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Summary of the mean value theorem of higher numbers

2022-07-07 05:21:00 Full stack o-jay

The mean value theorem

Generally used for proof questions , Analysis steps : Determine the interval 、 Determine auxiliary function 、 Determine the theorem used 、 Key point analysis .
f(x) stay [a,b] Continuous on ,

  1. Bounded and maximum theorem m ≤ f ( x ) ≤ M m \leq f(x)\leq M mf(x)M
  2. Intermediate value theorem m ≤ μ ≤ M , ∃ ϵ ∈ [ a , b ] , f ( ϵ ) = μ m\leq \mu \leq M, \exists \epsilon \in[a,b], f(\epsilon) = \mu mμM,ϵ[a,b],f(ϵ)=μ
  3. Mean value theorem a < x 1 < x 2 < ⋯ < x n < b , ∃ ϵ ∈ [ x 1 , x n ] , f ( ϵ ) = f ( x 1 ) + f ( x 2 ) + ⋯ + f ( x n ) n a<x_1<x_2<\cdots<x_n<b, \exists \epsilon \in[x_1,x_n], f(\epsilon) = \frac{f(x_1)+f(x_2)+\cdots+f(x_n)}{n} a<x1<x2<<xn<b,ϵ[x1,xn],f(ϵ)=nf(x1)+f(x2)++f(xn)
  4. Zero point theorem f ( a ) ⋅ f ( b ) < 0 , ∃ ϵ ∈ [ a , b ] , f ( ϵ ) = 0 f(a)\cdot f(b)<0, \exists \epsilon \in[a,b], f(\epsilon) = 0 f(a)f(b)<0,ϵ[a,b],f(ϵ)=0
  5. Fermat's theorem x 0 It's about can guide And by extremely value , f ′ ( x 0 ) = 0 x_0 Is differentiable and extremum , f'(x_0) = 0 x0 It's about can guide And by extremely value ,f(x0)=0
  6. Rolle's theorem [ a , b ) can guide , f ( a ) = f ( b ) , ∃ ϵ ∈ [ a , b ] , f ′ ( ϵ ) = 0 [a,b) Derivable , f(a) = f(b), \exists \epsilon \in[a,b], f'(\epsilon) = 0 [a,b) can guide ,f(a)=f(b),ϵ[a,b],f(ϵ)=0
  7. Lagrange mean value theorem ( a , b ) can guide , ∃ ϵ ∈ [ a , b ] , f ( b ) − f ( a ) = f ′ ( ϵ ) ( b − a ) (a,b) Derivable , \exists \epsilon \in[a,b], f(b) - f(a) = f'(\epsilon)(b-a) (a,b) can guide ,ϵ[a,b],f(b)f(a)=f(ϵ)(ba)
  8. Cauchy mean value theorem ( a , b ) can guide , g ′ ( x ) ≠ 0 , ∃ ϵ ∈ [ a , b ] , f ( b ) − f ( a ) g ( b ) − g ( a ) = f ′ ( ϵ ) g ′ ( ϵ ) (a,b) Derivable , g'(x)\neq 0, \exists \epsilon \in[a,b], \frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(\epsilon)}{g'(\epsilon)} (a,b) can guide ,g(x)=0,ϵ[a,b],g(b)g(a)f(b)f(a)=g(ϵ)f(ϵ)
  9. Taylor formula ( lagrange remainder ) f ( x ) = f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 ) + ⋯ + f ( n ) ( x 0 ) n ! ( x − x 0 ) n + f ( n + 1 ) ( ϵ ) ( n + 1 ) ! ( x − x 0 ) n + 1 f(x) = f(x_0) + f'(x_0)(x-x_0)+\cdots+ \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n +\frac{f^{(n+1)}(\epsilon)}{(n+1)!}(x-x_0)^{n+1} f(x)=f(x0)+f(x0)(xx0)++n!f(n)(x0)(xx0)n+(n+1)!f(n+1)(ϵ)(xx0)n+1
  10. Taylor formula ( Payano's remainder ) f ( x ) = f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 ) + ⋯ + f ( n ) ( x 0 ) n ! ( x − x 0 ) n + O ( ( x − x 0 ) n ) f(x) = f(x_0) + f'(x_0)(x-x_0)+\cdots+ \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n +O((x-x_0)^n) f(x)=f(x0)+f(x0)(xx0)++n!f(n)(x0)(xx0)n+O((xx0)n)
  11. Mean value theorem of integral ∃ ϵ ∈ [ a , b ] , ∫ a b f ( x ) d x b − a = f ( ϵ ) \exists \epsilon \in[a,b], \frac{\int_a^bf(x)dx}{b-a} = f(\epsilon) ϵ[a,b],baabf(x)dx=f(ϵ)

There is a memory skill , The most commonly used are seven ,“ Zero dielectric ferroratesi ”, If you see the proof question, you can try it one by one ,

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