I think this it:
You can show that any real solution of ##x^4+ax^3+bx^2+ax+1=0## is also a solution of ##x^4-ax^3+bx^2-ax+1=0##. Then you can combine these two equations and get that this solution must be a solution of ##x^4+bx^2+1=0##. But this is great, because you can show from here that...
Here is another way to think about it:
If ##x^4+ax^3+bx^2+ax+1## has at least one real root, then that means its graph must intersect the x-axis. Since this is a 4th degree equation with a positive leading coefficient the tails of the graph both go off towards positive infinity. Therefore, if...
I'm trying to understand the spinning egg phenomenon. That is, why does a hardboiled egg (or any solid ellipsoid) "stand on end" when spun at high velocities. Upon searching the web I found one site written by physicist Rod Cross in Sydney who tried to give a simple, intuitive explanation...
That's probably what you call it. Notice that the power reducing formulas are really just rewritten versions of double angle formulas - so I tend to call them all a "double angle formula" because I try to memorize as few formulas as possible.
Wolfram Alpha gave the following graph when asking for the tangent line to y=(x^2)/(2x+5) at x=-5. If your graph doesn't look like this then you have either told us the wrong function or mis-entered something into your calculator...
Your problem was in your substitution, you successfully substituted for ##x## but you did not substitute for " ## dx ## ".
If you do this you will get that: $$dx = -r \sin (\theta ) d\theta$$
Which will give the integral: $$\int_0^{\pi / 2} r^2 \sin^2 \theta$$
which you can solve using...
I found the expression \frac{d}{d V} \int_0^t{V(\tau)}d\tau a little surprising. I'm not sure how to interpret the \frac{d}{d V}. I would have expected to see \frac{d}{d t} \int_0^t{V(\tau)}d\tau instead. In that case it is exactly the fundamental theorem of calculus and the solution is...
I approached it a little differently than you - here was my thinking:
in the numerator distribute the 2's so that you get two products of only even numbers: ##\frac{(2*4*6*...*2n)^2}{2n!}## This would allow you to cancel out the even terms in the denominator and get...
Mathematicians do occasionally discuss "transfinite sequences" or sequences with an arbitrary index that may the uncountable. If the sequence is of real numbers and an uncountable number of the terms are nonzero, then the sum of the sequence necessarily diverges.
Thanks for the response - I really appreciate it.
let me see if I can clarify (or have clarified) some of these comments:
It sounds to me that you believe the axis about which an object will spin is most strongly determined by the stability of its axes. I was going to approach this...
I approached the problem as a triple integral in spherical coordinates and got $$\int_0^{aR} \int_0^{2\pi} \int_0^{\pi} f(\frac{r}{R})r^2 \sin (\phi) d\phi d\theta dr$$
Doing a simple substitution of ##x= \frac{r}{R}## and doing the two inner integrals gives the same result as you: $$4\pi R^3...
I would rephrase this to "suggesting that ∫ F * dr might not= 0" Note that IF the vector field is conservative then every closed path integral must be zero; however, the opposite is not necessarily true. There are many cases when the field is not conservative but a given closed path integral...
this is true, but I think if the curl is nonzero at some point then you could find a surface that makes the surface integral nonzero (you would need some sort of continuity assumption to prove this). Since this problem is talking about an arbitrary curve, L, I believe the only possibility is to...
I might note that ##\frac{1}{\frac{1}{x^n}}+\frac{1}{\frac{1}{y^n}}=\frac{1}{\frac{1}{z^n}}## is identical to writing ##x^n+y^n=z^n##. Then you can refer back to your earlier proof.
it has been a long time since I studied Ring Theory, but here is what I remember that might be relevant:
A unit is an element that has an inverse. So in order for ##x^2-1## to be a unit, there would have to exist an inverse of ##x^2-1## in your field. Your suggestion of ##\frac{1}{x^2-1}## is...