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: $$ (x+h)^{k+1}-x^{k+1}=x(x+h)^{k}-x \cdot x^{k}+h(x+h)^{k} . $$
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 $$ \left[\begin{array}{l} \hat{n} x \ \hat{n} y \ \hat{n} z \end{array}\right] $$
so $e^{y}-2 x-e^{-y}=0$ or, multiplying by $e^{y}$, $$ e^{2 y}-2 x e^{y}-1=0 $$ This is really a quadratic equation in $e^{y}$ : $$ \left(e^{y}\right)^{2}-2 x\left(e^{y}\right)-1=0 $$
证明 .
Solving by the quadratic formula, we get $$ e^{y}=\frac{2 x \pm \sqrt{4 x^{2}+4}}{2}=x \pm \sqrt{x^{2}+1} $$ Note that $e^{y}>0$, but $x-\sqrt{x^{2}+1}<0$ (because $x<\sqrt{x^{2}+1}$ ). Thus the minus sign is inadmissible and we have $$ e^{y}=x+\sqrt{x^{2}+1} $$ Therefore $$ y=\ln \left(e^{y}\right)=\ln \left(x+\sqrt{x^{2}+1}\right) $$
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, $$ D=\frac{\partial^{2} f}{\partial x^{2}}\left(\frac{1}{6}, \frac{1}{12}\right) \frac{\partial^{2} f}{\partial y^{2}}\left(\frac{1}{6}, \frac{1}{12}\right)-\left(\frac{\partial^{2} f}{\partial y \partial x}\left(\frac{1}{6}, \frac{1}{12}\right)\right)^{2}=\left(-6\left(\frac{1}{6}\right)\right)(-2)-1^{2}=1>0 $$
Suppose $a_{n} \rightarrow a$ and $b_{n} \rightarrow$ b as $n \rightarrow \infty$. Then the following results hold. (i) $a_{n}+b_{n} \rightarrow a+b$ as $n \rightarrow \infty$. (ii) $c a_{n} \rightarrow c$ a as $n \rightarrow \infty$ for any real number $c$. (iii) If $a_{n} \leq b_{n}$ for all $n \in \mathbb{N}$, then $a \leq b$. (iv) (Sandwich theorem) If $a_{n} \leq c_{n} \leq b_{n}$ for all $n \in \mathbb{N}$, and if $a=b$, then $c_{n} \rightarrow a$ as $n \rightarrow \infty$.
证明 .
Proof Let $\varepsilon>0$ be given. (i) Note that, for every $n \in \mathbb{N}$, $$ \begin{aligned} \left|\left(a_{n}+b_{n}\right)-(a+b)\right| &=\left|\left(a_{n}-a\right)+\left(b_{n}-b\right)\right| \ & \leq\left|a_{n}-a\right|+\left|b_{n}-b\right| \end{aligned} $$ Since $a_{n} \rightarrow a$ and $b_{n} \rightarrow b$, the above inequality suggests that we may take $\varepsilon_{1}=\varepsilon / 2$, and consider $N_{1}, N_{2} \in \mathbb{N}$ such that $$ \left|a_{n}-a\right|<\varepsilon_{1} \quad \forall n \geq N_{1} \text { and }\left|b_{n}-b\right|<\varepsilon_{1} \quad \forall n \geq N_{2} $$ so that $$ \left|\left(a_{n}+b_{n}\right)-(a+b)\right| \leq\left|a_{n}-a\right|+\left|b_{n}-b\right|<2 \varepsilon_{1}=\varepsilon $$ for all $n \geq N:=\max \left{N_{1}, N_{2}\right}$. (ii) Note that $$ \left|c a_{n}-c a\right|=|c|\left|a_{n}-a\right| \quad \forall n \in \mathbb{N} . $$
问题 2.
(Ratio test) Suppose $a_{n}>0$ for all $n \in \mathbb{N}$ such that $\lim {n \rightarrow \infty} \frac{a{n+1}}{a_{n}}=\ell$ for some $\ell \geq 0$. Then the following hold. (i) If $\ell<1$, then $a_{n} \rightarrow 0$. (ii) If $\ell>1$, then $a_{n} \rightarrow \infty$.
证明 .
Suppose $\ell<1$. Let $q$ be such that $\ell1$. Let $q$ be such that $1<q<\ell$. Then, taking for example the open interval $I$ containing $\ell$ as $I=(q, \ell+1)$, there exists $N \in \mathbb{N}$
Instructor: Emily Meyer Contact: Time & Place: [email protected] MWF 8:00-9:40 am, Chemistry 166 Office: MSB 3127 Office hours MWR 3:00-5:00 pm Course Website Homework, worksheets, updates and supplementary material will be online at http://math.ucdavis.edu/~emeyer/math12/~index.html, as well as on Canvas: http://canvas.ucdavis.edu/courses/282963. Textbook • Precalculus (Seventh Edition, 2012) by Cohen, Lee, and Sklar. Prerequisites/Placement Exam • It is expected that you have taken two years of high school algebra, plane geometry, and plane trigonometry. • You must have received a qualifying score (25 or more) on the UC Davis Math Placement Exam to take this course. If you have not yet taken the exam, you can take it during the testing session from 1pm July 30 to 1pm August 7, 2018. See department website for further details. NOTE: after completing this course, you are also required to take the Math Placement Exam again before you may enroll in calculus. There are two testing windows for fall quarter: 1pm September 5 to 1pm September 11, and 1pm September 26 to 1pm October 2. Course Information Grading Scheme • 40% final exam • 20% midterm exam 1 MAT 012 SSII18 • 20% quizzes • 20% homework and worksheets Class Expectations Class time will be interactive. I will not take attendance, but you are expected to be in class. Being present and engaged is crucial to your academic success. Homework and Quizzes • Homework will be in the form of worksheets given in class that you will have time to start in class and will be due at the beginning of the following class. These will be graded out of 2 points, and you can redo each one once for a regrade as long as your first version is complete and your redo is turned in within one week of the original due date. • Additional homework problems will be provided but will not be graded. It is not expected that you do all of the additional problems, but that you work through enough problems from each section to develop and solidify the skills and knowledge you have learned. This is the best way to study for exams! You are welcome to ask about these problems in office hours, or to ask for general qualitative feedback from the instructor on your work. • Quizzes will be given every Friday on which there is no exam (see Exams below). These are intended not to be especially difficult but to give you practice thinking about math in a similar environment to an exam. Exams There will be two exams: • Midterm on Friday, August 24 • Final on Friday, September 14 (last day of class) Each will be 100 minutes. Other • Make-ups and Absences: Make-up exams and quizzes will be given only in the case of documented emergencies. Your lowest quiz grade (out of four) will be dropped, so if you need to miss a quiz for a non-emergency reason, plan on using that as your dropped quiz. • Office hours: I will hold office hours (tentatively) 3-5pm Monday, Wednesday and Thursday. You are highly encouraged to attend! This is your opportunity to ask me to repeat or rephrase anything that did not make sense in class, to get individualized help on your homework, or to ask any lingering questions you might have. 2/3 MAT 012 SSII18 Academic Policies Academic Integrity and Honesty You are expected to comply with the UC Davis Code of Academic Conduct, which can be found at http://sja.ucdavis.edu/cac.html. Any evidence of misconduct will be reported to SJA and dealt with according to their policy. (Note that this does not include working with your classmates in class and/or on homework, which is fine – and recommended!) Accommodations for Disabilities If you need any accommodations, you are required to register with the Student Disability Center (http://sdc.ucdavis.edu). Once you have done this, the SDC will contact the instructor and/or the math department, and we will respond accordingly to meet your needs.
