Let y = (2x ^ 2 + X + 1) e ^ 2x, find the derivative of Y100? You'd better write the general term for me,

Let y = (2x ^ 2 + X + 1) e ^ 2x, find the derivative of Y100? You'd better write the general term for me,

The 100 order derivative of Y, which is the 100 order derivative of Y? Y ′ = (4x + 1) e ^ (2x) for the 100 order derivative of Y, the 100 order derivative of Y is (Y ′ = (4x + 1) e ^ (2x) (2x) (4x + 1) e ^ (2x) (2x (2x) (4x + 1) e \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\x & # 178; + 20
u=(2x^2+x+1)
u'=(4x+1)
u"=4
y^(100)=2^100×(2x^2+x+1)e^2x+100×2^99×(4x+1)e^2x+4950×2^100×e^2x
13 and 52 greatest common divisor and least common multiple 10 and 8 greatest common divisor and least common multiple
13 and 52 greatest common divisor and least common multiple
Because 13 is a multiple of 52,
So the greatest common divisor is 13,
The least common multiple is 52
The greatest common divisor and the least common multiple of 10 and 8
10=2×5
8=2×4
Greatest common divisor = 2
Least common multiple = 10 × 4 = 40
If the complex Z satisfies (Z-3) (2-I) = 5 (I is an imaginary unit), then the conjugate complex of Z. Z is ()
A. 2+iB. 2-iC. 5+iD. 5-i
∵ (Z-3) (2-I) = 5, ∵ Z-3 = 52 − I = 2 + I ∵ z = 5 + I, ∵. Z = 5-i
How to find the least common multiple and the greatest common divisor of a fraction, such as the least common multiple and the greatest common divisor of 5 / 24 and 3 / 28,
The least common multiple of the denominator is the denominator, and the greatest common divisor of the numerator is the numerator. This number is the greatest common divisor of the fraction. Therefore, the greatest common divisor of the above two numbers is 1 / 168
The greatest common divisor of the denominator is the denominator, and the least common multiple of the numerator is the numerator. This number is the least common multiple of the fraction. Therefore, the least common multiple of the above two numbers is 15 / 4
That's the proof of the matrix over the complex field
1. A square matrix over a complex field is similar to a Jordan square matrix (it is proved that we can see the textbook of linear algebra). That is to say, the linear transformation represented by a is a Jordan matrix for a matrix of a base. The Jordan matrix is lower triangular, while the matrix required by the title is upper triangular. Therefore, we consider making some transformations for the Jordan canonical form of A, We just need to change the order of the bases corresponding to the Jordan canonical form of A. the order of the bases is still the bases, and the matrix of a about the bases is upper triangular. You can also say that a is similar to a Jordan square matrix, that is, there is an invertible matrix C, such that a = C ^ (- 1) * J * C, where j is the Jordan canonical form of A, Let e = D * J * d ^ (- 1), then
A = C ^ (- 1) * d ^ (- 1) * e * D * C, where e is an upper triangular matrix
Let J be Jordan canonical form of a, then there exists invertible square matrix C such that a = C ^ (- 1) * J * C, then a ^ k = C ^ (- 1) * J ^ k * C. Furthermore, if f (x) is a polynomial function, then f (a) = C ^ (- 1) * f (J) * C, that is, f (a) is similar to f (J), So f (a) and f (J) have the same eigenvalues. Now consider J ^ K. J is a lower triangular matrix. The calculation shows that its k-th power is also a lower triangular matrix (Jordan matrix was written yesterday, sorry). Moreover, every diagonal element of J ^ k is the k-th power of the corresponding diagonal element of J. furthermore, its polynomial function is still a lower triangular matrix, that is, f (J) is a lower triangular matrix, Moreover, the diagonal elements of F (J) are exactly the polynomial functions of J corresponding to the diagonal elements, that is, f (λ 1), f (λ 2),..., f (λ n). Now we find the eigenvalue of F (J) and the lower triangle of F (J), so its characteristic polynomial is easy to calculate, and because its diagonal elements are
F (λ 1), f (λ 2),..., f (λ n), so its characteristic polynomial cannot have roots other than f (λ 1), f (λ 2),..., f (λ n), that is to say, all the eigenvalues of F (a) are f (λ 1), f (λ 2),..., f (λ n)
The greatest common divisor of 16.18 and 24 is () and the least common multiple is ()
Using short division
The greatest common divisor is (2), and the least common multiple is (72)
2 and 72
The greatest common divisor of 18 and 24 is (6), and the least common multiple is (72).
The greatest common divisor is (2), and the least common multiple is (72)
2 72
Solving the complex number proof problem
Z1, Z2, Z3 are three complex numbers, | Z1 | = | Z2 | = | Z3 | = 1, Z1 + Z2 + Z3 = 0, if | Z | = 3, find:
(1) It is proved that: | z-z1 | ^ 2 = 10 - (conjugate complex of Z * Z1 + conjugate complex of Z * z1)
(2) It is proved that: | z-z1 | ^ 2 + | z-z2 | ^ 2 + | z-z3 | ^ 2 = 30
Prove that (1) | z-z1 | ^ 2 = (z-z1) (conjugate complex of z-z1) = conjugate complex of Z * Z + conjugate complex of Z1 * Z1 - (conjugate complex of Z * Z1 + conjugate complex of Z * z1) = 9 + 1 -- (conjugate complex of Z * Z1 + conjugate complex of Z * z1) = 10 - (conjugate complex of Z * Z1 + conjugate complex of Z * z1); (2) prove that
The greatest common divisor of 24 and 18 is______ The least common multiple is______ .
24 = 2 × 2 × 2 × 318 = 2 × 3 × 3, so the greatest common factor of 24 and 18 is 2 × 3 = 6; the least common multiple of 24 and 18 is 2 × 2 × 2 × 3 × 3 = 72
It's a plural topic. Come on,
Z = (COS @ + isin @) quintic verification tan5 @ = 5tan @ - 10tan cubic @ + Tan quintic @, divided by 1-10tan square @ + 5tan quartic @
z=[cos@^5-10cos@^4sin@ +5cos@sin@ ^4]+i[5cos@^ 4sin@-10cos@ ^2sin@^3+sin@^5]
How to calculate the slope of 5 Tan @ - 10 Tan third power @ + Tan fifth power @, divided by 1-10 Tan square @ + 5 Tan fourth power@
Z is the plural of the unit;
Then z = (COS @ + isin @) ^ 5 = cos5 @ + isin5 @;
So the protagonist of Z is 5 @;
If you use a vector, the slope of the vector is tan5 @;
If we expand z = (COS @ + isin @) ^ 5,
We get z = [cos @ ^ 5-10cos @ ^ 4sin@ +5cos@sin@ ^4]+i[5cos@^ 4sin@-10cos@ ^2sin@^3+sin@^5]
So if we find the slope again, we will get 5tan @ - 10tan third power @ + Tan fifth power @, divided by 1-10tan square @ + 5tan fourth power@
The two are equal, so the equation holds;
The key to solve the problem is that the 5th power of the unit vector is to enlarge @ by 5 times;
What are the greatest common divisor and the least common multiple of 18, 24 and 42
6 and 504