1we can see that the following polynomial is satisfied for x1,x2,x3,...xm ;
but the polynomial is of degree (m-1)
since it has roots greater than its degree.
thus this polynomial is satisfied for all real numbers .
i.e it has infinite roots.
Based on the question of 29th August, read and try to find a solution to this one...
if x1, x2.... xm are distinct numbers, and f(x) is a polynomial of degree m-1
f(x)=f(x_1)\times\frac{(x-x_2)(x-x_3)...(x-x_m)}{(x_1-x_2)(x_1-x_3)...(x_1-x_m)}+f(x_2)\times\frac{(x-x_1)(x-x_3)...(x-x_m)}{(x_2-x_1)(x_2-x_3)...(x_2-x_m)}+...+f(x_m)\times\frac{(x-x_1)(x-x_2)...(x-x_{m-1})}{(x_m-x_1)(x_m-x_2)...(x_m-x_{m-1})}
What can you say about the function above?
PS: This question is open to students other than Aveek ;)
btw If you think this is way to tough, you wud be advised to read the [url=http://www.targetiit.com/iit-jee-forum/posts/29-08-09-another-quadratic-equation-question-10840.html]question[/url]
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18 Answers
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1Okie [1]
I actually thought that you gave the question like that on purpose
bcos we knew the value of f(x) (a M-1 degree polynomial) at M points which means (unless I am mistaken) that we have uniquely identified the function using langrange's interpolation formula (even here I may be mistaken).
now once we have identified the function , its not a big deal that you ask about its roots . however now i see that the roots of f(x) would not have been found without some more information
62@Philip You are right...
A slight justification.. (something that I may not have cleared nicely)
THe fact is that LHS is a n-1 degree polynomial. Ditto with the RHS.
LHS-RHS is a n-1 degree polynomial.
Hence it can have max n-1 roots.
BUt x1, x2.... xn are n roots that we can observe.
Hence, LHS-RHS = 0 everywhere....
Hence LHS = RHS for all values of x.
Sorry I did not frame the question correctly :(
I will try to fix the language
1can someone clarify my doubt...
i always thought root(s) of a function meant the value(s) of x for which the function equals zero....
and I can't see how f(x1) = f(x2) =......= f(xm)=0 either [2]
however i totally understand that a polynomial of degree 'n' cannot take same value more than 'n' times. but how to use that concept to find the roots of f(x) in this equation??
1Yes that may be clear that the equation written above is an identity and holds for all real values of x
but Nishant bhaiyya has asked us about the roots of f(x)
that means the values of x at which f(x) is zero
I dont see how ur explanation holds there ...
or am I too confused ?
1no i said that it has infinite real roots.
if we take x= x1,x21x3,...xm ;
it satisfies the given equation .
but the equation has (m-1) degree thus it should have at most (m-1) roots .
thus given polynomial is an identity and will be satisfied by all real numbers. thus infinite roots.
1@Anirudh
can u elaborate a bit
I dont think we can say that x1,x2,x3...xm can be said to be the roots of this function or can we?
62rickde.. there is a far simpler way to think.. and i did not completely understand your logic...
so the function we got after m-2 times will have one more root but as its differentiation has infinite roots this will have infinite roots
I dont think this statemet is true! THis is the converse of the arguement that n roots means n-1 roots of the derivative... Which is not true..
1we know that if f(x) has n roots then f'(x) has n-1 roots
if we differentiate the above function m-1 times we will get a constant function
but we see the RHS is zero at a point
as it is a constant and it is zero at a point it is zero at all points
so the function we got after m-2 times will have one more root but as its differentiation has infinite roots this will have infinite roots
going backwards like this we see that the given function has infinite roots
i.e. it is a identity
not a good proof innit?
62yeah.. i now realize that we had discussed this one.. but what is the proof of the same.. (not to say that this is difficult ;)
11 reference over here
http://www.targetiit.com/iit-jee-forum/posts/one-more-integral-3560.html
62Yes indeed it is that formula.. but giving names to simple things in life makes them seem more difficult :D
But yeah this one is a very easy one if you use the hint and the solution of the previous question :)
11This is a identity i have read about this it is langerange's interpolation Theorem
Rite sir?