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

• Notes
• 4

This preview shows page 3 - 4 out of 4 pages.

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]

Subscribe to view the full document.

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.
You've reached the end of this preview.
• Spring '12
• D.S.G.Pollock
• Taylor Series, Taylor’s Theorem, D.S.G. POLLOCK, stephen pollock

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern