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sql server reporting services barcode font iterations in Software
24 iterations Read QR In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Generate Denso QR Bar Code In None Using Barcode encoder for Software Control to generate, create QR Code ISO/IEC18004 image in Software applications. Equations
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FindRoot[(Exp[x] 1)2, {x, 2}, DampingFactor 2, EvaluationMonitor n++] Print[n," iterations"] {x 6.6703
1017 } 8 iterations
FindRoot can also be used to determine the solution of simultaneous equations.
FindRoot[equations, { var1, a1} , { var2, a2} ,...] attempts to solve equations using initial values a1, a2, . . . for var1, var2, . . . , respectively. The equations are enclosed in a list: {equation1, equation2,...}. Alternatively, the equations may be separated by && (logical and). Convergence of Newton s method for functions of several variables is much more sensitive to choice of starting values than its counterpart for single variables. Therefore, a good graph of the functions involved is quite helpful. EXAMPLE 23 Solve the system of equations sin x + cos y = 1
e x + ln y = 2
First we graph the equations. ContourPlot[{Exp[x] + Log[y] 2, Sin[x] + Cos[y] 1}, {x, 0, 2}, {y, 0, 3}, Frame False, Axes True] Equations
It appears that there is only one solution. We use x = 1, y = 1 for our initial guess. FindRoot[{Exp[x] + Log[y] 2, Sin[x] + Cos[y] 1}, {x, 1}, {y, 1}] 0.624295, y 1.14233} If the function in an equation is such that its evaluation is costly, particularly if high precision is desired, there is another procedure that may be beneficial. InterpolateRoot[lhs rhs, {x, a, b}] solves the equation lhs = rhs using initial values a and b.
Whereas FindRoot uses linear functions (straight lines) to approximate the root of the equation, InterpolateRoot uses polynomials of degree 3 or less. The result is that higher precision can be achieved with fewer function evaluations. InterpolateRoot is contained within the package FunctionApproximations` and must be loaded prior to use. As with FindRoot, the equation may be replaced by a function, in which case its zero is computed. This example computes the zero (between 2 and 3) of the Bessel function2 J0(x), using a working precision of 1000 significant digits. For comparison purposes, the Mathematica function Timing is used. The actual numerical approximation is suppressed to save space. As a result, the value Null is returned. Delete the semicolon and run the command to see the actual result of the calculation. EXAMPLE 24
FindRoot[BesselJ[0, x], {x, 2}, WorkingPrecision 1000]; //Timing
{0.219, Null} FunctionApproximations` InterpolateRoot[BesselJ[0, x], {x, 2, 3}, WorkingPrecision 1000]; //Timing {0.046, Null} SOLVED PROBLEMS
6.12 Solve the equation 5 cos x = 4 x 3 . Make sure you find all solutions.
SOLUTION
Since 5 cos x = 4 x 3 if and only if 5 cos x 4 + x 3 = 0 , we introduce the function f ( x ) = 5 cos x 4 + x 3 and look for xintercepts. (Although we could look for the intersection of two curves, it is easier to approximate where points intercept an axis.) f[x_] = 5 Cos [x] 4 + x3; Plot[f[x], {x, 1, 2}] 1.0 0.5 1 J0(x) is a solution of the differential equation x2 y" + x y ' + x2 y = 0.
Equations
It appears that there are three solutions, near 0.5, 0.8, and 1.6. FindRoot[f[x], {x, 0.5}] 0.576574} 0.797323} 1.61805} FindRoot[f[x], {x, 0.8}] FindRoot[f[x], {x, 1.6}] 6.13 Find a solution of the equation sin x = 2. (This problem may be omitted by those unfamiliar with functions of a complex variable.) SOLUTION
Since 1 sin x 1 for all real x, this problem has no real solutions. We can force FindRoot to search for a complex solution by using a complex initial guess. FindRoot[Sin[x] 2,{x, I}] 1.5708 + 1.31696 } 6.14 Find a 20 significant digit approximation to the equation x +  sin (x 1)  = 5.
SOLUTION
First we plot the function f(x) = x +  sin (x 1)  5. f[x_] = x + Abs[Sin[x 1]] 5; Plot[f[x], {x, 10, 10}] 5 5 It appears that the only solution lies between 4 and 5. FindRoot[f[x],{x,5}, AccuracyGoal 20, WorkingPrecision 25] 4.577640011987577295259374} To 20 significant digits, the solution is 4.5776400119875772953 (last digit rounded up).

