How do I solve for t?$$1.05^t>1+0.06t$$
Is there a way to solve for the exact value of $t$ or do I just use my calculator and estimate the answer?
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1Look up LambertW function – Andrei Aug 14 '20 at 06:27
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1Best you can do is estimation unless you want to attempt to make use of nonelementary functions, e.g. the Lambert W function. – PrincessEev Aug 14 '20 at 06:27
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I'm actually doing my finance homework, I suppose it'd be alright if I give an estimation? The question asked for a whole number of $t$ so estimation won't be hard. – Aug 14 '20 at 06:30
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Because I haven't come across this function before, so I don't know how to use it. – Aug 14 '20 at 06:31
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1Whether that's okay is entirely up to your professor; we wouldn't be the ones to know. – PrincessEev Aug 14 '20 at 06:31
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3$t<0$ or $t>8.204$, approximately – Kenta S Aug 14 '20 at 06:36
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Using Taylor series expansion around $x=8$, we get a good quadratic inequality which estimates the solution: $1.47746+0.0720853(x-8)+0.00175853(x-8)^2>1+0.06x$. The solution of this quadratic inequality gives an accurate solution for the original one. Hope this helps. – Hussain-Alqatari Aug 14 '20 at 06:48
1 Answers
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Working with whole numbers you want to find the zero of function $$f(t)=\left(\frac{21}{20}\right)^t-\frac{3 }{50}t-1$$ If you do not want to use Lambert function, you need a numerical method and then a "good" starting guess.
$t=0$ is an obvious solution. But, consider the first derivative $$f'(t)= \left(\frac{21}{20}\right)^t \log \left(\frac{21}{20}\right)-\frac{3}{50}$$ It cancels at $$t_*=-\frac{\log \left(\frac{50}{3} \log \left(\frac{21}{20}\right)\right)}{\log \left(\frac{21}{20}\right)}$$ This point corresponds to a minimum (the second derivative being positive $\forall t$). So, perform a Taylor expansion around $t_*$; this will give $$f(t)=f(t_*)+\frac 12 f''(t_*) (t-t_*)^2 + O\big[(t-t_*)^3\big]$$ Ignore the higher order terms to get the approximation $$t_0=t_* +\sqrt{ -2 \frac{f(t_*) }{f''(t_*) }}$$ This will give $t_0 \approx 8.33656$ while the exact solution is $8.20445$; not too bad.
Claude Leibovici
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