Contoh Soal Deret Taylor
Gaisssss mhonn bntu polinom deret taylor soal terlampirr
1. Gaisssss mhonn bntu polinom deret taylor soal terlampirr
Jawab:
[tex]\displaystyle f(x,y)=\frac{3x^3}{2-2x^2-2y^2}\\f(1,1)=\frac{3\cdot1^3}{2-2\cdot1^2-2\cdot1^2}\\f(x,y)=\frac{3x^3}{2-2x^2-2y^2}\\f(1,1)=\frac{3}{2-2-2}\\f(1,1)=-\frac{3}{2}[/tex]
[tex]\displaystyle f_x(x,y)=\frac{9x^2(2-2x^2-2y^2)-3x^3(-4x)}{(2-2x^2-2y^2)^2}\\f_x(x,y)=\frac{18x^2-18x^4-18x^2y^2+12x^4}{(2-2x^2-2y^2)^2}\\f_x(x,y)=\frac{(1-y^2)18x^2-6x^4}{(2-2x^2-2y^2)^2}\\f_x(1,1)=\frac{(1-1^2)18\cdot1^2-6\cdot1^4}{(2-2\cdot1^2-2\cdot1^2)^2}\\f_x(1,1)=\frac{0\cdot18-6}{(2-2-2)^2}\\f_x(1,1)=\frac{-6}{4}\\f_x(1,1)=-\frac{3}{2}[/tex]
[tex]\displaystyle f_y(x,y)=\frac{0(2-2x^2-2y^2)-3x^3(-4y)}{(2-2x^2-2y^2)^2}\\f_y(x,y)=\frac{12x^3y}{(2-2x^2-2y^2)^2}\\f_y(1,1)=\frac{12\cdot1^3\cdot1}{(2-2\cdot1^2-2\cdot1^2)^2}\\f_y(1,1)=\frac{12}{(2-2-2)^2}\\f_y(1,1)=\frac{12}{4}\\f_y(1,1)=3[/tex]
[tex]\displaystyle f_{xx}(x,y)=\frac{((1-y^2)36x-24x^3)(2-2x^2-2y^2)^2+((1-y^2)18x^2-6x^4)(8x)}{((2-2x^2-2y^2)^2)^2}\\f_{xx}(x,y)=\frac{(1-y^2)36x-24x^3}{(2-2x^2-2y^2)^2}+\frac{8x((1-y^2)18x^2-6x^4)}{(2-2x^2-2y^2)^4}\\f_{xx}(1,1)=\frac{(1-1^2)36\cdot1-24\cdot1^3}{(2-2\cdot1^2-2\cdot1^2)^2}+\frac{8\cdot1((1-1^2)18\cdot1^2-6\cdot1^4)}{(2-2\cdot1^2-2\cdot1^2)^4}\\f_{xx}(1,1)=\frac{(1-1)36-24}{(2-2-2)^2}+\frac{8((1-1)18-6)}{(2-2-2)^4}\\f_{xx}(1,1)=-\frac{24}{4}-\frac{48}{16}\\f_{xx}(1,1)=-6-3\\f_{xx}(1,1)=-9[/tex]
[tex]\displaystyle f_{xy}(x,y)=\frac{36x^2y(2-2x^2-2y^2)^2-12x^3y(-8x)}{((2-2x^2-2y^2)^2)^2}\\f_{xy}(x,y)=\frac{36x^2y(2-2x^2-2y^2)^2+96x^4y}{(2-2x^2-2y^2)^4}\\f_{xy}(1,1)=\frac{36\cdot1^2\cdot1(2-2\cdot1^2-2\cdot1^2)^2+96\cdot1^4\cdot1}{(2-2\cdot1^2-2\cdot1^2)^4}\\f_{xy}(1,1)=\frac{36(2-2-2)^2+96}{(2-2-2)^4}\\f_{xy}(1,1)=\frac{36\cdot4+96}{16}\\f_{xy}(1,1)=9+6\\f_{xy}(1,1)=15[/tex]
[tex]\displaystyle f_{yy}(x,y)=\frac{12x^3(2-2x^2-2y^2)^2-12x^3y(-8y)}{((2-2x^2-2y^2)^2)^2}\\f_{yy}(x,y)=\frac{12x^3(2-2x^2-2y^2)^2+96x^3y^2}{(2-2x^2-2y^2)^4}\\f_{yy}(1,1)=\frac{12(2-2-2)^2+96}{(2-2-2)^4}\\f_{yy}(1,1)=\frac{48+96}{16}\\f_{yy}(1,1)=3+6\\f_{yy}(1,1)=9[/tex]
[tex]\displaystyle f(x,y)=f(a,b)+\frac1{1!}((x-a)f_x(a,b)+(y-b)f_y(a,b))+\frac1{2!}(f_{xx}(a,b)(x-a)^2+2(x-a)(y-b)f_{xy}(a,b)+f_{yy}(a,b)(y-b)^2)\\f(x,y)=f(1,1)+(x-1)f_x(1,1)+(y-1)f_y(1,1)+\frac1{2}(f_{xx}(1,1)(x-1)^2+2(x-1)(y-1)f_{xy}(1,1)+f_{yy}(1,1)(y-1)^2)\\f(x,y)=-\frac32-\frac32(x-1)+3(y-1)+\frac1{2}(-9(x-1)^2+30(x-1)(y-1)+9(y-1)^2)[/tex]
Beberapa konsep yang dipakai:
[tex]\displaystyle \triangleright~f_x(x,y)=\frac{\partial}{\partial x}f(x,y)\\\triangleright~f_y(x,y)=\frac{\partial}{\partial y}f(x,y)\\\triangleright~f_{xx}(x,y)=\frac{\partial^2}{\partial x^2}f(x,y)\\\triangleright~f_{yy}(x,y)=\frac{\partial^2}{\partial y^2}f(x,y)\\\triangleright~f_{xy}(x,y)=\frac{\partial}{\partial y}(f_x(x,y))\\\triangleright~\frac{d}{dx}(y_1+y_2+\cdots+y_n)=y'_1+y'_2+\cdots+y'_n\\\triangleright~\frac{d}{dx}\left(\frac uv\right)=\frac{u'v-uv'}{v^2}\\\triangleright~\frac{d}{dx}(ax^n)=anx^{n-1}[/tex]
2. Tentukan deret Taylor Dari fungsi, dengan deret Taylor di a = 2 f (x) = x³ - 08
f(x) = x³ - 08
f(2) = (2)³ - (2)8
f(2) = 8 - 16
f(x) = 16 - 8
f(x) = 83. Manfaat deret taylor
Jawaban:
deret Taylor adalah representasi fungsi matematika sebagai jumlah tak hingga dari suku-suku yang nilai nya dihitung dari turunan fungsi tersebut di suatu titik
4. Tentukan deret taylor dari (x-a) sampai (x-a)³
saya tuliskan rumusnya...
fungsinya masih f(x), bila f(x) =e^x, maka cari f'(x), f"(x), f'"(x) dst...
5. Deretkan ke deret taylor F(x)= 1/4x+8 di x=2
Jawab:
Penjelasan dengan langkah-langkah:
6. Tentukan Deret Taylor dari fungsi, dengan deret taylor di a = 2 () = ³ − 4
Jawaban :
a = ( 4 - 2 )³
= ( 2 )³
= 2 × 2 × 2
= 4 × 2
=8
- Penyelesaian Soal - [tex] \: [/tex]Penyelesaian :a = 2 () = ³ − 4
a = (4 - 2)³a = ( 2 )³a = ( 2 × 2 × 2 )a = ( 4 × 2 ) a = 8Kesimpulana = 8iHope This Help7. Deret taylor f(x)=sin(3x)
• Trigonometri
-
Deret Taylor
[tex] \boxed{ \tt f(x) = \sin(3x) }[/tex]
Misal kita akan mencari Deret taylor di x = 0
Mencari Turunan Fungsi[tex] \: \: \: \: \tt f'(x) = 3 \cos(3x) \\ \tt \: \: f''(x) = - 9 \sin(3x) \\ \tt f'''(x) = - 27 \cos(3x) \\ \: \: \: \tt {f}^{4} (x) = 81 \sin(3x) \\ \: \: \: \tt {f}^{5} (x) = 243 \cos(3x) [/tex]
Mencari Nilai x = 0 tiap turunan[tex] \: \: \: \: \tt f'(0) = 3 \cos(0) = 3 \\ \tt \: \: f''(0) = - 9 \sin(0) = 0 \\ \tt f'''(0) = - 27 \cos(0) = - 27\\ \: \: \: \tt {f}^{4} (0) = 81 \sin(0) = 0 \\ \: \: \: \tt {f}^{5} (0) = 243 \cos(0) = 243[/tex]
Menentukan Deret Taylor[tex] \boxed{ \tt \sin(3x) = \sum\limits^{ \infty }_{n = 0} {f}^{n} x_{0} \frac{ {(x - x_{0})}^{n} }{n!} } [/tex]
Gunakan ekspansi diatas , diperoleh :
[tex] \tt \sin(3x) = f(0) + \frac{f'(0)}{1!} (x - 0)^{1} + \frac{f''(0)}{2!} (x - 0)^{2} + \frac{f'''(0)}{3!} (x - 0)^{3} + \frac{ {f}^{4}(0) }{4!} (x - 0)^{4} + \frac{ {f}^{5} (0)}{5!} (x - 0)^{5} + ... \\ \tt \sin(3x) =0 + 3 {x} + 0 - \frac{27}{6} {x}^{3} + 0 + \frac{243}{120} {x}^{5} \\ \tt \sin(3x) = 3x - \frac{9}{2} {x}^{3} + \frac{81}{40} {x}^{5} + ...[/tex]
•••
Jawaban:
• Deret Taylor
---------------------
Terlampir pada Gambar
8. cari deret taylor Pada fungsi f(x) = ex²
soal :
cari deret taylor Pada fungsi f(x) = ex².
jawab :
[tex]\mathbf{rumus\ taylor=}\sum_{n=0}^{\infty}\frac{f^{(n)}(a)(x-a)^n}{n!}[/tex]
[tex]n=0: \frac{f(0)(x-0)^{0}}{0 !}=\frac{e(0)^{2}(1)}{0 !}=\frac{0}{0 !}[/tex]
[tex]n=1: \frac{f^{\prime}(0)(x-0)^{1}}{1 !}=\frac{2 e(0)(x)}{1 !}=\frac{0}{1 !}[/tex]
[tex]n=2: \frac{f^{\prime \prime}(0)(x-0)^{2}}{2 !}=\frac{2 e x^{2}}{2 !}=\frac{2 e x^{2}}{2 !}[/tex]
[tex]n=3: \frac{f^{\prime \prime}(0)(x-0)^{3}}{3 !}=\frac{0 x^{3}}{3 !}=\frac{0}{3 !}[/tex]
[tex]n=4: \frac{f^{(4)}(0)(x-0)^{4}}{4 !}=\frac{0 x^{4}}{4 !}=\frac{0}{4 !}[/tex]
[tex]n=5: \frac{f^{(5)}(0)(x-0)^{5}}{5 !}=\frac{0 x^{5}}{5 !}=\frac{0}{5 !}[/tex]
[tex]n=6: \frac{f^{(6)}(0)(x-0)^{6}}{6 !}=\frac{0 x^{0}}{6 !}=\frac{0}{6 !}[/tex]
[tex]n=7: \frac{f^{f^{t h}(0)(x-0)^{r}}}{7 !}=\frac{0 x^{c}}{7 !}=\frac{0}{7 !}[/tex]
[tex]n=8: \frac{f^{(8)}(0)(x-0)^{8}}{8 !}=\frac{0 x^{8}}{8 !}=\frac{0}{8 !}[/tex]
[tex]n=9: \frac{f^{(9)}(0)(x-0)^{9}}{9 !}=\frac{0 x^{9}}{9 !}=\frac{0}{9 !}[/tex]
[tex]\textcolor{#2B7FBB}{\mathbf{sekian\ g\ trima\ gaji\ janlup\ leks\ }}[/tex]
9. 3. Dapatkan deret Taylor dari f(x) = 2/x di a = 2dan dari soal diatas tentukan juga deret maclaurin nya
Jawaban:
maaf kalau gak jelas
Penjelasan dengan langkah-langkah:
di folbackbya
10. Deret taylor f(x)=2sin(2x), pada x=phi/4 adalah...
Jawaban:
Ekspansi deret taylor di sekitar x = a:
$\begin{aligned}F(x)&=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+...\end{aligned}
\begin{gathered}$\begin{aligned}f(x)&=\sin(2x),&\ f(\frac{\pi}{2})&=0 \\ f'(x)&=2\cos(2x),&\ f'(\frac{\pi}{2})&=-2 \\ f''(x)&=-4\sin(2x),&\ f''(\frac{\pi}{2})&=0 \\ f'''(x)&=-8\cos(2x),&\ f'''(\frac{\pi}{2})&=8 \\ \text{dst, maka}\end{aligned}\end{gathered}
\begin{gathered}$\begin{aligned}F(x)&=0+(-2)(x - \frac{\pi}{2})+0+\frac{8}{3!}(x-\frac{\pi}{2})^3+0+\frac{(-32)}{5!}(x-\frac{\pi}{2})^5+... \\ &=-2(x-\frac{\pi}{2})+\frac{8}{3!}(x-\frac{\pi}{2})^3-\frac{32}{5!}(x-\frac{\pi}{2})^5+\frac{128}{7!}(x-\frac{\pi}{2})^7+...\end{aligned}\end{gathered}
Penjelasan dengan langkah-langkah:
maaf kalau salah
11. tentukan deret taylor f (x) = x e^x
Jawab:
Penjelasan dengan langkah-langkah:
12. Tentukan Deret Taylor dari fungsi, dengan deret taylor di a = 2 () = ³− 4
a = (4 - 2)³
a = (2)³
a = 2 × 2 × 2
a = 4 × 2
a = 8a = (4 - 2)³
a = (4 - 2)(4 - 2)(4 - 2)
a = 4(4 - 2) - 2(4 - 2)(4 - 2)
a = (16 - 8 - 8 + 4)(4 - 2)
a = (8 - 8 + 4)(4 - 2)
a = (0 + 4)(4 - 2)
a = 0(4 - 2) + 4(4 - 2)
a = 0 - 0 + 16 - 8
a = 0 + 16 - 8
a = 16 - 8
a =8
:)13. Ekspansikan f(x) = sin x , dalam deret taylor
Jawab:
x - x³/3! + x⁵/5! - x⁷/7! + ...
Penjelasan dengan langkah-langkah:
Deret Maclaurin dirumuskan
[tex]\displaystyle f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}~x^2+\frac{f'''(x)}{3!}~x^3+\frac{f^4(0)}{4!}~x^4+...+\frac{f^n(0)}{n!}x^n[/tex]
f(x) = sin x → f(0) = sin 0 = 0
f '(x) = cos x → f '(0) = cos 0 = 1
f ''(x) = -sin x → f ''(0) = -sin 0 = 0
f '''(x) = -cos x → f '''(0) = -cos 0 = -1
f ⁴(x) = sin x → f ⁴(0) = sin 0 = 0
Ingat turunan tingkat tinggi fungsi trigonometri
f ⁵(x) = f '(x) → f ⁵(0) = f '(0) = 1
f ⁶(x) = f ''(x) → f ⁶(0) = f ''(0) = 0
f ⁷(x) = f '''(x) → f ⁷(0) = f '''(0) = -1
f ⁸(x) = f ⁴(x) → f ⁸(0) = f ⁴(0) = 0
dan seterusnya
Maka ekspansi dari f(x) = sin x adalah
[tex]\displaystyle \sin x=0+1x+\frac{0}{2!}~x^2+\frac{-1}{3!}~x^3+\frac{0}{4!}~x^4+\frac{1}{5!}~x^5+\frac{0}{6!}~x^6+\frac{-1}{7!}~x^7+\frac{0}{8!}~x^8+...\\\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...[/tex]
x dalam radian.
14. Tentukan Deret Taylor dari fungsi f(x) = x^3 - 18, dengan deret taylor di a = 2
taylor series (deret taylor):
f(x) = f(a) + f'(a)•(x-a) + f''(a)/2! • (x-a)² + f"'(a)/3! • (x-a)³
(berhenti di 3 karena polinomnya berderajat 3).
f(x) = x³-18
f(2) = 2³-18 = -10
f'(x) = 3x²
f'(2) = 12
f"(x) = 6x
f"(2) = 12
f"'(x) = 6
f"'(2) = 6
berarti deret taylornya:
f(x) = f(2) + f'(2)•(x-2) + f"(2)/2! • (x-2)² + f"'(2)/3! • (x-2)³
f(x) = -10 + 12•(x-2) + 12/2 • (x-2)² + 6/6 • (x-2)³
f(x) = (x-2)³ + 6(x-2)² + 12(x-2) - 10
Jawab:
Penjelasan dengan langkah-langkah:
[tex]\displaystyle f(x) = x^3 - 18\\f(x) = x^3 - 3\cdot 2\cdot x^2 + 3\cdot 4\cdot x - 8 + 3\cdot 2\cdot x^2 - 3\cdot 4\cdot x + 8-18\\f(x) = (x-2)^3 + 6x^2 - 12x - 10 = (x-2)^3 + 3(x^2+x^2-4x) - 10\\f(x) = (x-2)^3 + 3(x^2-4x+4 - 4)+3x^2 - 10 \\f(x) = (x-2)^3 + 3(x-2)^2+3x^2 - 10 - 12\\f(x) = (x-2)^3 + 3(x-2)^2+3x^2-3\cdot 4\cdot x + 3\cdot 4\\ +3\cdot 4\cdot x - 3\cdot 4 - 22\\f(x) = (x-2)^3 + 6(x-2)^2 + 12x-34 \\[/tex]
[tex]f(x)= (x-2)^3 + 6(x-2)^2 + 12x-24 - 10 \\\\ \boxed{\boxed{f(x) = (x-2)^3 + 6(x-2)^2 + 12(x-2)-10}}\\\\[/tex]
15. Tentukan Deret Taylor dari fungsi, dengan deret taylor di a = 2f(x) = x^3 - 18
f(x) = 3/x ⇒f(1) = 3/1 = 3
f '(x) = -3/x² ⇒f '(1) = -3/1² = -3
f "(x) = 6/x³ ⇒f "(1) = 6/1³ = 6
f "'(x) = -18/x^4 ⇒f "'(1) = -18
f ""(x) = 72/x^5 ⇒f ""(1) = 72, dst
Maka,
f(x) = 3 + (-3)/1! + 6/2! (2-1) + (-18)/3! (2-1)² + 72/4! (2-1)³ + ....... ≈ 3/2 = 1,5
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