Taylor expansion formula of common functions

Taylor expansion formula of common functions

e^x = 1+x+x^2/2!+x^3/3!+…… +x^n/n!+……
ln(1+x)=x-x^2/2+x^3/3-…… +(-1)^(k-1)*(x^k)/k(|x|

Taylor formula McLaurin expansion can't remember? Is there any special memory method? And trigonometric function is also a headache

f(0)+f'(0)x+f''(0)x^2/(2!)+…… +The n-th derivative of F at 0 times the n-th power of X divided by the factorial plus remainder of n
The law is that the upper side is the n-th derivative multiplied by the n-th power of X, which is the factorial divided by n (do you see it? All are n). Needless to say, generally o (the n-th power of x). The Lagrange type remainder is: the N + 1-th derivative at thetax multiplied by the N + 1-th power of X, which is the factorial divided by N + 1, which is the law in front of it. It's too difficult to write, You'll find the McLaughlin formula easy to remember

For three numbers a, B and C, min | a, B and C | are used to represent the smallest of the three numbers, for example, min {- 1,2,3} = - 1,
Min {- 1,2, a} = a (a < = - 1) - 1 (a > = - 1), if min {2,2x + 2,4-2x} = 2, what is the value range of X?

2x+2≥2,x≥0
4-2x≥2,x≤1
So the value range of X is o ≤ x ≤ 1

Using Taylor's formula to get higher derivative
Let y = arcsinx,
(n)
Find y (0); (when x = 0, the N derivative of Y)

These heads are big,
Find the derivative of y = arcsinx, and then use Taylor formula directly,
Do you feel upset about finding the derivative of y = arcsinx

If the derivative function f '(x) of the differentiable function f (x) is known and satisfies XF' (x) > F (x), then when a > 1, the relationship between F (a) and AF (1) is?
It's better to have a detailed process! Urgent! Thank you

Let g (x) = f (x) / X
g'(x) = d(f(x)/x)/dx = [xf'(x) - f(x)]/x^2
Because XF '(x) > F (x), = > XF' (x) - f (x) > 0 = > when x is not 0, G '(x) > 0
Therefore, G (x) is an increasing function when x is not zero = > when a > 1, G (a) > G (1), that is, f (a) / a > F (1) / 1 = > F (a) > AF (1)

If symmetric matrix a satisfies a ^ 2 = 0, it is proved that a = 0

Because A2 = 0 and a is a symmetric matrix (i.e. a (I, J) = a (J, I)), any element in matrix a satisfies ∑ a (I, J)? J, I) = 0, so a (I, J) = 0. Because a (I, j) is arbitrary, so a = 0

The difference between in ten minutes and ten minutes later
I will be free________
A after ten minutes
B in ten minutes
C ten minutes later
D ten minutes after
The answer is B. explain why C can't. what's the difference between B and C?

In ten minutes, in ten minutes, in ten minutes, in ten minutes, in ten minutes, in ten minutes, in ten minutes, in ten minutes, in ten minutes, in ten minutes, in ten minutes, in ten minutes, in ten minutes, in ten minutes, in ten minutes

How to calculate ax - B = CX - D

ax - b = cx -d
ax-cx=b-d
（a-c）x=b-d
x=(b-d)/(a-c)

A large number of birds is singular and plural? What about the number of books?

A large number of birds are circling overhead. The number of books refers to the number of books. It is only a data, so it is singular

In the triangle ABC, ab = 8, BC = 10, AC = 13, then the shape of the triangle is -——
My question is: is it an acute triangle or an obtuse triangle? Because AB ^ 2 + BC ^ 2 < AC ^ 2

Therefore, if AC is the longest side of a triangle, then CoSb = [AB & # 178; + CB & # 178; - AC & # 178;] / (2 × ab × CB) = [64 + 100-169] / (160)