Question

using Newton raphson method find the real roots of equation
x^4-x-10=0 correct to three decimal places
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Answer 1

$x4−x−10=0⇒x4=x+10.x4−x−10=0⇒x4=x+10.</p><p></p><p>⇒x4≈10⇒x≈±10−−√4≈±1.75.⇒x4≈10⇒x≈±104≈±1.75.</p><p></p><p>The equation has only two real roots.</p><p></p><p>Using the Newton-Raphson method, we can solve this equation as under:</p><p></p><p>Let f(x)=x4−x−10⇒f′(x)=4x3−1.f(x)=x4−x−10⇒f′(x)=4x3−1.</p><p></p><p>We want to determine the value of xx for which f(x)=0.f(x)=0.</p><p></p><p>Let the first estimate of xx be x1.x1.</p><p></p><p>Then, the second and better estimate would be x2=x1−f(x1)f′(x1).x2=x1−f(x1)f′(x1).</p><p></p><p>The third and better estimate would be x3=x2−f(x2)f′(x2).x3=x2−f(x2)f′(x2).</p><p></p><p>We continue in this manner until the difference between two successive estimates is lesser than the tolerable error.</p><p></p><p>$

Answer 2

at x = 3 which looks a long way from the solution but the Newton Raphson process converged VERY quickly in only 3 iterations to 1.934 which is correct to 4 sig figs.

It is sometimes a little disappointing when the process converges so quickly if you like seeing it converge!

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Here is a nicer picture which I find more satisfying. It is the graph of y = f(x) so the solution of f(x) = 0 is the point where the graph crosses the x axis at x = α.

This diagram shows how the iterative process approaches the solution of the equation f(x) = 0.