Nevzat Eren Akkaya
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$x^p \equiv x \bmod p \Rightarrow x^{2p^2}=(x^p)^{2p} \equiv x^{2p} \equiv x^2 \bmod{p}$ $x^p \equiv x \bmod p \Rightarrow x^{p^2} \equiv x \bmod p$ $x^{2p^2}- \overline{2}x^{p^2} - x^p + \overline{2} ... View answer Accepted answer 3 votes$\begin{pmatrix} a &b \\0 &1 \end{pmatrix}\begin{pmatrix} c &d \\0 &1 \end{pmatrix}=\begin{pmatrix} ac &ad+b \\0 &1 \end{pmatrix}\Rightarrow\begin{pmatrix} a &...

$\frac{\dfrac{1}{\sqrt{2a-1}}}{\dfrac{1}{\sqrt{2a+1}}}=\frac{18\sqrt{2a+1}}{3(a^2-19)\sqrt{2a-1}}\Rightarrow \frac{6}{a^2-19}=1$ $\frac{6}{a^2-19}=1 \Rightarrow a^2=25$ which means $a=5$ or $a=-5$ but ...

Let $B=\{1,x,x^2\}$ then $\forall p \in P_2(R)$ $\exists c_0,c_1,c_2$: $p=c_0+c_1x+c_2x^2$ $T(p)=c_0T(1)+c_1T(x)+c_2T(x^2)$ $\Rightarrow$ $[T(p)]_B=A\begin{pmatrix} c_0 \\ c_1 \\ c_2 \end{pmatrix}$ $... View answer 2 votes Assume that$ABu=(\alpha+2)Bu[\pi(H),\pi(X)] = 2\pi(X) \Rightarrow AB-BA=2BAB-BA=2B \Rightarrow AB^2-BAB=2B^2AB^2-BAB=2B^2 \Rightarrow AB^2u-BABu=-(\alpha+2)B^2u+AB^2u=2B^2u-(\alpha+2)B^2u+...

Your method actually could work but there is an easier approach involving the Rotation matrix which is defined as R_\theta =\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta ... View answer 2 votes Note that $$\sum_{k=1}^{n+1}\frac{1}{(n+1)+k}= \sum_{k=1}^{n}\frac{1}{n+k}-\frac1{n+1}+\frac{1}{2n+1}+\frac1{2n+2}$$ \begin{align} -\frac1{n+1}+\frac{1}{2n+1}+\frac1{2n+2}&=\frac{1}{2n+1}-\frac{... View answer Accepted answer 2 votes You could use Fermats Little Theorem and Chinese Remainder Theorem together. Note that 12^{157} \equiv 0 \pmod{2} \Rightarrow 12^{157}=2k 12^4 \equiv 1 \pmod{5} \Rightarrow 12^{157} \equiv 2 \pmod{... View answer Accepted answer 2 votes Note that -1<\frac{\sin x}{x}<1 let u=\frac{\sin x}{x} then,\begin{align}\sum_{n=1}^{\infty}\left(\frac{\sin x}{x}\right)^n&=\sum_{n=1}^{\infty}(u)^n\\ &=\lim \limits_{n \to \... View answer Accepted answer 2 votesn=m=0 \Rightarrow \int T_{n}(x)\,T_{m}(x)\,{\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}=\int \frac{dx}{\sqrt{1-x^2}}=arcsin x\int \frac{dx}{\sqrt{1-x^2}}=arcsin x \Rightarrow \int_{-1}^{1} T_{n}(x)\,...

$u(x)=-1+x+\frac{x^2}{2}+2e^x-\int_0^xu(t)dt \Rightarrow \frac{du}{dx}=2x+2e^x+1-u$ $\frac{du}{dx}=2x+2e^x+1-u \Rightarrow e^x\frac{du}{dx}+e^xu=2e^xx+2e^{2x}+e^x$ $e^x\frac{du}{dx}+e^xu=2e^xx+2e^{2x}+... View answer Accepted answer 2 votes$A^{-1}$exists$\RightarrowDet(A)=-bc \neq 0$which means both$b$and$c$can not be$0A^{-1} = A \Rightarrow A^2=IA = \begin{pmatrix} a & b \\ c & 0 \end{pmatrix} \Rightarrow A^2=\...

let $p$ be a prime number $x^2\equiv 1 \pmod p$ $\Rightarrow$ $x^2-1\equiv (x-1)(x+1)\equiv 0 \pmod p$ $(x-1)(x+1)\equiv 0 \pmod p$ $\Rightarrow$ $x \equiv 1$ or $x \equiv -1$ Let $n=p_1p_2...p_n$ and ...

$Bu_3=u_4$ and $Bu_4=u_3$ $\Rightarrow$ $Bu_3+Bu_4=B(u_3+u_4)=u_4+u_3$ hence 1 is an eigenvalue for $u_3+u_4$ $Bu_3=u_4$ $\Rightarrow$ $-Bu_3=B(-u_3)=-u_4$ $B(-u_3)=-u_4$ and $Bu_4=u_3$ $\Rightarrow$ ...

$(I-Y)^2=I-Y$ $\Rightarrow$ $Y^2-2Y=-Y$ $\Rightarrow$ $Y^2=Y$ hence it is idempotent

We know that $\frac{\partial G}{\partial x}=\frac{\partial g}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial g}{\partial v}\frac{\partial v}{\partial x}$ $\frac{\partial G}{\partial y}=\frac{\... View answer Accepted answer 2 votes It is trivial actually.Lets say$v_1$,$v_2$and$v_3$are vectors . If$v_1=v_2 + v_3$, then$v_2 + v_3 - v_1=0$, so there exists a solution for$c_1v_1+c_2v_2+c_3v_3=0$which is different than$c_1=...

by the definition of the derivative, we know that $f^(a)=\lim_{x \to 0}\frac{f(x+a)-f(a)}{x}$ it can be observed that $f(0)=\ln(\cos(0))=0$ so according to definition $f^(0)=\lim_{x \to 0}\frac{\ln(\... View answer 1 votes Yes, since$A^T=-A$for all$v\in V\langle Av,v \rangle=0$Proof: Note that$\langle u,v\rangle=u^TvA^T=-A \Rightarrow \langle v,AV\rangle= v^TAv=-v^TA^Tv\langle -Av,v\rangle=-v^TA^Tv \...

Hint:Assume that $L>M$ and let $L-M=\epsilon$ than use $\epsilon-\delta$ definition of limit for given $\epsilon$

Let $v \in U$ $v \in U \Rightarrow f(v) \in im(f)$ $im(f) ⊂ ker(g) \Rightarrow f(v) \in ker(g)$ $f(v) \in ker(g) \Rightarrow g(f(v))=0$ $g(f(v))=0 \Rightarrow g ◦ f = 0$

$$\frac{d(-R\omega \sin(\theta))}{dt}=-R\frac{d(\omega sin(\theta))}{dt}$$ according to product rule and chain rule, \frac{d(\omega \sin(\theta))}{dt}=\frac{d\omega}{dt}\sin(\theta)+\omega (\frac{d\...

let $lim_{n \to \infty }b_n=b$ and $lim_{n \to \infty}(a_n+b_n)=u$ $lim_{n \to \infty }b_n=b \Rightarrow lim_{n \to \infty }-b_n=-b$ $lim_{n \to \infty }-b_n=-b$ and $lim_{n \to \infty}(a_n+b_n)=u$ $\... View answer Accepted answer 1 votes$y=ux$and$y(1 + xy) dx = x dy \Rightarrow uxdx+ux^3dx=uxdx+x^2duuxdx+u^2x^3dx=uxdx+x^2du \Rightarrow u^2x^3dx=x^2duu^2x^3dx=x^2du \Rightarrow x dx=\frac{du}{u^2}x dx=\frac{du}{u^2} \...

Limit does not exist in this case $\lim_{x\to 1^+}\frac{x^2+x+1}{x-1}=\lim_{x\to 1^+}(x+2+\frac{3}{x-1})=\infty$ $\lim_{x\to 1^-}\frac{x^2+x+1}{x-1}=\lim_{x\to 1^-}(x+2+\frac{3}{x-1})=-\infty$
$|x_{n+2}-x_{n+1}| \leq 0.2|x_{n+1}-x_n| \Rightarrow |x_{n+k}-x_{n}|\leq c^k |x_2-x_1|=c^k$ $s=\frac{1}{|c|}-1 \Rightarrow |c|=\frac{1}{1+s}$ and $Ns \geq \epsilon$ $|c|=\frac{1}{1+s} \Rightarrow |c|^... View answer 1 votes$x_1=c , x_2=c+4, x_3=c+6 \Rightarrow c+4=cr$and$c+6=cr^2\frac{c+4}{c}=\frac{c+6}{c+4} \Rightarrow 2c+16=0c=-8$hence$r=\frac12$View answer Accepted answer 1 votes$F(r(t))=(sin(t^3),cos(-t^2),t^4)$,$r(t)=(3t^2,-2t,1)dt \Rightarrow\int F \cdot dr=\int_0^1(3t^2sin(t^3)-2tcos(-t^2)+t^4)dt\int_0^1(3t^2sin(t^3)-2tcos(-t^2)+t^4)dt=sin(-1)-cos(1)+2$View answer 1 votes$a^2+2b^2=1 \Rightarrow 2b^2=1-a^2=(1-a)(1+a)2b^2 \geq 0 \Rightarrow -1 \leq a \leq 1$but we know the fact that$a$is an integer so$ -1 \leq a \leq 1 $and$a \in \Bbb Z \Rightarrow a=-1$or ... View answer 1 votes$(ln(u(x))^=\frac{u’(x)}{u(x)} y(x) = \frac{u’(x)}{u(x)r(x)}=\frac{d\ln u}{dx} \frac1{r(x)} \Rightarrow \frac{dy}{dx}=\frac{d^2(\ln u)}{dx^2}\frac1{r(x)}+\frac{d\ln u}{dx}\frac{-r^(x)}{r(x)^2}\$