pip install sympy

from sympy import Symbol, symbols

x = Symbol('x')
y, z = symbols('y z')

from sympy import Symbol

x = Symbol('x')
y = Symbol('y')

# 加法
expr1 = x + y
print(expr1)  # 輸出: x + y

# 乘法
expr2 = x * y
print(expr2)  # 輸出: x*y

# 減法
expr3 = x - y
print(expr3)  # 輸出: x - y

# 除法
expr4 = x / y
print(expr4)  # 輸出: x/y

from sympy import Symbol, evalf

x = Symbol('x')
expr = 5*x + 4

# 求值
y1 = expr.evalf(subs={x: 6})
print(y1)  # 輸出: 34.0000000000000

from sympy import Symbol, evalf

x, y = symbols('x y')
expr = x**2 + y**2

# 求值
result = expr.evalf(subs={x: 3, y: 4})
print(result)  # 輸出: 25.0000000000000

from sympy import Symbol, solve

x = Symbol('x')
# 求解方程 x^2 - 4 = 0
equation = x**2 - 4
solution = solve(equation, x)
print(solution)  # 輸出: [-2, 2]

from sympy import symbols, solve

x, y = symbols('x y')
# 定義方程組
a = 4*x + 7 - y
b = 5*y - x + 6
# 求解方程組
solutions = solve((a, b), (x, y))
print(solutions)  # 輸出: {x: 1, y: 3}

from sympy import Symbol, diff

x = Symbol('x')
f = 2*x**4 + 3*x + 6

# 對(duì) f 求導(dǎo)
df = diff(f, x)
print(df)  # 輸出: 8*x**3 + 3

# 偏導(dǎo)
y = Symbol('y')
f3 = 2*x**2 + 3*y**4 + 2*y
dfx = diff(f3, x)
dfy = diff(f3, y)
print(dfx)  # 輸出: 4*x
print(dfy)  # 輸出: 12*y**3 + 2

from sympy import Symbol, integrate

x = Symbol('x')
f = 2*x**3 + 3*x**2 + 1

# 對(duì) f 進(jìn)行不定積分
F = integrate(f, x)
print(F)  # 輸出: x**4 + x**3 + x

from sympy import Symbol, integrate

x = Symbol('x')
f = x**2

# 對(duì) f 在區(qū)間 [0, 1] 上進(jìn)行定積分
result = integrate(f, (x, 0, 1))
print(result)  # 輸出: 1/3

from sympy import Symbol, limit

x = Symbol('x')
expr = (x**2 - 9) / (x - 3)

# 計(jì)算 x 趨于 3 時(shí)的極限
limit_value = limit(expr, x, 3)
print(limit_value)  # 輸出: 6

from sympy import symbols, summation

n, i = symbols('n i')
# 計(jì)算前 n 項(xiàng)和 1 + 2 + ... + n
sum_n = summation(i, (i, 1, n))
print(sum_n)  # 輸出: n*(n + 1)/2

# 計(jì)算具體值，如 n = 10
sum_10 = sum_n.subs(n, 10)
print(sum_10)  # 輸出: 55

from sympy import symbols, sin, series

x = symbols('x')
expr = sin(x)

# 將 sin(x) 在 x = 0 處展開(kāi)到 x^5
series_expansion = series(expr, x, 0, 5)
print(series_expansion)
# 輸出: x - x**3/6 + O(x**5)

from sympy import Matrix

# 創(chuàng)建 2x2 矩陣
A = Matrix([[1, 2], [3, 4]])
print(A)

# 創(chuàng)建 3x1 矩陣（列向量）
v = Matrix([1, 2, 3])
print(v)

# 矩陣加法
B = Matrix([[5, 6], [7, 8]])
C = A + B
print(C)

# 矩陣乘法
D = A * B  # 或者使用 A.dot(B)
print(D)

# 矩陣求逆
A_inv = A.inv()
print(A_inv)

# 矩陣的轉(zhuǎn)置
A_T = A.T
print(A_T)

from sympy import symbols, Eq, solve

t = symbols('t')
g = 9.8  # 重力加速度，單位 m/s^2

# 定義速度公式
v = g * t

# 假設(shè)我們要求解在 t = 5s 時(shí)的速度
t_value = 5
v_value = v.subs(t, t_value)
print(f"在 t = {t_value}s 時(shí)的速度為: {v_value} m/s")

# 如果問(wèn)題是求解達(dá)到特定速度 v_target 時(shí)所需的時(shí)間，可以這樣設(shè)置并求解
v_target = 49  # 假設(shè)目標(biāo)速度為 49 m/s
equation = Eq(v, v_target)
solution = solve(equation, t)

print(f"達(dá)到 {v_target} m/s 所需的時(shí)間為: {solution[0]}s")

from sympy import symbols, Eq, solve

P = symbols('P')
r = 0.05  # 假設(shè)年利率為 5%
n = 10  # 假設(shè)投資期限為 10 年
A_target = 1500  # 假設(shè)目標(biāo)未來(lái)值為 1500

# 定義復(fù)利公式
A = P * (1 + r)**n

# 如果我們已知 P 和 n，要求解 A 的值
P_value = 1000  # 假設(shè)本金為 1000
A_calculated = A.subs({P: P_value, n: n})
print(f"本金為 {P_value} 元，年利率為 {r*100}%，投資期限為 {n} 年時(shí)，未來(lái)值為: {A_calculated} 元")

# 如果我們要求解達(dá)到特定未來(lái)值 A_target 所需的本金 P
equation = Eq(A, A_target)
solution = solve(equation, P)

print(f"為了達(dá)到 {A_target} 元的未來(lái)值，在年利率為 {r*100}% 和投資期限為 {n} 年的條件下，需要的本金為: {solution[0]} 元")

from sympy import symbols, Eq, Function, dsolve

x = symbols('x')
y = Function('y')(x)  # 定義一個(gè)關(guān)于x的函數(shù)y

# 定義微分方程：y'' - 2y' - 3y = 0
# 其中，y' 表示 y 關(guān)于 x 的一階導(dǎo)數(shù)，y'' 表示二階導(dǎo)數(shù)
equation = Eq(y.diff(x, 2) - 2*y.diff(x) - 3*y, 0)

# 求解微分方程
solution = dsolve(equation)

print(solution)

from sympy import Matrix, symbols

# 定義一個(gè)3x3矩陣
A = Matrix([[1, 2, 3], [4, 5, 6], [7, 8, 9]])

# 計(jì)算特征值和特征向量
eigenvals, eigenvecs = A.eigenvals_right(), A.eigenvects_right()

print("特征值:", eigenvals)
print("特征向量:", eigenvecs)

# 假設(shè)我們想要找到對(duì)應(yīng)于特征值λ的特征向量，其中λ是已知的
lambda_val = 2  # 注意：這里的2可能不是A的一個(gè)特征值，僅為示例
eigenvec = A.eigenvects_right(lambda_val)
if eigenvec:
    print(f"特征值 {lambda_val} 對(duì)應(yīng)的特征向量為: {eigenvec[0][2][0]}")
else:
    print(f"矩陣A沒(méi)有對(duì)應(yīng)于特征值 {lambda_val} 的特征向量。")

from sympy import symbols, factor, expand, simplify

x, y = symbols('x y')

# 因式分解
expr = x**2 - y**2
factored_expr = factor(expr)
print("因式分解:", factored_expr)

# 展開(kāi)
expr = (x + y)**2
expanded_expr = expand(expr)
print("展開(kāi):", expanded_expr)

# 簡(jiǎn)化
expr = (x**2 + 2*x*y + y**2) / (x + y)
simplified_expr = simplify(expr)
print("簡(jiǎn)化:", simplified_expr)

from sympy import symbols, summation, product

n, k = symbols('n k')

# 符號(hào)求和
sum_expr = summation(k**2, (k, 1, n))
print("求和:", sum_expr)

# 符號(hào)積（注意：這通常不是數(shù)學(xué)中的“積”概念，而是類似求和的連續(xù)乘法）
# 但我們可以模擬一個(gè)有限積的計(jì)算
product_expr = product(k, (k, 1, n))
print("有限積（連續(xù)乘法）:", product_expr)