Continuity is generally a big problem. Left continuity equals right continuity.

The formulas of definite integral and indefinite integral should be memorized well.

And formula derivation.

L'H?pital's law is a method to determine the indeterminacy under certain conditions by taking derivatives of the numerator and denominator respectively and then finding the limit.

set up

(1) When x→a, the functions f(x) and F(x) tend to zero;

(2) In the centripetal neighborhood of point A, both f'(x) and F'(x) exist, and F' (x) ≠ 0;

(3) When x→a, lim f'(x)/F'(x) exists (or is infinite), then

When x→a, lim f(x)/F(x)=lim f'(x)/F'(x).

rebuild

(1) When x→∞, both functions f(x) and F(x) tend to zero;

(2) f'(x) and F'(x) both exist when |x| > is n, and F' (x) ≠ 0;

(3) When x→∞, lim f'(x)/F'(x) exists (or is infinite), then

When x→∞, lim f(x)/F(x)=lim f'(x)/F'(x).

Using L'H?pital's law to find the limit of infinitive is one of the key points of differential calculus. When solving problems, we should pay attention to:

(1) before you start to seek the limit, you should first check whether you meet the 0/0 or ∞ /∞ indefinite formula, otherwise it is wrong to abuse L'H?pital's law. When it does not exist (excluding the ∞ case), it cannot be used. At this time, it is said that L'H?pital's law is not applicable, and other methods should be used to find the limit. For example, use Taylor formula to solve.

(2) If the conditions are met, the Lobida rule can be used repeatedly until the limit is found.

(3) The Lobida rule is an effective tool to find the limit of the indefinite form, but if only the Lobida rule is used, the calculation will often be very complicated, and other methods must be combined, such as separating the product factor of non-zero limit in time to simplify the calculation, and replacing the product factor with equivalent quantity. Taylor formula.

Taylor's mean value theorem: if the function f(x) has a derivative up to n+ 1 in the open interval (a, b), when the function is in this interval, it can be expanded into the sum of (x-x.) polynomials 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.

(Note: f(n) (x.) is the nth derivative of f (x.), not the product of f (n) and x. )

It is proved that we know f(x)=f(x.)+f'(x.)(x-x.)+α (the finite increment theorem derived from Lagrange's mean value theorem is lim δ x→ 0 f (x.+δ x)-f (x.) = f' (x.) δ x), where the error α is in. Therefore, we need a polynomial that is accurate enough to estimate the error:

p(x)=a0+a 1(x-x.)+a2(x-x.)^2+……+an(x-x.)^n

Approximate the function f(x) and write the specific expression of its error f(x)-P(x). Let the function P(x) satisfy p (x.) = f (x x.), p p' (x.) = f'' (x x.), p' (x.) = f'' (x.), ..., p (n) (x.) = f (n). Obviously, P(x.)=A0, so A0 = F (X.); P'(x.)=A 1，a 1 = f '(x .)； P''(x.)=2！ A2，A2=f''(x.)/2！ ……P(n) (x.)=n！ An，An=f(n)(x.)/n！ . So far, the coefficients of many terms have been calculated, which are: p (x) = f (x.)+f' (x.) (x-x.)+f'' (x.)/2! ? (x-x.)^2+……+f(n)(x.)/n！ ? (x-x.)^n.

Next, the specific expression of the error is needed. Let Rn(x)=f(x)-P(x), so rn (x) = f (x)-p (x) = 0. So it can be concluded that rn (x.) = rn' (x.) = rn'' (x.) = ... = rn (n) (x.) = 0. According to Cauchy mean value theorem, we can get rn (x)/(x-x.) (n+1) = (rn (x)-rn (x.)/(x-x.) (n+1)-0) = rn' (. Continue to use (rn' (ξ1)-rn' (X.))/((n+ 1) (ξ1-X.) n-0) = rn after continuous use of n+1times. Here ξ is between X. and X. But Rn (n+1) (x) = f (n+1) (x)-p (n+1) (x), because P(n)(x)=n! Ann, n! An is a constant, so P(n+ 1)(x)=0, so we get rn (n+1) (x) = f (n+1) (x). To sum up, the remainder rn (x) = f (n+1) (ξ)/(n+1)! ? (x-x.)^(n+ 1)。 Generally speaking, when expanding a function, it is for the need of calculation, so X often takes a fixed value, and then Rn(x) can also be written as Rn. Maclaurin expansion: If the function f(x) has a derivative of order up to 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 and a remainder:

f(x)=f(0)+f'(0)x+f''(0)/2！ ? x^2,+f'''(0)/3！ ? x^3+……+f(n)(0)/n！ ? x^n+Rn

Where rn = f (n+1) (θ x)/(n+1)! ? X (n+ 1), where 0

It is proved that if the polynomial p (x) = A0+A 1x+A2X 2+ is used to approximate the function f(x)...+ANX N and obtain the specific expression of its error, we can rewrite Taylor's formula into a simpler form, that is, a special form when x = 0:

f(x)=f(0)+f'(0)x+f''(0)/2！ ? x^2,+f'''(0)/3！ ? x^3+……+f(n)(0)/n！ ? x^n+f(n+ 1)(ξ)/(n+ 1)！ ? x^(n+ 1)

Since ξ is between 0 and x, it can be written as θx, 0.

1, expand trigonometric function y=sinx, y=cosx.

Solution: According to the derivative table: f (x) = sinx, f' (x) = cosx, f' (x) =-sinx, f' (x) =-cosx, f (4) (x) = sinx. ...

Thus, the periodic law is obtained. Calculate f(0)=0, f' (0) = 1, f'' (x) = 0, f'' (0) =- 1, and f (4) = 0. ...

Finally available: sinx = x-x 3/3! +x^5/5！ -x^7/7！ +x^9/9！ -(This is written in the form of infinite series. )

Similarly, y=cosx can also be expanded.

2. Calculate the approximate value e = lim x→∞ (1+1/x) x.

Solution: apply maclaurin expansion to exponential function y = e x and discard the remainder:

e^x≈ 1+x+x^2/2！ +x^3/3！ +……+x^n/n！

When x= 1, e≈ 1+ 1+ 1/2! + 1/3! +……+ 1/n！

If n= 10, the approximate value e≈2.7 1828 18 can be calculated.

3. Euler formula: e ix = cosx+isinx (I is the root of-1, that is, the imaginary unit).

Proof: This formula writes complex numbers as power exponents, which is actually proved by maclaurin expansion or McLaughlin series. I won't write the process in detail, but let's talk about the idea first: first expand the exponential function e z, and then write z in each term as ix. Because of the power periodicity of I, the term containing soil I in the coefficient can be written together by multiplication and division and distribution, and the remaining terms are also written together, which is exactly the expansion of cosx, sinx SINX. Then multiply sinx by the suggested I, and the Euler formula can be derived. You can prove it yourself if you are interested.

The discovery of Taylor expansion principle e began with differentiation. When h gradually approaches zero, the calculated value is infinitely close to a certain value of 2.7 1828 ..., and this fixed value is E. The first person to discover this value is the famous Swiss mathematician Euler. He named the irrational number after his name with the prefix lowercase e.

Calculate the derivative of logarithmic function, and get that when a=e, the derivative of is 0, so the logarithm based on e is more reasonable, which is called natural logarithm.

If the exponential function ex is Taylor expansion, then

Substitute x= 1 into the above formula.

This series converges quickly, and the value of E is approximately 40 decimal places.

When the exponential function ex is extended to the complex number z=x+yi, it is determined by the following formula.

Through this series of calculations, we can get

From this, Demol's theorem, the formula of sum and difference angles of trigonometric functions and so on can be easily deduced. For example, z 1 = x 1+y 1i, z2 = x2+y2i,

On the other hand,

So,

We can not only prove that e is an irrational number, but also a transcendental number, that is, it is not the root of any integer coefficient polynomial. This result was obtained by Hermite in 1873.

A) differences.

Consider a discrete function (sequence) R, whose value u(n) at n is denoted as un, and we usually write this function as OR (un). The difference of the sequence U is still a sequence, and its value in n is defined as

In the future, let's remember Jane

(Example): The difference sequence of sequence 1, 4, 8, 7, 6, -2, ... is 3, 4,-1, -8. ...

Note: We say "series" is "a function defined at discrete points". This statement sucks in high school, but it is appropriate here because it has a completely parallel analogy with continuous functions.

The nature of difference operator

(a) [collectively referred to as linear]

(ii) (Constant) [Basic Theorem of Difference Equation]

㈢

Where (n(k) is called a permutation sequence.

(4) called natural geometric series.

(iv) The difference sequence (i.e. "derivative function") of the' general exponential sequence (geometric sequence) rn is rn(r- 1).

(2). Sum integral

Give a series of numbers (un). The problem of summation is to calculate summation. How to calculate? We got the following important results:

Theorem 1 (basic theorem of difference and division) If we can find a sequence (vn), then

Sum and fraction also have linear properties:

A) differences

Given a function f, if the limit of Newton's quotient (or difference quotient) exists, then we call this limit value f the derivative of point x0 and write it as f'(x0) or Df(x), that is.

If the derivative of f exists at every point in the defined area, it is called a differentiable function. We call it the derivative function of f, not the differential operator.

Properties of differential operators:

(a) [collectively referred to as linear]

(ii) (Constant) [Basic Theorem of Difference Equation]

(3) Dxn=nxn- 1

Dex = ex

(iv)' The derivative function of the general exponential sequence ax is

(b) integration.

Let f be a function defined on [a, b], and the problem of integration is to calculate the shadow area. Our method is divided into [a, b]:

; Secondly, take a sample point [xi- 1, xi] for each small segment; Then find the approximate sum; Finally, take the limit (let the length of each segment approach 0).

If this limit exists, we will remember that the geometric meaning is the shadow area.

(In fact, continuity is also "almost" a necessary condition for the existence of integral. )

Integral operators also have linear properties:

Theorem 2 If F is a continuous function, it exists. (In fact, continuity is also a necessary condition for the existence of an integral. )

Theorem 3 (Basic Theorem of Calculus) Let f be a continuous function defined in the closed interval [a, b], and we want to get the integral. If we can find another function g that makes g'=f, then

Note: (1)(2) Although the two formulas are analogies, there is one difference, that is, be careful about the upper limit of the sum!

Theorem 1 and Theorem 3 above basically talk about difference and division, and differential and integral are two reciprocal operations, just as addition, subtraction, multiplication and division are reciprocal operations.

As we all know, the operation of difference and differential is much simpler than the operation of sum, fraction and integral. Theorem 1 and Theorem 3 above tell us that to calculate the sum of (un) and the integral of fraction and f, we only need to find another (vn) and g to satisfy G'= F (this is a problem of difference and differentiation), and then we can get the answer by substituting vn and g into the upper and lower limits. In other words, we can use something simpler.

A) Taylor expansion formula

There are discrete and continuous analogies. It is a special case of the important idea of approximation in mathematics. The idea of approximation is this: given a function F, we should study its behavior, but F itself may be very complicated and difficult to handle, so we try to find a simpler function G to make it "close" to F, and then we use G instead of F, which is to simplify the complexity.

Two questions: How to choose simple function and approximate scale?

(1) In the case of continuous world, the approximation idea of Taylor expansion is to select polynomial function as simple function and local tangency as approximation scale. More specifically, given a function f that can be differentiated to order n, we need to find a polynomial function g of order n so that it is "tangent" to f at point x0, that is, the answer is

This formula is called the n-order Taylor expansion of f at point x0.

G is very close to F near x0, so we use G to replace F locally, so we can use G to find some local qualitative behaviors of F, so Taylor expansion is only a local approximation. When F is a good enough function, that is, the so-called analytic function, F can be expanded into Taylor series, which is equal to F itself.

It is worth noting that in the special case of the first-order Taylor expansion, the graph of g(x)=f(x0)+f'(x0)(x-x0) is just a straight line that passes through the graph of points (x0, f(x0)) and is tangent to F. Therefore, the significance of the first-order Taylor expansion of f at point x0 is that we have used the point (x0).

Taylor expansion can help us do many things, such as judging the maximum and minimum values of functions, finding the approximate value of integrals, and making function tables (such as trigonometric function tables and logarithmic tables). In fact, we can "consistently" calculus with approximate ideas.

Many times, we notice that we choose polynomial function as simple approximation function for a simple reason: among many elementary functions, such as trigonometric function, exponential function, logarithmic function, polynomial function and so on. From the arithmetic point of view, polynomial function is the simplest, because to calculate the value of polynomial function, it only involves four operations of addition, subtraction, multiplication and division, and other functions are not so simple.

Of course, from other analysis angles, in some cases, there are other simple functions that are more useful and important. For example, trigonometric polynomials, combined with some approximation scales, give us Fourier series expansion, which plays an important role in applied mathematics. (In fact, Fourier series expansion is an approximation scale with the smallest variance, which often appears in higher mathematics and is also applied in statistics. )

Note: Take the special case of x0=0 as an example. At this time, Taylor expansion is also called Ma Kraulin expansion. However, as long as we can expand a special case and want a general Taylor expansion, it is good to do translation (or variable substitution). Therefore, Taylor expansion can only be done at the point of x=0 from the beginning.

(2) For discrete cases, Taylor expansion is:

Given a sequence, we need to find a polynomial sequence with degree n (gt) so that gt and ft have N-order "difference approximation" when t=0. The so-called zero-order n-order difference approximation refers to:

The answer is that this formula is Maclaurin formula in discrete case.

B) analogy between partial integral formula and Abel partial sum formula

(a) partial integral formula:

Let u (x) and v (x) be continuous on [a, b], then

(2) Abel partial sum formula:

Let (UN) and (v) be two series, so Sn = U 1+...+UN, then

The above two formulas are Leibniz's derivative formula D(uv)=(Du)v+u(Dv) and Leibniz's difference formula respectively. Note that one of the two Leibniz formulas is very symmetrical, while the other is asymmetrical.

(d) compound interest and continuous compound interest (this is also an analogy between discrete and continuous respectively)

(1) The compound interest problem is as follows: there is principal y0, annual interest rate R, and compound interest once a year. Ask the principal and interest after N years and yn= Obviously, this series satisfies the difference equation yn+ 1=yn( 1+r).

According to (2) of (c), we know that yn=y0( 1+r)n is a compound interest formula.

(2) If compound interest is considered m times a year, the sum of principal and interest after t years is

Order, you get the concept of continuous compound interest, and the sum of principal and interest at this time is y(t)= yoert.

In other words, the principal and interest of time t and y(t)= yoert are the solutions of the differential equation y'=ry.

As can be seen from the above, the discrete compound interest problem is described by difference equation, while the continuous compound interest problem is described by differential equation. For linear difference equations and differential equations with constant coefficients, the whole point of solving equations is the superposition principle, so the solving methods are completely parallel.

(e) Fubini's theorem of multiple sums and points and Fubini's theorem of multiple integrals (also an analogy between discrete and continuous).

(1) Fubini's double sum theorem: given a series of numbers (ars) with double exponents, we want to sum r= 1 to m and S = 1 to n (ars), then this sum can be obtained as follows: light sum r, and then sum s (and vice versa). In other words, we have.

(2) Fubini's multiple integral theorem: Let f(x, y) be an integrable function defined on, then

Of course, several variables are the same.

The concept of leberg integral

(1) Discrete case: Given a series (an), the sum should be estimated. Leberg's idea is that no matter what the order of the data indicators of this pile is, we only divide it into piles according to the size of the values, and then multiply a value from each pile by the number of the piles to get the overall sum.

(2) Continuity: Given a function f, we need to define the area enclosed by the curve y=f(x) and the X axis from A to B. 。

Leberg's idea is to divide the shadow domain of F:

X whose function value is between yi- 1 and yi is set together to make it 0, thus [a, b] is divided into sampling points and approximately summed.

If the limit of the above approximate sum exists, it is called Lebesgue integral of f on [a, b]. The remainder of Taylor's formula is f(x)=f(a)+f'(a)(x-a)/ 1! + f''(a)(x-a)^2/2！ + …… + f(n)(a)(x-a)^n/n！ +Rn(x)[ where f(n) is the nth derivative of f]

Taylor remainder can be written in the following different forms:

1. Peano remainder:

Rn(x) = o((x-a)^n)

2. Remaining projects of 2.Schlomilch-Roche:

rn(x)= f(n+ 1)(a+θ(x-a))( 1-θ)^(n+ 1-p)(x-a)^(n+ 1)/(n！ p)

[f(n+ 1) is the n+ 1 derivative of f, θ∈(0, 1)]

3. Lagrange remainder;

rn(x)= f(n+ 1)(a+θ(x-a))(x-a)^(n+ 1)/(n+ 1)！

[f(n+ 1) is the n+ 1 derivative of f, θ∈(0, 1)]

4. Cauchy remainder:

rn(x)= f(n+ 1)(a+θ(x-a))( 1-θ)^n(x-a)^(n+ 1)/n！

[f(n+ 1) is the n+ 1 derivative of f, θ∈(0, 1)]

5. Integer remainder:

Rn(X)=[f(n+ 1)(t)(X-t)n an integer from a to x ]/n!

[f(n+ 1) is the n+ 1 derivative of f]

Also known as Cauchy mean value theorem.

Let the functions f (x) and g (x) be continuous in [a, b], (a, b) be derivable, and g'(x)≠0(x∈(a, b)).

Then at least one point ξ∈(a, b) makes f' (ξ)/g' (ξ) = [f (a)-f (b)]/[g (a)-g (b)] hold.

Geometrically, if u = f (x) and v = g (x), this form can be understood as a parametric equation, [f(a)-f(b)]/[g(a)-g(b)] is the end slope of the connecting parametric curve, and f'(ξ)/g'(ξ

When g(x)=x in Cauchy mean value theorem, Cauchy mean value theorem is Lagrange mean value theorem.

Proof order f (x) = f (x)-[f (a)-f (b)] g (x)/[g (a)-g (b)]

∫F(a)= F(b)=[F(a)g(b)-F(b)g(a)]/[g(b)-g(a)]

According to Rolle's theorem, there exists ξ∈(a, b), which makes F'(ξ)=0.

f '(x)= f '(x)-[f(a)-f(b)]g '(x)/[g(a)-g(b)]

So f' (ξ)-[f (a)-f (b)] g' (ξ)/[g (a)-g (b)] = 0.

That is f' (ξ)/g' (ξ) = [f (a)-f (b)]/[g (a)-g (b)]

The proposition is proved.

Rolle theorem Rolle theorem explanatory diagram

If the function f(x) satisfies:

Continuous on the closed interval [a, b];

Derivable in the open interval (a, b);

Where a is not equal to b;

The function values at the end points of the interval are equal, that is, f(a)=f(b),

Then there is at least one point ξ (a

The intuitive significance of the three known conditions of Rolle's theorem is: f(x) on [a, b] continuously shows that the curve is seamless, including the endpoints; The derivation that (a, b) contains f(x) shows that the curve y=f(x) has a tangent at every point; F(a)=f(b) indicates that the secant (straight line AB) of the curve is parallel to the X axis. The intuitive significance of the conclusion of Rolle's theorem is that at least one point ξ can be found in (a, b), so that f'(ξ)=0, which means that the tangent slope of at least one point on the curve is 0, so that the tangent is parallel to the secant AB and also parallel to X.

good luck