\( P(x) \) is polynomial of degree 3 such that,

\( P(I)=1/(1+I) \),

Where \(I\) belongs to (1,2,3,4).

Then find the value of \(P(5)\).

\( P(x) \) is polynomial of degree 3 such that,

\( P(I)=1/(1+I) \),

Where \(I\) belongs to (1,2,3,4).

Then find the value of \(P(5)\).

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TopNewestThe answer should be \(\frac{2}{15}\). Use Lagrange Interpolation – Raushan Sharma · 11 months, 2 weeks ago

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– Naitik Sanghavi · 11 months, 2 weeks ago

Can u solve more easily? Or Please explain your solution!Log in to reply

– Raushan Sharma · 11 months, 2 weeks ago

Actually, in such questions where a polynomial \(f(x)\) follows a pattern for certain values in its domain, we generally use a trick like defining a new polynomial \(g(x)\) whose roots are known, and then calculating \(f(x)\). For instance, here we could have done this by defining \(g(x) = f(x) - \frac{1}{x+1}\). Then the roots of \(g(x)\) are \(1,2,3,4\). Then, \(g(x) = k(x-1)(x-2)(x-3)(x-4)\), giving us \(f(x) = k(x-1)(x-2)(x-3)(x-4) + \frac{1}{x+1}\). But then, the degree of \(f(x)\) will be \(4\), but we are given its degree to be \(3\). And in such cases, I think we have only one option, i.e. to use Lagrange Interpolation.Log in to reply

– Abdur Rehman Zahid · 11 months, 2 weeks ago

\(g(x)=f(x)-\dfrac{1}{x+1}\) is not a polynomialLog in to reply

– Raushan Sharma · 11 months, 2 weeks ago

Yes, but then, \(f(x) = g(x) + \frac{1}{x+1}\) is not a polynomial. So, there is some flaw in this method. That's why I didn't do it with this method, as I said earlier. A traditional method of Lagrange Interpolation, as I mentioned earlier is the best here, I think.Log in to reply

– Raushan Sharma · 11 months, 2 weeks ago

\(g(x)\) has roots \(1,2,3,4\). So, it can be expressed in the form \(g(x) = k(x-1)(x-2)(x-3)(x-4)\), which is a polynomial.Log in to reply

Let \(g(x)=(x+1)f(x)-1\) You have to express it in this form in order to make \(g(x)\) a polynomial.Anyways, \(g(x)=0\) for \(x=1,2,3,4\).So \((x+1)f(x)-1=g(x)=c(x-1)(x-2)(x-3)(x-4)\) for some constant \(c\).Putting \(x=-1\) we get: \[-1=c(-2)(-3)(-4)(-5)\implies c=\frac{-1}{120}\] Therefore \(g(x)=(x+1)f(x)-1=\frac{-1}{120}(x-1)(x-2)(x-3)(x-4)\) Now simply putting \(x=5\) gives: \[6\times f(5)-1=\frac{-1}{120} (4)(3)(2)(1)=\frac{-1}{5}\\ \implies \boxed{f(5)=\frac{2}{15}}\] – Abdur Rehman Zahid · 11 months, 2 weeks ago

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– Raushan Sharma · 11 months, 2 weeks ago

Ohh, that's really nice. So, this method also works here.... Thanks for your solution (+1). And definitely it's better than Interpolation in such cases, where there is a pattern given. Thanks once again!! :DLog in to reply

– Abdur Rehman Zahid · 11 months, 2 weeks ago

No problem. Glad to help :)Log in to reply

– Vighnesh Shenoy · 11 months, 1 week ago

Since you are evaluating for consecutive integers you could have used method of differences.Log in to reply

– Raushan Sharma · 11 months, 1 week ago

Yes, method of differences works in those cases, but actually there also I was facing the problem of degree.... :pLog in to reply