Chapter 8: Problem 18
\(f(x)=\sin x=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1) !} x^{2 k+1}\)
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Chapter 8: Problem 18
\(f(x)=\sin x=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1) !} x^{2 k+1}\)
These are the key concepts you need to understand to accurately answer the question.
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$$3 x^{2} y^{\prime \prime}+8 x y^{\prime}+(x-2) y=0$$
Duffing's Equation. In the study of a nonlinear spring with periodic forcing, the following equation arises: $$ y^{\prime \prime}+k y+r y^{3}=A \cos \omega t $$ Let \(k=r=A=1\) and \(\omega=10\). Find the first three nonzero terms in the Taylor polynomial approximations to the solution with initial values \(y(0)=0, y^{\prime}(0)=1\).
In applying the method of Frobenius, the following recurrence relation arose: \(a_{k+1}=15^{7} a_{k} /(k+1)^{9}\) \(k=0,1,2, \ldots\) (a) Show that the coefficients are given by the formula \(a_{k}=15^{7 k} a_{0} /(k !)^{9}, k=0,1,2, \ldots\) (b) Use the formula obtained in part (a) with \(a_{0}=1\) to compute \(a_{5}, a_{10}, a_{15}, a_{20},\) and \(a_{25}\) on your computer or calculator. What goes wrong? (c) Now use the recurrence relation to compute \(a_{k}\) for \(k=1,2,3, \ldots, 25,\) assuming \(a_{0}=1 .\) (d) What advantage does the recurrence relation have over the formula?
$$3 x^{2} y^{\prime \prime \prime}+8 x y^{\prime}+(x-2) y=0$$
$$x y^{\prime \prime}+(x-1) y^{\prime}-2 y=0$$
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