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seen Dec 12 at 1:41

Dec
3
revised Let $\lambda>1$ and show the equation $\lambda - z -e^{-z} = 0$ has exacly one solution in the half plane $\{z:Re(z)>0\}$
deleted 1 character in body
Dec
3
comment Let $\lambda>1$ and show the equation $\lambda - z -e^{-z} = 0$ has exacly one solution in the half plane $\{z:Re(z)>0\}$
Sorry, I meant to write "Rouche's Theorem"
Nov
21
comment Let $\lambda>1$ and show the equation $\lambda - z -e^{-z} = 0$ has exacly one solution in the half plane $\{z:Re(z)>0\}$
Ah yes, I am relying on that deliberately, that's a good point that you need to have the additive factor less than 1. Thank you for highlighting it.
Nov
20
reviewed Approve Nearest-neighbor interpolation
Nov
20
answered Some questions on basic number theory
Nov
20
answered Is the map, $ f:(0,1)⊂ \mathbb{R}$ → $(1,∞)⊂ \mathbb{R}$ : $x ↦ 1/x $continuous?
Nov
20
reviewed Approve Continuous limsup, Discrete limsup relation
Nov
20
answered Inequality: showing that $p(1-p)\leq \frac{1}{4}$ if $0<p<1$.
Nov
20
answered Chain rule using the expression F=150W^1/3
Nov
20
comment Generating a new password using each of (A-E) and (0-9) only one time randomly
Just go through and count how many was do what you are interested in, and divide by the $15!$
Nov
20
comment Generating a new password using each of (A-E) and (0-9) only one time randomly
You do each location separately and sum them. Each position for $A$ is equally likely.
Nov
20
answered Generating a new password using each of (A-E) and (0-9) only one time randomly
Nov
19
answered Let $\lambda>1$ and show the equation $\lambda - z -e^{-z} = 0$ has exacly one solution in the half plane $\{z:Re(z)>0\}$
Nov
19
comment Let $\lambda>1$ and show the equation $\lambda - z -e^{-z} = 0$ has exacly one solution in the half plane $\{z:Re(z)>0\}$
Oh gosh, I'm sorry. I was misreading this as the imaginary part. No, it's saying the real part, and obviously there are real numbers with positive real part.
Nov
19
comment Let $\lambda>1$ and show the equation $\lambda - z -e^{-z} = 0$ has exacly one solution in the half plane $\{z:Re(z)>0\}$
Maybe it says "Re(z)$\geq$ 0"? That would be more possible. To show that this function has a real root is a simple application of the intermediate value theorem.
Nov
19
comment Let $\lambda>1$ and show the equation $\lambda - z -e^{-z} = 0$ has exacly one solution in the half plane $\{z:Re(z)>0\}$
As phrased, this doesn't make a whole lot of sense, as there are no real points in the set $\{z:\Re{(z)}>0\}$
Nov
19
answered Entire function $\phi(z)$
Nov
19
answered Prove that $\zeta(4)=\pi^4/90$
Nov
19
reviewed Approve Trigonometric substitution in the integral $\int x^2 (1-x^2)^{\frac{9}{2}} \ \mathrm dx$
Nov
18
comment Breaking RSA if small subset invertible
^^ +1 Specifically, consider how the data you are given collapses the tree of possible prime factors and thus decreases the number of things you need to check.