pxc3110
  • Member for 7 years, 10 months
  • Last seen more than 1 year ago
Prove: $\kappa^2v^4=|\alpha^{''}|^2-(\frac{dv}{dt})^2.$
3 votes

$v\frac{dv}{dt}={\alpha'}\cdot{\alpha''}$ $\frac{dv}{dt}=\frac{\alpha'\cdot\alpha"}{|\alpha'|}$ $\kappa^2v^4=\frac{|\alpha'\times\alpha''|^2}{|\alpha'|^6}\cdot v^4=\frac{|\alpha'|^2|\alpha''|^2\sin^...

View answer
When does $(uv)'=u'v'?$
3 votes

Let $u$ and $v$ be functions of $t$. then $(uv)'=u'v'$ $\iff u'v+uv'=u'v'$ $\iff u'(v-v')+v'u=0$ $\iff u'+\frac {v'}{v-v'}u=0$ Let u be the unkown function, then multiply both sides by $e^{\int \frac{...

View answer
Prove that $(n-1)! \equiv -1 \pmod{n}$ iff $n$ is prime [Wilson's Theorem]
1 votes

Proof: Let p be an odd prime number. Consider the group $U_p=${equivalent classes of $a$|$p>a>0$, $gcd(a,p)=1$} (equivalent relation:$a\equiv b \pmod p$, binary operation:[a][b]=[ab]...

View answer
Solving differential equation (equation whose unknown is a function)
1 votes

$\int \frac1ydy=ln|y|+c=6x$, therefore $e^{ln|y|+c}=e^{6x}$, or $|y|e^c=e^{6x}$ or $|y|=e^{6x}e^c$ (we write $e^{-c}$ as $e^{c}$ since both $-c$ and $c$ represent an arbitrary constant.) or $y=\pm ...

View answer
Implicit Differentiation to find equation if the tangent line to a curve
1 votes

Differentiating both sides, we get: $2x+2yy'=2(2x^2+2y^2-x)*(4x+4yy'-1)$. Plugging in $x=0, y=\frac 12$, we get: $y'=2y'-1$, or $y'=1$. Therefore, $y-\frac 12=y'(0)(x-0)=x$. This will be transformed ...

View answer
Show that $G$ is a group if the cancellation law holds when identity element is not sure to be in $G$
1 votes

Using the pigeonhole principle: Consider the set $A=${$ab|b\in G$}, where $a\in G $ is fixed. If $a\notin A$, then since $a\in G$ and all elements in $A$ is in $G$, $A\subset G$. then consider the ...

View answer
Find the indefinite integral $\int \frac{2x^2 + 1}{(2x)^2}dx$
1 votes

$\int \frac{2x^2+1}{(2x)^2}dx=\int \frac12+\frac 1{(2x)^2}dx=\frac 12x+\frac14 \int x^{-2}dx=\frac 12 x-\frac 1{4x}+C$

View answer
Prove that $\log(x) < x$ for $x > 0$, $x\in \mathbb{N}$.
1 votes

(I) Let $x\in \mathbb R, x>0$. Maclaurin series for $e^x$: $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$, So $e^x-x=1+\frac{x^2}{2!}+\frac{x^3}{3!}+...>0$ or $e^x>x$ or $x>log(x)$. When $x\...

View answer
Solving the halting problem for *almost* all machines?
0 votes

It can be shown that neither $h(g,g)$ or $g(g)$ halt. Hence you can solve $g(g)$ by building a second halting machine $h'$ that returns $0$ if input is $(g,g)$ and calls $h$ otherwise. I don't have a ...

View answer
Constructive proof of $\forall n, m\in\mathbb N$ such that $\gcd (m,n)=1, $ the cyclic group $C_{nm}\cong C_n\times C_m$.
0 votes

Since cyclic groups are isomorphic to additive groups, set of all proofs of this theorem is "isomorphic", in a sense, to set of all proofs of CRT. Hence, find a constructive proof of CRT, which ...

View answer
The Proof of Wilson's Theorem using the auxiliary multiplicative modulous group
0 votes

Proof: Let p be an odd prime number. Consider the group $U_p=${equivalent classes of $a$|$p>a>0$, $gcd(a,p)=1$} (equivalent relation:$a\equiv b \pmod p$, binary operation:[a][b]=[ab]...

View answer
Simple Line Integral -- Just making sure I'm right
0 votes

As to the parametrization, I got the same answer as yours, and I can confidently say that your parametrization is correct. Obviously the first line integral is 0. For the second line integral, the ...

View answer
Error raising a complex number to a power
0 votes

You can save time by doing it another way: By Euler :$3+7i=e^{iarctan\frac 73}$, which means $(3+7i)^5=e^{i5arctan\frac 73}$ Now use a calculator to evaluate $5arctan \frac 73$. If we know that $...

View answer
Integration by substitution for line integrals
0 votes

Maybe you should use Green theorem to evaluate $I:=\int_{\gamma} P(x,y)dx+Q(x,y)dy$, and I seriously doubt the validity of your formula. I'm not sure if it is wrong though.

View answer
Show that the mapping is one to one iff $|f(S)|=|S|$, and the two definitions for being onto are equivalent.
0 votes

To show that the two definitions for being onto is equivalent: 1. If for $\forall t\in T$, $\exists s\in S|f(s)=t$, then $T\subseteq f(S)$, and for $\forall a\in f(S), a\in T$,so $f(S)\subseteq T$. ...

View answer