sad += (diff < 0)?(-diff):diff; c语言,

sad += (diff < 0)?(-diff):diff; c语言, 大智慧指标,错误line 5:C1001 :语法错误：“,”SQX:=(C-MA(CLOSE,13))*100;DIFF:EMA(MA(SQX,7),1),COLOR00FFFF;DEA:EMA(DIFF,3),COLORFF33FF;MACD:(DIFF-DEA)*2,COLORSTICK;低位金叉:=IF(CROSS(DIFF,DEA) AND DIFF=0));二次金叉:IF(CROSS(DIFF,DE matlab diff怎么不能求1/x^2的导数>> syms x>> f = 1/x^2 f = 1/x^2 >> diff('f',x) ans = 0 y=diff(u)/0.MATLAB matlab 关于 diff用法问题syms t x yx=sin(t);y=diff(x)想求 当t=(0:pi/8:2*pi)时,y(即cos(t)?),y的值程序应该怎么写呢? matlab解微分方程组并画图问题 方程组如下 diff(x,t,2)-a*diff(y,t)+b*x=0,diff(y,t,2)-a*diff(x,t)+b*y=0a=0.05 b=0.15刚学matlab的小白QAAAQ 完全不知道用什么命令解跪求指导QAAAAAAQx0=4 y0=4 Dx0=0 Dy0=0 绘图 t 由0到100 问一个偏微分方程的求解问题吧,MATLAB,MAPLE都可以,是2维的热传导问题的方程求解> PDE1 := piecewise(x > 0,diff(T(x,t),[`$`(t,1)]) = diff(T(x,t),[`$`(x,2)]));> PDE2 := piecewise(x = 0,diff(T(x,t),[`$`(x,1)]) = 1000);> PDE3 := MATLAB二重参数积分求最小值clearclear allsyms x y a b; Q = cos(0.212*y).*besselj(0,0.2405*x).*exp(-a*x.^2).*exp(-b*y.^2);Dx1 = diff(Q,x);Dy2 = diff(Q,y,2);DxxDx = diff(x.*Dx1,x);fxy = (Q.*(-19.*(1./x.*DxxDx+Dy2)-2654.*Q./sqrt(x.^2+y.^2)))./Q Matlab错误求改.急syms x1 x2 x3 x4 x5;e=0.001;c=[5;4;3;2;1]Q=[5 4 3 2 1;4 4 3 2 1;3 3 3 2 1;2 2 2 2 1;1 1 1 1 1];x=[x1;x2;x3;x4;x5]f=c'*x+0.5*x'*Q*xtidu=[diff(f,x1);diff(f,x2);diff(f,x3);diff(f,x4);diff(f,x5)]x1=0;x2=0;x3=0;x4=0;x5=0;flag=1;k=0;t matlab 求-120*x^3-30*x^4+18*x^5+5*x^6-30*xy^2的极值syms x yf='-120*x^3-30*x^4+18*x^5+5*x^6-30*xy^2';fx=diff(f,x)fy=diff(f,y)x0=roots([30,90,-120,-360])y0=roots([0])fxx=diff(f,x,2);fyy=diff(f,y,2);fxy=diff(f,x,y);A=(fxx)*(fyy);x=2;y=0;a1=eval(A)b1 有谁知道matlab里这句 tmp=([1;diff(x)]~ diff是什么意思 diff time sad SAD Sad 关于符号变量的分段函数function D = fcn(ut)syms rc1=0.5;c2=0.0014;c3=1;R=100;persistent a;if isempty(a)a=0.07;endif diff(ut)>0h=max(ut-r,a);elseif diff(ut) 0h=min(ut+r,a); %ut的极小值点a=h;elseif diff(ut,2) < 0h=max(ut-r,a); %ut的极 matlab 曲线求导syms x1 x2 x3 x4y1=50/2*(1-cos(2*x1));dy1=diff(y1) ; X1=0:pi/1000:pi/2;Dy1=subs(dy1,X1);y2=50;dy2=diff(y2) ; X2=pi/2:pi/1000:pi;Dy2=subs(dy2,X2);y3=50-50/2*(1-cos(2*x3));dy3=diff(y3) ; X3=pi:pi/1000:3*pi/2;Dy3=subs(dy3,X3);y4=0;dy4