import sympy as smp from matplotlib import pyplot as plt import numpy as nmp from scipy.integrate import solve_ivp p = 1.225 # air density m = 0.003 # mass of the plane a = (9.3/180)*nmp.pi # angle of attack ar = 0.86 # aspect ratio s = 0.017 # wing area g = 9.807 # acceleration of free fall cla = (nmp.pi*ar)/1+nmp.sqrt(1+(ar/2)**2) # lift slope derivative cl = cla*a # lift coefficient e = 0.9 # oswald efficiency factor ε = 1/(nmp.pi*e*ar) # induced darg factor cd = 0.02 + ε*(cl**2) def airplane(t,f): v, γ, h, r = f return [-cd(p*v**2)*s/(2*m)-g*nmp.sin(γ), ((cl*(p*v**2/2))-g*nmp.cos(γ))/v, v*nmp.sin(γ), v*nmp.cos(γ)] solution = solve_ivp(airplane, (0,1000),[3.7,((-11/180)*nmp.pi),2.0,0.0]) print(solution) 哪里有 type error

from sympy import * from math import * import numpy as np from matplotlib import pyplot as plt plt.rcParams['font.sans-serif'] = ['SimHei'] #用来正常显示中文标签 plt.rcParams['axes.unicode_minus'] = False #用来正常显示负号 import numpy as np import matplotlib.pyplot as plt from sympy import * # 用于求导积分等科学计算 # 一元一次函数图像 def fun_format(): plt.xlabel('x') plt.ylabel('y') plt.xlim((-10,10)) plt.ylim((-20,20)) plt.tight_layout() x,y = symbols('x y') # 引入x y变量 expr = log(x)# 计算表达式 x_value = [] # 用于保存x值 y_value = [] # 用于保存y值 y_value_dif = [] # 用于保存一阶导数值 expr_dif = diff(expr,x,1) for i in np.arange(-10,10,0.1): x_value.append(i) y_value.append(expr.subs('x',i)) # 将i值代入表达式 y_value_dif.append(expr_dif.subs('x',i)) # 将i值代入一阶求导表达式 fig=plt.figure() ax1=fig.add_subplot(2,1,1) # plt.title('f(x)='+str(expr)) fun_format() ax1.plot(x_value,y_value) # 画原函数图 ax2=fig.add_subplot(2,2,3) plt.title('f(x)_dot='+str(expr_dif)) fun_format() ax2.plot(x_value,y_value_dif) # 画一阶导数图

- 引入了 matplotlib 和 sympy 库。 - 定义了一个函数 fun_format()，用于设置坐标轴范围、标签等格式。 - 定义了变量 x 和 y。 - 定义了表达式 expr = log(x)，这是对数函数的表达式。 - 计算了 x 和 y 的值，...

import numpy as np import matplotlib.pyplot as plt import sympy from scipy.interpolate import interp1d gamma = 1.2 R = 8.314 T0 = 500 Q = 50 * R * T0 a0 = np.sqrt(gamma * R * T0) M0 = 6.216 P_P0 = sympy.symbols('P_P0') num = 81 x0 = np.linspace(0,1,num) t_t0 = np.linspace(0,15,num) x = x0[1:] T_T0 = t_t0[1:] h0 = [] h1 = []#创建拉姆达为1的空数组 r = [] t = [] c = [] s = [] i = 0 for V_V0 in x: n1 = sympy.solve(1 / (gamma-1) * (P_P0 * V_V0 - 1) - 0.5 * (P_P0 + 1) * (1 - V_V0)- gamma * 0 * Q / a0 ** 2,P_P0)#lamuda=0的Hugoniot曲线方程 n2 = sympy.solve(1 / (gamma-1) * (P_P0 * V_V0 - 1) - 0.5 * (P_P0 + 1) * (1 - V_V0)- gamma * 1 * Q / a0 ** 2,P_P0)#lamuda=1的Hugoniot曲线方程 n3 = sympy.solve(-1 * P_P0 + 1 - gamma * M0 ** 2 * (V_V0 - 1),P_P0)#Reyleigh曲线方程 n4 = 12.014556 / V_V0#等温线 n5 = sympy.solve((P_P0 - 1 / (gamma+1) )* (V_V0-gamma / (gamma + 1)) - gamma / ((gamma + 1) ** 2),P_P0)#声速线 n6 = 10.6677 / np.power(V_V0,1.2)#等熵线 h0.append(n1) h1.append(n2) r.append(n3) t.append(n4) c.append(n5) s.append(n6) i = i+1 h0 = np.array(h0) h1 = np.array(h1) r = np.array(r) t = np.array(t) c = np.array(c) s = np.array(s) plt.plot(x,r,label='Rayleigh') plt.plot(x,t,color='purple',label='isothermal') plt.plot(x,s,color='skyblue',label='isentropic') a = np.where(h0 < 0) b = np.where(c < 0) h0 = np.delete(h0,np.where(h0 < 0)[0],axis = 0)#去除解小于0的值 h1 = np.delete(h1,np.where(h1 < 0)[0],axis = 0)#去除解小于0的值 c = np.delete(c,np.where(c < 0)[0],axis = 0)#去除解小于0的值 x0 = np.delete(x,a,axis = 0)#对应去除x轴上错误值的坐标 x1 = np.delete(x,b,axis = 0) plt.plot(x0,h0,label='Hugoniot(lambda=0)') plt.plot(x0,h1,label='Hugoniot(lambda=1)') plt.plot(x1,c,color='yellow',label='soniclocus') plt.ylim((0,50)) plt.legend() # 显示图例 plt.xlabel('V/V0') plt.ylabel('P/P0') f1 = interp1d(x1, c.T, kind='cubic') f2 = interp1d(x,r.T,kind='cubic') f3 = interp1d(x, t.T, kind='cubic') epsilon = 0.0001 x0 = 0.56 y0 = f1(x0) - f2(x0) while abs(y0) > epsilon: df = (f1(x0 + epsilon) - f2(x0 + epsilon) - y0) / epsilon x0 -= y0 / df y0 = f1(x0) - f2(x0) plt.scatter(x0, y0, 50, color ='red') plt.show()

其中用到了一些科学计算库，比如 numpy、matplotlib、sympy 和 scipy.interpolate。这段代码还包括了一些数据处理的操作，比如删除数组中小于0的值、插值等等。最后，这段代码还用牛顿迭代法求解了两条曲线的交点。

#外点法(能运行出来) import math import sympy import numpy as np from matplotlib import pyplot as plt from mpl_toolkits.mplot3d import Axes3D plt.ion() fig = plt.figure() ax = Axes3D(fig) def draw(x,index,M): # F = f + MM * alpha # FF = sympy.lambdify((x1, x2), F, 'numpy') Z = FF((X, Y,M)) ax.plot_surface(X, Y, Z, rstride=1, cstride=1, cmap='rainbow',alpha=0.5) ax.scatter(x[0], x[1], FF((x[0],x[1],M)), c='r',s=80) ax.text(x[0], x[1], FF((x[0],x[1],M)), 'here:(%0.3f,%0.3f)' % (x[0], x[1])) ax.set_zlabel('F') # 坐标轴 ax.set_ylabel('X2') ax.set_xlabel('X1') plt.pause(0.1) # plt.show() # plt.savefig('./image/%03d' % index) plt.cla() C = 10 # 放大系数 M = 1 # 惩罚因子 epsilon = 1e-5 # 终止限 x1, x2 = sympy.symbols('x1:3') MM=sympy.symbols('MM') f = -x1 + x2 h = x1 + x2 - 1 # g=sympy.log(x2) if sympy.log(x2)<0 else 0 g = sympy.Piecewise((x2-1, x2 < 1), (0, x2 >= 1)) # u=lambda x: alpha = h 2 + g 2 F = f + MM alpha # 梯度下降来最小化F def GD(x,M,n): # F = f + M * alpha # delta_x = 1e-11 # 数值求导 # t = 0.0001 # 步长 e = 0.001 # 极限 # my_print(e) np.array(x) for i in range(15): t = sympy.symbols('t') grad = np.asarray( [sympy.diff(F, x1).subs([(x1, x[0]), (x2, x[1]),(MM,M)]), sympy.diff(F, x2).subs([(x1, x[0]), (x2, x[1]),(MM,M)])]) # print('g',grad) # print((x-tgrad)) # print(F.subs([(x1,(x-tgrad)[0]),(x2,(x-tgrad)[1])])) t = sympy.solve(sympy.diff(F.subs([(x1, (x - t grad)[0]), (x2, (x - t * grad)[1]),(MM,M)]), t), t) print('t',t) x = x - t * grad print('x', x) # print('mmm',M) draw(x,n*10+i,M) # my_print(np.linalg.norm(grad)) # print(type(grad)) if (abs(grad[0]) < e and abs(grad[1]) < e): # print(np.linalg.norm(grad)) print('g', grad) break return list(x) pass x = [-0.5, 0.2] X = np.arange(0, 4, 0.25) Y = np.arange(0, 4,

0.25) # 定义网格点坐标 X, Y = np.meshgrid(X, Y) # 生成网格点 FF = sympy.lambdify((x1, x2,MM), F, 'numpy') # 将表达式转化为可计算的函数 draw(x,0,M) # 绘制函数图像 for i in range(5): # 迭代次数为5次 x =...

import sympy from scipy.integrate import quad y=xsympy.atan(x)sympy.ln(sympy.exp(1+x**2)) result,error=quad(y,0,1) print(result)出错怎么办

from scipy.integrate import quad x = sympy.symbols('x') y = x*sympy.atan(x)*sympy.ln(sympy.exp(1+x**2)) result, error = quad(sympy.lambdify(x, y), 0, 1) print(result) 在这里，我们使用 sympy....

from sympy import simplify import numpy as np import sympy as sp from sympy import symbols x, y, z = symbols('x y z') F = (x2 + y2 - 1)(x2 + z2 - 1)(y2 + z2 - 1) - 1 F_x = F.diff(x) F_y = F.diff(y) F_z = x**2 * F.diff(x) - y * F.diff(y) F_x = simplify(F_x) F_y = simplify(F_y) F_z = simplify(F_z) G_x = - F_x/F_z G_y = - F_y/F_z G_x = simplify(G_x) G_y = simplify(G_y) Fi = F_x.diff(x) Fj = F_x.diff(y) Fk = F_y.diff(y) N = G_x.cross(G_y) # normale N = simplify(N/N.norm()) E = simplify(G_x.norm()2) F = simplify(G_x.dot(G_y)) G = simplify(G_y.norm()2) I = simplify(N.dot(Fi)) J = simplify(N.dot(Fj)) K = simplify(N.dot(FK)) x = Symbol('x') factor(det(Matrix([[I - xE, J - xF], [J - xF, K - xG]])).subs({x:0, y:0}))跑不出来，

从代码上看，这是一个使用 SymPy 执行符号计算的程序。其中，定义了一些变量和表达式，对这些表达式求导，然后求取法向量、曲率等数据。最后一行代码是要计算一个行列式，并且把其中的 $x$ 和 $y$ 分别代入为 $0$，...

import sympy as sp import math a = int(input("请输入a：")) n0= float(input("请输入n0：")) d = 2 * n0 * a / (1 - a ** 2) p = int(input("请输入p：")) q_sym, nx = sp.symbols('q_sym nx') # 定义符号变量 q1 = a * a * sp.cos(math.radians(p)) * sp.cos(math.radians(p)) * sp.cos(math.radians(q_sym)) ** 2 q2 = (d - a * sp.sin(math.radians(p))) ** 2 q3 = q1 / q2 eq = sp.Eq(q3, 1 - q_sym) # 构造方程 sol = sp.solve(eq, [q_sym, nx]) # 求解 for s in sol: print(sp.N(s)) # 将表达式转换为浮点型以输出结果

这段代码的作用是解决一个三棱锥的问题，其中需要输入参数a、n0、p，然后通过计算得到q_sym和nx的值。具体来说，这段代码首先导入了sympy和math库，然后让用户输入a、n0和p的值。接着，根据这些输入的值，计算出d的...

import sympy x=sympy.Symbol('x') y=xsympy.atan(x)sympy.ln(sympy.exp(1+x**2)) dx=sympy.N(sympy.integrate(y,x)) print(dx)没有输出怎么改

y = x*sympy.atan(x)*sympy.ln(sympy.exp(1+x**2)) dx = sympy.integrate(y, x, show=True) print(dx) 如果还是没有输出，那可能是你的计算机处理时间太长了，你可以尝试将 show=True 参数去掉，这样就不会...

#导入sympy库 import sympy as sp #定义变量符号 x,y =sp.symbols('x y') #定义被积函数 f=(x2+2y) #定义积分区域 D = sp.Interval(x=y2)sp.Interval(y=x-2) #计算符号解 result = sp.integrate(f,(x,1,4),(y,-1,2)) print(result)

这段代码使用了Python的符号计算库sympy来进行多重积分的计算。具体来说，它定义了两个符号变量x和y，然后定义了被积函数f为x的平方加2乘以y。接着，它定义了积分区域D为y的取值范围在[x-2,x]之间，x的取值范围在[1,...

from sympy import symbols,diff import sympy as sp from sympy.abc import beta,omega,u import math k1 = 150.0 k2 = 100.0 a = 0.6 b = 0.77 L = 1.37 step_size = 0.01 time = 10 m = 256 vx = 30.0 Iz = 0.1 vx_plot = [] omega_plot = [] vx = vx sigma = 0.5 eq1 = sp.Eq((k1+k2)beta+omega/u(ak1-bk2)-k1sigma -m(0.09+uomega),0) eq2 = sp.Eq((ak1-bk2)beta+omega/u(a2k1-b2k2)-ak1sigma -Izomega,0) eq3 = sp.Eq((u-vx/math.tan(beta)),0) sol = sp.solve((eq1,eq2,eq3),(beta,omega,u)) print(sol)不能正常运行

1. 在第12行中，您缺少了一个乘号(*)，应该将(k1+k2)和beta相乘，正确的代码应该是： eq1 = sp.Eq((k1+k2)*beta+omega/u*(a*k1-b*k2)-k1*sigma -m*(0.09+u*omega),0) 2. 在第13行中，您也缺少了一个乘号(*...

#导入sympy库 import sympy as sp #定义变量符号 x,y =sp.symbols('x y') #定义被积函数 f=(x**2+2*y) #定义积分区域 D = sp.Interval(x=y**2)*sp.Interval(y=x-2) #计算符号解 result = sp.integrate(f,(x,1,4),(y,-1,2)) print(result)

这段代码使用了 Python 中的 sympy 库来进行符号计算，实现了对被积函数 f 在积分区域 D 上的重积分计算。具体来说，首先通过 sp.symbols() 函数定义了变量符号 x 和 y，然后定义了被积函数 f，即 x^2 + 2y。接...

下列代码在运行中有哪些错误，并改正：import numpy as np import sympy as sym def ff(a,b,n=3): h=(b-a)/(n-1) x_k=[a+kh for k in range(n)] #x_k随着k取值不同，有很多个值，所以是列表形式[]# x=sym.Symbol('x') def l(k): terms=[(x-x_k[j])/(x_k[k]-x_k[j]) for j in range(n) if j!=k] return sym.prod(terms) A_k=[sym.integrate(l(k),(x,a,b)) for k in range(n)] C_k=[sym.simplify(A_k[k]/(b-a)) for k in range(n)] def f(x): return sym.integrate(sym.sqrt(x-1),(x,a,b)) return (b-a)np.sum([f(x_k[k])*C_k[k] for k in range(n)]) ff(a=-1,b=1)

import numpy as np import sympy as sym def ff(a, b, n=3): h = (b - a) / (n - 1) x_k = [a + k * h for k in range(n)] x = sym.Symbol('x') def l(k): terms = [(x - x_k[j]) / (x_k[k] - x_k[j]) for ...

这段代码为什么报错： Invalid limits given: ((-1.0, -1, 1),) 怎样修改？代码：import numpy as np import sympy as sym def ff(a,b,n=3): h=(b-a)/(n-1) x_k=[a+kh for k in range(n)] #x_k随着k取值不同，有很多个值，所以是列表形式[]# x=sym.Symbol('x') def l(k): terms=[(x-x_k[j])/(x_k[k]-x_k[j]) for j in range(n) if j!=k] return sym.prod(terms) A_k=[sym.integrate(l(k),(x,a,b)) for k in range(n)] C_k=[sym.simplify(A_k[k]/(b-a)) for k in range(n)] def f(x): return sym.integrate(sym.sqrt(x+1),(x,a,b)) return (b-a)np.sum([f(x_k[k])*C_k[k] for k in range(n)]) ff(a=-1,b=1,n=3)

import numpy as np import sympy as sym def ff(a, b, n=3): h = (b - a) / (n - 1) x_k = [a + k * h for k in range(n)] x = sym.Symbol('x') def l(k): terms = [(x - x_k[j]) / (x_k[k] - x_k[j]) for ...

将以下代码转化为matlab代码表示：import xlrd import sympy import numpy as np from scipy import linalg #%% queue = [ 0, 29, 17, 2, 1, 20, 19, 26, 18, 25, 14, 6, 11, 7, 15, 9, 8, 12, 27, 16, 10, 13, 5, 4, 3, 22, 28, 24, 23, 21, 0] def read_data_model(): data = xlrd.open_workbook("/Users/lzs/Downloads/2020szcupc/data/C2.xlsx") table = data.sheet_by_name("Sheet1") rowNum = table.nrows colNum = table.ncols consumes = [] for i in range(1, rowNum): # 忽略DC的消耗 if i == 1: pass else: consumes.append(0 if table.cell_value(i, 3) == '/' else table.cell_value(i, 3)) return consumes #%% 获得矩阵A def get_A_matrix(data): A = np.ones([29,29], dtype = float) diagonal = np.eye(29) for i in range(29): for j in range(29): A[i][j] = data['consumes'][j] / data['r'] A = A - diagonal return A #%% def get_b_maatrix(data): b = np.ones([29,1], dtype=float) for i in range(29): b[i][0] = -data['dst']data['consumes'][i]/data['velocity']+data['f'] for j in range(29): b[i][0] = b[i][0] + data['f']data['consumes'][i]/data['r'] return b #%% 数值解 def numerical(data): data['velocity'] = 50 data['dst'] = 11469 data['r'] = 200 data['f'] = 10 A = get_A_matrix(data) b = get_b_maatrix(data) x = linalg.solve(A, b) return x #%% 符号解决方案 def symbolic(data): data['velocity'] = sympy.symbols("v", integer = True) data['dst'] = 12100 data['r'] = sympy.symbols("r", integer = True) data['f'] = sympy.symbols("f", integer = True) # 获取矩阵A并转移到符号矩阵M A = np.ones([29,29], dtype = float).tolist() diagonal = np.eye(29).tolist() for i in range(29): for j in range(29): A[i][j] = data['consumes'][j] / data['r'] - diagonal[i][j] M = sympy.Matrix(A) # 得到矩阵b并转移到符号矩阵b b = np.ones([29,1], dtype=float).tolist() for i in range(29): b[i][0] = -data['dst']data['consumes'][i]/data['velocity']+data['f'] for j in range(29): b[i][0] = b[i][0] + data['f']data['consumes'][i]/data['r'] b = sympy.Matrix(b) # LU solver x = M.LUsolve(b) return x #%% 主功能 if name == 'main': data = {} data['consumes'] = read_data_model() options = {"numerical":1, "symbolic":2} option = 1 if option == options['numerical']: x = numerical(data) print(x) elif option == options['symbolic']: x = symbolic(data) print(x) else: print("WARN!!!")

import numpy as np from scipy import linalg %% 读取数据模型 function consumes = read_data_model() data = xlsread('/Users/lzs/Downloads/2020szcupc/data/C2.xlsx'); consumes = []; for i = 2:size(data...

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#导入sympy库 import sympy as sp #定义变量符号 x,y =sp.symbols('x y') #定义被积函数 f=(x**2+2*y) #定义积分区域 D = sp.Interval(x=y**2)*sp.Interval(y=x-2) #计算符号解 result = sp.integrate(f,(x,1,4),(y,-1,2)) print(result)

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