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1(课本习题2)
做因式分解 f ( x ) = x 4 − 5 x 3 + 5 x 2 + 5 x − 6 f(x) = x^{4} - 5x^{3} + 5x^{2} + 5x - 6 f(x)=x4−5x3+5x2+5x−6.
代码:
syms x;
f = x^4-5*x^3+5*x^2+5*x-6;
factor(f)
输出:
( x − 1 x − 2 x − 3 x + 1 ) (\begin{matrix} x - 1 & x - 2 & x - 3 & x + 1 \end{matrix}) (x−1x−2x−3x+1)
2(课本习题3)
求矩阵 A = ( 1 2 2 a ) A = \begin{pmatrix}1 & 2 \\ 2 & a \end{pmatrix} A=(122a)的逆和特征值.
代码:
syms a;
A = [1 2;2 a];
iA = inv(A)
e = eig(A)
答:
该矩阵的逆(iA)为 ( a a − 4 − 2 a − 4 2 a − 4 1 a − 4 ) (\begin{matrix} \frac{a}{a - 4} & - \frac{2}{a - 4} \\ \frac{2}{a - 4} & \frac{1}{a - 4} \end{matrix}) (a−4aa−42−a−42a−41),
特征值(e)为 ( a 2 − a 2 − 2 a + 17 2 + 1 2 a 2 + a 2 − 2 a + 17 2 + 1 2 ) (\begin{array}{r} \frac{a}{2} - \frac{\sqrt{a^{2} - 2\, a + 17}}{2} + \frac{1}{2} \\\ \frac{a}{2} + \frac{\sqrt{a^{2} - 2\, a + 17}}{2} + \frac{1}{2} \end{array}) (2a−2a2−2a+17+21 2a+2a2−2a+17+21)。
3(课本习题4)
计算极限
lim x → ∞ ( 3 x + 9 x ) 1 x , lim y → 0 + lim x → 0 + ln ( 2 x + e − y ) x 3 + y 2 , lim x → ∞ ln ( 1 + 1 x ) a r c c o t x , lim x → 0 1 − 1 − x 2 e x − cos x \lim_{x \rightarrow \infty}\left( 3^{x} + 9^{x} \right)^{\frac{1}{x}},\lim_{y \rightarrow 0^{+}}{\lim_{x \rightarrow 0^{+}}\frac{\ln{(2x + e^{- y})}}{\sqrt{x^{3} + y^{2}}}},\\ \lim_{x \rightarrow \infty}\frac{\ln{(1 + \frac{1}{x})}}{arccot \, x},\lim_{x \rightarrow 0}\frac{1 - \sqrt{1 - x^{2}}}{e^{x} - \cos x} x→∞lim(3x+9x)x1,y→0+limx→0+limx3+y2ln(2x+e−y),x→∞limarccotxln(1+x1),x→0limex−cosx1−1−x2
代码:
syms x y;
ans1 = limit((3^x+9^x)^(1/x),x,inf)
temp1 = limit(log(2*x+exp(-y))/(sqrt(x^3+y^2)-1),x,0);
ans2 = limit(s1,y,0)
ans3 = limit(log(1+1/x)/acot(x),x,inf)
ans4 = limit((1-sqrt(1-x^2))/(exp(x)-cos(x)),x,inf)
输出:
ans1 = 9
ans2 = 0
ans3 = 1
ans4 = 0
4(课本习题5)
计算
∑ k = 1 n k 2 , ∑ k = 1 ∞ 1 k 2 , ∑ n = 0 ∞ 1 ( 2 n + 1 ) ( 2 x + 1 ) 2 n + 1 . \sum_{k = 1}^{n}k^{2},\ \sum_{k = 1}^{\infty}\frac{1}{k^{2}},\ \sum_{n = 0}^{\infty}\frac{1}{(2n + 1)(2x + 1)^{2n + 1}}. k=1∑nk2, k=1∑∞k21, n=0∑∞(2n+1)(2x+1)2n+11.
代码:
syms k n x;
s1=symsum(k^2,k,1,n);
s2=symsum(k^(-2),k,1,inf);
s3=symsum(1/(2*n+1)/(2*x+1)^(2*n+1),n,0,inf);
s1=simplify(s1)
s2=simplify(s2)
s3=simplify(s3)
输出:
s1 = n ( 2 n + 1 ) ( n + 1 ) 6 \frac{n\,(2\, n + 1)\,(n + 1)}{6} 6n(2n+1)(n+1)
s2 = π 2 6 \frac{\pi^{2}}{6} 6π2
s3 = a t a n h ( 1 2 x + 1 ) if 1 < ∥ 2 x + 1 ∥ atanh(\frac{1}{2x + 1}) \text{ if }1 < \|2x + 1\| atanh(2x+11) if 1<∥2x+1∥
5(课本习题6)
求 ∂ 3 ∂ x 2 ∂ y sin ( x 2 y z ) ∥ x = 1 , y = 1 , z = 3 \left. \ \frac{\partial^{3}}{\partial x^{2}\partial y}\sin\left( x^{2}y z \right) \right\|_{x = 1,y = 1,z = 3} ∂x2