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The Maclaurin series of the hyperbolic cosine is $\cosh x=1+x^2/2+x^4/24+O(x^6)$. The binomial series with exponent $-1/2$ gives $(1+u{)}^{-1/2}=1-u/2+3u^2/8+O(u^3)$. Substituting $u=x^2/2+x^4/24+O(x^6)$ yieldsThese are the first three nonzero terms; the remainder is written in Big O notation.

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 $x\to 0$.

For the odd extension, the cosine integral vanishes and the sine integral doubles. Integration by parts givesThe value at $k=0$ is zero by the odd function symmetry, or by the Maclaurin series of the sine. The transform is purely imaginary and odd.

Write the Maclaurin series coefficients as $a_n=f^{(n)}(0)/n!$. When the successive ratios exist eventually and have a limit, the ratio test gives the displayed radius of convergence, including zero and infinity. The unconditional formula is $R^{-1}={\mathrm{limsup}}_{n\to \infty }|a_n{|}^{1/n}$, from the Cauchy-Hadamard theorem. Endpoints must be checked separately.

The existence and convergence of this power series do not by themselves imply that it equals $f$: equality requires analyticity or control of the Taylor remainder. The smooth function $f(x)=e^{-1/x^2}$ for $x\ne 0$, $f(0)=0$, has all derivatives zero at zero, so its Maclaurin series is identically zero despite the nonzero values away from zero.

For a function with derivatives of every order at $a$, its Taylor series is the power seriesIts partial sums are Taylor polynomials. At $a=0$ it is called a Maclaurin series. Forming the coefficients, proving that its partial sums form a convergent sequence, and proving that their limit equals $f(x)$ are separate tasks. A smooth function provides all the coefficients, but a real analytic function is additionally represented locally by the resulting power series.

For the exponential function, every derivative of $e^x$ at zero equals one, givingHere equality holds for every real $x$: the Taylor remainder is bounded by $e^{|x|}|x{|}^{N+1}/(N+1)!$, which tends to zero for each fixed $x$. At $x=1$, the degree-five Taylor polynomial is $163/60\approx 2.71666667$, and the absolute difference from $e$ is less than $e/720$. The factorial denominator eventually overwhelms every fixed power, allowing local derivative information to reconstruct this function throughout the real line.

The geometric series shows why the region of validity matters. For $f(x)=1/(1-x)$, the Taylor series at zero is ${\sum }_{k\ge 0}{x}^k$, with radius of convergence one. For $x\ne 1$, its finite Taylor polynomial satisfies the exact identityAt $x=1/2$ and $N=5$, the remainder is $1/32$. At $x=0.9$, the same degree-five polynomial has value $4.68559$ and remainder $5.31441$, although the Taylor series still converges to $10$ there. Convergence of the infinite expression does not guarantee accuracy from a particular small number of terms. For $|x|>1$, even the terms fail to tend to zero, so this power series cannot represent the function. The standard expansions and the conditions for convergent series are discussed in OpenStax, working with Taylor series.

A stronger caution comes from the flat functionIt is a smooth function with $h^{(n)}(0)=0$ for every $n$, so every Taylor polynomial at zero is zero and its Taylor series converges everywhere to zero. Nevertheless $h(1/2)=e^{-4}>0$. Off zero, repeated differentiation produces a polynomial in $1/x$ times $e^{-1/x^2}$; the exponential decay makes each required limit at zero vanish. This explains how all the local coefficients can miss a positive function arbitrarily close to the centre. Real analyticity supplies the additional condition that rules out this behavior.