MHB High school inequality 2 |x^2y-a^2b|<A

AI Thread Summary
The discussion focuses on finding a suitable B>0 given A>0, such that the inequality |x^2y - a^2b| < A holds under specified conditions for x, y, a, and b. The initial step involves manipulating the expression |x^2 - a^2| to establish a bound using the triangle inequality. It concludes that |x^2y - a^2b| can be bounded by 3B, leading to the recommendation to set B = 1/3A. This choice ensures that the inequality |x^2y - a^2b| remains valid and less than A. The findings emphasize the relationship between the variables and the constraints necessary for the inequality to hold.
solakis1
Messages
407
Reaction score
0
Given A>0, find a B>0 such that :

for all x,y,a,b : if 0<x<1,0<y<1,0<a<1,0<b<1,and |x-a|<B,|y-b|<B,then $$|x^2y-a^2b|<A$$
 
Mathematics news on Phys.org
solakis said:
Given A>0, find a B>0 such that :

for all x,y,a,b : if 0<x<1, 0<y<1, 0<a<1, 0<b<1, and |x-a|<B, |y-b|<B, then $$|x^2y-a^2b|<A$$
[sp]First step: $|x^2-a^2| = |(x+a)(x-a)| = |x+a|x-a| \leqslant 2|x-a|$.

Next (using the triangle inequality), $|x^2y - a^2b| = |x^2y-a^2y + a^2y - a^2b| \leqslant |x^2y-a^2y| + |a^2y - a^2b| = |x^2-a^2|y + a^2|y-b| \leqslant 2|x-a| + |y-b| < 3B.$

So take $B = \frac13A$. Then $|x^2y - a^2b| < 3B = A$.[/sp]
 
Last edited:
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top