TAYLOR SERIES, Schemes and Mind Maps of Signals and Systems

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TAYLOR SERIES

CHALLENGE!^2 3 Construct a polynomial Sounds hard right? But luckily, the predictability of the differentiation ofpolynomials helps out here!

(^40 1 2 3 ) ( ) P x^ a^ a x^ a x^ a x^

a x = + + + + with the following behavior at x = 0:

(0)^1 P^ = ′^ (0)^2 P = ′′^ (0)^3 P = ′′′^ (0)^4 P = (4)(0)^5 P =

POLYNOMIAL APPROXIMATION

ln(1^ ) x +

th^ Construct a 4degree polynomialthat matches the behavior of at x = 0 through its first 4 derivatives. ln(1^ ) x +

(^2 3 4) ( ) P x a a x a x a x^ a x = + + +^ + 0 1 2 3 4

POLYNOMIAL APPROXIMATION

ln(1^ ) x + What did we just do?th^ • We constructed the 4^ order Taylor^ ( )^ ln(1^ ) @Polynomial for • If we kept going, we would improve theapproximation near • The series is called a Taylor Series.

0 f x x x = + = 0 x =

SERIES FOR SIN(X) AND COS(X)How do we come up with the explicit form to represent the series?^ Let’s start by noticing the alternating signs…

3 5 7 9 ...^0

??? ∞ x x x xx − + − + + = ∑3! 5! 7! 9! n = Next, let’s try to find a pattern in those exponents in terms of n.Can you find an expression in terms of n that explains that pattern?

th n term^ Degree 0 1 1 3 2 5 3 7 2 1^ n^ +∞ xn ( 1) (^) − (^) ∑(2 1)! n + (^0) n = Note: 3 terms is sufficient to establish a pattern.

GROUP IT UP!!!th Construct the 6order Taylor polynomial and the Taylor series forcos(x) at x = 0.Compare your method with that of the other groups. Is there a shortcut?

th n term^ Degree 0 1 2 3

APPROXIMATING A FUNCTIONNEAR ZERO th^ Find the 4^ order Taylor polynomial that approximates^ There are really two ways we can handle this…^ (^ )^ n^ nP^ x! n

near x = 0. cos 2 y^ x = Using the definition^

Intuitively How good is this approximation?

TAYLOR SERIESGENERATED AT X = A We can match a power series with^ f^ in the same way at ANY value

x = a , provided we can take the derivatives. Actually, this is where we apply allthose “horizontal transformations”!^ Let^ f^ be a function with derivatives of all orders throughout some openinterval containing^ a. Then the

Taylor series generated by

f^ at^ x = a^ is ( ) (^ ) 2

0 ( )^ ( )

( )^ ( )(^ )^

(^ )^ ...^ (^

)^ ...^ (^

2!^!^

n^

k n k k f^ a^ f^

a^ f^

a f a^ f^ a^ x^ a^

x^ a^ x^

a^ x^

a n^ ∞ k = ′′′+ − + −^ +^ +^

−^ +^ =^

−∑ The partial sum shown here, iscalled the^ Taylor polynomial oforder^ n^ for^ f^ at^ x = a.

(^ )^ kn ( ) f ak ( ) (^ ) P x x^ a = − n ∑! k (^0) k =

OR, FOR EXAMPLE… Find the third order Taylor polynomial for

(^3 2) ( ) (^2 3 4 5) f x x x^ x = −^ +^ − (a) At x = 0^

(b) At x = 1

COMBINING TAYLOR SERIES^ On the intersection of their interval of convergence, Taylor series maybe added, subtracted, and multiplied by constants and powers of

x , and the results are once again Taylor series. The Taylor series for

f(x) + g(x)^ is the sum of the Taylor series for

f(x)^ and the Taylor series for

g(x) because the nth derivative of

(n)^ (n) f + g is f+ g^ , and so on. We can obtain the Maclaurin series for

by substituting^ 2x^ in the Maclaurin series for

cos x , adding 1, and dividing the result by 2. The Maclaurin series for

sinx + cos x^ is the term-by-term sum of the series for

sin x^ and^ cos x. We obtain the Maclaurin series for^ xsinx^ by multiplying all the terms of theMaclaurin series for^ sin x^ by^

(1^ cos 2^ )^ x +^2 x.

Series you mightwant to know(not as vital though!) 2 3 1

(^11) 3 5 2 1

2 1 1

0 ln(1^ )^ ...

( 1)^ ...^ ( 1)

( 1^ 1)

(^2 3) tan ...^ ( 1)^

...^ ( 1)^ (^

n^ nx x x xn n n n n n n^2 1 n x^ x^

x n^ n x x x x x^ x^

x ∞− − = + +∞ n n −

= +^ =^ −^ +^ −^ + −

+^ =^ −^

−^ ≤^ ≤

=^ −^ +^ −^ + −^

+^ =^ −^

2004 FORM BBC2^ (Parts a, b, and c only)

2005 FORM BBC3^ (Parts a and b only)

2005 FORM BBC3 ANSWERS