In Exercises 7 – 12, find a formula for the nth term of the Taylor series of f(x), centered at c, by finding the coefficients of the first few powers of x and looking for a paƩern. (The formulas for several of these are found in Key Idea 8.8.1; show work verifying these formula.)
Solution:
Let’s take the derivative of f(x) a couple of times and set x = π/2.
f(x) = cos(x), cos(π/2) = 0 f’(x) = -sin(x), -sin(π/2) = -1 f’‘(x) = -cos(x), -cos(π/2) = 0 f’’‘(x) = sin(x), sin(π/2) = 1 f’’’‘(x) = cos(x), cos(π/2) = 0 f’’’’‘(x) = -sin(x), -sin(π/2) = -1 f’’’’’’(x) = -cos(x), -cos(π/2) = 0
Notice the pattern, you get a 0 followed by alternating signs of 1.
The Taylor Series of f(x) centered at π/2 is \(\sum _{ n=0 }^{ \infty }{ \frac { { f }^{ (n) }(\frac { \pi }{ 2 } ) }{ n! } } { (x-\frac { \pi }{ 2 } ) }^{ n }\)
Now we plug in the derivatives at π/2:
\(\sum _{ n=0 }^{ \infty }{ \frac { { f }^{ (n) }(\frac { \pi }{ 2 } ) }{ n! } } { (x-\frac { \pi }{ 2 } ) }^{ n }\) =
\({ \frac { { f }^{ (0) }(\frac { \pi }{ 2 } ) }{ 0! } } { (x-\frac { \pi }{ 2 } ) }^{ 0 }+{ \frac { { f }^{ (1) }(\frac { \pi }{ 2 } ) }{ 1! } } { (x-\frac { \pi }{ 2 } ) }^{ 1 }+{ \frac { { f }^{ (2) }(\frac { \pi }{ 2 } ) }{ 2! } } { (x-\frac { \pi }{ 2 } ) }^{ 2 }+{ \frac { { f }^{ (3) }(\frac { \pi }{ 2 } ) }{ 3! } } { (x-\frac { \pi }{ 2 } ) }^{ 3 }+{ \frac { { f }^{ (4) }(\frac { \pi }{ 2 } ) }{ 4! } } { (x-\frac { \pi }{ 2 } ) }^{ 4 }+{ \frac { { f }^{ (5) }(\frac { \pi }{ 2 } ) }{ 5! } } { (x-\frac { \pi }{ 2 } ) }^{ 5 }+{ \frac { { f }^{ (6) }(\frac { \pi }{ 2 } ) }{ 6! } } { (x-\frac { \pi }{ 2 } ) }^{ 6 }+...\) =
\({ \frac { 0 }{ 0! } } { (x-\frac { \pi }{ 2 } ) }^{ 0 }+{ \frac { -1 }{ 1! } } { (x-\frac { \pi }{ 2 } ) }^{ 1 }+{ \frac { 0 }{ 2! } } { (x-\frac { \pi }{ 2 } ) }^{ 2 }+{ \frac { 1 }{ 3! } } { (x-\frac { \pi }{ 2 } ) }^{ 3 }+{ \frac { 0 }{ 4! } } { (x-\frac { \pi }{ 2 } ) }^{ 4 }+{ \frac { -1 }{ 5! } } { (x-\frac { \pi }{ 2 } ) }^{ 5 }+{ \frac { 0 }{ 6! } } { (x-\frac { \pi }{ 2 } ) }^{ 6 }+...\) =
\(0- { (x-\frac { \pi }{ 2 } ) }^{ 1 }+{ \frac { 1 }{ 6 } } { (x-\frac { \pi }{ 2 } ) }^{ 3 }+0-{ \frac { 1 }{ 120 } } { (x-\frac { \pi }{ 2 } ) }^{ 5 }+0+...\) =
\(- { (x-\frac { \pi }{ 2 } ) }^{ 1 }+{ \frac { 1 }{ 6 } } { (x-\frac { \pi }{ 2 } ) }^{ 3 }-{ \frac { 1 }{ 120 } } { (x-\frac { \pi }{ 2 } ) }^{ 5 }+...\)
We can see that when n is even the term ends up equal to 0, so we are left with the terms when n is odd.
From this pattern a formula for the nth term of the Taylor Series of cos(x) centered at π/2 is \(\sum _{ n=0 }^{ \infty }{ \frac { { f }^{ (n) }(\frac { \pi }{ 2 } ) }{ n! } } { (x-\frac { \pi }{ 2 } ) }^{ n }=- { (x-\frac { \pi }{ 2 } ) }^{ 1 }+{ \frac { 1 }{ 6 } } { (x-\frac { \pi }{ 2 } ) }^{ 3 }-{ \frac { 1 }{ 120 } } { (x-\frac { \pi }{ 2 } ) }^{ 5 }+...\)
This is my attempt to simplify the formula: \(\sum _{ n=0 }^{ \infty }-{ \frac { (-1+{ (-1) }^{ n }){ (x-\frac { \pi }{ 2 } ) }^{ n } }{ 2n! } }\)