Prove that   $$\int_0^x dx_1 \int_0^{x_1}dx_2 \cdots \int_0^{x_{n-1}}f(x_n) \, dx_n =\frac{1}{(n-1)!}\int_0^x (x-t)^{n-1}f(t) \, dt$$  I'm trying induction over $n$. The case $n=1$ is trivial.   When $n=2$   \begin{align}\int_0^x dx_1 \int_0^{x_1}f(x_{2})\,dx_2 = &amp; \int_0^x\int_0^{x_1}f(x_2) \, dx_2 \, dx_1 \\ = &amp; \int_0^x \int_{x_2}^{x} f(x_2) \, dx_1 \, dx_2 \\ =&amp; \int_0^x f(x_2)(x_1-x_2) \, dx_2 = \frac{1}{(2-1)!}\int_0^x (x-t)f(t) \, dt\end{align}  I couldn't take the induction step though. Any thoughts? I appreciate the help.
According to two questions  and  asked on this site earlier there exists a nice relation:  $$\frac1{n!} \left(\int_{0}^t\mathrm dt \; f(t)\right)^n = \int_{0}^t\mathrm dt_1 \int_{0}^{t_1}\mathrm dt_2 \cdots \int_{0}^{t_{n-1}}\mathrm dt_n\; f(t_1)\,f(t_2) \ldots f(t_n). $$  Unfrotunatly I am more interested in integrals of the form  $$\int_{0}^t\mathrm dt_1 \int_{0}^{t_1}\mathrm dt_2 \cdots \int_{0}^{t_{n-1}}\mathrm dt_n\; f(t_n). $$  Can it be somehow brought to an unary integral?  All I have figured out by now is that  $$\int_{0}^t\mathrm dt_1 \int_{0}^{t_1}\mathrm dt_2 \cdots \int_{0}^{t_{n-1}}\mathrm dt_n\; t_n^m = \frac{m!}{(m+n)!} t^{m+n} .$$
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Question is :     Prove that the polynomial $p(x)=x^4-4x^2+8x+2$ is irreducible over the quadratic field $F=\mathbb{Q}(\sqrt{-2})$. [Hint : first use the method of proposition $11$ for the U.F.D $\mathbb{Z}[\sqrt{-2}]$(cf.Exercise $8$, Section $8.1$) to show that if $\alpha \in \mathbb{Z}[\sqrt{-2}]$ is a root of $p(x)$ then $\alpha$ is a divisor of $2$ in $\mathbb{Z}[\sqrt{-2}]$ . Conclude that $\alpha$ must be $\pm 1,\pm \sqrt{-2}$ or $\pm 2$ and hence show that $p(x)$ has no linear factor over $F$. Show similarly that $p(x)$ is not the product of quadratics with coefficients in $F$.]   What I have done so far is :  Suppose $\alpha \in\mathbb{Z}[\sqrt{-2}]$ is a root of $p(x)=x^4-4x^2+8x+2$ we would then have :  $p(\alpha)=\alpha^4-4\alpha^2+8\alpha+2=0\Rightarrow 2=\alpha(-\alpha^3+4\alpha-8)$  i.e., $\alpha$ is a divisor of $2$ in $\mathbb{Z}[\sqrt{-2}]$.  so, I have used the method of proposition $11$ for the U.F.D $\mathbb{Z}[\sqrt{-2}]$(cf.Exercise $8$, Section $8.1$) to show that if $\alpha \in \mathbb{Z}[\sqrt{-2}]$ is a root of $p(x)$ then $\alpha$ is a divisor of $2$ in $\mathbb{Z}[\sqrt{-2}]$  I do not understand why does he mentioned that $\mathbb{Z}[\sqrt{-2}]$ is U.F.D and all... I do not use that at all... It is unnecessarily confusing me or i am unnecessarily getting confused.. hint is actually misleading me :(  Now, I have to prove that $\alpha$ must be $\pm 1,\pm \sqrt{-2}$ or $\pm 2$  i.e., suppose I have $2=ab$ in $\mathbb{Z}[\sqrt{-2}]$ then, $N(2)=N(ab)\Rightarrow 4=N(a)N(b)$   i.e.,$N(a)=1\text{ or }2\text{ or } 4$ i.e., $p^2+2q^2=1\text{ or }2\text{ or } 4$ for $a=p+\sqrt{-2}q$   $p^2+2q^2=1\Rightarrow p=\pm 1 \Rightarrow a=\pm 1$ $p^2+2q^2=2\Rightarrow q=\pm 1\Rightarrow a=\pm\sqrt{-2}$  $p^2+2q^2=4\Rightarrow p=\pm 2\Rightarrow a=\pm 2$    Once I prove that those are the only divisors then I would consider :    $p(1)=(1)^4-4(1)^2+8(1)+2\neq 0$ $p(-1)=(-1)^4-4(-1)^2+8(-1)+2\neq 0$ $p(\sqrt{-2})=(\sqrt{-2})^4-4(\sqrt{-2})^2+8(\sqrt{-2})+2\neq 0$ $p(-\sqrt{-2})=(-\sqrt{-2})^4-4(-\sqrt{-2})^2+8(-\sqrt{-2})+2\neq 0$ $p(2)=(2)^4-4(2)^2+8(2)+2\neq 0$ $p(-2)=(-2)^4-4(-2)^2+8(-2)+2\neq 0$   So, no divisor of $2$ is a root..  Thus $p(x)$ do not have a root in $\mathbb{Z}[\sqrt{-2}]$  suppose I have something like :  $$x^4-4x^2+8x+2=(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(b+d+ac)x^2+(ad+bc)x+bd$$  But then, $bd=2\Rightarrow \text { b,d  are  a divisors of 2 in $\mathbb{Z}\sqrt{-2}$}$  But then we have seen that only divisors of $2$ in $\mathbb{Z}\sqrt{-2}$ are $\pm 1,\pm \sqrt{-2}$ or $\pm 2$  So, only possibilities are   $x^4-4x^2+8x+2=(x^2+ax\pm 1)(x^2+cx\mp 2)$ $x^4-4x^2+8x+2=(x^2+ax\pm \sqrt{-2})(x^2+cx\mp \sqrt{-2})$   But these are not possible...  This only tells that $p(x)$ is irreducible in $\mathbb{Z}[\sqrt{-2}]$ but then how do i show this is irreducible in $\mathbb{Q}(\sqrt{-2})$ I was expecting gauss lemma to help but it only works for integers and rationals...  So, please help me to clear this..  Thank you...   P.S : Proposition $11$ is Rational root theorem and Exercise $8$ is that ring of integers of $\mathbb{Q}(\sqrt{-2})$ is an Euclidean domain. $M$ = mass of the Sun  $m$ = mass of the Earth  $r$ = distance between the Earth and the Sun  The sun is converting mass into energy by nuclear fusion.   $$F = \frac{GMm}{r^2} = \frac{mv^2}{r} \rightarrow r = \frac{GM}{v^2}$$  $$\Delta E = \Delta M c^2 = (M_{t} - M_{t+\Delta t}) c^2 \rightarrow \Delta M = \Delta E / c^2$$  $$\rightarrow \frac{\Delta r}{\Delta t} = \frac{G}{v^2 c^2}.\frac{\Delta E}{\Delta t}$$  Sun radiates $3.9 × 10^{26}~\mathrm W = \Delta E/\Delta t$  Velocity of the earth $v = 29.8 \mathrm{km/s}$  There is nothing that is stopping the earth from moving with the same velocity so for centripetal force to balance gravitational force $r$ must change.   Is $r$ increasing? ($\Delta r/ \Delta t = 3.26070717 × 10^{-10} \mathrm{m/s} $) Kinetic energy is simple, right?! from the equation it seems as though if you know the speed and mass you'll know the work. Let's try to prove it. let's assume we know the acceleration and distance for now. $$w=mda$$ So the distance is $$d= 1/2 *at^2$$ Then  $$t = \sqrt{2d/a}$$ So $a*t$, which is the speed, is  $$a*\sqrt{2d/a} = a* \sqrt{2d}/\sqrt{a} = \sqrt{2da} = \sqrt{2w/m}$$ So regardless of the distance or acceleration, if you know the work and the mass, you can calculate the speed which is  $$v=\sqrt{2w/m}$$  What if you know the speed and the mass? You can easily calculate the work. $$v = \sqrt{2w/m}$$ Then  $$v^2 = 2w/m$$  That would mean  $$m*1/2*v^2=W $$ Rearrange it and  $$W= 1/2 *m *v^2$$ would be the equation.  What if you have an initial speed and a new speed and want calculate the amount of work done after an initial speed to get the new speed? You'll just subtract the total work from the initial speeds work.  $$1/2mv^2 - 1/2ms^2$$  where s is  the initial speed and v is the new speed. However if you do it the other way which is  $$d= 1/2at^2 + vt$$ and solve it the answer would be $$1/2ms^2  + 1/2msv$$ Why aren't they equal? Google Drive search for part of a file name Is there any way I can search based on the last three digits of a file name?  I have a file with many invoices that are easy to reference with the last three numbers. Is there any way I can search something like *765 to search for ends with 765  Another example.. sometimes files are prefixed with a 0, sometimes not.. 093241251 vs 93241251 and if I just search 93241251 it will not find the 0 prefix version Can I install Microsoft Office? If so, how? What are WPF Preview Events? How do I change NEI to cheat mode? Can noindex tags on several pages affect the overall SEO or ranking on Google? 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