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12h
revised Extract $A+B+C$ from $A^{\frac{3}{2}}+B^{\frac{3}{2}}+C^{\frac{3}{2}}= R*D^{\frac{3}{2}}$
deleted 1 character in body
12h
comment Extract $A+B+C$ from $A^{\frac{3}{2}}+B^{\frac{3}{2}}+C^{\frac{3}{2}}= R*D^{\frac{3}{2}}$
Iredi: By your last comment do you mean you do not want the $3/2$ to be a power in the equation? [Why would you want to "remove" it?] Or is the last comment referring to your wanting to find a valid math technique to remove the powers? If that's is, there is none which works over sums of separate terms. (See my answer below, I think your equation does not allow determination of the sum A+B+C.)
13h
comment Extract $A+B+C$ from $A^{\frac{3}{2}}+B^{\frac{3}{2}}+C^{\frac{3}{2}}= R*D^{\frac{3}{2}}$
Also it seems you want to have $R,D$ be the same for all cases and are viewing $A,B,C$ as variables. Is that right?
13h
answered Extract $A+B+C$ from $A^{\frac{3}{2}}+B^{\frac{3}{2}}+C^{\frac{3}{2}}= R*D^{\frac{3}{2}}$
13h
comment Extract $A+B+C$ from $A^{\frac{3}{2}}+B^{\frac{3}{2}}+C^{\frac{3}{2}}= R*D^{\frac{3}{2}}$
Are $A,B,C,D,R$ restricted to be integers, positive integers, or what restrictions are there on them if any?
1d
comment Doesn't $x^3+2y^3+3z^3=0$ give a surface in $R^3$?
What regular value has inverse image $S$ minus $0$?
Jul
27
revised Prove that given a triangle satisfying $8\prod \sin\frac{A}{2}=\prod \cos(A-B)$ then that triangle is equilateral.
inserted better explanation of why example not equilateral
Jul
26
revised This inequality $a+b^2+c^3+d^4\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}+\frac{1}{d^4}$
added 1 character in body
Jul
26
answered Prove that given a triangle satisfying $8\prod \sin\frac{A}{2}=\prod \cos(A-B)$ then that triangle is equilateral.
Jul
26
comment Prove that given a triangle satisfying $8\prod \sin\frac{A}{2}=\prod \cos(A-B)$ then that triangle is equilateral.
Suggest you re-word the question to something like "prove that given a triangle satisfying (your equation) then that triangle is equilateral." In your question you already have said it holds for equilaterals, but it seems you want it to hold only for equilaterals.
Jul
25
comment Find k for which the equation has equal roots.
It's easier to just use the quadratic equation. Generally $ax^2+bx+c=0$ has (one or two) real roots iff its discriminate $b^2-4ac\ge 0.$ You can apply that hereto your second rearrangement and then simplify.
Jul
22
comment End-point as point of tangency
$8x^{3/2}$ has domain $[0,\infty)$ as you say, which at $0$ means a left hand domain endpoint. In this case the slope of the tangent line (if it exists) is typically defined as the right hand derivative. For your function that will be $0$.
Jul
22
comment Find the number of sets of $(a,b,c)$ for $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{29}{72}$
So only 4 satisfy all requirements, which I got also by another long-ish (not clever) method. +1, still wonder if there is some clever way without trying so many cases...
Jul
21
comment Notation for matrix that is partially unknown.
If you want to give variable names to the unknown positions there's not a better way than you have. Without naming them, some writers just place asterisks (i.e. *) where the unknown entries are.
Jul
19
comment I need to show a union is countable (most of proof is done)
Your union is over a subset of $\mathscr{J}^\circ_{\text{rat}}$ and the latter is a countable set. Since any subset of a countable set is (at most) countable, it seems you already know the union is a countable (or a finite) union, without further analysis.
Jul
18
comment What function satisfies $F'(x) = F(2x)$?
@TheoBendit Where's the $x$ in the series, i.e. did you mean to multiply the $n$th term by $x^n$ (and sum over $n \ge 0$)?
Jul
18
revised This inequality $a+b^2+c^3+d^4\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}+\frac{1}{d^4}$
removed "improvement" of argument which wasn't one.
Jul
17
revised This inequality $a+b^2+c^3+d^4\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}+\frac{1}{d^4}$
simpler finish to proof added at end
Jul
17
revised This inequality $a+b^2+c^3+d^4\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}+\frac{1}{d^4}$
minor typos
Jul
17
answered This inequality $a+b^2+c^3+d^4\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}+\frac{1}{d^4}$