Hot answers tagged

8

I think $-\infty $ make sense. Indeed, let $P$ a polynomial of degree $\geq 1$. Then, you have that $$\deg(PQ)=\deg(P)+\deg(Q),$$ for every polynomial $Q$. Now, if you define $\dim(0)=0$, you'll get $$\deg(0\cdot P)=0+\deg(P)>0,$$ which is not compatible with the degree formula. The only way to give a sense to this formula is to define $\deg(0)=-\infty $. ...


8

Well, it depends. Mathematical practice shows that sometimes it is useful to define the degree of the zero polynomial to be zero, sometimes to define it to be $-\infty$ and sometimes to leave is undefined. Which option one chooses depends on what one is trying to do. This is quite different with what happens with the degree of all other polynomials, which ...


8

Hint: Between any two roots of $f$ there is a root of $f'$


6

One can check that $1$ and $-1$ are both roots of this polynomial. After factoring out $x-1$ and $x+1$, you're left with a quadratic polynomial. As a general rule, the rational root theorem is a good place to start for questions like these.


5

Defining it as $-\infty$ makes the most sense. As mentioned in Surb's answer and comments, some properties of degrees are kept intact this way, e.g. $\deg(PQ)=\deg P+\deg Q$ If $\deg P>\deg Q$ then $\deg(P+Q)=\deg P$ It also starts making more sense if you consider expressions that can take on negative powers as well. That is, instead of ...


5

There is no nice formula to get the roots of $P+Q$ from the roots of $P$ and of $Q$. For example, the roots of $x^5$ and $2x+1$ are easy to find, but the sum of these polynomials is $x^5 + 2 x + 1$, an irreducible quintic whose roots can't be expressed in radicals.


4

When you need to develop Taylor series around $x_0\neq 0$, it is generally simpler to make a change of variable $x=y+x_0$ and consider the development around $y=0$ using more standard formulae. In your case, for $f=\sqrt x$ around $x=1$, this gives $f=\sqrt{1+y}$ around $y=0$. Using Taylor series or the generalized binomial theorem, this will give ...


4

Let $P_k(x)=P(P(...(x))$ with $k$ iterations of $P$. Suppose (for contradiction) that $P(x)$ has a non-positive root $s$ and let $r$ be a positive root of $P(x)$ so that $P(x)=(x-r)(x-s)f(x)$. Then $P_2(x)= (P(x)-r)(P(x)-s)g(x)$. Note that $P(x)$ must have arbitarily large negative values or arbit. large positive values to the left of $s$; thus $P(x)=r$ or ...


4

Here is a way. Suppose that we want to visualize the roots of the cubic equation $$ z^3+z+1=0 $$ Write $z=x+iy$ and expand: \begin{align} (x+iy)^3+(x+iy)+1&=0\\ x^3+3ix^2y-3xy^2-iy^3+x+iy+1&=0\\ \end{align} Taking real and imaginary parts, we get \begin{align} x^3-3xy^2+x+1&=0\\ 3x^2y-y^3+y&=0 \end{align} Plotting the solution sets of ...


4

We assume that $deg P(x) \geq 2$ and that $a_n \neq 0$. Since $P(1) = P(2) = 0$, we have $a_n2^n + a_{n - 1}2^{n-1} + .. + a_12 + a_0 = 0$ and $a_n + a_{n - 1} + .. + a_1 + a_0 = 0$. Together, we have $(a_{n - 1} + .. + a_1 + a_0)2^n = a_{n - 1}2^{n - 1} + .. + a_12 + a_0$. If we assume that $a_k \geq -1$ for $k \leq n - 1$, then $a_{n - 1}2^{n - 1} + .. + ...


3

Hint: $$f'(x)=101x^{100}-100^2x^{99}=x^{99}\left(101x-100^2\right)$$ has two solutions. $x_{\max}=0, x_{\min}=\frac{100^2}{101}$ Then $$x_1<x_{\max}=0<x_2<x_{\min}<x_3$$ Then $x_2,x_2>0$


3

I think you might be mixing up which parts to evaluate. Namely, you want to evaluate the coefficients at $x_0=1$, but to get a polynomial (i.e. a function), you don't want to evaluate the expression $(x-x_0)=(x-1)$. I.e. what the answer is: $$P_2(x)=f(x_0)+\displaystyle\frac{f^{(1)}(x_0)}{1!} (x-x_0)^1 + \displaystyle\frac{f^{(2)}(x_0)}{2!}(x-x_0)^2 \\ = ...


3

Single Variable Taylor Polynomial Formula: $$P_{n}(x)=\sum_{n=0}^{n}\frac{f^{n}(c)}{n!}(x-c)^n$$ Given: $$f(x)= \sqrt x , c=1$$ $$f(1)=1$$ Derivatives: $$f'(x)=\frac{1}{2\sqrt x} , f'(1)=\frac{1}{2}$$ $$f''(x)=\frac{-1}{4x^\frac{3}{2}} , f''(1)=\frac{-1}{4}$$ Second Degree Taylor Polynomial: $$P_{2}(x)=1+\frac{1}{2}(x-1)-\frac{1}{8}(x-1)^2$$


3

A polynomial is separable (has distinct roots) if it shares no zeroes with its formal derivative. If $ P(X) - w $ is a polynomial of degree $ n $ for $ w $ a constant, then its formal derivative $P'(X) $ is a polynomial of degree at most $ n-1 $ and does not depend on $ w $. Let the distinct zeroes of the formal derivative be $ z_1, z_2, \ldots, z_r $ where ...


3

Suppose there are repeated roots. Then $f$ and $f'$ would have at least one common root ($f'$ is the formal derivative). Because $f' = nx^{n-1}$ and $p$ does not divide $n$ the only root of $f'$ is $0$. But $0$ is not a root for $f$ because $p$ does not divide $a$


2

A difference of squares is always factorable into two binomial factors. This immediately tells you that $9x^2 - 16y^2$ can be factored further. A sum of squares cannot be factored into two binomials, and in your case $9x^2 + 16y^2$ has no common factor to pull out, nor can it be factored in any other way. As Lubin points out below, there are sums of ...


2

Yes. There are two factors which affect whether or not a polynomial $P(x)$ can be split or not: The polynomial $P(x)$ itself. The field $F$ that the polynomial is over. For example, the polynomial $P(x) = x^2 + 1$ does not split over $\mathbb{R}$, but it does split into $P(x) = (x - i)(x + i)$ over $\mathbb{C}$. Furthermore, if $P(x)$ is a polynomial ...


2

Let $m=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ and $n=\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$. Then $\frac{a}{b},\frac{b}{c},\frac{c}{a}$ are the roots of $x^3-mx^2+nx-1=0$. Hence we have $a^3-ma^2b+nab^2-b^3=0$, where $a,b,m,n$ are all integers, with $a,b$ positive. Let $p^r$ be the highest power of $p$ dividing $a$, and $p^s$ be the highest power of $p$ ...


2

For monomials, this is not to difficult to solve: In particular, consider $x_1^{n_1}x_2^{n_2}\cdots x_k^{n_k}$ Then any sequence $(a_1,a_2,\ldots,a_k) \in \{0,1,...,n_1\}\times\{0,1,...,n_2\}\times \cdots \{0,1,\ldots,n_k\}$ Map it to the monomial $\dfrac{d^{a_1+a_2+\cdots+a_k}}{dx_1^{a_1} dx_2^{a_2}\cdots dx_k^{a_k}}x_1^{n_1}x_2^{n_2}\cdots x_k^{n_k}$ ...


2

I coded up the formula in the Wikipedia page and got the right results. Perhaps you made a mistake in your program. The intermediate values, using Wikipedia's variable names, are: $p = -14.375$ $q = -16.875$ $\Delta_0 = 268$ $\Delta_1 = 7040$ $\sqrt{\Delta_1^2 - 4\Delta_0^3} = 5237.721642i$ $Q^3 = 3520 + 2618.860821i$ $Q = 16 + 3.464102i$ $S = 2.25$ ...


2

If you choose $g(z) = -5z^4+ 3z^2$ and $f(z) = z^6-1$, then $|f(z)| \le 2$ and $|g(z)| \ge 2$ on $|z| = 1$. Moreover, on $|z|=1$ we have $|g(z)| = 2$ only at $\pm 1$ and $|f(1)| = |f(-1)| = 0 < 2$. Thus, Rouché's theorem applies and you need only count the roots of $z^2(-5z^2+3)$ within $|z|=1$.


2

Note that the polynomial is positive at $0$, negative at (what?) and positive for large $x$, so there area least two positive roots. Now take the derivative. Where is it zero?


1

Descartes' rule of signs indicates that $P(x)=x^{101}-100x^{100}+100$ has either zero or two positive roots. But $P(0)>0$ and $P(2)<0$ so $P(x)$ has at least one positive root, hence it has exactly two positive roots.


1

You should simply study the variations of the function $f$ defined by $f(x) = x^{101}-100x^{100}+100$ on $\mathbb R$.


1

We have $$x^4-3x^3-11x^2+3x+10$$ Using rational root test: $$(x-1)(x^3-2x^2-13x-10)$$ And rational root test again: $$(x-1)(x+1)(x^2-3x-10)$$ Factoring quadratic: $$(x-1)(x+1)(x-5)(x+2)$$ Thus, roots are $x=\pm1,\; x=-2,\; x=5$.


1

The presence of multiple roots is detected by the Discriminant which is expressible in terms of the coefficient of the polynomial. So if the coefficients are rational, the presence of multiple roots is decidable. Edit: I didn't get the OP's question properly in the first place. Concerning the computation of multiplicities once a complete list of zeros is ...


1

The question is whether the set $U$ of all $(a,b,c,d)\in{\mathbb R}^4$ satisfying $$(a+b)(c+d)=0$$ forms a subspace of ${\mathbb R}^4$. The condition looks so strange that we are lead to the conjecture that the answer is no. Therefore we try to find a counterexample. If some ${\bf x}=(a,b,c,d)$ satisfies the condition then so does any $\lambda{\bf x}$. ...


1

For the polynomial $p+q=\left(a+e\right)+\left(b+f\right)t+\left(c+g\right)t^2+\left(d+h\right)t^3$, you now have that $a' = a+e, b' = b+f, c' = c + g,$ and $d' = d+ h$. When you multiply out $$ (a' + b')(c'+d') $$ the terms $(a+b)(c+d)$ and $(e+f)(g+h)$ will show up both equaling $0$, however, there will also be many other terms, like $ag, ah, \dots$, a ...


1

As I said above, given any polynomial $f$ that's bound correctly, we can construct infinitely more by writing $x^n f(x)$. Moreover, given $f,g$ that are both bound correctly, their product $fg$ will also be bound correctly. Now, to see that there isn't some finite set $\{f_0,f_1,...,f_n\}$ such that all other polynomials bound this way are a product of ...


1

If you define the degree of a polynomial as the degree of the highest non-zero power, then if the polynomial is zero, the degree is undefined. You could define it by convention, to make sense of some general rules, and then as the other answers explain, you get different results. Now, think of why $x^0 =1$. The proof is by using the laws of exponents, ...



Only top voted, non community-wiki answers of a minimum length are eligible