Sichuan University 2009 years of Mathematical analysis of postgraduate examination questions

One, limit (7 points per small question, total 28 points)

1.$\displaystyle \lim\limits_{n\to \infty} \frac{1}{n^{2}}\sum\limits_{n=1}^{\infty}\ln \tbinom{n}{k}$

2.$\displaystyle \lim\limits_{n\to \infty} \sin ^{2}\left (\pi \sqrt{n^{2}+n}\right) $

3.$\displaystyle \lim\limits_{x\to 0^{+}}\frac{\displaystyle \int_{0}^{x^{2}}\sin ^{\frac{3}{2}}tdt}{\displaystyle \int_{0}^{x}t (T-\sin t) dt}$

4.$\displaystyle \lim\limits_{x \to 0}\frac{e^{x}-1-x}{\sqrt{1-x}-\cos \sqrt{x}}$

Second, (10 points for each sub-question, a total of 40) calculate the following points.

1.$\displaystyle \iint\limits_{d}\left| \frac{x+y}{2}-x^{2}-y^{2}\right|dxdy$, where $\displaystyle d=\{(x, y) \in r^{2}\mid x^{2}+y^{2}\le 1\}$.

2.$\displaystyle \int_{l}yzds$, wherein the curve $l$ is the intersection of the spherical $\displaystyle x^{2}+y^{2}+z^{2}=a^{2}$ and the plane $z+y+z=1$.

3. Set $f (x) $ within $ (-\infty,+\infty) $ for continuous conduction function, quadrature $\displaystyle \int\limits_{l}\frac{1+y^{2}f (XY)}{y}dx+\frac{x}{y^{2}}[ Y^{2}f (XY) -1]dy$, where $l$ is a straight segment from the point $ a\left (3,\frac{2}{3}\right) $ to $ b\left (1,2\right) $.

4.$\displaystyle \iint\limits_{s}\frac{xdydz+ydzdx+zdxdy}{\sqrt{(x^{2}+y^{2}+z^{2}) ^{3}}}$, where $s$ is a parabolic $\ Displaystyle 1-\frac{z}{7}=\frac{(x-2) ^{2}}{25}+\frac{(y-1) ^{2}}{16} (Z\GE0) $ on the upper side.

Third, (the subject 10 points) set $f (x, y) $ in the bounded closed area $d$ has a continuous second derivative, and $\displaystyle f ' _{xx}+f ' _{yy}=0,f ' _{xy}\not =0$.

Proof: The maximum minimum value of $z =f (x, y) $ can only be obtained on the boundary of the zone $d$.

Four, (12 points) prove: Under the transformation $\displaystyle u=\frac{x}{y},v=x,w=xz-y$, the equation

$ $y \frac{\partial ^{2} z}{\partial y^{2}}+2\frac{\partial z}{\partial y}=\frac{2}{x}$$ can become $$\frac{\partial ^{2} w}{\ Partial u^{2}}=0$$

V. (12 points) (12 points) prove: $\displaystyle \sum\limits_{n=1}^{\infty} (1-x) \frac{x^{n}}{1-x^{2n}}\sin nx$ in $\displaystyle \ Left (\frac{1}{2},1\right) $ within uniform convergence.

Six, (12 points) set $f (x) $ on $[a,b]$ continuous, and $f (a) =f (b) $, proving that: $\displaystyle \forall n \in \mathbb{z}^{+},\exists\ \xi\in (A, b) $, making $$\ Displaystyle f\left (\xi +\frac{b-a}{n}\right) =f (\xi) $$.

Seven, (12 points) set $f (x) $ on $[0,1]$ continuous, in $ (0,1) $ can be guided, and $f ' (x) >0,f (0) =0$, proving:

$\displaystyle \exists \xi, \eta \in (0,1) $ makes $\xi +\eta =1$ and $\displaystyle \frac{f ' (\xi)}{f (\xi)}=\frac{f ' (\eta)}{f (\ ETA)}$.

Eight, (the subject 12 points) Set function $f (x) $ on $[a,b]$ continuous, Proof: $$\int_{a}^{b}\left|f (x) \right|dx\le\max \left\{(b-a) \int_{a}^{b}\left|f (x) \ RIGHT|DX, \left|\int_{a}^{b}f (x) dx\right| \right\}.$$

Nine, (the subject 12 points) to any real number $a>0$, the function $f (x) $ on the $[0,a]$ integrable, and $\displaystyle \lim\limits_{x\to +\infty}f (x) =b$ ($B $ limited).

Proof: $\displaystyle \lim\limits_{t\to 0^{+}}t\int_{0}^{+\infty}e^{-tx}f (x) dx=b$.

Sichuan University 2009 years of Mathematical analysis of postgraduate examination questions