How do you review the median value theorem for high numbers?

Linear algebra and probability theory and mathematical statistics in advanced mathematics are more difficult than those who are new to it.

Linear Algebra includes determinants, matrices, systems of linear equations, vector spaces and linear transformations, eigenvalues and eigenvectors, diagonalization of matrices, quadratic forms and application problems.

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The median value theorem is an important theorem that reflects the connection between functions and derivatives, and is also the theoretical basis of calculus. It plays an important role in many ways, and has many applications in the derivation of some formulas and theorem proofs.

A function and its derivative are two different functions; The derivative, on the other hand, is only a local feature of the function at a point; If we want to understand the overall behavior of the absolute basis solution function in its defined domain, it is necessary to establish a connection between the derivative and the function, and this is the case with the differential median value theorem. Differential median theorem, including Roll's theorem, Lagrange's theorem, Cauchy's theorem, Taylor's theorem.

It is a bridge between the value of the derivative and the value of the function, and it is a tool for inferring the overall properties of the function by using the local properties of the derivative. A set of median theorems consisting of Roll's theorem, Lagrangian median theorem and Cauchy's median theorem is the theoretical basis of the whole code macro differential calculus. The lagrangian delay value theorem establishes a quantitative relationship between the value of the function and the derivative value, so the median value theorem can be used to study the behavior of the function through the derivative.

Practical application of the median value theorem:

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First of all, the three median theorems are premised on closed-interval continuity.

1 The essence of Roll's theorem is that if the two endpoints on the closed interval are equal, then there must be such a point on the function, and its derivative value is zero.

That is, if the two endpoints are equal, that is, the tangent of one point is a horizontal horizontal line (parallel to the x-axis).

2 The Lagrangian median theorem means that two endpoints are connected by a line segment, which is the string of the curve function. There must be a point, a point, on the function, whose tangent is parallel to the strings.

Using Rolle's theorem to prove the Lagrangian, construct a function, that is, the function of the curve minus its string, which geometrically is equivalent to pulling the curve parallel to the x-axis. That is, the endpoint values are equal. Then Roll's theorem exists a little bit.

The essence of this proof process is that any curve function can be pulled into a function that is horizontal to the x-axis, which satisfies Roll's theorem.

3 Cauchy's median theorem is to treat two functions as parametric equations, and there is a point on this parametric equation, and its tangent is parallel to the string of the equation curve, which is the Lagrangian median theorem at a higher level. This is actually the case of g(x)=x. It is required that the first derivative of g(x) is not equal to 0, and we know that the derivative equal to 0 must be a constant function...

You know. 4 Taylor's formula is a formula that connects functions and series, and there are two forms, which are actually different remainders.

The implication is that if a function is continuously derivable of n order in an interval, then the function can be a series of any point in the interval, that is, the sum of many formulas related to n, the point chosen. But no matter what point you choose, the final sum is still the same as the original function.

That's why we chose to be at zero (i.e., McLaughlin series) – it's the easiest way to do the math because it doesn't matter at which point.

Math is all about meaning, not a pair of silly 2b formulas. Although it looks quite similar. If you grasp the mathematical ideas, you can learn them well, but if you rush those formulas, you will perish. Hope it helps.

Look at the example problems, grasp the key points of the example problems, and it is best to know how the theorem came about.

First, using the integral median theorem, we can get f(1)=2 (3 2 2)f(x)dx=2f(c)(2-3 2)=f(c)where c is a median between 3, 2, and 2. Then, because f(x) is derivable in [1,2], the result can be obtained directly using Rawl's theorem.

Since f(x) is continuous on [3 2,2], there is 1 [3 2,2] such that f( 1) 2=f( 1)(2-3 2)= f(x)dx (integral range [3 2,2]).

Thus f(1)=2*f( 1) 2=f( 1) Since f(x) is continuous on [1, 1] and derivable on (1, 1), there exists (1, 1), such that f'( ) = 0 Note: Proof of the step of the integration.

Since f(x) is continuous on [3 2,2], there is m,m [3 2,2] such that for any x [3 2,2], f(m) f(x) f(m) so f(m) 2=(2-3 2)f(m) f(x)dx (integral range [3 2,2]) 2-3 2)f(m)=f(m) 2

And because f(x) is continuous on [3 2,2] and f(m) f(x) f(m) f(m) there is 1 [3 2,2] such that f( 1) 2= f(x)dx (integration range [3 2,2]).

Take the midpoint of the interval [a,b] (a+b) 2 According to the Lagrangian median theorem, there exists (a,(a+b) 2 such that f'( )=[f((a+b) 2)-f(a)] [(a+b) 2-a]=2[f((a+b) 2)-f(a)] (b-a) Let g(x)=x 2, then according to Cauchy's median theorem, there exists (a+b) 2,b), such that f'(η)/g'(η)=[f(b)-f((a+b)/2)]/[g(b)-g((a+b)/2)] f'( ) 2 =[f(b)-f((a+b) 2)] [b 2-(a+b) 2 4]=4[f(b)-f((a+b) 2)] (3b+a)(b-a) so f'(ξ)/(3b+a)+f'(η)/4η =2[f((a+b)/2)-f(a)]/(b-a)(3b+a)+2[f(b)-f((a+b)/2)]/(3b+a)(b-a) =2[f(b)-f(a)]/(b-a)(3b+a) =0

Take the midpoint of interval [a,b] (a+b) 2

According to the Lagrangian median theorem, there is (a,(a+b) 2), such that.

f'(ξ)=[f((a+b)/2)-f(a)]/[(a+b)/2-a]=2[f((a+b)/2)-f(a)]/(b-a)

Let g(x)=x 2, then according to Cauchy's median theorem, exist (a+b) 2,b), such that.

f'(η)/g'(η)=[f(b)-f((a+b)/2)]/[g(b)-g((a+b)/2)]

f'(η)/2η=[f(b)-f((a+b)/2)]/[b^2-(a+b)^2/4]=4[f(b)-f((a+b)/2)]/(3b+a)(b-a)

So f'(ξ)/(3b+a)+f'(η)/4η

2[f((a+b)/2)-f(a)]/(b-a)(3b+a)+2[f(b)-f((a+b)/2)]/(3b+a)(b-a)

2[f(b)-f(a)]/(b-a)(3b+a)=0

f'( 1)=f(1)-f(0) is 2 1=1 1=1 2f'( 2)=f(1)-f(0) so 3 2 2=1 2=root number 1 3

Verify Cauchy's median theorem.

So f'(ξ3)/f'(ξ3)=f(1)-f(0)/f(1)-f(0)=1

That is, 3 2 3=1 is trapped by 3=2 3

The function g(x)=f(x)e x is considered because f(x) is continuous at [a,b] and derivable at (a,b), and f(a)=f(b)=0

Therefore, g(x) is argued in [a,b] continuous bush hall, and is derivable in (a,b), and g(a)=g(b)=0

For g(x) we can apply Rohr's median theorem to (a,b).

At least a little bit of t in (a,b) belongs to (a,b) such that g'(t)=0, i.e. e x =0 because e x >0 so f(t)+f'(t)=0

In the problem, a point d is given, and the slope of the ab line is k, i.e., k=(f(a)-f(b)) (a-b), and d is on ab, so k=(f(a)-f(b)) (a-b)=(f(a)-f(d)) (a-d)=(f(d)-f(b)) (d-b).

From the Lagrangian median theorem:

f'(ξ)=(f(a)-f(d))/(a-d)f'(η)=(f(d)-f(b))/(d-b)f'(ξ)=f'(η)=k

From the Rohr median value theorem:

f''(c)=(f'(ξ)f'(η)/(ξ-