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A power series is a sum of terms . You can differentiate or integrate it term-by-term inside its interval of convergence. If , then and . This lets you build new series from known ones, such as , , and . For an alternating series whose terms decrease in magnitude to zero, the remainder after stopping at a term is bounded by the absolute value of the next term.
Practice
483Endless problems made on demand, each with an exactly computed answer. Every click is a new one.
From open sources
16Complete problems only, with worked solutions first. Show everything
| # | Title | Difficulty | |
|---|---|---|---|
| 134548 | By estimating … prove that the Taylor series for … around … | Medium | |
| 134549 | (Proof of the remainder formula when …) (a) At … find | Medium | |
| 134550 | All derivatives of … are zero at …, including | Medium | |
| 134552 | Show that the binomial coefficients in … are | Medium | |
| 134554 | Change the dummy variable and add lower limits to make | Medium | |
| 134555 | In … the coefficient of … is the number of groups of… | Medium | |
| 134556 | (a) What is the coefficient of … when … multiplies | Medium | |
| 134558 | Use the binomial series to compute 33–36 with error less than | Medium | |
| 134560 | Medium | ||
| 134561 | Medium | ||
| 134562 | Medium | ||
| 134566 | (Composition of series) If … and … find the … coefficients of… | Medium | |
| 134551 | (a) Find the first three terms in the binomial series for | Medium | |
| 134553 | For … and … find nice formulas for the binomial coefficients | Medium | |
| 134559 | Medium | ||
| 134563 | From … find the Taylor series of … up to | Medium |