Solving a Nonlinear System

\begin{align} (1/300)a + (-1/200)b &= 5\\ (-1/300)a + ((-1/300) + (1/200))b + (-1/200)c &= -e^b\\ (-1/200)b + (1/200)c &= -e^c \end{align} how do I solve for $a, b$ and $c$? Thanks!

I know if I derive an equation that isolates a variable, like $kx + e^x = 0$ I can use Newton's method to approximate x. But still can't figure it out.

(I am just a high school student - so if you can make your answers as easy to understand as possible)

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Newton's method also works for systems, as you can read here, for example. –  martini Oct 10 '11 at 15:57

$$\left\{ \begin{array}{c} \frac{1}{300}a-\frac{1}{200}b=5 \\ -\frac{1}{300}a+\left( -\frac{1}{300}+\frac{1}{200}\right) b-\frac{1}{200} c=-e^{b} \\ -\frac{1}{200}b+\frac{1}{200}c=-e^{c}\tag{1} \end{array} \right.$$

is equivalent to

$$\left\{ \begin{array}{c} 2a-3b=3000 \\ -2a+b-3c=-600e^{b} \\ -b+c=-200e^{c}\tag{2} \end{array} \right.$$

and to

$$\left\{ \begin{array}{c} a=\frac{3}{2}\left( c+200e^{c}\right) +1500 \\ -5c-400e^{c}-3000+600\text{ exp}(c+200e^{c})=0 \\ \tag{3} b=c+200e^{c}. \end{array} \right.$$

The second equation has two solutions (computation in WolframAlpha): $c\approx -600.000$ and $c\approx -3.64058$.

The Newton's method applied to $$f(c)=-5c-400e^{c}-3000+600\text{ exp}(c+200e^{c})\tag{4}$$

consists of the following iterations

$$c_{k+1}=c_{k}-\frac{f(c_{k})}{f^{\prime }(c_{k})},\qquad k=1,2,\dots\tag{5}$$

with

$$f'(c)=-5-400e^{c}+600\left( 1+200e^{c}\right) \text{ exp}(c+200e^{c})\tag{6}$$

Starting with e.g. $c_{1}=-500$, we get $$\begin{eqnarray*} f(c_{1}) &=&-5c_{1}-400e^{c_{1}}-3000+600\text{ exp}(c_{1}+200e^{c_{1}}) \\ f(-500) &=&5\times 500-400e^{-500}-3000+600\text{ exp}(-500+200e^{-500}) \\ &\approx &-500.0 \end{eqnarray*}$$

and $$\begin{eqnarray*} f^{\prime }(c_{1}) &=&-5-400e^{c_{1}}+600\left( 1+200e^{c_{1}}\right) \text{ exp}(c_{1}+200e^{c_{1}}) \\ f^{\prime }(-500) &=&-5-400e^{-500}+600\left( 1+200e^{-500}\right) \text{ exp}(-500+200e^{-500}) \\ &\approx &-5.0. \end{eqnarray*}$$

And so, $$c_{2}=c_{1}-\frac{f(c_{1})}{f^{\prime }(c_{1})}\approx -500-\frac{-500.0}{ -5.0}\approx -600.0,$$

which is already a good approximation.

For $c_{1}=-3.6$, we get successively $$\begin{eqnarray*} c_{2} &\approx &-3.6-\frac{879.64}{25019.0}\approx -3.6352 \\ c_{3} &\approx &-3.6352-\frac{71.578}{19406.}\approx -3.6389 \\ c_{4} &\approx &-3.6389-\frac{1.\,417\,3}{18902.}\approx -3.6390 \\ c_{5} &\approx &-3.6390-\frac{29.\,615}{18889.}\approx -3.6406 \\ c_{6} &\approx &-3.6406-\frac{-0.435\,93}{18676}\approx -3.6406. \end{eqnarray*}$$

From $(3)$ for $c\approx -3.6406$, we get the solution $(a,b,c)\approx (1502.4,1.6067,-3.6406)$ and for $c\approx -600.0$ the solution $(a,b,c)\approx (600.0,-600.0,-600.0)$.

Plot of $f(c)$ for $c=-700$ to $c=-3.62$

Plot of $f(c)$ for $c=-3.7$ to $c=-3.6$

Added: You have perhaps in mind the general method for solving a nonlinear system, applied to the present case. Let's denote $x_{1}=a,x_{2}=b,x_{3}=c$. The system

$$\begin{pmatrix} f_{1}(x_{1},x_{2},x_{3}) \\ f_{2}(x_{1},x_{2},x_{3}) \\ f_{3}(x_{1},x_{2},x_{3}) \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix},\tag{7}$$

where

$$\begin{pmatrix} f_{1}(x_{1},x_{2},x_{3}) \\ f_{2}(x_{1},x_{2},x_{3}) \\ f_{3}(x_{1},x_{2},x_{3}) \end{pmatrix} = \begin{pmatrix} \frac{1}{300}x_{1}-\frac{1}{200}x_{2}-5 \\ -\frac{1}{300}x_{1}+\left( -\frac{1}{300}+\frac{1}{200}\right) x_{2}-\frac{1 }{200}x_{3}+e^{x_{2}} \\ -\frac{1}{200}x_{2}+\frac{1}{200}x_{3}+e^{x_{3}} \end{pmatrix},\tag{8}$$

has the Jacobian matrix

$$\begin{eqnarray*} J\left( x\right) &=& \begin{pmatrix} \frac{\partial f_{1}(x_{1},x_{2},x_{3})}{\partial x_{1}} & \frac{\partial f_{1}(x_{1},x_{2},x_{3})}{\partial x_{2}} & \frac{\partial f_{1}(x_{1},x_{2},x_{3})}{\partial x_{3}} \\ \frac{\partial f_{2}(x_{1},x_{2},x_{3})}{\partial x_{1}} & \frac{\partial f_{2}(x_{1},x_{2},x_{3})}{\partial x_{2}} & \frac{\partial f_{2}(x_{1},x_{2},x_{3})}{\partial x_{3}} \\ \frac{\partial f_{3}(x_{1},x_{2},x_{3})}{\partial x_{1}} & \frac{\partial f_{3}(x_{1},x_{2},x_{3})}{\partial x_{2}} & \frac{\partial f_{3}(x_{1},x_{2},x_{3})}{\partial x_{3}} \end{pmatrix} \\ &=& \begin{pmatrix} \frac{1}{300} & -\frac{1}{200} & 0 \\ -\frac{1}{300} & \frac{1}{600}+e^{x_{2}} & -\frac{1}{200} \\ 0 & -\frac{1}{200} & \frac{1}{200}+e^{x_{3}} \end{pmatrix} . \end{eqnarray*}\tag{9}$$

The Newton's method consists of starting with an approximation $x^{(1)}$ and find successively

$$x^{(k+1)}=x^{(k)}+\Delta x^{(k)},\qquad k=1,2,\dots\tag{10}$$

where $\Delta x^{(k)}$ is a solution of

$$J\left( x^{(k)}\right) \Delta x^{(k)}=-f\left( x^{(k)}\right) ,\tag{11}$$

which can be found by Gaussian elimination.

Notation: $x^{(k)}$ is the vector $\left( x_{1}^{(k)},x_{2}^{(k)},x_{3}^{(k)}\right) ^{T}$, $x^{(k+1)}$ is the vector $% \left( x_{1}^{(k+1)},x_{2}^{(k+1)},x_{3}^{(k+1)}\right) ^{T}$, $J\left( x^{(k)}\right)$ is the Jacobian matrix evaluated at $\left( x_{1}^{(k)},x_{2}^{(k)},x_{3}^{(k)}\right)$ and $f\left( x^{(k)}\right)$ is the vector column $(8)$ of $f$ evaluated at $\left( x_{1}^{(k)},x_{2}^{(k)},x_{3}^{(k)}\right)$.

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The first equation is used to substitute all $a \rightarrow \frac{3}{2}b+1500$. The two remaining equations can be collected into a 2x1 vector $f = 0$

$$f = \begin{pmatrix} \hat{e}^b-\frac{b}{300}-\frac{c}{200}-5 \\ \hat{e}^c-\frac{b}{200}+\frac{c}{200} \end{pmatrix}$$

The derivatives with respect to $b$ and $c$ for each part are

$$f\,' = \begin{pmatrix} \hat{e}^b-\frac{1}{300} & -\frac{1}{200} \\ -\frac{1}{200} & \hat{e}^c+\frac{1}{200} \end{pmatrix}$$

Newton raphson with vectors is $(b,c) \rightarrow (b,c) - {f\,'}^{-1} f$

$$\begin{pmatrix}b\\c\end{pmatrix} \rightarrow \begin{pmatrix}b\\c\end{pmatrix} - \begin{pmatrix} \hat{e}^b-\frac{1}{300} & -\frac{1}{200} \\ -\frac{1}{200} & \hat{e}^c+\frac{1}{200} \end{pmatrix}^{-1} \begin{pmatrix} \hat{e}^b-\frac{b}{300}-\frac{c}{200}-5 \\ \hat{e}^c-\frac{b}{200}+\frac{c}{200} \end{pmatrix}$$

with an initial guess of $\begin{pmatrix}b = 1\\c = 1 \end{pmatrix}$ I get the following iterations

$$\begin{bmatrix} 1&1\\1.841661&0.003381327\\1.634514&-0.9835724\\1.608907&-1.936174\\1.607725&-2.783955\\1.607126&-3.380392 \end{bmatrix}$$

So in the end we have $a= \frac{3}{2} 1.607126 + 1500$, $b = 1.607126$, $c = -3.380392$.

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