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Worked problem · Calculus II · Unit 4B

Taylor Polynomial From Derivatives: Worked Example and Solution

A complete taylor or power-series construction example with method choice, derivation, verification, and a common wrong approach.

1 problem8 min estimated timeIntermediate to advanced progression

Problem

  1. If f^(n)(0)=2^n, find the Maclaurin series.

Complete worked solutions

Every problem has a source-matched answer and independently reviewed derivation.

01

Problem 1: If f^(n)(0)=2^n, find the Maclaurin series.

Answer: ∑n=0∞2nxn/n!=e2x\sum_{n=0}^{\infty}2^nx^n/n!=e^{2x}

Why this method: Taylor or power-series construction matches the mathematical structure before any algebraic cleanup.

  1. The Maclaurin coefficient is f(n)(0)/n!=2n/n!f^{(n)}(0)/n!=2^n/n!.
  2. Thus the series is ∑n=0∞2nxn/n!=e2x\sum_{n=0}^{\infty}2^nx^n/n!=e^{2x}.

What is included

See a concise answer and full derivation for if f^(n)(0)=2^n, find the maclaurin series.

Skills assessed

  • Taylor or power-series construction

Prerequisites

  • derivatives
  • infinite series
Taylor Polynomial From Derivatives instructional sequence
A complete taylor or power-series construction example with method choice, derivation, verification, and a common wrong approach. The numbered labels and written sequence preserve meaning without relying on color.
Long description

Read the diagram from top to bottom. Each numbered box names one decision or mathematical operation. Arrows show the required order; the text labels remain the complete interpretation in print, dark mode, and nonvisual reading.

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Common errors

  • Transforming the formula but forgetting to transform the interval, endpoints, factorial, or remainder.

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Supporting steps