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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
8 months ago

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The answer should be $$\frac{2}{15}$$. Use Lagrange Interpolation · 8 months ago

Can u solve more easily? Or Please explain your solution! · 7 months, 4 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. · 7 months, 3 weeks ago

$$g(x)=f(x)-\dfrac{1}{x+1}$$ is not a polynomial · 7 months, 3 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. · 7 months, 3 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. · 7 months, 3 weeks ago

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: $6\times f(5)-1=\frac{-1}{120} (4)(3)(2)(1)=\frac{-1}{5}\\ \implies \boxed{f(5)=\frac{2}{15}}$ · 7 months, 3 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!! :D · 7 months, 3 weeks ago

No problem. Glad to help :) · 7 months, 3 weeks ago

Since you are evaluating for consecutive integers you could have used method of differences. · 7 months, 3 weeks ago