New answers tagged

0

The proof works up to line 5 as user21820 explained. The assumption made on your line 6 would not be easy to discharge later. Rather take the derived line 5 and continue with it. Here is a completed proof using a Fitch-style proof checker which you may use to check this and other proofs. On lines 7 to 9, I derived $\neg\neg R$ to use modus tollens on ...


0

If you want to avoid using partial fractions, then you could take this approach. We are given that $$\frac{dx}{dt}=3x(x-5)$$ which rearranges to (observing that $x\equiv 0$ and $x \equiv 5$ are solutions to the ODE but not the IVP) $$\frac{1}{x(x-5)}dx=3dt \implies-\frac{1}{5}\Big(\frac{(x-5)-x}{x(x-5)}\Big)dx=3dt\implies -\frac{1}{5}\Big(\frac{1}{x}-\...


1

HINT \begin{align*} \frac{1}{3x(x-5)} = \frac{1}{3}\times\frac{1}{x(x-5)} = \frac{1}{15}\times\frac{x - (x-5)}{x(x-5)} = \frac{1}{15(x-5)} - \frac{1}{15x} \end{align*}


2

Before giving you a complete solution I want to point out the mistakes in your solution. Your partial fractions are correct. As mentioned in the comments $\int \frac{1}{ax} dx \ne \ln | a x |$. Instead we have $$ \int \frac{1}{ax} dx = \frac{1}{a} \int \frac{1}{x} dx = \frac{1}{a} \ln | x| $$ for $a \ne 0$. We have $e^{\ln(a) - \ln(b)} \ne a - b$. This is ...


0

$$ 1 = \frac{A}{3x} + \frac{B}{x-5} $$ if and only if $$ (3x)(x-5) = A(x-5) + B(3x) \text{,} $$ with $x \not\in \{0,5\}$. How do we get "$15xB$"?


0

There's a mistake. Let $\alpha = \sup(\mathbb{N})$ which exists by the reasons you mentiond. It is true that this means that for all $\varepsilon>0$ there exists $n\in\mathbb{N}$ such that $\alpha-\varepsilon <n\leq \alpha$. From this you conclude that $\alpha<n+\varepsilon$ which is fine. However this does not mean that $\alpha\leq n$. You ...


0

You may have the right idea, but for the proof to be complete and rigorous you need to more clearly justify the following crucial inference: When we square a number, we merely repeat its factors, therefore $A^2$ and $B^2$ must also not share any factors". As it stands, your justification "when we square a number, we merely repeat its factors" could be ...


1

$$ \sum\limits_{n = 1}^{ + \infty } {\left( { - 1} \right)^n } \frac{1} {{\sqrt {4n + 1} }} $$ is convergent by Leibniz. The same is true for $$ \sum\limits_{n = 1}^{ + \infty } {\left( { - 1} \right)^n } \frac{1} {{\sqrt {5n + 1} }} $$ thus the given series is convergent because is the difference of two convergent series.


1

Your proof is fine. You showed that the sequence is decreasing and bounded below (thus convergent). You might elaborate that the equation $a = (a+1/a)/2$ has two solutions ($a= \pm 1$), but only $a=1$ can be the limit. Alternatively observe that $$ 0 \le x_{n+1} - 1 = \frac{(x_n-1)^2}{2x_n} \le \frac{(x_n-1)^2}{2} $$ which also implies convergence $x_n \...


3

I will mention one (easily corrected) logical error and one stylistic piece of advice that could make the proof more readable. But the upshot is that this is a well-argued proof by any standard, and especially impressive for a first effort. When you said that $A^2$ and $B^2$ share no factors aside from 1, that does not imply that $\frac{A^2}{B^2}$ is not ...


1

If one doesn't count multiplicities of roots $$ f(x)=36x^4-76x^3+42x^2+1 $$ has one point at which $f(x)=1$, two points at which $f(x)=2$, three points at which $f(x)=3$, and four points at which $f(x)=4$.


2

Since $f(x) - 1$ has only one root and all coeficients are in $\mathbb{R}$, we have only two cases to consider: Case 1. $f(x) - 1 = a(x - \alpha)^4$ In this case, the function obtained is convex (if $a > 0$) and concave otherwise. In particular, it does not have any horizontal line that crosses it $3$ times. Case 2. $f(x) - 1 = a(x - \alpha)^2 g(x)$ ...


1

Consider $f$ a polynomial of degree $4$, then its derivative is a polynomial of degree $3$, so it has at most three critical points. There are only two cases. First case: localmin-localmax-localmin (or max-min-max). The plot you have to imagine is this one. If you count the roots from the global minimum upwards you find $1 \rightarrow 2 \rightarrow 3 \...


3

It is correct. I would have used Abel's test to justify the convergence of $\sum_{n=1}^\infty\frac{a_n}n$ without the assumption that $(\forall n\in\mathbb N):a_n\geqslant0$. It allows you to deduce that, say, $\sum_{n=1}^\infty\frac{na_n}{n+1}$ also converges.


0

By the physics rules, I have that: $l=\frac{1}{2}at^2$. Substituing: $l=\frac{1}{2}g\cdot \sin(\alpha)\cdot t^2$. Also: $\sin(\alpha)=\frac{h}{l}$. From this: $2l^2=ght^2$ and so: $t=l\sqrt{\frac{2}{hg}}$.


1

I think your mistake is in deriving the constant of integration $C$. $s$ measures distance along the plane, not vertical distance. So you should have $C=l$, not $C=h$. If the acceleration due to gravity is $g$ then you should get $t=l\sqrt{\frac{2}{gh}}$ If $l$ is in feet and you take $g$ to be $32$ feet per second$^2$ then you have $t=\frac{l}{4\sqrt{h}...


1

There is an error in your calculation. The acceleration $32\sin A$ is along the slanted plane, not vertically. Thus, the object travels a distance of $l$, not $h$. So, the correct equation after your derivation is instead, $$-16t^2\sin A + l = 0,$$ which leads to the correct answer.


2

You know that $\alpha: X \to P$ satisfies $$\forall j: \beta_j \circ \alpha = \text{pr}_j\tag{1}$$ and $\beta: P \to X$ satisfies $$\forall j : \text{pr}_j \circ \beta= \beta_j \tag{2}$$ now using $(1)$ and $(2)$ we get that for any $j$: $$\text{pr}_j \circ (\beta \circ \alpha) = (\text{pr}_j \circ \beta) \circ \alpha = \beta_j \circ \alpha = \text{pr}_j\...


0

Anothet approach is to use the theorem that A is connected iff for all continuous f:A -> {0,1} with the discrete topology, f is constant. To prove your problem and the generalization: connected A and A subset B subset $\bar A$ implies B is connected, one can use the above theorem and the fact that for continuous f, f($\bar A$) subset $\overline {f(A)}$.


1

1) Depending on your definition of monotonically increasing, being $p_{n} > p_{n-1}$ or $p_{n} ≥ p_{n-1}$. The first case is easily disprovable, just pick $a_{n} = 2$, and $p_{1}$ and $p_{2}$ are both 2. For the second case, since $p_{n}|a_{n} \rightarrow p_{n}|a_{n} + p_{n} \rightarrow p_{n}|a_{n+1}$, so $p_{n+1}$ is at least $p_{n}$. So depending on ...


3

It's better to start with an element $n\in N$ then show that $n$ commutes with any $g\in G$, like so. Since $N$ is normal and $g$ is arbitrary in $G$, we have $gng^{-1}\in N$. Since $N$ is a subgroup of $G$, we have $$h:=\underbrace{gng^{-1}}_{\in N}n^{-1}\in N.$$ But $h$ is the commutator $[g, n]$ of $g$ with $n$, so $h\in N\cap G'=\{e_G\}$, whence $h=gng^{...


2

Well, this is a consequence of the famous Steinitz exchange lemma: If $\{v_{1},\dots ,v_{m}\}$ is a set of $m$ linearly independent vectors in a vector space V, and $\{w_{1},\dots ,w_{n}\}$ spans $V$, then $m\leq n$ and after reordering $\{v_{1},\dots ,v_{m},w_{m+1},\dots ,w_{n}\}$ spans $V$. Here you take $\{v_{1},\dots ,v_{m}\}$ as a basis of $W_1$ and $\...


0

you can just say: Let $\{w_1, \ldots, w_n\}$ be a base for $W_1$, and let $\{w_1, \ldots, w_m \}$ be a base for $W_2$. Because $W_2 \subseteq W_1 \to Sp(W_2) \subseteq Sp(W_1) \to \{w_1, \ldots, w_m \} \subseteq \{w_1, \ldots, w_n \}$, hence $m \le n$ (if $W_1 \subseteq W_2 \to m=n$, and $W_1 = W_2$).


0

If the function is of exponential order, then by definition it follows that we can find constants $a$ and $M>0$ such that $$|f(t-t_0)|\le Me^{a(t-t_0)}$$ whenever $t>t_0.$ Since this function is continuous it follows that as $t_0\to 0,$ we must have $f(t-t_0)\to f(t).$ Therefore, in this limit, we have $$|f(t)|\le Me^{at},$$ since the exponential ...


5

There is no such thing as "more correct" when talking about proofs. A proof can only be one of two things: correct or incorrect. A half correct proof is incorrect. A 99.999999% correct proof is incorrect. Both your proofs are correct, and there is nothing better or worse about either of them. Personally, I think the second is a little easier to understand, ...


3

The mistake is in thinking that $x \in E_n$ iff $\frac 4 {10^{n}} \leq x \leq \frac 5 {10^{n}}$. In fact $E_n$ is a union of $10^{n-1}$ disjoint intervals of length $\frac 1 {10^{n}}$ each, so $m(E_n)=\frac 1 {10}$. For each $n$. Using Borel - Cantelli Lemmas we can show that $m(E)=1$.


-1

Disregarding the validity of its mathematical correctness, here is my thought on how you can structure it in the form of a mathematical statement and proof. Club the $key$ and $formula$ section together as $Keys$, starting with the formula section, ending with your key section. Then put $X=E$ as the $Formula$ and the remaining part of the proof as its ...


5

If $F\colon \mathcal{C}\to \mathbf{Set}$ is any functor and $f$ is an iso in $\mathcal{C}$, then $Ff$ has inverse $F(f^{-1})$. This is an inverse in $\mathbf{Set}$, as you've noticed, and the isos in that category are precisely the bijections. However, what I suspect you want to ask is if in every concrete category, an arrow is an iso iff its underlying ...


2

If different linear combinations give the same vector $v$, we have $v=a_1v_1+\dots+a_nv_n=b_1v_1+\dots +b_nv_n$ and then subtracting, we get $(a_1-b_1)v_1+\dots+(a_n-b_n)v_n=0$, without all $a_i-b_i=0$. This contradicts linear independence of the $v_i$.


1

By contradiction assume otherwise, so there exists scalars $a_i$ and $b_i$ such that $a_j\neq b_j$ for some $j$ and $\sum a_i v_i=\sum b_i v_i$, then $$\sum (a_i-b_i)v_i=0$$ can you take it from here?


1

Here is a proof of the result by cases using a proof checker: Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/


2

$$\frac{dx}{dt}=3x(x-5)\implies\frac{dx}{x(x-5)}=3dt\implies$$ $$\int\frac{dx}{x(x-5)}=\int3dt\implies\int\frac{1}{5}\left(\frac{1}{x-5}-\frac{1}{x}\right)dx=3t+c\implies$$ $$\frac{1}{5}\ln|x-5|-\frac{1}{5}\ln|x|=\frac{1}{5}\ln\left|\frac{x-5}{x}\right|=3t+c$$ Taking the exponential of both sides, $$\left(\frac{x-5}{x}\right)^{\frac{1}{5}}=e^{3t+c}=e^{3t}e^...


0

Put more formally, prove that, for $L, K, n \in \mathbb{Z}$, $n \ge 1$, $$ \tag{1}\label{proposition} (L < K) \wedge (L < m \le K) \wedge (K/n \text{ is upper bound}) \wedge (L/n \text{ is no upper bound}) \Rightarrow (\exists m) (m/n \text{ is upper bound}) \wedge ((m-1)/n \text{ is no upper bound}) $$ Proof is by induction over $K - L$. Base case: ...


0

Yes, just a bit more is needed. Assume $x\to y$ as a premise. Assume $y\to z$ as a premise. Assume $x$. Derive $y$ from 1 and 3 via modus ponens. Derive $z$ from 2 and 4 via modus ponens. Deduce $x\to z$ from subproof 3-5. $\blacksquare~x\to z$ is derivable from $x\to y, y\to z$


2

$$y(x)=\sum_{n=0}^{\infty}a_{n}x^{n}$$ So $$y^{'}(x)=\sum_{n=0}^{\infty}na_{n}x^{n-1}$$ import these power series to main differential equation $y^{'}(x)=1+xy(x)$to receive to $$\sum_{n=0}^{\infty}na_{n}x^{n-1}=1+\sum_{n=0}^{\infty}a_{n}x^{n+1}$$ by simplifying this equation: $$a_{1}=1 , a_{3}=\frac{1}{3}, a_{5}=\frac{1}{15}, \cdot\cdot\cdot,a_{2n-1}=...


1

If the conditions of the Picard–Lindelöf theorem are satisfied then the function sequence $(y_n)$ defined iteratively by $$ y_0(x) = 0 \, , \\ y_{n+1}(x) = Ty(x) = x + \int_0^x t y_n(t) \, dt $$ converge to a solution of the initial value problem. This is called Picard-iteration. The first iterates are $$ \begin{align} y_0(x) &= 0 \\ y_1(x) &= x + ...


2

$(\Rightarrow)$ Your proof is correct, but you start with $p$, then you start calling it $n$ and in the end it's $p$ again. Don't do that. $(\Leftarrow)$ There is the same problem here. Besides, when you wrote $1<a<n$, you should have written $1\leqslant a<b$. Finally, you should explain how you passed from $a\nmid p$ to the assertion that $\mathbb ...


1

I have a possibly more rigorous proof for the $\Leftarrow$ proof: Assume $p$ is prime. Now, consider $ab \equiv 0 \pmod p$. This means, $ab=np$ for some $n \in \Bbb{Z}$. Thus, $p \mid ab$, so by the definition of a prime element, either $p \mid a$ or $p \mid b$. Therefore, either $a \equiv 0 \pmod p$ or $b \equiv 0 \pmod p$. This proves that $\Bbb{Z}_p$ ...


1

Your partial fraction set up is \begin{align*} \frac{1-v^2}{v(1+v^2)}&=\frac{A}{v}+\frac{B\color{red}{v}+C}{1+v^2} \end{align*} Then you got the values as $A=1, B=-2$ and $C=0$. This means \begin{align*} \frac{1-v^2}{v(1+v^2)}&=\frac{1}{v}-\frac{2\color{red}{v}}{1+v^2} \end{align*} Now when you integrate you get \begin{align*} \int \frac{1-v^2}{v(1+...


0

Let $d=\gcd(a,b)$ with $d>1$. Because of the proprieties of $\gcd(a,b)$, I have: $(d\mid a) \land (d\mid b)$. Squaring, I obtain: $(d^2\mid a^2) \land (d^2\mid b^2)$. $d$ is the $\gcd(a,b)$ so also $d^2$ is the $\gcd(a^2,b^2)$. Thus $d^2>1$ and $\gcd(a^2,b^2)>1$.


0

You don't need a proof by contradiction for such a statement: if $d=\gcd(a,b)$, $d$ divides a, hence it trivially dids $a^2$. Similarly, it divides $b^2$, hence, it divides $\gcd(a^2, b^2)$, and it is greater than $1$…


2

All that you have proved was that if your limit exists, then it cannot be greater than $0$. You can prove that it is $0$ using the fact that$$\left\lvert(x^2-y^2)\sin\left(\frac1{x^2+y^2}\right)\right\rvert\leqslant x^2+y^2.$$


0

Here are three ways to see that if $X\to Y$ and $Y\to Z$ then $X\to Z$. Consider how this might look in a diagram. If $X$ is a sufficient condition for something else, say, $Y$, then whenever we have $X$ we have $Y$. That could be represented as one circle containing everything in $X$ being fully contained in a circle containing everything in $Y$. Here is ...


0

There is an easier way: $4^n+6n-1\iff (3+1)^n+6n-1=(9M+3n+1)+6n-1=9(M+n)$


1

And now for something totally different: your condition implies that $G$ has unique Sylow subgroups and hence is nilpotent. This reduces the problem to $p$-groups. Now maximal subgroups of $p$-groups all have index $p$, hence have the same order. Again, the condition now implies $G$ to have a unique maximal subgroup $M$. Pick a $g \in G$ with $g \notin M$, ...


0

Schur's inequality: $$a^3+b^3+c^3+3abc \ge a^2(b+c)+b^2(c+a)+c^2(a+b) \iff \\ (a+b+c)^3+9abc\ge 4(a+b+c)(ab+bc+ca) \Rightarrow \\ 1+9abc\ge 4(ab+bc+ca) \quad (1)$$ Rearrangement: $$a+b+c=1 \Rightarrow a^2+b^2+c^2=1-2(ab+bc+ca)\ge ab+bc+ca \Rightarrow \\ 1\ge 3(ab+bc+ca) \quad (2)$$ Now add $(1)$ and $(2)$.


1

What you have written is actually how most people would prove this in their heads. But if you are looking for a valid mathematical proof, you have to go backwards. What I mean by going backwards is something like this: Proof. Let $M \gt 0$ be arbitrary. Let $N = M$. Then for all $n \gt N$ we have: $$a_n = n+1 \gt n \gt N = M.$$ Thus we have found the ...


0

Yes, your proof is correct. Without returning to the definition, you could also say that $$ a_n = n+1 \rightarrow + \infty $$ when $n \rightarrow + \infty$.


0

A proof by SOS: $$2+9abc-7(ab+ac+bc)=2(a+b+c)^3+9abc-7(a+b+c)(ab+ac+bc)=$$ $$=\sum_{cyc}(2a^3+6a^2b+6a^2c+4abc+3abc-7a^2b-7a^2c-7abc)=$$ $$=\sum_{cyc}(a^3-a^2b-ab^2+b^3)=\sum_{cyc}(a-b)^2(a+b)\geq0.$$ Also, $uvw$ kills it immediately.


2

Your proof looks good. Here's an unsophisticated alternative proof, using only elementary algebra . . . We don't even need $a,b,c$ to be nonnegative. As shown below, if $a,b,c\;$are real numbers such that $a+b+c=1$, and if at least one of $a,b,c\;$is between $-1$ and ${\large{\frac{7}{9}}}$ inclusive, then the inequality holds. Without loss of generality,...


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