10.3E: Mazoezi ya Sehemu ya 10.3

Taylor Polynomials

Katika mazoezi ya 1 - 8, tafuta polynomials ya Taylor ya shahada mbili zinazokadiria kazi iliyotolewa katikati katika hatua fulani.

1)\( f(x)=1+x+x^2\) katika\( a=1\)

2)\( f(x)=1+x+x^2\) katika\( a=−1\)

Jibu

\( f(−1)=1;\;f′(−1)=−1;\;f''(−1)=2;\quad p_2(x)=1−(x+1)+(x+1)^2\)

3)\( f(x)=\cos(2x)\) katika\( a=π\)

4)\( f(x)=\sin(2x)\) katika\( a=\frac{π}{2}\)

Jibu

\( f′(x)=2\cos(2x);\;f''(x)=−4\sin(2x);\quad p_2(x)=−2(x−\frac{π}{2})\)

5)\( f(x)=\sqrt{x}\) katika\( a=4\)

6)\( f(x)=\ln x\) katika\( a=1\)

Jibu

\( f′(x)=\dfrac{1}{x};\; f''(x)=−\dfrac{1}{x^2};\quad p_2(x)=0+(x−1)−\frac{1}{2}(x−1)^2\)

7)\( f(x)=\dfrac{1}{x}\) katika\( a=1\)

8)\( f(x)=e^x\) katika\( a=1\)

Jibu

\( p_2(x)=e+e(x−1)+\dfrac{e}{2}(x−1)^2\)

Theorem ya Taylor

Katika mazoezi 9 - 14, hakikisha kwamba uchaguzi uliotolewa wa\(n\) makadirio ya salio\( |R_n|≤\dfrac{M}{(n+1)!}(x−a)^{n+1}\), wapi\(M\) thamani ya\( ∣f^{(n+1)}(z)∣\) juu ya muda kati\(a\) na hatua iliyoonyeshwa, mavuno\( |R_n|≤\frac{1}{1000}\). Pata thamani ya polynomial\( p_n\) ya Taylor ya\( f\) katika hatua iliyoonyeshwa.

9) [T]\( \sqrt{10};\; a=9,\; n=3\)

10) [T]\( (28)^{1/3};\; a=27,\; n=1\)

Jibu

\( \dfrac{d^2}{dx^2}x^{1/3}=−\dfrac{2}{9x^{5/3}}≥−0.00092…\)wakati\( x≥28\) hivyo makadirio salio inatumika kwa makadirio linear\( x^{1/3}≈p_1(27)=3+\dfrac{x−27}{27}\), ambayo inatoa\( (28)^{1/3}≈3+\frac{1}{27}=3.\bar{037}\), wakati\( (28)^{1/3}≈3.03658.\)

11) [T]\( \sin(6);\; a=2π,\; n=5\)

12) [T]\( e^2; \; a=0,\; n=9\)

Jibu

Kutumia makadirio\( \dfrac{2^{10}}{10!}<0.000283\) tunaweza kutumia upanuzi wa Taylor wa utaratibu wa 9 ili kukadiria\( e^x\)\( x=2\) katika. kama\( e^2≈p_9(2)=1+2+\frac{2^2}{2}+\frac{2^3}{6}+⋯+\frac{2^9}{9!}=7.3887\)... wakati\( e^2≈7.3891.\)

13) [T]\( \cos(\frac{π}{5});\; a=0,\; n=4\)

14) [T]\( \ln(2);\; a=1,\; n=1000\)

Jibu

Tangu\( \dfrac{d^n}{dx^n}(\ln x)=(−1)^{n−1}\dfrac{(n−1)!}{x^n},R_{1000}≈\frac{1}{1001}\). Moja ina\(\displaystyle p_{1000}(1)=\sum_{n=1}^{1000}\dfrac{(−1)^{n−1}}{n}≈0.6936\) wakati\( \ln(2)≈0.6931⋯.\)

Kukadiria Integrals uhakika Kutumia Taylor Series

15) Unganisha makadirio\(\sin t≈t−\dfrac{t^3}{6}+\dfrac{t^5}{120}−\dfrac{t^7}{5040}\) tathmini katika\( π\) t kwa takriban\(\displaystyle ∫^1_0\frac{\sin πt}{πt}\,dt\).

16) Unganisha makadirio\( e^x≈1+x+\dfrac{x^2}{2}+⋯+\dfrac{x^6}{720}\) tathmini katika\( −x^2\) takriban\(\displaystyle ∫^1_0e^{−x^2}\,dx.\)

Jibu

\(\displaystyle ∫^1_0\left(1−x^2+\frac{x^4}{2}−\frac{x^6}{6}+\frac{x^8}{24}−\frac{x^{10}}{120}+\frac{x^{12}}{720}\right)\,dx =1−\frac{1^3}{3}+\frac{1^5}{10}−\frac{1^7}{42}+\frac{1^9}{9⋅24}−\frac{1^{11}}{120⋅11}+\frac{1^{13}}{720⋅13}≈0.74683\)wakati\(\displaystyle ∫^1_0e^{−x^2}dx≈0.74682.\)

Zaidi Taylor salifu Theorem Matatizo

Katika mazoezi ya 17 - 20, tafuta thamani ndogo zaidi ya\(n\) vile makadirio ya salio\( |R_n|≤\dfrac{M}{(n+1)!}(x−a)^{n+1}\), wapi\(M\) thamani ya\( ∣f^{(n+1)}(z)∣\) juu ya muda kati\(a\) na hatua iliyoonyeshwa,\( |R_n|≤\frac{1}{1000}\) huzaa wakati ulioonyeshwa.

17)\( f(x)=\sin x\) juu\( [−π,π],\; a=0\)

18)\( f(x)=\cos x\) juu\( [−\frac{π}{2},\frac{π}{2}],\; a=0\)

Jibu

Tangu\( f^{(n+1)}(z)\) ni\(\sin z\) au\(\cos z\), tuna\( M=1\). Tangu\( |x−0|≤\frac{π}{2}\), tunatafuta\(n\) ndogo zaidi\( \dfrac{π^{n+1}}{2^{n+1}(n+1)!}≤0.001\). Thamani ndogo zaidi ni\( n=7\). Makadirio yaliyobaki ni\( R_7≤0.00092.\)

19)\( f(x)=e^{−2x}\) juu\( [−1,1],a=0\)

20)\( f(x)=e^{−x}\) juu\( [−3,3],a=0\)

Jibu

Kwa kuwa\( f^{(n+1)}(z)=±e^{−z}\) mtu ana\( M=e^3\). Tangu\( |x−0|≤3\), moja inataka ndogo\(n\) kama hiyo\( \dfrac{3^{n+1}e^3}{(n+1)!}≤0.001\). Thamani ndogo zaidi ni\( n=14\). Makadirio yaliyobaki ni\( R_{14}≤0.000220.\)

Katika mazoezi 21 - 24, upeo wa upande wa kulia wa makadirio\( |R_1|≤\dfrac{max|f''(z)|}{2}R^2\) ya salio\( [a−R,a+R]\) hutokea saa\(a\) au\( a±R\). Tathmini thamani ya juu ya\(R\) vile kwamba\( \dfrac{max|f''(z)|}{2}R^2≤0.1\) juu\( [a−R,a+R]\) ya kupanga upeo huu kama kazi ya\(R\).

21) [T]\( e^x\) inakadiriwa na\( 1+x,\; a=0\)

22) [T]\( \sin x\) inakadiriwa na\( x,\; a=0\)

Jibu

Kwa kuwa\( \sin x\) ni kuongezeka kwa wadogo\( x\) na tangu\( \frac{d^2}{dx^2}\left(\sin x\right)=−\sin x\), makadirio inatumika wakati wowote\( R^2\sin(R)≤0.2\), ambayo inatumika hadi\( R=0.596.\)

23) [T]\( \ln x\) inakadiriwa na\( x−1,\; a=1\)

24) [T]\( \cos x\) inakadiriwa na\( 1,\; a=0\)

Jibu

Kwa kuwa derivative pili ya\( \cos x\) ni\( −\cos x\) na tangu\( \cos x\) ni kupungua mbali na\( x=0\), makadirio inatumika wakati\( R^2\cos R≤0.2\) au\( R≤0.447\).

Taylor Series

Katika mazoezi 25 - 35, tafuta mfululizo wa Taylor wa kazi iliyotolewa katikati ya hatua iliyoonyeshwa.

25)\(f(x) = x^4\) saa\( a=−1\)

26)\(f(x) = 1+x+x^2+x^3\) saa\( a=−1\)

Jibu

\( (x+1)^3−2(x+1)^2+2(x+1)\)

27)\(f(x) = \sin x\) katika\( a=π\)

28)\(f(x) = \cos x\) katika\( a=2π\)

Jibu

Maadili ya derivatives ni sawa na kwa\( x=0\) hivyo\(\displaystyle \cos x=\sum_{n=0}^∞(−1)^n\frac{(x−2π)^{2n}}{(2n)!}\)

29)\(f(x) = \sin x\) saa\( x=\frac{π}{2}\)

30)\(f(x) = \cos x\) saa\( x=\frac{π}{2}\)

Jibu

\( \cos(\frac{π}{2})=0,\;−\sin(\frac{π}{2})=−1\)hivyo\(\displaystyle \cos x=\sum_{n=0}^∞(−1)^{n+1}\frac{(x−\frac{π}{2})^{2n+1}}{(2n+1)!}\), ambayo pia\( −\cos(x−\frac{π}{2})\).

31)\(f(x) = e^x\) saa\( a=−1\)

32)\(f(x) = e^x\) katika\( a=1\)

Jibu

Derivatives ni\( f^{(n)}(1)=e,\) hivyo\(\displaystyle e^x=e\sum_{n=0}^∞\frac{(x−1)^n}{n!}.\)

33)\(f(x) = \dfrac{1}{(x−1)^2}\) katika\( a=0\) (ladha: Tofauti Taylor Series kwa\( \dfrac{1}{1−x}\).)

34)\(f(x) = \dfrac{1}{(x−1)^3}\) saa\( a=0\)

Jibu

\(\displaystyle \frac{1}{(x−1)^3}=−\frac{1}{2}\frac{d^2}{dx^2}\left(\frac{1}{1−x}\right)=−\sum_{n=0}^∞\left(\frac{(n+2)(n+1)x^n}{2}\right)\)

35)\(\displaystyle F(x)=∫^x_0\cos(\sqrt{t})\,dt;\quad \text{where}\; f(t)=\sum_{n=0}^∞(−1)^n\frac{t^n}{(2n)!}\) katika =0 (Kumbuka:\( f\) ni mfululizo wa Taylor\(\cos(\sqrt{t}).)\)

Katika mazoezi 36 - 44, compute mfululizo wa Taylor wa kila kazi kote\( x=1\).

36)\( f(x)=2−x\)

Jibu

\( 2−x=1−(x−1)\)

37)\( f(x)=x^3\)

38)\( f(x)=(x−2)^2\)

Jibu

\( ((x−1)−1)^2=(x−1)^2−2(x−1)+1\)

39)\( f(x)=\ln x\)

40)\( f(x)=\dfrac{1}{x}\)

Jibu

\(\displaystyle \frac{1}{1−(1−x)}=\sum_{n=0}^∞(−1)^n(x−1)^n\)

41)\( f(x)=\dfrac{1}{2x−x^2}\)

42)\( f(x)=\dfrac{x}{4x−2x^2−1}\)

Jibu

\(\displaystyle x\sum_{n=0}^∞2^n(1−x)^{2n}=\sum_{n=0}^∞2^n(x−1)^{2n+1}+\sum_{n=0}^∞2^n(x−1)^{2n}\)

43)\( f(x)=e^{−x}\)

44)\( f(x)=e^{2x}\)

Jibu

\(\displaystyle e^{2x}=e^{2(x−1)+2}=e^2\sum_{n=0}^∞\frac{2^n(x−1)^n}{n!}\)

Maclaurin Series

[T] Katika mazoezi 45 - 48, kutambua thamani ya\(x\) vile kwamba mfululizo uliopewa\(\displaystyle \sum_{n=0}^∞a_n\) ni thamani ya mfululizo Maclaurin wa\( f(x)\) saa\( x\). Takriban thamani ya\( f(x)\) kutumia\(\displaystyle S_{10}=\sum_{n=0}^{10}a_n\).

45)\(\displaystyle \sum_{n=0}^∞\frac{1}{n!}\)

46)\(\displaystyle \sum_{n=0}^∞\frac{2^n}{n!}\)

Jibu

\( x=e^2;\quad S_{10}=\dfrac{34,913}{4725}≈7.3889947\)

47)\(\displaystyle \sum_{n=0}^∞\frac{(−1)^n(2π)^{2n}}{(2n)!}\)

48)\(\displaystyle \sum_{n=0}^∞\frac{(−1)^n(2π)^{2n+1}}{(2n+1)!}\)

Jibu

\(\sin(2π)=0;\quad S_{10}=8.27×10^{−5}\)

Katika mazoezi 49 - 52 tumia kazi\( S_5(x)=x−\dfrac{x^3}{6}+\dfrac{x^5}{120}\) na\( C_4(x)=1−\dfrac{x^2}{2}+\dfrac{x^4}{24}\) kuendelea\( [−π,π]\).

49) [T] njama\(\sin^2x−(S_5(x))^2\) juu ya\( [−π,π]\). Linganisha tofauti ya kiwango cha juu na mraba wa makadirio ya salio ya Taylor\( \sin x.\)

50) [T] njama\(\cos^2x−(C_4(x))^2\) juu ya\( [−π,π]\). Linganisha tofauti ya kiwango cha juu na mraba wa Taylor salio makadirio ya\( \cos x\).

Jibu

Tofauti ni ndogo juu ya mambo ya ndani ya muda lakini inakaribia\( 1\) karibu na mwisho. Makadirio yaliyobaki ni\( |R_4|=\frac{π^5}{120}≈2.552.\)

51) [T] njama\( |2S_5(x)C_4(x)−\sin(2x)|\) juu ya\( [−π,π]\).

52) [T] Linganisha\( \dfrac{S_5(x)}{C_4(x)}\)\( [−1,1]\) na\( \tan x\). Kulinganisha hii na Taylor salio makadirio ya makadirio ya\( \tan x\) na\( x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}\).

Jibu

Tofauti ni juu ya utaratibu wa\( 10^{−4}\) juu ya\( [−1,1]\) wakati Taylor makadirio makosa ni karibu\( 0.1\) karibu\( ±1\). Curve juu ni njama ya\(\tan^2x−\left(\dfrac{S_5(x)}{C_4(x)}\right)^2\) na chini dashed njama inaonyesha\( t^2−\left(\dfrac{S_5}{C_4}\right)^2\).

53) [T] Plot\( e^x−e_4(x)\) ambapo\( e_4(x)=1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}+\dfrac{x^4}{24}\) juu ya\( [0,2]\). Linganisha hitilafu ya kiwango cha juu na makadirio ya Taylor

54) (Taylor makadirio na mizizi kutafuta.) Kumbuka kwamba njia ya Newton\( x_{n+1}=x_n−\dfrac{f(x_n)}{f'(x_n)}\) inakaribia ufumbuzi wa\( f(x)=0\) karibu na pembejeo\( x_0\).

a Kama\( f\) na\( g\) ni kazi inverse, kueleza kwa nini ufumbuzi wa\( g(x)=a\) ni thamani\( f(a)\) ya\( f\).

b Hebu\( p_N(x)\) kuwa\( N^{\text{th}}\) shahada Maclaurin polynomial ya\( e^x\). Tumia njia ya Newton kwa ufumbuzi wa\( p_N(x)−2=0\) takriban\( N=4,5,6.\)

c Eleza kwa nini mizizi ya takriban\( p_N(x)−2=0\) ni maadili ya takriban ya\(\ln(2).\)

Jibu

a Majibu yatatofautiana.
b. zifuatazo ni\( x_n\) maadili baada\( 10\) iterations ya njia Newton kwa makadirio mzizi wa\( p_N(x)−2=0\): kwa\( N=4,x=0.6939...;\)\( N=5,x=0.6932...;\) ajili ya\( N=6,x=0.69315...;.\) (Kumbuka:\( \ln(2)=0.69314...\))
c. majibu zitatofautiana.

Kutathmini mipaka kutumia Taylor Series

Katika mazoezi 55 - 58, tumia ukweli kwamba ikiwa\(\displaystyle q(x)=\sum_{n=1}^∞a_n(x−c)^n\) hujiunga katika kipindi kilicho na\( c\), kisha kutathmini kila kikomo\(\displaystyle \lim_{x→c}q(x)=a_0\) kwa kutumia mfululizo wa Taylor.

55)\(\displaystyle \lim_{x→0}\frac{\cos x−1}{x^2}\)

56)\(\displaystyle \lim_{x→0}\frac{\ln(1−x^2)}{x^2}\)

Jibu

\( \dfrac{\ln(1−x^2)}{x^2}→−1\)

57)\(\displaystyle \lim_{x→0}\frac{e^{x^2}−x^2−1}{x^4}\)

58)\(\displaystyle \lim_{x→0^+}\frac{\cos(\sqrt{x})−1}{2x}\)

Jibu

\(\displaystyle \frac{\cos(\sqrt{x})−1}{2x}≈\frac{(1−\frac{x}{2}+\frac{x^2}{4!}−⋯)−1}{2x}→−\frac{1}{4}\)