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  2. AP Calculus BC
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AP Calculus BC Flashcards: Finding Taylor Polynomial Approximations Of Functions

Study Finding Taylor Polynomial Approximations Of Functions in AP Calculus BC with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

This deck focuses on Finding Taylor Polynomial Approximations Of Functions, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus BC.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

AP Calculus BC Flashcards: Finding Taylor Polynomial Approximations Of Functions

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QUESTION

What is the nthn^{th}nth degree Taylor polynomial for f(x)=(1+x)kf(x) = (1+x)^kf(x)=(1+x)k centered at 000?

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ANSWER

1+kx+k(k−1)x22!+⋯+k(k−1)...(k−n+1)xnn!1 + kx + \frac{k(k-1)x^2}{2!} + \cdots + \frac{k(k-1)...(k-n+1)x^n}{n!}1+kx+2!k(k−1)x2​+⋯+n!k(k−1)...(k−n+1)xn​. Binomial series with binomial coefficients.

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Flashcard 1: What is the nthn^{th}nth degree Taylor polynomial for f(x)=(1+x)kf(x) = (1+x)^kf(x)=(1+x)k centered at 000?

Answer: 1+kx+k(k−1)x22!+⋯+k(k−1)...(k−n+1)xnn!1 + kx + \frac{k(k-1)x^2}{2!} + \cdots + \frac{k(k-1)...(k-n+1)x^n}{n!}1+kx+2!k(k−1)x2​+⋯+n!k(k−1)...(k−n+1)xn​. Binomial series with binomial coefficients.

Flashcard 2: Determine the Taylor polynomial of degree 1 for f(x)=ln⁡(x)f(x) = \ln(x)f(x)=ln(x) at x=1x = 1x=1.

Answer: 0+(x−1)0 + (x-1)0+(x−1). f(1)=0f(1)=0f(1)=0, f′(1)=1f'(1)=1f′(1)=1 gives linear approximation.

Flashcard 3: Identify the convergence condition for a Taylor series.

Answer: Converges if ∣x−a∣<R|x-a| < R∣x−a∣<R (radius of convergence RRR). Series converges within its radius of convergence RRR.

Flashcard 4: State the Taylor series for cos(x)\text{cos}(x)cos(x) centered at 000.

Answer: cos(x)=1−x22!+x44!−⋯\text{cos}(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdotscos(x)=1−2!x2​+4!x4​−⋯. Alternating signs with even powers only.

Flashcard 5: What is the 1st1^{st}1st degree Taylor polynomial for f(x)=e2xf(x) = \text{e}^{2x}f(x)=e2x at x=0x = 0x=0?

Answer: 1+2x1 + 2x1+2x. Replace xxx with 2x2x2x in exponential series.

Flashcard 6: What is the nthn^{th}nth degree Taylor polynomial of f(x)=ln(1+x)f(x) = \text{ln}(1+x)f(x)=ln(1+x) centered at 000?

Answer: x−x22+x33−⋯+(−1)n−1xnnx - \frac{x^2}{2} + \frac{x^3}{3} - \cdots + (-1)^{n-1}\frac{x^n}{n}x−2x2​+3x3​−⋯+(−1)n−1nxn​. Alternating signs with terms (−1)n−1xnn\frac{(-1)^{n-1}x^n}{n}n(−1)n−1xn​.

Flashcard 7: Find the 3rd3^{rd}3rd degree Taylor polynomial for f(x)=ln(1+x)f(x) = \text{ln}(1+x)f(x)=ln(1+x) centered at 000.

Answer: x−x22+x33x - \frac{x^2}{2} + \frac{x^3}{3}x−2x2​+3x3​. First three terms of natural log series.

Flashcard 8: Calculate the 2nd2^{nd}2nd degree Taylor polynomial for f(x)=exf(x) = \text{e}^xf(x)=ex centered at 000.

Answer: 1+x+x221 + x + \frac{x^2}{2}1+x+2x2​. First three terms of exponential series.

Flashcard 9: What is the Taylor polynomial approximation for f(x)=sin(x)f(x) = \text{sin}(x)f(x)=sin(x) at x=0x = 0x=0 up to x2x^2x2?

Answer: xxx. Only first term survives since x2x^2x2 term has coefficient zero.

Flashcard 10: What is the remainder term R3(x)R_3(x)R3​(x) for Taylor polynomial of f(x)=exf(x) = e^xf(x)=ex at x=0x = 0x=0?

Answer: ecx44!\frac{e^c x^4}{4!}4!ecx4​ for some c∈(0,x)c \in (0,x)c∈(0,x). Next derivative term after cubic polynomial.

Flashcard 11: What is the 1st1^{st}1st degree Taylor polynomial for f(x)=exf(x) = \text{e}^xf(x)=ex centered at 111?

Answer: e+e(x−1)e + e(x-1)e+e(x−1). e1=ee^1=ee1=e with slope eee at x=1x=1x=1.

Flashcard 12: Calculate the 3rd3^{rd}3rd degree Taylor polynomial for f(x)=cos(x)f(x) = \text{cos}(x)f(x)=cos(x) at x=πx = \pix=π.

Answer: −1+(x−π)2-1 + (x-\pi)^2−1+(x−π)2. cos⁡(π)=−1\cos(\pi)=-1cos(π)=−1, derivatives give quadratic term.

Flashcard 13: Find the Taylor polynomial of degree 2 for f(x)=x3f(x) = x^3f(x)=x3 at a=1a = 1a=1.

Answer: 1+3(x−1)+3(x−1)21 + 3(x-1) + 3(x-1)^21+3(x−1)+3(x−1)2. Taylor expansion of x3x^3x3 around x=1x=1x=1.

Flashcard 14: What is the 4th4^{th}4th degree Taylor polynomial for f(x)=x2exf(x) = x^2 \text{e}^xf(x)=x2ex at x=0x = 0x=0?

Answer: 0+0+x2+x3+x420 + 0 + x^2 + x^3 + \frac{x^4}{2}0+0+x2+x3+2x4​. Product of x2x^2x2 and exponential series.

Flashcard 15: What is the radius of convergence for the Taylor series of exe^xex?

Answer: Radius of convergence is ∞\infty∞. Exponential function converges everywhere.

Flashcard 16: Find the Taylor polynomial of degree 2 for f(x)=e−xf(x) = \text{e}^{-x}f(x)=e−x at x=0x = 0x=0.

Answer: 1−x+x221 - x + \frac{x^2}{2}1−x+2x2​. Replace xxx with −x-x−x in exponential series.

Flashcard 17: What is the general form of the nthn^{th}nth degree Taylor polynomial of f(x)f(x)f(x) centered at aaa?

Answer: Pn(x)=f(a)+f′(a)(x−a)+⋯+f(n)(a)n!(x−a)nP_n(x) = f(a) + f'(a)(x-a) + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^nPn​(x)=f(a)+f′(a)(x−a)+⋯+n!f(n)(a)​(x−a)n. Uses successive derivatives divided by factorials with powers of (x−a)(x-a)(x−a).

Flashcard 18: What is the Taylor series expansion of exe^xex centered at 000?

Answer: ex=1+x+x22!+x33!+⋯e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdotsex=1+x+2!x2​+3!x3​+⋯. Each term is xnn!\frac{x^n}{n!}n!xn​ for the exponential function.

Flashcard 19: Find the first three non-zero terms of the Taylor series for arctan(x)\text{arctan}(x)arctan(x) centered at 000.

Answer: x−x33+x55x - \frac{x^3}{3} + \frac{x^5}{5}x−3x3​+5x5​. Pattern follows (−1)nx2n+12n+1\frac{(-1)^n x^{2n+1}}{2n+1}2n+1(−1)nx2n+1​.

Flashcard 20: What is the Taylor series expansion for ln(1−x)\text{ln}(1-x)ln(1−x) centered at 000?

Answer: −x−x22−x33−⋯-x - \frac{x^2}{2} - \frac{x^3}{3} - \cdots−x−2x2​−3x3​−⋯. Replace xxx with −x-x−x in ln⁡(1+x)\ln(1+x)ln(1+x) series.

Flashcard 21: What is the 3rd3^{rd}3rd degree Taylor polynomial for f(x)=sin(2x)f(x) = \text{sin}(2x)f(x)=sin(2x) centered at 000?

Answer: 2x−8x33!2x - \frac{8x^3}{3!}2x−3!8x3​. Replace xxx with 2x2x2x in sine series.

Flashcard 22: What is the 2nd2^{nd}2nd degree Taylor polynomial for f(x)=sin(x)f(x) = \text{sin}(x)f(x)=sin(x) centered at π\piπ?

Answer: 0−(x−π)0 - (x-\pi)0−(x−π). sin⁡(π)=0\sin(\pi)=0sin(π)=0, sin⁡′(π)=−1\sin'(\pi)=-1sin′(π)=−1 gives linear term.

Flashcard 23: What is the 3rd3^{rd}3rd degree Taylor polynomial for f(x)=arcsin(x)f(x) = \text{arcsin}(x)f(x)=arcsin(x) at x=0x = 0x=0?

Answer: x+x36x + \frac{x^3}{6}x+6x3​. Arcsine has pattern (2n)!x2n+14n(n!)2(2n+1)\frac{(2n)!x^{2n+1}}{4^n(n!)^2(2n+1)}4n(n!)2(2n+1)(2n)!x2n+1​.

Flashcard 24: Find the 4th4^{th}4th degree Taylor polynomial for f(x)=ln(x)f(x) = \text{ln}(x)f(x)=ln(x) centered at 111.

Answer: (x−1)−(x−1)22+(x−1)33−(x−1)44(x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4}(x−1)−2(x−1)2​+3(x−1)3​−4(x−1)4​. Natural log series shifted to center at x=1x=1x=1.

Flashcard 25: Calculate the 2nd2^{nd}2nd degree Taylor polynomial for f(x)=cos(3x)f(x) = \text{cos}(3x)f(x)=cos(3x) at x=0x = 0x=0.

Answer: 1−9x221 - \frac{9x^2}{2}1−29x2​. Replace xxx with 3x3x3x in cosine series.

Flashcard 26: What is the 4th4^{th}4th degree Taylor polynomial for f(x)=cos(x)f(x) = \text{cos}(x)f(x)=cos(x) centered at 000?

Answer: 1−x22+x4241 - \frac{x^2}{2} + \frac{x^4}{24}1−2x2​+24x4​. Even powers with alternating signs through degree 4.

Flashcard 27: What is the formula for the remainder term Rn(x)R_n(x)Rn​(x) in Taylor polynomial?

Answer: Rn(x)=f(n+1)(c)(n+1)!(x−a)n+1R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}Rn​(x)=(n+1)!f(n+1)(c)​(x−a)n+1. Lagrange form with ccc between aaa and xxx.

Flashcard 28: State the first four terms of the Taylor series for cosh⁡(x)\cosh(x)cosh(x) centered at 000.

Answer: 1+x22!+x44!+x66!1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!}1+2!x2​+4!x4​+6!x6​. Even powers only, like cosine but hyperbolic.

Flashcard 29: What is the Taylor series for sin(x)\text{sin}(x)sin(x) centered at 000?

Answer: sin(x)=x−x33!+x55!−⋯\text{sin}(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdotssin(x)=x−3!x3​+5!x5​−⋯. Alternating signs with odd powers only.

Flashcard 30: What is the general form of the nthn^{th}nth degree Taylor polynomial of f(x)f(x)f(x) centered at aaa?

Answer: Pn(x)=f(a)+f′(a)(x−a)+⋯+f(n)(a)n!(x−a)nP_n(x) = f(a) + f'(a)(x-a) + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^nPn​(x)=f(a)+f′(a)(x−a)+⋯+n!f(n)(a)​(x−a)n. Uses successive derivatives divided by factorials with powers of (x−a)(x-a)(x−a).