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5

You have found that $p(x)$ is a rational function. It is zero if and only if its numerator is zero. Its numerator is a polynomial of the form $Ax^{n+2}+Bx^{n+1}+C$, where $A,B,C$ are constants. The derivative of that numerator is of the form $Dx^{n+1}+Ex^n=x^n(Dx+E)$ for some constants $D,E$. It's easy to see how many real roots that derivative has. Then ...


5

Answer to the title question: Yes! There exists such a function. For example, consider the function $$f(x)=\frac{d(x,A)}{d(x,A)+d(x,B)}$$ where $A$ is the fat cantor set in $[0,1]$ and $B$ is any closed singleton set $\{b\}$ where $b \notin A$. Then $f$ is continuous whose zero set is $A$, which is nowhere dense in $[0,1]$ of positive measure! Edit: The ...


3

One way to see $a+2$ is an eigenvalue is that $$A\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}a+2\\a+2\\a+2\end{bmatrix}.$$ Then you can use the fact that $x-(a+2)$ divides the characteristic polynomial. More generally: if all the rows of $A$ add up to $\lambda$, then $\lambda$ is an eigenvalue.


3

use the following way $$x=\sqrt{5+\sqrt{5 + x} }$$ $$x=\sqrt{5+\sqrt{5 + \sqrt{5+\sqrt{5 + \sqrt{5+\sqrt{5 + ....} }} }} } $$ or $$x=\sqrt{5+x }$$ $$x^2-x-5=0$$ $$x=\frac{1}{2}\pm\frac{\sqrt{21}}{2}$$ now use long division to get the other roots and then check which which one satisfies the original equation


2

Note that if $x = \sqrt{x+5}$ then $x = \sqrt{\sqrt{x+5}+5}$. So, try solving $x = \sqrt{x+5}$. This is a quadratic.


2

Hint: Both the real and imaginary parts of $P(ki)$ are $0$. Consider the real part and use (a).


2

Hint: try to use the triangle inequality. If $|z|\leq 1$, then what can you say about $|z^3+3z|$?


1

We know that $ki$ is a solution to $P(x)=0$ so in particular, $$\begin{split} (ki)^4+a(ki)^3+b(ki)^2+c(ki)+d&=0 \\ \Leftrightarrow k^4-ak^3i-bk^2+cki+d&=0 \\ \Leftrightarrow k^4-bk^2+d+ki(c-ak^2)&=0 \end{split}$$ Using part (a), this gives $k^4-bk^2+d=0$. Now add $bk^2$ on both sides and multiply through with $a^2$ to get $$a^2k^4+a^2d=a^2bk^2=...


1

Basically, you need to solve $(a-\lambda)^3-3(a-\lambda)+2 =0$ for $\lambda$. Don't expand the brackets, instead denote: $t=a-\lambda$. Then: $$t^3-3t+2=0 \Rightarrow (t-1)^2(t+2)=0 \Rightarrow \\ t_1=1 \Rightarrow a-\lambda =1 \Rightarrow \lambda_1 =a-1\\ t_2=-2\Rightarrow a-\lambda =-2 \Rightarrow \lambda_2=a+2.$$


1

Hint: if $I$ denotes the identity matrix, then the eigenvalues of $A+cI$ are easily obtained from the eigenvalues of $A$: $$ (A+cI)v=\lambda v \iff Av=(\lambda-c)v $$ What if you take $c=1-a$?


1

There is no guarantee that there are "$10$ roots with the smallest real part". That is, it is quite possible that there is an infinite sequence of roots with real parts positive but approaching $0$, and no roots with real part $0$.


1

You can take $p(x)=x^5-4x-2$. It is irreducible in $\mathbb Q[x]$, by Eisenstein's criterion. And it is easy to deduce from the fact that $p'(x)=5x^4-4$ and from the intermediate value theorem that it has $3$ and only $3$ real roots.


1

$$ p(x)=1+2x+3x^2+\ldots+(n+1)x^n;\\ xp(x)=x+2x^2+3x^3+\ldots+(n+1)x^{n+1};\\ (1-x)p(x)=1+x+x^2+\ldots+x^n-(n+1)x^{n+1}=\frac{1-x^{n+1}}{1-x}-(n+1)x^{n+1}\\ =\frac{1-x^{n+1}-(1-x)(n+1)x^{n+1}}{1-x}=0.\\ \Rightarrow1-x^{n+1}-(1-x)(n+1)x^{n+1}=0 $$ Note that $x=1$ is clearly not a solution of the original equation. As a result, the above equation is actually ...


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