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Taylor series

Taylor coefficients come from derivatives at a point; equality of a function to its series requires additional convergence and remainder conditions.

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Different ways to understand it

• Taylor Series and Maclaurin Series - Calculus 2

  The Organic Chemistry Tutor · English · Explanation
  Connection evidence

  For ln x around $x=1$, derivative values give ($x-1$)−($x-1$)²/2+($x-1$)³/3−⋯. Equality with the function holds for |$x-1$|<1; endpoints require separate checking. At zero, the even-power coefficients of sine vanish and the odd-power terms alternate: $x-x$³/3!+x⁵/5!−⋯. This series converges to sine for every real x. Termwise differentiation gives the cosine series; replacing x by x² then gives $1-x$⁴/2!+x⁸/4!−⋯ for cos(x²). These operations are justified within the power series’ convergence range.

• Intro to Taylor Series: Approximations on Steroids

  Dr. Trefor Bazett · English · Explanation
  Connection evidence

  This 155-second introductory calculus segment motivates Taylor series by approximating $e^x$ at $x=0$. It begins with the numerical question of how to evaluate $e^{0.2}$, then builds a sequence of increasingly accurate polynomial approximations: the constant $y=1$, the tangent line $y=1+x$, the quadratic $y=1+x+\frac{x^2}{2}$, and finally the cubic $y=1+x+\frac{x^2}{2}+\frac{x^3}{6}$. Animated vertical gap markers visualize approximation error, showing why higher-degree terms improve the local fit. This introductory calculus segment motivates Taylor series by asking how to build polynomial approximations, whether they are optimal, and how accurate they are. Using the graph of $y=e^x$ together with lower- and higher-degree polynomial approximations, the lecturer observes that increasing degree improves the local fit near $x=0$. He then reviews linearization: the tangent line must match both the function value and the slope at the base point. Working with a generic line $c_0+c_1x$, he substitutes $x=0$ to get $c_0=1$, differentiates to match slopes and obtains $c_1=1$, concluding $e^x \approx 1+x$. Finally he generalizes this to $f(x) \approx f(0)+f$'(0)x and signals the next step toward quadratic approximations. This 155-second calculus lecture segment develops a quadratic Taylor-style approximation for $e^x$ at $x=0$ by matching value, slope, and concavity. The board and narration determine $c_0=1$, $c_1=1$, and $c_2=1/2$, yielding $e^x\approx 1+x+\frac12x^2$. The speaker then generalizes the method to a power series $f(x)=c_0+c_1x+c_2x^2+c_3x^3+\cdots$, displaying the first and second term-by-term derivatives before the clip ends. This 155-second lecture segment derives the coefficient formula for a power series centered at 0 by differentiating term by term, observing that the nth derivative leaves the constant contribution n!·$c_n$, and then substituting $x=0$ to isolate $c_n = f^{(n)}(0)/n$!. It rewrites the first few coefficients in unified factorial notation, states the Maclaurin series definition, generalizes the same pattern to an arbitrary center a to obtain the Taylor series formula $c_n = f^{(n)}(a)/n$!, and begins applying the result to $f(x)=e^x$ with $a=0$. The clip ends during the setup of that example, before the simplified series for $e^x$ is displayed. This 143-second excerpt explains Taylor series through one algebraic example and one graphical example. First, the instructor states that if $f(x)$=sum $c_n(x-a)^n$, then $c_n=f^{(n)}(a)/n$!, and applies this to $f(x)=e^x$ with $a=0$. Because every derivative of $e^x$ is $e^x$ and $e^0=1$, the series simplifies to $e^x$=sum $x^n/n$!. The second half graphs $\cos (x)$ against its first two Taylor polynomials at $a=0$, namely 1 and $1-x^2/2$, to show that the approximation is strong near the center but poor far away. The clip then shifts the center to a=pi, producing a new quadratic approximation -1+(x-pi)^$2/2$, reinforcing that Taylor series encode local information at a chosen point.

• Best Explanation of Taylor Series on Internet!

  Brain Station Advanced · English · Explanation
  Connection evidence

  A first-order approximation near p is $f(p)+f$′(p)($x-p$). It matches both the function value and slope at the center. Adding f″(p)($x-p$)²/2! matches the second derivative as well. Further coefficients use higher derivatives divided by the corresponding factorials. At $p=0$ this is the Maclaurin construction. For sine it yields $x-x$³/3!+x⁵/5!−⋯; in general, matching all derivatives alone does not guarantee that an infinite Taylor series equals the function away from its center.

Knowledge connections

• Series · Prerequisite · Map context
• Derivatives · Prerequisite · Map context

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