It follows that for x 1 we have 1 1 x 1 x x 2 x 3 x 4

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It follows that, for | x | < 1, we have 1 1 x = 1 + x + x 2 + x 3 + x 4 + · · · . The last of these examples is a familiar one which may have been en- countered for the first time in the context of the summation of a geometric progression. There is a wide variety of alternative ways of demonstrating this expansion. Amongst these is simple long division: 1 + x + x 2 + · · · 1 x 1 1 x x x x 2 x 2 x 2 x 3 We can also proceed in the opposite direction. That is to say, we can evaluate S = { 1 + x + x 2 + · · ·} to show that S = (1 x ) 1 . The calculation is as follows: S = 1 + x + x 2 + · · · xS = x + x 2 + · · · S xS = 1 . D.S.G. Pollock: stephen [email protected]
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TAYLOR’S THEOREM Then S (1 x ) = 1 immediately implies that S = 1 / (1 x ). There is also a formula for the partial sum of the first n terms of the series, which is S n = 1 + x + · · · + x n 1 . Consider the following subtraction: S = 1 + x + · · · + x n 1 + x n + x n +1 + · · · x n S = x n + x n +1 + · · · S x n S = 1 + x + · · · + x n 1 . This shows that S (1 x n ) = S n , whence S n = 1 x n 1 x . Example. An annuity is a sequence of regular payments, made once a year, until the end of the n th year. Usually, such an annuity may be sold to another holder; and, almost invariably, its outstanding value can be redeemed from the institution which has contracted to make the payments. There is clearly a need to determine the present value of the annuity if it is to be sold or redeemed. The principle which is applied for this purpose is that of discounting.
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  • Spring '12
  • D.S.G.Pollock
  • Taylor Series, Taylor’s Theorem, D.S.G. POLLOCK, stephen pollock

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