For $\mathbf{r}(t)=(t^2,t{)}^T$, the tangent vector is $(2t,1{)}^T$ and $\mathbf{F}(\mathbf{r}(t))=(t^3,t^4+t{)}^T$. ThusThe unequal values for the same endpoints already rule out a conservative vector field.

With $\mathbf{r}(t)=(t,t^2{)}^T$, the tangent vector is $(1,2t{)}^T$ and $\mathbf{F}(\mathbf{r}(t))=(t^3,2t^2{)}^T$. The line integral is

In the nonsingular case the unique solution isGeometrically this is a single selected point of the straight line through $\mathbf{a}$ parallel to $\mathbf{c}$.

When $\mathbf{b}\cdot \mathbf{c}=1$ and $\mathbf{b}\cdot \mathbf{a}=0$, the solution set is that entire straight line, $\mathbf{a}+\lambda \mathbf{c}$. The remaining case has empty solution set. In matrix terms the equation is $(I-\mathbf{c}{\mathbf{b}}^T)\mathbf{x}=\mathbf{a}$, a rank-one matrix update of the identity matrix.

Substitute $\mathbf{x}=\mathbf{a}+\lambda \mathbf{c}$ into the original equation and cancel $\mathbf{a}$. Because $\mathbf{c}$ is nonzero, the scalar condition isIf $\mathbf{b}\cdot \mathbf{c}\ne 1$, there is one value, $\lambda =(\mathbf{b}\cdot \mathbf{a})/(1-\mathbf{b}\cdot \mathbf{c})$. If $\mathbf{b}\cdot \mathbf{c}=1$ and $\mathbf{b}\cdot \mathbf{a}=0$, every real $\lambda$ works. If $\mathbf{b}\cdot \mathbf{c}=1$ and $\mathbf{b}\cdot \mathbf{a}\ne 0$, the scalar equation is inconsistent and there is no solution.

Thus “undefined” in the last case means that no parameter value solves the equation; it must not be used to conceal the distinct case with infinitely many solutions.

Take the cross product with $\mathbf{c}$. Since $\mathbf{c}\times \mathbf{c}=0$, the equation impliesAs $\mathbf{c}\ne 0$, the displacement is parallel to $\mathbf{c}$. ThereforeThis is a necessary condition; substitution must still determine which points on this straight line satisfy the full equation.

For the positive terms $u_n=2^n/n!$, the consecutive ratio is $u_{n+1}/u_n=2/(n+1)\to 0$, so the ratio test proves convergence. The exponential series gives $e^2={\sum }_{n=0}^{\infty }{2}^n/n!$. Removing the $n=0$ term yields

The summands repeat the six-term patternIn particular they do not tend to zero: the subsequence with $n=6k+1$ is always $\sqrt{3}/2$. The term test for divergence therefore shows that the series diverges. Although each block of six sums to zero, that only makes a subsequence of partial sums zero; it does not make the full sequence of partial sums convergent.

The inner geometric series has common ratio $2/3$, soThe outer geometric series therefore converges as well, with valueAll summands are nonnegative, so there is no cancellation or conditional-convergence issue in evaluating the nested sums.

For every integer $n\ge 1$, $\sin ((2n-1)\pi )=0$ and $\cos ((2n-3)\pi )=-1$. Thus the summand simplifies to $2/n$. The resulting series is twice the harmonic series, so it diverges to $+\infty$.

Start with the supplied Gaussian integral $I(0)=\sqrt{\pi }/2$ and part (i)'s $I(1)=1/2$. Repeated use of the Gaussian moment recurrence givesThe even indices inherit the $\sqrt{\pi }$ factor from $I(0)$, while the odd indices inherit rational values from $I(1)$.

For $n\ge 2$, write $x^n{e}^{-x^2}=x^{n-1}(x{e}^{-x^2})$ and apply integration by parts, using the antiderivative $-e^{-x^2}/2$ for the second factor:The boundary term is zero: at zero, $n-1>0$, and at infinity Gaussian decay dominates every power. Consequently the Gaussian moment recurrence is

Use the substitution rule $u=x^2$, $du=2xdx$. The convergent improper integral is

The difference and sum of the two antiderivatives in part (i) giveandTheir derivatives also verify the signs immediately.

Write $I=\int e^x\sin xdx$ and $J=\int e^x\cos xdx$. Two applications of integration by parts, first taking the sine and then the cosine as the differentiated factor, giveThus $I+J=e^x\sin x$ and $J-I=e^x\cos x$, in each case up to an arbitrary constant of integration. Keeping that freedom avoids identifying independently chosen antiderivatives too rigidly.

Here $t_0=0$ and $t_2=t_1^2$, so the constant and quadratic terms vanish, while the linear coefficient remains $t_1\ne 0$. A further derivative givesConsequently $f^{(3)}(0)=t_3-3t_1{t}_2+2t_1^3=t_3-t_1^3$. Since $t_3$ and $t_1$ have opposite signs, $t_3$ and $t_1^3$ also have opposite signs, and their difference is nonzero. The first two nonzero terms are thereforeThis is where the full sign condition $t_3{t}_1<0$ rules out a further cancellation in the Taylor expansion.

The second derivative chain rule givesWith $t_0=0$, the constant term vanishes, $f^{\prime }(0)=t_1\ne 0$, and $f^{\prime \prime }(0)=t_2-t_1^2\ne 0$ by the specified case. Thus the first two nonzero terms in the Taylor expansion are

Write $t_j=t^{(j)}(0)$ and $s=1+t_0>0$. The chain rule givesThe condition $t_3{t}_1<0$ ensures $t_1\ne 0$. Since $t_0\ne 0$, also $s\ne 1$ and $\ln s\ne 0$. Hence the first two nonzero terms are the constant and linear terms:The inequality $s>0$, together with continuity of $t$, makes the logarithm defined on a neighbourhood of zero. The remainder is written in Big O notation.

First express the function in the original variables:The chain rule givesTaking its partial derivative with respect to $y$ yieldsThe numerator in the second fraction reduces to $4xy$, reproducingBoth logarithms are defined for every real $x,y$ because their arguments are at least one.

For $f(u,v)=\ln u+\ln v$, the needed partial derivatives areSubstituting these into part (ii) givesSince $v-x^2{y}^2=1$, the last two terms combine. Thus

Apply the product rule and the chain rule once more to each expression in part (i). The second derivative chain rule givesandFor example, the coefficient of $f_{uv}$ in the mixed partial derivative is $u_x{v}_y+v_x{u}_y=4x^{3}y+4x{y}^3$, while the final term is $v_{xy}{f}_v=4xy{f}_v$. The symmetry $f_{uv}=f_{vu}$ is used in combining the mixed terms under the usual twice-continuously-differentiable interpretation.