# Tutorial 5 - 2010 - Solution/Question 2

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# Question 2 [1]

1. Derive a $$g(x) = x$$ function to use in the fixed-point algorithm.
2. Show the first 3 iterations of using the fixed-point algorithm, starting with an initial guess of $$T = 380$$ K.
3. Will the fixed-point method converge for this problem, using your $$g(x)$$?

## Solution

1. The goal of fixed-point iterative methods is to locate "fixed-point(s)" associated with a given function. A fixed-point is special case of a function where in it maps a value $$x$$ to itself (what this means in layman's terms is if you put a value $$x$$ into a function you get $$x$$ out, i.e. $$x = g(x)$$). Knowing that our objective is to find the root(s) of a known objective function (we could have just had a black box function...) we may then construct $$g(x)$$ in way that gives it the greatest chance of converging. Let us start by looking at our objective function:

$f(T) = 0 = - 24097 - 0.26T + 1.69\text{x}10^{-3}T^{2} + \frac{1.5\text{x}10^{5}}{T} - (-23505)$

Likely the first choice of g(x) that popped to everyone's mind was

$T = \frac{- 24097 + 1.69\text{x}10^{-3}T^{2} + \frac{1.5\text{x}10^{5}}{T} - (-23505)}{0.26} = g(T)_{1}$

Given that we have a function $$f(T) = 0$$, another good choice would be one of $$g(T) = T + f(T)$$ or $$g(T) = T - f(T)$$. Let's say we were trying to identify the lower of the two roots. In this case we would want $$f(T)$$ to yield a negative value when $$T > T_{root}$$ and a positive value when $$T < T_{root}$$. Looking at the plot generated in Question 1 we see that the function is positive when $$T$$ is smaller than the lower root and greater than the upper root and is negative when $$T$$ is between the two roots. Therefore if we wished to locate the lower root $$g(T) = T - f(T)$$ would locally satisfy the desired properties and if we wished to find the upper root $$g(T) = T + f(T)$$ would locally satisfy the desired properties (Note that these are gross generalizations and do not prove stability or ensure convergence. They are merely educated guesses as to potentially good functions). Therefore we will also investigate both:

$g(T)_{2} = T + (- 24097 - 0.26T + 1.69\text{x}10^{-3}T^{2} + \frac{1.5\text{x}10^{5}}{T} - (-23505))$
$g(T)_{3} = T - (- 24097 - 0.26T + 1.69\text{x}10^{-3}T^{2} + \frac{1.5\text{x}10^{5}}{T} - (-23505))$
2. Starting with a value of $$T^{(0)} = 380 K$$:

FUNCTION $$g(T)_{1}$$

$\begin{split}\begin{array}{rl} \Delta H_{r}^{0}\left(T^{(0)}\right) &= - 24097 - 0.26\left(T^{(0)}\right) + 1.69\text{x}10^{-3}\left(T^{(0)}\right)^{2} + \frac{1.5\text{x}10^{5}}{\left(T^{(0)}\right)}\\ \Delta H_{r}^{0}\left(T^{(0)}\right) &= - 24097 - 0.26\left(380\right) + 1.69\text{x}10^{-3}\left(380\right)^{2} + \frac{1.5\text{x}10^{5}}{\left(380\right)}\\ \Delta H_{r}^{0}\left(T^{(0)}\right) &= -23557.02715 \end{array}\end{split}$

ITERATION 1

$\begin{split}\begin{array}{rl} T^{(1)} &= \frac{- 24097 + 1.69\text{x}10^{-3}(T^{(0)})^{2} + \frac{1.5\text{x}10^{5}}{T^{(0)}} - (-23505)}{0.26}\\ T^{(1)} &= \frac{- 24097 + 1.69\text{x}10^{-3}(380)^{2} + \frac{1.5\text{x}10^{5}}{(380)} - (-23505)}{0.26}\\ T^{(1)} &= 179.8955\\ & \\ \Delta H_{r}^{0}\left(T^{(1)}\right) &= - 24097 - 0.26\left(T^{(1)}\right) + 1.69\text{x}10^{-3}\left(T^{(1)}\right)^{2} + \frac{1.5\text{x}10^{5}}{\left(T^{(1)}\right)}\\ \Delta H_{r}^{0}\left(T^{(1)}\right) &= - 24097 - 0.26\left(179.8955\right) + 1.69\text{x}10^{-3}\left(179.8955\right)^{2} + \frac{1.5\text{x}10^{5}}{\left(179.8955\right)}\\ \Delta H_{r}^{0}\left(T^{(1)}\right) &= -23255 \end{array}\end{split}$
$\begin{split}\begin{array}{rl} \epsilon_{tol,x}^{(1)} &= \left| \frac{T^{(1)}-T^{(0)}}{T^{(0)}} \right| < \epsilon_{tol} ? \\ \epsilon_{tol,x}^{(1)} &= \left| \frac{\left(179.8955\right)-\left(380\right)}{\left(380\right)} \right| < \left(10^{-6}\right) ? \\ \epsilon_{tol,x}^{(1)} &= 0.5266 < \left(10^{-6}\right) ? \;\; \rightarrow \;\; \text{No, therefore keep going}. \\ & \\ \epsilon_{tol,f}^{(1)} &= \left| \frac{\Delta H_{r}^{0}\left(T^{(1)}\right)-\Delta H_{r}^{0}\left(T^{(0)}\right)}{\Delta H_{r}^{0}\left(T^{(0)}\right)} \right| < \epsilon_{tol} ? \\ \epsilon_{tol,f}^{(1)} &= \left| \frac{\left(-23255\right)-\left(-23557.02715\right)}{\left(-23557.02715\right)} \right| < \left(10^{-6}\right) ? \\ \epsilon_{tol,f}^{(1)} &= 0.0128 < \left(10^{-6}\right) ? \;\; \rightarrow \;\; \text{No, therefore keep going}. \end{array}\end{split}$

ITERATION 2

$\begin{split}\begin{array}{rl} T^{(2)} &= \frac{- 24097 + 1.69\text{x}10^{-3}(T^{(1)})^{2} + \frac{1.5\text{x}10^{5}}{T^{(1)}} - (-23505)}{0.26}\\ T^{(2)} &= \frac{- 24097 + 1.69\text{x}10^{-3}(179.8955)^{2} + \frac{1.5\text{x}10^{5}}{(179.8955)} - (-23505)}{0.26}\\ T^{(2)} &= 1140.4\\ & \\ \Delta H_{r}^{0}\left(T^{(2)}\right) &= - 24097 - 0.26\left(T^{(2)}\right) + 1.69\text{x}10^{-3}\left(T^{(2)}\right)^{2} + \frac{1.5\text{x}10^{5}}{\left(T^{(2)}\right)}\\ \Delta H_{r}^{0}\left(T^{(2)}\right) &= - 24097 - 0.26\left(1140.4\right) + 1.69\text{x}10^{-3}\left(1140.4\right)^{2} + \frac{1.5\text{x}10^{5}}{\left(1140.4\right)}\\ \Delta H_{r}^{0}\left(T^{(2)}\right) &= -22064 \end{array}\end{split}$
$\begin{split}\begin{array}{rl} \epsilon_{tol,x}^{(2)} &= \left| \frac{T^{(2)}-T^{(1)}}{T^{(1)}} \right| < \epsilon_{tol} ? \\ \epsilon_{tol,x}^{(2)} &= \left| \frac{\left(1140.4\right)-\left(179.8955\right)}{\left(179.8955\right)} \right| < \left(10^{-6}\right) ? \\ \epsilon_{tol,x}^{(2)} &= 5.3394 < \left(10^{-6}\right) ? \;\; \rightarrow \;\; \text{No, therefore keep going}. \\ & \\ \epsilon_{tol,f}^{(2)} &= \left| \frac{\Delta H_{r}^{0}\left(T^{(2)}\right)-\Delta H_{r}^{0}\left(T^{(1)}\right)}{\Delta H_{r}^{0}\left(T^{(1)}\right)} \right| < \epsilon_{tol} ? \\ \epsilon_{tol,f}^{(2)} &= \left| \frac{\left(-22064\right)-\left(-23255\right)}{\left(-23255\right)} \right| < \left(10^{-6}\right) ? \\ \epsilon_{tol,f}^{(2)} &= 0.0512 < \left(10^{-6}\right) ? \;\; \rightarrow \;\; \text{No, therefore keep going}. \end{array}\end{split}$

ITERATION 3

$\begin{split}\begin{array}{rl} T^{(3)} &= \frac{- 24097 + 1.69\text{x}10^{-3}(T^{(2)})^{2} + \frac{1.5\text{x}10^{5}}{T^{(2)}} - (-23505)}{0.26}\\ T^{(3)} &= \frac{- 24097 + 1.69\text{x}10^{-3}(1140.4)^{2} + \frac{1.5\text{x}10^{5}}{(1140.4)} - (-23505)}{0.26}\\ T^{(2)} &= 6682.6\\ & \\ \Delta H_{r}^{0}\left(T^{(3)}\right) &= - 24097 - 0.26\left(T^{(3)}\right) + 1.69\text{x}10^{-3}\left(T^{(3)}\right)^{2} + \frac{1.5\text{x}10^{5}}{\left(T^{(3)}\right)}\\ \Delta H_{r}^{0}\left(T^{(2)}\right) &= - 24097 - 0.26\left(6682.6\right) + 1.69\text{x}10^{-3}\left(6682.6\right)^{2} + \frac{1.5\text{x}10^{5}}{\left(6682.6\right)}\\ \Delta H_{r}^{0}\left(T^{(3)}\right) &= 49659 \end{array}\end{split}$
$\begin{split}\begin{array}{rl} \epsilon_{tol,x}^{(3)} &= \left| \frac{T^{(3)}-T^{(1)}}{T^{(2)}} \right| < \epsilon_{tol} ? \\ \epsilon_{tol,x}^{(3)} &= \left| \frac{\left(6682.6\right)-\left(1140.4\right)}{\left(1140.4\right)} \right| < \left(10^{-6}\right) ? \\ \epsilon_{tol,x}^{(3)} &= 4.8598 < \left(10^{-6}\right) ? \;\; \rightarrow \;\; \text{No, therefore keep going}. \\ & \\ \epsilon_{tol,f}^{(3)} &= \left| \frac{\Delta H_{r}^{0}\left(T^{(3)}\right)-\Delta H_{r}^{0}\left(T^{(2)}\right)}{\Delta H_{r}^{0}\left(T^{(2)}\right)} \right| < \epsilon_{tol} ? \\ \epsilon_{tol,f}^{(3)} &= \left| \frac{\left(49659\right)-\left(-22064\right)}{\left(-22064\right)} \right| < \left(10^{-6}\right) ? \\ \epsilon_{tol,f}^{(3)} &= 3.2507 < \left(10^{-6}\right) ? \;\; \rightarrow \;\; \text{No, therefore keep going}. \end{array}\end{split}$

FUNCTION $$g(T)_{2}$$

$\begin{split}\begin{array}{rl} \Delta H_{r}^{0}\left(T^{(0)}\right) &= - 24097 - 0.26\left(T^{(0)}\right) + 1.69\text{x}10^{-3}\left(T^{(0)}\right)^{2} + \frac{1.5\text{x}10^{5}}{\left(T^{(0)}\right)}\\ \Delta H_{r}^{0}\left(T^{(0)}\right) &= - 24097 - 0.26\left(380\right) + 1.69\text{x}10^{-3}\left(380\right)^{2} + \frac{1.5\text{x}10^{5}}{\left(380\right)}\\ \Delta H_{r}^{0}\left(T^{(0)}\right) &= -23557.02715 \end{array}\end{split}$

ITERATION 1

$\begin{split}\begin{array}{rl} T^{(1)} &= T^{(0)} - 24097 - 0.26T^{(0)} + 1.69\text{x}10^{-3}(T^{(0)})^{2} + \frac{1.5\text{x}10^{5}}{T^{(0)}} - (-23505)\\ T^{(1)} &= (380) - 24097 - 0.26(380) + 1.69\text{x}10^{-3}(380)^{2} + \frac{1.5\text{x}10^{5}}{(380)} - (-23505)\\ T^{(1)} &= 283.2531\\ & \\ \Delta H_{r}^{0}\left(T^{(1)}\right) &= - 24097 - 0.26\left(T^{(1)}\right) + 1.69\text{x}10^{-3}\left(T^{(1)}\right)^{2} + \frac{1.5\text{x}10^{5}}{\left(T^{(1)}\right)}\\ \Delta H_{r}^{0}\left(T^{(1)}\right) &= - 24097 - 0.26\left(283.2531\right) + 1.69\text{x}10^{-3}\left(283.2531\right)^{2} + \frac{1.5\text{x}10^{5}}{\left(283.2531\right)}\\ \Delta H_{r}^{0}\left(T^{(1)}\right) &= -23505.49146 \end{array}\end{split}$
$\begin{split}\begin{array}{rl} \epsilon_{tol,x}^{(1)} &= \left| \frac{T^{(1)}-T^{(0)}}{T^{(0)}} \right| < \epsilon_{tol} ? \\ \epsilon_{tol,x}^{(1)} &= \left| \frac{\left(283.2531\right)-\left(380\right)}{\left(380\right)} \right| < \left(10^{-6}\right) ? \\ \epsilon_{tol,x}^{(1)} &= 0.080 < \left(10^{-6}\right) ? \;\; \rightarrow \;\; \text{No, therefore keep going}. \\ & \\ \epsilon_{tol,f}^{(1)} &= \left| \frac{\Delta H_{r}^{0}\left(T^{(1)}\right)-\Delta H_{r}^{0}\left(T^{(0)}\right)}{\Delta H_{r}^{0}\left(T^{(0)}\right)} \right| < \epsilon_{tol} ? \\ \epsilon_{tol,f}^{(1)} &= \left| \frac{\left(-23505.49146\right)-\left(-23557.02715\right)}{\left(-23557.02715\right)} \right| < \left(10^{-6}\right) ? \\ \epsilon_{tol,f}^{(1)} &= 0.0022 < \left(10^{-6}\right) ? \;\; \rightarrow \;\; \text{No, therefore keep going}. \end{array}\end{split}$

ITERATION 2

$\begin{split}\begin{array}{rl} T^{(2)} &= T^{(1)} - 24097 - 0.26T^{(1)} + 1.69\text{x}10^{-3}(T^{(1)})^{2} + \frac{1.5\text{x}10^{5}}{T^{(1)}} - (-23505)\\ T^{(2)} &= (283.2531) - 24097 - 0.26(283.2531) + 1.69\text{x}10^{-3}(283.2531)^{2} + \frac{1.5\text{x}10^{5}}{(283.2531)} - (-23505)\\ T^{(2)} &= 282.7616\\ & \\ \Delta H_{r}^{0}\left(T^{(2)}\right) &= - 24097 - 0.26\left(T^{(2)}\right) + 1.69\text{x}10^{-3}\left(T^{(2)}\right)^{2} + \frac{1.5\text{x}10^{5}}{\left(T^{(2)}\right)}\\ \Delta H_{r}^{0}\left(T^{(2)}\right) &= - 24097 - 0.26\left(282.7616\right) + 1.69\text{x}10^{-3}\left(282.7616\right)^{2} + \frac{1.5\text{x}10^{5}}{\left(282.7616\right)}\\ \Delta H_{r}^{0}\left(T^{(2)}\right) &= -23504.913333 \end{array}\end{split}$
$\begin{split}\begin{array}{rl} \epsilon_{tol,x}^{(2)} &= \left| \frac{T^{(2)}-T^{(1)}}{T^{(1)}} \right| < \epsilon_{tol} ? \\ \epsilon_{tol,x}^{(2)} &= \left| \frac{\left(282.7616\right)-\left(283.2531\right)}{\left(283.2531\right)} \right| < \left(10^{-6}\right) ? \\ \epsilon_{tol,x}^{(2)} &= 0.0017 < \left(10^{-6}\right) ? \;\; \rightarrow \;\; \text{No, therefore keep going}. \\ & \\ \epsilon_{tol,f}^{(2)} &= \left| \frac{\Delta H_{r}^{0}\left(T^{(2)}\right)-\Delta H_{r}^{0}\left(T^{(1)}\right)}{\Delta H_{r}^{0}\left(T^{(1)}\right)} \right| < \epsilon_{tol} ? \\ \epsilon_{tol,f}^{(2)} &= \left| \frac{\left(-23504.913333\right)-\left(-23505.49146\right)}{\left(-23505.49146\right)} \right| < \left(10^{-6}\right) ? \\ \epsilon_{tol,f}^{(2)} &= 0.000026 < \left(10^{-6}\right) ? \;\; \rightarrow \;\; \text{No, therefore keep going}. \end{array}\end{split}$

ITERATION 3

$\begin{split}\begin{array}{rl} T^{(3)} &= T^{(2)} - 24097 - 0.26T^{(2)} + 1.69\text{x}10^{-3}(T^{(2)})^{2} + \frac{1.5\text{x}10^{5}}{T^{(2)}} - (-23505)\\ T^{(3)} &= (282.7616) - 24097 - 0.26(282.7616) + 1.69\text{x}10^{-3}(282.7616)^{2} + \frac{1.5\text{x}10^{5}}{(282.7616)} - (-23505)\\ T^{(2)} &= 282.8483\\ & \\ \Delta H_{r}^{0}\left(T^{(3)}\right) &= - 24097 - 0.26\left(T^{(3)}\right) + 1.69\text{x}10^{-3}\left(T^{(3)}\right)^{2} + \frac{1.5\text{x}10^{5}}{\left(T^{(3)}\right)}\\ \Delta H_{r}^{0}\left(T^{(2)}\right) &= - 24097 - 0.26\left(282.8483\right) + 1.69\text{x}10^{-3}\left(282.8483\right)^{2} + \frac{1.5\text{x}10^{5}}{\left(282.8483\right)}\\ \Delta H_{r}^{0}\left(T^{(3)}\right) &= -23505.0156 \end{array}\end{split}$
$\begin{split}\begin{array}{rl} \epsilon_{tol,x}^{(3)} &= \left| \frac{T^{(3)}-T^{(1)}}{T^{(2)}} \right| < \epsilon_{tol} ? \\ \epsilon_{tol,x}^{(3)} &= \left| \frac{\left(282.8483\right)-\left(282.7616\right)}{\left(282.7616\right)} \right| < \left(10^{-6}\right) ? \\ \epsilon_{tol,x}^{(3)} &= 0.00031 < \left(10^{-6}\right) ? \;\; \rightarrow \;\; \text{No, therefore keep going}. \\ & \\ \epsilon_{tol,f}^{(3)} &= \left| \frac{\Delta H_{r}^{0}\left(T^{(3)}\right)-\Delta H_{r}^{0}\left(T^{(2)}\right)}{\Delta H_{r}^{0}\left(T^{(2)}\right)} \right| < \epsilon_{tol} ? \\ \epsilon_{tol,f}^{(3)} &= \left| \frac{\left(-23505.0156\right)-\left(-23504.913333\right)}{\left(-23504.913333\right)} \right| < \left(10^{-6}\right) ? \\ \epsilon_{tol,f}^{(3)} &= 0.0000044 < \left(10^{-6}\right) ? \;\; \rightarrow \;\; \text{No, therefore keep going}. \end{array}\end{split}$

FUNCTION $$g(T)_{3}$$

$\begin{split}\begin{array}{rl} \Delta H_{r}^{0}\left(T^{(0)}\right) &= - 24097 - 0.26\left(T^{(0)}\right) + 1.69\text{x}10^{-3}\left(T^{(0)}\right)^{2} + \frac{1.5\text{x}10^{5}}{\left(T^{(0)}\right)}\\ \Delta H_{r}^{0}\left(T^{(0)}\right) &= - 24097 - 0.26\left(380\right) + 1.69\text{x}10^{-3}\left(380\right)^{2} + \frac{1.5\text{x}10^{5}}{\left(380\right)}\\ \Delta H_{r}^{0}\left(T^{(0)}\right) &= -23557.02715 \end{array}\end{split}$

ITERATION 1

$\begin{split}\begin{array}{rl} T^{(1)} &= T^{(0)} - \left(- 24097 - 0.26T^{(0)} + 1.69\text{x}10^{-3}(T^{(0)})^{2} + \frac{1.5\text{x}10^{5}}{T^{(0)}} - (-23505) \right)\\ T^{(1)} &= (380) - \left(- 24097 - 0.26(380) + 1.69\text{x}10^{-3}(380)^{2} + \frac{1.5\text{x}10^{5}}{(380)} - (-23505) \right)\\ T^{(1)} &= 432.0272\\ & \\ \Delta H_{r}^{0}\left(T^{(1)}\right) &= - 24097 - 0.26\left(T^{(1)}\right) + 1.69\text{x}10^{-3}\left(T^{(1)}\right)^{2} + \frac{1.5\text{x}10^{5}}{\left(T^{(1)}\right)}\\ \Delta H_{r}^{0}\left(T^{(1)}\right) &= - 24097 - 0.26\left(432.0272\right) + 1.69\text{x}10^{-3}\left(432.0272\right)^{2} + \frac{1.5\text{x}10^{5}}{\left(432.0272\right)}\\ \Delta H_{r}^{0}\left(T^{(1)}\right) &= -23547 \end{array}\end{split}$
$\begin{split}\begin{array}{rl} \epsilon_{tol,x}^{(1)} &= \left| \frac{T^{(1)}-T^{(0)}}{T^{(0)}} \right| < \epsilon_{tol} ? \\ \epsilon_{tol,x}^{(1)} &= \left| \frac{\left(432.0272\right)-\left(380\right)}{\left(380\right)} \right| < \left(10^{-6}\right) ? \\ \epsilon_{tol,x}^{(1)} &= 0.1369 < \left(10^{-6}\right) ? \;\; \rightarrow \;\; \text{No, therefore keep going}. \\ & \\ \epsilon_{tol,f}^{(1)} &= \left| \frac{\Delta H_{r}^{0}\left(T^{(1)}\right)-\Delta H_{r}^{0}\left(T^{(0)}\right)}{\Delta H_{r}^{0}\left(T^{(0)}\right)} \right| < \epsilon_{tol} ? \\ \epsilon_{tol,f}^{(1)} &= \left| \frac{\left(-23547\right)-\left(-23557.02715\right)}{\left(-23557.02715\right)} \right| < \left(10^{-6}\right) ? \\ \epsilon_{tol,f}^{(1)} &= 0.00044 < \left(10^{-6}\right) ? \;\; \rightarrow \;\; \text{No, therefore keep going}. \end{array}\end{split}$

ITERATION 2

$\begin{split}\begin{array}{rl} T^{(2)} &= T^{(1)} - \left(- 24097 - 0.26T^{(1)} + 1.69\text{x}10^{-3}(T^{(1)})^{2} + \frac{1.5\text{x}10^{5}}{T^{(1)}} - (-23505) \right)\\ T^{(2)} &= (432.0272) - \left(- 24097 - 0.26(432.0272) + 1.69\text{x}10^{-3}(432.0272)^{2} + \frac{1.5\text{x}10^{5}}{(432.0272)} - (-23505) \right)\\ T^{(2)} &= 473.7196\\ & \\ \Delta H_{r}^{0}\left(T^{(2)}\right) &= - 24097 - 0.26\left(T^{(2)}\right) + 1.69\text{x}10^{-3}\left(T^{(2)}\right)^{2} + \frac{1.5\text{x}10^{5}}{\left(T^{(2)}\right)}\\ \Delta H_{r}^{0}\left(T^{(2)}\right) &= - 24097 - 0.26\left(473.7196\right) + 1.69\text{x}10^{-3}\left(473.7196\right)^{2} + \frac{1.5\text{x}10^{5}}{\left(473.7196\right)}\\ \Delta H_{r}^{0}\left(T^{(2)}\right) &= -23524 \end{array}\end{split}$
$\begin{split}\begin{array}{rl} \epsilon_{tol,x}^{(2)} &= \left| \frac{T^{(2)}-T^{(1)}}{T^{(1)}} \right| < \epsilon_{tol} ? \\ \epsilon_{tol,x}^{(2)} &= \left| \frac{\left(473.7196\right)-\left(432.0272\right)}{\left(432.0272\right)} \right| < \left(10^{-6}\right) ? \\ \epsilon_{tol,x}^{(2)} &= 0.0965 < \left(10^{-6}\right) ? \;\; \rightarrow \;\; \text{No, therefore keep going}. \\ & \\ \epsilon_{tol,f}^{(2)} &= \left| \frac{\Delta H_{r}^{0}\left(T^{(2)}\right)-\Delta H_{r}^{0}\left(T^{(1)}\right)}{\Delta H_{r}^{0}\left(T^{(1)}\right)} \right| < \epsilon_{tol} ? \\ \epsilon_{tol,f}^{(2)} &= \left| \frac{\left(-23524\right)-\left(-23547\right)}{\left(-23547\right)} \right| < \left(10^{-6}\right) ? \\ \epsilon_{tol,f}^{(2)} &= 0.00095 < \left(10^{-6}\right) ? \;\; \rightarrow \;\; \text{No, therefore keep going}. \end{array}\end{split}$

ITERATION 3

$\begin{split}\begin{array}{rl} T^{(3)} &= T^{(2)} - \left(- 24097 - 0.26T^{(2)} + 1.69\text{x}10^{-3}(T^{(2)})^{2} + \frac{1.5\text{x}10^{5}}{T^{(2)}} - (-23505) \right)\\ T^{(3)} &= (473.7196) - 24097 - 0.26(473.7196) + 1.69\text{x}10^{-3}(473.7196)^{2} + \frac{1.5\text{x}10^{5}}{(473.7196)} - (-23505)\\ T^{(2)} &= 492.9904\\ & \\ \Delta H_{r}^{0}\left(T^{(3)}\right) &= - 24097 - 0.26\left(T^{(3)}\right) + 1.69\text{x}10^{-3}\left(T^{(3)}\right)^{2} + \frac{1.5\text{x}10^{5}}{\left(T^{(3)}\right)}\\ \Delta H_{r}^{0}\left(T^{(2)}\right) &= - 24097 - 0.26\left(492.9904\right) + 1.69\text{x}10^{-3}\left(492.9904\right)^{2} + \frac{1.5\text{x}10^{5}}{\left(492.9904\right)}\\ \Delta H_{r}^{0}\left(T^{(3)}\right) &= -23510 \end{array}\end{split}$
$\begin{split}\begin{array}{rl} \epsilon_{tol,x}^{(3)} &= \left| \frac{T^{(3)}-T^{(1)}}{T^{(2)}} \right| < \epsilon_{tol} ? \\ \epsilon_{tol,x}^{(3)} &= \left| \frac{\left(492.9904\right)-\left(473.7196\right)}{\left(473.7196\right)} \right| < \left(10^{-6}\right) ? \\ \epsilon_{tol,x}^{(3)} &= 0.0407 < \left(10^{-6}\right) ? \;\; \rightarrow \;\; \text{No, therefore keep going}. \\ & \\ \epsilon_{tol,f}^{(3)} &= \left| \frac{\Delta H_{r}^{0}\left(T^{(3)}\right)-\Delta H_{r}^{0}\left(T^{(2)}\right)}{\Delta H_{r}^{0}\left(T^{(2)}\right)} \right| < \epsilon_{tol} ? \\ \epsilon_{tol,f}^{(3)} &= \left| \frac{\left(-23510\right)-\left(-23524\right)}{\left(-23524\right)} \right| < \left(10^{-6}\right) ? \\ \epsilon_{tol,f}^{(3)} &= 0.00060 < \left(10^{-6}\right) ? \;\; \rightarrow \;\; \text{No, therefore keep going}. \end{array}\end{split}$
3. It would appear that $$g(T)_{1}$$ is rapidly diverging while $$g(T)_{2}$$ is converging to the lower root and $$g(T)_{3}$$ is converging to the upper root.