If the required hyperbolic equation is the same as the asymptote of the known hyperbola, how to set the hyperbolic equation
Asymptote equation is the multiplication of two linear equations, and hyperbolic equation is to change the right 0 to any constant which is not 0 after multiplication. Line 1: A1 x + B1 Y - C1 = 0 line 2: A2 x + B2 Y - C2 = 0 asymptote equation: (A1 x + B1 Y - C1) * (A2 x + B2 Y - C2) = 0 hyperbolic
The equiaxed hyperbolic equation can be set as (), and its asymptote equation is ()
x^2-y^2=k
Asymptote: y = x, y = - x
Just think about it
Finding the general solution of differential equation y '' - 4Y '+ 4Y = 2 ^ 2x + e ^ x + 1
The characteristic equation is R & # 178; - 4R + 4 = 0, with a pair of multiple roots r = 2
The general solution of the corresponding homogeneous equation is y = (C1 + C2 · x) · e ^ (2x)
C1 and C2 are arbitrary constants
Let f (x) = 2 ^ 2x + e ^ x + 1
Let f (d) = 4-4d + D & # 178;, then a special solution of the original differential equation is y * = [1 / F (d)] f (x)
=[1/F(D)](2^2x+e^x+1)
=[1/F(D)]2^2x + [1/F(D)]e^x + 1/(4-4D+D²)
=[1/F(D)]e^((2ln2)x) + [1/F(D)]e^x + (1/4 +(1/4)D +………… )
=[1/F(2ln2)]e^((2ln2)x) + [1/F(1)]e^x + 1/4
=e^((2ln2)x)/(2ln2 -2)² + e^x/(1 -2)² + 1/4
=4^x /(2ln2 -2)² + e^x + 1/4
Then the general solution of the original differential equation is
y=Y+y*
=(C1+C2·x)·e^(2x) + 4^x /(2ln2 -2)² + e^x + 1/4
Finding the general solution of the differential equation (2x + 1) y ″ + 4xy ′ - 4Y = [(2x + 1) ^ 2] (e ^ x)
1/9*exp(x)*(6*x+1)+C1*x+C2*exp(-2*x),
Exp (x) is e ^ X
It is proved that the following sequence limits exist and their values are obtained: (1) A1 = √ C (c > 0), a (n + 1) = √ [C + a (n)], n = 1,2,3
(2) A (n) = n power of C / N! (c > 0), n = 1,2
(1) In this paper, we prove that {a (n)} is an increasing sequence of numbers and has an upper bound of 1 + √ C, which can be proved by induction, and then we can get the existence of {a (n)} limit by monotone boundedness theorem, which is denoted as a; then we can get a * a-a-c = 0 by finding the limit on both sides of the equation a (n + 1) = √ [C + a (n)], and we can get the positive root a = {1 + √ [1 + 4 * C]} / 2 by solving quadratic equation
For any positive integer n, if a1 + A2 +.. + an = n ^ 3, LIM (1 / (A2-1) + 1 / (a3-1) +. 1 (an-1))=
For any positive integer n, if a1 + A2 +.. + an = n ^ 3, LIM (1 / (A2-1) + 1 / (a3-1) +. 1 (an-1)) = find out the general term of an, and then do nothing
A1 + A2 +.. + an = n ^ 3 (1) a1 + A2 +.. + an + a (n + 1) = (n + 1) ^ 3 (2) from (2) - (1), a (n + 1) = 3N ^ 2 + 3N + 1 = 3N (n + 1) + 1, that is, an = 3 (n-1) n + 1, so 1 / (an-1) = 1 / 3N (n-1) = 1 / 3 [1 / (n-1) - 1 / N] LIM (1 / (A2-1) + 1 / (a3-1) +. 1 (an-1)) = 1 / 3 {(1-1 / 2) + 1 / 2-1 / 3. (1 / (...)
Given the sequence {an}, we define its mean as VN = (a1 + A2 +...) +an)/n ,n∈N*.
(1) If the mean of sequence {an} VN = 2n + 1, find an
(2) If the sequence {an} is an equal ratio sequence with the first term of 1 and the common ratio of 2, and its average is VN, it is constant for any n ∈ n *, (VN + 1 / N) * k > = 3, the value range of real number k is obtained
(1)
Sn =a1+a2+..+an
Vn = 2n+1
Sn/n = 2n+1
Sn = n(2n+1) (1)
S(n-1) = (n-1)(2n-1) (2)
(1)-(2)
an = n(2n+1) -(n-1)(2n-1)
= 2n^2 +n -(2n^2-3n+1)
=4n-1
(2)
an = 2^(n-1)
Sn = a1+a2+..+an
= 2^n-1
Vn = (2^n-1)/n
(Vn + 1/n).k >=3
(2^n). k >=3
k >= 3. 2^(-n)
What does log mean
I'm a junior high school student. Someone asked me what log means
Log is the sign of logarithm. Logarithm seeks the exponent in the power operation. For example, a ^ B = C reads: the power of B of a is equal to C. here, a is called the base number, B is called the exponent, C is called the power operation, C is called the power operation, C is called the root number, a logarithm seeks the exponent B. It is written in the form of log (a) C = B
log
For
Activity log in for each backup script
Extended system log for troubleshooting
What does log logistic mean
Log means record
Logistics, supply, supply
log-logistic(s)
Record of logistics supply