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Polynoial Tactics 2!

\( P(x) \) 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 \(P(5)\).

Note by Naitik Sanghavi
5 months, 2 weeks ago

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The answer should be \(\frac{2}{15}\). Use Lagrange Interpolation Raushan Sharma · 5 months, 2 weeks ago

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@Raushan Sharma Can u solve more easily? Or Please explain your solution! Naitik Sanghavi · 5 months, 2 weeks ago

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@Naitik Sanghavi 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. Raushan Sharma · 5 months, 2 weeks ago

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@Raushan Sharma \(g(x)=f(x)-\dfrac{1}{x+1}\) is not a polynomial Abdur Rehman Zahid · 5 months, 2 weeks ago

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@Abdur Rehman Zahid 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. Raushan Sharma · 5 months, 2 weeks ago

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@Abdur Rehman Zahid \(g(x)\) 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. Raushan Sharma · 5 months, 2 weeks ago

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@Raushan Sharma No,this question can still be done by Remainder Factor Theorem.

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 · 5 months, 2 weeks ago

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@Abdur Rehman Zahid 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!! :D Raushan Sharma · 5 months, 2 weeks ago

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@Raushan Sharma No problem. Glad to help :) Abdur Rehman Zahid · 5 months, 2 weeks ago

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@Naitik Sanghavi Since you are evaluating for consecutive integers you could have used method of differences. Vighnesh Shenoy · 5 months, 1 week ago

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@Vighnesh Shenoy Yes, method of differences works in those cases, but actually there also I was facing the problem of degree.... :p Raushan Sharma · 5 months, 1 week ago

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