Newton Raphson Method for Solving

This is the graph of y=6×3+7×2-9x-7. I am going to find the f(x) =0 for all roots. The same process will be used for three roots. The first root I am going to find out is between the interval [1, 2]. The first approximation was 1. The table worked out the Newton Raphson automatically. As you can see the line from x=1 touches the curve, it then measure the tangent from that point. The second point is when the tangent cuts the x axis. The second x value is 1. 1304 f(1. 1304) = 0. 4376. The next tangent hits at 1. 1157. f(1. 1157) = 0. 005055151.

This shows the root is going towards f(x) = 0. The third tangent hits the x axis on the value of 1. 1155. Other iteration gives the same value. As the value is so accurate the software cannot proceed. There is a sign of convergence because the differences between two numbers are decreasing. This shows a change of sign; it indicates that there is a root between 1. 1155 and 1. 11555. The error bounds are ±0. 00005. So the final answer is 1. 1155 the solution bounds are (1. 11545? x ? 1. 11555). This is the second root which is between the interval 0 and -1.

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The first guess was x = 1. The second value is when the tangent of the curve passes the x-axis at -0. 4. The difference between -1 and -0. 4 shows the accuracy increased to 60%. The rate of convergence is fast at this point. The picture to the right shows the second value where x = -0. 6273. As you can see the accuracy has increased. This mean the value is going to the root. Further analysis shows that the other X value is -0. 63486. Other iteration shows the same value this is because the value is precise so the software can’t proceed.

The final value is x= -0. 63488 because it is the last number showing in the table. I substituted x= -0. 63488 to the equation 6×3+7×2-9x-7 the f(x) = 0. 000011852. f(-0. 634875) = -0. 000041312 and f(-0. 634885)= 0. 000065017. The error bounds are ±0. 000005 and the solution bounds are (-0. 634875? x ? -0. 634885). I am going to find out the third root which is located between the intervals -1 and -2. The f(-2) = -9. This will be the last root for this equation. The error bounds and the final answer will be stated in this root.

The first approximation is -2. The tangent at this point on the curve passes the x – axis at -1. 7429. The f(-1. 7429) = -1. 81645. Further analysis shows that the second tangent hits the x-axis at -1. 6575. The f(-1. 6575) = -0. 17332. This is the next approximation. The rate of convergence is now faster because the differences between two values are now smaller 0. 25714 to 0. 085333 indicates the level of accuracy has increased.

The next X value is -1. 6475 where the f(-1. 6475) = -0. 00313 The fourth value is -1. 6473 where the f(-1. 6473) = 0.00023 this is the final answer for this root. Two other iteration gives the same answer because the levels of stability have reached at this point. The error bounds are ±0. 00005 and the solution bounds are (-1. 64725? x ? -1. 64735). So the final answer to 5 significant figure is x=-1. 6473. The error bounds shows the possible root is between x= -1. 64725 and -1. 6473 because the f(x) is positive. The table values indicate that the rate of convergence is high because the level of accuracy has increased from -1. 7429 to -1. 6473.

The equation of y=1.7ln(x-3)+3 shows the limitations of Newton Raphson where failure can be experienced because the function is discontinuous. My first guess was x=4. The tangent at that point on the curve hits the x-axis at 2. 2353 the graph does not meet at that point so therefore the method is stopped. It has asymptotes at three so the second value cannot touch the graph this shows that any values lower than three is not defined for that equation. Overflow is shown after the second value this indicates that no solution cannot be found after x=2. 2353.