Polynoial Tactics 2!

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

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

Where II belongs to (1,2,3,4).

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

Note by Naitik Sanghavi
3 years, 8 months ago

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The answer should be 215\frac{2}{15}. Use Lagrange Interpolation

Raushan Sharma - 3 years, 8 months ago

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Can u solve more easily? Or Please explain your solution!

naitik sanghavi - 3 years, 8 months ago

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Actually, in such questions where a polynomial f(x)f(x) follows a pattern for certain values in its domain, we generally use a trick like defining a new polynomial g(x)g(x) whose roots are known, and then calculating f(x)f(x). For instance, here we could have done this by defining g(x)=f(x)1x+1g(x) = f(x) - \frac{1}{x+1}. Then the roots of g(x)g(x) are 1,2,3,41,2,3,4. Then, g(x)=k(x1)(x2)(x3)(x4)g(x) = k(x-1)(x-2)(x-3)(x-4), giving us f(x)=k(x1)(x2)(x3)(x4)+1x+1f(x) = k(x-1)(x-2)(x-3)(x-4) + \frac{1}{x+1}. But then, the degree of f(x)f(x) will be 44, but we are given its degree to be 33. And in such cases, I think we have only one option, i.e. to use Lagrange Interpolation.

Raushan Sharma - 3 years, 7 months ago

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@Raushan Sharma g(x)=f(x)1x+1g(x)=f(x)-\dfrac{1}{x+1} is not a polynomial

Abdur Rehman Zahid - 3 years, 7 months ago

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@Abdur Rehman Zahid g(x)g(x) has roots 1,2,3,41,2,3,4. So, it can be expressed in the form g(x)=k(x1)(x2)(x3)(x4)g(x) = k(x-1)(x-2)(x-3)(x-4), which is a polynomial.

Raushan Sharma - 3 years, 7 months 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)1g(x)=(x+1)f(x)-1 You have to express it in this form in order to make g(x)g(x) a polynomial.Anyways, g(x)=0g(x)=0 for x=1,2,3,4x=1,2,3,4.So (x+1)f(x)1=g(x)=c(x1)(x2)(x3)(x4)(x+1)f(x)-1=g(x)=c(x-1)(x-2)(x-3)(x-4) for some constant cc.Putting x=1x=-1 we get: 1=c(2)(3)(4)(5)    c=1120-1=c(-2)(-3)(-4)(-5)\implies c=\frac{-1}{120} Therefore g(x)=(x+1)f(x)1=1120(x1)(x2)(x3)(x4)g(x)=(x+1)f(x)-1=\frac{-1}{120}(x-1)(x-2)(x-3)(x-4) Now simply putting x=5x=5 gives: 6×f(5)1=1120(4)(3)(2)(1)=15    f(5)=2156\times f(5)-1=\frac{-1}{120} (4)(3)(2)(1)=\frac{-1}{5}\\ \implies \boxed{f(5)=\frac{2}{15}}

Abdur Rehman Zahid - 3 years, 7 months 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 - 3 years, 7 months ago

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@Raushan Sharma No problem. Glad to help :)

Abdur Rehman Zahid - 3 years, 7 months ago

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@Abdur Rehman Zahid Yes, but then, f(x)=g(x)+1x+1f(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 - 3 years, 7 months ago

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Since you are evaluating for consecutive integers you could have used method of differences.

A Former Brilliant Member - 3 years, 7 months ago

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@A Former Brilliant Member Yes, method of differences works in those cases, but actually there also I was facing the problem of degree.... :p

Raushan Sharma - 3 years, 7 months ago

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