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Advanced calculus theorem
Rolle's theorem: If the function f(x) is continuous in the closed interval [a, b] and derivable in the open interval (a, b), and the function values at the end points of the interval are equal, that is, f(a)=f(b), there is at least one point ξ (a).

Lagrange theorem: Lagrange theorem in calculus (Lagrange mean value theorem)

Let the function f(x) satisfy the condition:

(1) is continuous on the closed interval [a, b];

(2) Derivable in the open interval (a, b);

Then at least a little ε∈( a, b), so

f(b) - f(a)

F' (ε) = - or

b-a

f(b)=f(a) + f(ε)'(b - a)

Roberta's Law: If there is 1 for the infinitive of type 0/0, the functions f(x) and F(x) tend to zero when x tends to a; 2. In the centripetal neighborhood of point A, both F'(x) and F'(x) exist, and F'(x) is not zero; 3. If the limit of f'(x)/F'(x) exists, then the limit of f(x)/F(x) is equal to f'(x).

Taylor theorem: Taylor mean value theorem: If the function f(x) has a derivative of order n+ 1 in the open interval (a, b), when the function is in this interval, it can be expanded into the sum of a polynomial about (x-x.) and a remainder;

f(x)= f(x .)+f '(x .)(x-x .)+f ' '(x .)/2! ? (x-x.)^2,+f'''(x.)/3! ? (x-x.)^3+……+f(n)(x.)/n! ? (x-x.)^n+Rn

Where rn = f (n+1) (ξ)/(n+1)! ? (x-X.) (n+ 1), where ξ is between x and x, and this remainder is called Lagrange remainder.

Mclaughlin formula: f(x)=f(0)+f'(0)*x+f''(x)/2! *x^2+...+f(n)(0)/n! * x n (McLaughlin formula, where n in the last term represents the n-order derivative)

Fermat's last theorem: Fermat's last theorem;

When the integer n > 2, the indefinite equation about x, y and z.

x^n + y^n = z^n.

((x, x, y) = (x, z) = (y, z) =1[n+0 (n is an odd prime number) x > 0, y>0, z>0, and xyz≠0) has no integer solution.

Cauchy theorem: Cauchy mean value theorem

If the functions f(x) and F(x) satisfy:

(1) is continuous on the closed interval [a, b];

(2) Derivable in the open interval (a, b);

(3) For any x∈(a, b), F'(x)≠0,

Then there is at least one zeta in (a, b), which constitutes the equation.

[f (b)-f (a)]/[f (b)-f (a)] = f' (zeta)/f' (zeta) holds.