Tugas Kelompok Pengantar Fungsi Khusus

Tugas Kelompok II Pengantar Fungsi Khusus Oleh : Kelompok 3

1. Febri Heryo Yudanto M0119032 2. Narisya Zahrani M0121055 3. Gohan Lasuli Silvia M0121077

1. Use the recursion relation Γ(𝑧 + 1) = 𝑧Γ(𝑧), and if needed, equation Γ(𝑧 + 1) =

∫ 𝑡₀^{∞ 𝑧}𝑒^−𝑡𝑑𝑡 to simplify a. ^Γ(

2 3) Γ(⁸

3)

b. ^Γ(4)Γ(

3 4) Γ(⁷

4)

2. Express each of the following integrals as a Γ function a. ∫ 𝑥

2 3𝑒^−𝑥𝑑𝑥

∞ 0

b. ∫ √ln(𝑥)₀¹³ 𝑑𝑥

3. Express the following integrals as 𝐵 functions, and then, by 𝐵 =^{Γ(𝑝)Γ(𝑞)}

Γ(𝑝+𝑞), in terms of Γ functions. When possible, use Γ function formulas to write an exact answer in terms of 𝜋, √2, etc. Compare your answers with computer results and reconcile any discrepancies.

a. ∫ 𝑥₀^{1 2}(1 − 𝑥²)³²𝑑𝑥 b. ∫ √sin^{3 3}𝑥 cos 𝑥𝑑𝑥

𝜋 2 0

4. Soal

a. By computer, plot graphs of 𝐽_𝑝(𝑥) for 𝑝 = 4,5 and 𝑥 from 0 to 15.

b. From the graphs in problem 4a, state the number and location of the zeros.

5. Soal

a. Express 7𝑥⁴− 3𝑥 + 1 as linear combinations of Legendre polynomials. Hint: Start with the highest power of 𝑥 and work down in finding the correct combination.

b. Find the best (in the least squares sense) second-degree polynomial approximation to 𝑥⁴ over the interval −1 < 𝑥 < 1.

Pembahasan : 1.

a. Dengan menggunakan relasi rekrusi Γ(𝑧 + 1) = 𝑧Γ(𝑧) dan persamaan Γ(𝑧 + 1) =

∫ 𝑡₀^{∞ 𝑧}𝑒^−𝑡𝑑𝑡, maka diperoleh Γ (2

3) Γ (8

3)

= Γ (2 3) Γ (5

3+ 1)

= Γ (2 3) 5 3Γ (5

3)

= Γ (2 3) 5

3Γ (2 3+ 1)

= Γ (2 3) (5

3) (2 3) Γ (2

3)

= 1 10

9

=1 × 9 10

= 0,9 Jadi, nilai dari ^Γ(

2 3) Γ(⁸

3)= 0,9.

b. Dengan menggunakan relasi rekrusi Γ(𝑧 + 1) = 𝑧Γ(𝑧) dan persamaan Γ(𝑧 + 1) =

∫ 𝑡₀^{∞ 𝑧}𝑒^−𝑡𝑑𝑡, maka diperoleh

Γ(4)Γ (3 4) Γ (7

4)

= Γ(4)Γ (3 4) Γ (3

4+ 1)

= Γ(4)Γ (3 4) 3

4Γ (3 4)

= Γ(4) 3 4

= Γ(3 + 1) 3 4

= ∫ (𝑡₀^{∞ 3}𝑒^−𝑡)𝑑𝑡 3 4

= 6 3 4

= 6 × 4 3

= 12 Jadi, nilai dari ^Γ(4)Γ(

3 4) Γ(⁷

4) = 12.

2.

a. Diketahui bahwa Γ(𝑧) = ∫ 𝑡₀^{∞ 𝑧−1}𝑒^−𝑡𝑑𝑡, maka

∫ 𝑥²³𝑒^−𝑥𝑑𝑥

∞

0

= ∫ 𝑥⁵³⁻¹𝑒^−𝑥𝑑𝑥

∞

0

= Γ (5 3)

Jadi, ∫ 𝑥

2 3𝑒^−𝑥𝑑𝑥

∞

0 = Γ (⁵

3).

b. Diketahui bahwa Γ(𝑧) = ∫ 𝑡₀^{∞ 𝑧−1}𝑒^−𝑡𝑑𝑡, maka

Dimisalkan ln(𝑥) = −𝑡, 𝑥 = 𝑒^−𝑡, dan 𝑑𝑥 = −𝑒^−𝑡 𝑑𝑡. Hal ini berakibat batas atas dan batas bawah limit akan berubah, saat 𝑥 → 0 ⟹ 𝑡 → ∞ dan saat 𝑥 → 1 ⟹ 𝑡 → 0 sehingga

∫(−𝑡)¹³(−𝑒^−𝑡)𝑑𝑡

0

∞

= − ∫ (𝑡)¹³(𝑒^−𝑡)𝑑𝑡

∞

0

= − ∫ (𝑡)⁴³⁻¹(𝑒^−𝑡)𝑑𝑡

∞

0

= −Γ (4 3)

= −Γ (1 3+ 1)

= − (1 3) Γ (1

3) Jadi, ∫ √ln(𝑥)₀¹³ 𝑑𝑥 = − (¹

3) Γ (¹

3).

3.

a. Misal, 𝑢 = 𝑥² ⟶ 𝑑𝑢 = 2𝑥 𝑑𝑥 dan 𝑥 = 𝑢¹² Diperoleh,

∫ 𝑥²(1−𝑥²)^3/2 𝑥𝑥

1 0

= ∫ x²

2x(1−𝑥)^3/2 𝑥𝑥

1 0

= 1

2∫ 𝑥^1/2(1−𝑥)^3/2 𝑥𝑥

1 0

1

2𝑥(𝑥,𝑥) =1

2∫ 𝑥^𝑥−1(1−𝑥)^𝑥−1 𝑥𝑥

1

0

𝑥−1=1/2 ⟶ 𝑥=3/2 𝑥−1=3/2 ⟶ 𝑥=5/2

∫ 𝑥²(1−𝑥²)^3/2 𝑥𝑥

1 0

= (1 2)𝑥(3

2,5 2)

Dari

𝑥(𝑥, 𝑥) =

^{𝑥(𝑥)𝑥(𝑥)}

𝑥(𝑥+𝑥) diperoleh,

∫ 𝑥²(1−𝑥²)^3/2 𝑥𝑥

1 0

= (1 2)𝑥(3

2,5 2)

∫ 𝑥²(1−𝑥²)^3/2 𝑥𝑥

1 0

= (1

2) 𝑥(3/2)𝑥(5/2) 𝑥((3/2) + (5/2)) = (¹

2)(1/2)(3/2)𝑥(1/2)𝑥(3/2) 𝑥(4)

= (¹

2)(3/4)(1/2)𝑥(1/2)𝑥(1/2) 3!

= (¹

2)(3/8)𝑥(1/2)𝑥(1/2) 3!

= (¹

2) (³

8) (¹

3!)𝑥(1/2)𝑥(1/2) = (¹

32)𝑥(1/2)𝑥(1/2) = (¹

32) (√𝑥)² =𝑥/32 Dapat disimpulkan bahwa,

∫ 𝑥²(1−𝑥²)^3/2 𝑥𝑥

1 0

= 𝑥/32

Dengan menggunakan mathematica nilai pendekatan dari 𝑥/32 adalah 0.098174770424.

b. ∫₀^𝑥/2√𝑥𝑥𝑥³𝑥 𝑥𝑥𝑥 𝑥 𝑥𝑥= ∫₀^𝑥/2(𝑥𝑥𝑥³𝑥 𝑥𝑥𝑥 𝑥)^1/2 𝑥𝑥 = ∫₀^𝑥/2(𝑥𝑥𝑥³𝑥)^1/2 (𝑥𝑥𝑥 𝑥)^1/2 𝑥𝑥 =¹

2(2∫₀^𝑥/2(𝑥𝑥𝑥 𝑥)^3/2 (𝑥𝑥𝑥 𝑥)^1/2 𝑥𝑥) 1

2𝑥(𝑥,𝑥) =1

2(2∫ (𝑥𝑥𝑥 𝑥)^2𝑥−1(𝑥𝑥𝑥 𝑥)^2𝑥−1 𝑥𝑥

𝑥/2 0

) 2𝑥−1 =3/2 ⟶ 𝑥= 5/4

2𝑥−1 =1/2 ⟶ 𝑥= 3/4

1

2

𝑥(𝑥, 𝑥) =

¹

2

𝑥(𝑥)𝑥(𝑥) 𝑥(𝑥 + 𝑥) =

¹

2

𝑥

(

5 4 , 3

4

⁾

1 2𝑥(5

4,3 4) =1

2

𝑥(5/4)𝑥(3/4) 𝑥((5/4) + (3/4))= 1

2

(1/4)𝑥(1/4)𝑥(3/4) 𝑥(2)

= (1 2) (1

4)𝑥(1/4)𝑥(3/4) = (1

8)𝑥(1/4)𝑥(3/4) Jadi,

∫ √𝑥𝑥𝑥³𝑥 𝑥𝑥𝑥 𝑥 𝑥𝑥

𝑥/2 0

= (1

8)𝑥(1/4)𝑥(3/4) 4.

a. Diperoleh grafik sebagai berikut :

b. from the graphs in problem 4a, state the number and location of the zeros number of zeros: 4 (disekitar 0.5, 8, 12, dan 14)

location of zeros : (disekitar 0.5, 8, 12, dan 14)

5.

a. Express 7𝑥⁴− 3𝑥 + 1 as linear combinations of Legendre polynomials. Hint:

Start with the highest power of 𝑥 and work down in finding the correct combination.

 Polinomial Legendre:

𝑃₀(𝑥) = 1 (1)

𝑃₁(𝑥) = 𝑥 (2)

𝑃₂(𝑥) =¹

2(3𝑥²− 1) (3)

𝑃₃(𝑥) =¹

2(5𝑥²− 3𝑥) (4)

𝑃₄(𝑥) = ¹

8(35𝑥⁴− 30𝑥²+ 3) (5)

 Misalkan ruang vektor (𝑉), karena polinomial berderajat 𝑛 ≤ 4. Maka basis standar dari 𝑉:

𝐴 = {1, 𝑥, 𝑥², 𝑥³, 𝑥⁴}

 Sehingga polinomial dapat dituliskan sebagai 𝑝(𝑥) = 𝑎₀+ 𝑎₁𝑥 + 𝑎₂𝑥 + 𝑎₃𝑥 + 𝑎₄𝑥 ∈ 𝑉 dan dikatakan sebagai coordinat relative vector terhadap 𝐴:

[𝑃(𝑥)]_𝐴= [𝑎₀ 𝑎₁ 𝑎₂ 𝑎₃ 𝑎₄]^𝑇

 Jadi, coordinat relative vector untuk setiap polynomial Legendre-nya:

[𝑃(𝑥)]_𝐴= [1, 0,0,0,0]^𝑇 [𝑃(𝑥)]_𝐴= [0, 1,0,0,0]^𝑇 [𝑃(𝑥)]_𝐴= [−1

2, 0,3 2, 0,0]^𝑇 [𝑃(𝑥)]_𝐴= [0, −3

2, 0,5 2, 0]^𝑇

[𝑃(𝑥)]_𝐴= [3

8, 0, −30 8 , 0,35

8]^𝑇

 Express 𝑃(𝑥) = 7𝑥⁴− 3𝑥 + 1 sebagai kombinasi linear. Maka koordinatif linear dari 𝑃(𝑥):

[𝑝(𝑥)]_𝐴= [1 − 3 0 0 7]^𝑇

 Mencari koefisien dari 𝑎₀, 𝑎₁, 𝑎₂, 𝑎₃, 𝑎₄∈ ℝ dengan rumus:

𝑝(𝑥) = 𝑎₀𝑝₀(𝑥) + 𝑎₁𝑝₁(𝑥) + 𝑎₂𝑝₂(𝑥) + 𝑎₃𝑝₃(𝑥) + 𝑎₄𝑝₄(𝑥) Untuk memudahkan perhitungan, digunakan perkalian matriks untuk

𝑝(𝑥) = 𝑎₀𝑝₀(𝑥) + 𝑎₁𝑝₁(𝑥) + 𝑎₂𝑝₂(𝑥) + 𝑎₃𝑝₃(𝑥) + 𝑎₄𝑝₄(𝑥)

[

1 0 −1

2 0 3

8 0 1 0 −3

2 0

0 0 3

2 0 −30 8

0 0 0 5

2 0

0 0 0 0 −35 8 ]

[ 𝑎₀ 𝑎₁ 𝑎₂ 𝑎₃ 𝑎₄]

= [

1

−3 0 0 7 ]

 Menggunakan rumus 𝐴𝑋 = 𝐵 → 𝐴⁻¹𝐵

𝑃(𝑥)⁻¹=

[

1 0 −1

2 0 3

8 0 1 0 −3

2 0

0 0 3

2 0 −30 8

0 0 0 5

2 0

0 0 0 0 −35 8

||

1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 ]

2 3𝐵3

[

1 0 −1

2 0 3

8 0 1 0 −3

2 0

0 0 3

2 0 −30 8

0 0 0 5

2 0

0 0 0 0 −35 8

|

|

1 0 0 0 0 0 1 0 0 0 0 0 2

3 0 0 0 0 0 1 0 0 0 0 0 1 ]

1 3𝐵3 + 𝐵1

[

1 0 0 0 −7 8 0 1 0 −3

2 0

0 0 1 0 −10 8 0 0 0 5

2 0

0 0 0 0 35 8

|

| 1 0 1

3 0 0 0 1 0 0 0 0 0 2

3 0 0 0 0 0 1 0 0 0 0 0 1 ]

2 5𝐵4

[

1 0 0 0 −7 8 0 1 0 −3

2 0

0 0 1 0 −10 8

0 0 0 1 0

0 0 0 0 35 8

|

|

| 1 0 1

3 0 0 0 1 0 0 0 0 0 2

3 0 0 0 0 0 2

5 0 0 0 0 0 1]

3 2𝐵4 + 𝐵2

[

1 0 0 0 −7 8 0 1 0 0 0 0 0 1 0 −10

4 0 0 0 1 0 0 0 0 0 35 8

|

|

| 1 0 1

3 0 0 0 1 0 3

5 0 0 0 2

3 0 0 0 0 0 2

5 0 0 0 0 0 1]

8 35𝐵5

[

1 0 0 0 −7 8 0 1 0 0 0 0 0 1 0 −10

4 0 0 0 1 0 0 0 0 0 1 |

|

| 1 0 1

3 0 0 0 1 0 3

5 0 0 0 2

3 0 0 0 0 0 2

5 0 0 0 0 0 8 35]

7 8𝐵5 + 𝐵1

[

1 0 0 0 0 0 1 0 0 0 0 0 1 0 −10

4 0 0 0 1 0 0 0 0 0 1

|

|

| 1 0 1

3 0 7 35 0 1 0 3

5 0 0 0 2

3 0 0 0 0 0 2

5 0 0 0 0 0 8 35]

10 4𝐵5 + 𝐵3

[

1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1

|

|

| 1 0 1

3 0 7 35 0 1 0 3

5 0 0 0 2

3 0 20 35 0 0 0 2

5 0 0 0 0 0 8 35]

→ 𝑝(𝑥)⁻¹=

[ 1 0 1

3 0 7 35 0 1 0 3

5 0 0 0 2

3 0 20 35 0 0 0 2

5 0 0 0 0 0 8 35]

 Solusinya adalah:

[ 𝑎₀ 𝑎₁ 𝑎₂ 𝑎₃ 𝑎₄]

=

[ 1 0 1

3 0 7 35 0 1 0 3

5 0 0 0 2

3 0 20 35 0 0 0 2

5 0 0 0 0 0 8 35]

[ 1

−3 0 0 7 ]

=

[ 12

−35 0 08 5 ]

 Jadi, ekspansi dari polinomial Legendre 𝑃(𝑥) = 7𝑥⁴− 3𝑥 + 1 adalah 7𝑥⁴− 3𝑥 + 1 = 12

5 𝑃₀(𝑥) − 3𝑃₁(𝑥) + 4𝑃₂(𝑥) +8

5𝑃₅(𝑥)

b. Find the best (in the least squares sense) second-degree polynomial approximation to 𝑥⁴ over the interval −1 < 𝑥 < 1.

 We want to find coefficients 𝑎₀, 𝑎₁, 𝑎₂ such that

𝑥⁴≈ 𝑎₀ 𝑝₀(𝑥) + 𝑎₁ 𝑝₁(𝑥) + 𝑎₂ 𝑝₂(𝑥)

 Where 𝑝_𝑛’s are normalized Legendre polynomials:

𝑝_𝑛(𝑥) = √2𝑛 + 1 2 𝑝_𝑛(𝑥) Then,

𝑎₀ = ∫ 𝑥⁴

1

−1

𝑝₀(𝑥)𝑑𝑥 = √1 2∫ 𝑥⁴

1

−1

𝑑𝑥 = 2

5√2=√2 5

𝑎₁= ∫ 𝑥⁴

1

−1

𝑝₁(𝑥)𝑑𝑥 = √3 2∫ 𝑥⁵

1

−1

𝑑𝑥 = 0

𝑎₂= ∫ 𝑥⁴

1

−1

𝑝₂(𝑥)𝑑𝑥 =1 2√5

2∫ 𝑥⁴(3𝑥²− 1)

1

−1

𝑑𝑥 = 0

=1 2√5

2∫ (3𝑥⁶− 𝑥⁴)

1

−1

𝑑𝑥 =1 2√5

2[3 7𝑥⁷−1

5𝑥⁵]

−1 1

=1 2√5

2 [6 7−2

5] =4 7√2

5

 Thus, we get the least-squares quadratic aproximation of 𝑥⁴:

 Figure

𝑥⁴≈√2

5 𝑝₀(𝑥) +4 7√2

5𝑝₂(𝑥)