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
4 years, 11 months 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 $\displaystyle\sum_{k=1}^{n} P(k) = \sum_{k=1}^{n} (3k^{2} - 3k + 1) =$

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

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

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

- 4 years, 11 months ago

Thanks for you solution, sir.

- 4 years, 11 months ago

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

- 4 years, 11 months ago

I used method of difference to find the polynomial and then did the second part the same as you.

- 4 years, 9 months 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.

- 4 years, 11 months ago

- 4 years, 11 months ago

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Why? Reason?

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Ok. I will take care of it.

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