How to solve $\\e^{-u}+\frac{u}{5}=1\\$ for $\\u\\$ ? What is the method to solve it without using graph.

Question

Solving $\\e^{-u}+\frac{u}{5}=1\\$ without using graph.

From graphing line $y(u)=1$ intersects with $\\e^{-u}+\frac{u}{5}\\$ at two points (0,1) & (4.97,1) so that gives $\\u = 0\\$ or $\\4.97\\$. But how can I solve it analytically ?

[[graph]

Answer 1

As mentioned elsewhere, the Lambert W function is what you need to use to solve this equation exactly. The Lambert W function is actually more correctly described as a family of functions on the complex plane, functions which solve the equation $W(x) \cdot \exp(W(x)) = x$.

Here is how this is going to go: you want to start by isolating the exponential expression. So you want to rewrite $\exp(-u) + \frac{u}{5} = 1$ as $\exp(-u) = 1 - \frac{u}{5}$. With the exponential expression isolated, now you want to multiply both sides by the inverse of the exponential expression, so that you get $1$ on the other side. This will result in $1 = (1 - \frac{u}{5})\exp(u)$. Now, you want the coefficient of $u$ in the polynomial to be the same as the coefficient of $u$ in the exponent. The exponential has a coefficient of $1$ in the exponent, but in the polynomial, the coefficient is $-\frac{1}{5}$. Hence, you need to multiple both sides of the equation by $-5$. Therefore, $-5 = (-5 + u)\exp(u)$. Next, you want the constant in the exponent to match that of the polynomial. In the polynomial, we have $-5$ as the constant, but since the exponent is just $u = u + 0$, the constant in the exponent is $0$. So we need to multiple both sides of the equation by $\exp(-5)$, so that the exponent can have this constant too. Hence, $-5\exp(-5) = (-5 + u)\exp(-5 + u)$. Now, we are ready to use the Lambert W function(s). Since you only want the real solutions to the equation, we actually only need the functions $W_0 : [-\frac{1}{e}, \infty) \rightarrow [-1, \infty)$ and $W_{-1} : [-\frac{1}{e}, 0) \rightarrow (-\infty, -1]$. The idea is that $W_0(x\exp(x)) = x$ if and only if $x \geq -1$, while $W_{-1}(x\exp(x)) = x$ if and only if $x \leq -1$. With this, we have that $-5\exp(-5) = (-5 + u)\exp(-5 + u)$ iff $-5 + u = W_0(-5\exp(-5))$ or $-5 + u = W_{-1}(-5\exp(-5))$. Since $-5 \leq -1$, we know $W_{-1}(-5\exp(-5)) = -5$, while $W_0(-5\exp(-5))$ cannot be simplified further in exact form. Thus, $-5 + u = -5$ or $-5 + u = W_0(-5\exp(-5))$. Therefore, $u = 0$ or $u = 5 + W_0(-5\exp(-5))$. Now, if you type $5 + W_0(-5\exp(-5))$ into Wolfram Alpha for a numerical approximation, you will obtain the same answer you obtained, so $5 + W_0(-5\exp(-5))$ is indeed the other exact solution. But be careful: Wolfram Alpha uses different notation for this. What you need to do is type "5 + ProductLog(-5e^(-5))" into the bar. It can also be seen that $5 + W_0(-5\exp(-5))$ is correct, because your solution is between $4$ and $5$, and $-1 \lt W_0(-5\exp(-5)) \lt 0$ by its defining properties, so $4 \lt 5 + W_0(-5\exp(-5)) \lt 5$ is expected.

Answer 2

You can show by standard method of calculus that there are 2 solutions. To find the first one we we write $$e^u=1+P(u)$$ where the power series $P(u)$ has only powers greater than $0$, thus: $P(u)+\dfrac u5=0$, and it follows that one of the roots is exactly $0$. We then guess that the other solution is near $u=5$ and write: $$ e^{-5}e^{-(u-5)}+\dfrac u5=1 $$ and expanding for small $|u-5|$ obtain the approximate equation: $$ e^{-5}(6-u)+\dfrac u5=1 $$ This is a linear equation whose solution is: $$ u\approx 4.965 $$ which is close enough for the effort expended.

Answer 3

I have come up with an "algebrical" approximate solution of the curve shown in function $\\y(u)=e^{-u}+\frac{u}{5}-1=0\\$ which is as follows:

As can be noticed and observed from graph, the curve in excess of $\\u>7.7\\$ becomes a straight line.

Hence first we take the derivative of the function $\\y(u)\\$:

$\\y(u)=e^{-u}+\frac{u}{5}-1\\\tag{1}$

$\frac{\mathrm{d(y(u)} )}{\mathrm{d} u}=-e^{-u}+\frac{1}{5}$ = slope

Now the equation of straight line as mentioned above will be:

$\\y=mu+c\\$ and $\\y'(u)\\$ becomes,

$\\y=(-e^{-u}+\frac{1}{5})u+c\\\tag{2}$

Now original equation Eq. (1) in this form will be,

$\\y=(u.e^{-u}+\frac{u}{5u})u-1\\\tag{3}$

Considering Eq. (2) & (3), it is can deduced that,

$c=(-1)$

Now put $c=(-1)$ and $y=0$ for $u-intercept$ in Eq. (2), we get:

$\\0=(e^{-u}+\frac{1}{5})u-1\\$

$\\1=u.e^{-u}+\frac{u}{5}\\$

$\\e^{-u}=\frac{1}{u}(1-\frac{u}{5})=\frac{(5-u)}{5u}\\$

Now put value of $e^{-u}$ and $y=0$ in Eq. (2) again, we get:

$\\0=(-\frac{(5-u)}{5u}+\frac{1}{5})u-1\\$ simplifying will get,

$\\5=-(5-u)+u\\$

$\\10=2u\\$

$\\u=\frac{10}{2}=5\\$

Hence $u=5$ an approximate solution.

Putting value of $u$ in Eq. (1):

$\\e^{-5}+\frac{5}{5}-1=0\\$

$\\e^{-5} = 0\\$

$\\(0.00673) \approx 0\\$

[The idea of doing all excercise is to see if there is an algebric solution available or not]