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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.

Use $2{\cos }^{2}x=1+\cos 2x$ to reduce the equation to $\sin 2x=\cos 2x$. Equivalently, $\sqrt{2}\sin (2x-\pi /4)=0$, so $x=\pi /8+k\pi /2$ for an integer $k$. Restricting to $[0,\pi ]$ gives

The known root gives a factor $x-1$ by the factor theorem. Polynomial division leaves $2x^2-7x+6=(2x-3)(x-2)$, soConsequently all the roots areThe three linear factors account for the full degree of the polynomial, so there are no further complex roots.

The inequality is equivalent to $|\cos \theta |>1/\sqrt{2}$. The cosine exceeds $1/\sqrt{2}$ close to zero and $2\pi$, and is less than $-1/\sqrt{2}$ close to $\pi$. Within the specified interval this givesThe four equality angles are excluded because the inequality is strict; zero and $2\pi$ satisfy it and are included.

The logarithm must be expanded about two, not about one. FactorApplying the supplied logarithm expansion to the bracket gives $\log 2+(x/2+x^2/4)-\frac{1}{2}(x/2{)}^2+O(x^3)$. Thus, through second order,In fact the cubic coefficient vanishes, since $\log (1+e^x)=x/2+\log [2\cosh (x/2)]$ and the second term is even; the Maclaurin series remainder can be sharpened to $O(x^4)$.

At $x=1$, the natural logarithm has value zero and derivative $1/x=1$. Its tangent line is therefore

Expanding gives $u=2x^2+2c^2{t}^2$, so $u_{tt}=4c^2$ and $u_{xx}=4$. Hence $u_{tt}=c^2{u}_{xx}$, proving that it solves the one-dimensional wave equation. Equivalently its two summands are oppositely travelling profiles, as in D'Alembert's formula.

Factorization gives $y=(x-2{)}^2(x+1)$, so the zeros are $x=-1$ and $x=2$, with a double root at two. The derivatives are $y^{\prime }=3x(x-2)$ and $y^{\prime \prime }=6(x-1)$. ThusThe second derivative changes sign at one, confirming the inflection point. The curve rises from $-\infty$, crosses at $-1$, reaches its local maximum, falls to touch the axis at two, then rises to $+\infty$. Neither local extremum is a global extremum on the whole real line.

Zero is an eigenvalue exactly when the determinant vanishes. Here $\det A=1-{\alpha }^2$, soFor either value the other eigenvalue is two, as follows from the trace.

The gradients are $\nabla f=(1,1,1)$ and $\nabla g=(2x,2y,2z)$. Equality of their three components therefore gives the unique point

For $\lambda \ge 0$, the exponential power series gives the normalization of the Poisson distribution:For $\lambda =0$, the limiting law assigns probability one to $k=0$ and zero to every positive integer $k$. Negative $\lambda$ would not define the stated probability distribution.

If $y\ne 0$, the equation would imply $b=-(x/y)a$, contradicting the assumed nonparallel directions. Hence $y=0$. Then $xa=0$ and $a\ne 0$ force $x=0$. This proves that the two vectors are linearly independent.

The line has equation $2x-y+2=0$, with normal vector $(2,-1)$. The distance from a point to a line is the absolute value of the normal component of the displacement:The perpendicular foot is $(4,-6)-(16/5)(2,-1)=(-12/5,-14/5)$, which lies on the line and realizes this shortest distance.