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October 29th, 2019, 09:10 AM   #1
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Math Focus: Elementary Math
Need help to verify the number of solutions

Given equation $\displaystyle x^2 +x +\lambda =0 \; ,x\in \mathbb{R}$.

Verify N - the number of solutions, using any software (matlab, calculators... etc.)

$\displaystyle N_{\lambda } =\frac{\displaystyle 1 + \lim_{s\rightarrow \infty} \left[-2\left(1+e^{-2s(-1-4\lambda)}\right)+1\right]}{2}\lim_{s\rightarrow \infty} \left[-2\left(1+e^{-2s(-1-4\lambda)}\right)+1\right]\cdot \left\lceil \frac{Re\{ \sqrt{-1-4\lambda}\}+1}{Re \{ \sqrt{-1-4\lambda}\}+2} \right\rceil.$
Re - real part of imaginary number (if it exists).

Last edited by skipjack; October 30th, 2019 at 02:43 PM.
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October 29th, 2019, 09:29 AM   #2
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x is a real number. not an imaginary or complex number.
x^2 +x +L = 0
x = (1 +/- sqrt(1 -4*L))/2

Last edited by phillip1882; October 29th, 2019 at 10:00 AM.
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October 29th, 2019, 10:58 AM   #3
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For example $\displaystyle \lambda =1 $ ; $\displaystyle D=i\sqrt{5}$ ; $\displaystyle Re(D)=0$.
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Last edited by idontknow; October 29th, 2019 at 11:18 AM.
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October 29th, 2019, 01:21 PM   #4
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Why is this question repeated?
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October 30th, 2019, 12:48 PM   #5
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it specifically states in the first sentence, x is a real number
also, if L = 1;
(1 +/- sqrt(1-4*1))/2
(1 +/- sqrt(-3))/2
1+/- i*sqrt(3))/2
x = 1/2 +i*sqrt(3)/2
or
x = 1/2 -i*sqrt(3)/2
so the real part is 1/2.
and will be 1/2 for any L value > 1/4
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Last edited by phillip1882; October 30th, 2019 at 12:52 PM.
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October 30th, 2019, 01:10 PM   #6
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I fixed the exponents now: $\displaystyle N_{\lambda } = \frac{\displaystyle 1 + \lim_{s\rightarrow \infty} \left[-2\left(1+e^{-2s(-1-4\lambda)}\right)^{-1}+1\right]}{2}\lim_{s\rightarrow \infty} \left[-2\left(1+e^{-2s(-1-4\lambda)}\right)^{-1}+1\right]\cdot \left\lceil \frac{Re\{ \sqrt{-1-4\lambda}\}+1}{Re \{ \sqrt{-1-4\lambda}\}+2} \right\rceil.$

Last edited by skipjack; October 30th, 2019 at 02:47 PM.
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November 2nd, 2019, 07:54 AM   #7
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Just plug the value of $\displaystyle \lambda $ into $\displaystyle N(\lambda )$ and it gives the number of solutions .
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