nLab circumference of a circle





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infinitesimal cohesion

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id id fermionic bosonic bosonic Rh rheonomic reduced infinitesimal infinitesimal & étale cohesive ʃ discrete discrete continuous * \array{ && id &\dashv& id \\ && \vee && \vee \\ &\stackrel{fermionic}{}& \rightrightarrows &\dashv& \rightsquigarrow & \stackrel{bosonic}{} \\ && \bot && \bot \\ &\stackrel{bosonic}{} & \rightsquigarrow &\dashv& \mathrm{R}\!\!\mathrm{h} & \stackrel{rheonomic}{} \\ && \vee && \vee \\ &\stackrel{reduced}{} & \Re &\dashv& \Im & \stackrel{infinitesimal}{} \\ && \bot && \bot \\ &\stackrel{infinitesimal}{}& \Im &\dashv& \& & \stackrel{\text{étale}}{} \\ && \vee && \vee \\ &\stackrel{cohesive}{}& \esh &\dashv& \flat & \stackrel{discrete}{} \\ && \bot && \bot \\ &\stackrel{discrete}{}& \flat &\dashv& \sharp & \stackrel{continuous}{} \\ && \vee && \vee \\ && \emptyset &\dashv& \ast }


Lie theory, ∞-Lie theory

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The circumference of a circle (the length of the curve that is the circle) as discussed in geometry.

Definition/proposition and proof


Depending on which circle constant you use, given a radius rr of a circle 𝒞\mathcal{C} in the Euclidean plane 2\mathbb{R}^2, the circumference of a circle is either C(r)=τrC(r) = \tau r or C(r)=2πrC(r) = 2 \pi r.

Proof by integration


In this proof, we are using the circle constant τ=2π\tau = 2 \pi.

Given any Euclidean plane 2\mathbb{R}^2, one could select an orthonormal basis on 2\mathbb{R}^2 by postulating an origin 00 at the center of the circle 𝒞\mathcal{C} and two orthonormal vectors i^\hat{i} and j^\hat{j}. The circle 𝒞\mathcal{C} could be parameterized by a function r:[0,τ] 2\overrightarrow{r}:[0, \tau] \to \mathbb{R}^2 defined as

r(θ)rcos(θ)i^+rsin(θ)j^\overrightarrow{r}(\theta) \coloneqq r \cos(\theta) \hat{i} + r \sin(\theta) \hat{j}

Then the circumference of 𝒞\mathcal{C} is given by the following integral:

C(r)= 0 τ|r(θ)|dθC(r) = \int_{0}^{\tau} \vert \overrightarrow{r}(\theta) \vert d \theta

which evaluates to

C(r)= 0 τ|rcos(θ)i^+rsin(θ)j^|dθ= 0 τr((cos(θ)) 2+(sin(θ)) 2)dθ= 0 τrdθ=τrC(r) = \int_{0}^{\tau} \vert r \cos(\theta) \hat{i} + r \sin(\theta) \hat{j} \vert d \theta = \int_{0}^{\tau} r((\cos(\theta))^2 + (\sin(\theta))^2) d \theta = \int_{0}^{\tau} r d \theta = \tau r

Proof by limits of regular polygons


In this proof, we are using the circle constant τ=2π\tau = 2 \pi.

The perimeter of a regular polygon 𝒫 n\mathcal{P}_n (strictly speaking, a regular polygonal line) with nn sides and circumradius rr is given by the sequence of functions P 𝒫:()P_\mathcal{P}:\mathbb{N} \to (\mathbb{R} \to \mathbb{R})

P 𝒫(n)(r)=r(2n)sin(τ2n)P_\mathcal{P}(n)(r) = r (2 n) \sin\left(\frac{\tau}{2n}\right)

which embeds in the +\mathbb{R}_+-action P 𝒫 : +()P_\mathcal{P}^\prime:\mathbb{R}_+ \to (\mathbb{R} \to \mathbb{R}), defined as

P 𝒫 (n)(r)=r(2n)sin(τ2n)P_\mathcal{P}^\prime(n)(r) = r (2 n) \sin\left(\frac{\tau}{2 n}\right)

The limit of P P^\prime as nn goes to infinity is the circumference of a circle with radius rr:

C(r)=lim nP 𝒫 (n)(r)=lim nr(2n)sin(τ2n)=rlim m0sin(τm)m=rlim m0 msin(τm) mm=rlim m0τcos(τm)1=τrC(r) = \lim_{n \to \infty} P_\mathcal{P}^\prime(n)(r) = \lim_{n \to \infty} r (2 n) \sin\left(\frac{\tau}{2 n}\right) = r \lim_{m \to 0} \frac{\sin(\tau m)}{m} = r \lim_{m \to 0} \frac{\partial_m \sin(\tau m)}{\partial_m m} = r \lim_{m \to 0} \frac{\tau \cos(\tau m)}{1} = \tau r

See also

Last revised on May 17, 2022 at 01:02:20. See the history of this page for a list of all contributions to it.