\(函数f(x,y)=\ln(x+y+1)在点(0,0)处的带佩亚诺余项的二阶泰勒公式为(\,)\)
A: \(x+y-\frac{1}{2}(x+y)+o(x^2+y^2)\)
B: \(x+y-\frac{1}{2}(x+y)^2+o(\sqrt{x^2+y^2})\)
C: \(x+y-\frac{1}{2}(x+y)^2+o(x^2+y^2)\)
D: \(x+y-\frac{1}{2}(x+y)^3+o(x^2+y^2)\)
A: \(x+y-\frac{1}{2}(x+y)+o(x^2+y^2)\)
B: \(x+y-\frac{1}{2}(x+y)^2+o(\sqrt{x^2+y^2})\)
C: \(x+y-\frac{1}{2}(x+y)^2+o(x^2+y^2)\)
D: \(x+y-\frac{1}{2}(x+y)^3+o(x^2+y^2)\)
举一反三
- \(函数f(x,y)=\ln(x+y+1)在点(0,0)处的带佩亚诺余项的三阶泰勒公式为(\,)\) A: \(x+y-\frac{1}{2}(x+y)^2+\frac{1}{6}(x+y)^3+o((\sqrt{x^2+y^2})^3)\) B: \(x+y-\frac{1}{2}(x+y)^2+\frac{1}{3}(x+y)^3+o((\sqrt{x^2+y^2})^3)\) C: \(x+y-\frac{1}{3}(x+y)^2+\frac{1}{6}(x+y)^3+o((\sqrt{x^2+y^2})^3)\) D: \(x+y-\frac{1}{2}(x+y)^2-\frac{1}{3}(x+y)^3+o((\sqrt{x^2+y^2})^3)\)
- 下列函数在点$(0,0)$的重极限存在的是 A: $f(x,y)=\frac{y^2}{x^2+y^2}$ B: $f(x,y)=(x+y)\sin\frac{1}{x}\sin\frac{1}{y}$ C: $f(x,y)=\frac{x^2y^2}{x^2y^2+(x-y)^2}$ D: $f(x,y)=\frac{x^2y^2}{x^3+y^3}$
- 函数$f(x,y)=\sin x\cdot \ln (1+y)$在点$(0,0)$处带有Peano型余项的3阶Taylor公式为$f(x,y)=$ A: $xy+\frac{1}{2}x{{y}^{2}}+o({{(\sqrt{{{x}^{2}}+{{y}^{2}}})}^{3}})$ B: $xy-\frac{1}{2}x{{y}^{2}}+o({{(\sqrt{{{x}^{2}}+{{y}^{2}}})}^{3}})$ C: $xy-x{{y}^{2}}+o({{(\sqrt{{{x}^{2}}+{{y}^{2}}})}^{3}})$ D: $xy+x{{y}^{2}}+o({{(\sqrt{{{x}^{2}}+{{y}^{2}}})}^{3}})$
- $(0,0)$是以下函数$f(x,y)$的定义域的内点是 A: $f(x,y)=\sqrt{x} \ln(x+y)$ B: $f(x,y)=\frac{x+y}{x^2+y^2 }$ C: $f(x,y)=\arcsin \frac{x}{y}$ D: $f(x,y)=\ln (1-x^2-y^2)$
- 4.已知二元函数$z(x,y)$满足方程$\frac{{{\partial }^{2}}z}{\partial x\partial y}=x+y$，并且$z(x,0)=x,z(0,y)={{y}^{2}}$，则$z(x,y)=$（ ） A: $\frac{1}{2}({{x}^{2}}y-x{{y}^{2}})+{{y}^{2}}+x$ B: $\frac{1}{2}({{x}^{2}}{{y}^{2}}+xy)+{{y}^{2}}+x$ C: ${{x}^{2}}{{y}^{2}}+{{y}^{2}}+x$ D: $\frac{1}{2}({{x}^{2}}y+x{{y}^{2}})+{{y}^{2}}+x$