9.3E: Mazoezi ya Sehemu ya 9.3

Matatizo ya mtihani wa Tofauti

Fikiria mlolongo wa kila mfululizo katika mazoezi 1 - 14, ikiwa mtihani wa tofauti unatumika, ama hali ambayo\(\displaystyle \lim_{n→∞}a_n\) haipo au kupata\(\displaystyle \lim_{n→∞}a_n\). Ikiwa mtihani wa kutofautiana hautumiki, sema kwa nini.

1)\(\displaystyle \sum_{n=1}^∞ \dfrac{n}{n+2}\)

2)\(\displaystyle \sum_{n=1}^∞\dfrac{n}{5n^2−3}\)

Jibu

\(\displaystyle \lim_{n→∞}a_n=0\). Mtihani wa Divergence hautumiki.

3)\(\displaystyle \sum_{n=1}^∞\dfrac{n}{\sqrt{3n^2+2n+1}}\)

4)\(\displaystyle \sum_{n=1}^∞\dfrac{(2n+1)(n−1)}{(n+1)^2}\)

Jibu

\(\displaystyle \lim_{n→∞}a_n=2\). Hivyo mfululizo unatofautiana na Mtihani wa\(n^{\text{th}}\) -Term kwa Divergence.

5)\(\displaystyle \sum_{n=1}^∞\dfrac{(2n+1)^{2n}}{(3n^2+1)^n}\)

6)\(\displaystyle \sum_{n=1}^∞\dfrac{2^n}{3^{n/2}}\)

Jibu

\(\displaystyle \lim_{n→∞}a_n=∞\)(haipo). Hivyo mfululizo unatofautiana na Mtihani wa\(n^{\text{th}}\) -Term kwa Divergence.

7)\(\displaystyle \sum_{n=1}^∞\dfrac{2^n+3^n}{10^{n/2}}\)

8)\(\displaystyle \sum_{n=1}^∞e^{−2/n}\)

Jibu

\(\displaystyle \lim_{n→∞}a_n=1.\)Hivyo mfululizo unatofautiana na Mtihani wa\(n^{\text{th}}\) -Term kwa Divergence.

9)\(\displaystyle \sum_{n=1}^∞\cos n\)

10)\(\displaystyle \sum_{n=1}^∞\tan n\)

Jibu

\(\displaystyle \lim_{n→∞}a_n\)haipo. Hivyo mfululizo unatofautiana na Mtihani wa\(n^{\text{th}}\) -Term kwa Divergence.

11)\(\displaystyle \sum_{n=1}^∞\dfrac{1−\cos^2(1/n)}{\sin^2(2/n)}\)

12)\(\displaystyle \sum_{n=1}^∞\left(1−\dfrac{1}{n}\right)^{2n}\)

Jibu

\(\displaystyle \lim_{n→∞}a_n=1/e^2.\)Hivyo mfululizo unatofautiana na Mtihani wa\(n^{\text{th}}\) -Term kwa Divergence.

13)\(\displaystyle \sum_{n=1}^∞\dfrac{\ln n}{n}\)

14)\(\displaystyle \sum_{n=1}^∞\dfrac{(\ln n)^2}{\sqrt{n}}\)

Jibu

\(\displaystyle \lim_{n→∞}a_n=0.\)Mtihani wa Divergence hautumiki.

\(p\)-Matatizo ya mfululizo & Matatizo ya mtihani wa Integral

Katika mazoezi ya 15 - 20, sema kama\( p\) mfululizo uliopewa unajiunga.

15)\(\displaystyle \sum_{n=1}^∞\frac{1}{\sqrt{n}}\)

16)\(\displaystyle \sum_{n=1}^∞\frac{1}{n\sqrt{n}}\)

Jibu

Mfululizo hujiunga, tangu\( p=3/2>1\).

17)\(\displaystyle \sum_{n=1}^∞\frac{1}{\sqrt[3]{n^2}}\)

18)\(\displaystyle \sum_{n=1}^∞\frac{1}{\sqrt[3]{n^4}}\)

Jibu

Mfululizo hujiunga, tangu\( p=4/3>1.\)

19)\(\displaystyle \sum_{n=1}^∞\frac{n^e}{n^π}\)

20)\(\displaystyle \sum_{n=1}^∞\frac{n^π}{n^{2e}}\)

Jibu

Mfululizo hujiunga, tangu\( p=2e−π>1.\)

Katika mazoezi 21 - 27, tumia mtihani muhimu ili uone kama kiasi kinachofuata kinajiunga.

21)\(\displaystyle \sum_{n=1}^∞\frac{1}{\sqrt{n+5}}\)

22)\(\displaystyle \sum_{n=1}^∞\frac{1}{\sqrt[3]{n+5}}\)

Jibu

mfululizo diverges na Integral Test tangu\(\displaystyle ∫^∞_1\frac{dx}{(x+5)^{1/3}}\) inaweza kuonyeshwa kuachana.

23)\(\displaystyle \sum_{n=2}^∞\frac{1}{n\ln n}\)

24)\(\displaystyle \sum_{n=1}^∞\frac{n}{1+n^2}\)

Jibu

mfululizo diverges na Integral Test tangu\(\displaystyle ∫^∞_1\frac{x}{1+x^2}\,dx\) inaweza kuonyeshwa kuachana.

25)\(\displaystyle \sum_{n=1}^∞\frac{e^n}{1+e^{2n}}\)

26)\(\displaystyle \sum_{n=1}^∞\frac{2n}{1+n^4}\)

Jibu

Mfululizo hujiunga na Mtihani wa Integral tangu\(\displaystyle ∫^∞_1\frac{2x}{1+x^4}\,dx\) hujiunga.

27)\(\displaystyle \sum_{n=2}^∞\frac{1}{n\ln^2n}\)

Eleza kiasi katika mazoezi 28 - 31 kama\( p\) -mfululizo na kuamua kama kila hujiunga.

28)\(\displaystyle \sum_{n=1}^∞2^{−\ln n}\)\(\quad\Big(\) Kidokezo:\( 2^{−\ln n}=\dfrac{1}{n^{\ln 2}}.\Big)\)

Jibu

\( 2^{−\ln n}=1/n^{\ln 2}.\)Tangu\(p=\ln 2<1\), mfululizo huu unatofautiana na mtihani wa\( p\) mfululizo.

29)\(\displaystyle \sum_{n=1}^∞3^{−\ln n}\)\(\quad\Big(\) Kidokezo:\( 3^{−\ln n}=\dfrac{1}{n^{\ln 3}}.\Big)\)

30)\(\displaystyle \sum_{n=1}^n2^{−2\ln n}\)

Jibu

\( 2^{−2\ln n}=1/n^{2\ln 2}.\)Tangu\(p = 2\ln 2−1<1\), mfululizo huu unatofautiana na mtihani wa\( p\) mfululizo.

31)\(\displaystyle \sum_{n=1}^∞n3^{−2\ln n}\)

Katika mazoezi 32-35, kutumia makadirio ya\(\displaystyle R_N≤∫^∞_Nf(t)\,dt\) kupata amefungwa kwa ajili ya salio\(\displaystyle R_N=\sum_{n=1}^∞a_n−\sum_{n=1}^Na_n\) ambapo\( a_n=f(n).\)

32)\(\displaystyle \sum_{n=1}^{1000}\frac{1}{n^2}\)

Jibu

\(\displaystyle R_{1000}≤∫^∞_{1000}\frac{dt}{t^2}=\lim_{b\to ∞}−\frac{1}{t}\bigg|^b_{1000}=\lim_{b\to ∞}\left(−\frac{1}{b}+\frac{1}{1000}\right)=0.001\)

33)\(\displaystyle \sum_{n=1}^{1000}\frac{1}{n^3}\)

34)\(\displaystyle \sum_{n=1}^{1000}\frac{1}{1+n^2}\)

Jibu

\(\displaystyle R_{1000}≤∫^∞_{1000}\frac{dt}{1+t^2}=\lim_{b\to ∞} \left(\tan^{−1}b−\tan^{−1}(1000)\right)=π/2−\tan^{−1}(1000)≈0.000999\)

35)\(\displaystyle \sum_{n=1}^{100}\frac{n}{2^n}\)

[T] Katika mazoezi 36 - 40, kupata thamani ya chini ya\( N\) vile kwamba makadirio salio\(\displaystyle ∫^∞_{N+1}f(x)\,dx<R_N<∫^∞_N f(x)\,dx\) dhamana kwamba\(\displaystyle \sum_{n=1}^Na_n\) makadirio\(\displaystyle \sum_{n=1}^∞a_n,\) sahihi kwa ndani ya makosa fulani.

36)\( a_n=\dfrac{1}{n^2},\) hitilafu\( <10^{−4}\)

Jibu

\(\displaystyle R_N<∫^∞_N\frac{dx}{x^2}=1/N,\;\text{for}\;N>10^4\)

37)\( a_n=\dfrac{1}{n^{1.1}},\) hitilafu\( <10^{−4}\)

38)\( a_n=\dfrac{1}{n^{1.01}},\) hitilafu\( <10^{−4}\)

Jibu

\(\displaystyle R_N<∫^∞_N\frac{dx}{x^{1.01}}=100N^{−0.01},\;\text{for}\;N>10^{600}\)

39)\( a_n=\dfrac{1}{n\ln^2n},\) hitilafu\( <10^{−3}\)

40)\( a_n=\dfrac{1}{1+n^2},\) hitilafu\( <10^{−3}\)

Jibu

\(\displaystyle R_N<∫^∞_N\frac{dx}{1+x^2}=π/2−\tan^{−1}(N),\;\text{for}\;N>\tan(π/2−10^{−3})≈1000\)

Katika mazoezi 41 - 45, pata thamani ya\( N\) vile ambavyo\( R_N\) ni ndogo kuliko kosa linalohitajika. Compute jumla sambamba\(\displaystyle \sum_{n=1}^Na_n\) na kulinganisha na makadirio ya kutolewa ya mfululizo usio.

41)\( a_n=\dfrac{1}{n^{11}},\) hitilafu\( \displaystyle <10^{−4}, \sum_{n=1}^∞\frac{1}{n^{11}}=1.000494…\)

42)\( a_n=\dfrac{1}{e^n},\) hitilafu\(\displaystyle <10^{−5}, \sum_{n=1}^∞\frac{1}{e^n}=\frac{1}{e−1}=0.581976…\)

Jibu

\(\displaystyle R_N<∫^∞_N\frac{dx}{e^x}=e^{−N},\;\text{for}\;N>5\ln(10),\)sawa kama\(\displaystyle N=12;\sum_{n=1}^{12}e^{−n}=0.581973....\) Makisio anakubaliana na\( 1/(e−1)\) kwa maeneo tano decimal.

43)\( a_n=\dfrac{1}{e^{n^2}},\) hitilafu\(\displaystyle <10^{−5}, \sum_{n=1}^∞\dfrac{1}{e^{n^2}}=0.40488139857…\)

44)\( a_n=\dfrac{1}{n^4},\) hitilafu\(\displaystyle <10^{−4}, \sum_{n=1}^∞\dfrac{1}{n^4}=\frac{π^4}{90}=1.08232...\)

Jibu

\(\displaystyle R_N<∫^∞_N\dfrac{dx}{x^4}=4/N^3,\;\text{for}\;N>(4.10^4)^{1/3},\)sawa kama\( N=35\);\(\displaystyle \sum_{n=1}^{35}\dfrac{1}{n^4}=1.08231….\) Makisio anakubaliana na jumla ya maeneo manne decimal.

45)\( a_n=\dfrac{1}{n^6}\), hitilafu\(\displaystyle <10^{−6}, \sum_{n=1}^∞\frac{1}{n^6}=\frac{π^6}{945}=1.01734306...,\)

46) Kupata kikomo kama\( n→∞\) ya\( \dfrac{1}{n}+\dfrac{1}{n+1}+⋯+\dfrac{1}{2n}\). \(\quad\Big(\)Kidokezo: Linganisha na\(\displaystyle ∫^{2n}_n\frac{1}{t}\,dt.\Big)\)

Jibu

\( \ln(2)\)

47) Kupata kikomo kama\( n→∞\) ya\( \dfrac{1}{n}+\dfrac{1}{n+1}+⋯+\dfrac{1}{3n}\)

Mazoezi machache yafuatayo yanalenga kutoa hisia ya maombi ambayo kiasi cha sehemu ya mfululizo wa harmonic hutokea.

48) Katika baadhi ya maombi ya uwezekano, kama vile kinachojulikana Watterson estimator kwa ajili ya utabiri wa viwango vya mabadiliko katika genetics idadi ya watu, ni muhimu kuwa na makadirio sahihi ya idadi\( H_k=(1+\frac{1}{2}+\frac{1}{3}+⋯+\frac{1}{k})\). Kumbuka kwamba\( T_k=H_k−\ln k\) ni kupungua. Compute\(\displaystyle T=\lim_{k→∞}T_k\) kwa maeneo manne decimal.

\(\quad\Big(\)Kidokezo:\(\displaystyle \frac{1}{k+1}<∫^{k+1}_k\frac{1}{x}\,dx.\Big)\)

Jibu

\( T=0.5772...\)

49) [T] Kamili sampuli na badala, wakati mwingine huitwa tatizo Coupon mtoza, ni maneno kama ifuatavyo: Tuseme una vitu\( N\) kipekee katika bin. Katika kila hatua, kipengee kinachaguliwa kwa random, kutambuliwa, na kurejeshwa kwenye bin. Tatizo linauliza ni nambari gani inayotarajiwa ya hatua\( E(N)\) ambazo inachukua kuteka kila kitu cha kipekee angalau mara moja. Inageuka kuwa\( E(N)=N.H_N=N\left(1+\frac{1}{2}+\frac{1}{3}+⋯+\frac{1}{N}\right)\). Kupata\( E(N)\) kwa\( N=10,20,\) na\( 50\).

50) [T] Njia rahisi ya shuffle kadi ni kuchukua kadi ya juu na kuingiza katika nafasi random katika staha, aitwaye juu random insertion, na kisha kurudia. Sisi kufikiria staha kuwa nasibu shuffled mara moja ya kutosha juu kuingizwa random yamefanywa kuwa kadi awali chini umefikia juu na kisha nasibu kuingizwa. Ikiwa staha ina\( n\) kadi, basi uwezekano kwamba kuingizwa itakuwa chini ya kadi awali chini (piga kadi hii\( B\)) ni\( 1/n\). Hivyo idadi inatarajiwa ya kuingizwa juu random kabla\( B\) ni tena chini ni\( n\). Mara baada ya kadi moja ni chini\( B\), kuna maeneo mawili chini\( B\) na uwezekano kwamba kadi nasibu kuingizwa kuanguka chini\( B\) ni\( 2/n\). idadi inatarajiwa ya kuingizwa juu random kabla ya hii hutokea ni\( n/2\). kadi mbili hapa chini\( B\) ni sasa katika utaratibu random. Kuendelea kwa njia hii, kupata formula kwa idadi inatarajiwa ya kuingizwa juu random zinahitajika kufikiria staha kuwa nasibu shuffled.

Jibu

idadi inatarajiwa ya insertions random\( B\) kupata juu ni\( n+n/2+n/3+⋯+n/(n−1).\) Kisha moja kuingizwa zaidi unaweka\( B\) nyuma katika random. Hivyo, idadi inatarajiwa ya shuffles kwa randomize staha ni\( n(1+1/2+⋯+1/n).\)

51) Tuseme pikipiki inaweza kusafiri\( 100\) km kwenye tank kamili ya mafuta. Kutokana kwamba mafuta yanaweza kuhamishwa kutoka pikipiki moja hadi nyingine lakini inaweza tu kufanyika katika tank, sasa utaratibu ambayo itawezesha moja ya scooters kusafiri\( 100H_N\) km, ambapo\( H_N=1+1/2+⋯+1/N.\)

52) Onyesha kwamba kwa makadirio ya salio kuomba juu\( [N,∞)\) yake ni ya kutosha kwamba\( f(x)\) kupungua juu ya\( [N,∞)\), lakini\( f\) haja ya kuwa na kupungua kwa\( [1,∞).\)

Jibu

Kuweka\( b_n=a_{n+N}\) na\( g(t)=f(t+N)\) vile kwamba\( f\) ni kupungua\( [t,∞).\)

53) [T] Tumia makadirio yaliyobaki na ushirikiano na sehemu ili takriban\(\displaystyle \sum_{n=1}^∞\frac{n}{e^n}\) ndani ya kosa ndogo kuliko\( 0.0001.\)

54) Je,\(\displaystyle \sum_{n=2}^∞\frac{1}{n(\ln n)^p}\) hujiunga ikiwa\( p\) ni kubwa ya kutosha? Ikiwa ndivyo, kwa\( p\) nini?

Jibu

Mfululizo hujiunga\( p>1\) na mtihani muhimu kwa kutumia mabadiliko ya kutofautiana.

55) [T] Tuseme kompyuta inaweza jumla ya maneno milioni moja kwa sekunde ya mfululizo tofauti\(\displaystyle \sum_{n=1}^N\frac{1}{n}\). Tumia mtihani muhimu ili takriban sekunde ngapi itachukua ili kuongeza maneno ya kutosha kwa jumla ya sehemu ya kuzidi\( 100\).

56) [T] Kompyuta ya haraka inaweza jumla ya maneno milioni moja kwa sekunde ya mfululizo tofauti\(\displaystyle \sum_{n=2}^N\frac{1}{n\ln n}\). Tumia mtihani muhimu ili takriban sekunde ngapi itachukua ili kuongeza maneno ya kutosha kwa jumla ya sehemu ya kuzidi\( 100.\)

Jibu

\( N=e^{e^{100}}≈e^{10^{43}}\)masharti zinahitajika.