Error In Taylor Approximations: Lagrange Remainder & Alternating Series Bounds

1) What is the coefficient of (x-1)⁴ in the Taylor series for ln(x) centered at x=1?

-1/4

1/4

-1/3

1/3

The general term for n ≥ 1 in the Taylor series of ln(x) centered at 1 is (-1)ⁿ⁺¹ (x-1)ⁿ / n. For n=4: (-1)^{5} / 4 = -1/4.

Explanation

The general term for n ≥ 1 in the Taylor series of ln(x) centered at 1 is (-1)ⁿ⁺¹ (x-1)ⁿ / n. For n=4: (-1)^{5} / 4 = -1/4.

2) What is the coefficient of (x−2)⁵ in the Taylor series for f(x)=cos(x) centered at x=2?

Cos(2)/120

−cos(2)/120

Sin(2)/120

−sin(2)/120

The general coefficient for the Taylor series centered at a is f^{(n)}(a)/n!. Here a=2 and we want n=5. The derivatives of cos(x) cycle every four steps: cos(x) → −sin(x) → −cos(x) → sin(x) → cos(x) → −sin(x) → −cos(x) → …. So the 5th derivative is −sin(x). Therefore f^{(5)}(2)=−sin(2) and the coefficient of (x−2)⁵ is f^{(5)}(2)/5! = −sin(2)/120.

Explanation

The general coefficient for the Taylor series centered at a is f^{(n)}(a)/n!. Here a=2 and we want n=5. The derivatives of cos(x) cycle every four steps: cos(x) → −sin(x) → −cos(x) → sin(x) → cos(x) → −sin(x) → −cos(x) → …. So the 5th derivative is −sin(x). Therefore f^{(5)}(2)=−sin(2) and the coefficient of (x−2)⁵ is f^{(5)}(2)/5! = −sin(2)/120.

3) What is T3(x) for f(x)=cos(x) centered at x=π?

-1 + 0·(x-π) - (x-π)²/2 - 0·(x-π)³/6

-1 - (x-π)²/2

-1 + (x-π)²/2

1 - (x-π)²/2

f(π) = -1, f'(π) = 0, f''(π) = -cos(π) = -(-1) = 1, f'''(π) = sin(π) = 0. So T3(x) = -1 + 0*(x-π) + (1)/2 (x-π)² + 0 = -1 + (x-π)²/2.

Explanation

f(π) = -1, f'(π) = 0, f''(π) = -cos(π) = -(-1) = 1, f'''(π) = sin(π) = 0. So T3(x) = -1 + 0*(x-π) + (1)/2 (x-π)² + 0 = -1 + (x-π)²/2.

4) Given that the Taylor series for e^x centered at x=0 is 1 + x + x²/2! + ..., what is the Taylor series for the function f(x) = e^(x-2) centered at x=2?

1 + (x-2) + (x-2)²/2! + ...

1 + x + x²/2! + ...

E² + e²(x-2) + e²(x-2)²/2! + ...

1 - (x-2) + (x-2)²/2! - ...

We can use the substitution u = x-2. Since we want the series centered at x=2, we are looking for powers of (x-2). We know e^u = 1 + u + u²/2! + ... Substituting u = x-2 directly gives 1 + (x-2) + (x-2)²/2! + ...

Explanation

We can use the substitution u = x-2. Since we want the series centered at x=2, we are looking for powers of (x-2). We know e^u = 1 + u + u²/2! + ... Substituting u = x-2 directly gives 1 + (x-2) + (x-2)²/2! + ...

5) The coefficient of x⁴ in the Taylor series centered at 0 for sin(3x) is

0

-81/24

81/24

-27/8

The series for sin(3x) contains only odd powers of x because sin is an odd function and 3x scales the argument. The general term is (-1)^k (3x)^{2k+1} / (2k+1)! = (-1)^k 3^{2k+1} x^{2k+1} / (2k+1)!. All even powers have coefficient 0.

Explanation

The series for sin(3x) contains only odd powers of x because sin is an odd function and 3x scales the argument. The general term is (-1)^k (3x)^{2k+1} / (2k+1)! = (-1)^k 3^{2k+1} x^{2k+1} / (2k+1)!. All even powers have coefficient 0.

6) Which function has a Taylor series with only even powers when centered at x=0?

Cos(x)

Eˣ

Ln(1+x)

Sin(x)

cos(x) is an even function, so only even powers appear. sin(x) is odd (only odd powers), eˣ and ln(1+x) have both.

Explanation

cos(x) is an even function, so only even powers appear. sin(x) is odd (only odd powers), eˣ and ln(1+x) have both.

7) The general coefficient in the Taylor series for f(x) centered at a is:

F^{(n)}(a)/n!

F^{(n)}(a)·n!

F(a)/n!

F^{(n)}(a)/(n+1)!

By definition, the Taylor series is ∑ [f^{(n)}(a)/n!] (x-a)ⁿ from n=0 to ∞. The coefficient of the nth term is exactly f^{(n)}(a)/n!.

Explanation

By definition, the Taylor series is ∑ [f^{(n)}(a)/n!] (x-a)ⁿ from n=0 to ∞. The coefficient of the nth term is exactly f^{(n)}(a)/n!.

8) The remainder term in Taylor's theorem tells us:

Exactly how far the polynomial is from the function

That the error goes to zero as n increases inside the interval of convergence

Both A and B for analytic functions

The error is always zero

For functions that are equal to their Taylor series (analytic functions), the remainder → 0 as n → ∞ in the interior of the interval of convergence.

Explanation

For functions that are equal to their Taylor series (analytic functions), the remainder → 0 as n → ∞ in the interior of the interval of convergence.

9) Why can we use the alternating series error bound for sin(x), cos(x), and ln(1+x) approximations?

Because their series alternate and decrease

Because they are polynomials

Because the error is always positive

Only for sin(x)

Each of these series satisfies the conditions of the alternating series test for |x| small enough, so the error is less than the first omitted term.

Explanation

Each of these series satisfies the conditions of the alternating series test for |x| small enough, so the error is less than the first omitted term.

10) To apply the Lagrange error bound when using Tₙ(x) centered at a to approximate f(x), you need to know:

A bound on |f^{(n+1)}(t)| for t between a and x

Value of f^{(n+1)}(c)

Previous term in series

Nothing

The Lagrange remainder is |Rₙ(x)| ≤ M ( |x-a|ⁿ⁺¹ ) / (n+1)! where M is a bound on the (n+1)th derivative on the interval between a and x.

Explanation

The Lagrange remainder is |Rₙ(x)| ≤ M ( |x-a|ⁿ⁺¹ ) / (n+1)! where M is a bound on the (n+1)th derivative on the interval between a and x.

11) For the function f(x) = eˣ, the Taylor series centered at any point a equals eˣ everywhere because:

Remainder → 0

Eˣ is its own derivative

Radius infinite

All of the above

eˣ is analytic everywhere, its derivatives grow but factorials grow faster, radius infinite, remainder → 0 for all real x.

Explanation

eˣ is analytic everywhere, its derivatives grow but factorials grow faster, radius infinite, remainder → 0 for all real x.

12) A common strategy to bound the error when using a partial sum of a Taylor series is to:

Use the next term (for alternating series) or the Lagrange form with a bound on the next derivative

Just take more terms without checking

Assume error zero

Use linear approx only

These are the two standard rigorous methods taught in high school calculus for estimating or bounding Taylor series approximation error.

Explanation

These are the two standard rigorous methods taught in high school calculus for estimating or bounding Taylor series approximation error.

13) Why does the alternating series error bound work particularly well for sin(x) and cos(x) approximations near x=0?

Factorial makes terms small

Series not alternating

Derivatives bounded by 1

Both A and C

The terms contain n! in the denominator (grows very fast) and the derivatives of sin/cos are ±sin or ±cos, so |f^{(n)}(x)| ≤ 1 for all x and n, guaranteeing the terms get smaller and the alternating series test applies.

Explanation

The terms contain n! in the denominator (grows very fast) and the derivatives of sin/cos are ±sin or ±cos, so |f^{(n)}(x)| ≤ 1 for all x and n, guaranteeing the terms get smaller and the alternating series test applies.

14) If the limit as x approaches 2 of [f(x) - 3] / (x-2) = 5, what must be true about the Taylor polynomial T1(x) for f(x) centered at x=2?

3 + 5(x-2)

5 + 3(x-2)

5(x-2)

3

If the limit of [f(x) - 3] / (x-2) as x approaches 2 is 5, this is the definition of the derivative f'(2) = 5, and it implies f(2) = 3 (for the numerator to be 0). The first-degree Taylor polynomial is f(2) + f'(2)(x-2), which is 3 + 5(x-2).

Explanation

If the limit of [f(x) - 3] / (x-2) as x approaches 2 is 5, this is the definition of the derivative f'(2) = 5, and it implies f(2) = 3 (for the numerator to be 0). The first-degree Taylor polynomial is f(2) + f'(2)(x-2), which is 3 + 5(x-2).

15) When you increase the degree of a Taylor polynomial centered at the same point, the approximation at points close to the center generally:

Gives smaller error

Larger error

Same

Exact only at center

Adding higher-degree terms makes the polynomial match more derivatives at the center point, so near the center the graphs stay closer together for a wider range, reducing the error in the approximation.

Explanation

Adding higher-degree terms makes the polynomial match more derivatives at the center point, so near the center the graphs stay closer together for a wider range, reducing the error in the approximation.

Error in Taylor Approximations: Lagrange Remainder & Alternating Series Bounds