19this question was given by anant sir in one of his class tests , and i solved it using one of my " sleight of the hand " tricks(i had the options for the answer there because the question was an mcq!)
We have seen so many polynomial questions.. But I am still giving this one because I cant resist from feeling how many of you have really understood some of what we discussed in thsoe questions..
P(x) is a polynomial of degree n such that p(k) = k/(k+1) for k=0,1 ,2, ... n
find the value of P(k+1)
Found this link in a paper of one of my friends who is Caltech.. and completed his PHd in maths and represented India 2 times at IMO :)
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9 Answers
19
3i feel dat it is like dat previous one,but i fear that it is an olympaid one also its a qod.
1why it is very much the same thing ..
http://targetiit.com/iit-jee-forum/posts/nice-one-on-polynomials-11705.html
62Yes it is indeed very very similar to that one.. and that is why i commented that before even teh statement fo the question..
But what is the complete solution?
1g(x)=(x+1)f(x)-x....
it is a poynomial of degree n+1 having n+1 roots as
0,1,2,3........n
so....
g(x)=ax(x-1)(x-2)......(x-n)
g(-1)=a(-1)n(n+1)!=1
a=1/(-1)n(n+1)!
g(n+1)=(n+2)f(n+1)-(n+1)
g(n+1)=a(n+1)!
f(n+1)=(a(n+1)!+(n+1))/n+2
= (-1)n+n+1 /n+2
62Awesome :)
Good I am happy that some of you have learnt something from that post :)
62damn my firefox crashed... I have to rewirte the whole thing again...
There is a very important question to be answered.. How did we get to the function g(x) as constructed by xyz? THe construction may seem very difficult at the first look but is not so difficult. The solution is far simpler.. The method to do this is that f(k) = k/(k+1) for x= 0, 1, 2, ... n
Now we should also realize that it is far more difficult to handle f(x)/g(x) than to handle f(x) alone.. so if we can somehow reduce the information in the quesiton to a polynomial, we will be in a very good shape to handle these questions....
so we can observe from f(k) = k/(k+1) that it is same as saying that f(k).(k+1) = k or f(k) (k+1) - k = 0 for k = 0, 1, , 2, ... n
and if we write g(x) = f(x)(x+1)-x then g(x) is zero for x = 0, 1, .... n
hence the motivation for writing g(x) the way it has been written by xyz