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Taylor series | Chapter 11, Essence of calculus thumbnail

Taylor series | Chapter 11, Essence of calculus

3Blue1Brown·
6 min read

Based on 3Blue1Brown's video on YouTube. If you like this content, support the original creators by watching, liking and subscribing to their content.

TL;DR

Taylor series convert derivative information at a single point into polynomial approximations that match the function’s value and derivatives there.

Briefing

Taylor series turn local derivative information at a single point into accurate polynomial approximations nearby—often so accurate that, when enough terms are added, the approximation becomes an infinite series equal to the original function. That “derivatives at one point → values near that point” pipeline matters because it replaces hard-to-handle functions (like trig and logs) with polynomials that are easier to compute, differentiate, and integrate, which is why Taylor methods show up across physics and engineering.

The core example builds a quadratic approximation to cos(x) near x = 0. Start with the most basic requirement: the polynomial must match the function’s value at the expansion point. Since cos(0) = 1, the constant term must be c0 = 1. Next comes the slope. The derivative of cos(x) is −sin(x), which is 0 at x = 0, so the quadratic’s linear coefficient must be chosen so the approximation also has a flat tangent there. That forces c1 = 0. Finally, the curvature: cos(x) has a negative second derivative at x = 0 because the second derivative of cos(x) is −cos(x), giving −1 at the origin. Matching the quadratic’s second derivative (which equals 2c2) yields c2 = −1/2. The result is the small-angle approximation cos(x) ≈ 1 − x^2/2, which already predicts cos(0.1) extremely well.

The same logic scales to higher-degree polynomials by matching higher derivatives. Adding a cubic term c3x^3 would require the third derivative at x = 0 to match that of cos(x). But the third derivative of cos(x) is sin(x), which equals 0 at x = 0, so c3 must be 0—meaning the quadratic is also the best possible cubic approximation. Adding a quartic term works differently: the fourth derivative of cos(x) is cos(x), which equals 1 at x = 0, so the coefficient must be 1/24. That produces 1 − x^2/2 + x^4/24, a polynomial that stays very close to cos(x) for small x.

A key structural takeaway is why factorials appear. When taking n derivatives of x^n, power-rule cascades produce factors 1·2·…·n, so the coefficient of x^n must be the nth derivative value divided by n! to “cancel out” that growth. Another crucial property: adding higher-order terms doesn’t disturb the lower-order matching at x = 0, because every extra term contains powers of x that vanish when evaluating derivatives at the expansion point. This is why each derivative order at x = 0 is controlled by exactly one coefficient.

General Taylor polynomials formalize the pattern: for an expansion about x = a, the coefficient of (x − a)^n is f^(n)(a)/n!. The video then connects the second-order term to geometry via the fundamental theorem of calculus by approximating how the area-under-a-curve function changes, where the quadratic term corresponds to a triangular area involving the second derivative. Stopping after finitely many terms gives Taylor polynomials; adding infinitely many terms gives Taylor series. Some series converge to the function everywhere (e^x, and also sine/cosine), while others converge only within a limited interval (ln x around x = 1), defining a radius of convergence. The unifying intuition is that derivative data at one point can be “propagated” into an approximation around that point—sometimes indefinitely, sometimes only up to a boundary.

Cornell Notes

Taylor series build polynomials that match a function’s value and successive derivatives at a chosen point. For cos(x) near x = 0, matching cos(0)=1 fixes the constant term, matching the flat slope (−sin(0)=0) fixes the linear term, and matching curvature (second derivative −cos(0)=−1) fixes the x^2 coefficient, giving cos(x) ≈ 1 − x^2/2. Extending to higher orders works by matching higher derivatives; factorials appear because differentiating x^n repeatedly produces factors 1·2·…·n, so coefficients must use f^(n)(a)/n!. Adding infinitely many terms yields a Taylor series, which may converge everywhere (e^x, sine, cosine) or only within a radius of convergence (ln x about 1).

How does matching derivatives at x = 0 determine the coefficients of a quadratic approximation to cos(x)?

Assume a quadratic c0 + c1 x + c2 x^2. Value matching: cos(0)=1 forces c0=1. Slope matching: (cos x)' = −sin x, and −sin(0)=0, so the quadratic’s derivative c1 + 2c2 x must equal 0 at x=0, giving c1=0. Curvature matching: (cos x)'' = −cos x, so at x=0 it equals −1. The quadratic’s second derivative is 2c2, so 2c2 = −1, hence c2 = −1/2. The approximation becomes 1 − x^2/2.

Why do factorials show up in Taylor series coefficients?

Differentiating x^n repeatedly produces a cascading product. After n derivatives, the factor becomes 1·2·3·…·n = n!. So if the nth derivative of the target function at the expansion point is f^(n)(a), the coefficient of (x−a)^n must be f^(n)(a)/n! to counteract that growth. In the cos example, the fourth derivative at 0 is 1, but the polynomial’s fourth derivative of c4 x^4 is 24c4, so c4 must be 1/24 = 1/4!.

Why doesn’t adding a higher-order term ruin the earlier matching at x = 0?

Because higher-order terms contain powers of x. When evaluating derivatives at x=0, any term with an extra factor of x vanishes after enough differentiation. More specifically, the derivative of order k of a term like x^m at x=0 is zero whenever m>k, so each coefficient is effectively “responsible” for exactly one derivative order at the expansion point. That’s why introducing an x^4 term doesn’t change the second derivative at 0.

What changes when the expansion point is x = a instead of 0?

The polynomial must be written in powers of (x−a) so that plugging in x=a plays the same role as plugging in x=0 did before. Then the coefficient of (x−a)^n becomes f^(n)(a)/n!, ensuring the value and all matched derivatives occur at x=a. Without shifting to (x−a), the cancellation that makes derivative matching clean would be lost.

How does the fundamental theorem of calculus connect to the second-order term?

Let A(x) be the area under a graph from a fixed left point to a variable right point x. The graph of A'(x) is the original function (height). When x moves from a to x, the change in area can be approximated by a rectangle plus a small triangular correction. The triangle’s area is (1/2)·(x−a)·(slope of the height function at a)·(x−a), which becomes (1/2)A''(a)(x−a)^2. That matches the quadratic term in the Taylor polynomial for A(x).

Why do some Taylor series converge only within a limited interval?

Convergence depends on how the infinite sum behaves as more terms are added. For e^x, the Taylor series converges to e^x for any x, so the series equals the function everywhere. For ln x expanded around x=1, the series converges only when x stays within a certain range (between 0 and 2 in the described example). Outside that range, adding more terms makes the partial sums oscillate wildly instead of approaching ln x, so the series diverges. The boundary distance defines the radius of convergence.

Review Questions

  1. For cos(x) near x=0, which derivative order fixes the x^2 coefficient, and what value does that derivative take at x=0?
  2. In general Taylor polynomials about x=a, what is the formula for the coefficient of (x−a)^n, and why does n! appear?
  3. What does the radius of convergence mean in terms of where the Taylor series partial sums approach the target function?

Key Points

  1. 1

    Taylor series convert derivative information at a single point into polynomial approximations that match the function’s value and derivatives there.

  2. 2

    For cos(x) near x=0, enforcing value, slope, and curvature matching yields cos(x) ≈ 1 − x^2/2, a small-angle approximation used in physics.

  3. 3

    Higher-degree Taylor polynomials come from matching higher derivatives; in the cos example, the cubic coefficient is forced to 0 because the third derivative at 0 is sin(0)=0.

  4. 4

    Factorials arise because differentiating x^n repeatedly produces a factor of n!; coefficients must divide by n! to align with the desired derivative values.

  5. 5

    Adding higher-order terms doesn’t disturb lower-order matching at the expansion point because powers of (x−a) vanish appropriately when evaluating derivatives at x=a.

  6. 6

    Taylor series may converge everywhere (e^x, sine, cosine) or only within a radius of convergence (ln x about x=1), where partial sums stop approaching the function outside the interval.

Highlights

Matching cos(x) at x=0 requires c0=1, c1=0, and c2=−1/2, producing the classic approximation cos(x) ≈ 1 − x^2/2.
The coefficient of x^n in a Taylor polynomial is f^(n)(a)/n!, because n derivatives of x^n generate the product 1·2·…·n = n!.
Higher-order terms don’t break earlier derivative matching at the expansion point since derivatives of extra powers vanish at x=a.
The second-order term can be visualized geometrically as a triangular area correction involving (1/2)·f''(a)(x−a)^2, tied to the fundamental theorem of calculus.
Taylor series convergence is not guaranteed: ln x around x=1 converges only within a finite interval, defining a radius of convergence.

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