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Multiply the Taylor series $\cos x=1-x^2/2+O(x^4)$ by $(1-x{)}^{-1}=1+x+x^2+O(x^3)$. Counting the constant as the first nonzero term gives

Write $\cos x=1+u$ with $u=-x^2/2+x^4/24-x^6/720+O(x^8)$. Substitution into $\ln (1+u)=u-u^2/2+u^3/3+O(u^4)$ givesThus the first three nonzero terms of this even Taylor series areThe real logarithm is well defined near zero because $\cos x>0$ there.

In the Cauchy product, $a_j{x}^j$ combines with $b_{i-j}{x}^{i-j}$ to give a term of degree $i$. HenceThis coefficient identity holds for formal power series. For convergent Taylor series it also gives the function product throughout the common open interval of convergence, where absolute convergence justifies regrouping.

The Taylor series $\sin x=x-x^3/6+O(x^5)$ gives ${\sin }^{2}x=x^2(1-x^2/3+O(x^4))$. Taking its reciprocal and cancelling the singular term givesThus the apparent singularity at zero is removable, and the first nonzero term is the constant $1/3$.

The Taylor series of the sine and arctangent give $\sin x-\arctan x=x^3/6+O(x^5)$, while $x^2\ln (1+x)=x^3(1-x/2+O(x^2))$. Dividing the two series givesThe first nonzero term after cancellation is the constant $1/6$.

Factor the numerator as $(e^x-1{)}^2=x^2+O(x^3)$. The Taylor series of the cosine gives $\cos 3x-2\cos 2x+\cos x=(-9/2+4-1/2)x^2+O(x^4)=-x^2+O(x^4)$. HenceThe first nonzero term is the constant $-1$.

Taylor's theorem expands about $a$ in powers of $h=x-a$. With the usual differentiability assumptions, its finite form with Lagrange remainder iswhere $\xi$ lies between $a$ and $x$. If the remainder tends to zero as $n\to \infty$, this yields the Taylor series $f(x)={\sum }_{k\ge 0}{f}^{(k)}(a)(x-a{)}^k/k!$. Smoothness alone does not guarantee that infinite-series equality.

Put $h=x-1$. The Taylor series of the natural logarithm gives the first four nonzero terms:The corresponding infinite series converges for $|h|<1$.

The lower-order terms cancel, so the numerator must be expanded through sixth order. The Taylor series give $\sin (\sin x)=x-x^3/3+x^5/10+O(x^7)$ and ${\sin }^{2}x=x^2-x^4/3+2x^6/45+O(x^8)$. ThereforeThe first nonzero term is $x^2/18$, with the quotient continuously extended to zero if needed.

For $|t|<1$, the geometric series gives $(1+t{)}^{-1}={\sum }_{n\ge 0}(-t{)}^n$. Its uniform convergence on every closed interval strictly inside $(-1,1)$ permits termwise integration from zero to $x$. ThereforeSubtracting cancels every even power and doubles every odd power, giving the Taylor series

The first four terms form the cubic Taylor polynomial about $x_0$:With a continuous fourth derivative nearby, Taylor's theorem gives a remainder $O((x-x_0{)}^4)$. Continuous derivatives through third order suffice for the weaker remainder $o((x-x_0{)}^3)$. These are local approximations; equality with an infinite Taylor series requires additional convergence information.

The cubic Taylor polynomial about $a$ isIf $f$ has a continuous fourth derivative near $a$, Taylor's theorem gives $f(x)$ equal to this polynomial plus $O((x-a{)}^4)$. With continuous derivatives only through third order, the corresponding remainder is $o((x-a{)}^3)$. An equality with an infinite Taylor series would require additional convergence information, rather than just the existence of these four terms.

Factor $1+e^x=2e^{x/2}\cosh (x/2)$. This givesThe last function is even. Its Taylor series begins $\ln \cosh (x/2)=x^2/8+O(x^4)$, since $\cosh (x/2)=1+x^2/8+O(x^4)$ and $\ln (1+w)=w+O(w^2)$. ThereforeIn particular, the cubic coefficient is zero; the identity explains its cancellation before any lengthy differentiation.

Near $x=2$, use $f(x)=2\ln x$, so $f^{\prime }(2)=1$ and $f^{\prime \prime }(2)=-1/2$. The Taylor series therefore beginsThe constant term is one of the requested three terms. The local expression $2\ln x$ is legitimate because this neighbourhood has $x>0$.

Put $h=x-\pi /4$. The trigonometric identities give $\cos x\sin x=\frac{1}{2}\sin 2x=\frac{1}{2}\cos 2h$. Expanding cosine at zero givesThe linear and cubic terms vanish, so these are the first two nonzero Taylor series terms.

With sufficiently many derivatives, Taylor's theorem gives a finite approximation and a controlled remainder. For example, if $f$ is $N+1$ times continuously differentiable between $a$ and $a+\delta$, thenThus derivatives at the nearby base point determine the polynomial approximation; a bound on the next derivative bounds its error. The smallness of $\delta$ makes this useful, provided the derivatives remain controlled.

The statement's bare assumption of differentiability alone guarantees only $f(a+\delta )=f(a)+f^{\prime }(a)\delta +o(\delta )$. Higher-order formulas need higher regularity. Even smoothness does not by itself guarantee equality with an infinite Taylor series; that stronger conclusion requires the remainder to tend to zero.

Multiply the standard Taylor series through quadratic order:The linear term is $x$ and the quadratic coefficient is $1-1/2$, hence

For $n\ge 3$, Taylor's theorem after $n$ terms, counting the constant term, isUnder the usual hypothesis that $f$ is $n-1$ times continuously differentiable on the interval joining $a$ and $x$, and $n$ times differentiable in its interior, the Taylor theorem with Lagrange remainder givesThis is a finite expansion with a controlled remainder; it does not assume an infinite Taylor series converges. For $x=a$ the remainder is zero.

The Taylor series of the logarithm givesThe first nonzero term is therefore $\boxed{x^2}$. It is positive on either side of the origin, so the stationary point is a strict local minimum. In fact $\log (1+x^2)\ge 0$, with equality only at zero, making it the global minimum as well.

In a multivariable Taylor series, order means total degree. The only nonzero monomial of total degree at most three is $xy$; there are no constant, linear, or cubic terms. Thusnear the origin.