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I was wondering how to solve the following exponential equation for a by hand:

$$-e^{-a/2}-\frac{1}{2}ae^{-a/2} = -0.95$$

I'm able to find a solution with computer software, but am stuck when it comes to solving it algebraically by hand. Any help would be appreciated!

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    $\begingroup$ It is not possible to solve it algebraically by hand! $\endgroup$ Commented Mar 18, 2018 at 13:50

5 Answers 5

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We can use the Lambert $W$ function to solve the equation "by hand".

Notice that your equation can be written as: $$2+a-1.9e^{\frac a2}=0 \Rightarrow \color{red}{1.9e^{\frac a2}=2+a}$$ Now we let $-2x=2+a$ and $a=-2x-2$ and thus we have: $$-2x=1.9\exp\Big(\frac{-2x-2}{2}\Big)$$ $$-2x=1.9\exp(-x-1)\Rightarrow\color{red}{-2x=\frac{1.9}{e^x\cdot e}}$$ The last equation can be rewritten as: $$x\cdot \color{red}{e^x}=\frac{1.9}{\color{red}{-2}\cdot e}$$ And now we have the function $xe^x=k$, which can be written as $x=W(k)$, and thus we have: $$\begin{align} x&=W\Big(-\frac{1.9}{2e}\Big)\qquad{\mathrm{recall \ \ that\ -2x=2+a}}\\ 2+a&=-2W\Big(-\frac{1.9}{2e}\Big)\\ \end{align}$$

$$\therefore\bbox[8px, border:2px solid black]{a=-2-2W\Big(-\frac{1.9}{2e}\Big)}$$

Now, it is debatable whether the Lambert-$W$ function constitutes an explicit solution. However, we can solve this numerically thru NR, using the recurrence relation: $${a_{n+1}=-2+a_n-\frac{a_n}{1-0.95e^{\frac a2}}}$$ For approximately close $a_0$ and after sufficient iterations, the solution over $\mathbb{R}$ would be:

$$\bbox[8px, border: 2px solid black]{a=\begin{cases} -0.574097\\ 0.710723\\ \end{cases}}$$

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As already said, numerical methods would be required.

However, consider that you look for the zero's of $$f(a)=e^{-\frac a 2}\left(1+\frac{1}{2}a\right)-\frac{19}{20}$$ Using Taylor series built around $a=0$ you would get $$f(a)=-\frac{1}{20}+\frac{a^2}{8}-\frac{a^3}{24}+\frac{a^4}{128}+O\left(a^5\right)$$ which can be solved using (ugly) radicals.

The solutions will be $\approx -0.574459$ and $\approx 0.709325$ while the exact solutions would be $\approx -0.574097$ and $\approx 0.710723$ which is not too bad.

Sooner or later, you will learn that, better than with Taylor series, functions can be approximated using Padé approximants. Using a $[2,2]$ one, we should have $$f(a)=\frac{-\frac{1}{20}-\frac{1}{60}a+\frac{353 }{2880}a^2}{1+\frac{1}{3}a+\frac{7}{144}a^2}$$ and the numerator cancels for $$a_ \pm=\frac{12}{353} \left(2\pm\sqrt{357}\right)$$ that is to say $\approx -0.574315$ and $\approx 0.710293$. We did not need to solve an ugly quartic polynomial to have something quite good.

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you can write $$-e^{-a/2}\left(1+\frac{1}{2}a\right)=-\frac{95}{100}$$ and then a numerical method will help you!

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You did the right thing to use your calculator to solve $$ -e^{-a/2}-\frac{1}{2}ae^{-a/2} = -0.95$$

Our current algebra is not sufficient to solve such equations by hand.

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If you want a solution ''by hand'' this can be only an approximate solution and the approximation depends on the work you are willing to do (and that a calculator can do in a much shorter time) .

With a little work (and using your knowledge of elementary continuous functions):

Plot $y_1=1+\frac{1}{2}x$ and $y_2=0.95e^{\frac{x}{2}}$

You can easily see that for $x\ge 2$ or $x\le -2$ we have $y_2>y_1$

but, in a sub-interval of $(-2 ,2)$ the value of $y_2$ becomes less than the value of $y_1$ because the two functions are continuous and $y_1(0)=1<0.95=y_2(0)$

So you can conclude that the equation $-e^{-a/2}-\frac{1}{2}ae^{-a/2} = -0.95$ has one solution in the interval $(-2,0)$ and another solution in $(0,2)$.

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