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Don't have much time these days...


Nov
5
comment simple functional inequality
$\alpha(u) = u$ for $u > 0$ seems to be a counterexample.
Nov
5
comment Work out the values of a and b from the identity $x^2 - ax + 144 = (x-b)^2$
@Josh: This answer and the comment should have been part of the question.
Nov
5
comment Work out the values of a and b from the identity $x^2 - ax + 144 = (x-b)^2$
@Djaian was right (he deleted his comment). THere are multiple solutions. For instance a = -24 and b = -12.
Nov
5
comment Solving the equation $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$
@J.M: Yes, I gathered that (and hence changed my answer to reflect that) :-)
Nov
5
revised Solving the equation $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$
added 359 characters in body; added 2 characters in body
Nov
5
comment Solving the equation $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$
@user0: It is not about cube roots of unity, but about the definition of cuberoot itself! You can pick one of 3...
Nov
5
comment Solving the equation $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$
It depends on the definition of (5x-x^3)^(1/3). Given that this is a high school level problem, I presume we are only talking of the real values of the three possible values. I am not sure what is standard, though. Apparently, Mathematica seems to take the one with positive imaginary part.
Nov
5
comment Solving the equation $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$
cuberoot of an arbitrary real R number is well defined: the unique real root of x^3 = R. In this case 5x-x^3 becomes negative and your software seems to be getting unnecessarily "complex".
Nov
5
comment Solving the equation $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$
@user02: Yes, I verified with the actual (as should be evident from the figure -2.37688... I quoted).
Nov
5
awarded  sequences-and-series
Nov
5
awarded  analysis
Nov
5
revised Solving the equation $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$
added 222 characters in body; deleted 9 characters in body
Nov
5
comment Solving the equation $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$
@Alexsander: $\frac{1}{{12}}\left( {17 + \sqrt {97} } \right)$ is also a root. I am not sure why wolframalpha says it isn't. I compute LHS = RHS = -2.3768876790113715. (My guess is it is not able to handle cuberoots of negative numbers).
Nov
5
revised Sufficient conditions for convergence of functions of random variables
edited title
Nov
5
revised Boolean Simplification of A'B'C'+AB'C'+ABC'
edited title
Nov
5
comment Repeating Tiled Piecewise Function Help
It is not really clear what you are asking. Perhaps explain what you mean by "equation of pink etc in range 0-2"? What is x_1, x_2 etc
Nov
5
revised Solving the equation $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$
added 3 characters in body; edited title
Nov
5
comment Solving the equation $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$
I am not sure if "at least one real solution is spurious" is correct. Did you actually verify?
Nov
5
comment Solving the equation $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$
@Tom: Knowing that is was a factor (based on user02138's solution) helped. I am not sure exactly how I came up with it. Sorry.
Nov
5
comment Solving the equation $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$
@J.M, Ross: Maybe the downvoters did this because the statement is false? btw, I don't see any downvotes!