For $h \neq 0$ and $x+h$ in the domain of $f$, $$ f(x+h)-f(x)=\frac{f(x+h)-f(x)}{h} \cdot h $$ With $f$ differentiable at $x$, $$ \lim {h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=f^{\prime}(x) $$ Since $\lim {h \rightarrow 0} h=0$, we have $$ \lim {h \rightarrow 0}[f(x+h)-f(x)]=\left[\lim {h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\right] \cdot\left[\lim _{h \rightarrow 0} h\right]=f^{\prime}(x) \cdot 0=0 . $$
Show that if $f(x)=x^{3}$, then $f^{\prime}(x)=3 x^{2}$. Prove by induction that for each positive integer $n$, $f(x)=x^{n} \quad$ has derivative $\quad f^{\prime}(x)=n x^{n-1} .$ HINT: $$ (x+h)^{k+1}-x^{k+1}=x(x+h)^{k}-x \cdot x^{k}+h(x+h)^{k} . $$
Given a series $\Sigma_{m=1}^{m} a_{n}=a_{1}+a_{2}+a_{s}+\cdots$, let $s_{a}$ denote its rth partial sum: $$ s_{n}=\sum_{i=1}^{n} a_{i}=a_{1}+a_{2}+\cdots+a_{n} $$
证明 .
If the sequence $\left{s_{\mathrm{n}}\right}$ is convergent and $\lim {\mathrm{a} \rightarrow \mathrm{m}} s{\mathrm{a}}=s$ exists as a real number, then the series $\Sigma a_{n}$ is called convergent and we write $$ a_{1}+a_{2}+\cdots+a_{n}+\cdots=s \quad \text { or } \quad \sum_{n=1}^{\infty} a_{n}=s $$ The number $s$ is called the sum of the series. Otherwise, the series is called divergent.
If $r=1$, then $s_{n}=a+a+\cdots+a=n a \rightarrow \pm \infty$. Since $\lim {n \rightarrow-} s{n}$ doesn’t exist, the geometric series diverges in this case. If $r \neq 1$, we have $$ \begin{aligned} &s_{\mathrm{a}}=a+a r+a r^{2}+\cdots+a r^{\mathrm{n}-1} \ &r s_{\mathrm{a}}=\quad a r+a r^{2}+\cdots+a r^{\mathrm{n}-1}+a r^{\mathrm{n}} \end{aligned} $$ Subtracting these equations, we get $$ \begin{array}{r} s_{\mathrm{a}}-r s_{\mathrm{a}}=a-a r^{n} \ s_{\mathrm{n}}=\frac{a\left(1-r^{\mathrm{n}}\right)}{1-r} \end{array} $$ If $-1<r<1$, we know from $(11.1 .9)$ that $r^{n} \rightarrow 0$ as $n \rightarrow \infty$, so $$ \lim {n \rightarrow \infty} s{n}=\lim {n \rightarrow \infty} \frac{a\left(1-r^{n}\right)}{1-r}=\frac{a}{1-r}-\frac{a}{1-r} \lim {n \rightarrow \infty} r^{n}=\frac{a}{1-r} $$
This is $$ \left[\left[\begin{array}{c} R \circ g \ -Q \circ g \ P \circ g \end{array}\right] \cdot\left[\left[\frac{\partial g}{\partial u}\right] \times\left[\frac{\partial g}{\partial v}\right]\right]\right] d u \wedge d v $$ (The permutation of the $P, Q, R$ (and the minus sign) come from the way the $d x \wedge d y$ acts on a piece of surface normal to the $(d) z$ direction.)
证明 .
We can rewrite this as $$ \left|\frac{\partial g}{\partial u} \times \frac{\partial g}{\partial v}\right|\left[\left[\begin{array}{c} R \circ g \ -Q \circ g \ P \circ g \end{array}\right] \cdot \hat{\boldsymbol{n}}[u, v]\right] \quad d u \wedge d v $$ where $\hat{\boldsymbol{n}}[u, v]$ is the unit normal to the surface at $g\left[\begin{array}{l}u \ v\end{array}\right]$, and $$ \left|\frac{\partial g}{\partial u} \times \frac{\partial g}{\partial v}\right| $$ is the “area stretching factor”. We have that $$ \int_{g\left(I^{2}\right)} \omega $$
is the limit of the sums of values of $\omega$ on small elements of the surface $g\left(I^{2}\right)$. Suppose $g$ takes a rectangle $\triangle u \times \triangle v$ in $I^{2}$ to a (small) piece of the surface. $\omega$ at $g\left[\begin{array}{l}u \ v\end{array}\right]$ is, say, $$ P d x \wedge d y+Q d x \wedge d z+R d y \wedge d z $$ and the unit normal to the surface is $\hat{\boldsymbol{n}}[u, v]$ (located at $\left.g\left[\begin{array}{l}u \ v\end{array}\right]\right)$. Write $\hat{\boldsymbol{n}}[u, v]$ as $$ \left[\begin{array}{l} \hat{n} x \ \hat{n} y \ \hat{n} z \end{array}\right] $$
Johannes Kepler’s work Stereometrica Doliorum formed the basis of integral calculus.Kepler developed a method to calculate the area of an ellipse by adding up the lengths of many radii drawn from a focus of the ellipse.
微积分课后作业代写
The first equation yields $y=3 x^{2}$, substituting that into the second equation yields $x-6 x^{2}=0$, which has the solutions $x=0$ and $x=\frac{1}{6}$. So $x=0 \Rightarrow y=3(0)=0$ and $x=\frac{1}{6} \Rightarrow y=3\left(\frac{1}{6}\right)^{2}=\frac{1}{12}$. So the critical points are $(x, y)=(0,0)$ and $(x, y)=\left(\frac{1}{6}, \frac{1}{12}\right)$. To use Theorem 2.6, we need the second-order partial derivatives: $$ \frac{\partial^{2} f}{\partial x^{2}}=-6 x, \quad \frac{\partial^{2} f}{\partial y^{2}}=-2, \quad \frac{\partial^{2} f}{\partial y \partial x}=1 $$ So $$ D=\frac{\partial^{2} f}{\partial x^{2}}(0,0) \frac{\partial^{2} f}{\partial y^{2}}(0,0)-\left(\frac{\partial^{2} f}{\partial y \partial x}(0,0)\right)^{2}=(-6(0))(-2)-1^{2}=-1<0 $$ and thus $(0,0)$ is a saddle point. Also, $$ D=\frac{\partial^{2} f}{\partial x^{2}}\left(\frac{1}{6}, \frac{1}{12}\right) \frac{\partial^{2} f}{\partial y^{2}}\left(\frac{1}{6}, \frac{1}{12}\right)-\left(\frac{\partial^{2} f}{\partial y \partial x}\left(\frac{1}{6}, \frac{1}{12}\right)\right)^{2}=\left(-6\left(\frac{1}{6}\right)\right)(-2)-1^{2}=1>0 $$
