2023/5/27 20:08 : 9:00~12:00 this morning, I took the National Mathematics Competition. Guangzhou is really hot! It's only 17 degrees in Xi'an, and it's still raining lightly. After arriving in Guangzhou, the bright sun was completely useless for me. I walked for 20 minutes on the way to the hotel, and I was really dying.
When I got the test paper, I collapsed. Use the orthogonal matrix surface equation as the standard equation (the second question)? The Gram-Schmidt I am most afraid of is orthogonalized. After calculating the eigenvalues of that matrix for a long time, I first guessed -2, which was wrong, and then guessed 2, yes, this is all right, the remaining two eigenvalues can be obtained only by using the trace and determinant. Then I realized that I didn't seem to have learned the standard equation? It seems that the line substitute teacher skipped this section in class, and I didn't review it? After crashing, I wrote the standard and normative forms of the quadratic form, and deducted 1 point for the last question.
Then there is the third question. What the hell? First you can think of total differential, and then there is a differential equation. What? Higher order linear differential equations? I didn't review this! I thought I didn't take the exam! I haven't passed the exam in previous sessions! Then just make it up, it took a lot of effort to make f ( x ) + 4 f ′ ′ ( x ) = 0 f(x)+4f''(x)=0f(x)+4f _′′(x)=0 , the second-order linear homogeneous differential equation that can be reduced, was solved, and the last question was only given 4.5 points.
The answer to the fourth question was wrong, and 2.5 points were deducted.
The first two questions of the fifth question were awarded points, and the third question was blind. I wrote it directly according to the idea of the second question. . . I just want to go back in time and slap myself twice. . .
The sixth question, the most amazing question, I don't know how to prove it, but I only wrote the shortest line between two points, and I actually gave it 11.5 points? ! This is something I didn't expect.
The seventh question, the proof of convergence is very good, but if the sum is used, two ln \lnWhat the hell is ln multiplication? Never seen it before! Helpless! After thinking for half an hour, I couldn't figure it out, so I gave up. Scored 2.5 points.
Summary : After preparing for a month, all the exams were unprepared. This time it will be considered as a public-funded trip, woo woo woo. Shushu from the barren land in the northwest came to Guangzhou, an international metropolis, just like Grandma Liu entered the Grand View Garden. The Mulianzhuang Hotel where I stayed was quite luxurious.
2023/5/27 20:23 : At present, the roll noodles are divided into 74, and the national two are suspended. . .
2023/5/27 20:38 : Summarize the lessons. The basic knowledge is still not solid, I was a little flustered in the process of writing the paper, the handwriting was poor, and the steps were not written in detail. . . Eat a fall and gain a wisdom.
The complete answers to the questions are as follows:
The picture comes from the official public account article .
Fill-in-the-blank question (4): Originally, the fill-in-the-blank question was a sub-question (I finished it in 10 minutes), but some people actually did this question wrong! Be sure to pay attention to the positive definiteness of the Hesse matrix after finding out the extreme points!
The third question, if you know f ( x ) + 4 f ′ ′ ( x ) = 0 f(x)+4f''(x)=0f(x)+4f _′′(x)=The general solution for 0 is f ( x ) = C 1 sin 2 x + C 2 cos 2 xf(x)=C_1\sin 2x+C_2\cos 2xf(x)=C1sin2x _+C2cos2x is pretty easy. (It's a pity that I didn't review it enough at the time...
For the fourth question, it is not acceptable to use the definite integral directly, and it must be transformed into a triple integral under spherical coordinates. I have a variable substitution not divided by 2 \sqrt{2}2, the car overturned, huh~
Fifth question, note: t sin x ≥ sin txt\sin x\ge\sin txtsinx≥sint x is just atx ∈ [ 0 , π ] x\in[0,\pi]x∈[0,π]， t ∈ [ 0 , 1 ] t\in[0,1] t∈[0,1 ] when established. The first question I used Jensen's inequality to prove, the second question gave a sub-question, and the third question fell into a pit... I was wondering whysin u \sin usinU wants to add the absolute value, it is in[ 0 , π ] [0,\pi][0,π ] is not a non-negative number? In short, this question should be considered in sections, eachπ \piDivide π into one segment, then divide the remaining tail into another segment, and then combine each segment.
The sixth question, in fact, I have tried the answer method, but I think this method cannot be done. The use of this Cauchy inequality is really too mysterious. But this question did not deduct any points, I am still very happy.
The seventh question, I can't think of an = a 2 n + a 2 n + 1 a_n=a_{2n}+a_{2n+1}an=a2 n+a2 n + 1this kind of thing. . . Simple push: an = ln ( 1 + 1 n ) a 2 n + a 2 n + 1 = ln ( 1 + 1 2 n ) + ln ( 1 + 1 2 n + 1 ) = ln ( 1 + 1 2 n + 1 2 n + 1 + 1 ( 2 n ) ( 2 n + 1 ) ) a_n=\ln\left(1+\frac{1}{n}\right)\\ a_{2n}+ a_{2n+1}=\ln\left(1+\frac{1}{2n}\right)+\ln\left(1+\frac{1}{2n+1}\right)=\ln\ left(1+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{(2n)(2n+1)}\right)an=ln(1+n1)a2 n+a2 n + 1=ln(1+2 n1)+ln(1+2 n+11)=ln(1+2 n1+2 n+11+( 2 n ) ( 2 n+1)1)其中1 ( 2 n ) ( 2 n + 1 ) = 1 2 n − 1 2 n + 1 \frac{1}{(2n)(2n+1)}=\frac{1}{2n}-\frac {1}{2n+1}( 2 n ) ( 2 n + 1 )1=2 n1−2 n + 11，故 a 2 n + a 2 n + 1 = ln ( 1 + 1 2 n + 1 2 n + 1 + 1 2 n − 1 2 n + 1 ) = ln ( 1 + 1 n ) = a n a_{2n}+a_{2n+1}=\ln\left(1+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n}-\frac{1}{2n+1}\right)=\ln\left(1+\frac{1}{n}\right)=a_n a2 n+a2 n + 1=ln(1+2 n1+2 n+11+2 n1−2 n+11)=ln(1+n1)=anIt's so fucking mysterious! Then, for the summation of this kind of non-geometric sequence, it is generally to use some means of subtraction and cancellation of adjacent two terms. I thought of a split item at the time, but it didn't come out, but the overall idea is similar. He means, let bn = ∑ k = n 2 n − 1 ak 2 b_n=\sum\limits_{k=n}^{2n-1}a_k^2bn=k=n∑2n − 1 _ak2, bn − bn + 1 = an 2 − a 2 n 2 − a 2 n + 1 2 b_n-b_{n+1}=a_n^2-a_{2n}^2-a_{2n+1}^2bn−bn+1=an2−a2 n2−a2 n + 12； because = a 2 n + a 2 n + 1 a_n=a_{2n}+a_{2n+1}an=a2 n+a2 n + 1, so after subtraction it becomes bn − bn + 1 = 2 anan + 1 b_n-b_{n+1}=2a_na_{n+1}bn−bn+1=2a _nan+1. Then because bn b_nbn趋于0， ∑ n = 1 ∞ a 2 n a 2 n + 1 = 1 2 lim N → ∞ ∑ n = 1 N ( b n − b n + 1 ) = b 1 2 = ln 2 2 2 \sum\limits_{n=1}^{\infty}a_{2n}a_{2n+1}=\frac{1}{2}\lim\limits_{N\to\infty}\sum\limits_{n=1}^N(b_n-b_{n+1})=\frac{b_1}{2}=\frac{\ln^2 2}{2} n=1∑∞a2 na2 n + 1=21N→∞limn=1∑N(bn−bn+1)=2b1=2ln22。
Absolutely amazing is all I can say! The multiplication of two terms can be constructed using the sum of squares - the sum of squares, and this difference is obtained by the difference between adjacent two terms of another sequence. I can't think of such a solution in eight lifetimes.
2023/5/28 7:51 : I'm infused! 74 points country two! The score line for non-numerical categories is 76.5, and the score line for National Two is 62, which is only 2.5 points away from National One! I feel that he changed the third question unfairly. My solution should not only be given 4.5 points. If you give a little more points, you can pass the country. But I still blame myself for being too bad. After all, a person with a really high level will not lose Guoyi because the teacher corrects the papers too strictly. Keep up the good work!
2023/5/28 22:19 : Send!
The column of shame belongs to yes.