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Solve for $t$ in the equation: $$(1+t)e^{-t}=0.05$$ This is where I have reached. $$(1+t)e^{-t}=0.05$$ $$1+t=0.05e^t$$ $$\ln(1+t)=\ln0.05e^t$$ $$\ln1+\ln t=\ln0.05+\ln e^t$$ $$\ln t=\ln0.05+t$$ I am stucked

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    $\begingroup$ Please use MathJax. $\endgroup$ – José Carlos Santos Sep 23 '17 at 10:37
  • $\begingroup$ is this $$e^{-T(1+T)}=0.05$$? $\endgroup$ – Dr. Sonnhard Graubner Sep 23 '17 at 10:59
  • $\begingroup$ No. I have made the necessary corrections. It's e^ (-T) $\endgroup$ – Ashalley Samuel Sep 23 '17 at 11:09
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    $\begingroup$ the logarithm of the sum is not the sum of the logarithms $\endgroup$ – M. Van Sep 23 '17 at 11:27
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    $\begingroup$ there is no closed solution for this, use numerical methods $\endgroup$ – Vasya Sep 23 '17 at 11:28
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Solution in terms of the Lambert W function. $$ (1+t)e^{-t} = \frac{1}{20} \\ (-1-t)e^{-t} = \frac{-1}{20} \\ (-1-t)e^{-1-t} = \frac{-1}{20e} \\ -1-t = W\left(\frac{-1}{20e}\right) \\ 1+t = -W\left(\frac{-1}{20e}\right) \\ t = -W\left(\frac{-1}{20e}\right)-1 $$ All complex solutions are obtained by taking all branches of W. The real solutions are $$ -W_0\left(\frac{-1}{20e}\right)-1 \approx -.981258037995 \\ -W_{-1}\left(\frac{-1}{20e}\right)-1 \approx 4.74386451839 $$

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  • $\begingroup$ (+1) nice use of Lambert function, its kinda obvious, but i think not to the OP. $\endgroup$ – Ahmad Sep 23 '17 at 11:52
  • $\begingroup$ So how do you get the figures in using Lambert W function. I understand only to where the figures were obtained $\endgroup$ – Ashalley Samuel Sep 29 '17 at 13:51
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If you do not use the beautiful Lambert function, just consider the most general problem where you look for the zero(s) of function $$f(t)=(1+t)e^{-t}-a$$ $$f'(t)=-te^{-t}$$ $$f''(t)=(t-1)e^{-t}$$ So, the function increases when $t<0$, goes through a maximum when $t=0$ (by the second derivative test), decreases when $t>0$.

At the maximum, we have $f(0)=1-a$; so, if $a<1$, there will be two real roots (one negative and one positive). If $a=1$, a double root $t=0$ and if $a>1$, no root at all.

If you graph the function for $a=0.05$, you will see that there is one root close to $t=-1$ and another one close to $t=5$. You then have all the elements to start Newton method which will give the following iterates $$\left( \begin{array}{cc} n & t_n \\ 0 & -1.00000000000000000000 \\ 1 & -0.98160602794142788392 \\ 2 & -0.98125816020529694931 \\ 3 & -0.98125803799504305042 \\ 4 & -0.98125803799502797244 \end{array} \right)$$

$$\left( \begin{array}{cc} n & t_n \\ 0 & 5.0000000000000000000 \\ 1 & 4.7158684089742339658 \\ 2 & 4.7435578246859564746 \\ 3 & 4.7438644812774523864 \\ 4 & 4.7438645183905778323 \\ 5 & 4.7438645183905783759 \end{array} \right)$$ These are the solutions for twenty significants figures.

Concerning the estimates of the solutions (for $a<1$), we can gnerate estimates approximating the function $f(x)$ using, say, a $[2,3]$ Padé approximant. Solving for the roots of numerator, we should get $$t_{1,2}=\frac{2 \left(42-24a-18 a^2\pm\sqrt{-131 a^4-3386 a^3-4356 a^2+3194 a+4679}\right)}{13 a^2+114 a+53}$$ For $a=0.05$, this will give as estimates $t_1=-0.978$ and $t_2=3.754$.

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  • $\begingroup$ Am trying to use the numerical methods approach. But how do i get the initial value for "t"? $\endgroup$ – Ashalley Samuel Sep 29 '17 at 15:15
  • $\begingroup$ @AshalleySamuel. See my edit. $\endgroup$ – Claude Leibovici Sep 30 '17 at 3:13

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