一类四阶两点边值问题对称正解的单调迭代法

利用单调算子的迭代方法,给出了一类四阶两点边值问题x(4)(t)+f(t,x(t))=0(0≤t≤1),x(0)=x’(0)=x(1)=x’(1)=0的对称正解.     

抽象函数解读

Ⅰ.正比例函数f(x)=kx(k≠0,x∈R)的抽象函数的特征式为:(1)f(x+y)=f(x)+f(y);(2)f(x-y)=f(x)-f(y);(3)f(xy)=k1f(x)f(y),特别地当k=1时,有f(xy)=f(x)f(y).例1:定义在R上的函数f(x),恒有f(x+y)=f(x)+f(y),若f(16)=4,那么f(2003)=.解法1(基本解法):令x=y=0,得f(0)=2f(0),∴f(0)=0.令y=-x,得f(x-x)=f(x)+f(-x),即f(-x)=-f(x),∴f(x)是奇函数.令y=x,得f(2x)=2f(x),f(22x)=f(2·2x)=2f(2x)=22f(x),…,f(2nx)=2nf(x).又∵f(16)=4,∴f(1)=41.∵f(2003)=f(211-25-23-22-1),∴f(2003)=f(211)-f(25)-f(23)-f(22)-f(1)=(211-25-23-22-1)·f(1)=20403.…     

导数的活用

一、活用导数求极限【例1】求(1)li mx→0ex-1x(2)lix→m0sixnx解:(1)令f(x)=ex,则f′(x)=ex,f(0)=1∴li mx→0ex-1x=lix→m0f(x)x--0f(1)=f′(0)=1(2)li mx→0sinxx=lix→m0sinxx--0sin0=(sinx)′|x=0=1二、活用导数解决函数的单调性问题【例2】已知:函数f(x)=x2cosθ 2xsinθ     

函数解析式求解面面观

一、直接法例1已知f(x)=x2(x≥0)x(x<0),g(x)=x(x≥0)-x2(x<0),则x<0时,f[g(x)]为()(A)-x(B)-x2(C)x(D)x2解:当x<0时,g(x)=-x2<0,所以f[g(x)]=g(x)=-x2,选(B).求复合函数的解析式,先求内层函数,再求外层函数,另外,分段函数要注意变量的范围.二、换元法例2已知f(1-cosx)=sin2x,求f(x).解:令1-cosx=t则cosx=1-t,-1≤1-t≤1,所以0≤t≤2.所以f(t)=1-(1-t)2=-t2+2t(0≤t≤2)所以f(x)=-x2+2x(0≤x≤2)三、配方法例3f(x-1x)=x2+x12.求f(x).解:f(x-1x)=x2+x12=(x-1x)2+2,所以f(x)=x2+2.四、待定系数法例4已知f(x)=3x-1,f[h(x)]=g(x)=2x+3,h(x)为x…     

曲线的和与积

一、二曲线的和系定义1:在实数域内,设有二曲线 f_1(x、y)=0,f_2(x、y)=0,称曲线系mf_1(x、y)+nf_2(x、y)=0为曲线f_1、f_2的和系.m、n是不为0的实参数.令λ=n/m,则曲线f_1、f_2的和系可以写成: f_1(x、y)+λf_2(x、y)=0,当f_1=f_2时,规定λ≠—1。性质1:当二曲线f_1(x、y)=0与f_2(x、y)=0有公共点时,二曲线的和系f_1(x、y)+λf_2(x、y)=0为过f_1、f_2公共点的曲线系。性质2:除曲线f_1(x、y)=0与f_2(x、y)=0的公共点以外,二曲线的和系f_1(x、y)+λf_2(x、y)=0与曲线f_1或f_2没有其他的公共     

不定型1∞的一种简捷求法

利用连续函数ex,给出不定型1∞求极限的结论:lim(1 α(x))M(x)=eλ,其中limα(x)=0(α(x))≠0),limαx→x0(x)M(x)=λ(|λ|＜ ∞).     

聚焦以拐点处切线为背景的高考压轴题

<正>1考情新动向题1(2018年高考全国3卷理科)已知函数f(x)=2(+x+ax²)ln(1+x)-2x.(1)若a=0,证明:当-10时,f(x)>0;⑵略.命题组给出的标准答案如下:(1)当a=0时,f(x)=2(+x)ln(1+x)-2x,f′(x)=ln(1+x)-x/1+x.设函数g(x)=f′(x)=ln(1     

二次函数迭代的一个问题的探究

文献[1]~[3]对二次函数f(x)=x2+bx+c的迭代进行了探讨,其中文献[2]、[3]得到了关于方程f2(x)=x在特殊情形下根的一个结论:设f(x)=x2+bx+c,记Δ0=(b-1)2-4c,若方程f(x)=x有2个不等实根,则1)当0<Δ0<4时,f2(x)=x只有2个不等实根;2)当Δ0>4时,f2(x)=x有4个不等实根.方程f2(x)=x中的f2(x)为f2(x)=f(f(x)),一般地有fn(x)=f(fn-1(x)).本文将考虑一般二次函数f(x)=ax2+bx+c(其中a≠0且a,b,c∈R)的迭代,用初等方法给出     

对一道数学题的探究性学习与反思

原题已知函数f(x)的定义域为(0, ∞),且对于任意的正实数x、y都有f(xy)=f(x) f(y).当x>1时,f(x)>0,f(4)=1.(1)求证:f(1)=0.(2)求f(116).(3)解不等式:f(x) f(x-3)≤1.一、教学过程老师:如何证明f(1)=0?学生1:令x=1,y=1,得f(1)=f(1×1)=f(1) f(1)=2f(1),∴f(1)=0.学生2:令x=4,y=1     

一类特殊方程的无解性的判别法

中学代数中,有些较为特殊的方程,在实数范围内无解,若依照一般解法,不但演算过程复杂,而且很难判定它们在实数范围内是否无解。本文试图给出这类无解方程的两个判定定理,可以简化解题过程,省时省力。定理1:若方程f(x)=0可表示成f_1[g(x)]=0,且f_1(y)=0无实数根,则方程f(x)=0无实数根。(其中f(x),g(x),f_1(y)均为代数函数,下面定理2假设相同。)。证明:设f(x)=0有实数根x_0,则有: f_1[g(x_0)]=0。令 y_0=g(x_0),则f_1(y_0)=0 即y_0是方程f_1(y)=0的实数根,与题设相矛盾。从而方程f(x)=0无实数根。定理2:若f(x)=0可表示成f_1[g(x)]=0,且f_1(y)=0有实数根y_1,y_2,…,y_n,但对于每一个y_i(1≤i≤n),方程g(x)=y_i都无实数根,则方程f(x)=0无实根。     

卡思卡特城堡

It is worthy of noting that, whilst Crookston Castle witnessed the earlier and happier portion of Mary's variegated life,     

不同的吃喝玩乐

一、吃和喝吃苹果 eat an apple, 吃药 take medicine,吃糖 have some sweets,吃饭 have one's meals,吃馆子 dine out,吃惊 be surprised/     

漏洞百出的教育

Many college freshmen arrive woefully unprepared to do college work, and as disadvantaged populations continue to grow, the share of the American work force that has made it through college is expected to plummet. Many experts blame that educational failure not just on high schools but also on colleges. School & College, a special report by The Chronicle, looks at efforts to fix the system. What reforms would better prepare students for college? What should schools and colleges be doing differently? How should state and federal officials help?     

只要努力,不要后悔

There are numbers of crossroads on our long and unpredictable life journey where we totally have no idea about which direction to choose. No matter what our decision is, we should not turn back, but face the music and go ahead instead. I am this kind of girl who always does try without regretting, one example is how I dealt with my love.     

浅谈移民问题

Migration occurs behind a variety of reasons and has a great effect on the whole world. People may migrate in order to improve their economic situation, or in order to escape civil strife, persecution, and environmental disasters. The impact of migration is complex, bringing both benefits anddisadvantages. This paper briefly talks about the causes of migration, the allocation of benefits, and the ways in which individual countries and the international community deal with this important subject.     

Sister Carrie, an Adherent of Desires

Sister Carrie is one of the most controversial characters in American literature.Thought as a "fallen woman" firstly,she was defined as a "new woman" by some critics later. However, by digging into the motivaton behind the whole process of Carrie's "success", the relationship between Carrie and her creator (the author), the social conditions of then American, it can be found that Carrie has never been free-standing on her thought and she has never found her real-sdf even after becoming a famous actress. In a society dominated by mass consumerism Carrie is only an adherent of her own desires. She also is a representative of all those country girls flooded into cities, a symbol and a sacrifice of the urbanization of America in a time countryside was overcome by cities.     

Improvement of one-cycle controller by use of proportional integral differential controller

1.IntroductionOne-cyclecontrolmethod,whichwasproposedaboutonedecadeago[1],hasbecomeanattractivemethodinspecialfieldssuchaspowerfactorcorrection[2-6],switchingamplifiers[7,8],etc.Themainideaofthiscontrollerisbasedonintegrationofdiodevoltageinone-cycleandforcesittobeexactlyequaltothereferencevalue.Themainadvantageofthiscontrollerisitsrealtimeabilitytorejectthevariationofinputvoltage[1].Despitethisgreatability,ithasnogoodperformancesinrejectingofloaddisturbanceandfollowingreferencecommands.Espec…     

库克斯顿城堡

城堡, 是欧洲典型的建筑之一, 每个城堡都有一个美丽的传说或是一段历史。从本期开始, 我们将推出“城堡之旅”, 带各位读者去浏览、观赏欧洲各地的城堡, 开始一段文化与历史的旅途!     

The Reader Response Approach to Teaching of English Literature

During interview with English majors, it was found that a majority of students expressed disappointment with their experience of English literature classes. Students' dislike of learning English literature appears to be the major problem of teaching English literature in China. Reader response approach is a good way to solve this problem. In this paper, I explain the rationale behind the reader response approach, such as its origin, definition, features, assumptions, and strategies. Then I illustrate how reader response approach works as a teaching strategy by presenting several teaching models. At last I evaluate its usefuiness for teachers and present the advantages of reader response approach in Chinese context.