fabian s. klinke

S-Tier Acrobatics Calculus That Physicists Do

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S-Tier Acrobatics Calculus That Physicists Do

Source: S-Tier Acrobatics Calculus That Physicists Do, Ron & Math, 1:54, uploaded 2024-03-03, category Mathematics, playlist index 1130.

The 1:54 video starts with a useful memory failure. If the Taylor expansion of (e^x) has gone missing, begin with the property that (e^x) is its own derivative and its own antiderivative. Ron & Math turns that fact into a piece of formal operator algebra and lets the algebra produce the series

ex=1+x+x22!+x33!+⋯ .e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots.

The integral as an operator

The usual antiderivative statement is written as

ex=∫ex dx.e^x = \int e^x\,dx.

The video drops the bounds and then drops (dx), treating the integral sign as an operation that acts on a function. Moving the resulting term to the left gives

(1−∫)ex=0.(1 - \int)e^x = 0.

The next move treats (1-\int) like an algebraic object with an inverse:

ex=(1−∫)−10.e^x = (1 - \int)^{-1}0.

The inverse has the familiar geometric expansion

(1−∫)−1=1+∫+∫2+∫3+⋯ .(1 - \int)^{-1} = 1 + \int + \int^2 + \int^3 + \cdots.

That makes the answer a repeated-integration problem. Applying the expansion to zero gives

ex=0+∫0+∫20+∫30+⋯ .e^x = 0 + \int 0 + \int^2 0 + \int^3 0 + \cdots.

The first term is zero. The second term needs the video’s change in viewpoint. Here the integral sign names an operation. A definite integral fixes a value from a zero integrand. An antiderivative of zero is a constant. Ron & Math chooses that constant as (1), which gives the next terms by repeated integration:

∫0=1,\int 0 = 1, ∫20=∫1=x,\int^2 0 = \int 1 = x,

and

∫30=∫x=x22!.\int^3 0 = \int x = \frac{x^2}{2!}.

The same step continues. Integrating (x^2/2!) gives (x^3/3!), and each further integration adds a power of (x) together with the next factorial. The geometric expansion therefore recovers the Taylor series of (e^x) without starting from Taylor’s theorem.

The freedom inside the notation

The calculation works because the video uses integration in the loose physics convention that the opening joke announces. An indefinite integral carries an arbitrary constant, and the choice (\int 0 = 1) supplies the first non-zero term. The inverse of (1-\int) also needs a defined space of functions and a convention for which antiderivative the operator returns. An infinite geometric expansion needs its own convergence conditions.

Ron & Math flags this at the end by promising proper definitions, then says that the preceding calculation remains legitimate in a calculus class, provided the physics professor has approved it. The caption track records the qualification without spelling out a base point, operator domain, or convergence argument. The video therefore establishes a compact formal derivation and a way to remember the series. Its strict proof status depends on definitions that the source leaves outside the one-minute demonstration.

Further reading / references

  • Professor Stewart’s Casebook of Mathematical Mysteries, the book named in the video description as the inspiration for the example.

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