# Questions tagged [absolute-value]

For questions about or involving the absolute value function also known as modulus function.

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### Trying to solve $|2x-15| = -x^2 - 5x -8$

My first instinct was to take the positive and negative of the right hand side, resulting in $2x-15 = -x^2 - 5x - 8$, and $2x-15 = x^2 + 5x + 8$, which results in the first giving me two real answers ...
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### Squaring complex number equation with absolute values

I don't understand how you go from the first to the second line in this problem : $$|(a-k)+i(7-2a)|=|(a-2)+i(9-2a)|$$ $$(a-k)^2+(7-2a)^2=(a-2)^2+(9-2a)^2.$$ Firstly, squaring i should make it -1 I ...
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### Show that if $0 < \frac{1}{j},\frac{1}{k} \leq \frac{1}{N} \leq 1$, then $|\frac{1}{j}-\frac{1}{k}| \leq \frac{1}{N}$. ($j,k,N \in \mathbb{N}^+$))

I am working through Tao's analysis I book and I am trying to prove that the sequence $a_n = \frac{1}{n}$ is a Cauchy sequence (Proposition 5.1.11). I understand the proof, but I am having trouble ...
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### If $|x|+|y|>|x+z|$, then is $y > z, y<z$, or $y=z$, or not determinable?

If $|x|+|y|>|x+z|$, then is $y > z, y<z$, or $y=z$, or not determinable? I thought about cases, $x>0,x<0,y>0,y<0,z>0,z<0.$ Plugging numbers is unconvincing to me.
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### minimum value of sum of absolute diferences

i was thinking about finding a $x$ that minimize the function $f(x)=\sum_{i=0}^{n}|x-a_i|$. I plot this function in geogebra (here is the url) and i saw that the function have an x that minimize it ...
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### Why isn't there any Anti-Modulus Function?

Why isn't there anything which works like anti modulus? That is, a function which gives negative of absolute values of the number? Simply, if modulus function |x| is: when x≥0 then |x|= x and when x&...
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### Inequality searching for an upper-bound

Suppose that there exists a constant $C$ such that the following relation holds for all $G$: \begin{equation*} \vert T(F)-T(G) \vert \le C \sup_y \vert F(y)-G(y) \vert \end{equation*} Suppose that ...
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The question is to solve the inital-value problem $xy'=y+x^2\sin{x}, y\left(\pi\right)=0$. I got to the same answer as the textbook except I have an x in absolute value. Here's what I did: y'-\frac{...