Recent content by lurflurf

What is partial integration? You mean like this? $$\int \mathrm{f}(x,y,z) \partial{}x$$ or this $$\int u\, dv=u\, v-\int v\, du$$ maybe your friend should work on his communication skills. The student probably did not know what he was on about. What is forth semester physics these days? Is that...

The initial condition is part of the problem statement. You do not figure out the initial condition, it is given to you. Keep in mind that Euler's method while simple is very unstable. I don't know what you are doing with f(x,y) the example shows how to find it.

To find the implied domain of a composite function we need to consider the inverse image of the outer function. So we If we want 0<arctan(x)<pi we must have 0<=x as the domain. Next we consider values of cos(arctan(x)) that are possible. That will be the range. 0 excluded from the range of the...

^positive derivative suffices I found an interesting paper https://www.researchgate.net/publication/292674112_Fast_and_flexible_methods_for_monotone_polynomial_fitting that states the general monotonic polynomial is $$a+b\int_0^x \bigg(p_1(u)^2+p_2(u)^2\bigg) du$$ and that it may be proved...

Yes they do, in particular they say given two sets A and B the relation R is a subset of their Cartesian product $$R\subset A\times B$$ the inverse of R is defined by $$R^{-1}=\{(b,a)|(a,b)\in R\}$$ If you are more interested in functions it is true that not all functions have full inverses...

$$x^2-16$$ has an inverse. The complete inverse is multivalued, so you can chose branches if you like. That is like saying sine or exp don't have inverses, my pocket calculator says they do. I would say $$f^{-1}(0)=\{-4,4\} $$ If want a single valued inverse pick a branch.

^I think the question is if the particular example is easier than the general case. This example is easier, it is a sum of cubes. I don't know what you mean by polynomials don't have an inverse, as they clearly do. The inverse is not a function so we need to choose branches. In this example we...

$$f(x)=3x^3 -18x^2 +36x$$ complete the cube $$f(x)=3(x-2)^3 +24$$ There will be three inverses For real values other than two, two preimages are complex

That function is not analytic. Its not equal to its Taylor series so you cannot substitute the series for the function.

I liked it because I liked it. Sometimes I like things that are not perfect. Clearly a simple typo. We all make them. Naming variables conveniently not a the most important technique, but is helpful at times. 4x,4x+4,4x+8,4x+12 is nice I would probably use x-6,x-2,x+2,x+6 instead but it does...

Well x is a multiple of 2 was used x is a multiple of 4 is not needed

I would let t=arg(z) then z=|z|exp(i t)=exp(i t) $$\mathrm{arg}\left( \frac{1+z^2}{1+\bar{z}^2 }\right)=\mathrm{arg}\left( \frac{1+\exp(2i t)}{1+\exp(-2i t) }\right)$$ $$\mathrm{arg}\left( \frac{1+z^2}{1+\bar{z}^2 }\right)=\mathrm{arg}\left(\exp(2i t)\right)$$

In short you need more functions. For example for the quintic https://en.wikipedia.org/wiki/Generalized_hypergeometric_function https://en.wikipedia.org/wiki/Theta_function https://en.wikipedia.org/wiki/Bring_radical see https://en.wikipedia.org/wiki/Quintic_function

It is natural to substitute the contents of the square root. $$u=x^b+3$$ In 3 and 6 we can finish the integral easily. The others we can use elliptic integrals as mathwonk recommends or hypergeometric functions. For example use $$\frac{1}{x^{a+1}\sqrt{x^b+3}}=...

Your proof would be conditional. It would become valid once the conjecture is proven or false if the conjecture is disproven.