Correct, it's pointless because it equals ## 0 ## anyways. I just feel relieved a lot more since I don't have to include the ## sec(x) ## function on part b), and it makes the proof a lot shorter than before.
a) Observe that ## \frac{\partial}{\partial z}F(y, z)=y^{n-1}\cdot \frac{2z}{2\sqrt{y^2+z^2}}=\frac{zy^{n-1}}{\sqrt{y^2+z^2}} ##.
This means ## G(y, z)=\frac{z^2\cdot y^{n-1}}{\sqrt{y^2+z^2}}-y^{n-1}\cdot \sqrt{y^2+z^2}=\frac{z^2\cdot...
Yes, because some students haven't taken Algebra 2 yet but they have already entered college, and this is why colleges label the name of Algebra 2 as 'College Algebra' to sound more difficult, but in reality, it's the same as Algebra 2 in high school. I don't know too much about other countries...
Okay, here's my revised proof on part (i).
Consider the second-order linear differential equation ## \frac{d^2u}{dx^2}+\frac{fu}{2}=0, f=f(x) ##.
Then ## u_{i}''+\frac{fu}{2}=0\implies u_{i}''=-\frac{f}{2}u ##, so ## -\frac{2}{f}u_{i}''=u_{i} ##.
Given that ## w=\frac{u_{2}}{u_{1}} ##, we have...
I see where you got the ## -\frac{2}{f}u_{i}''=u_{i} ## from. But I don't understand why/how does ## w=\frac{u_{2}}{u_{1}}=\frac{u_{2}''}{u_{1}''} ##. And by starting both differentiations of ## \omega ##, do you mean to take up to second derivatives of ## \omega ##?
I noticed the issue. Since both of the functions ## u_{1}(x), u_{2}(x) ## were incorrect, then how should I solve this differential equation and find these correct functions?
Proof:
(i) Consider the second-order linear differential equation ## \frac{d^2u}{dx^2}+\frac{fu}{2}=0, f=f(x) ##.
Then ## u''+\frac{f}{2}u=0\implies r^2+\frac{f}{2}=0 ##, so ## r=\pm \sqrt{\frac{f}{2}}i ##.
This implies ## u_{1}=c_{1}cos(\sqrt{\frac{f}{2}}x) ## and ##...
Note that ## \frac{\partial F}{\partial x}=\frac{2x}{2\sqrt{x^2+y'^2}}=\frac{x}{\sqrt{x^2+y'^2}}, \frac{\partial F}{\partial y}=0, \frac{\partial F}{\partial y'}=\frac{2y'}{2\sqrt{x^2+y'^2}}=\frac{y'}{\sqrt{x^2+y'^2}} ##.
Now we have ## \frac{dF}{dx}=\frac{\partial F}{\partial x}+\frac{\partial...
I don't know either. So what should the book normally express these primes then, instead? Also, if ## exp(y') ## mean ## e^{y'} ##. Then the expression for ## dF/dy' ## is ## dF/dy'=e^{y'} ##?
a) ## dF/dy'=\frac{1}{4}(1+y'^2)^{\frac{-3}{4}}\cdot 2y' ##
b) ## dF/dy'=cos (y') ##
I just took the derivatives above and found out these expressions, but may anyone please check/verify to see if these expressions for ## dF/dy' ## are correct? Also, I do not understand part c). What does 'exp'...
By the Euler's equation of the functional, we have
## J(\mathrm u)=\int ((\mathrm{u})^{2}+e^{\mathrm{u}}) \, dx ##.
Then ## J(\mathrm{u}+\epsilon\eta)=\int ((\mathrm{u}'+\epsilon\eta')^{2}+e^{\mathrm{u}+\epsilon\eta}) \, dx=\int...
But how should I verify that ## N\equiv 1\pmod {6} ##? And I think I made some mistakes in my previous proof attempts, because ## p\mid N ## and ## p\mid (2p_{1}\dotsb p_{n})^{2} ## implies that ## p\mid (N-(2p_{1}\dotsb p_{n})^{2}) ##, so ## p\mid 3 ##.
Because ## 3\nmid 2p_{1}p_{2}\dotsb p_{n} ##. And ## p\mid N, p\mid 3 ## implies ## p\mid (N-3) ##, so ## p\mid (2p_{1}p_{2}\dotsb p_{n}) ##. Also, how should I show that ## N ## has an odd prime divisor of the form ## 6k+1 ##? At first, I thought this is so because ## N ## itself is odd. But it...
Okay, so I revised this proof:
Suppose for the sake of contradiction that the only primes of the form ## 6k+1 ## are ## p_{1}, p_{2}, ..., p_{n} ##.
Consider the integer ## N=4p_{1}^{2}p_{2}^{2}\dotsb p_{n}^{2}+3=(2p_{1}p_{2}\dotsb p_{n})^{2}+3 ##.
Since ## N ## is odd, it follows that ## N ##...
Proof:
Suppose that the only prime numbers of the form ## 6k+1 ## are ## p_{1}, p_{2}, ..., p_{n} ##,
and let ## N=4p_{1}^{2}p_{2}^{2}\dotsb p_{n}^{2}+3 ##.
Since ## N ## is odd, ## N ## is divisible by some prime ## p ##,
so ## 4p_{1}^{2}\dotsb p_{n}^{2}\equiv -3\pmod {p} ##.
That is, ##...
So now we have that ## \chi(k)^{2}=\chi(k)\cdot \chi(k)=\chi(k\cdot k)=\chi(k^{2})=\chi(1)=1 ## for all ## k\{8, 13, 20\} ## and ## \chi(k)^{6}=\chi(k^{6})=\chi(1)=1 ## for all ## k\{2, 4, 5, 10, 11, 16, 17, 19\} ##. And this implies that ## \chi(n)=8, 13, 20 ## can either be ## -1, 1 ##. But...
Since ## \varphi(21)=\varphi(3)\varphi(7)=2\cdot 6=12 ##, there are ## 12 ## elements such that ## G=\{1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20\} ##. So ## G ## can be generated by order ## 2 ## or ## 6 ##. And we have ## \chi(k)^{2}=\chi(k)\cdot \chi(k)=\chi(k\cdot k)=\chi(k^{2})=\chi(1)=1 ##...
No. I do not know the structure of finite abelian groups. I do not know the characters of cyclic groups are. You said that my group of ## G ## is isomorphic to ## C_{6}\times C_{2} ##, which are the two cyclic groups of order ## 6 ## and ## 2 ##. But how did you get these?
This is my question, too. I do not know what they mean, I just posted them under the relevant equation(s) just because my book has these definitions. Since these definitions are preventing people to make sense of my question, then please ignore them. How should I find those values then, starting...
I wish there are more details in this question, but no. The question states: "Write an account of prime numbers in arithmetic progressions. Your account should be in the form of an essay of 500-1000 words."
How should I write an account of prime numbers in arithmetic progressions? Assuming this account should be in the form of an essay of at least ## 500 ## words. Should I apply the formula ## a_{n}=3+4n ## for ## 0\leq n\leq 2 ##? Can anyone please provide any idea(s)?
After breaking down into smaller pieces, I got the following:
\begin{align*}
&(y'(x)+\tau\psi'(x))^{2}=y'^{2}(x)+2y'(x)\tau\psi'(x)+\tau^{2}\psi'^{2}(x)\\
&\omega^{2}(y(x)+\tau\psi(x))^{2}=\omega^{2}y^{2}(x)+2\tau\psi(x)\omega^{2}+\omega^{2}\tau^{2}\psi^{2}(x)\\...