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3) Let $$P(x)$$ be second degree polynomial such that : $$P(1)=1$$ $$P(2)=7$$ $$P(3)=19$$ show that $$P(1) + P(2) + P(3) + ...... + P(n) = n^3$$

4) Find all pairs of polynomials $$P(x)$$ and $$Q(x)$$ with real coefficients such that : $$P(x^2 + 1) = Q(x)^2 + 2x$$ and $$Q(x^2 + 1) = P(x)^2$$

Different approaches will be appreciated

Note by Dev Sharma
1 year ago

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3) With $$P(x) = ax^{2} + bx + c$$ we have the system

$$(i) a + b + c = 1, (ii) 4a + 2b + c = 7, (iii) 9a + 3b + c = 19.$$

Now (ii) - (i) yields $$3a + b = 6,$$ and (iii) - (i) yields $$8a + 2b = 18.$$

Thus $$(8a + 2b) - 2*(3a + b) = 18 - 2*6 = 6 \Longrightarrow 2a = 6 \Longrightarrow a = 3, b = -3, c = 1.$$

So $$\displaystyle\sum_{k=1}^{n} P(k) = \sum_{k=1}^{n} (3k^{2} - 3k + 1) =$$

$$\displaystyle 3*\sum_{k=1}^{n} k^{2} - 3*\sum_{k=1}^{n} k + \sum_{k=1}^{n}1 =$$

$$3*\dfrac{n(n + 1)(2n + 1)}{6} - 3*\dfrac{n(n + 1)}{2} + n =$$

$$\dfrac{n}{2}((n + 1)(2n + 1) - 3(n + 1) + 1) = \frac{n}{2}(2n^{2} + 3n + 1 - 3n - 3 + 1) = n^{3}.$$ · 1 year ago

I used method of difference to find the polynomial and then did the second part the same as you. · 11 months, 1 week ago

Thanks for you solution, sir. · 1 year ago

I solved 3rd very easily and my solution was exactly like @Brian Charlesworth. Now I will try 4th. · 1 year ago

The polynomials $$P(x)$$ and $$Q(x)$$ have the same degree and are monic. This is easily inferred by assuming degrees to be $$m$$ and $$n$$ and then equating the degree of the polynomials thus formed in the second equation. Now it is important to realize that all the terms in the polynomial $$P(x^2 +1)$$ will be ones with an even power of $$x$$. Thus so must $$Q(x)^2 + 2x$$. Thus $$Q(x)^2$$ must have a $$- 2x$$ in its expansion. Also it must be the only term with a odd power of $$x$$. Such a term in the expansion indicates that the coefficient of $$x$$ and the constant term in $$Q(x)$$ are non-zero. I'll give a rough sketch as to why should these be the only nonzero coefficients. Assume $$\displaystyle Q(x) = \sum_{i=0}^n a_i x^i$$ such that $$a_0 , a_1 ,a_n \neq 0$$. Consider the coefficient of $$x^{2n-1}$$ in the expansion of $$Q(x)^2$$. This will have to be zero for $$n >1$$. It implies $$a_{n-1} =0$$. With a similar argument for $$x^{2n-3}$$, we can conclude that $$a_{n-3} = 0$$. This can be continued until $$a_0$$ or $$a_1$$ is reached depending upon $$n$$, which would force either $$a_n$$ to be $$0$$ or $$a_0 / a_1$$ to be $$0$$. Since the latter condition is more important for the given condition to hold true, all the conditions, when imposed together will end up with $$n=1$$, i.e., $$Q(x)$$ is linear. After this, it is easy to realize that $$P(x) = x$$ and $$Q(x) = x - 1$$. I think these are the only possible polynomials.

Note: I haven't given the complete proof. It is rough sketch which can be made more rigorous and complete with more logical flow of arguments all of which I have mentioned. If you think I have missed a step, let me know I'll make myself more clear. · 1 year ago

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