Observe that if and then and for it follows that

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Unformatted text preview: d % , we have . . That is, . The four- and five-term polynomial approximations are ________________________________________________________________________ page 186 —————————————————————————— CHAPTER 5. —— . . The four-term approximation on the interval . appears to be reasonably accurate within 20. Two linearly independent solutions of Airy's equation about Applying the ratio test to the terms of lim are , lim Similarly, applying the ratio test to the terms of lim Hence both series converge absolutely for all 21. Let % , lim . . Then and ________________________________________________________________________ page 187 —————————————————————————— CHAPTER 5. —— Substitution into the ODE results in . First write We then obtain Setting the coefficients equal to zero, it follows that for . Note that the indices differ by two, so for and Hence the linearly independent solutions of the Hermite equation about are . Based on the recurrence relation ________________________________________________________________________ page 188 —————————————————————————— CHAPTER 5. —— , the series solution will terminate as long as is a nonnegative even integer. If , then one or the other of the solutions in Part will contain at most terms. In particular, we obtain the polynomial solutions corresponding to . Observe that if , and , then and for . It follows that the coefficient of , in and , is for for Then by definition, for for Therefore the first six Hermite polynomials are 23. The series solution is given by ________________________________________________________________________ page 189 —————————————————————————— CHAPTER 5. —— 24. The series solution is given by . 25. The series solution is given by . ________________________________________________________________________ page 190 —————————————————————————— CHAPTER 5. —— 26. The series solution is given by 27. The series solution is given by . ________________________________________________________________________ page 191 —————————————————————————— CHAPTER 5. —— 28. Let . Then and Substitution into the ODE results in . After appropriately shifting the indices, it follows that . We find that for Since and . Writing out the individual equations, and , the remaining coefficients satisfy the equations ________________________________________________________________________ page 192 —————————————————————————— CHAPTER 5. —— That is, , , of the initial value problem is , , . Hence the series solution ________________________________________________________________________ page 193 ——————————————————...
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