66@Qwerty
How do you define a polynomial?
First look when i saw this question, I got bowled.. Then the feeling of solving this one was good.. [1] .. I thought THis is a nice question that you guys should try
F(x) is a polynomial of degree n such that f(k)=1/k for k=1, 2, .. n+1
find the value of f(n+2)
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22 Answers
3thanq prophet sir for that hint,it was in the rite time.Atleast from this problem i wont forget these three points ever.
3lastly i am going to add wat i have learnt from the above discussions.
g(x)=xf(x)-1
from the qn g(x) has roots 1,2,3,....n+1.
so we have g(x)=a(x-1).....(x-n-1)
g(0)=a(-1)n+1(n+1)!=-1
a=(-1)n/(n+1)!
g(n+2)=(-1)n
f(n+2)=1+(-1)n/n+2
341you have in your hand all the roots of the polynomial. Why dont you take option 2?
341friends, when it comes to polynomials, it is a helpful to have three ways of thinking about it in your RAM:
(1) Regular representation as a sum of terms - i would call it the coefficient representation because here the coeffs get prominence
(2) as a product of linear terms a \prod_{i=1}^n (x-r_i) where ri are the roots - the roots representation
(3) As a curve on a graph
Now, perhaps the next step is clear
62now that you have g(x) then what are the roots of g(x)
Now what?
3still didn get ne answers from the hint sir
neways this is wat i got.
g(x)=xf(x)-1
from the qn g(x) has roots 1,2,3,....n+1.
so g(n+2)> or < 0
which gives us f(n+2) >1/n+2 or <1/n+2
62One hint: Define a new polynomial.. from all the discussion above! and try to analyze that..
1No one ??!?!
Try to observe the properties of the polynomial
xf(x)-1 .....
66No....the answer is 1+(-1)n/(n+2)
I hope Nishant sir won't mind me giving the answer.
62hmm.. sorry slightly wrong hint :D :P
Notice that the new polynomail that you gave is of degree n+1 now! so it is allowed to have n+1 real roots!
1Oh shit ! I messed up
xF(x) - 1 is a n+1 degree so we can't say whatever rubbish i said
Im so sorry .
62to put kaymant sir's question in a slightly different way,
f(x) = 0 for x=1, 2, ... n+1
then does it mean that f(x) = 0 for all x? ? [7]