# numerical solution examples

/Parent 5 0 R >> /Parent 3 0 R /Type /Page >> /ProcSet [/PDF /Text /ImageB] /Resources << /T1_4 93 0 R For Euler's Method, we just take the first 2 terms only. /CropBox [0 0 612 792] /GS0 82 0 R /T1_4 93 0 R /ProcSet [/PDF /Text] >> /T1_4 98 0 R /T1_6 93 0 R endobj /Type /Page 53 0 obj /Resources << /Resources << /Type /Pages 13 0 obj >> /T1_7 102 0 R The function file /T1_3 99 0 R >> /ProcSet [/PDF /Text /ImageB] /T1_2 97 0 R /Title <507265666163650D> /ExtGState << >> /Font << /T1_0 83 0 R /Contents 121 0 R /GS0 54 0 R /Resources << /Parent 5 0 R >> /Font << >> /Count 4 >> /T1_8 114 0 R /MediaBox [0 0 612 792] /ExtGState << << /T1_7 97 0 R /T1_5 99 0 R /Prev 7 0 R /Count 40 /T1_5 101 0 R /ProcSet [/PDF /Text /ImageB] /T1_6 98 0 R /Contents 141 0 R /T1_3 97 0 R /CropBox [0 0 612 792] /CropBox [0 0 612 792] /Contents 96 0 R An analytical solution involves framing the problem in a well-understood form and calculating the exact solution. /MediaBox [0 0 612 792] As a result, we need to resort to using numerical methods for solving such DEs. /T1_2 97 0 R /ProcSet [/PDF /Text] >> MATLAB Examples Hans-Petter Halvorsen Numerical Differentiation. 39 0 obj /GS0 82 0 R /T1_3 102 0 R /Im0 79 0 R /T1_9 93 0 R /Contents [151 0 R 152 0 R 153 0 R 154 0 R 155 0 R] /Type /Page /T1_4 98 0 R /T1_6 101 0 R /T1_4 99 0 R This means the approximate value of the solution when `x=2.1` is `2.8540959`. /T1_3 93 0 R /Type /Page >> /T1_0 89 0 R /MediaBox [0 0 612 792] /T1_4 93 0 R /T1_2 97 0 R endobj /T1_6 101 0 R /MediaBox [0 0 612 792] /T1_4 102 0 R << /T1_2 83 0 R (There's no final `dy/dx` value because we don't need it. /T1_4 93 0 R endobj /T1_0 89 0 R /T1_5 131 0 R /T1_0 85 0 R We obtain the formula. >> /ExtGState << /T1_2 93 0 R >> /T1_1 89 0 R /Resources << So it's a little more steep than the first 2 slopes we found. /Parent 6 0 R /GS0 82 0 R We have: Once again, we substitute our current point and the derivative we just found to obtain the next point along. All numerical solutions are approximations, some better than others, depending on the context of the problem and the numerical method used. endobj 32 0 obj A Student Study Guide for the Ninth Edition of Numerical Analysis is also avail-able and the solutions given in the Guide are generally more detailed than those in the Instructor’s Manual. In the next graph, we see the estimated values we got using Euler's Method (the dark-colored curve) and the graph of the real solution `y = e^(x"/"2)` in magenta (pinkish). You will ﬁnd them here. /T1_1 89 0 R endobj >> << /T1_1 85 0 R or /T1_0 89 0 R /MediaBox [0 0 612 792] /Type /Catalog /ExtGState << Note that the right hand side is a function of `x` and `y` in each case. >> /T1_6 120 0 R 8 0 obj /Length 2597 The simplest example of a one-step method for the numerical solution of the initial value problem (1–2) is Euler’s method4. << /Contents 117 0 R /T1_4 93 0 R /T1_6 99 0 R /GS0 82 0 R Example of Numerical Instability Example of Unstable Sequence Another Example of Unstable Sequence Solution of Nonlinear Equations 3.1 Bisection (Interval Halving) Method Example of Bisection Method 3.2 Newton's Method /Last 8 0 R /T1_3 93 0 R << /T1_3 99 0 R /ProcSet [/PDF /Text /ImageB] >> We continue this process for as many steps as required. << /T1_3 93 0 R endobj /T1_10 101 0 R /Parent 6 0 R endobj /ProcSet [/PDF /Text /ImageB] endobj /Next 50 0 R /T1_3 97 0 R You could use an online calculator, or Google search. /MediaBox [0 0 612 792] endobj /Parent 5 0 R /T1_4 99 0 R /ExtGState << >> >> >> /GS0 82 0 R >> /CropBox [0 0 612 792] /T1_10 83 0 R >> /ProcSet [/PDF /Text] endobj Next value: To get the next value `y_2`, we would use the value we just found for `y_1` as follows: `y_2` is the next estimated solution value; `f(x_1,y_1)` is the value of the derivative at the current `(x_1,y_1)` point. /T1_6 132 0 R /Type /Page 23 0 obj /T1_6 102 0 R We proceed for the required number of steps and obtain these values: In the next section, we see a more sophisticated numerical solution method for differential equations, called the Runge-Kutta Method. endobj /Contents 123 0 R >> << /MediaBox [0 0 612 792] /CropBox [0 0 612 792] /T1_3 99 0 R /Creator (alibabadownload.com) /ProcSet [/PDF /Text /ImageB] >> /Type /Page /T1_2 93 0 R /Font << /TT1 56 0 R /ProcSet [/PDF /Text /ImageB] /T1_5 101 0 R /MediaBox [0 0 612 792] endobj The following question cannot be solved using the algebraic techniques we learned earlier in this chapter, so the only way to solve it is numerically. /T1_4 93 0 R /Resources << << >> /T1_7 98 0 R >> /T1_2 97 0 R /CropBox [0 0 612 792] /Parent 6 0 R << endobj 46 0 obj /T1_4 102 0 R We'll do this for each of the sub-points, `h` apart, from some starting value `x=a` to some finishing value, `x=b`, as shown in the graph below. 26 0 obj /Font << /Font << >> endobj << For instance, you can approximate an integral numerically by subdividing the domain in little pieces and summing the area (or volume) they cover “under the curve”. ], Differential equation: separable by Struggling [Solved! endobj /Alternate /DeviceRGB /ExtGState << A numerical example may clarify the mechanics of principal component analysis. /Resources << >> /CropBox [0 0 612 792] >> /CropBox [0 0 612 792] >> /T1_8 120 0 R /Contents 87 0 R /T1_5 101 0 R /T1_7 101 0 R /T1_5 99 0 R >> endobj