that you have a simple recursion equation with 2 previous members of the sequence
f(x+2)=root3 f(x+1) - f(x) which can be rewritten thus

Answer 25

f(x+1) = root3 f(x) - f(x-1)

Answer 26

So just jump forward 1 index, 2 indexes, 3 indexes - you will quickly discover a regularity

Answer 27

Or , you can look up "z-transform" + "recursive equation" in ggl

Answer 28

recursive equation will not work since it requiers 2 given condition

Answer 29

i m not getting it xactly plz help me

Answer 30

\[f(x)+f(x+2)=\sqrt{3} \ f(x+1)\]let x=x+1\[f(x+1)+f(x+3)=\sqrt{3} \ f(x+2) \ \ (1)\]let x=x-1\[f(x-1)+f(x+1)=\sqrt{3} \ f(x) \ \ (2)\]
add (1) and (2)\[f(x-1)+f(x+3)+2f(x+1)=\sqrt3 \ (f(x)+f(x+2))=\sqrt3 \ (\sqrt{3} \ f(x+1))\]\[=3f(x+1)\]\[f(x-1)+f(x+3)=f(x+1)\]check till here

Answer 31

\[f(x-1)+f(x+3)=f(x+1) \ \ (1)\]let x=x+2\[f(x+1)+f(x+5)=f(x+3) \ \ (2)\]
add (1) and (2)\[f(x-1)+f(x+3)+f(x+1)+f(x+5)=f(x+1)+f(x+3)\]\[f(x-1)+f(x+5)=0\]

Answer 32

\[f(x-1)+f(x+5)=0\]let x=x+1\[f(x+6)+f(x)=0\]

Answer 33

\[f(x+6)=-f(x)\]let x=x+6\[f(x+12)=-f(x+6)=-(-f(x))=f(x)\]\[f(x+12)=f(x)\]so f is a periodic function with \(T=12\)

Answer 34

\[f(5+12r)=f(5)=3 \ \ \ r=0,1,2,...\]

Answer 35

i hope its clear :)