級(jí)數(shù)求和公式

級(jí)數(shù)求和公式張金偉一、引言我們經(jīng)常遇到諸如1+2+3+....+100、1/2+1/4+1/8+...+1/1000之類的對(duì)一列有規(guī)律的數(shù)求和的問題，即所謂的級(jí)數(shù)求和，但常常是這樣的一列數(shù)太長，使我們畏怯于枯燥的計(jì)算，或者由于結(jié)果累計(jì)誤差太大而失去計(jì)算的意義。級(jí)數(shù)求和曾給人帶來麻煩事，有這樣一個(gè)傳說，古印度有個(gè)國王對(duì)印度象棋頗有興趣，某日招象棋發(fā)明人進(jìn)宮，欲予賞賜，國王問發(fā)明人所欲何物。發(fā)明人想得到一車稻谷，但為了顯示自己的斯文與機(jī)智，便說：大王只須在棋盤的第1格中放1粒稻谷，在第2格中放2粒稻谷，在第3格中放4粒稻谷，在第4格中放8粒稻谷，以此類推放滿整個(gè)棋盤的64格，卑人足已。國王聽罷大笑，說：智者索物的方式亦高人一籌，朕一定滿足你的要求。便咐左右為其發(fā)賞不得馬虎?？扇旌髱焖静潘愠鼋Y(jié)果，貼耳告訴國王，把全國的所以庫糧加在一起也不夠發(fā)。國王覺得發(fā)明人當(dāng)眾愚弄了他，便使人暗中殺了發(fā)明人，獎(jiǎng)賞兌現(xiàn)的事自然也就不了了之?，F(xiàn)在中學(xué)生都能很快計(jì)算這個(gè)問題了，其稻谷數(shù)總和為264-1=1064lg2-1>1019(粒)，這豈止是一車稻谷啊，至少是100萬億噸！哎，發(fā)明人和國王不會(huì)計(jì)算，誤會(huì)釀禍，聰明反被聰明誤，實(shí)在可悲！或許你也碰到過奇特的級(jí)數(shù)求和問題，當(dāng)然不會(huì)象上面提到的故事那么嚴(yán)重，不過我相信它會(huì)給你帶來煩惱與喜悅那絕對(duì)是真的。這里給大家介紹一個(gè)級(jí)數(shù)求和公式，我在學(xué)泰勒級(jí)數(shù)時(shí)獨(dú)立地發(fā)現(xiàn)并推導(dǎo)了它，當(dāng)時(shí)很讓我高興了一陣，現(xiàn)在將其要點(diǎn)列出，與各位共賞。一堆曰二、推導(dǎo)設(shè)一級(jí)數(shù)的第i項(xiàng)為f(i),S(i)是第x項(xiàng)之前包括第x項(xiàng)的所以各項(xiàng)之和，則有S(x)-S(0)=乂f(x)=f(1)+f(2)+f(3)+...+f(i)+...+f(x) (1)i=1所以對(duì)通項(xiàng)f(t+1)有f(t+1)=S(t+1)-S(t) ……………(2)用泰勒公展開S(t+1)得：f(t+1)=S'(t)+1/2!*S“(t)+1/3!*S“(t)+1/4!*S“'(t)+1/5!*S““(t)+...+1/n!*S(n)(t+0) (3)其中0v0<l;可以把此式看著是S(t)高階微分方程式，若其各項(xiàng)逐次減小，且級(jí)數(shù)收斂，就可以設(shè)其解為如下形式：S(t)=k1f(t+1)dt+k2f(t+1)+k3f'(t+1)+k4f''(t+1)+k5f'''(t+1)+.knf(n)(t+1)+.+C (4)其中k］、k2、…、kn、…、C都是恒常數(shù)，為求解k,將⑷式代回(3)式得k1f(t+1)+ k2f'(t+1)+ k3f''(t+1)+ k4f'''(t+1)+ k5f''''(t+1)+ k6f'''''(t+1)+.1/2!*kf'(t+1)+1/2!*k2f“(t+1)+1/2!*k3f“'(t+1)+1/2!*k4f“'(t+1)+1/2!*k5f““(t+1)+...1/3!*k1f"(t+1)+1/3!*￡「迫+1)+1/3!*k3f'-(t+1)+1/3!*k4f"(t+1)+...1/4!*k1f'''(t+1)+1/4!*k2f''''(t+1)+1/4!*k3f'''''(t+1)+...1/5!*k1f''''(t+1)+1/5!*k2f'''''(t+1)+...+

=f(t+1)由于對(duì)于一切=f(t+1)由于對(duì)于一切t,上式均應(yīng)成立故其系數(shù)k須滿足下列方程k1=1k2+1/2!*k1=0k3+1/2!*k2+1/3!*k1=0k4+1/2!*k3+1/3!*k2+1/4!*k1=0k5+1/2!*k4+1/3!*k3+1/4!*k2+1/5!*k1=0解此代數(shù)方程組得:K1=1k2=-1/2k3=1/12k4=0k5=-1/720k6=0k7=1/30240k8=0k9=-1/1209600k10=0k11=1/47900160k12=0k13=-691/1307674368000k13=0n，n對(duì)于更多的kn，見后面的計(jì)算機(jī)計(jì)算結(jié)果表。可以看到kn是很有趣的，當(dāng)n>2時(shí)，偶數(shù)的n,kn，n奇數(shù)的n,kn的正負(fù)號(hào)交替變化；叱絕對(duì)值逐次減小，當(dāng)n->-時(shí)，kn/kn+2-〉(2n)2?，F(xiàn)在將kn代會(huì)(i),可以得到計(jì)算公式無f(t)=Jf(t無f(t)=Jf(t+1)dt-2f(t+1)+12廣(t+1)-72of(t+D+…t=n-1(5)t=0t=-1送f(t)屮f(t)譏2f(t)t=0+12廣(t)-72o廣'(t)+…t=n(6)t=0等式兩端同時(shí)加f(n)整理得乞f(t)乞f(t)=Jnf(t)dt+j(f(n)+f(0))+ f'(t)-t=0 0 2 L12127I0f'''(t)+…t=0(7)三、應(yīng)用下面我們來應(yīng)用所得公式(7)1. 計(jì)算S=04+14+24+34+...+n4.解：通項(xiàng)為f(t)=t4,所以f(-1)=1/5*t5,F=4t3,F'=12t2,F，，=24t,f4=24,f(5)=0,...代入(7)式得S=o4+14+24+34+.+n4=1/5*n5+1/2n4+1/12*4n3-1/72o*24n=1/5*n5+1/2*n4+1/3*n3-1/3o*n=n(n2(6n2+15n+1o)-1)/3o2.計(jì)算I=100.1*100.2*100.3*.*(100+0.1i)*.200.解：InI=ln100.1+ln100.2+ln100.3+...ln(100+0.1i)+...ln200

通項(xiàng)為f(t)=ln(100+0.1t)所以f(-1)=(100+0.1t)(ln(100+0.1t)-1),f'=0.1/(100+0.1t),f''=-0.01/(100+0.1t)2,f'''=0.002/(100+0.1t)3,…取前三項(xiàng)代入(7)式得1(100+1(100+o.”)(ln(i00+)-1)+2(ln(i00+0.”)+20012(100+0.1t)」100二4.99641624921758e+03該計(jì)算結(jié)果與精確計(jì)算結(jié)果的相對(duì)誤差E小于1.0e-15.所以：I=10exp(lnl0*4.99641624921758e+03)=8.24150121404295*102169.相對(duì)誤差E小于3.0e-ll.四、余項(xiàng)7)式可以寫成代余項(xiàng)的形式迅f(t)=1nf(t)dt+；(f(n)+f(0))+\1f'(t)--Lf'''(t)+...+kf(n)(n9)］「"0 2 12 720 n I t=0 」 t=0其中0<e<1.這就是著名的“歐拉—馬克勞林求和公式”，更多知識(shí)，感興趣者只有查閱級(jí)數(shù)專著1.20000000000000e+01-6.00000000000000e+01-4.20000000000001e+01-3.99999999999995e+01-3.96000000000004e+01-3.95079594790159e+01-3.94857142857141e+01-3.94802322366602e+01-3.94788702213512e+01-3.94785306440354e+01-3.94784458516140e+01-3.94784246647685e+01-3.94784193693046e+01-3.94784180455771e+01-3.94784177146606e+01-3.94784176319330e+01-3.94784176112510e+01-3.94784176060810e+01-3.94784176047882e+01-3.94784176044651e+01-3.94784176043844e+01-3.94784176043643e+01-3.94784176043591e+01-3.94784176043581e+01-3.94784176043573e+011.20000000000000e+01-6.00000000000000e+01-4.20000000000001e+01-3.99999999999995e+01-3.96000000000004e+01-3.95079594790159e+01-3.94857142857141e+01-3.94802322366602e+01-3.94788702213512e+01-3.94785306440354e+01-3.94784458516140e+01-3.94784246647685e+01-3.94784193693046e+01-3.94784180455771e+01-3.94784177146606e+01-3.94784176319330e+01-3.94784176112510e+01-3.94784176060810e+01-3.94784176047882e+01-3.94784176044651e+01-3.94784176043844e+01-3.94784176043643e+01-3.94784176043591e+01-3.94784176043581e+01-3.94784176043573e+01五、系數(shù)k表nk1=1.00000000000000e+00k3=8.33333333333333e-02k5=-1.38888888888889e-03k7=3.30687830687830e-05k9=-8.26719576719585e-07k11=2.08767569878681e-08k13=-5.28419013868749e-10k15=1.33825365306847e-11k17=-3.38968029632260e-13k19=8.58606205627782e-15k21=-2.17486869855807e-16k23=5.50900282836021e-18k25=-1.39544646858125e-19k27=3.53470703962946e-21k29=-8.95351742703749e-23k31=2.26795245233766e-24k33=-5.74479066887214e-26k35=1.45517247561486e-27k37=-3.68599494066528e-29k39=9.33673425709499e-31k41=-2.36502241570062e-32k43=5.99067176248211e-34k45=-1.51745488446828e-35k47=3.84375812545416e-37k49=-9.73635307264657e-39k2=-5.00000000000000e-01k4=0.00000000000000e+00k6=2.16840434497101e-19k8=3.04931861011548e-20k10=9.52912065661088e-22k12=-3.30872245021211e-24k14=-1.55096364853693e-25k16=-1.61558713389263e-27k18=3.78653234506086e-28k20=-1.57772181044202e-30k22=3.82104500966428e-31k24=-6.54816181091660e-33k26=2.04630056591144e-34k28=-6.39468926847324e-36k30=1.58691737361009e-37k32=-2.57139389242375e-39k34=5.73971850987445e-42k36=-1.16588032231825e-42k38=1.40129846432482e-44k40=-2.62743462060903e-46k42=1.36845553156720e-47k44=-5.55935059699177e-49k46=6.68191177523049e-51k48=-2.92333640166334e-52k50=-1.95759134039956e-54六、附系數(shù)k計(jì)算程序n/*programedbyzhangjinwei2003/6/30*/#include"stdio.h"#include<conio.h>doublemulti(intn);main(){inti,j,max=50;doublek[300],klk;FILE*fp;k[0]=1;k[1]=1;fp=fopen("d:\\kk.txt","w");for(i=2;i<=150;i++){k[i]=0;for(j=1;j<i;j++){k[i]=k[i]-k[j]/multi(i-j+1);}if(k[i]!=0)klk=k[i-2]/k[i];elseklk=1;}for(i=1;i<=50;i+=2){printf("\nk%d=\t\b\b\b%21.15e"