简单计算一下即可，答案如图所示
证明(1)必要性设f(x,y,z)是k次齐次函数,即f(tx,ty,tz)=t^kf(x,y,z) 令u=tx,v=ty, ω=tz 将上式两边对t求导数,由复合函数微分法,得xf_k'+yf'_o+zf_∞'=kt^(k-1)f(x,y,z) 将上式两边乘以t,得t[f_u+tyf_v+tzf_w']=kt^kf(x,y,z) 因为f(x,y,z)是k次齐...
你前面提到的方程是一个二元方程,而后面的微分方程中,y是x的函数,如果y是齐次的,那么必定就会有那第二个方程
u=tx,v=ty.f(tx,ty)=t^3f(x,y)对t求导: xf'u(tx,ty) yf'v(tx,ty)=3t^2f(x,y).令t=1,x=1,y=2,代入可得f(1,2)。是隐函数，t是对x,y的函数。函数（function）的定义通常分为传统定义和近代定义，函数的两个定义本质是相同的，只是叙述概念的出发点不同，传统定义是从运动变...
y*F2'(tx,ty,tz)+z F3'(tx,ty,tz);同时dG/dt= kt^(k-1)*F(x,y,z);所以联立上三式得:x*F1'(tx,ty,tz)+y*F2'(tx,ty,tz)+z*F3'(tx,ty,tz)= kt^(k-1)*F(x,y,z)= kt^(-1)*F(tx,ty,tz);换元,令tx=x,ty=y,tz=z(符号无关),得:xFx'+yFy'+zFz'=kF(x,...
f(tx,ty)=(tx)²+(ty)²-tx·tytan(tx/ty)=t²x²+t²y²-t²xytanxy =t²(x²+y²-xytanxy)=t²f(x,y)
如果函数F(x,y)满足F(tx,ty)=t^kF(x,y),称F(x,y)为k次齐次函数。如果k=0,那么零次齐次函数F(x,y)一定可以表示为f(y/x)的形式。两种方程没有任何关系。 结果一 题目 若代数式2xay3zc与-(1,2x4ybz2是同类项,则( ) A. a=4,b=2,c=3 B. a=4,b=4,c=3 C. a=4,b=3,...
(tx,ty)\ge 1. \end{aligned}$$ definition 1.23 [ 30 ] let \((x,g)\) be a \(g\) -metric space, \(t\) be a self-mapping on \(x\) and \(\alpha : x^3 \rightarrow [0,+\infty )\) be a function. we say that \(t\) is an \(g\) - \(\alpha \) -admissible ...
8.若f(x,y,z)满足关系式f(tx,ty,tz)=tnf(x,y,z),其中t为任意实数,则称f(x,y,z)为n次齐次函数.证明:任意一个可微的n次齐次函数均满足
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