If for some $\alpha , \beta \in \mathbb{R}$ , the integration formula   $\int_{0}^{2} p(x) dx = p(\alpha) + p(\beta)$ , holds for all polynomials $p(x)$ of degree atmost $3$, then the valueof $3(\alpha - \beta)^2 = ?$.  How to approach this question? i thought of taking Polynomial as $x^3$ and i am getting the answer but is there any formal way of doing it without explicitly substituting any polynomial of degree atmost $3$ ?.  Any suggestion , hints , solution ?
121648
Let $f$ be a polynomial of degree $n$ at most $3$ such that there exists some $\alpha,\beta$ satisfying $\int_0^2 f(x)dx=f(\alpha) +f(\beta)$ for all such $f$. Find the value of $3(\alpha-\beta)^2$  Attempt: We need to find the value of $3(\alpha-\beta)^2$. Now, this expression reminds of the function $f(x) = (x-\beta)^3$. Differentiating this function at $x=\alpha$ will give us the desired result.  Hence, we just need to know the derivative of $k(x)=(x-\beta)^3$ at every point.  However, I am unable to move forward with this intuition. Could anyone please help me to move ahead in this problem.  Thank you very much for your help in this regard.
1226
$f:\mathbb{R}^3\to\mathbb{R}, f(rx)=r^{\alpha}f(x)$ for some $\alpha&gt;0$, $\forall x\in\mathbb{R}$, for any $r\in\mathbb{R}$, could any one tell me which of the following is true?  $1.$ If $f(x)=f(y)$ whenever $||x||=||y||=\beta&gt;0$, then $f(x)=\beta ||x||^{\alpha}$  $2.$ If $f(x)=f(y)$ whenever $||x||=||y||=1$, then $f(x)= ||x||^{\alpha}$  $3.$ If $f(x)=f(y)$ whenever $||x||=||y||=1$, then $f(x)=c||x||^{\alpha}$ for some constant $c$  $4.$ If $f(x)=f(y)$ whenever $||x||=||y||$, then $f(x)$ must be a constant function. Thank you.
60447
I want to solve the integral $\int 1/(a^2+x^2)^n dx$. I guess this can be solved using the reduction formula as in the case of $\int x^2/(a^2+x^2)^n dx$. However, I am not able to do it.
116448
Given a polynomial $f(x)$ of n degree such that $$f(x)+f\left(\frac{1}{x}\right)=f(x)\cdot f\left(\frac{1}{x}\right)$$ Find the polynomial  I've tried considering $f(x)=\sum \limits_{i=1}^{n} a_{i}x^{i}$ and after a number of steps I finally arrived at the answer.However I am not convinced.Is there a better way to solve this functional equation. Thanks in advanced!
121582
Fix a nonnegative integer $n$. Let $P(x)$ be a polynomial of degree $n$ (over the real numbers) such that for all $k\in\left\{0,1,\ldots,n\right\}$, we have $P(k)=\dfrac{k}{k+1}$. Find $P(n+1)$.   My attempt:  I consider a new polynomial $Q(x)=(x+1)P(x)-x$. This $Q$ is a polynomial of degree $\leq n+1$ satisfying: $Q(k)=0\quad\forall k\in\{0,1,2,\ldots,n\}$. Therefore $Q(x)=\lambda x(x-1)\cdots(x-n)$ for a certain real $\lambda$. Hence, $Q(-1)=\lambda \cdot (-1)^{n+1}(n+1)!$, but also $Q(-1)=(-1+1)P(-1)+1=1$. Comparing these, we get $\lambda \cdot (-1)^{n+1} (n+1)! = 1$, thus $\lambda=\dfrac{(-1)^{n+1}}{(n+1)!}$. Since $Q(n+1)=\lambda(n+1)!$ then $Q(n+1)=(-1)^{n+1} $, and from $P(x) = \dfrac{Q(x)+x}{x+1}$ we deduce: $$P(n+1)=\dfrac{(-1)^{n+1}+n+1}{n+2}.$$ I doubted that the answer were $P(n+1)=\frac{n+1}{n+2}$ but the term $(-1)^{n+1}$ wasn’t expected. Is there any mistake here? Does someone have an alternative proof?
47939
If we have an algebraic number $\alpha$ with (complex) absolute value $1$, it does not follow that $\alpha$ is a root of unity (i.e., that $\alpha^n = 1$ for some $n$).  For example, $(3/5 + 4/5 i)$ is not a root of unity.  But if we assume that $\alpha$ is an algebraic integer with absolute value $1$, does it follow that $\alpha$ is a root of unity?    I know that if all conjugates of $\alpha$ have absolute value $1$, then $\alpha$ is a root of unity by the argument below:  The minimal polynomial of $\alpha$ over $\mathbb{Z}$ is $\prod_{i=1}^d (x-\alpha_i)$, where the $\alpha_i$ are just the conjugates of $\alpha$.  Then $\prod_{i=1}^d (x-\alpha_i^n)$ is a polynomial over $\mathbb{Z}$ with $\alpha^n$ as a root.  It also has degree $d$, and all roots have absolute value $1$.  But there can only be finitely many such polynomials (since the coefficients are integers with bounded size), so we get that $\alpha^n=\sigma(\alpha)$ for some Galois conjugation $\sigma$. If $\sigma^m(\alpha) = \alpha$, then $\alpha^{n^m} = \alpha$.  Thus $\alpha^{n^m - 1} = 1$.
88183
This question is realy hard for me and I'd like to get some advises:  $$f(x) =   \begin{cases}      x^{\alpha}(1-\cos(1-x))^{\beta} &amp;  x\in (0,1)\\[6px]     \dfrac{1}{x^{\alpha}+x^{\beta}}  &amp;   x\ge 1   \end{cases}$$  I need to find the values of $\alpha$ and $\beta$ that make $\int_{0}^{\infty}f(x)\,dx$ converge.  Now I think I need to split the integral to $\int_{0}^{1}f(x)\,dx$ and  $\int_{1}^{\infty}f(x)\,dx$ and check the converge for each integral. But I'm stuck here.
116589
Let $P(x)=a_nx^n+a_{n-1}x^{n-1}+\dots +a_0$ be a real polynomial such that $a_0+\frac{a_1}{2}+\frac{a_2}{3}+\dots + \frac{a_n}{n+1}=0$.  Show that $P$ has a root in $[0,1]$. I would like proof verification and alternative proofs.  My solution:      notice that the antiderivative of $P(x)$ is $\frac{a_nx^{n+1}}{n+1}+\frac{a_{n-1}x^n}{n}+\dots + \frac{a_1x^2}{2}+a_0x+C$. From here $\int_{0}^1P(x)dx=a_0+\frac{a_1}{2}+\frac{a_2}{3}+\dots + \frac{a_n}{n+1}=0$. If $P$ had no roots then $P$ would be a strictly positive(or negative) continuous function, so its integral would be strictly positive ( or negative), a contradiction.
104661
We're learning Riemann integrals, and I'm trying to see whteher the following indefinite integral $\int _{1}^{\infty}\frac{p(x)}{e^x}$  converges, when p(x) is a polynomial. I don't know where to start this question - does it depends on p(x)? What theorem should I use?
48253
I had an example in the book given as follows:        Find the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ .       Solution:  $~~~~~$$(\sqrt{2}+\sqrt{3})^2=5+2 \sqrt6$      $(\sqrt{2}+\sqrt{3})^4=49+20 \sqrt6$       Then $(\sqrt{2}+\sqrt{3})^4-10(\sqrt{2}+\sqrt{3})^2+1=0.$       Thus $a=\sqrt{2}+\sqrt{3}$ satisfies $f(x)=x^4-10x^2+1$ over $\mathbb Q$.       Let $p(x)$ be the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$.Then $(\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})$ are also roots of $p(x).$   So degree of $p(x)$ is atleast $4$.      But $f(a)=0 $ and $f(x) \in \mathbb Q[x] \implies p(x)$ divides $f(x)$ .      So $f(x)$ is minimal polynomial of $\sqrt{2}+\sqrt{3}.$   But I can't get the step  how are $(\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})$ also roots of $p(x).$   Kindly help with this.
10631
Let $A\in M_3(\Bbb R)$ which is not a diagonal matrix. Let $P$ be a polynomial (in one variable), with real coefficients and of degree 3 such that $P(A) = 0$. Pick out the true statements:  a. $P = cP_A$ where $c \in\Bbb R$ and $P_A$ is the characteristic polynomial of $A$;  b. if $P$ has a complex root (i.e. a root with non-zero imaginary part), then $P = cP_A$, with $c$ and $P_A$ as above;  c. if $P$ has a complex root, then $A$ is diagonalizable over $\Bbb C$.  I guess b and c are true although I am unable to disprove a. Because A could satisfy $(X-A)$ but this is not a polynomial. Now if $P$ has a complex root, then its conjugate too is its root as the polynomial has  real coefficients. As we all know An odd degree polynomial has always a real root, what we have is three distinct roots. Since $A$ satisfies the polynomial the only difference between the characteristic polynomial and $P$ has to be a constant.
8894
The question that I am trying to answer is :  Suppose is $p(x)$ is a polynomial with integer coefficients. Show that if $p(a) = 1$ for some integer a then $p(x)$ has at most two integer roots.  I have no idea how to get started. Any help would be awesome!
51577
I don't think I fully understand.  Let's say there is a root $x_0 \in K=\mathbb{F}_p(t)$, where $p$ is a prime number.  Then $x_0 = \frac{P(t)}{Q(t)}$ for some polynomials $P,Q \in \mathbb{F}_p[t]$. We can assume $\gcd(P,Q)=1$  and $x_0^p-t= \frac{(P(t))^p}{(Q(t))^p}-t= \frac{P(t)^p-tQ(t)^p}{Q(t)^p} =0$, so coefficients of $P(t)^p, tQ(t)^p$ must be identical, which contradicts $\gcd(P,Q)=1$,   hence such $x_0$ does not exist and the polynomial has no root in $K$.   am I about right?    anyway, I would appreciate an explanation about $\mathbb{F}_p(t)$, what is $t$? what is the meaning of a variable which does not belong to any specific "world"? I cannot use $t$ as if it was a member of $\mathbb{F}_p$ and hence cannot assume $t^{p-1} = 1 \pmod p$...
5078
Find all polynomials $p(x)$ such that for all $x$, we have $$(x-16)p(2x)=16(x-1)p(x)$$  I tried working out with replacing $x$ by $\frac{x}{2},\frac{x}{4},\cdots$, to have $p(2x) \to p(0)$ but then factor terms seems to create a problem.  Link:
75159
Question:  If $p(x) \in F[x]$ is of degree $3$, and $p(x)=a_0+a_1x+a_2x^2+a_3x^3$, show that $p(x)$ is irreducible over $F$ if there is no element $r\in F$ such that $a_0+a_1r+a_2r^2+a_3r^3 =0$.  My understanding: If $p(x)$ is reducible, then there exists $ax + b$ such that $a, b \in F$ and $a\ne 0$. And $p(x) = (ax + b)(cx^2 + dx + e)$. Then an $r$ exists such that $p(r) = 0$. Is that how you're supposed to tackle it? I understand that I am supposed to first explain the argument for a polynomial of degree 1, but I do not understand how to go about that, and why that helps.
110254
Let $f(z)$ be analytic in a closed set $A$ and bounded by contour $C$. Let $z_1,z_2,...,z_n$ be different arbitrary point in the interior of $C$ and let $w_n(z)=(z-z_1)(z-z_2)...(z-z_n) $. Prove the integral $$P(z)=\frac{1}{2\pi i} \int_C \frac{f(x)}{w_n(x)} \frac{w_n(x)-w_n(z)}{x-z} dx$$ is a polynomial of degree $(n-1)$ that is the same as $f(z)$ at $z_1,z_2,...,z_n$ points.  Note: P(z) is the Lagrange interpolating polynomial.  I think I have an idea for the proof, but I don't know how to write it formally.  Idea I think this formula $$f^n(z_0)=\frac{n!}{2\pi i} \int_J \frac{f(x)}{(x-z_0)^{n+1}} dx $$ will be useful because practically is telling us that P$(z)$ must be 'equal to' $\frac{f^n(z_0)}{n!} \left(\frac{w_n(x)-w_n(z)}{x-z} \right)$.  Now the rest is trying to proof that $\frac{f^n(z_0)}{n!} \left(\frac{w_n(x)-w_n(z)}{x-z} \right)$ must be something very similar to the Lagrange polynomial.  Is this a good idea? Could be demonstrated in another way? Thanks for any suggestions/hint/tip/help.
1542
Calculating with Mathematica, one can have $$\int_0^{\pi/2}\frac{\sin^3 t}{\sin^3 t+\cos^3 t}\,\mathrm dt=\frac{\pi}{4}.$$   How can I get this formula by hand? Is there any simpler idea than using $u = \sin t$?     Is there a simple way to calculate $$ \int_0^{\pi/2}\frac{\sin^n t}{\sin^n t+\cos^n t}\,\mathrm dt $$ for $n&gt;3$? Could anyone come up with a reference for this exercise?