33Now got sumthing, actually i have never encountered recurrence relations... but now read something about it... that basic i think is enough....
Thanxxxx
It is given that :
f(x+2)-5f(x+1)+6f(x)=0
Find f(x)
??????????????
-
UP 0 DOWN 0 0 19
19 Answers
33
62abhishek i think ur last term should have been zero.. or should have had a dy/dx.
in your first step
33assuming it dx...
differentiating it wrt x
we get
yf(xy)=xf(x)+0∫xf(x)dx
Putting y=1...
f(x)-xf(x)=0∫xf(x)dx
again differentiating wrt x....
f'(x)-xf'(x)-f(x)=f(x)
f'(x){1-x}=2f(x)
(dy/dx)(1-x)=2y
dy/2y=dx/(1-x)
is this correct ??
62dude what is the variable of integration?
0∫xyf(xy) = x(0∫x f(x)) + y(0∫yf(y))
0∫xyf(xy)(dx/dy/dxy) = x(0∫x f(x)(dx/dy/dxy)) + y(0∫yf(y)(dx/dy/dxy))
1ok nishant i've understood the logic
Now plz solve this one
0 ∫ xy f(xy) = x ( 0 ∫ x f(x) dx) + y ( 0 ∫ y f(y)dy )
Given that f(1) = 3
62Abhishek, as such there is no logic! (Atleast i dont know of any!)
basically this is a standard method which gives a solution! and the solution is very satisfactory.
Have u solved some problems where u assume function to be of polynomial kind. and then find the corresponding polynomial?
basically what u are asking is why do we solve it the way we do...
i think it is bcos we get a solution this way :)
33U can refer:
http://en.wikipedia.org/wiki/Recurrence_relation
1nishant or abhishek priyam anyone, anyone
Can u plz explain why the f() whwre relatd to quadratic coefficients .and plz explain whaT WAS The logic of solving .
62f(x+2)=5f(x+1)-6f(x)
Well for any two starting values of f(0) and f(1)
it will give the values for all the further x's
neither will it be a continuous function!
I mean unless u want a solution for integer values... (i mean x in Z or N)
62f(x+2)-5f(x+1)+6f(x)=0
take the related quadratic!
x2-5x+6=0
the roots are 2 and 3
so the general soln will be of the form
f(x)=a.2n+b.3n
substitute this in the above equation..
u will have two values f(0) and f(1) as constants or some other way to find two constants a and b
btw u should try to get the solution of the general term of the fibonacci series 1,1,2,3,5,8....
the recurrence relation will be f(n+2)=f(n+1)+f(n)
btw dont get worried about the complex looking roots.. they will work as the way these worked here :)
62yes the answer is an exponential form.. i will tell u the proof as well :)
33Don't know the answer but one of my frnd gave me said it is from some book named JPNP or so.....
just overheard its answer is some exponential or might this question is not complete....
62btw this method that i am talking is the same that is used to find the general term of the fibonacci series!
62Having said that, there is a standard metod by assuming a form for f(x)
this is for recurrence relations!
we first convert this to a quadratic form then take the roots and then the general soluton is found by taking the sum of powers of the roots mulitplied by a coefficient!
That is a standard method..
if that is the solution u are looking for then i will post it! I dont see anything brilliant about the solution.. (unfortunately!)