问:
How to proceed $\int \frac{dx}{(x-2) \left(1+\sqrt{7 x-10-x^2}\right)} $

$$I=\int \frac{dx}{(x-2) \left(1+\sqrt{7 x-10-x^2}\right)} $$

I have taken the second term of denominator (in square root) as $(x - 2)t$. But cannot go further. Please suggest how to proceed or any alternate method to solve it.

答:
If you want to use integration by substitution, you'll need to determine what you're substituting for, so that you can actually follow through with the substitution.

Added: In this case, it seems that you're wanting to rewrite $\sqrt{7x-10-x^2}$ in the form $(x-2)t,$ which turns out to be workable, unlike what I thought you were trying to do, and turns out to be much simpler than what I ended up doing instead.

Since $7x-10-x^2=(5-x)(x-2)$ is nonnegative exactly when $x\in[2,5],$ then $$\sqrt{7x-10-x^2}=\sqrt{5-x}\sqrt{x-2}=\sqrt{\frac{5-x}{x-2}}(x-2)$$ except at the single point $x=2,$ which doesn't alter the antiderivative. Hence, we want $$t=\sqrt{\frac{5-x}{x-2}}.$$ Now, for $x\in(2,5],$ we have $$(x-2)t=\sqrt{5-x}\sqrt{x-2}\\(x-2)^2t^2=(5-x)(x-2)\\(x-2)t^2=5-x\\xt^2-2t^2=5-x\\x+xt^2=5+2t^2\\x\left(1+t^2\right)=5+2t^2\\x=\frac{5+2t^2}{1+t^2}.$$ From there, we can see that $$x-2=\frac{5+2t^2}{1+t^2}-\frac{2+2t^2}{1+t^2}=\frac{3}{1+t^2},$$ whence $$1+\sqrt{7x-10-x^2}=1+(x-2)t=1+\frac{3t}{1+t^2}=\frac{1+3t+t^2}{1+t^2}.$$ Finally, applying the quotient rule shows us that $$dx=\frac{4t\left(1+t^2\right)-2t\left(5+2t^2\right)}{\left(1+t^2\right)^2}\,dt=\frac{-6t}{\left(1+t^2\right)^2}\,dt.$$

At this point, we're ready to follow through on the substitution, finding that $$I=\int\frac1{x-2}\cdot\frac1{1+\sqrt{7x-10-x^2}}\,dx\\I=\int\frac{1+t^2}{3}\cdot\frac{1+t^2}{1+3t+t^2}\cdot\frac{-6t}{\left(1+t^2\right)^2}\,dt\\I=-\int\frac{2t}{1+3t+t^2}.$$ This is definitely nicer looking!

Since $$3t+t^2=(3+t)t=\left(\frac32+t+\frac32\right)\left(\frac32+t-\frac32\right)=\left(\frac32+t\right)^2-\left(\frac32\right)^2=\left(\frac32+t\right)^2-\frac94,$$ then $$1+3t+t^2=\left(\frac32+t\right)^2-\frac94=\left(\frac32+t\right)^2-\left(\frac{\sqrt 5}2\right)^2=\left(\frac32+t+\frac{\sqrt 5}2\right)\left(\frac32+t-\frac{\sqrt 5}2\right),$$ and so $$I=-\int\frac{2t}{\left(\frac32+t+\frac{\sqrt 5}2\right)\left(\frac32+t-\frac{\sqrt 5}2\right)}\\I=-\int\frac{2t}{\left(t+\frac12\left(3+\sqrt 5\right)\right)\left(t+\frac12\left(3-\sqrt 5\right)\right)}.$$

The next step I'd use is called partial fraction decomposition, in which we suppose that $$\frac{2t}{\left(t+\frac12\left(3+\sqrt 5\right)\right)\left(t+\frac12\left(3-\sqrt 5\right)\right)}=\frac{A}{t+\frac12\left(3+\sqrt 5\right)}+\frac{B}{t+\frac12\left(3-\sqrt 5\right)}\tag{$\diamondsuit$}$$ for some constants $A$ and $B,$ then determine what we can about those constants. Multiplying both sides by the left-hand side's denominator gives us $$2t=A\left(t+\frac12\left(3-\sqrt 5\right)\right)+B\left(t+\frac12\left(3+\sqrt 5\right)\right)\\2t+0=(A+B)t+\frac32(A+B)+\frac{\sqrt 5}2(B-A).$$ As both sides are polynomials, the only way for equality to hold is if the corresponding coefficients are equal. Hence, we need $A+B=2,$ meaning that $A=2-B.$ Substituting back in gives us $$2t+0=2t+3+\frac{\sqrt 5}2(2B-2)\\2t+0=2t+3+(B-1)\sqrt 5,$$ so we also need $$0=3+(B-1)\sqrt 5\\-3=(B-1)\sqrt 5\\-\frac3{\sqrt 5}=B-1\\1-\frac3{\sqrt 5}=B\\B=-\frac{3-\sqrt 5}{\sqrt 5},$$ so $$A=2-B\\A=2+\frac{3-\sqrt 5}{\sqrt 5}\\A=\frac{3+\sqrt 5}{\sqrt 5}.$$ Thus, by $(\diamondsuit),$ we find that $$I=-\frac{3+\sqrt 5}{\sqrt 5}\int\frac{1}{t+\frac12\left(3+\sqrt 5\right)}\,dt+\frac{3-\sqrt 5}{\sqrt 5}\int\frac{1}{t+\frac12\left(3-\sqrt 5\right)}\,dt\\I=-\frac{3+\sqrt 5}{\sqrt 5}\ln\left|t+\frac12\left(3+\sqrt 5\right)\right|+\frac{3-\sqrt 5}{\sqrt 5}\ln\left|t+\frac12\left(3-\sqrt 5\right)\right|+C.$$ Noting that $t$ is nonnegative on $(2,5]$ and that $3-\sqrt 5>0,$ we see then that $$I=-\frac{3+\sqrt 5}{\sqrt 5}\ln\left(t+\frac12\left(3+\sqrt 5\right)\right)+\frac{3-\sqrt 5}{\sqrt 5}\ln\left(t+\frac12\left(3-\sqrt 5\right)\right)+C.$$ Finally, we recall our logarithm rules to get $$I=-\ln\left(\left(t+\frac12\left(3+\sqrt 5\right)\right)\left(t+\frac12\left(3-\sqrt 5\right)\right)\right)+\frac{3}{\sqrt 5}\ln\left(\frac{t+\frac12\left(3-\sqrt 5\right)}{t+\frac12\left(3+\sqrt 5\right)}\right)+C\\I=-\ln\left(t^2+3t+1\right)+\frac{3}{\sqrt 5}\ln\left(\frac{2t+3-\sqrt 5}{2t+3+\sqrt 5}\right)+C\\I=-\ln\left(\frac{1}{t^2+3t+1}\right)+\frac{3}{\sqrt 5}\ln\left(\frac{2t+3-\sqrt 5}{2t+3+\sqrt 5}\right)+C\\I=-\ln\left(\frac13\cdot\frac{3}{1+t^2}\cdot\frac{1+t^2}{t^2+3t+1}\right)+\frac{3}{\sqrt 5}\ln\left(\frac{2t+3-\sqrt 5}{2t+3+\sqrt 5}\right)+C\\I=-\ln\left(\frac13\cdot\frac{x-2}{1+\sqrt{7x-10-x^2}}\right)+\frac{3}{\sqrt 5}\ln\left(\frac{2t+3-\sqrt 5}{2t+3+\sqrt 5}\right)+C\\I=-\ln\left(\frac{x-2}{3+3\sqrt{7x-10-x^2}}\right)+\frac{3}{\sqrt 5}\ln\left(\frac{2t+3-\sqrt 5}{2t+3+\sqrt 5}\right)+C.$$ One last substitution gets us $$I=\ln\left(\frac{x-2}{3+3\sqrt{7x-10-x^2}}\right)+\frac3{\sqrt 5}\ln\left(\frac{2\sqrt{\frac{5-x}{x-2}}+3-\sqrt 5}{2\sqrt{\frac{5-x}{x-2}}+3+\sqrt 5}\right)+C\\I=\ln\left(\frac{x-2}{3+3\sqrt{7x-10-x^2}}\right)+\frac3{\sqrt 5}\ln\left(\frac{2\sqrt{5-x}+(3-\sqrt 5)\sqrt{x-2}}{2\sqrt{5-x}+(3+\sqrt 5)\sqrt{x-2}}\right)+C.$$

While we might be able to massage that a bit to make it nicer, that's probably enough time spent on this one problem.

Everything after this point was based on my initial misunderstanding of the OP. While accurate (thanks to a few later corrections), it's very long, and a little embarassing, so feel free to ignore it.

For example, if you want the expression under the radical to have the form $(x-2)t,$ that turns out to be doable, since factoring shows that $$7x-10-x^2=(x-2)(5-x).$$ That means that you'll need $$t=5-x,$$ so that $$x=5-t$$ and $$dx=-dt.$$ Following through with this substitution will then yield $$I=\int\frac{-dt}{(5-t-2)\left(1+\sqrt{(5-t-2)t}\right)}=-\int\frac{dt}{(3-t)\left(1+\sqrt{(3-t)t}\right)}.$$

That's not looking much better, but let's see what we can do with it.

Multiplying the numerator and denominator by $1-\sqrt{(3-t)t}$ will give us $$I=-\int\frac{(1-\sqrt{(3-t)t})dt}{(3-t)\bigl(1-(3-t)t\bigr)},$$ but that isn't a lot better, either. We could try another substitution, say $u=3-t,$ so that $t=3-u$ and $dt=-du,$ which would give us $$I=-\int\frac{-(1-\sqrt{u(3-u)})du}{u\bigl(1-u(3-u)\bigr)}=\int\frac{(1-\sqrt{(3-u)u})du}{u\bigl(1-(3-u)u\bigr)},$$ but while we've made the denominator a little simpler, we've done nothing at all for the numerator, and it still isn't obvious how to proceed.

To go any further, we could be a bit clever, and rewrite the bit under the radical as a difference of two squares. This will allow us to make a special substitution called a trigonometric substitution.

Let's take it back to the original form $$I=\frac{dx}{(x-2)\left(1+\sqrt{7x-10-x^2}\right)}.$$ Now, observe that $$7x-x^2=-(x^2-7x)=-(x-7)x=-\left(x-\frac72-\frac72\right)\left(x-\frac72+\frac72\right).$$ Recalling that $(a-b)(a+b)=a^2-b^2,$ then letting $a=x-\frac72$ and $b=\frac72$ shows us that $$7x-x^2=-\left(\left(x-\frac72\right)^2-\left(\frac72\right)^2\right)=\left(\frac72\right)^2-\left(x-\frac72\right)^2=\frac{49}4-\left(x-\frac72\right)^2.$$ Consequently, $$7x-10-x^2=\frac{49}4-\left(x-\frac72\right)^2-10=\frac94-\left(x-\frac72\right)^2=\left(\frac32\right)^2-\left(x-\frac72\right)^2.$$

We're not quite there, yet, but we're getting closer. Note that $$x-\frac72=\frac32\left(\frac23x-\frac73\right),$$ so that $$7x-10-x^2=\left(\frac32\right)^2\left(1-\left(\frac23x-\frac73\right)^2\right).$$ Here's where we're ready for the clever bit: we want to replace $\frac23x-\frac73$ with $\sin t,$ giving us $$7x-10-x^2=\left(\frac32\right)^2(1-\sin^2t)=\left(\frac32\cos t\right)^2$$ by the Pythagorean Identity, and so $$\sqrt{7x-10-x^2}=\frac32|\cos t|.$$ That means we'll need $$2x-7=3\sin t\\2x=7+3\sin t\\x=\frac72+\frac32\sin t,$$ and so by the chain rule, $$dx=\frac32\cos t\,dt.$$ Also, $$x-2=\frac32+\frac32\sin t=\frac32(1+\sin t).$$ Following through with the substitution, we get $$I=\int\frac{\frac32\cos t\,dt}{\frac32(1+\sin t)\left(1+\frac32|\cos t|\right)}=\int\frac{2\cos t}{(1+\sin t)\bigl(2+3|\cos t|\bigr)}\,dt.$$

But why are we even allowed to do this? Couldn't this substitution make the bit under the radical a negative number? We'll, let's find out if that's possible. Since we require $7x-10-x^2\ge 0,$ then $(x-2)(5-x)\ge0,$ meaning either that $x=2,$ that $x=5,$ or that $x-2$ and $5-x$ have the same sign. Since $x-2$ and $5-x$ only have the same sign when $2<x<5,$ then we need $$2\le x\le 5\\2\le\frac72+\frac32\sin t\le 5\\4\le 7+3\sin t\le 10\\-3\le 3\sin t\le 3\\-1\le\sin t\le1,$$ which is always true! Excellent! Our substitution is safe, after all.

I see that another answer has appeared already, but it doesn't really offer much insight as to why it works or what steps were taken to get there (assuming it was worked out by hand and not just computed with software). With that in mind, I will go ahead and continue my answer.

Unfortunately, the absolute value seems to make things problematic, but let's see what we can do.  We'd really like to make sure that $\cos t\ge 0$ with our initial substitution. One way to make it happen is to assume that $t\in[-\pi/2,\pi/2],$ which is fine, since in that interval, $\sin t$ can take on any of its possible values, so we aren't preventing our antiderivative from being valid. Good news! That lets us go with $$I=\int\frac{2\cos t}{(1+\sin t)\bigl(2+3\cos t\bigr)}\,dt.$$

Our next step involves what's known as partial fraction decomposition. In this case, that means trying to find constants $A,$ $B,$ $C,$ $D,$ $E,$ and $F$ so that $$\frac{2\cos t}{(1+\sin t)\bigl(2+3\cos t\bigr)}=\frac{A\cos t+B\sin t+C}{1+\sin t}+\frac{D\cos t+E\sin t+F}{2+3\cos t}.\tag{$\heartsuit$}$$ Multiplying both sides of $(\heartsuit)$ by $(1+\sin t)\bigl(2+3\cos t\bigr)$ gives us $$2\cos t=(A\cos t+B\sin t+C)(2+3\cos t)+(D\cos t+E\sin t+F)(1+\sin t),$$ or $$0=(A\cos t+B\sin t+C)(2+3\cos t)+(D\cos t+E\sin t+F)(1+\sin t)-2\cos t.\tag{$\star$}$$ Note that if we can find values of those constants to make $(\star)$ hold, then they will make $(\heartsuit)$ hold, as well. Unfortunately, that right-hand side is a mess, but expanding it out yields $$3A\cos^2t+E\sin^2t+(3B+D)\cos t\sin t+(2A+3C+D-2)\cos t+(2B+E+F)\sin t+2C+F,$$ which is a bit friendlier. Now we can try out some different values of $t$ to see if any of them simplify things for us. One nice one is $t=-\frac\pi 2,$ which makes $\sin t=-1$ and $\cos t=0,$ so that $(\star)$ becomes $$0=E-(2B+E+F)+2C+F$$ or $$0=2C-2B.$$ Thus, we need $C=B,$ yielding $$\frac{2\cos t}{(1+\sin t)\bigl(2+3\cos t\bigr)}=\frac{A\cos t}{1+\sin t}+\frac{D\cos t+E\sin t+F}{2+3\cos t}+B\tag{$\heartsuit_1$}$$ and making the right-hand side of $(\star)$ equal to $$3A\cos^2t+E\sin^2t+(3B+D)\cos t\sin t+(2A+3B+D-2)\cos t+(2B+E+F)\sin t+2B+F,$$ which can be factored somewhat, allowing us to rewrite $(\star)$ as $$0=\bigl((3\cos t+2)A-2\bigr)\cos t+\bigl((3B+D)\cos t+E\sin t+2B+F\bigr)(1+\sin t).\tag{$\star_1$}$$ Note that we could not have made the substitution $t=-\frac\pi2$ into $(\heartsuit),$ since it would have made two of the denominators equal to $0.$ To simplify $(\star_1)$ further, we actually want to find a way for the other denominator in $(\heartsuit)$ to be made equal to $0$--that is, we want some $t$ such that $\cos t=-\frac23.$ There are many such $t,$ and for such $t$ we have by Pythagorean identity that either $\sin t=\frac{\sqrt 5}3$ or $\sin t=-\frac{\sqrt 5}3.$ Making the substitutions $\cos t=-\frac23$ and $\sin t=\pm\frac{\sqrt 5}3$ into $(\star_1)$ gives us $$0=-\frac23\bigl(0A-2\bigr)+\left(-\frac23(3B+D)+2B+F\pm\frac{\sqrt5}3E\right)\left(1\pm\frac{\sqrt 5}3\right)\\0=\frac43+\left(-\frac23D+F\pm\frac{\sqrt5}3E\right)\left(1\pm\frac{\sqrt 5}3\right)\\0=\frac43+\frac13\left(-2D+3F\pm E\sqrt5\right)\left(1\pm\frac{\sqrt 5}3\right)\\0=4+\left(-2D+3F\pm E\sqrt5\right)\left(1\pm\frac{\sqrt 5}3\right)\\0=4+\frac13\left(-2D+3F\pm E\sqrt5\right)\left(3\pm\sqrt 5\right)\\0=12+\left(-2D+3F\pm E\sqrt5\right)\left(3\pm\sqrt 5\right)\\0=12-6D+9F+E(\pm\sqrt5)^2\pm(-2D+3F+3E)\sqrt 5\\0=12-6D+5E+9F\pm(-2D+3E+3F)\sqrt 5.$$ Now, this is technically not one equation, but two, depending on whether we substituted $\sin t=-\frac{\sqrt 5}3$ or $\sin t=\frac{\sqrt 5}3,$ meaning that we have both $$0=12-6D+5E+9F\pm(-2D+3E+3F)\sqrt 5\tag{1}$$ and $$0=12-6D+5E+9F\pm(-2D+3E+3F)\sqrt 5\tag{2}$$ but the only way for both $(1)$ and $(2)$ to be true is if $-2D+3E+3F=0$--meaning that $D=\frac32(E+F)$--which turns both $(1)$ and $(2)$ into $$0=12-9(E+F)+5E+9F\\0=12-4E.$$ Hence, we need $E=3$ and $D=\frac32(3+F),$ which yields $$\frac{2\cos t}{(1+\sin t)\bigl(2+3\cos t\bigr)}=\frac{A\cos t}{1+\sin t}+\frac{\frac92\cos t+3\sin t}{2+3\cos t}+B+\frac12F\tag{$\heartsuit_2$}$$ and lets us rewrite $(\star_1)$ as follows: $$0=\bigl((3\cos t+2)A-2\bigr)\cos t+\left(\left(3B+\frac92+\frac32F\right)\cos t+3\sin t+2B+F\right)(1+\sin t)$$ $$0=\bigl((6\cos t+4)A-4\bigr)\cos t+\bigl((6B+3F+9)\cos t+6\sin t+4B+2F\bigr)(1+\sin t).\tag{$\star_2$}$$ Next, we can try $t=\frac\pi2,$ which makes $\sin t=1$ and $\cos t=0,$ so that $$0=0+\bigl(0+6+4B+2F\bigr)(2)\\0=12+8B+4F\\F=-2B-3.$$ Thus, we have $$\frac{2\cos t}{(1+\sin t)\bigl(2+3\cos t\bigr)}=\frac{A\cos t}{1+\sin t}+\frac{\frac92\cos t+3\sin t}{2+3\cos t}-\frac32\tag{$\heartsuit_3$}$$ and we can rewrite $(\star_2)$ as follows: $$0=\bigl((6\cos t+4)A-4\bigr)\cos t+\bigl((6B+3(-2B-3)+9)\cos t+6\sin t+4B+2(-2B-3)\bigr)(1+\sin t)$$ $$0=\bigl((6\cos t+4)A-4\bigr)\cos t+(6\sin t-6)(1+\sin t)$$ $$0=\bigl((6\cos t+4)A-4\bigr)\cos t+6\sin^2 t-6$$ $$0=\bigl((6\cos t+4)A-4\bigr)\cos t-6(1-\sin^2 t)$$ $$0=\bigl((6\cos t+4)A-4\bigr)\cos t-6\cos^2 t$$ $$0=\bigl((6\cos t+4)A-6\cos t-4\bigr)\cos t$$ $$0=(6\cos t+4)(A-1)\cos t.$$ Substituting $t=0$ so that $\cos t=1,$ we can immediately conclude that $A=1,$ and so $$\frac{2\cos t}{(1+\sin t)\bigl(2+3\cos t\bigr)}=\frac{\cos t}{1+\sin t}+\frac{\frac92\cos t+3\sin t}{2+3\cos t}-\frac32\\\frac{2\cos t}{(1+\sin t)\bigl(2+3\cos t\bigr)}=\frac{\cos t}{1+\sin t}+\frac{\frac92\cos t+3\sin t}{2+3\cos t}-\frac{\frac32(2+3\cos t)}{2+3\cos t}\\\frac{2\cos t}{(1+\sin t)\bigl(2+3\cos t\bigr)}=\frac{\cos t}{1+\sin t}+\frac{\frac92\cos t+3\sin t}{2+3\cos t}-\frac{\frac92\cos t+3}{2+3\cos t}\\\frac{2\cos t}{(1+\sin t)\bigl(2+3\cos t\bigr)}=\frac{\cos t}{1+\sin t}+\frac{3\sin t}{2+3\cos t}-\frac{3}{2+3\cos t}.$$

At this point, an integral table would really be your friend, but let's see if we can get around it. By the work above, we know that $$I=\int\frac{\cos t}{1+\sin t}\,dt+\int\frac{3\sin t}{2+3\cos t}\,dt-\int\frac{3}{2+3\cos t}\,dt.$$ The first integral is easily found by noting that if $u=1+\sin t,$ then $du=\cos t,$ so we get $$I=\int\frac1u\,du+\int\frac{3\sin t}{2+3\cos t}\,dt-\int\frac{3}{2+3\cos t}\,dt\\I=\ln|u|+\int\frac{3\sin t}{2+3\cos t}\,dt-\int\frac{3}{2+3\cos t}\,dt\\I=\ln|1+\sin t|+\int\frac{3\sin t}{2+3\cos t}\,dt-\int\frac{3}{2+3\cos t}\,dt.$$ The second integral is easily found by noting that if $v=2+3\cos t,$ then $dv=-3\sin t\,dt$ so we similarly get $$I=\ln|1+\sin t|-\ln|2+3\cos t|-\int\frac{3}{2+3\cos t}\,dt.$$ The last integral is more of a pain. To prepare, we can use the double angle identity $\cos 2\theta=2\cos^2\theta-1$ with $\theta=\frac t2$ to show that $$I=\ln|1+\sin t|-\ln|2+3\cos t|-\int\frac{3}{-1+6\cos^2\left(\frac t2\right)}\,dt\\I=\ln|1+\sin t|-\ln|2+3\cos t|-\int\frac{\frac12\sec^2\left(\frac t2\right)}{1-\frac16\sec^2\left(\frac t2\right)}\,dt.$$ Next, we can use the Pythagorean identity $\sec^2\theta=1+\tan^2\theta$ to see that $$I=\ln|1+\sin t|-\ln|2+3\cos t|-\int\frac{\frac12\sec^2\left(\frac t2\right)}{\frac56-\frac16\tan^2\left(\frac t2\right)}\,dt.$$ Next, we make the substitution $w=\tan\frac t2,$ so that $dw=\frac12\sec^2\frac t2\,dt,$ and so $$I=\ln|1+\sin t|-\ln|2+3\cos t|-\int\frac1{\frac56-\frac16w^2}\,dw\\I=\ln|1+\sin t|-\ln|2+3\cos t|-\frac65\int\frac1{1-\frac15w^2}\,dw\\I=\ln|1+\sin t|-\ln|2+3\cos t|-\frac65\int\frac1{1-\left(\frac w{\sqrt 5}\right)^2}\,dw\\I=\ln|1+\sin t|-\ln|2+3\cos t|-\frac6{\sqrt 5}\int\frac{\frac1{\sqrt 5}}{1-\left(\frac w{\sqrt 5}\right)^2}\,dw.$$ A final substitution of $y=\frac w{\sqrt 5}$ then gives us $$I=\ln|1+\sin t|-\ln|2+3\cos t|-\frac6{\sqrt 5}\int\frac1{1-y^2}\,dy\\I=\ln|1+\sin t|-\ln|2+3\cos t|-\frac6{\sqrt 5}\tanh^{-1} y+C\\I=\ln|1+\sin t|-\ln|2+3\cos t|-\frac6{\sqrt 5}\tanh^{-1}\left(\frac{w}{\sqrt 5}\right)+C\\I=\ln|1+\sin t|-\ln|2+3\cos t|-\frac6{\sqrt 5}\tanh^{-1}\left(\frac{\tan\frac t2}{\sqrt 5}\right)+C.$$

At last, we let remember that by our earlier assumptions, $t=\sin^{-1}\left(\frac23 x-\frac 73\right).$ Immediately, $\sin t=\frac23 x-\frac 73,$ so that $$I=\ln\left|\frac23 x-\frac43\right|-\ln|2+3\cos t|-\frac6{\sqrt 5}\tanh^{-1}\left(\frac{\tan\frac t2}{\sqrt 5}\right)+C,$$ and recalling that $x\ge 2$ we see that $$I=\ln\left(\frac23 x-\frac43\right)-\ln|2+3\cos t|-\frac6{\sqrt 5}\tanh^{-1}\left(\frac{\tan\frac t2}{\sqrt 5}\right)+C.$$ Also, with a little Pythagorean work, we can show that $\cos\bigl(\sin^{-1}(z)\bigr)=\sqrt{1-z^2}$ for any $z\in[-1,1],$ so that $$\cos t=\sqrt{1-\left(\frac23 x-\frac73\right)^2}=\sqrt{-\frac49 x^2+\frac{28}9x-\frac{40}9}=\frac23\sqrt{7x-10-x^2},$$ and so $$I=\ln\left(\frac23 x-\frac43\right)-\ln\left|2+2\sqrt{7x-10-x^2}\right|-\frac6{\sqrt 5}\tanh^{-1}\left(\frac{\tan\frac t2}{\sqrt 5}\right)+C\\I=\ln\left(\frac23 x-\frac43\right)-\ln\left(2+2\sqrt{7x-10-x^2}\right)-\frac6{\sqrt 5}\tanh^{-1}\left(\frac{\tan\frac t2}{\sqrt 5}\right)+C.$$ With quite a bit more work, we can show that $\tan\frac{\sin^{-1}(a)}2=\frac{a}{1+\sqrt{1-a^2}}$ for any $a\in[-1,1],$ so that $$\tan\frac t2=\frac{\frac23 x-\frac73}{1+\sqrt{1-\left(\frac23 x-\frac73\right)^2}}=\frac{\frac13(2x-7)}{1+\frac23\sqrt{7x-10-x^2}}=\frac{2x-7}{3+2\sqrt{7x-10-x^2}},$$ and so $$I=\ln\left(\frac23 x-\frac43\right)-\ln\left(2+2\sqrt{7x-10-x^2}\right)-\frac6{\sqrt 5}\tanh^{-1}\left(\frac{\frac{2x-7}{3+2\sqrt{7x-10-x^2}}}{\sqrt 5}\right)+C\\I=\ln\left(\frac23 x-\frac43\right)-\ln\left(2+2\sqrt{7x-10-x^2}\right)-\frac6{\sqrt 5}\tanh^{-1}\left(\frac{(2x-7)\sqrt{5}}{15+10\sqrt{7x-10-x^2}}\right)+C.$$

The moral to the story is that sometimes it's not really worth doing it all out by hand. :-P


问:
How does the Cauchy integral formula follow from the Taylor expansion?
In my book (in German) there is a theorem stated the following:

Every $f\in H(U)$ can be expressed as a power series, with $H(U)$ the set of all holomorphic functions on the subset $U\subset \mathbb{C}$. With $B_r(z_0)\subset U$ and $z\in B = B_p(z_0)(0<p<r)$ it follows:
$$
f(z) = \sum_{n=0}^{\infty}a_n(z-z_0)^n,\quad a_n =\frac{1}{2\pi i} \int_{\partial B}^{} \frac{f(\zeta)}{(\zeta-z_0)^{n+1}}\, d\zeta
$$

and also $$f^{(n)}(z) = \frac{n!}{2\pi i}\int_{\partial B}^{} \frac{f(\zeta)}{(\zeta-z)^{n+1}}\, d\zeta$$

I really don't understand how he gets this formula for the derivatives. When I define the $n$-th derivative on the "normal" way I get the following expression:

$$f^{(n)}(z) = \sum_{k=n}^{\infty}(k\cdot \ldots \cdot (k-n+1))\;a_k\;(z-z_0)^{k-n}$$

I just don't understand why this two expressions should be the same or how the author even got his formula(the third equation).

答: Since$$f(z)=\frac1{2\pi i}\int_{\partial B}\frac{f(\zeta)}{\zeta-z}\,\mathrm d\zeta,$$you have\begin{align}f'(z)&=\frac1{2\pi i}\int_{\partial B}\frac{\mathrm d}{\mathrm dz}\left(\frac{f(\zeta)}{\zeta-z}\right)\,\mathrm d\zeta\\&=\frac1{2\pi i}\int_{\partial B}\frac{f(\zeta)}{(\zeta-z)^2}\,\mathrm d\zeta.\end{align}Then you have\begin{align}f''(z)&=\frac1{2\pi i}\int_{\partial B}\frac{\mathrm d}{\mathrm dz}\left(\frac{f(\zeta)}{(\zeta-z)^2}\right)\,\mathrm d\zeta\\&=\frac2{2\pi i}\int_{\partial B}\frac{f(\zeta)}{(\zeta-z)^2}\,\mathrm d\zeta,\end{align}and so on.

ContentLicense: CC BY-SA 4.0


问:
Definition of a dominating function and the Dominated Convergence Theorem.
I apologise if this is a rather simplistic or even silly question, but I am confused with the word "dominated" in Lebesgue's Dominated Convergence Theorem (DCT) since I can find no definition of a dominating function in the textbook I am following, rather I can only find a definition in relation to measures.

In the DCT it is assumed we have a sequence $\{f_{n}\}$ of real measurable functions where $|f_{n}|\leq g$ almost everywhere for all $n$, and where $g$ is integrable. Now I can clearly see from various discussions on the DCT that $g$ is considered to be the dominating function, but I cannot find any definition to confirm this. The only definition I can find is related to measures and absolute continuity:

The measure $v$ is absolutely continuous with respect to the measure $\mu$ if for all $A\in\mathscr{F}$, $\mu(A)=0$ implies $v(A)=0$. In the text I am following this is written as $v<<\mu$ which means $v$ is dominated by $\mu$.

Am I right in assuming there is nothing linking these two uses of the word dominating?

答:
Given two positive functions $f$ and $g$ such that $f$ dominates $g$, let
$$\nu(A) = \int_A g d\mu, \qquad \eta(A) = \int_A f d\mu.$$
We note that $\nu(A) \leq \eta(A)$ for all measurable sets $A$.  In particular, $\nu(A)=0$ for all $\eta$-null sets $A$, so that $\nu \ll \eta$.  In this sense,

Domination of functions implies absolute continuity of the induced measures.

Of course, the converse is wildly false.  For example, the measures induced by $f=2$ and $g=1$ are absolutely continuous with respect to each other, whereas $f$ dominates $g$ (and not conversely).

I suppose you could use the Radon-Nikodym theorem to further analyze this situation.

ContentLicense: CC BY-SA 3.0


问:
find the number of possible positive integer solutions when the inequality is $a \times b \times c \lt 180$
As most of you know there is a classical question in elementary combinatorics such that if $a \times b  \times c = 180$ , then how many possible positive integer solution are there for the equation $?$

The solution is easy such that $180=2^2 \times 3^2 \times 5^1$ and so , for $a=2^{x_1} \times 3^{y_1} \times 5^{z_1}$ ,   $b=2^{x_2} \times 3^{y_2} \times 5^{z_2}$ ,   $c=2^{x_3} \times 3^{y_3} \times 5^{z_3}$ .

Then: $x_1+x_2+x_3=2$ where $x_i \geq0$ ,  and $y_1+y_2+y_3=2$ where $y_i \geq0$ and $z_1+z_2+z_3=1$ where $z_i \geq0$.

So , $C (4,2) \times C(4,2) \times C(3,1)=108$.

Everything is clear up to now.However , i thought that how can i find that possible positive integer solutions when the equation is $a \times b  \times c \lt 180$  instead of $a \times b  \times c = 180$

After , i started to think about it. Firstly , i thought that if i can calculute the possible solutions for $x_1+x_2+x_3 \lt2$ where $x_i \geq0$ ,  and $y_1+y_2+y_3 \lt 2$ where $y_i \geq0$ and $z_1+z_2+z_3 \lt1$ where $z_i \geq0$ , then i can find the solution.However , there is a problem such that when i calculate the solution , i do not include the prime numbers and their multiplicites which is in $180$.

For example , my solution does not contain $1 \times 1 \times 179 \lt 180$

My question is that how can we solve these types of question . Is there any $\color{blue}  {\text{TRICK}}  $ for include all possible ways ? Moreover ,this question can be generalized for  $a \times b  \times c \leq 180$ , then what would happen for it ?

Thanks for helps..

答:
We can start with listing all the factors of $180$  so we have

$x,y,z|180: x\land y\land z\in\big\{
1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30,36,45,60,90, 180 \big\}$

We cannot use such as $(x,y,z)\in\big\{(180,1,1),\space (90,2,1)\big\}$ because $x\cdot y\cdot z<180$ but we can us such as $\space (x,y,z)=(90,1,1)\space $ because $\space 90 \cdot 1\cdot 1 < 180.\space $
Likewise

$$(60\cdot 3\cdot 1)= 180\implies\\(60\cdot 1\cdot 1) < (60\cdot 2\cdot 1)< 180$$

$$(45\cdot 4\cdot 1)= 180\implies\\(45\cdot 1\cdot 1) < (45\cdot 2\cdot 1)<(45\cdot 3\cdot 1)< 180$$

$$(36\cdot 5\cdot 1)= 180\quad \implies\\
(36\cdot 1\cdot 1) < (36\cdot 2\cdot 1)<(36\cdot 3\cdot 1)<(36\cdot 4\cdot 1)< 180\\
\land \quad (36\cdot 2\cdot 2)<180$$

$$(30\cdot 6\cdot 1) = 180\quad \implies\\
(30\cdot 1\cdot 1) < (30\cdot 2\cdot 1)<(30\cdot 3\cdot 1)<(30\cdot 4\cdot 1)< (30\cdot 5\cdot 1< 180\\
\land \quad (30\cdot 2\cdot 2)<180$$

$$(20\cdot 9\cdot 1) = 180\quad \implies\\
(20\cdot 1\cdot 1) < (20\cdot 2\cdot 1)<(20\cdot 3\cdot 1)<(20\cdot 4\cdot 1)\\< (20\cdot 5\cdot 1< 180
<(20\cdot 6\cdot 1)< (20\cdot 7\cdot 1<(20\cdot 8\cdot 1) < 180\\
\land \quad (20\cdot 2\cdot 2)<(20\cdot 3\cdot 2)< (20\cdot 4\cdot 2)<180$$

If we continue this process through $x=1$, we will have all of the combinations and $\frac16$ of the permutations of $x,y,z$ that satisfy the equation.


问:
Puzzling solution for $1/x = 3 - 2\sqrt{x}$
What is the solution for

$$\frac{1}{x} = 3 - 2\sqrt{x}$$

When I plot $1/x$ and $3-2\sqrt{x}$ separately, it meets at $x = 1$.

However when I solve the equation $1/x = 3 - 2\sqrt{x}$ algebraically, I get two solutions $-1/4$ and $1$.

What am I missing?

答:
At some point in your calculation you are squaring both sides.  This introduces extraneous solutions.  If you have

$$x=-2$$

and square both sides you get

$$x^2=4$$

which has two solutions.


问:
Generalizing Ex6.14 from Atiyah-Macdonald: is $S^{-1}(M/N)\cong S^{-1}M/ S^{-1}N$?
In doing exercise 6.14 from Atiyah-McDonald

Let $M$ be an $A$-module, and $a$ and ideal fo $A$. Suppose that $M_\mathfrak{m}=0$ for all maximal ideals $\mathfrak{m}\supset a$. Then $M=\mathfrak{a}M$.

We need the following fact:

$(M/\mathfrak{a}M)_\mathfrak{p}=M_\mathfrak{p}/(\mathfrak{a}M)_\mathfrak{p}$

This is not hard too hard to prove as $M/\mathfrak{a}M\cong M\otimes_A A/\mathfrak{a}$ and since localizing commutes with the tensor product we
$(M/\mathfrak{a}M)_\mathfrak{p}\cong M_\mathfrak{p}\otimes_{A_\mathfrak{p}} (A/\mathfrak{a})_\mathfrak{p}\cong M_\mathfrak{p}/(\mathfrak{a}M)_\mathfrak{p}$.

How much does this fact generalise? Is the following true?

Let $M$ and $A$-module. Let $N$ be an $A$-submodule of $M$. Then $S^{-1}(M/N)\cong S^{-1}M/ S^{-1}N$.

答: Yes: the functor $M\mapsto S^{-1}M$ is exact so that the image of the short exact sequence $$0\to N\longrightarrow M\longrightarrow M/N\to 0$$ under it is a short exact sequence $$0\to S^{-1}N\longrightarrow S^{-1}M\longrightarrow S^{-1}(M/N)\to 0$$ This shows that $S^{-1}(M/N)\simeq S^{-1}M/S^{-1}N$.


问:
Probability that a randomly chosen eldest child is from a one-child family; a four-child family
The probability of having a one-child family is $0.1$; the probability of having a two child family is $0.25$; the probability of having a three- child family is $0.35$ and the probability of having a four-child family is $0.3$. If a randomly chosen child is the eldest child in his/her family, calculate the probability that he/she is from:

(a) a one-child family;

(b) a four-child family

my thoughts for part a) 0.1/10 = 0.01 or 1% and for part b) 0.3/10=0.03 or 3%. Unless I am missing something here...

Body:
Use Bayes' rule, where $e$ represents "eldest is chosen":

$$P(1|e) = \frac{P(e|1)P(1)}{P(e)}
= \frac{P(e|1)P(1)}{\sum\limits_{i=1}^4 P(e|i)P(i)} \\
= \frac{1 \cdot 0.1}{0.1 \cdot 1 + 0.25 \cdot 1/2 + 0.35 \cdot 1/3 + 0.3 \cdot 1/4} \\
= \frac{0.1}{0.416667} = 0.24 .$$


问:
Need help on understanding a proof of the formula for calculating the $\zeta(2n)$
In this thread, Jack D'Aurizio provided a succinct proof for the formula of calculating the values of the Zeta function $\zeta(2n)$

$$ \coth z-\frac{1}{z} = \sum_{n\geq 1}\frac{4^n\,B_{2n}}{(2n)!}z^{2n-1}.\tag{2}$$$$\begin{eqnarray*} \coth z -\frac{1}{z} &=& \color{red}{\sum_{n\geq 1}}\frac{d}{dz}\log\left(1+\frac{z^2}{n^2 \pi^2}\right)\\&=&\sum_{n\geq 1}\frac{2z}{\pi^2 n^2+z^2}\\&=&\color{red}{\sum_{n\geq 1}\sum_{m\geq 1}}\frac{2\color{red}{(-1)^{m-1}z^{2m-1}}}{\pi^{2m}\color{red}{n^{2m}}} \\&=&\sum_{m\geq 1}\frac{2\,\zeta(2m)}{\pi^{2m}}(-1)^{m-1}z^{2m-1}\tag{3}\end{eqnarray*}$$

and we have the claim by comparing the coefficients in the RHSs of (2) and (3).

I just have a couple of questions regarding this derivation.

First, the sigma notation. I thought that the infinite product for $\sinh(x)$ is $\displaystyle\prod_{n=1}^{+\infty}\color{red}{z}\left(1+\dfrac{z^2}{n^2\pi^2}\right)$. So why there is a sum here, shouldn't it be a product? Also where does the $\color{red}z$ go?

How to prove, or how do we know that $\coth(z)-\dfrac{1}{z}=\displaystyle\sum_{n=1}^{+\infty}\dfrac{d}{dz}\log\left(1+\dfrac{z^2}{n^2\pi^2}\right)$. This is my guess. The derivative of $\log(\sinh(z))$ is $\coth(z)$. So that is why it is, am I correct?

I don't understand anything at all about the double sigma notation. I am always in fear of double sigma notation because I haven't learnt how to manipulate them well. So what was Jack doing here? Suddenly there are $(-1)^{2m-1}$, $z^{2m-1}$, $\pi^{2m}$

I don't understand anything when he says "comparing the coefficients in the RHSs of (2) and (3)." What should I suppose to do to understand this?

答:
$$\sinh z=z\prod_{n=1}^\infty\left(1+\frac{z^2}{n^2\pi^2}\right) \implies \log(\sinh{z}) = \log(z)+ \log\prod_{n=1}^\infty\left(1+\frac{z^2}{n^2\pi^2}\right) \\ \implies
\log(\sinh{z})-\log(z) =  \log\prod_{n=1}^\infty\left(1+\frac{z^2}{n^2\pi^2}\right) = \sum_{n=1}^\infty\log\left(1+\frac{z^2}{n^2\pi^2}\right)$$

Differentiating both sides we get $$\coth z -\frac{1}{z} = {\sum_{n\geq 1}}\frac{d}{dz}\log\left(1+\frac{z^2}{n^2 \pi^2}\right) = \sum_{n\geq 1}\frac{2z}{\pi^2 n^2+z^2}$$

Next, we write the summand as an infinite series. Since $\displaystyle \frac{1}{1+t} = \sum_{m=1}^{\infty}(-1)^{m-1}t^{m-1}$ we have $$\frac{2z}{\pi^2 n^2+z^2} = \frac{2z}{\pi^2 n^2} \cdot \frac{1}{1+\left(\frac{z}{\pi n}\right)^2} = \frac{2z}{\pi^2 n^2} \cdot\sum_{m=1}^{\infty}(-1)^{m-1} \left(\frac{z}{\pi n}\right)^{2m-2}$$

Which is equal to  $\displaystyle \frac{2z}{\pi^2 n^2} \cdot\sum_{m=1}^{\infty}(-1)^{m-1} \frac{z^{2m-2}}{\pi^{2m-2}n^{2m-2}} = \sum_{m=1}^{\infty}(-1)^{m-1} \frac{z^{2m-1}}{\pi^{2m}n^{2m}}.$

So we have $$\coth z -\frac{1}{z} ={\sum_{n\geq 1}\sum_{m\geq 1}}\frac{2{(-1)^{m-1}z^{2m-1}}}{\pi^{2m}{n^{2m}}} $$

And summing over $n$ first we have
$${\sum_{n\geq 1}\sum_{m\geq 1}}\frac{2{(-1)^{m-1}z^{2m-1}}}{\pi^{2m}{n^{2m}}} = {\sum_{m\geq 1}\sum_{n\geq 1}}\frac{2{(-1)^{m-1}z^{2m-1}}}{\pi^{2m}{n^{2m}}} ={\sum_{m\geq 1}}\frac{2{(-1)^{m-1}z^{2m-1}}}{\pi^{2m}} \bigg(\sum_{n \ge 1} \frac{1}{n^{2m}}\bigg)$$

Since $$\displaystyle \sum_{n \ge 1} \frac{1}{n^{2m}}= \zeta(2m)$$

We have

$$\displaystyle\coth z -\frac{1}{z} = \sum_{m\geq 1}\frac{2\,\zeta(2m)}{\pi^{2m}}(-1)^{m-1}z^{2m-1}.$$

So, on the one hand

$$\displaystyle \coth z-\frac{1}{z} = \sum_{n\geq 1}\frac{4^n\,B_{2n}}{(2n)!}z^{2n-1}$$
but on the other $$\displaystyle \coth z-\frac{1}{z} = \displaystyle \sum_{m\geq 1}\frac{2\,\zeta(2m)}{\pi^{2m}}(-1)^{m-1}z^{2m-1}$$

So $$\sum_{n\geq 1}\frac{4^n\,B_{2n}}{(2n)!}z^{2n-1} = \displaystyle \sum_{m\geq 1}\frac{2\,\zeta(2m)}{\pi^{2m}}(-1)^{m-1}z^{2m-1}$$

Reindexing the RHS  with $n$

$$\sum_{n\geq 1}\frac{4^n\,B_{2n}}{(2n)!}z^{2n-1} = \displaystyle \sum_{n\geq 1}\frac{2\,\zeta(2n)}{\pi^{2n}}(-1)^{n-1}z^{2n-1}$$

Then comparing the coefficient of $z^{2n-1}$ on both sides

$$\frac{4^n\,B_{2n}}{(2n)!} =\frac{2\,\zeta(2n)}{\pi^{2n}}(-1)^{n-1} $$

And finally solving for $\zeta(2n)$ gives:

$$\zeta(2n) = \frac{4^n\,B_{2n} (-1)^{n-1}\pi^{2n}}{2(2n)! } =\frac{\,B_{2n} (-1)^{n-1}(2\pi)^{2n}}{2(2n)! }$$


问:
Why are Line Integrals of Vector Fields independent of Magnitude of $\vec{r}'$?
If we have a line integral of a vector field $\vec{F}$ along a space curve $\vec{r}(t)$
$$\int\limits_{t=a}^{t=b}\vec{F}(\vec{r}(t))\cdot\vec{r}'(t)\,dt$$
Why doesn't it matter that $\vec{r}'$ doesn't always have unit length? For example, in this visualization of a line integral (https://en.wikipedia.org/wiki/File:Line_integral_of_vector_field.gif), it seems that $r'$ always has unit length, while it usually doesn't. Why does the length/magnitude of $\vec{r}'$ not matter?

答:
The line integral is expressed more univerally as $$\int_C \vec F\cdot d\vec r$$
where $C$ is the curve of integration. This can be considered at the limit of a Riemann sum, where the $P = \{\vec r_i\}_{i=0}^N$ are partition points of  on the curve:
$$\int_C \vec F\cdot d\vec r = \lim_{|P|\to 0} \sum_{i=1} \vec F(\vec r_i)\cdot (\vec r_i - \vec r_{i-1})$$
When we move to a paramtetrization, $\vec r = \vec r(t)$, then by the vector version of the chain rule, $d\vec r = \frac {d\vec r}{dt}dt$.

The fact that $\frac {d\vec r}{dt}$ is not restricted to be $1$ is exactly what is needed. If we change to a parameter that zooms through the curve twice as fast, $u = 2t$, then $du = 2dt$, so we would have $d\vec r = \frac {d\vec r}{dt}\frac {du}2$. But this works because by the chain rule
$$\frac {d\vec r}{dt} = \frac {d\vec r}{du}\frac {du}{dt} = 2\frac {d\vec r}{du}$$ so the differential becomes
$$d\vec r = \left(2\frac {d\vec r}{du}\right)\left(\frac {du}2\right) = \frac {d\vec r}{du}du$$
which is the exactly the same form as $t$. If they did not have the same form, then this expression for $d\vec r$ would not be true for any arbitrary smooth parametrization.

Basically, when the parameter runs through the curve faster, we need $\vec r'(t)$ to be larger to compensate for $dt$ being smaller to cover the same ground.


问:
Identify all $a \mod 35$ such that $35$ is a pseudoprime to base $a$

Identify all $a \mod 35$ such that $35$ is a pseudoprime to base $a$.

By defintion, $\gcd(a,n) = 1$ for $a,n \in \mathbb{Z}$, then $a^{n-1} \equiv 1 \mod n$ means that $n$ is a pseudoprime to base $a$.

$G_{35}$ is the group of elements all coprime to $35$. So I have said that

$$G_{35} \cong G_5 \times G_7.$$

The order of the two isomporhic groups are $4$ and $6$ and so the order of $G_{35} = \mathrm{lcm}(4,6) = 12$. That means all divisors of $12$ will be potential orders of $a$. I want all the $a$ such that $a^{34} = 1 \mod 35$. By Eulers theorem, this tells me that the order of this element is $\gcd(12,34) = 2$. So I want all elements of order $2$ in $G_5$ and $G_7$. Clearly, this is only $\pm 1 \mod 5$ and $\pm 1 \mod 7$ and so all values of $a$ are of the form $\pm1 \mod 7, \pm 1 \mod 5$?

Have I done this correctly? I think I might have cheated a little with the Euler theorem bit, because normally I would say $34 \mid 12$ but seeing as $34$ doesn't divide $12$ and $12$ doesn't divide $34$, I got a little stuck.

EDIT: Actually, I don't think I've finished the question as the answer should be $\mod 35$. So would the final answer just be $\pm 1 \mod 35$?

答:
We will use standard elementary number theory terminology, not group terminology, but that makes no difference.

We want to characterize those  $a$ such that $a^{34}\equiv 1\pmod{35}$.  For such an $a$, the order of $a$ modulo $35$ must divide $34$. It must also divide $\varphi(5)$ and $\varphi(7)$, giving the possibilities $1$ or $2$. And it is clear that if $a$ has order $1$ or $2$ then $a^{34}\equiv 1\pmod{35}$.

So we want to find all elements of order $1$ or $2$ modulo $35$. So we want to solve the $4$ systems of congruences $x\equiv \pm 1\pmod{5}$, $x\equiv \pm 1\pmod{7}$.

If the signs match, the solutions are obvious. For $x\equiv 1\pmod{5}$, $x\equiv -1\pmod{7}$, a short search gives $x\equiv 6\pmod{35}$. The remaining congruence then has solution $x\equiv -6$.


问:
$\text{Sat}_{\Delta_0}(x, y)$ is $\Delta_1(\textbf{PA})$ (Kaye's book)
In Kaye's "Models of Peano Arithmetic" in chapter 9 on satisfaction a lot of effort is expanded to prove that $\text{Sat}_{\Delta_0}(x, y)$, representing truth of $\Delta_0$-sentences, is a $\Delta_1$-sentence under $\textbf{PA}$ (Theorem 9.13 in Kaye, p. 125). However, I'm unsure how this is used. Here is the theorem:

Kaye then concludes:

I guess the latter is a result of the properties listed in Theorem 9.13 plus induction in the metatheory on the complexity of formulas. I presume it is here that $\text{Sat}_{\Delta_0}(x, y)$ is $\Delta_1(\textbf{PA})$ is used, because somewhat earlier in the chapter Kaye has a similar strategy. He proves:

And then mentions a similar result where instead the $\Delta_1$-ness of the formula $\text{term}(x)$ is used.

However, I also don't see how $\mathbb{N} \models \text{term}(m) \Leftrightarrow \textbf{PA} \models \text{term}(m)$ is used here. Furthermore, isn't it enough that $\text{term}(x)$ is $\Sigma_1$ to show this?

答:
FYI there seems to be a typo in Kaye: in the claimed ${\bf PA}$-theorem scheme of the "in particular" part, the first block of $v$s should be a block of $x$s.

I don't think the $\Delta_1$-ness of $\mathsf{Sat}_{\Delta_0}$ is in fact used in the "in particular" passage. Rather, Theorem $9.13$ proves two things:

a complexity bound on $\mathsf{Sat}_{\Delta_0}$, modulo ${\bf PA}$; and

some nice ${\bf PA}$-provable properties of $\mathsf{Sat}_{\Delta_0}$.

The "in particular"  refers only to this second point; at a glance, the metatheoretic induction taking place (your guess is correct re: this) only uses the fact that ${\bf PA}$ proves enough to "break down" instances of $\mathsf{Sat}_{\Delta_0}$. Now granted, this second bulletpoint is vastly simpler than the first and, as its proof indicates, it (and consequently the "in particular" bit) could have been stated right after the Explanation of the Definition of $\mathsf{Sat_{\Delta_0}}$ in the first place.

But nonetheless the $\Delta_1$-ness of the formula $\mathsf{Sat}_{\Delta_0}$ (as always, modulo ${\bf PA}$) is interesting in its own right. In general, keeping track of the "provable complexity" of a formula is a useful thing to do. Note that this type of analysis is not subsumed by the computability-theoretic interpretation of the arithmetic hierarchy: while knowing that $\mathsf{Sat}_{\Delta_0}$ is computable is enough to conclude that it is $\Delta_1$ in the structure $\mathbb{N}$, that is not enough to conclude that it is $\Delta_1$ modulo ${\bf PA}$. So this sort of analysis is nontrivial.

Meanwhile, as to your last question, note that the $\Sigma_1$-ness of $\mathsf{term}(x)$ gives us $$\mathbb{N}\models\mathsf{term}(m)\implies{\bf PA}\models\mathsf{term}(m)$$ (conflating a number with its numeral, as usual). However, for the converse implication we need that $\mathsf{term}(x)$ is $\Pi_1$. For example, take $\mathsf{term}(x)$ to be "${\bf PA}$ is inconsistent" (so it ignores $x$ completely). Then - unless we grant more than mere consistency of ${\bf PA}$ - it could be the case that ${\bf PA}\models \mathsf{term}(0)$ (say) while $\mathbb{N}\not\models\mathsf{term}(0)$.


问:
How do we know the uniqueness of the transport equation?
I'm learning PDE and have a basic question.

In Evans book on PDE, in order to solve $\partial_t u+\langle b, Du\rangle=0$ with the initial condition $u(0,\cdot)=g$, I first have to use the auxillary function $w(s)=u(t+s,x+sb)$ and realize that $u(t+s,x+sb)=u(t,x)$.  Now we can plug in $s=-t$ and use the initial condition to derive that the solution is of the form $u(x,t)=g(x-tb)$. But here, since we are using some "auxillary function," analytically speaking I don't know the uniquenss, right? Can somebody give me some insight on how to achieve uniqueness in the simplest way? Thanks!

答:
If $\partial_tu + \langle b,\nabla u\rangle = 0$ for some $b \in \Bbb R^n$, and we wish to solve for $u$ (defined on some subset of $\Bbb R^{n+1}$), we try to reconstruct the graph of $u$ inside $\Bbb R^{n+2}$, by rewriting the PDE as $$\langle (\partial_tu, \nabla u, -1),(1,b,0)\rangle = 0.$$Note that $(\partial_tu, \nabla u, -1)$ would be the normal direction to the graph of $u$. The flow of the vector field $V(t,x,u)=(1,b,0)$ (which will be tangent to the graph) is given by $$(t,x,u)=\Phi_V(s,(t_0,x_0,u_0))=(t_0+s,x_0+sb, u_0).$$So, saying that $u(0,x)=g(x)$ is saying that for $t_0=0$, we have that $u_0=g(x_0)$. Now, since $$x_0 = x-sb \quad\mbox{and}\quad s = t,\quad\mbox{and}\quad u=u_0$$we have $x_0 = x-tb$. Thus, the general solution is of the form $u(t,x)=g(x-tb)$. If you do not want to make reference to $g$, write $u(t,x) = u(0,x-tb)$, this is what the transport equation gave us.

In the same way when solving an ODE you get a constant of integration which depends on initial conditions, for PDEs you have not a constant, but a function which depends on initial conditions. So, changing the initial conditions changes this function as well and you don't quite have uniqueness.


问:
Simple connected eulerian graph $G = (V, E)$ cannot be bipartite if $(V, E - \{e_1,e_2,e_3\})$ is also eulerian
Let $G = (V, E)$ be a simple connected eulerian graph, I was asked to prove that it cannot be bipartite if $(V, E - \{ e_1,e_2,e_3 \})$ is also eulerian.

While processing my proof I came up with this simple, connected and eulerian graph: $$v_1-v_2-v_3-v_4-v_5-v_6-v_1$$
$G$ is also bipartite e.g. let $A = \{v_1,v_3,v_5\}$,$B = \{v_2,v_4,v_6\}$. If I would remove the edges $v_1v_2,v_2v_3,v_3v_4$, $$v_4-v_5-v_6-v_1, v_2, v_3$$ is still eulerian, what am I missing?

Edit: I now noticed that an eulerian graph is defined as a graph that conatins an eulerian cycle my example only has an eulerian path therefore the graph is not eulerian.

答: Since $G$ is Eulerian, each vertex has even degree, and the sum of the degrees in each of its two parts is even. Since $G$ is bipartite, all of the edges are between the two parts, so removing $3$ of them reduces the sum of the degrees of the vertices in each part by $3$, leaving an odd total. This is possible only if at least one vertex in each part now has odd degree, in which case the new graph cannot be Eulerian. (In fact the only way to remove $3$ edges from an Eulerian graph and not produce a vertex of odd degree is to remove a triangle of edges, which does not exist in a bipartite graph.)


问:
Circle of radius 1 is inscribed in an equilateral triangle PQR. The point of contacts of C with PQ, QR, RP are D,E,F....

Circle of radius 1 is inscribed in an equilateral triangle PQR. The point of contacts of C with PQ, QR, RP are D,E,F respectively. If PQ is $\sqrt 3 x + y-6=0$ and D is $(\frac{\sqrt 3}{2} , \frac 32)$ and origin and centre of the C are on the same side, find equation of C

Now according to the answer, the center of the given circle is $(\sqrt 3,1)$

Let the center be $(h,k)$

Then
$$-\sqrt 3 \times \frac{k-\frac 32}{h-\frac{\sqrt 3}{2}}=-1$$
But the answer doesn’t seem to satisfying $(h,k)$ in this equation.

I feel like this question will end up embarrassing me for being too simple, but I really don’t know why this formula is not working

答:
Well, since the point $D$ is definitively not on $PQ$, we will "slightly change" the statement, so that the expected answer is an answer, according to the following picture:

Problem:
$\Delta PQR$ is an equilateral triangle with points in the cartesian plane $\Bbb R^2$, so that the equation of the line $PQ$ is $\sqrt 3 x +y-6=0$. The incircle $\odot (I)$ of $\Delta PQR$ touches the sides of this triangle opposite to $P,Q,R$ in $D,E,F$ respectively.
The point $D$ has coordinates $\displaystyle\left(\frac {\sqrt 3}2,\frac 32\right)$.

Find the equation of the incircle $\odot(I)=\odot(DEF)$.

Note: I could not understand the "same side" story.

Note:
I am against formulations of the shape <<If $PQ$ is ... and $D$ is ... and ... find equation of the circle $\odot(I)$>>

Why? Since this "if-then sentence" is not giving anything, it is an implication. Even if we understand the if-condition, and would be polite to assume it, it would be false to solve the problem till we can show that the assumption is true. But we do not have enough information to decide. I would not recommend to do this in an exam, as i did, except for the case of an ultima ratio. Well, i had luck, the professor accepted my logic.

Solution:
Since $\Delta PQR$ is equilateral, the triangle $\Delta DEF$ is also equilateral.
Its circumcircle $\odot(I)$ has radius $1$, so the two sides of equal length $1$ in $\Delta IDF$ are building an angle of $120^\circ$, which gives
$$
DF^2=1^2+1^2-2\cdot 1\cdot 1\cdot \cos 120^\circ
=1+1+1=3\ .
$$
So $F\in PQ$ is a point on the line $PQ$ with known equation, at known distance $\sqrt 3$ from the known point $D$. Very quickly the "two chances for $F$", let us denote them by $F_1$ and $F_2$ are
$$
F_1 =
\left(
\frac{3\sqrt 3}2,\frac 32
\right) \ ,
\qquad
F_2=(\sqrt 3,3)\ .
$$
We consider the two chances separately.

For $F=F_1$, since the point $Q$ makes $\Delta DFQ$ equilateral, and $Q$ satisfies the given equation of $PQ$, we have only the chance $Q=F_2$. The homothety centered in $Q$ and factor $2$ brings $\Delta QDF$ into $\Delta QRP$, so the other two vertices of the latter are $R(0,0)$ and $P(2\sqrt 3,0)$. The mid point of $PR$ is thus $E=(\sqrt 3,0)$. (Or we could compute $E$ as the reflection of $Q$ in $DF$.) The median $QE$ between $Q(\sqrt3,3)$ and $E(\sqrt 3,0)$ has length $3$, and contains the point $I$, which is also the centroid, placed at distances in proportion $2:1$ from $Q$ and $E$, so
$$
I=(\sqrt 3,1)\ .
$$
(Well, this point and the origin both satisfy $\sqrt 3 x +y-6<0$, so they are on the same side of $PQ$, but the next point has the same property.)

It remains to write down the equation of the circle centered in this point $I$ having radius one, $(x-\sqrt 3)^2+(y-1)^2=1^2$.

For $F=F_2$, with the same argument, $Q$ is the other "$F$-choice", $Q=F_1$, and we construct similarly $I$. The reflection of $Q$ w.r.t. $DF$ is the point $E(0,3)$, so the second choice of $I$ is $\displaystyle\left(\frac{\sqrt 3}2,\frac 52\right)$.


答:
If we let $y = \sqrt{x}$, we get
\begin{align*}
\frac{1}{y^{2}} = 3 - 2y & \Longleftrightarrow 1 = 3y^{2} - 2y^{3}\\
& \Longleftrightarrow 2y^{3} - 3y^{2} + 1 = 0\\\\
& \Longleftrightarrow (2y^{3} - 2y^{2}) - (y^{2} - 1) = 0\\\\
& \Longleftrightarrow 2y^{2}(y-1) - (y-1)(y+1) = 0\\\\
& \Longleftrightarrow (2y^{2} - y - 1)(y-1) = 0\\\\
& \Longleftrightarrow \left(y + \frac{1}{2}\right)(y-1)^{2} = 0\\\\
& \Longleftrightarrow \left(y = -\frac{1}{2}\right)\vee(y = 1)
\end{align*}

Since $y = \sqrt{x} > 0$, there is only one possible solution, which is $x = 1$.

Hopefully this helps!

Title: Puzzling solution for $1/x = 3 - 2\sqrt{x}$

Body:
What is the solution for

$$\frac{1}{x} = 3 - 2\sqrt{x}$$

When I plot $1/x$ and $3-2\sqrt{x}$ separately, it meets at $x = 1$.

However when I solve the equation $1/x = 3 - 2\sqrt{x}$ algebraically, I get two solutions $-1/4$ and $1$.

What am I missing?


问:
Formalizing picking the nth element of each row of a matrix, being n determined by a vector
I've struggled so far to formalize the following in mathematical notation.

Let $h_{i}$ be a variable containing a natural number between 1 and n. Let S be the following matrix of size $m x n$

$$
S =
\begin{pmatrix}
s_{11} & \cdots & s_{1n} \\
\vdots & \ddots & \vdots \\
s_{i1} & \cdots & s_{in}
\end{pmatrix}
$$

I would like to be able to formalize the following mathematically. For each row of the matrix, I want to pick the $n^{th}$ element, being this element determined by vector $\vec h$.

In other words. Let's assume the following.

$$
h=[3,2,1]
$$

$$
S=
\begin{pmatrix}
4 & 3 & 5 \\
2 & 9 & 7 \\
5 & 8 & 1
\end{pmatrix}
$$

I would like to end up with the following vector $\vec z$

$$
z = [5, 9, 5]
$$

In the first row I pick the third element ($h_{1}=3$). In the second row I pick the second element ($h_{2}=2$), and I pick the first in the last ($h_{3}=1$).

As I've said I would like to formalize this mathematically, any hints will be greatly appreciated.

答:
There are typos in your S matrix, the lower left entry should be $S_{m1}$, not $S_{i1}$ and the lower right should be $S_{mn}$ not $S_{in}$. Let $$S=\left(\begin{matrix}S_{11}&S_{12}&...&S_{1n}\\S_{21}&S_{22}&...&S_{2n}\\
...&&...&\\S_{m1}&S_{m2}&...&S_{mn}\end{matrix}\right), h=\left(\begin{matrix}h_1\\h_2\\...\\h_m\end{matrix}\right)$$

Then z with $z_i=S_{i h_i}$ for $i=1,2,...,m$ chooses the "ith entry of h"th element of the ith row of S and puts it in the ith entry of z. That is,

$$z=\left(\begin{matrix}S_{1h_1}\\S_{2h_2}\\...\\S_{mh_m}\end{matrix}\right)$$


答:
First, for fixed $z\in B=B_\rho(z_0)$, the geometric series converges uniformly on $\partial B$:
$$
\frac{1}{\zeta-z}=\sum_{n=0}^{\infty} \frac{\left(z-z_{0}\right)^{n}}{\left(\zeta-z_{0}\right)^{n+1}}
$$

By the uniform convergence, you can switch the integral and summation sign if you apply the Cauchy integral formula (Theorem 2.5.1 in the book):
$$
f(z)=\frac{1}{2 \pi i} \int_{\partial B} \frac{f(\zeta)}{\zeta-z} d \zeta
=\frac{1}{2 \pi i} \int_{\partial B} {f(\zeta)} \sum_{n=0}^{\infty} \frac{\left(z-z_{0}\right)^{n}}{\left(\zeta-z_{0}\right)^{n+1}} d \zeta\\
=\sum_{n=0}^{\infty}\left(\frac{1}{2 \pi i} \int_{\partial B} {f(\zeta)}  \frac{1}{\left(\zeta-z_{0}\right)^{n+1}} d \zeta\right)\left(z-z_{0}\right)^{n}
 \quad(z \in B)
$$
the n-th coefficient of which gives you the formula:
$$
f^{(n)}(z_0)=\frac{n !}{2 \pi i} \int_{\partial B} \frac{f(\zeta)}{(\zeta-z_0)^{n+1}} d \zeta\tag{1}
$$

For $z\in B\setminus\{z_0\}$, the argument above shows that

$$
f^{(n)}(z)=\frac{n !}{2 \pi i} \int_{\partial B_\rho(z)} \frac{f(\zeta)}{(\zeta-z)^{n+1}} d \zeta\tag{2}
$$

But Cauchy's theorem allows you to shift the contour to conclude that
$$
f^{(n)}(z)=\frac{n !}{2 \pi i} \int_{\partial B_\rho(z)} \frac{f(\zeta)}{(\zeta-z)^{n+1}} d \zeta
=\frac{n !}{2 \pi i} \int_{\partial B} \frac{f(\zeta)}{(\zeta-z)^{n+1}} d \zeta
\tag{2'}
$$

The original proof in your book.

Title: How does the Cauchy integral formula follow from the Taylor expansion?

Body:
In my book (in German) there is a theorem stated the following:

Every $f\in H(U)$ can be expressed as a power series, with $H(U)$ the set of all holomorphic functions on the subset $U\subset \mathbb{C}$. With $B_r(z_0)\subset U$ and $z\in B = B_p(z_0)(0<p<r)$ it follows:
$$
f(z) = \sum_{n=0}^{\infty}a_n(z-z_0)^n,\quad a_n =\frac{1}{2\pi i} \int_{\partial B}^{} \frac{f(\zeta)}{(\zeta-z_0)^{n+1}}\, d\zeta
$$

and also $$f^{(n)}(z) = \frac{n!}{2\pi i}\int_{\partial B}^{} \frac{f(\zeta)}{(\zeta-z)^{n+1}}\, d\zeta$$

I really don't understand how he gets this formula for the derivatives. When I define the $n$-th derivative on the "normal" way I get the following expression:

$$f^{(n)}(z) = \sum_{k=n}^{\infty}(k\cdot \ldots \cdot (k-n+1))\;a_k\;(z-z_0)^{k-n}$$

I just don't understand why this two expressions should be the same or how the author even got his formula(the third equation).


问:
Proof using Squeeze Theorem
Prove that if

then if

then

I know that the squeeze theorem states that there are three real number sequences such that if two of them converge to L then the third converges to L.  In this case -1/n and 1/n are both converging to 0.  Therefore can we say that

 ?

Therefore,

Can someone let me know where I am off?

答:
You can make use of the fact that the sine function is bounded:
\begin{align*}
\left|\frac{\sin(n)}{n}\right| \leq \frac{1}{n}
\end{align*}

Then, if you apply the squeeze theorem, you get that
\begin{align*}
\lim_{n\to\infty}\left|\frac{\sin(n)}{n}\right| \leq \lim_{n\to\infty}\frac{1}{n} = 0 \Rightarrow \lim_{n\to\infty}\frac{\sin(n)}{n} = 0
\end{align*}
where the last relation holds because a sequence $a_{n}$ converges to zero iff $|a_{n}|$ converges to zero.

Hopefully this helps!

Title: Proof using Squeeze Theorem


答:
Rewrite the equation as
$$
2\sqrt{x}=3-\frac{1}{x}
$$
This shows that there is an implied condition, besides $x>0$ necessary in order that the equation makes sense: indeed you need
$$
3-\frac{1}{x}\ge0
$$
which becomes, taking into account that $x>0$, $3x-1\ge0$, so $x\ge1/3$.

Now you can safely square and get
$$
4x=9-\frac{6}{x}+\frac{1}{x^2}
$$
and therefore
$$
4x^3-9x^2+6x-1=0
$$
This factors as $(x-1)^2(4x-1)=0$. Since $1/4<1/3$, this is a spurious solution that must be discarded. Hence the only solution is $x=1$.


问:
Finding the sum of a Taylor expansion
I want to find the following sum:

$$
\sum\limits_{k=0}^\infty (-1)^k \frac{(\ln{4})^k}{k!}
$$

I decided to substitute $x = \ln{4}$:

$$
\sum\limits_{k=0}^\infty (-1)^k \frac{x^k}{k!}
$$

The first thing I noticed is that this looks an awful lot like the series expansion of $e^x$:

$$
e^x = \sum\limits_{k=0}^\infty \frac{x^k}{k!}
$$

The only obstacle is the $(-1)^k$ term. I tried getting rid of it by rewriting:

\begin{align*}
\sum\limits_{k=0}^\infty (-1)^k \frac{x^k}{k!} &= 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \frac{x^5}{5!} + \dots\\
&= (1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots) - (x + \frac{x^3}{3!} + \frac{x^5}{5!} + \dots)\\
&= \sum\limits_{k=0}^\infty \frac{x^{2k}}{(2k)!} - \sum\limits_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}
\end{align*}

These sums look a lot like the series expansions of $\sin(x)$ and $\cos(x)$:

\begin{align*}
\sin(x) &= \sum\limits_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}\\
\cos(x) &= \sum\limits_{k=0}^\infty (-1)^k \frac{x^{2k}}{(2k)!}
\end{align*}

However, these sums do have a $(-1)^k$ term, just when I got rid of it! So now I'm stuck. Can someone help me in the right direction?

Title: Finding the sum of a Taylor expansion

Body: Note that $$e^{-x}=\sum_{k=0}^\infty \frac{(-x)^k}{k!}=\sum_{k=0}^\infty  (-1)^k\frac{x^k}{k!}$$


问:
$n^k\alpha^n \to 0$
For $k \geq 1$ and $0 < \alpha < 1$, I have to prove that $n^k\alpha^n \to 0$ when $n \to \infty$. Intuitively, I think I have to prove that $\alpha^n$ is going faster to $0$ than $n^k$ is going to $\infty$ when $n \to \infty$. I think I have to use the binomial theorem, but it is very unclear right now. Can you give me a hint how to solve this?

Note: I can't use the ratio test here.

答:
What about the good old ratio test?
$$\frac{(n+1)^k\alpha^{n+1}}{n^k\alpha^n}=(1+\frac{1}{n})^k\cdot\alpha\to\alpha<1 $$

After your edit:

What about the good old root test?

$$(n^k\alpha^n)^\frac{1}{n}=\alpha\cdot n^\frac{k}{n}\to\alpha<1$$

Even if you can't use that, see this as the values $f(n)$ for $f(x)=x^k\cdot \alpha^x$ and use L' Hospital rule to compute the limit of $f$ as $x\to\infty$.

Title: $n^k\alpha^n \to 0$


答:
You started fine, but then you got sidetracked:

$$\sum\limits_{k\ge 0}(-1)^k \frac{(\ln{4})^k}{k!}=\sum_{k\ge 0}\frac{(-\ln 4)^k}{k!}=e^{-\ln 4}=\frac1{e^{\ln 4}}=\frac14\;.$$

Title: Finding the sum of a Taylor expansion

Body:
I want to find the following sum:

$$
\sum\limits_{k=0}^\infty (-1)^k \frac{(\ln{4})^k}{k!}
$$

I decided to substitute $x = \ln{4}$:

$$
\sum\limits_{k=0}^\infty (-1)^k \frac{x^k}{k!}
$$

The first thing I noticed is that this looks an awful lot like the series expansion of $e^x$:

$$
e^x = \sum\limits_{k=0}^\infty \frac{x^k}{k!}
$$

The only obstacle is the $(-1)^k$ term. I tried getting rid of it by rewriting:

\begin{align*}
\sum\limits_{k=0}^\infty (-1)^k \frac{x^k}{k!} &= 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \frac{x^5}{5!} + \dots\\
&= (1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots) - (x + \frac{x^3}{3!} + \frac{x^5}{5!} + \dots)\\
&= \sum\limits_{k=0}^\infty \frac{x^{2k}}{(2k)!} - \sum\limits_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}
\end{align*}

These sums look a lot like the series expansions of $\sin(x)$ and $\cos(x)$:

\begin{align*}
\sin(x) &= \sum\limits_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}\\
\cos(x) &= \sum\limits_{k=0}^\infty (-1)^k \frac{x^{2k}}{(2k)!}
\end{align*}

However, these sums do have a $(-1)^k$ term, just when I got rid of it! So now I'm stuck. Can someone help me in the right direction?


问:
Given vectors v = [5, −1], b1 = [1, 1], and b2 = [1, −1] all written in the standard basis, what is v in the basis defined by b1 and b2?

Given vectors v= [5, −1], b1 = [1, 1], and b2 = [1, −1] all written in the standard basis, what is v in the basis defined by b1 and b2?

The solution is [2, 3], but I need to know how.

v in the same basis as b1 and b2 is [5,-1]

but if v has to be defined in terms of b1 and b2, then

vb = 2b1, 3b2 = [2, 3] because this vector v is now being shown with b1
and b2 as new basis vectors.

I understand b1, but get confused how projection b2 gets calculated the way it does.

The solution provided completed projection b2 as follows:

Proj. b2 = $\frac{v.b2}{|b2|^2}$ = $\frac{(5x1)+(-1x-1)}{-1^2 + (-1^2)}$ = $\frac{6}{2}$ = 3

How does b2 [1, -1] become [-1, -1] in the calculation?

And even still, shouldn't that equal -2, not 2? Giving answer -3, not 3?

I seem to calculate  1^2, + (-1^2) = 0, and arrive at 6/0, which obviously isn't right. Where am I going wrong?

答:
Given vectors v = [5, −1], b1 = [1, 1], and b2 = [1, −1] all written in the standard basis, what is v in the basis defined by b1 and b2?

Suppose $v = r\times b_1 + s\times b_2.$

Then, looking first at the 1st component of the vectors $b_1, b_2$ and $v$ and then looking at the 2nd component, you have:

$r(1) + s(1) = 5$ and

$r(1) + s(-1) = -1.$

Two linear equations in two unknowns.  Solve for $r$ and $s$.


答:
Note what happens to $n^k$ and $\alpha^n$ if you double $n$:

$n^k$ gets multiplied by $2^k$

$\alpha^n$ gets squared

Once $n$ is large enough that $\alpha^n<2^{-k}$, the product $n^k\alpha^n$ clearly decreases as you double $n$. And the next time you double $n$ the product decreases even more. And next time, more.

Noting that $n^k$ and $\alpha^n$ are both monotonous, while it doesn't by itself show that the product is monotonous, does let you bound the product and show that it goes to $0$.

Title: $n^k\alpha^n \to 0$

Body:
For $k \geq 1$ and $0 < \alpha < 1$, I have to prove that $n^k\alpha^n \to 0$ when $n \to \infty$. Intuitively, I think I have to prove that $\alpha^n$ is going faster to $0$ than $n^k$ is going to $\infty$ when $n \to \infty$. I think I have to use the binomial theorem, but it is very unclear right now. Can you give me a hint how to solve this?

Note: I can't use the ratio test here.


问:
Proving the Bounded Inverse Theorem
Let $X$, $Y$ be Banach spaces. and let $T: X \rightarrow Y$ be a bijective bounded linear map.

So I'm trying to prove that Bounded Inverse Theorem.

I already showed that for each $y \in Y$, there exists a $x\in X$ such that $Tx=y$ and $\| x\| \leq M\| y\|$, for some constant $M>0$.

Hence, we have $\| T^{-1}y\| = \|x \| \leq M\| y\|$.

Am I done here? If not, how would I finish proving this?

I'm worried that the boundedness of $\| T^{-1}y\|$ is really depending on $y$...

Thank you.

答:
I think the tricky part here is that the Bounded Inverse Theorem is equivalent to other strong results such as the Open Mapping Theorem and the Closed Graph Theorem (see the first paragraph in the Wikipedia page of this theorem https://en.wikipedia.org/wiki/Bounded_inverse_theorem).

So, if you have or want to find an "easy" proof of the BIT then, in some way, you can produce an "easy" proof of the other two theorems which, if I may add, I don't think there is one (for instance, the proof I know of the OMT needs the Baire Category Theorem).

With that said, one thing I can do for you is to show you a way to prove the BIT assuming the OMT, and give you a reference where you can check the proof of the OMT. Recall that a function $f:X\to Y$ between topological spaces is called open if and only if every open subset $U$ of $X$ satisfies that $f[U]$ is an open subset of $Y$.

Open Mapping Theorem: Let $X$ and $Y$ be Banach spaces and the linear operator $T:X\to Y$ be continuous. Then its image $T[X]$ is a closed subspace of $Y$ if and only if the operator $T$ is open.

Fact: A linear operator between normed linear spaces is continuous if and only if is bounded.

Assuming the OMT and the fact, the proof of the BIT is quite easy. From the fact we deduce that, in order to check that $T^{-1}$ is bounded, it is enough to prove that $T^{-1}$ is continuous. Now, observe that $T^{-1}$ is continuous if and only if $T$ is an open map. So, if we take a look at the OMT, the only thing we have to show is that $T[X]$ is a closed subspace of $Y$. Lastly, since $T$ is a surjective function, we have the equality $T[X]=Y$; so, trivially, $T[X]$ is closed in $Y$ and the result follows.

The material discussed in this answer can be found in Royden & Fitzpatrick - Real Analysis, Fourth Edition, Chapter 13. Particularly, check Theorem 1, Theorem 8 and Corollary 9.

Hope this helps.

Title: Proving the Bounded Inverse Theorem


问:
Modify the rules of Gomoku (Five-in-a-row) or Connect Four type games to enforce the fairness among players
One colleague and me were discussing this problem during lunch today, and I did a little bit digging for several hours after returning to my office.

Fact: For an $(m,n,k)$-game, there does not exist a strategy that assures that the second player will win. For example, in Five-in-a-row, first player wins with perfect play.

Now our question is: How to modify the rules to make the game fair for both players?

I googled a bit more, found an MSE question that addresses the fairness of this kind of games.

I think the fairness in that question can be strenghthen by the following definition: A game is fair if and only if that two players play with perfect strategy, the game will always be a draw. This is fair in that this game favors player with less mistakes. For example, Tic-tac-toe is fair.

Several proposals to modify the rules to enforce the fairness:

Impose more restrictions on the player who plays first. For example, Five-in-a-row bans black to play "three and three", "four and four", and "overlines".

Change the $(m,n,k)$-game to an $(m,n,k,p,q)$-game: $k$-in-a-row on an $(m\times n)$-board, first player put $p$ stones on board, in subsequent moves, players put $q$ stones on board. For example, [Connect 6].

Go to higher dimension. For example, Five-in-a-row played in a $19\times 19\times 19$ "board".

I am no expert in combinatorial game theory or computational complexity in board games, but it is always nice to learn some new things in summer time. My question is: Any papers or treatises dealing the fairness of this kind of games? or more specifically, any proof of the fairness using above three modifications of rules? especially, for Five-in-a-row, do those additional rules make the game fair?

Any analysis of fairness on simpler cases like Connect 4 is welcome too.

Lastly just out of curiousity, as an avid Go player myself, I wonder is there any mathematical analysis on addressing the fairness using komi in the Go game? Or Go is just too complex to analyze...

答:
One other option is to modify that game so that there are two stages:

Randomly determine who is the first player

Play the game as usual

This skirts the question.


问:
Sum over (squares of) Laguerre Polynomials
I'm looking for a closed form of the sum \begin{equation}
\sum_{n=0}^\infty \frac{n!}{(n+k)!} (L_n^k(x))^2 t^n,
\end{equation}
where $L_n^k(x)$ are the Laguerre Polynomials. I have been looking for some time and only found equations that are pretty close, but not exactly what I need (like here).

答: Go the wikipedia article for Laguerre polynomials and search for 'Hardy-Hille' formula near the end of it.  Put $\alpha=k$ and $y=x.$


问:
complex and real spectral theorem for matrices
I am studying the spectral theorem for matrices, and the book says that if a $nxn$ matrix A is real and symmetric then its diagonalizable over $\mathbb{R}$. And that this fact is a corollary of the Spectral Theorem for the complex case of normal matrices.

Although I agree that since $A$ is symmetric then $A$ is normal hence it implies that $A$ is diagonalizable over $\mathbb{C}$, and moreover it is easy to prove that all eigenvectors are real. But how can I see that all eigenvectors are also real?

Thanks in advance.

答:
It's mostly an addition to Jacky Chong's answer. Even though he left a comment about why the real eigenvectors constructed by taking $v + \bar v$ are ortgogonal, I really fail to see how taking the real component helps. I mean it's easy to come up with orthogonal complex vectors such that their real components aren't orthogonal: $(1, i)$ and $(1, -i)$, for example. Maybe I'm just missing his point. So here's my proof. Let $v$ and $w$ be eigenvectors with $\lambda$ and $\mu$ as corresponding eigenvalues (here $\lambda \neq \mu$) Then
$$ \lambda \langle v, w \rangle = \langle \lambda v, w \rangle = \langle Av, w\rangle = \langle v, A^T w \rangle = \langle v, Aw \rangle = \langle v, \mu w \rangle = \mu \langle v, w \rangle $$
If $\langle v, w \rangle \neq 0$, then we can divide by it and get $\lambda = \mu$, so it must be zero.


问:
prove $\cos \left(\alpha \right)+\cos \left(\beta \right)+\cos \left(\gamma \right)$ equals.... if $\alpha +\beta +\gamma =\pi $
I am looking at the solution of this problem:
 $$\cos \left(\alpha \right)+\cos \left(\beta \right)+\cos \left(\gamma \right)=4\sin \left(\frac{\alpha }{2}\right)\sin \left(\frac{\beta }{2}\right)\sin \left(\frac{\gamma }{2}\right)$$ and it says this in the solution as the first step: $$\cos \left(\alpha \right)+\cos \left(\beta \right)+\cos \:\left(\gamma \:\right)=2\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)-2\cos \left(\frac{\alpha \:+\beta \:}{2}\right)^2+1$$

I understand that the $\cos \alpha +\cos \beta=2\cos \left({\frac {\alpha +\beta }{2}}\right)\cos \left({\frac {\alpha -\beta }{2}}\right)$ is used, however how is the $cos \:\left(\gamma \:\right)$ transformed? I assume I should use the $\alpha +\beta +\gamma =\pi $ to replace $\gamma$ but can't seem to figure it out?
Any help is very much appreciated. x

答:
We have
$$ \cos \gamma = \cos(\pi-(\alpha+\beta)) = -\cos(\alpha+\beta) = -2 \cos^2\frac{\alpha+\beta}{2} + 1$$
Two formulas are used here:
$$ \cos (\pi-x) = -\cos x $$
and
$$ \cos (2x) = 2 \cos^2 x -1$$
Later you'll also need
$$ \cos(\frac{\pi}{2} -x) = \sin x $$


问:
Determinant Identity: Elegant Solution Please.
Can anyone provide a proof of this identity:
$$\begin{vmatrix}
(b+c)^2 &a^2&a^2\\
b^2&(a+c)^2&b^2\\
c^2&c^2&(a+b)^2\\
\end{vmatrix}=2abc(a+b+c)^3 ?$$

Ideally I would like a series of row column operations resulting in the formula.
No brute force calculations. Note that letting $a=0$ we get two rows equal. So $abc$ is a factor. But I have unable to produce this by a simple series of operations, I mean to get one row or column all of whose entries are divisible by $a$. Getting divisibility by $(a+b+c)^2$ is also not hard just subtract one column from the other two. After that however I have been unable to make any further progress. Hopefully someone here is smarter than I.

答:
I appreciate the idea of computing the determinant by using its polynomial properties, and symmetry, thus not computing it really. I will try to go in this direction. Let $F(a,b,c)$ be the function given by the determinant from the OP. We know it is a symmetric polynomial of degree $2+2+2=6$ in $a,b,c$.
(And want to show it is $2abc(a+b+c)^3$.)

Let $x$ be some new transcendental variable, and let us compute (generically)
$$
\begin{aligned}
&F(x-b-c,b,c)
\\
&=
\begin{vmatrix}
(b+c)^2 &(b+c-x)^2&(b+c-x)^2\\
b^2&(x-b)^2&b^2\\
c^2&c^2&(x-c)^2\\
\end{vmatrix}
=
\begin{vmatrix}
2bc & 2(b-x)c&2(c-x)b\\
b^2&(b-x)^2&b^2\\
c^2&c^2&(c-x)^2\\
\end{vmatrix}
\\
&=
2b^3c^3
\begin{vmatrix}
1 & (1-x/b)&(1-x/c)\\
1&(1-x/b)^2&1\\
1&1&(1-x/c)^2\\
\end{vmatrix}
\\
&=
2b^3c^3
\begin{vmatrix}
1 & -x/b&-x/c\\
1&-x/b(2-x/b)&0\\
1&0&-x/c(2-x/c)\\
\end{vmatrix}
\\
&=
2b^2c^2x^2
\begin{vmatrix}
1 & 1& 1\\
1& 2-x/b&0\\
1&0& 2-x/c\\
\end{vmatrix}\ .
\end{aligned}
$$
This is a point where i would like to pause.
Now we have some possibilities.

The first one is further calculating, well, not in the spirit of the OP, but easy to bring to a good end the last determinant. We subtract the first line from the other two. We immediately get $\begin{vmatrix}1-x/b & -1\\-1&1-x/c\end{vmatrix}=(1-x/b)(1-x/c)-1$ and so on.

But i would like to stop here the calculus, and observe that making $x=0$ in the last determinant we get $\begin{vmatrix}1 & 1& 1\\1 & 2&0\\1&0&2\end{vmatrix}=0$.
(The sum of the last two rows depends linearly on the first row.) So we can finally isolate the factor $x^3$. So we know that $F(a,b,c)$ of homogeneous degree $6$ has the factors $a,b,c$, and $(a+b+c)^3$. So we need only to determine the constant involved, e.g. from $F(-1,1,1)
=\begin{vmatrix}4 & 1& 1\\1 & 0&1\\1&1&0\end{vmatrix}
=\begin{vmatrix}2 & 0& 0\\1 & 0&1\\1&1&0\end{vmatrix}=-2$.

$\square$


Title: prove $\cos \left(\alpha \right)+\cos \left(\beta \right)+\cos \left(\gamma \right)$ equals.... if $\alpha +\beta +\gamma =\pi $

Body:
I am looking at the solution of this problem:
 $$\cos \left(\alpha \right)+\cos \left(\beta \right)+\cos \left(\gamma \right)=4\sin \left(\frac{\alpha }{2}\right)\sin \left(\frac{\beta }{2}\right)\sin \left(\frac{\gamma }{2}\right)$$ and it says this in the solution as the first step: $$\cos \left(\alpha \right)+\cos \left(\beta \right)+\cos \:\left(\gamma \:\right)=2\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)-2\cos \left(\frac{\alpha \:+\beta \:}{2}\right)^2+1$$

I understand that the $\cos \alpha +\cos \beta=2\cos \left({\frac {\alpha +\beta }{2}}\right)\cos \left({\frac {\alpha -\beta }{2}}\right)$ is used, however how is the $cos \:\left(\gamma \:\right)$ transformed? I assume I should use the $\alpha +\beta +\gamma =\pi $ to replace $\gamma$ but can't seem to figure it out?
Any help is very much appreciated. x


问:
Essentially self adjoint operators: a verification procedure
$\langle \ .,. \rangle : \scr H \! \times \! \scr H \rightarrow \mathbb C$ is the inner product.

Let $T : {\cal D}(T) \rightarrow \scr H $ be a linear symmetric
operator, so:

${\cal D} (T)$ is dense in $\scr H$

$\langle \phi, T\psi \rangle = \langle T\phi, \psi \rangle \quad \forall \phi, \psi \in {\cal D}(T)$

The following properties are equivalent:

$T$ is essentially self-adjoint: $\ T^* = \overline T $

${\rm Ker} (T^* \pm \lambda  I ) \equiv \{ 0\} $

If I have understood correctly, point 2. of the theorem is telling us that $(T^* \pm \lambda I)\psi = 0 \Longleftrightarrow \psi =0 $, or, to put it in other terms, the only eigenvector satisfying $$\ T^* \psi = \mp \lambda \psi $$ is the null vector.

My professor: "It's therefore evident that if we want to check if any symmetric operator $T$ admits a unique self-adjoint extension - its closure - we just have to make sure $\pm i$ are not eigenvalues of $T^*$"

Is this really enough? And what in the world has it to do with the kernel triviality?

答: Your $\ \lambda\ $ here should be $\ i\ $.  That $\ T\ $ is essentially self-adjoint if and only if both $\ T^*+iI\ $ and $\ T^*-iI\ $ have trivial kernels is a well-known result of functional analysis. See Corollary $9.22$ in Hall's Quantum Theory for Mathematicians, for instance.


问:
Linear independence and pairwise angles
Given $p+1$ vectors with pairwise angles $> \pi/2$ in $\mathbb{R}^n$ then every $p$ of them are linearly independent. How to prove this fact?

The motivation for this question is the idea of extending the notion of "linear independence" for points on tangent cones on some metric spaces (Alexandrov spaces), on which we do not have a vector space structure, but have a good notion of angle.

答:
The same idea as Dmitry K's answer yields the following non-inductive argument. Suppose you had a linear dependence among $p$ of the given $p+1$ vectors. As in Dmitry's answer, transpose all the terms with negative coefficients, so the dependence relation looks like
$\sum_{v\in B}\beta_v v=\sum_{w\in C}\gamma_w w$, where $B$ and $C$ are disjoint subsets of your set of $p$ dependent vectors, and where all the $\beta$'s and $\gamma$'s are positive.  Write $x$ for the common value of $\sum_{v\in B}\beta_v v$ and $\sum_{w\in C}\gamma_w w$.  On the one hand, the inner product $x\cdot x$ of a vector with itself is non-negative.  On the other hand, this inner product is $\sum_{v\in B, w\in C}\beta_v\gamma_w(v\cdot w)$, in which every term is strictly negative, because the $\beta$'s and $\gamma$'s are positive while the inner products $(v\cdot w)$ are negative. The only way for such a sum to be non-negative is for it to be the empty sum.  That is, $B$ or $C$ must be empty, and our original linear dependence relation had coefficients of only one sign; without loss of generality, suppose that sign is positive, so $C$ is empty and $\sum_{v\in B}\beta_v v=0$.  Now use the assumption that there is a $(p+1)$-st vector $y$ whose inner product with all $v\in B$ is negative.  Then since $\sum_{v\in B}\beta_v v=0$, we have $0=\sum_{v\in B}\beta_v (y\cdot v)$.  But every term in this sum is strictly negative, so $B$ has to be empty too, and our original alleged linear dependence relation has disappeared.</p>

Title: Linear independence and pairwise angles


问:
Approximating bounded functions $f \in B([0,1], \mathbb{R})$ by simple functions (not necessarily measurable)
If we have any arbitrary bounded function $f: [0,1] \to \mathbb{R}$, is it possible to approximate this by a sequence of simple functions, i.e. functions of the form $\sum_{j=1}^{n} a_j \chi_{A_j}$ (where $\chi_{A}$ denotes the characteristic function on $A$)?

$f$ may not be measurable, all we are given is that it is in $B([0,1], \mathbb{R})$, which we are considering as a normed vector space with the sup-norm.

答:
Since $f$ is bounded, set $K=\overline{f([0,1])}$. Then $K$ is a compact set. Let $E=\{r_n\}$ be a countable, dense subset of $K$ and for $\varepsilon>0$ consider the open cover $\mathcal{U}_\varepsilon=\{U_{n,\varepsilon}\}_{n=1}^\infty$, where $U_{n,\varepsilon}=(r_n-\varepsilon,r_n+\varepsilon)$. Now for any $\varepsilon>0$ this cover reduces to a finite subcover, so there exists an integer $M_\varepsilon$ and a function $q:\{1,\dots, M_\varepsilon\}\to\mathbb{N}$ (just a fancy way of picking $M_\varepsilon$-many indexes) so that
$$K\subset\bigcup_{i=1}^{M_\varepsilon}U_{q(i),\varepsilon}$$

For $n\geq1$ set $M_n=M_{\frac{1}{n}}$ and $V_{i,n}=f^{-1}(U_{q(i),\frac{1}{n}})$ for all $i=1,\dots, M_n$. Define
$$f_n(x)=\sum_{i=1}^{M_n}r_{q(i)}\cdot\chi_{V_{i,n}}$$
for $x\in[0,1]$. These are simple functions. Now let $x\in[0,1]$. Then $f(x)\in K$ so for each $n\geq1$ we have that $f(x)$ belongs to some $U_{q(i),\frac{1}{n}}$, so $f_n(x)=r_{q_i}$ while $f(x)\in U_{q(i),\frac{1}{n}}$, so $|f_n(x)-f(x)|<\frac{1}{n}$.

This shows that $f_n\to f$ uniformly, where $f_n$ are simple functions. Of course the sets can be extremely crazy, but you put no such restriction so there is no problem with that.


问:
How to solve $\int_0^\infty f(x) \sin(wx) \, dx=\frac a {a^2+w^2}$ with using Fourier transform?
$\int_0^\infty f(x) \sin(wx) \, dx=\frac{a}{a^2+w^2}$
I thought it is Fourier sine transform of $f(x)$

since  $\ F_s(f(x))=F_s(k)=\sqrt{\frac{2}{\pi}} \int_0^\infty f(x)\sin(wx) \, dx$

so $\sqrt{\frac{\pi}{2}} F_s(w)=\frac{a}{a^2+w^2}$. by taking invers sin transform of this

$$\sqrt{\frac{\pi}{2}}f(x)=\int_0^\infty \frac{a}{a^2+w^2} \sin(wx)\,dx$$

答:
This will be clear to those who have dealt with Laplace transforms. The Laplace transform of $\sin wt$ is given by ${w \over w^2 + s^2}$. In other words
$$\int_0^{\infty} e^{-st} \sin wt \,dt = {w \over w^2 + s^2}$$
Thus in your notation, the answer to your question is ${a \over w} e^{-ax}$.


问:
Difficulty with question about countable product of copies of $\mathbb{N}$
I had a lot of trouble trying to figure this problem out. Can someone explain how to properly do this?

Problem:Assume that $\mathbb{N}$ has the usual order. Let $\mathbb{N}^{\omega}$ denote the Cartesian product of a countable number of copies of the space $\mathbb{N}$. It can be endowed with the dictionary order in a natural way. Show that $\mathbb{N}^{\omega}$ with the dictionary order topology is uncountable, is not well ordered, and any set that does not have a least element does have a limit point.

Edit:I realized I showed a misunderstanding for the first part of the problem since for some reason I assumed each $s_{i,j} \in \{1,...,9\}$ instead of $\mathbb{N}$.

Attmempt:Suppose that $\{S_1,S_2,...\}$ is an enumeration of the elements of $\mathbb{N}^{\omega}$.Where $f:\mathbb{N} \rightarrow \mathbb{N}^{\omega}$ given by $f(n)=S_n$, is bijective.We will produce an element in $\mathbb{N}^{\omega}$ that is not on this list. Then there is no surjection from $\mathbb{N}$ onto $\mathbb{N}^{\omega}$, and our assumption that $f$ is bijective is false.
A list of elements in the enumeration is given by:
$$S_1=(s_{1,1},s_{1,2},...)$$
$$S_2=(s_{2,1},s_{2,2},...)$$
$$\vdots$$
$$S_n=(s_{n,1},s_{n,2},...)$$
$$\vdots$$
Where $s_{i,j} \in \mathbb{N}$ for all $i,j \in \mathbb{N}$.
Consider the element $A=(a_1,a_2,...) \in \mathbb{N}^{\omega}$ such that $a_k=7$ if $s_{k,k} \in \{1,2,3,4,5,6\}$, and $a_k=3$ if $s_{k,k} \in \{7,8,9\}$. $f$ is bijective so $A \in \mathbb{N}^{\omega}$ must appear on the list. However, the $k$'th component of $a_k$ of $A$ and $s_{k,k}$ of $S_k$ differ by between $4$ and $7$, so $A$ does not occur on this list and $\mathbb{N}^{\omega}$ is uncountable.
We will construct a set of elements that has no least element.  Let $S=\{S_1,S_2,...\}$ be the infinite set with elements given by:
$$S_1=(1,2,3,4,5,6,7,...)$$
$$S_2=(1,1,2,3,4,5,6,...)$$
$$S_3=(1,1,1,2,3,4,5,...)$$
$$S_4=(1,1,1,1,2,3,4,...)$$
$$S_5=(1,1,1,1,1,2,3,...)$$
$$\vdots$$
Then the succesor of every element in the set is less than it's predecessor in the lexicographical ordering. Since there is no smallest element in this nonempty subset of $\mathbb{N}^{\omega}$,$\mathbb{N}^{\omega}$ is not well ordered. Let $S$ be any subset of elements in $\mathbb{N}^{\omega}$ that does not have a limit point.

Note for the part below $(\downarrow$).I ran out of time, and had to turn in but knew it was wrong. I felt like it was wrong, because for a decreasing sequence of elements (in the dictionary order) must have reach a point where all the components are identical, up to a point where, the the components become eventually constant, to a point. However I had difficulty expressing this mathematically.

Any subset $S$ of $\mathbb{N}^{\omega}$ without a minimum contains a sequence of elements of the form $(a_1,a_2,a_3,...)$, where $a_i=a_j$ for $i \neq j$ holds for all terms after some index $t$ for all indices $n$ with $1\leq t \leq n<k$, and $a_k>a_n$. Consider the element in $\mathbb{N}^{\omega}$ having each entry the element $a_n \in \mathbb{N}$ after $a_t$ . Then any open set in $\mathbb{N}^{\omega}$ containing this element, contains elements in $S$. Since $S$ has no minimum there is an element $(a_1,a_2,...) \in S$ where $a_{t+1}=\dots=a_n$ for each natural number $n$, so that any open interval containing the element with all entries $a_n$ must contain an element of this form.

答:
Your diagonal argument for the first part is correct, though you don’t have to assume that $f$ is a bijection: you have in fact demonstrated that no function from $\Bbb N$ to $\Bbb N^\omega$ is a surjection, so in particular none of them is a bijection. In other words, there’s no need to make it a proof by contradiction.

Your construction of a set with no least element is also fine, though you could improve it slightly be giving an explicit definition of the sequences $S_n$: for $n,k\in\Bbb N$ let

$$s_{n,k}=\begin{cases}
1,&\text{if }k\le n\\
k-n+1,&\text{if }k>n\,,
\end{cases}$$

and let $S_n=\langle s_{n,k}:k\in\Bbb N\rangle$. You can even go on to argue that if $m<n$, then $s_{m,k}=1=s_{n,k}$ for $k\le m$, and $s_{m,m+1}=2>1=s_{n,m+1}$, so $S_n\prec S_m$, where $\prec$ is the lexicographic order on $\Bbb N^\omega$.

For the last part, suppose that $\mathscr{S}\subseteq\Bbb N^\omega$ has no least element. Let $\mathscr{S}_0=\mathscr{S}$. Given $\mathscr{S}_n\subseteq\mathscr{S}$ for some $n\ge 0$, let $a_{n+1}=\min\{\pi_{n+1}(S):S\in\mathscr{S}_n\}$, and let

$$\mathscr{S}_{n+1}=\{S\in\mathscr{S}_n:\pi_{n+1}(S)=a_{n+1}\}\,.$$

In this fashion we recursively construct a sequence $T=\langle a_n:n\in\Bbb N\rangle$; I claim that $T=\inf\mathscr{S}$ and hence that $T$ is a limit point of $\mathscr{S}$.

It’s straightforward to check that $T$ is a lower bound for $\mathscr{S}$, and I’ll leave that to you. It’s not much harder to check that if $T\prec T'$ for some $T'\in\Bbb N^\omega$, then there is an $S\in\mathscr{S}$ such that $S\prec T'$. Specifically, if $T\prec T'=\langle a_n':n\in\Bbb N\rangle$, let $m$ be minimal such that $a_m<a_m'$. Pick any $S\in\mathscr{S}_m$: $\pi_m(S)=a_m<a_m'$, and $\pi_k(S)=a_k=a_k'$ for $k<m$, so $S\prec T'$, as desired. Thus, $T'$ is not a lower bound for $\mathscr{S}$, and $T$ is therefore the greatest lower bound of $\mathscr{S}$. And in the process of showing this, we’ve also shown that $T$ is a limit point of $\mathscr{S}$: every open interval around $T$ contains an open interval $(T,T')_\prec$ for some $T'\succ T$, and we just showed that $(T,T')_\prec\cap\mathscr{S}\ne\varnothing$.


问:
Why is a cofinal set called "cofinal?"
Let $(A, \leq)$ be a poset. A set $B \subseteq A$ is said to be cofinal in $A$ if every $a \in A$ has a $b\in B$ with $a \leq b$.

Why is a cofinal set called "cofinal?" Is there a related notion of a "final" set? The "co" doesn't seem to be related to the word "complement," as the complement of a cofinal set can also be cofinal; such sets don't seem to be "final" in any natural sense of the word I can imagine.

答:
I'm not sure who came up with the term. For me it suggest that the set is "coming with you to the end", co- as in compansion, co-traveller; -final for end, the larger elements are towards the end in an order, intuitively.
So whatever direction you go towards larger elements, there will always be elements from this cofinal set too..

But one should find the first usage of the term and maybe it's explained there? Not sure if Cantor used e.g.

Title: Why is a cofinal set called "cofinal?"


问:
$\cot 30^\circ = \cot 20^\circ - \operatorname{cosec} 80^\circ$ (?)
Prove that,
$\cot 30^\circ = \cot 20^\circ - \operatorname{cosec} 80^\circ$

I could not do that by trying heart and soul. Please solve it.

答:
We are required to prove that (TPT)
$$\frac{\cos 30^{\circ}}{\sin 30^{\circ}} = \frac{\cos 20^{\circ}}{\sin 20^{\circ}} - \frac{1}{\cos 10^{\circ}}$$

i.e., TPT

$$\sin 20^{\circ} \cos 10^{\circ} \cos 30^{\circ} = \sin 30^{\circ} \cos 10^{\circ} \cos 20^{\circ} - \sin 30^{\circ} \sin 20^{\circ}$$

i.e., TPT

$$\sin 30^{\circ} \sin 20^{\circ} = \cos 10^{\circ}(\sin 30^{\circ}\cos 20^{\circ} - \sin 20^{\circ} \cos 30^{\circ})$$

i.e., TPT

$$\sin 30^{\circ} \sin 20^{\circ} = \cos 10^{\circ}\sin 10^{\circ}$$

which is always true since $$\sin 30^{\circ} \sin 20^{\circ} = \frac{1}{2} (2 \sin 10^{\circ} \cos 10^{\circ})$$

Hence the proof.


问:
Factoring quadratics with certain property
I am looking into which quadratics factor with the following property:
$x^2±ax±b$ factors e.g. all the following factor for $a=5, b=6$

$x^2+5x+6=(x+2)(x+3)$

$x^2-5x+6=(x-2)(x-3)$

$x^2+5x-6=(x+6)(x-1)$

$x^2-5x-6=(x-6)(x+1)$

I have found $x^2+10x+24, x^2+20x+96, ..., x^2+5 \cdot 2^{n-1}x+6\cdot 4^{n-1}$ all have this property. I am looking for others or is this the only family of MONIC quadratics that have this property?

答: You're looking for $a^2\pm 4b$ to be perfect squares, say$$a^2\pm 4b=(c\pm d)^2=c^2+d^2\pm 2cd\iff a^2=c^2+d^2,\,b=cd/2.$$So this reduces to Pythagorean triples viz.$$c=2klm,\,d=(k^2-l^2)m,\,a=(k^2+l^2)m,\,b=klm^2(k^2-l^2).$$In your example $k=2,\,l=m=1$.


问:
Tangent vector of geodesic on sphere trough points $(a,b,c)$ and $(d,e,f)$
I am seeking the unit vector $v=(x,y,z)$ tangent at $p$ to the geodesic on sphere connecting $p=(a,b,c)$ and $q=(d,e,f)$. The curve is orientated from $p$ to $q$.

Once the tangent vector is ortogonal to the point vectors, so I must have

$$\begin{cases}ax+by+cz=0\\
x^2+y^2+z^2=1\end{cases}$$

I need more two equations to get the vector, could you help me?

The aim is to prove that the application $(p,q)\mapsto v$ is continuous.

Thank you in advance.

答:
Another way to get $v$ is to first consider the vector $w: = q-p$, which points in the right direction, but it's not tangent. One can check that its projection onto the tangent space $T_p \mathbb{S}^2$
$$ v := w - (w \cdot p) p $$
is the tangent vector we want.

The outcome is the same if we just consider the projection of $q$ onto $T_p \mathbb{S}^2$: $v = q - (q \cdot p) p$.


问:
Prove that $\exists a,g,h\in G\colon h=aga^{-1}, g\neq h ,gh=hg$ in a finite non-abelian group $G$.
Let $G$ be a finite and non-abelian group.  How do I prove the following statement? $$\exists a,g,h\in G \colon\quad h=aga^{-1},\ g\neq h ,\ gh=hg.$$

Thanks in advance.

答:
Using the induction by $|G|$, we can suppose that all subgroups of $G$ are Abelian, i.e. $G$ is a Miller–Moreno group.  The structure of these groups are well-known (see What can we say of a group all of whose proper subgroups are abelian?). They are of 2 kinds:

1) $|G|=p^m$. Let $H<G, |H|=p^{m-1}$. Since $H$ is invariant and is not in the centre, it contains conjugate elements.

2) $|G|=p^mq^n$. Then, e.g., $P$ with $|P|=p^m$ is invariant of the type $(1,1,\ldots, 1)$ and $Q$ with $|Q|=q^n$ is cyclic, so $P$ is not in the centre and the proof is similar to above one.


答:
The term cofinal is a compound of the prefix co- and the adjective final. Here the meaning of the prefix is ‘together; mutually; jointly’: if $B$ is cofinal in $A$, the two sets reach the ‘end’ together, so to speak. The prefix is from Latin co-, a variant of con-, and indeed one occasionally sees confinal in place of cofinal. The Latin prefix con- in turn is derived from the preposition cum ‘with, along with’.

The OED does not yet have entries for either cofinal or confinal, so I can’t say when either came into use in the mathematical sense.

Title: Why is a cofinal set called "cofinal?"

Body:
Let $(A, \leq)$ be a poset. A set $B \subseteq A$ is said to be cofinal in $A$ if every $a \in A$ has a $b\in B$ with $a \leq b$.

Why is a cofinal set called "cofinal?" Is there a related notion of a "final" set? The "co" doesn't seem to be related to the word "complement," as the complement of a cofinal set can also be cofinal; such sets don't seem to be "final" in any natural sense of the word I can imagine.


问:
$F[x_1,...,x_n]$, $F(x_1,...,x_n)$, $F(\alpha_1, ..., \alpha_n)$ and $F[\alpha_1, ..., \alpha_n]$
I am having a hard time understanding the difference between $F[x_1,...,x_n]$ and $F[\alpha_1, ..., \alpha_n]$ (let $F$ be a subfield of a field extension $K$). If I understand this correctly, the first is the field $F$ with $n$ variables while the latter is the same field evaluated at $\alpha_1, ..., \alpha_n$.

Then $F(\alpha_1, ..., \alpha_n)$ is also defined as:

$$F(\alpha_1, ..., \alpha_n) = \bigg\{ \frac{\alpha}{\beta} | \alpha,\beta \in F[\alpha_1, ..., \alpha_n], \beta \neq0 \bigg\} $$

Then there is a theorem that states $F(\alpha_1, ..., \alpha_n)$ is the smallest subfield of the field $K$ that contains $F$ and $\alpha_1, ..., \alpha_n$. Why $F[\alpha_1, ..., \alpha_n]$ is not the smallest subfield with those properties?

In genenal, what are the difference between $F[x_1,...,x_n]$, $F(x_1,...,x_n)$, $F[\alpha_1, ..., \alpha_n]$ and $F(\alpha_1, ..., \alpha_n)$? Some concrete examples that shows the differences would be much appreciated. This question helped a bit, but here I am asking for a comparison between all these four fields.

答:
If F is a field, $F[x_1,...,x_n]$ is the ring of polynomials over F (with coefficients in F that is) in the n variables $x_1,...,x_n$, while $F(x_1,...,x_n)$ is defined just like the definition you gave for $F(\alpha_1,...,\alpha_n)$ but with $F[x_1,...,x_n]$ instead of $F[\alpha_1,...,\alpha_n]$. Since $F[\alpha_1,...,\alpha_n]$ is just the ring of polynomials over $\alpha_1,...,\alpha_n$, notice how $F(\alpha_1,...,\alpha_n)$ is just the smallest field containging F and $\alpha_1,...,\alpha_n$ (it is the closure under multiplicative inverses for nonzero elements), while $F[\alpha_1,...,\alpha_n]$ is the smallest ring containing F and $\alpha_1,...,\alpha_n$ (it need not be a field because it might not be closed under multiplicative inverses, e.g. $\frac{1}{x}\notin \mathbb{R}[x]$). But also, sometimes $F[\alpha_1,...,\alpha_n]$ coincides with $F(\alpha_1,...,\alpha_n)$. For instance, consider the ring $\mathbb{Q}(i)$, it can be shown that this is actually the same as $\mathbb{Q}[i]$ (you can take an arbitrary fraction in $\mathbb{Q}(i)$ and multiply by the conjugate of the denominator to show you really just have an element of $\mathbb{Q}[i]$). In particular:

$\mathbb{Q}(i) = \{x+yi\mid x,y\in \mathbb{Q}\}$

replace the above equality with $\sqrt{2}$ and it is still true. Note that always:

$F[x]\subseteq F(x)$,

so $F[x] = F(x)$ iff $F[x]$ is a field (same with $F(x_1,...,x_n)$ and $F[x_1,...,x_n]$).

The other glaring difference between $F[\alpha_1,...,\alpha_n]$ and $F[x_1,...,x_n]$ is that the latter is a polynomial ring in some n variables, and the former is over these elements $\alpha_i$ which are presumed to be in an extension field of F (and not formal indeterminates) (e.g. they could be roots of polynomials over F). Also, the latter (polynomial ring in n variables) will always be a transcendental extension (since the variables won't be roots of nonzero polynomials) while the former might be algebraic or even just F possibly (e.g. $\mathbb{Q}[\sqrt{2}]$ is algebraic over $\mathbb{Q}$, $\mathbb{Q}[7] = \mathbb{Q}$, while $\mathbb{Q}(\pi)$ is transcendental over $\mathbb{Q}$, but $F(x,y)$ is transcendental over F and even over $F(y)$, which is itself transcendental over F).


问:
integration wrt to a function
I was checking some papers and I learned that the following expression is valid.

$\int_{0}^{\infty} f(x)d(g(x)) = \int_{0}^{\infty} f(x) g'(x) dx $.

It seems to help solving integrals when complicated terms involved.

But I am not sure whether the following is correct.

$\int_{0}^{\infty} f(x)d(g(x)) = - \int_{0}^{\infty} g(x) d(f(x)) $

Can anyone enlighten me about it? Thank you.

答: No, it is not correct. Take $f(x)=1$ for all $x$ and $g(x)=e^{-x}$ for a counter-example.

Title: integration wrt to a function


问:
What is the higher dimensional analogy of $x^3, x^4$, etc.?
We know that a paraboloid, $x_1^2 + x_2^2 = x^T x$ is the generalization of a quadratic $x^2$.

Is there some function similar to $x^Tx$ that can be used to represent cubic, quartic, quintic...polynomials?

I was thinking of $x x^Tx$ for $x^3$ but obviously that doesn't work. Or perhaps $x_1^3 + x_2^3$ is the natural generalization that you simply cannot be put into a dot product between two vectors.

答:
One possible generalization is $x$, $x^{\otimes 2}$, $x^{\otimes 3}$, ... where we take the $n$- fold tensor power. This gives rise to the general version of Taylor's theorem for functions $\mathbb R^{n}\to \mathbb R^{m}$:

$$\begin{aligned}f(x)
&=f(x_{0})+(Df(x_{0}))\cdot (x-x_{0})+{}{\tfrac {1}{2}}(D^{2}f(x_{0}))\cdot (x-x_{0})^{\otimes 2}+\cdots
\\ &\qquad \cdots +{\tfrac {1}{k!}}D^{k}f(x_{0})\cdot (x-x_{0})^{\otimes k}+{\tfrac {1}{(k+1)!}}R_{k+1}(x)\cdot (x-x_{0})^{\otimes (k+1)}
\end{aligned}$$

Or in other words the Taylor series is $T_{f,x_0}(x+x_0)=\sum_{k\ge 0}\tfrac{1}{k!}D^kf(x_0)\cdot x^{\otimes k}$.  Note that here, in the term $D^{k}f(x_{0})\cdot (x-x_{0})^{\otimes k}$, the "$\cdot$"  denotes the appropriate tensor contraction. For example, a case you are probably familiar with is the Taylor polynomial of degree $2$ for a scalar, multivariate function $\mathbb R^{n}\to\mathbb R$, which can be expressed in terms of the Jacobian and Hessian:

$$ f(x+ x_0)\approx f(x_0) + Jf(x_0)\cdot x + \tfrac{1}{2}x^THf(x_0)x $$

here, note that the Hessian term can be equivalently be expressed in many ways, including the Frobenius inner product on $\mathbb R^{n\times n}$, which is equivalent to the induced inner product on the tensor product of Hilbert spaces $\mathbb R^{n}\otimes\mathbb R^{n}$ or, finally, simply the corresponding tensor contraction

$$\begin{aligned}
\tfrac{1}{2}x^THf(x_0)x
= \tfrac{1}{2}\operatorname{tr}(x^T Hf(x_0)x)
&= \tfrac{1}{2}\operatorname{tr}(Hf(x_0) xx^T)
\\&= \tfrac{1}{2}\langle Hf(x_0), xx^T\rangle_F
\\&= \tfrac{1}{2}\langle Hf(x_0), xx^T\rangle_{\mathbb R^{n}\otimes\mathbb R^n}
\\&= \tfrac{1}{2} Hf(x_0)\cdot x^{\otimes 2}
\end{aligned}$$

Example: $f(x)= x_1^3 + x_2^3$

Since this is a polynomial function of degree 3, it should be equal to its own Taylor expansion of degree 3. Taking $x_0=0$ we have:

$D^0 f$ is the rank-0 tensor $\Big(f\Big) = x_1^3 + x_2^3$
$\leadsto D^0 f(x_0) =0$

$D^1 f$ is the rank-1 tensor $\Big(\frac{\partial f}{\partial x_i}\Big)_i =    \begin{bmatrix}3x_1^2 & 3 x_2^2 \end{bmatrix}$ $\leadsto D^1 f(x_0) =
\begin{bmatrix} 0 & 0 \end{bmatrix}$

$D^2 f$ is the rank-2 tensor $\Big(\frac{\partial f}{\partial x_i x_j}\Big)_{ij} =    \begin{bmatrix}6 x_1 & 0  \\  0 & 6 x_2 \end{bmatrix}$ $\leadsto D^2 f(x_0) = \begin{bmatrix} 0 & 0 \\ 0 & 0\end{bmatrix}$

$D^3 f$ is the rank-3 tensor $\Big(\frac{\partial f}{\partial x_i x_j x_k}\Big)_{ijk} =    \begin{bmatrix}\begin{bmatrix}6 & 0  \\  0 & 0 \end{bmatrix}, \begin{bmatrix}0 & 0  \\  0 & 6 \end{bmatrix}\end{bmatrix}$ $= D^3 f(x_0)$

And on the other hand:

$x^{\otimes 0}$ is the rank-0 tensor $(\text{empty product}) = 1$

$x^{\otimes 1}$ is the rank-1 tensor $\Big(x_i\Big)_i =    \begin{bmatrix} x_1 & x_2\end{bmatrix}$

$x^{\otimes 2}$ is the rank-2 tensor $\Big(x_i x_j\Big)_{ij} =    \begin{bmatrix} x_1^2 & x_1x_2 \\ x_2 x_1 & x_2^2\end{bmatrix}$

$x^{\otimes 3}$ is the rank-3 tensor $\Big(x_i x_j x_k\Big)_{ijk} =    \begin{bmatrix}\begin{bmatrix} x_1  x_1 x_1& x_1 x_1 x_2  \\  x_1 x_2 x_1 & x_1 x_2 x_2\end{bmatrix}, \begin{bmatrix} x_2 x_1 x_1 &  x_2 x_1 x_2  \\  x_2 x_2 x_1 & x_2 x_2 x_2 \end{bmatrix}\end{bmatrix}$, which simplifies, thanks to commutativity, to
$x^{\otimes 3} = \begin{bmatrix}\begin{bmatrix} x_1^3 & x_1^2 x_2  \\  x_1^2 x_2 & x_1x_2^2\end{bmatrix}, \begin{bmatrix} x_1^2 x_2 &  x_1x_2^2 \\  x_1x_2^2& x_2^3 \end{bmatrix}\end{bmatrix}$

And the tensor contraction in each case, since $f$ is a scalar function, is simply the total sum over the multi-index

$$D^k f(x_0) \cdot x^{\otimes k} = \sum_{\substack{\mathbf{i}=(i_1,\ldots, i_k) \\ i_j \in\{1, \ldots, n\}}} [D^k f(x_0)]_{\mathbf{i}} \cdot  [x^{\otimes k}]_{\mathbf{i}}$$

So

$$ \tfrac{1}{3!}D^3 f(0) \cdot x^{\otimes 3} = \tfrac{1}{6}\sum_{ijk} \Big(\frac{\partial f}{\partial x_i x_j x_k}\Big)_{ijk} x_ix_jx_k = \tfrac{1}{6} (6x_1^3 + 6x_2^3) = x_1^3 + x_2^3$$

Remark:

The symmetric tensors that Vercassivelaunos talked about you can think of in the rank $2$ case: every rank-2 tensor ($\leftrightsquigarrow$ matrix) of the form

$$
T = \sum_{i} \Big( (u_i \otimes v_i) + (v_i \otimes u_i)\Big)
\leftrightsquigarrow \sum_{i} \Big( u_iv_i^T + v_i u_i^T\Big)
$$

I.e. the symmetric matrices you are familiar with. And in the general rank-$n$ case everything that can be written as a linear combination of symmetric pure tensors, i.e. $\displaystyle  T=\sum_{i}\sum_{\sigma\in S_n} \bigotimes_{j=1}^n u_{\sigma(j)}^{(i)}$


问:
Group of mappings containing an injective map is a subset of symmetric group

Let $G$ be a group of mappings on a set $X$ with respect to function composition. Show that if $G$ contains some injective function, then $G\subseteq \text{Sym}(X)$.

What I did: If $X$ is finite, then the injective mapping $g\in G$ is also bijective. Suppose the identity $i\in G$ is not bijective. Then $gi=g$ is not bijective, impossible, so $i$ must be bijective. Then any $h\in G$ must be bijective because of the existence of $h^{-1}$ such that $hh^{-1}=i$. So $G\subseteq\text{Sym}(X)$ indeed.

But when $X$ is infinite, the injective mapping is no longer necessarily bijective, and the whole argument breaks down.

答:
Here's a recipe for how you can do it:

Prove that $i$ is injective

Prove every $h\in G$ is injective

Prove that $i$ is surjective (consider $i=i^2$)

Prove every $h\in G$ is surjective

At each stage, employ proof by contradiction. Spoiler:

 Suppose $g\in G$ is injective. Suppose that $i$ is not injective. Then $gi=g$ is not injective, absurd, so $i$ is injective. Suppose $h\in G$ is not injective. Then $h^{-1}h=i$ is not injective, absurd, so every element of $G$ is an injective function. Suppose $i\in G$ is not surjective. Then $\exists x\in X$ such that $x\not\in i(X)$. But then $i(x)\in i(X)$, but $i(x)=i^2(y)\implies x=i(y)$ hence $i(x)\not\in i^2(X)$, hence $i\ne i^2$, a contradiction. Hence $i$ is surjective. Suppose now that $h\in G$ is not surjective. Then $h h^{-1}=i$ is not surjective, a contradiction. Hence every element is surjective as well as injective.