Can anyone tell me about calculus? 50 points

Calculus is the branch of mathematics in advanced mathematics that studies the differentiation and integration of functions, as well as related concepts and applications. It is a fundamental subject of mathematics. The content mainly includes limits, differential calculus, integral science and their applications.

Differential calculus consists of the operation of finding derivatives and is a set of theories about the rate of change. It makes it possible to discuss functions, velocities, accelerations, and slopes of curves in a common set of notations. Integralism, including the operation of finding integrals, provides a general set of methods for defining and calculating area, volume, etc.

If you want to learn, it's best to buy a textbook.

I can't explain it clearly in a few words.

If you want to study before going to university, if you don't want to do scientific research, it's not necessary.

Of course, if you are interested, you can, after all, this thing is very useful, in advanced science, but it is very complicated.

Look for advanced math books, they say it very clearly! Simply put, it is the inverse process of finding the derivative ......

First, according to the differential equation y given in the problem'+y = e x, we can solve it using the solution of a first-order linear ordinary differential equation.

Deform the original equation:

y' =e^x - y

It is then normalized to Y for early onset'+p(x)y = q(x), where p(x) =1 and q(x) =e x, get:

y' +y = e^x

The special solution is solved using the constant variation method. First, write the homogeneous equation y'+y = 0 for general solution:

y_h = c*e^(-x)

where c is a constant.

Then, let the constant c be a function of x, c(x), i.e., c=c(x), and substitute it into the nonhomogeneous equation to get :

c'(x)e^(-x) +c(x)*(e^(-x)) c(x)*e^x = e^x

Simplification yields: c'(x)*e^(-x) =e^(2x)

Solution: c(x) = 1 2*e x + a

where a is a constant.

Therefore, the special interpretation is:

y_p = 1/2)e^x + ae^(-x)

Substituting the initial note x = 0, y = 2 yields:

a = y_0 + 1/2)*e^x_0 = 2+1/2)*1 = 5/2

Therefore, the special interpretation is:

y = 1/2)*e^x + 5/2)*e^(-x) +2

After examination, it can be found that y = 1 2)*e x + 5 2)*e (-x) +2 satisfies the original differential equation and the initial conditions.

According to Taylor's formula, for any x>0, t>0, there is.

f(x+t)=f(x)+f'(x)*t+(1/2)*f''(t 2, where (x,x+t).

then f'(x)*t=f(x+t)-f(x)-(1/2)*f''(t^2f'(x)*t|=|f(x+t)-f(x)-(1/2)*f''(t^2|

f(x+t)|+f(x)|+1/2)*|f''(t^2=a+a+(1/2)*b*t^2

2a+(b/2)*t^2

f'(x)|<2a/t+bt/2

Because the above inequality is true for any x>0, |f'(x)|=2 (ab), and if and only if t=2 (a b), the equal sign holds.

So |f'(x)|<2√(ab)

You don't quite understand what the purpose of this definition is.

This definition directly refers to the need for anything that is small

n can always be found, which is that the absolute value of the difference between the term after an and a is less than , which can also be understood as, when n, n, > n existsan-a|The absolute value of is always infinitely decreasing as n increases, and infinitely close to 0

The key to the definition is that the latter term is closer to 0 than the previous term is 0 from the value of a, which is a kind of recursive definition, similar to the mathematical induction method of an establishment and the introduction of an+1 is established, which can ensure that all are true.

I say this just to help you understand the definition, that is, the latter term is closer to a than the previous one, this is a definition in itself, not a proposition, the definition itself is unprovable, just an explanatory statement, and if it can be proved, it is a theorem, not a definition.

Calculus, calculus, so irritable, so irritable.

Arbitrary is that no matter how you take it, it can be infinitely small... The results are all true...

This is the most basic limit idea in high mathematics, just understand it.

For example, if there is a point, as long as the center of the circle is at the origin, then no matter how small the radius of the circle is, this point is in the circle, then this point can only be the origin.

How do you know that you have taken it all.

Due to the continuity of real numbers, this is generally described by a sequence of real numbers, such as 1 n, 3 n 2, and so on.

The method is shown in the figure below, please check carefully, I wish you a happy study and academic progress!