This book is a math review book specially written for the candidates who took part in the 20 17 entrance examination for Mathematics I, covering all the contents stipulated in the examination syllabus. The book is divided into three parts: advanced mathematics, linear algebra and probability statistics, and each part has chapters. Each chapter module is generally: outline selection, basic review module, knowledge extension module, key problem analysis, test questions and test questions reference answers, etc.
Among them, the knowledge extension module and the analysis of key issues are the highlights of this book. The knowledge expansion module is a special module with great difficulty and high test frequency. For example, the knowledge expansion module in the chapter of the mean value theorem mainly explains the application of the mean value theorem and the construction of auxiliary functions, while the analysis of key questions is explained through a large number of examples, and it is also divided according to the test sites, and there are ideas to analyze and summarize.
As shown in the picture, who can help me solve the postgraduate entrance examination, advanced mathematics, mathematical analysis and science and engineering, and ask for detailed steps. The sum of the elements in each row of the n-order matrix A is zero, which means that (1, 1, ...,1) t (the n-column vector of1) is the solution of Ax=0. Since the rank of A is: n- 1, the dimension of the basic solution system is:. 1) t is a solution of the equation, not 0, so the general solution of Ax=0 is: k( 1, 1, …,1) t.
Illustrations of postgraduate entrance examination, advanced mathematics, and science and engineering are as follows: Why are there n-r+ 1 corresponding homogeneous equations AX=0 and n-r unsolvable vectors? Suppose AX=b has 1 nonzero solutions, and then superimpose the aforementioned irrelevant solutions respectively, and then include itself. * * There are n-r+ 1 irrelevant solutions.
How can this prove how the contracts for postgraduate entrance examination, advanced mathematics and secondary science and engineering disciplines are contracted for XTAX and XTBX? This is the definition.
If contract a and contract b
Then call the quadratic x'Ax and x'Bx contracts (x' stands for x transposition).
(The contract is not just a matrix contract, but a quadratic contract. )
As shown in the figure, why is the quadratic form X'AX after X= CY still the A in the middle of the new quadratic form? It's not a in the middle of advanced mathematics, postgraduate entrance examination and science and engineering, but C TAC, and the quadratic form is y t (c tac) y.
Draw a circle to answer? Please write down the detailed process ~ Thank you (Advanced Mathematics and Science) 18. The matrix of quadric surface is A =
[ 1 b 1]
[b a 1]
[ 1 1 1]
The eigenvalue of a is 1, 4, 0,
Then1+a+1=1+4+0, a = 3；; ;
|A| =
|0 b- 1 0|
|b a 1|
| 1 1 1|
| a | = -(b- 1)^2 = 1 * 4 * 0 = 0，b = 1。
The eigenvalues of A 1, 4,0 are all independent eigenvalues, and their eigenvectors are orthogonal.
For eigenvalue 1, E-A =
[ 0 - 1 - 1]
[- 1 -2 - 1]
[- 1 - 1 0]
The basic transformation of rows is
[ 1 1 0]
[0 1 1]
[0 0 0]
The eigenvector is (1,-1, 1) t, and the unit is (1/√ 3,-1/√ 3) t;
For eigenvalue 4, 4E-A =
[ 3 - 1 - 1]
[- 1 1 - 1]
[- 1 - 1 3]
The basic transformation of rows is
[ 1 - 1 1]
[0 -2 4]
[0 2 -4]
The basic transformation of rows is
[ 1 0 - 1]
[0 1 -2]
[0 0 0]
The eigenvector is (1, 2, 1) t, and the unit is (1/√ 6,2/√ 6,1/√ 6) t;
For eigenvalue 0, 0E-A =
[- 1 - 1 - 1]
[- 1 -3 - 1]
[- 1 - 1 - 1]
The basic transformation of rows is
[ 1 1 1]
[0 -2 0]
[0 0 0]
The basic transformation of rows is
[ 1 0 1]
[0 1 0]
[0 0 0]
The feature vector is (1, 0,-1) t, and the unit is (1√ 2,0,-1√ 2) t.
Orthogonal matrix P =
[ 1/√3, 1/√6, 1/√2]
[- 1/√3, 2/√6, 0]
[ 1/√3, 1/√6, - 1/√2]
For the circled solution, please write (detailed process) (advanced mathematics and science) thank you ... solution: ∫ dx/[x (n-x)] = (1/n) [1(n-x)+1/x.
∴∫dx/[x(n-x)]=( 1/n)∫[ 1/(n-x)+ 1/x)]dx =( 1/n)
∴ln 丨 x/(N-x) 丨 =Nkt+lnc, ∴ x/(n-x) = CE (NKT).
In addition, ∫t = 0, c=x0/(N-x0), x = n [ce (NKT)]/[1+ce (NKT)] = (nx0) e (NKT)]/[n-x0+.
For reference.
Downstream: ∫ < 0,2 > x 2dx/9 = [x 3/27] < 0，2 & gt= 8/27
Upstream: ∫ < 2,3 > x 2dx/9+∫ <; 1，y & gtx^2dx/9
=[x^3/27]<； 2，3 & gt+[x^3/27]<； 1，y & gt
= 1-8/27+y^3/27- 1/27 =(y^3+ 18)/27
Postgraduate entrance examination, advanced mathematics, science and engineering assume that the eigenvalue of A is S, why is the eigenvalue of E-A 1+s? Hello! You wrote it wrong. The eigenvalue of E-A is 1-s, and if x is the eigenvector, then Ax=sx, so (e-a) x = ex-ax = x-sx = (1-s) X. The economic mathematics team will help you solve the problem, so please adopt it in time. thank you
What is the equivalent infinitesimal used in addition and subtraction of advanced mathematics, science and engineering and postgraduate entrance examination? Hello, viceroy Wen at your service. The purpose of finding the limit with equivalent infinitesimal is to reduce the power of x whose denominator tends to 0.
If the denominator is a polynomial of x, the key is whether the denominator tends to 0 when x→0. If so, then you have to raise the power of x (for example, k times), and then you have to turn the molecule into k times.
Generally, the higher order shall prevail.
The problem of two subtractions of a molecule must be a whole to be replaced, but if it is a product, it can be replaced separately.