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A Euclidean motion has the form $E(x)=Ax+b$ with $A\in O(3)$. It preserves Euclidean distance, angles, and all extrinsic geometric quantities transformed with their orientations.

This is the dimension-three case $E(3)=O(3)⋉{\mathbb{R}}^3$ of the Euclidean group.

For a sufficiently smooth unit-speed curve with $\kappa >0$, the Frenet frame is the positively oriented orthonormal frame

Writing $\mathcal{F}=n_e/n_B$ couples the two equilibrium equations:

For a frame $L$, an $L$-valued matrix $f:A⇝B$ is a function $A\times B\to L$. Composition is matrix multiplication with join as addition and meet as multiplication.

An integral ideal is a fractional ideal of an integral domain $A$ that is contained in $A$. For a number field, take $A$ to be its ring of integers ${\mathcal{O}}_K$.

The cutoffs ${\theta }_n(\xi )=(1-|\xi |/n{)}_{+}$ increase pointwise to one, while their Fourier transforms are nonnegative and have integral $2\pi$ under the angular-frequency convention.

The squared sinc function is nonnegative and integrable; it occurs as the Fourier transform of a triangular function.

It is the pointwise square of the sinc function, with its normalization inherited from that function.

The triangular functionhas angular-frequency Fourier transform $4{\sin }^2(\xi /2)/{\xi }^2$.

For $y^{\prime }=i\omega y$ with real nonzero $\omega$, explicit Euler has amplification factor $1+i\omega h$, whose modulus $\sqrt{1+{\omega }^2{h}^2}$ exceeds one for every $h>0$.

If $C$ is Hermitian positive definite and $D$ is Hermitian, every eigenvalue of $CD$ is real. Equivalently, $CD$ is similar to the Hermitian matrix $C^{1/2}D{C}^{1/2}$.

If a real Fourier polynomial $c(x)$ is strictly positive, then $C_{nm}={\hat{c}}_{n-m}$ is Hermitian positive definite becausefor every nonzero $z$.

The degree-$d$ moment curve is $\gamma (t)=(t,t^2,\dots ,t^d)$. Tangent or radial directions from separated parameter intervals are quantitatively transverse by the Vandermonde determinant.

Kakeya inequalities bound overlaps of long thin tubes with controlled directions.

These tube estimates are used to study Kakeya sets and their dimension.

Constructive interference occurs when many oscillatory phases nearly agree, making their exponential terms add with comparable arguments and producing a large value of the sum.

A function whose Fourier support has widths $N$ in spatial-frequency directions and $N^2$ in a time-frequency direction varies only on the dual scales $N^{-1}$ and $N^{-2}$. Convolution with an adapted Schwartz kernel makes this quantitative.

A decoupling inequality controls an $L^p$ norm of a function with Fourier support near a curved set by an ${\ell }^2$ sum of norms of pieces supported near smaller caps.

For an integrable function and the angular-frequency convention,so a translation multiplies each frequency component by a factor of modulus one. The amplitude of every component stays the same, while its wave phase changes by $-ka$. For example, delaying $\cos (2\pi \nu t)$ by $a$ produces $\cos (2\pi \nu t-2\pi \nu a)$: faster oscillations accumulate a larger wave phase shift for the same delay. Substituting $y=x-a$ in the defining Fourier transform is the entire proof. Conversely, this multiplier in the Fourier transform produces a translation after Fourier inversion.

For the angular-frequency convention,Consequently cosine transforms to the half-sum of the two shifted deltas, while sine transforms to their signed difference divided by $2i$.

For a Schwartz function, with the angular-frequency convention,More generally, the identity holds when integration by parts is justified and the boundary term vanishes. Repeated differentiation gives $\widehat{f^{(m)}}(k)=(ik{)}^m\hat{f}(k)$. The short proof transfers the derivative from $f$ to $e^{-ikx}$ by integration by parts. High frequencies receive larger multipliers, so rapid ripples contribute more strongly to a derivative. For the heat equation, the second derivative becomes multiplication by $-k^2$, directly explaining the faster decay of fine spatial structure.

When $\hat{f}\in L^1$, its inverse Fourier integral is continuous by dominated convergence. If $f$ is also continuous and Fourier inversion identifies the two functions almost everywhere, then they agree everywhere.