题目描述 求解方程ax2+bx+c=0的根。要求a, b, c由用户输入,并且可以为任意实数。 输入 输入只有一行,包括三个系数,之间用空格格开。 输出...
函数文件 quadratic.m 将包含的主要 quadratic 函数和子函数 disc 来计算判别。...在MATLAB中建立一个函数文件 quadratic.m 并输入下述代码: function [x1,x2] = quadratic(a,b,c) %this function returns the...可以从命令提示符调用上述函数为: [a,b] = quadratic(2,4,-4) MATLAB执行上面的语句,返回以下结果: >> [a,b] = quadratic(2,4,-4) a =...在MATLAB中建立一个函数文件 quadratic2.m,并输入下述代码: function [x1,x2] = quadratic(a,b,c) d = disc(a,b,c); function...可以从命令提示符调用上面的函数为: [a,b] = quadratic2(2,4,-4) MATLAB执行上面的语句,返回以下结果: >> [a,b] = quadratic(2,4,-4) a
= PolynomialFeatures(degree=2) X_train_quadratic = quadratic_featurizer.fit_transform(X_train) X_test_quadratic...= quadratic_featurizer.transform(X_test) print("原特征\n",X_train) print("二次项特征\n",X_train_quadratic)...regressor_quadratic = LinearRegression() regressor_quadratic.fit(X_train_quadratic, y_train) xx_quadratic...= quadratic_featurizer.transform(xx.reshape(xx.shape[0], 1)) plt.plot(xx, regressor_quadratic.predict...(X_test_quadratic, y_test)) ?
(100,100,60,math.pi/6) #其实这里的返回值是一个tuple print(x,y) print("practice-------------") import math def quadratic...-b+math.sqrt(b*b-4*a*c))/(2*a) x2=(-b-math.sqrt(b*b-4*a*c))/(2*a) return x1,x2 #test print('quadratic...(2, 3, 1) =', quadratic(2, 3, 1)) print('quadratic(1, 3, -4) =', quadratic(1, 3, -4)) if quadratic(2...= (-0.5, -1.0): print('测试失败') elif quadratic(1, 3, -4) !
(X_train_quadratic, y_train)xx_quadratic = quadratic_featurizer.transform(xx.reshape(xx.shape[0], 1))...(degree=2)X_train_quadratic = quadratic_featurizer.fit_transform(X_train)X_test_quadratic = quadratic_featurizer.transform...(X_test)regressor_quadratic = LinearRegression()regressor_quadratic.fit(X_train_quadratic, y_train)xx_quadratic...(degree=2)X_train_quadratic = quadratic_featurizer.fit_transform(X_train)X_test_quadratic = quadratic_featurizer.transform...(X_test)regressor_quadratic = LinearRegression()regressor_quadratic.fit(X_train_quadratic, y_train)xx_quadratic
add to the x-coordinate of the last point on * this contour, for the control point of a quadratic...add to the y-coordinate of the last point on * this contour, for the control point of a quadratic...* * @param x1 The x-coordinate of the control point on a quadratic curve * @param y1 The y-coordinate...of the control point on a quadratic curve * @param x2 The x-coordinate of the end point on a quadratic...curve * @param y2 The y-coordinate of the end point on a quadratic curve */ public void quadTo(float
1.先回忆一下ax2+bx+c=0这个一元二次方程的数学解法 2.python实现 在我们知道求根公式后,我们用python来实现一下: def my_quadratic(a,b,c): if...+ math.sqrt(d))/(2*a) x2 = (-b - math.sqrt(d))/(2*a) return x1,x2 print(my_quadratic...(2,3,1)) # (-0.5, -1.0) print(my_quadratic(1,3,-4)) # (1.0, -4.0) print(my_quadratic(2,2,5)) #
题目: 请定义一个函数 ’quadratic(a,b,c)‘,接收三个参数,返回一元二次方程: ax² + bx + c = 0 的两个解。...(提示:计算平方根可以调用math.sqrt()函数) import math def quadratic(a, b, c): if not isinstance(a, (int, float)...derta)) / (2 * a) x2 = (-b - math.sqrt(derta)) / (2 * a) return x1, x2 print(quadratic...(2, 3, 1)) print(quadratic(1, 3, -4)) 发布者:全栈程序员栈长,转载请注明出处:https://javaforall.cn/156154.html原文链接:https
2 方法 定义一个函数quadratic接收三个参数,运用数学计算∆的方法赋值给变量s,调用计算平方根的方法算出x1、x2的值 代码清单 def quadratic(a,b,c): #定义一个函数接受三个参数...x2=(-b-math.sqrt(s))/(2*a) return x1,x2 #求解该方程 else: return 'unsolvable' #无解 print(quadratic
解决方案 以math.sqrt()函数为例 定义一个函数,quadratic(a,b,c),接收三个参数,返回一元二次方程ax²+bx+c=0的两个解。...import math def quadratic(a,b,c): if a == 0: raise TypeError('a不能为0') if not isinstance...无实根' x1= (math.sqrt(delta)-b)/(2*a) x2= (math.sqrt(delta)+b)/(2*a) return x1,x2 print(quadratic...(1,3,1)) print(quadratic(2,3,-4)) 上面的函数会输出以下结果 ?
2 方法 定义一个函数,quadratic(a,b,c),接受三个参数 插将一元二次方程转换为形如a+b+c=0 插入“import math” 引用“math.sqrt 利用if语句得到最终解或显示“...代码清单 1 def quadratic(a,b,c): n=b*b-4*a*c import math if n>=0: x1=(-b+math.sqrt(n))/(2...*a) x2=(-b-math.sqrt(n))/(2*a) return x1,x2 else: return('该一元二次方程无解') print(quadratic
1 问题 请定义一个函数,quadratic(a,b,c),接受三个参数,返回一元二次方程ax2+bx+c=0的两解。...代码清单 1 #quadratic(a,b,c),接受三个参数 #math.sqrt()函数计算平方根 import math def quadratic(a,b,c): m = b**2 - 4*a*
dqm.set_linear(i, linear_term) # Quadratic term for node pairs that do *not* share edges for p0,...p1 in nx.non_edges(G): quadratic_term = ......dqm.set_quadratic(p0, p1, {(c, c): quadratic_term for c in range(num_partitions)}) # Quadratic term...for node pairs which have edges between them for p0, p1 in G.edges: quadratic_term = ......dqm.set_quadratic(p0, p1, {(c, c): quadratic_term for c in range(num_partitions)}) # solve!
1 问题 定义一个函数,quadratic(a,b,c),接收三个参数,返回一元二次方程ax2+bx+c=0的两个解。...代码清单 1 def quadratic(a,b,c): for x in range(-1000,1000): if a*x*x+b*x+c==0: print...(x) a,b,c=map(int,input().split()) quadratic(a,b,c) 4 结语 针对编写函数求解一元二次方程的问题,提出def语句定义函数进行求解的方法,证明该方法是有效的
x_val', [h_grid;v_grid;d_scores], 'Q'); %% Part 2 & 3: Linear and Quadratic Regression for i = 1:length...*x(2,:))' (x(2,:).^2)']; % quadratic parameters initial_theta_Q = zeros(6, 1); label = double...% Quadratic: plot training data and trained classifier plot_training_data(label, i, x_Q...*x_val(2,:))' (x_val(2,:).^2)']; decision_Q = test_set_Q*theta_Q >= 0; % Quadratic: plot...length(vGrid)); for i = 1:length(hGrid) for j = 1:length(vGrid) % Map to a quadratic
) # eg:定义函数,返回方程 ax2 + bx + c = 0 的根 print('----------求解ax2 + bx + c = 0 的根------------') def my_quadratic...+ math.sqrt(d))/(2*a) x2 = (-b - math.sqrt(d))/(2*a) return x1,x2 print(my_quadratic...(2,3,1)) # (-0.5, -1.0) print(my_quadratic(1,3,-4)) # (1.0, -4.0) print(my_quadratic(2,2,5)) #
float x2, float y2,float x3, float y3) quadTo()方法,可以从源码的注释看出(x1,y1)是控制点坐标,(x2,y2)是终点坐标 /** * Add a quadratic...* * @param x1 The x-coordinate of the control point on a quadratic curve * @param y1 The y-coordinate...of the control point on a quadratic curve * @param x2 The x-coordinate of the end point on a quadratic...curve * @param y2 The y-coordinate of the end point on a quadratic curve */ public void quadTo
use quickdiv::DivisorU64; fn is_quadratic_residue(q: u64, modulus: u64) -> bool { // Initializing...(is_quadratic_residue(152, 169)); assert!(!...is_quadratic_residue(51, 111)); ReadMore: https://github.com/dtrifuno/quickdiv
can be efficiently solved with a number of methods, for example the finite-horizon iterative Linear Quadratic...u1:T⋆=argminx1:T∈X,u1:T∈U∑t=1TCt(xt,ut)subjecttoxt+1=f(xt,ut)x1=xinit, Our code currently supports a quadratic...cost function CC (non-quadratic support coming soon!)...Internally we solve a sequence of quadratic programs More details on this are in the box-DDP paper that
retro quadratic hypothesis nonlinear transform price on nonlinear transform structured hypothesis sets...quadratic hypothesis 如果线性不可分,虽然dvcd_{vc}可以保证generation,但是EinE_{in}很大也不行啊,这时候可以通过非线性变换将XX空间变换到ZZ空间,维度可能增加
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