For example, let $f= \sum_{i=0}^n a_ix^i$. If every $a_i\not\in U(A)$, then $\exists b_i\in A$ with $b_i \neq 0$ for every $i$ such that $a_ib_i=0$, so I can take $a = \prod b_i$ reaches what I need.  I tried using $f|0$ saying that exists a $g$ such that $fg\equiv0$, but the only thing I've reached with this is this is that the independant term of $fg$ is $0$.   Another thing I noticed is that $Im(f) \in A-U(A)$
22707
Let $R$ be commutative ring with no (nonzero) nilpotents. If $f(x) = a_0+a_1x+\cdots+a_nx^n$ is a zero divisor in $R[x]$, how do I show there's an element $b \ne 0$ in $R$ such that $ba_0=ba_1=\cdots=ba_n=0$?
7044
I understand intuitively that this is true, but I'm embarrassed to say I'm having a hard time constructing a rigorous proof that $|a+b| \leq |a|+|b|$. Any help would be appreciated :)
75223
Suppose $f(x)$ and $g(x)$ are continuous functions on $[a,b]$ with $f$ monotone increasing. Assume there exists a sequence $x_n \in [a, b]$ such that for all $n \in N$ , $g(x_n) = f(x_{n+1})$. Show that there exists $x_0 \in [a,b]$ such that $g(x_0) = f(x_0)$.  Can someone provide an example of functions that fulfills this condition?
22884
Assuming $a&lt;b$, function f(x) is continous at [a,b], and we have  $\int_a^bf(x)dx=\int_a^bxf(x)dx=\int_a^bx^2f(x)dx=0$  Prove that $\exists \ x_1,x_2,x_3\text {(different from each other)}  \in(a,b)$ satisfying $f(x_1)=f(x_2)=f(x_3)=0$.  I have proved the existence of one zero point by differential mean value theorem, but have no idea about going on.  Thanks for any solution or hint.
111403
I'm reading through How to Prove It by Velleman and I'm having trouble with this exercise in the section about Existence and Uniqueness proofs. Here is the exercise:     Suppose $A$ is a set and for every family of sets $\mathcal{F}$, if $\cup \mathcal{F}=A$ then $A \in \mathcal{F}$. Prove that $A$ has exactly one element.    He hints that for both the existence and uniqueness parts of the proof it would be a good idea to use contradiction. I've been playing around with this proof for a while but I can't seem to make any substantial progress.   I current idea is considering some cases  where for some family of sets $\mathcal{G}$, $A \in \mathcal{G}$ and $A \notin \mathcal{G}$. I thought if I could show that if $A= \varnothing$ lead to contradictions, I could at least say that there is something in $A$, and try to prove that it is unique from there. I haven't been able to make any progress with this though. Any help with this problem would be greatly appreciated!
69087
Give:  $fn(S)=\prod_{x\in S}x$  what is:  $fn(\emptyset)$  I can see reason that it would be defined as 1 or 0.    Note: I thought about restricting the domain of $S$ but that would make the problem less general. If there is no general answer then answers for restricted domains would be valid.
39046
I managed to prove it by rewriting the partial products in terms of $n!$ and then applying Stirling's approximation. But I was wondering if there is a more elementary method to show that the limit is zero, without using such heavy machinery.  Also, is it true that in general $$\prod_{n=0}^\infty \frac{an+b}{an+c}=0, \qquad a&gt;0\ \textrm{and}\ c&gt;b$$
113667
Suppose $\{a_k\}$ ,$\{b_k\}$ and $\{\xi_k\}$ are non negative, and for all $k\ge 0$ , we have $$a_{k+1}^2\le(a_k+b_k)^2-\xi_k^2$$   Prove   1.$$\sum_{i=1}^{k}\xi_i^2\le(a_1+\sum_{i=0}^kb_i)^2$$   2. If $\{b_k\}$ also satisfy $\sum_{k=0}^{\infty}b_k^2\lt+\infty $, then $$\lim_{k\to\infty}\frac{1}{k}\sum_{i=1}^k\xi_i^2=0$$   After doing this transformation:$$\xi_k^2\le(a_k+b_k)^2-a_{k+1}^2$$ $\Rightarrow$ $$\sum_{i=1}^{k}\xi_k^2\le(a_1+b_1-a_2)(a_1+b_1+a_2)+(a_2+b_2-a_3)(a_2+b_2+a_3)+\cdots(a_k+b_k-a_{k+1})(a_k+b_k+a_{k+1})$$ then I can't move forward any more.
6285
$$\lim _{x\rightarrow a} f(x)+g(x) = f(a)+g(a)$$ Let $\epsilon&gt;0$ be given Since $f$ and $g$ are continuous, $|f(x)-f(a)|&lt; \epsilon$ when $0&lt;|x-a|&lt; \delta_f $ and $|g(x)-g(a)|&lt; \epsilon$ $\ $ when $0&lt;|x-a|&lt; \delta_g$ Let  $\delta_h$ be defined as $\min(\delta_g ,\delta_f)$ and $h(x)$ be defined as $f(x)+g(x)$ $$\ |f(x)-f(a)|+|g(x)-g(a)|&lt; 2\epsilon\tag{1}$$ $$\ |f(x)-f(a) + g(x)-g(a)|&lt; 2\epsilon$$ $$\ |h(x)-h(a)|&lt; 2\epsilon$$ We can replace $\delta_f$  and  $\delta_g$ by $𝛿_h$ to get $|f(x)-f(a)| &lt; \epsilon$ when $0&lt;|x-a|&lt;\delta_h$ and $\ |g(x)-g(a)|&lt; \epsilon$ when $0&lt;|x-a|&lt;\delta_h$ Since (1) is true when $0&lt;|x-a|&lt;\delta_h$, we can find a $\delta_h$ for every value of $2ε$ for every $2\epsilon&gt;0$ there is a $\delta&gt;0$ such that $|x−a|&lt;\delta\implies ∣h(x)−h(a)∣&lt;2\epsilon$ and for every $2\epsilon$ there exists an $\epsilon$ so for every $\epsilon$ there exists a $\delta$. Is the proof logically correct and is it worded correctly?
65114
Let $ Y = \{ (1,1),(0,0\} \subset \mathbb A_k^2$. Find the $ \mathbb I(Y)$.   As if $f(x,y) \in \mathbb I(Y)$ then as $f(0,0)=0$ hence constant term of $f$ is zero and as $f(1,1)=0$ therefore sum of the coefficients of $f(x,y)$ is zero. But how can we describe this set properly?
115427
I've been stuck on this problem over the weekend so I decided to ask for some direction. The problem reads:  "The multiplication theorem for series requires that the two series be absolutely convergent; if this condition is not met, their product may be divergent. Show that the series $\sum_0^∞ \frac{(-1)^i}{\sqrt{1 + i}}$ gives an example: it is conditionally convergent, but its product with itself is divergent. (Estimate the size of the odd terms $c_{2n+1}$ in the product.)"  Someone earlier suggest I show the Cauchy product and show that it diverges. If I understand correctly, the two series must converge absolutely. If I can show that two do not converge, then  I have shown what the problem ask for.   So I tried and had this in mind. I reckon the prove is not correct because I do not fully understand how to prove Cauchy product, but it's an idea.   Consider $a_n$ = $(-1)^n$, $b_n = \frac{1}{\sqrt{n+1}}$. The sum of their product, $C_n$ = $\sum_0^{\infty} c_i$ = $\sum_0^{\infty}\frac{(-1)^i}{\sqrt{1 + i}}$ conditional converges. Thus $\forall$$\epsilon$ > 0, $\exists$N s.t. n, m $\geq$ N $\rightarrow$ |$\sum_n^m c_i$| &lt; $\epsilon$.    We assume without loss of generality that the series of $a_n$ and the series of $b_n$ does not converge absolutely.   |$\sum_0^{\infty} (-1)^i$| $\leq$ $\sum_0^{\infty} |(-1)^i|$ $\leq$ $\sum_0^{\infty} 1^i$  As n approaches infinity, the $\sum_0^{\infty} 1^i$ = $1 + 1 + ... + 1 = \infty$, thus diverging.   |$\sum_0^{\infty} \frac{1}{\sqrt{i+1}}$| = $\sum_0^{\infty} |\frac{1}{\sqrt{i+1}}|$ = $\sum_0^{\infty} \frac{1}{\sqrt{i+1}}$.   As n approaches infinity, the $\sum_0^{\infty} \frac{1}{\sqrt{i+1}}$ = $\infty$, thus diverging.   So I understand the idea, I just don't know how to go about finishing up the proof using cauchy.   Thank you for taking the time to read this and thanks in advance for commenting.
32979
Let $f(z) = z + g(z)$ where $g$ is holomorphic. Suppose that $|\operatorname{Im} g(z)| &lt; 1$ for $z ∈ [−1 − i, 1 − i]∪[−1 + i, 1 + i]$ and $|\operatorname{Re} g(z)| &lt; 1$ for $z ∈ [−1 − i, −1 + i] ∪ [1 − i, 1 + i]$.   Show that $f$ has exactly one zero on the square $Q =$ {$x + iy ∈ \Bbb C : |x| &lt; 1, |y| &lt; 1$}.  My attempt:  I let $h(z) = z$. Then, I want to compare $|g(z)|$ and $|h(z)|$ because if $|g(z)| &lt; |h(z)|$ then by Rouché's theorem, $h$ and $h+g$ have the same number of zeros, and $h$ has in fact one zero. But then $h+g = f$ and thus $f$ would also have the same number of zeros as $h+g$ which has one zero.  This is what I could come up with: $|g(z)| = |u(z) + iv(z)|$. Then for $z \in Q$, we have $|g(z)| \leq |u(z)| + |v(z)| &lt; 1 + 1 = 2$ (since $|\operatorname{Im} g(z)| &lt; 1$ for $z ∈ [−1 − i, 1 − i]∪[−1 + i, 1 + i]$ and $|\operatorname{Re} g(z)| &lt; 1$ for $z ∈ [−1 − i, −1 + i] ∪ [1 − i, 1 + i]$)  But I don't know how to continue from here. Any help please?
38631
My Approach. To prove the given statement it is sufficient to show that $$\exists \epsilon &gt; 0, \exists x \in [0,1], \forall N \in \mathbb{N}, \exists n &gt; N $$ $$ \Rightarrow | f_n(x) - f(x)| &gt; \epsilon$$    Let $\epsilon = \frac{1}{10}, x = \frac{1}{n}, n =N+1$. Hence $| f_n(x) - f(x)| = \frac{1}{2} &gt; \epsilon = \frac{1}{10}$. Hence, we have shown for every $N \in \mathbb{N}$, how to choose $n$ such that $| f_n(x) - f(x)| &gt; \epsilon$. Thus, $f_n(x)$ does not converge uniformly on $[0,1]$.    Is there any logical flaw or wrong steps in my proof. Any help would be appreciated.
73031
Let $(a_n)_{n \in \mathbb{N}}$ a sequence of real numbers with $0 \leq a_n \leq 1$ for all $n \in \mathbb{N}$.  I want to prove the following inequality using mathematical induction:  $\prod \limits_{k=0}^{n}(1-a_k) \geq  1- \sum\limits_{k=0}^{n}a_k$.  It's obviously true for $n=0$.  $n \rightarrow n+1$  Assume that it's true for n. Then one has:  $\prod \limits_{k=0}^{n+1}(1-a_k) = (1- a_{n+1}) \prod \limits_{k=0}^{n}(1-a_k) \geq (1- a_{n+1}) (1- \sum\limits_{k=0}^{n}a_k) = 1- \sum\limits_{k=0}^{n}a_k - a_{n+1} + a_{n+1} \sum\limits_{k=0}^{n}a_k = 1- \sum\limits_{k=0}^{n+1}a_k + a_{n+1} \sum\limits_{k=0}^{n}a_k=...$  How to go on?
76270
If we knew $\sum_{i=1}^{n} f(i)=S$  ($n$ can be $\infty$), is there anything we could say about $\sum_{i=1}^{n} \dfrac {1}{f(i)}$ in terms of $S$?  The only thing I was able to find on the web is . If there is no known relationship, is it possible for there to exist one that we just haven't discovered?
19698
I have found this exercise.  Let $f, g$ be entire functions.  a) Show that there exists entire $f_{1}, g_{1}$ and $h$ such that $f=hf_{1}$, $g = hg_{1}$ and $f_{1}$ and $g_{1}$ have no common zeros.  b) For $f$, $g$, and $h$ as above, show that there exists entire functions $A$ and $B$ such that $Af+Bg = h$.  Any idea ? How to manage the case ord($f$) = $\infty$ and ord($g$) = $\infty$ ?
103459
Suppose $f:\mathbb{R}\to\mathbb{R}$ has period 1, and for some $q\in(0,1)$:  $$|f(x)-f(y)|\leq q|x-y|\quad \forall x,y$$  Let $g(x)=x+f(x)$, for any $a\in\mathbb{R}$,define the following sequence:  $$a_1=a,\quad a_2=g(a_1),\quad a_3=g(a_2),\quad \dots,\quad a_{n+1}=g(a_n)$$  Show that  $$\lim_{n\to\infty}\frac{a_1+a_2+\dots+a_n}{n^2}$$  exists and does not depend on the choice of $a$.  I don't even know what is the possible approach, any hints? Thanks!