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Mar
15
comment Absolute value of complex number
Assuming all other symbols are real numbers, it might help to first multiply top and bottom by the complex conjugate of the denominator, then expand the denominator. This will give you a complex number of the form $x+jy$, which you should then be able to find the modulus.
Mar
6
comment How to do large number of arithmetic operations
Try to be concise and efficient in the steps you take.
Mar
3
revised Why is $1 - \frac{1}{1 - \frac{1}{1 - \ldots}}$ not real?
deleted 2 characters in body
Mar
3
answered Weird Inequality that seems to be true
Mar
3
comment Weird Inequality that seems to be true
Afraid not: Try $x=2.5$ and $y=0.5$.
Mar
2
comment How to calculate $\sum_{k=0}^{\infty}\int_k^{k+\frac{1}{2}} e^{-st} dt $
Try evaluating the integral, then summing, or take the sum inside the integral if you can, then evaluate, then integrate.
Feb
28
answered why does this $2\sin^{2}\frac{1}{2n}\leq2(\frac{1}{2n})^2$ hold?
Feb
25
revised Is this inequality for integrals correct: $\left|\int_{\mathbb{R}}\frac{f(x)}{g(x)}dx\right|>\left|\int_{\mathbb{R}}\frac{f(x)}{g(x)^2}dx\right|.$
added 537 characters in body
Feb
25
accepted Is this inequality for integrals correct: $\left|\int_{\mathbb{R}}\frac{f(x)}{g(x)}dx\right|>\left|\int_{\mathbb{R}}\frac{f(x)}{g(x)^2}dx\right|.$
Feb
25
comment Is this inequality for integrals correct: $\left|\int_{\mathbb{R}}\frac{f(x)}{g(x)}dx\right|>\left|\int_{\mathbb{R}}\frac{f(x)}{g(x)^2}dx\right|.$
Yes, thought so. Thanks for the counter-example :-)
Feb
25
comment Is this inequality for integrals correct: $\left|\int_{\mathbb{R}}\frac{f(x)}{g(x)}dx\right|>\left|\int_{\mathbb{R}}\frac{f(x)}{g(x)^2}dx\right|.$
That seems to work after your edit, but the left integral is zero. Presumably I can construct similar examples where the left integral is non-zero. I'll try to construct one.
Feb
25
comment Is this inequality for integrals correct: $\left|\int_{\mathbb{R}}\frac{f(x)}{g(x)}dx\right|>\left|\int_{\mathbb{R}}\frac{f(x)}{g(x)^2}dx\right|.$
Thanks for your answer. Not sure your counter-example works out though? I get $2>0.75$, which agrees with the inequality.
Feb
25
revised Is this inequality for integrals correct: $\left|\int_{\mathbb{R}}\frac{f(x)}{g(x)}dx\right|>\left|\int_{\mathbb{R}}\frac{f(x)}{g(x)^2}dx\right|.$
edited body
Feb
25
revised Is this inequality for integrals correct: $\left|\int_{\mathbb{R}}\frac{f(x)}{g(x)}dx\right|>\left|\int_{\mathbb{R}}\frac{f(x)}{g(x)^2}dx\right|.$
edited body
Feb
25
revised Is this inequality for integrals correct: $\left|\int_{\mathbb{R}}\frac{f(x)}{g(x)}dx\right|>\left|\int_{\mathbb{R}}\frac{f(x)}{g(x)^2}dx\right|.$
added 95 characters in body
Feb
25
asked Is this inequality for integrals correct: $\left|\int_{\mathbb{R}}\frac{f(x)}{g(x)}dx\right|>\left|\int_{\mathbb{R}}\frac{f(x)}{g(x)^2}dx\right|.$
Feb
24
comment Does $\mid x-y\mid>0,x\neq-y$ imply $\mid\mid x\mid-\mid y\mid\mid>0$?
I think my answer would be correct for $x,y\in\mathbb{R}$, but as has been pointed out in the answers, my proof is incorrect for $x,y\in\mathbb{C}$.
Feb
24
accepted Does $\mid x-y\mid>0,x\neq-y$ imply $\mid\mid x\mid-\mid y\mid\mid>0$?
Feb
24
comment Does $\mid x-y\mid>0,x\neq-y$ imply $\mid\mid x\mid-\mid y\mid\mid>0$?
Thanks, so my problem lies here: $x\neq y\implies |x|\neq |y|$ since complex $x,y$ on the same "complex circle" will have the same moduli.
Feb
24
revised Does $\mid x-y\mid>0,x\neq-y$ imply $\mid\mid x\mid-\mid y\mid\mid>0$?
edited body