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I need help with solving this difficult fluid dynamic expression. I have tried using rules of logs, symbolab algebra calculator and Wolfram Alpha calculator, and I have got no solution.

How would you solve the following expression USING the NEWTON-RAPHSON method for $x$?
$$e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$$

When solving this USING the NEWTON-RAPHSON method, the solution is: $x=2.2418sqrt{vt}$

I want to know how you could solve the first expression using the NEWTON-RAPHSON method to get the solution. So could someone please provide a step-by-step solution, by using this method please?

Note: This question was answered, however it was NOT answered using NEWTON-RAPHSON method. It was answered using the Lambert W function, which is a very long and complicated process as compared to the Newton-Raphson method.

Let $u = frac{x^2}{4 v t}$ so that
$$ e^u - 1 - 2u = 0. $$

Defining $g(u) = e^u - 1 - 2u, $ we iterate
$$ u mapsto u - frac{g(u)}{g'(u)} ,$$
$$ u mapsto frac{1 +(u-1)e^u}{e^u - 2} $$

? u = 1.25

%9 = 1.250000000000000000000000000

? u = ( 1 + (u-1) * exp(u)    ) / (exp(u)-2 ) 

%10 = 1.256479745141752637179209827

? u = ( 1 + (u-1) * exp(u)    ) / (exp(u)-2 ) 

%11 = 1.256431211361022818343348929

? u = ( 1 + (u-1) * exp(u)    ) / (exp(u)-2 ) 

%12 = 1.256431208626169685666336003

? u = ( 1 + (u-1) * exp(u)    ) / (exp(u)-2 ) 

%13 = 1.256431208626169676982737617

? u = ( 1 + (u-1) * exp(u)    ) / (exp(u)-2 ) 

%14 = 1.256431208626169676982737617

? 

We find that
$$ frac{x^2}{4vt} = 1.256431208626169676982737617 $$ so
$$ x^2 = 5.025724834504678707930950466 ; vt $$
$$ x = 2.241812845557068063953533471 ; sqrt {vt} $$

Well, we have:

$$expleft(frac{x^2}{4cdottext{v}cdot t}right)=1+frac{x^2}{2cdottext{v}cdot t}tag1$$

Now, we know that we can write:

$$expleft(alpharight)=sum_{text{n}=0}^inftyfrac{alpha^text{n}}{text{n}!}=frac{alpha^0}{0!}+frac{alpha^1}{1!}+frac{alpha^2}{2!}+dots=$$
$$1+alpha+frac{alpha^2}{2}+dotstag2$$

So, for equation $(1)$ we can write:

$$1+frac{x^2}{4cdottext{v}cdot t}+frac{1}{2}cdotleft(frac{x^2}{4cdottext{v}cdot t}right)^2+dots=1+frac{x^2}{2cdottext{v}cdot t}tag3$$

Using the aproximation of three terms we have:

$$1+frac{x^2}{4cdottext{v}cdot t}+frac{1}{2}cdotleft(frac{x^2}{4cdottext{v}cdot t}right)^2approx1+frac{x^2}{2cdottext{v}cdot t}spaceLongleftrightarrowspace$$
$$xapprox0spaceveespace xapproxpm2sqrt{2}cdotsqrt{text{v}cdottext{t}}tag4$$

First express this in terms of a single variable: letting $s = x/sqrt{vt}$, the equation becomes
$$ e^{s^2/4} = 1 + s^2/2$$
Now with $f(s) = exp(s^2/4) - (1 + s^2/2)$, $f'(s) = s exp(s^2/4)/2 - s$, and the Newton iteration is
$$ s_{n+1} = s_n - frac{f(s_n)}{f'(s_n)}$$
Note that $s=0$ is also a solution, so you don't want to start too close to that.
Starting with, say, $s_0 = 2$, you just iterate until the numbers get close enough to each other.

$s_1 = 2 - f(2)/f'(2) = 2.392211192$

$s_2 = 2.392211192 - f(2.392211192)/f'(2.392211192) = 2.269512712$

etc.

I find that $s_5$ and $s_6$ differ only in the $9$'th decimal place.

We put $0 le y=x^2/(4vt)$ and look for the non-negative zeros of the function $f(y)$
$$
left{ matrix{
f(y) = e^{,y} - 2y - 1 = 0 hfill cr
f'(y) = e^{,y} - 2quad Rightarrow quad min f(y):;y = ln 2 hfill cr
0 < f''(y) = e^{,y} hfill cr} right.
$$

Clearly, $f(y)$ is convex, has a negative minimum at $y=ln2$, thus it has two zeros.

One of them is at $y=0$ and the other will be past the minimum.

Since for $ln2 < y$ the function is increasing we can apply Newton-Raphson
method to find the second zero, provided that the starting point $y_0$ be to the right of the minimum.

We can choose $y_0=2$ for instance, and then start the recursion
$$
eqalign{
& {{f(y_0 )} over {y_1 - y_0 }} = f'(y_0 )quad Rightarrow quad cr
& Rightarrow quad y_1 = y_0 + {{f(y_0 )} over {f'(y_0 )}} = y_0 + {{e^{,y_0 } - 2y_0 - 1} over {e^{,y_0 } - 2}}quad Rightarrow cr
& Rightarrow quad y_{n + 1} = y_n + {{e^{,y_{,n} } - 2y_n - 1} over {e^{,y_{,n} } - 2}} cr}
$$

Of course, once found a satisfactory value for $y$, you can easily
convert it back to $x$

Just as Will Jagy did, let$u = frac{x^2}{4 v t}$ to make the equation $e^u - 1 - 2u = 0$.

So, let consider that you look for the zero's of function
$$f(u)=e^u - 1 - 2u $$ for which
$$f'(u)=e^u - 2 qquad text{and} qquad f''(u)=e^u > ,,forall u$$

The first derivative cancels when $u=log(2)$. You can get an estimate of the root builiding the Taylor series at this point. This would give
$$e^u - 1 - 2u =(1-2 log (2))+(u-log (2))^2+Oleft((u-log (2))^3right)$$ Ignoring the gigher order terms, you then have as an estimate
$$u_0=log(2)+sqrt{2 log (2)-1}approx 1.31467$$ With this estimate, you can now use Newton method
$$u_{n+1}=u_n-frac{f(u_n)}{f'(u_n)}=frac{e^{u_n} (u_n-1)+1}{e^{u_n}-2}$$ and get, for twelve significant figures, the following iterates
$$left(
begin{array}{cc}
n & u_n \
0 & 1.31467301359 \
1 & 1.26002526328 \
2 & 1.25644611685 \
3 & 1.25643120888 \
4 & 1.25643120863
end{array}
right)$$

One of the important points when you use Newton method is to get a "reasonable" estimates.

You will notice in the table that, at no time, we overshoot the solution because we started at a point whe $f(x_0) times f''(x_0) > 0$ (Darboux theorem).

Starting instead with $x_0=1$, the iterates would have been
$$left(
begin{array}{cc}
n & u_n \
0 & 1.00000000000 \
1 & color{red}{1.39221119118} \
2 & 1.27395717022 \
3 & 1.25677778598 \
4 & 1.25643134800 \
5 & 1.25643120863
end{array}
right)$$