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Put $h=x-\pi /2$. The sine addition identity gives $\sin x=\cos h=1-h^2/2+h^4/24+O(h^6)$. Applying the exponential series givesThus the first three nonzero terms in the Taylor series about $\pi /2$ are

Set $\ell =\ln 2$. Applying the exponential series and the Maclaurin series of the sine givesMultiply the two power series, retaining every contribution through degree four:Since $0<\ln 2<1$, the coefficient of $x^4$ is nonzero, so these are indeed the first two nonzero terms. The remainder is in Big O notation.

Use the exponential series to obtain the Maclaurin series of the hyperbolic sine and hyperbolic cosine:Dividing these power series gives the hyperbolic tangent expansionThe displayed remainder uses Big O notation as $x\to 0$.

Picard iteration replaces a tentative trajectory by the accumulated integral of its proposed rate. For an initial value problem $y^{\prime }=f(t,y)$, $y(t_0)=y_0$, define the Picard map and its starting function byHere $y_0(t)$ is the constant starting trajectory; the same symbol without its argument denotes the initial state. A fixed point of $\mathcal{P}$ solves this integral equation, and the fundamental theorem of calculus gives the original ordinary differential equation.

For $y^{\prime }=y$, $y(0)=1$, successive integrals giveAt $t=1$, the first five values are $1,2,2.5,2.66666667,2.70833333$, while $e\approx 2.71828183$. The fourth polynomial misses by about $0.00994850$. On $0\le t\le 1$, every omitted term of the exponential series is nonnegative and increasing, so the largest error occurs at $t=1$. The tail tends to zero, giving uniform convergence to the exponential function on this interval.

The beauty of Picard iteration is that each proposed trajectory supplies a better rate history for the next one. For this example, one integration adds exactly the next term of the exponential series. In general, the integrals may be difficult to compute, but the construction still explains local existence and uniqueness.

The short contraction mapping argument needs two independent controls. Choose a time-state cylinder $[t_0-a,t_0+a]\times {\overline{B}}_r(y_0)\subset U$ on whichThe second bound is the uniform local Lipschitz condition. Set $J=[t_0-h,t_0+h]$ with $0<h\le a$, $Mh\le r$ and $Lh<1$. In the function space of continuous functions $J\to {\mathbb{R}}^d$, use the supremum norm $\Vert u{\Vert }_{\infty }={\sup }_{t\in J}\Vert u(t)\Vert$. On its closed ball $\Vert u-y_0{\Vert }_{\infty }\le r$, direct integral estimates giveThus the map preserves the closed ball and is a contraction mapping there. This closed ball is a complete metric space, so the contraction mapping theorem gives convergence to exactly one fixed point. Any other local solution of a differential equation stays in the same ball after shortening its interval, and hence agrees there.

For $y^{\prime }=y$, take $r=1$, so $|y|\le 2$ in the state ball and $M=2$, $L=1$. Choosing $h=1/2$ gives $Mh=1$ and $Lh=1/2$, with the actual solution satisfying $|e^{1/2}-1|\approx 0.64872127<1$. The plotted interval $[0,1]$ is longer than this simple guarantee: the explicitly computed exponential series converges there anyway. The condition $Lh<1$ is sufficient for this one-step contraction mapping argument, rather than a necessary lifespan bound.

For a real angle $\alpha$ and a unit vector $\mathbf{n}$, the rotation about $\mathbf{n}$ sendsThe vector triple product identity gives the equivalent expression $\mathbf{x}+\sin \alpha \mathbf{n}\times \mathbf{x}+(1-\cos \alpha )\mathbf{n}\times (\mathbf{n}\times \mathbf{x})$. To see why it is a rotation, decompose $\mathbf{x}$ into its parallel and perpendicular components relative to $\mathbf{n}$. The parallel component is fixed, while the perpendicular component becomes $\cos \alpha {\mathbf{x}}_{\perp }+\sin \alpha \mathbf{n}\times {\mathbf{x}}_{\perp }$, the usual plane rotation. It is also the matrix exponential of the linear operator $H\mathbf{x}=\alpha \mathbf{n}\times \mathbf{x}$: the identity $H^3=-{\alpha }^{2}H$ reduces its exponential series to these three terms.

For $a>0$, this integral is an entire function of $x$: the exponential series converges uniformly on bounded subsets after multiplication by the integrable bound $e^{|x|}{u}^{a-1}$. ThusFor positive real $x$, a change of variables gives $F_a(x)=x^{-a}\gamma (a,x)$, where here $\gamma (a,x)={\int }_0^x{e}^{-t}{t}^{a-1}dt$ denotes the lower incomplete gamma function. The singular-looking factor $x^{-a}$ is removed by the integral over $[0,1]$. Applying integration by parts to the derivative of $u^a{e}^{-xu}$ gives $x{F}_a^{\prime }+a{F}_a=e^{-x}$; differentiating and eliminating $e^{-x}$ proves $x{F}_a^{\prime \prime }+(a+1+x)F_a^{\prime }+a{F}_a=0$.

Multiply the two exponential series in $e^{-(x-a{)}^2}=e^{-a^2}{e}^{2ax}{e}^{-x^2}$ to obtain the displayed Taylor expansion. Its first three nonzero terms depend on the real parameter $a$. For $a\ne 0,\pm 1/\sqrt{2}$, they end at degree two. At $a=0$, odd coefficients vanish and the first three terms are $1-x^2+x^4/2$. At $a=\pm 1/\sqrt{2}$, the quadratic coefficient vanishes, leaving $e^{-1/2}(1+2ax-4a{x}^3/3)$ as the first three nonzero terms. Parameter-dependent vanishing coefficients must be checked when a question counts nonzero terms rather than prescribing an expansion order.