Welcome to Part 20 of 21: Taylor and Maclaurin Series.

The Taylor and Maclaurin series are representation of the function f(x) by using an infinite series. If we use a finite number of terms, the series can (I stress can), but a good approximation f(x).

A Taylor series of f(x) is centered on a focus point x = a. Generally, approximations are best when x is around a, and gets worse the further x gets from a.

Taylor Series:

About the point x = a:

f(x) = f(a) + f'(a) * (x - 1) + f''(a) * (x - a)^2 / 2! + f'''(a) * (x - a)^3 / 3! + ....

If the series is cut off at n terms, the final term of the Taylor series is:

f^(n+1)(t) / (n+1)! * (x - a)^(n+1). This is known as the error term.

Maclaurin Series:

The Taylor series with a = 0.

f(x) = f(0) + f'(0) * x + f''(0) * x^2/2! + f'''(0) *x^3/3! + ...

With error term f^(n+1)(0) / (n+1)! * x^(n+1)

Some famous series:

e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + .... + x^n/n! + ....

sin x = x - x^3/3! + x^5/5! - x^7/7! + .... + ( (-1)^n * x^(2n+1) ) / (2n+1)! + ....

cos x = 1 - x^2/2! + x^4/4! - x^6/6! + ... + ( (-1)^n * x^(2n) ) / (2n)! + ....

Problems

1. Find a Macluarin series for

f(x) = ln(x + 1) to four terms

f(x) = ln(x + 1)

f(0) = ln(0 + 1) = 0

f'(x) = 1/(x+1)

f'(0) = 1/1 = 1

f''(x) = -1/(x+1)^2

f''(0) = -1/(1^2) = -1

f'''(x) = 2/(x+1)^3

f'''(0) = 2/(1^3) = 2

Then:

ln(x + 1) = 0 + 1 * x + (-1) * x^2/2! + 2 * x^3/3! + ....

= x - x^2/2! + 2x^3/3! + ....

2. Find the Macluarin series for

f(x) = e^(-x^2) to three *nonzero* terms. Approximate its integral.

f(x) = e^(-x^2)

f(0) = 1

f'(x) = e^(-x^2) * (-2x)

f'(0) = 0

f''(x) = e^(-x^2) * (4x^2 - 2)

f''(0) = -2

f'''(x) = e^(-x^2) * (-8x^3 + 12x)

f'''(0) = 0

f''''(x) = e^(-x^2) * (16x^4 - 48x^2 + 12)

f''''(0) = 12

Then:

e^(-x^2) = 1 - 2 * x^2/2! + 12 * x^4/4! + ....

= 1 - x^2 + x^4/2 + ...

∫ e^(-x^2) dx = x -x^3/3 + x^5/10 + .... + C

3. Find the Taylor series for cos x at a = π/4 to three *nonzero* terms.

f(x) = cos x

f(π/4) = √2/2

f'(x) = - sin x

f'(π/4) = -√2/2

f''(x) = -cos x

f''(π/4) = -√2/2

Then:

cos x = √2/2 - √2/2 * (x - π/4) + √2/2 * (x - π/4)^2/2! + ....

That concludes our section on Taylor and Maclaurin series. The next and last entry will be our "catch all" entry.

Thank you as always,

Eddie

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