Show that the Fourier coefficients of the triangular wave of Figure 16.7 are zero for n even and ( − 1 ) m 32 ( 2 m + 1 ) 2 π 2 ( 1 − cos ( 2 m + 1 ) π 8 ) for n odd.

Question

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Answer 1

Question

Chapter 16, Problem 16.10P

To determine

Show that the Fourier coefficients of the triangular wave of Figure 16.7 are zero for $n$ even and $(−1)m32(2m+1)2π2(1−cos(2m+1)π8)$ for $n$ odd.

Expert Solution & Answer

Answer to Problem 16.10P

Showed that the Fourier coefficients of the triangular wave of Figure 16.7 are zero for $n$ even and $(−1)m32(2m+1)2π2(1−cos(2m+1)π8)$ for $n$ odd.

Explanation of Solution

Consider the Figure shown below.

Classical Mechanics, Chapter 16, Problem 16.10P , additional homework tip 1

From the Figure, it is clear that, $u0(x)=0$ for $0<x<3&5<x<8$ , $u0(x)$ is increasing in the interval $x=3 to 4$ from $0 to1$ , and decreasing from $1 to 0$ in the interval $x=4 to 5$ .

Write the equation 16.33.

$Bn=2L∫0Lu0(x)sinnπxLdx$ (I)

Divide the integral into four parts, use $8$ for $L$ , and solve.

$Bn=28[∫03u0(x)sinnπx8dx+∫34u0(x)sinnπx8dx+∫45u0(x)sinnπx8dx+∫58u0(x)sinnπx8dx]=14[0+∫34u0(x)sinnπx8dx+∫45u0(x)sinnπx8dx+0]=14[∫34u0(x)sinnπx8dx+∫45u0(x)sinnπx8dx]$ (II)

Consider the enlarged view of triangle in the Figure 1 is shown below.

Classical Mechanics, Chapter 16, Problem 16.10P , additional homework tip 2

Let D is a point on straight line CA, the coordinate of this point is $(x,uo(x))$ , E is the foot of the perpendicular of D on CB, then the triangle CAH of CDE are similar triangles. So the ratio of corresponding sides of the two triangles are equal.

$DEAH=CECH$ (III)

From the Figure 2, use $u0(x)$ for $DE$ , $1$ for $AH$ , $x−3$ for $CE$ , and $1$ for $CH$ .

$u0(x)1=x−31u0(x)=x−3 for 3<x<4$ (IV)

Let F is a point on straight line AB, the coordinate of this point is $(x,uo(x))$ , G is the foot of the perpendicular of F on CB, then the triangle AHB of FGB are similar triangles. So the ratio of corresponding sides of the two triangles are equal.

$FGAH=GBHB$ (V)

From the Figure 2, use $u0(x)$ for $FG$ , $1$ for $AH$ , $5−x$ for $CE$ , and $1$ for $CH$ .

$u0(x)1=5−x1u0(x)=5−x for 4<x<5$ (VI)

Use equation (IV) and (VI) in the equation (II).

$Bn=14[∫34(x−3)sinnπx8dx+∫45(5−x)sinnπx8dx]$ (VII)

Integrate the first part of the equation (VII).

$∫34(x−3)sinnπx8dx=∫34xsinnπx8dx−3∫34sinnπx8dx={[x(−cosnπx8)nπ8]34−∫341⋅(−cosnπx8)nπ8}−3[(−cosnπx8)nπ8]34=−8nπ(4cos4nπ8−3cos3nπ8)+8nπ∫34cosnπx8dx+38nπ(cos4nπ8−cos3nπ8)$ (VIII)

Solve the equation (VIII).

$∫34(x−3)sinnπx8dx=−32nπcosnπ2+24nπcos3nπ8+8nπ(sin(nπx8)nπ8)34+24nπcosnπ2−24nπcos3nπ8=−8nπcosnπ2+(8nπ)2(sin4nπ8−sin3nπ8)=−8nπcosnπ2+(8nπ)2(sinnπ2−sin3nπ8)$ (IX)

Integrate the second part of the equation (VII).

$∫34(5−x)sinnπx8dx=∫455sinnπx8dx−∫45xsinnπx8dx=[(−cosnπx8)nπ8]45−{[x⋅(−cosnπx8)nπ8]45−∫451⋅(−cosnπx8)nπ8dx}=−40nπ(cos5nπ8−cos4nπ8)+8nπ(5cos5nπ8−4cos4nπ8)−8nπ∫45cosnπx8dx$ (X)

Solve the equation (X).

$∫34(5−x)sinnπx8dx=−40nπcos5nπ2+40nπcosnπ2+40nπcos5nπ8−32nπcosnπ2−8nπ(sin(nπx8)nπ8)45=8nπcosnπ2−(8nπ)2(sin5nπ8−sin4nπ8)=8nπcosnπ2−(8nπ)2(sin5nπ2−sinnπ2)$ (XI)

Use equation (X) and (XI) in (VII).

$Bn=14[−8nπcosnπ2+(8nπ)2(sinnπ2−sin3nπ8)+8nπcosnπ2−(8nπ)2(sin5nπ2−sinnπ2)]=14[(8nπ)2(sinnπ2−sin3nπ8−sin5nπ2+sinnπ2)]=16n2π2(2sinnπ2−sin3nπ8−sin5nπ2)=16n2π2(2sinnπ2−[sin3nπ8+sin5nπ2])$ (XII)

Solve equation (XII).

$Bn=16n2π2(2sinnπ2−[sin3nπ8+sin5nπ2])=16n2π2(2sinnπ2−2sin(5nπ2+3nπ82)cos(5nπ2−3nπ82))=16n2π2(2sinnπ2−2sinnπ2cosnπ8)=16n2π22sinnπ2(1−cosnπ8)=32n2π2sinnπ2(1−cosnπ8)$ (XIII)

If $n$ is even, $sinnπx2=0$ , and if $n$ is odd, $sinnπx2=(−1)m where m=0,1,2..$ . So equation (XIII) becomes zero for even $n$ . Then the equation (XIII) can be written as for odd $n$

$Bn=(−1)m32(2m+1)2π2[1−cos(2m+1)π8]$ .

Conclusion:

Therefore, the Fourier coefficients of the triangular wave of Figure 16.7 are zero for $n$ even and $(−1)m32(2m+1)2π2(1−cos(2m+1)π8)$ for $n$ odd.

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Answer 2

Question

Chapter 16, Problem 16.10P

To determine

Show that the Fourier coefficients of the triangular wave of Figure 16.7 are zero for $n$ even and $(−1)m32(2m+1)2π2(1−cos(2m+1)π8)$ for $n$ odd.

Expert Solution & Answer

Answer to Problem 16.10P

Showed that the Fourier coefficients of the triangular wave of Figure 16.7 are zero for $n$ even and $(−1)m32(2m+1)2π2(1−cos(2m+1)π8)$ for $n$ odd.

Explanation of Solution

Consider the Figure shown below.

Classical Mechanics, Chapter 16, Problem 16.10P , additional homework tip 1

From the Figure, it is clear that, $u0(x)=0$ for $0<x<3&5<x<8$ , $u0(x)$ is increasing in the interval $x=3 to 4$ from $0 to1$ , and decreasing from $1 to 0$ in the interval $x=4 to 5$ .

Write the equation 16.33.

$Bn=2L∫0Lu0(x)sinnπxLdx$ (I)

Divide the integral into four parts, use $8$ for $L$ , and solve.

$Bn=28[∫03u0(x)sinnπx8dx+∫34u0(x)sinnπx8dx+∫45u0(x)sinnπx8dx+∫58u0(x)sinnπx8dx]=14[0+∫34u0(x)sinnπx8dx+∫45u0(x)sinnπx8dx+0]=14[∫34u0(x)sinnπx8dx+∫45u0(x)sinnπx8dx]$ (II)

Consider the enlarged view of triangle in the Figure 1 is shown below.

Classical Mechanics, Chapter 16, Problem 16.10P , additional homework tip 2

Let D is a point on straight line CA, the coordinate of this point is $(x,uo(x))$ , E is the foot of the perpendicular of D on CB, then the triangle CAH of CDE are similar triangles. So the ratio of corresponding sides of the two triangles are equal.

$DEAH=CECH$ (III)

From the Figure 2, use $u0(x)$ for $DE$ , $1$ for $AH$ , $x−3$ for $CE$ , and $1$ for $CH$ .

$u0(x)1=x−31u0(x)=x−3 for 3<x<4$ (IV)

Let F is a point on straight line AB, the coordinate of this point is $(x,uo(x))$ , G is the foot of the perpendicular of F on CB, then the triangle AHB of FGB are similar triangles. So the ratio of corresponding sides of the two triangles are equal.

$FGAH=GBHB$ (V)

From the Figure 2, use $u0(x)$ for $FG$ , $1$ for $AH$ , $5−x$ for $CE$ , and $1$ for $CH$ .

$u0(x)1=5−x1u0(x)=5−x for 4<x<5$ (VI)

Use equation (IV) and (VI) in the equation (II).

$Bn=14[∫34(x−3)sinnπx8dx+∫45(5−x)sinnπx8dx]$ (VII)

Integrate the first part of the equation (VII).

$∫34(x−3)sinnπx8dx=∫34xsinnπx8dx−3∫34sinnπx8dx={[x(−cosnπx8)nπ8]34−∫341⋅(−cosnπx8)nπ8}−3[(−cosnπx8)nπ8]34=−8nπ(4cos4nπ8−3cos3nπ8)+8nπ∫34cosnπx8dx+38nπ(cos4nπ8−cos3nπ8)$ (VIII)

Solve the equation (VIII).

$∫34(x−3)sinnπx8dx=−32nπcosnπ2+24nπcos3nπ8+8nπ(sin(nπx8)nπ8)34+24nπcosnπ2−24nπcos3nπ8=−8nπcosnπ2+(8nπ)2(sin4nπ8−sin3nπ8)=−8nπcosnπ2+(8nπ)2(sinnπ2−sin3nπ8)$ (IX)

Integrate the second part of the equation (VII).

$∫34(5−x)sinnπx8dx=∫455sinnπx8dx−∫45xsinnπx8dx=[(−cosnπx8)nπ8]45−{[x⋅(−cosnπx8)nπ8]45−∫451⋅(−cosnπx8)nπ8dx}=−40nπ(cos5nπ8−cos4nπ8)+8nπ(5cos5nπ8−4cos4nπ8)−8nπ∫45cosnπx8dx$ (X)

Solve the equation (X).

$∫34(5−x)sinnπx8dx=−40nπcos5nπ2+40nπcosnπ2+40nπcos5nπ8−32nπcosnπ2−8nπ(sin(nπx8)nπ8)45=8nπcosnπ2−(8nπ)2(sin5nπ8−sin4nπ8)=8nπcosnπ2−(8nπ)2(sin5nπ2−sinnπ2)$ (XI)

Use equation (X) and (XI) in (VII).

$Bn=14[−8nπcosnπ2+(8nπ)2(sinnπ2−sin3nπ8)+8nπcosnπ2−(8nπ)2(sin5nπ2−sinnπ2)]=14[(8nπ)2(sinnπ2−sin3nπ8−sin5nπ2+sinnπ2)]=16n2π2(2sinnπ2−sin3nπ8−sin5nπ2)=16n2π2(2sinnπ2−[sin3nπ8+sin5nπ2])$ (XII)

Solve equation (XII).

$Bn=16n2π2(2sinnπ2−[sin3nπ8+sin5nπ2])=16n2π2(2sinnπ2−2sin(5nπ2+3nπ82)cos(5nπ2−3nπ82))=16n2π2(2sinnπ2−2sinnπ2cosnπ8)=16n2π22sinnπ2(1−cosnπ8)=32n2π2sinnπ2(1−cosnπ8)$ (XIII)

If $n$ is even, $sinnπx2=0$ , and if $n$ is odd, $sinnπx2=(−1)m where m=0,1,2..$ . So equation (XIII) becomes zero for even $n$ . Then the equation (XIII) can be written as for odd $n$

$Bn=(−1)m32(2m+1)2π2[1−cos(2m+1)π8]$ .

Conclusion:

Therefore, the Fourier coefficients of the triangular wave of Figure 16.7 are zero for $n$ even and $(−1)m32(2m+1)2π2(1−cos(2m+1)π8)$ for $n$ odd.

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Answer 3

Question

Chapter 16, Problem 16.10P

To determine

Show that the Fourier coefficients of the triangular wave of Figure 16.7 are zero for $n$ even and $(−1)m32(2m+1)2π2(1−cos(2m+1)π8)$ for $n$ odd.

Expert Solution & Answer

Answer to Problem 16.10P

Showed that the Fourier coefficients of the triangular wave of Figure 16.7 are zero for $n$ even and $(−1)m32(2m+1)2π2(1−cos(2m+1)π8)$ for $n$ odd.

Explanation of Solution

Consider the Figure shown below.

Classical Mechanics, Chapter 16, Problem 16.10P , additional homework tip 1

From the Figure, it is clear that, $u0(x)=0$ for $0<x<3&5<x<8$ , $u0(x)$ is increasing in the interval $x=3 to 4$ from $0 to1$ , and decreasing from $1 to 0$ in the interval $x=4 to 5$ .

Write the equation 16.33.

$Bn=2L∫0Lu0(x)sinnπxLdx$ (I)

Divide the integral into four parts, use $8$ for $L$ , and solve.

$Bn=28[∫03u0(x)sinnπx8dx+∫34u0(x)sinnπx8dx+∫45u0(x)sinnπx8dx+∫58u0(x)sinnπx8dx]=14[0+∫34u0(x)sinnπx8dx+∫45u0(x)sinnπx8dx+0]=14[∫34u0(x)sinnπx8dx+∫45u0(x)sinnπx8dx]$ (II)

Consider the enlarged view of triangle in the Figure 1 is shown below.

Classical Mechanics, Chapter 16, Problem 16.10P , additional homework tip 2

Let D is a point on straight line CA, the coordinate of this point is $(x,uo(x))$ , E is the foot of the perpendicular of D on CB, then the triangle CAH of CDE are similar triangles. So the ratio of corresponding sides of the two triangles are equal.

$DEAH=CECH$ (III)

From the Figure 2, use $u0(x)$ for $DE$ , $1$ for $AH$ , $x−3$ for $CE$ , and $1$ for $CH$ .

$u0(x)1=x−31u0(x)=x−3 for 3<x<4$ (IV)

Let F is a point on straight line AB, the coordinate of this point is $(x,uo(x))$ , G is the foot of the perpendicular of F on CB, then the triangle AHB of FGB are similar triangles. So the ratio of corresponding sides of the two triangles are equal.

$FGAH=GBHB$ (V)

From the Figure 2, use $u0(x)$ for $FG$ , $1$ for $AH$ , $5−x$ for $CE$ , and $1$ for $CH$ .

$u0(x)1=5−x1u0(x)=5−x for 4<x<5$ (VI)

Use equation (IV) and (VI) in the equation (II).

$Bn=14[∫34(x−3)sinnπx8dx+∫45(5−x)sinnπx8dx]$ (VII)

Integrate the first part of the equation (VII).

$∫34(x−3)sinnπx8dx=∫34xsinnπx8dx−3∫34sinnπx8dx={[x(−cosnπx8)nπ8]34−∫341⋅(−cosnπx8)nπ8}−3[(−cosnπx8)nπ8]34=−8nπ(4cos4nπ8−3cos3nπ8)+8nπ∫34cosnπx8dx+38nπ(cos4nπ8−cos3nπ8)$ (VIII)

Solve the equation (VIII).

$∫34(x−3)sinnπx8dx=−32nπcosnπ2+24nπcos3nπ8+8nπ(sin(nπx8)nπ8)34+24nπcosnπ2−24nπcos3nπ8=−8nπcosnπ2+(8nπ)2(sin4nπ8−sin3nπ8)=−8nπcosnπ2+(8nπ)2(sinnπ2−sin3nπ8)$ (IX)

Integrate the second part of the equation (VII).

$∫34(5−x)sinnπx8dx=∫455sinnπx8dx−∫45xsinnπx8dx=[(−cosnπx8)nπ8]45−{[x⋅(−cosnπx8)nπ8]45−∫451⋅(−cosnπx8)nπ8dx}=−40nπ(cos5nπ8−cos4nπ8)+8nπ(5cos5nπ8−4cos4nπ8)−8nπ∫45cosnπx8dx$ (X)

Solve the equation (X).

$∫34(5−x)sinnπx8dx=−40nπcos5nπ2+40nπcosnπ2+40nπcos5nπ8−32nπcosnπ2−8nπ(sin(nπx8)nπ8)45=8nπcosnπ2−(8nπ)2(sin5nπ8−sin4nπ8)=8nπcosnπ2−(8nπ)2(sin5nπ2−sinnπ2)$ (XI)

Use equation (X) and (XI) in (VII).

$Bn=14[−8nπcosnπ2+(8nπ)2(sinnπ2−sin3nπ8)+8nπcosnπ2−(8nπ)2(sin5nπ2−sinnπ2)]=14[(8nπ)2(sinnπ2−sin3nπ8−sin5nπ2+sinnπ2)]=16n2π2(2sinnπ2−sin3nπ8−sin5nπ2)=16n2π2(2sinnπ2−[sin3nπ8+sin5nπ2])$ (XII)

Solve equation (XII).

$Bn=16n2π2(2sinnπ2−[sin3nπ8+sin5nπ2])=16n2π2(2sinnπ2−2sin(5nπ2+3nπ82)cos(5nπ2−3nπ82))=16n2π2(2sinnπ2−2sinnπ2cosnπ8)=16n2π22sinnπ2(1−cosnπ8)=32n2π2sinnπ2(1−cosnπ8)$ (XIII)

If $n$ is even, $sinnπx2=0$ , and if $n$ is odd, $sinnπx2=(−1)m where m=0,1,2..$ . So equation (XIII) becomes zero for even $n$ . Then the equation (XIII) can be written as for odd $n$

$Bn=(−1)m32(2m+1)2π2[1−cos(2m+1)π8]$ .

Conclusion:

Therefore, the Fourier coefficients of the triangular wave of Figure 16.7 are zero for $n$ even and $(−1)m32(2m+1)2π2(1−cos(2m+1)π8)$ for $n$ odd.

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Answer 4

Question

Chapter 16, Problem 16.10P

To determine

Show that the Fourier coefficients of the triangular wave of Figure 16.7 are zero for $n$ even and $(−1)m32(2m+1)2π2(1−cos(2m+1)π8)$ for $n$ odd.

Expert Solution & Answer

Answer to Problem 16.10P

Showed that the Fourier coefficients of the triangular wave of Figure 16.7 are zero for $n$ even and $(−1)m32(2m+1)2π2(1−cos(2m+1)π8)$ for $n$ odd.

Explanation of Solution

Consider the Figure shown below.

Classical Mechanics, Chapter 16, Problem 16.10P , additional homework tip 1

From the Figure, it is clear that, $u0(x)=0$ for $0<x<3&5<x<8$ , $u0(x)$ is increasing in the interval $x=3 to 4$ from $0 to1$ , and decreasing from $1 to 0$ in the interval $x=4 to 5$ .

Write the equation 16.33.

$Bn=2L∫0Lu0(x)sinnπxLdx$ (I)

Divide the integral into four parts, use $8$ for $L$ , and solve.

$Bn=28[∫03u0(x)sinnπx8dx+∫34u0(x)sinnπx8dx+∫45u0(x)sinnπx8dx+∫58u0(x)sinnπx8dx]=14[0+∫34u0(x)sinnπx8dx+∫45u0(x)sinnπx8dx+0]=14[∫34u0(x)sinnπx8dx+∫45u0(x)sinnπx8dx]$ (II)

Consider the enlarged view of triangle in the Figure 1 is shown below.

Classical Mechanics, Chapter 16, Problem 16.10P , additional homework tip 2

Let D is a point on straight line CA, the coordinate of this point is $(x,uo(x))$ , E is the foot of the perpendicular of D on CB, then the triangle CAH of CDE are similar triangles. So the ratio of corresponding sides of the two triangles are equal.

$DEAH=CECH$ (III)

From the Figure 2, use $u0(x)$ for $DE$ , $1$ for $AH$ , $x−3$ for $CE$ , and $1$ for $CH$ .

$u0(x)1=x−31u0(x)=x−3 for 3<x<4$ (IV)

Let F is a point on straight line AB, the coordinate of this point is $(x,uo(x))$ , G is the foot of the perpendicular of F on CB, then the triangle AHB of FGB are similar triangles. So the ratio of corresponding sides of the two triangles are equal.

$FGAH=GBHB$ (V)

From the Figure 2, use $u0(x)$ for $FG$ , $1$ for $AH$ , $5−x$ for $CE$ , and $1$ for $CH$ .

$u0(x)1=5−x1u0(x)=5−x for 4<x<5$ (VI)

Use equation (IV) and (VI) in the equation (II).

$Bn=14[∫34(x−3)sinnπx8dx+∫45(5−x)sinnπx8dx]$ (VII)

Integrate the first part of the equation (VII).

$∫34(x−3)sinnπx8dx=∫34xsinnπx8dx−3∫34sinnπx8dx={[x(−cosnπx8)nπ8]34−∫341⋅(−cosnπx8)nπ8}−3[(−cosnπx8)nπ8]34=−8nπ(4cos4nπ8−3cos3nπ8)+8nπ∫34cosnπx8dx+38nπ(cos4nπ8−cos3nπ8)$ (VIII)

Solve the equation (VIII).

$∫34(x−3)sinnπx8dx=−32nπcosnπ2+24nπcos3nπ8+8nπ(sin(nπx8)nπ8)34+24nπcosnπ2−24nπcos3nπ8=−8nπcosnπ2+(8nπ)2(sin4nπ8−sin3nπ8)=−8nπcosnπ2+(8nπ)2(sinnπ2−sin3nπ8)$ (IX)

Integrate the second part of the equation (VII).

$∫34(5−x)sinnπx8dx=∫455sinnπx8dx−∫45xsinnπx8dx=[(−cosnπx8)nπ8]45−{[x⋅(−cosnπx8)nπ8]45−∫451⋅(−cosnπx8)nπ8dx}=−40nπ(cos5nπ8−cos4nπ8)+8nπ(5cos5nπ8−4cos4nπ8)−8nπ∫45cosnπx8dx$ (X)

Solve the equation (X).

$∫34(5−x)sinnπx8dx=−40nπcos5nπ2+40nπcosnπ2+40nπcos5nπ8−32nπcosnπ2−8nπ(sin(nπx8)nπ8)45=8nπcosnπ2−(8nπ)2(sin5nπ8−sin4nπ8)=8nπcosnπ2−(8nπ)2(sin5nπ2−sinnπ2)$ (XI)

Use equation (X) and (XI) in (VII).

$Bn=14[−8nπcosnπ2+(8nπ)2(sinnπ2−sin3nπ8)+8nπcosnπ2−(8nπ)2(sin5nπ2−sinnπ2)]=14[(8nπ)2(sinnπ2−sin3nπ8−sin5nπ2+sinnπ2)]=16n2π2(2sinnπ2−sin3nπ8−sin5nπ2)=16n2π2(2sinnπ2−[sin3nπ8+sin5nπ2])$ (XII)

Solve equation (XII).

$Bn=16n2π2(2sinnπ2−[sin3nπ8+sin5nπ2])=16n2π2(2sinnπ2−2sin(5nπ2+3nπ82)cos(5nπ2−3nπ82))=16n2π2(2sinnπ2−2sinnπ2cosnπ8)=16n2π22sinnπ2(1−cosnπ8)=32n2π2sinnπ2(1−cosnπ8)$ (XIII)

If $n$ is even, $sinnπx2=0$ , and if $n$ is odd, $sinnπx2=(−1)m where m=0,1,2..$ . So equation (XIII) becomes zero for even $n$ . Then the equation (XIII) can be written as for odd $n$

$Bn=(−1)m32(2m+1)2π2[1−cos(2m+1)π8]$ .

Conclusion:

Therefore, the Fourier coefficients of the triangular wave of Figure 16.7 are zero for $n$ even and $(−1)m32(2m+1)2π2(1−cos(2m+1)π8)$ for $n$ odd.

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Answer 5

Chapter 16, Problem 16.10P

To determine

Show that the Fourier coefficients of the triangular wave of Figure 16.7 are zero for $n$ even and $(−1)m32(2m+1)2π2(1−cos(2m+1)π8)$ for $n$ odd.

Expert Solution & Answer

Answer to Problem 16.10P

Showed that the Fourier coefficients of the triangular wave of Figure 16.7 are zero for $n$ even and $(−1)m32(2m+1)2π2(1−cos(2m+1)π8)$ for $n$ odd.

Explanation of Solution

Consider the Figure shown below.

Classical Mechanics, Chapter 16, Problem 16.10P , additional homework tip 1

From the Figure, it is clear that, $u0(x)=0$ for $0<x<3&5<x<8$ , $u0(x)$ is increasing in the interval $x=3 to 4$ from $0 to1$ , and decreasing from $1 to 0$ in the interval $x=4 to 5$ .

Write the equation 16.33.

$Bn=2L∫0Lu0(x)sinnπxLdx$ (I)

Divide the integral into four parts, use $8$ for $L$ , and solve.

$Bn=28[∫03u0(x)sinnπx8dx+∫34u0(x)sinnπx8dx+∫45u0(x)sinnπx8dx+∫58u0(x)sinnπx8dx]=14[0+∫34u0(x)sinnπx8dx+∫45u0(x)sinnπx8dx+0]=14[∫34u0(x)sinnπx8dx+∫45u0(x)sinnπx8dx]$ (II)

Consider the enlarged view of triangle in the Figure 1 is shown below.

Classical Mechanics, Chapter 16, Problem 16.10P , additional homework tip 2

Let D is a point on straight line CA, the coordinate of this point is $(x,uo(x))$ , E is the foot of the perpendicular of D on CB, then the triangle CAH of CDE are similar triangles. So the ratio of corresponding sides of the two triangles are equal.

$DEAH=CECH$ (III)

From the Figure 2, use $u0(x)$ for $DE$ , $1$ for $AH$ , $x−3$ for $CE$ , and $1$ for $CH$ .

$u0(x)1=x−31u0(x)=x−3 for 3<x<4$ (IV)

Let F is a point on straight line AB, the coordinate of this point is $(x,uo(x))$ , G is the foot of the perpendicular of F on CB, then the triangle AHB of FGB are similar triangles. So the ratio of corresponding sides of the two triangles are equal.

$FGAH=GBHB$ (V)

From the Figure 2, use $u0(x)$ for $FG$ , $1$ for $AH$ , $5−x$ for $CE$ , and $1$ for $CH$ .

$u0(x)1=5−x1u0(x)=5−x for 4<x<5$ (VI)

Use equation (IV) and (VI) in the equation (II).

$Bn=14[∫34(x−3)sinnπx8dx+∫45(5−x)sinnπx8dx]$ (VII)

Integrate the first part of the equation (VII).

$∫34(x−3)sinnπx8dx=∫34xsinnπx8dx−3∫34sinnπx8dx={[x(−cosnπx8)nπ8]34−∫341⋅(−cosnπx8)nπ8}−3[(−cosnπx8)nπ8]34=−8nπ(4cos4nπ8−3cos3nπ8)+8nπ∫34cosnπx8dx+38nπ(cos4nπ8−cos3nπ8)$ (VIII)

Solve the equation (VIII).

$∫34(x−3)sinnπx8dx=−32nπcosnπ2+24nπcos3nπ8+8nπ(sin(nπx8)nπ8)34+24nπcosnπ2−24nπcos3nπ8=−8nπcosnπ2+(8nπ)2(sin4nπ8−sin3nπ8)=−8nπcosnπ2+(8nπ)2(sinnπ2−sin3nπ8)$ (IX)

Integrate the second part of the equation (VII).

$∫34(5−x)sinnπx8dx=∫455sinnπx8dx−∫45xsinnπx8dx=[(−cosnπx8)nπ8]45−{[x⋅(−cosnπx8)nπ8]45−∫451⋅(−cosnπx8)nπ8dx}=−40nπ(cos5nπ8−cos4nπ8)+8nπ(5cos5nπ8−4cos4nπ8)−8nπ∫45cosnπx8dx$ (X)

Solve the equation (X).

$∫34(5−x)sinnπx8dx=−40nπcos5nπ2+40nπcosnπ2+40nπcos5nπ8−32nπcosnπ2−8nπ(sin(nπx8)nπ8)45=8nπcosnπ2−(8nπ)2(sin5nπ8−sin4nπ8)=8nπcosnπ2−(8nπ)2(sin5nπ2−sinnπ2)$ (XI)

Use equation (X) and (XI) in (VII).

$Bn=14[−8nπcosnπ2+(8nπ)2(sinnπ2−sin3nπ8)+8nπcosnπ2−(8nπ)2(sin5nπ2−sinnπ2)]=14[(8nπ)2(sinnπ2−sin3nπ8−sin5nπ2+sinnπ2)]=16n2π2(2sinnπ2−sin3nπ8−sin5nπ2)=16n2π2(2sinnπ2−[sin3nπ8+sin5nπ2])$ (XII)

Solve equation (XII).

$Bn=16n2π2(2sinnπ2−[sin3nπ8+sin5nπ2])=16n2π2(2sinnπ2−2sin(5nπ2+3nπ82)cos(5nπ2−3nπ82))=16n2π2(2sinnπ2−2sinnπ2cosnπ8)=16n2π22sinnπ2(1−cosnπ8)=32n2π2sinnπ2(1−cosnπ8)$ (XIII)

If $n$ is even, $sinnπx2=0$ , and if $n$ is odd, $sinnπx2=(−1)m where m=0,1,2..$ . So equation (XIII) becomes zero for even $n$ . Then the equation (XIII) can be written as for odd $n$

$Bn=(−1)m32(2m+1)2π2[1−cos(2m+1)π8]$ .

Conclusion:

Therefore, the Fourier coefficients of the triangular wave of Figure 16.7 are zero for $n$ even and $(−1)m32(2m+1)2π2(1−cos(2m+1)π8)$ for $n$ odd.

Answer 6

Chapter 16, Problem 16.10P

To determine

Show that the Fourier coefficients of the triangular wave of Figure 16.7 are zero for $n$ even and $(−1)m32(2m+1)2π2(1−cos(2m+1)π8)$ for $n$ odd.

Expert Solution & Answer

Answer to Problem 16.10P

Showed that the Fourier coefficients of the triangular wave of Figure 16.7 are zero for $n$ even and $(−1)m32(2m+1)2π2(1−cos(2m+1)π8)$ for $n$ odd.

Explanation of Solution

Consider the Figure shown below.

Classical Mechanics, Chapter 16, Problem 16.10P , additional homework tip 1

From the Figure, it is clear that, $u0(x)=0$ for $0<x<3&5<x<8$ , $u0(x)$ is increasing in the interval $x=3 to 4$ from $0 to1$ , and decreasing from $1 to 0$ in the interval $x=4 to 5$ .

Write the equation 16.33.

$Bn=2L∫0Lu0(x)sinnπxLdx$ (I)

Divide the integral into four parts, use $8$ for $L$ , and solve.

$Bn=28[∫03u0(x)sinnπx8dx+∫34u0(x)sinnπx8dx+∫45u0(x)sinnπx8dx+∫58u0(x)sinnπx8dx]=14[0+∫34u0(x)sinnπx8dx+∫45u0(x)sinnπx8dx+0]=14[∫34u0(x)sinnπx8dx+∫45u0(x)sinnπx8dx]$ (II)

Consider the enlarged view of triangle in the Figure 1 is shown below.

Classical Mechanics, Chapter 16, Problem 16.10P , additional homework tip 2

Let D is a point on straight line CA, the coordinate of this point is $(x,uo(x))$ , E is the foot of the perpendicular of D on CB, then the triangle CAH of CDE are similar triangles. So the ratio of corresponding sides of the two triangles are equal.

$DEAH=CECH$ (III)

From the Figure 2, use $u0(x)$ for $DE$ , $1$ for $AH$ , $x−3$ for $CE$ , and $1$ for $CH$ .

$u0(x)1=x−31u0(x)=x−3 for 3<x<4$ (IV)

Let F is a point on straight line AB, the coordinate of this point is $(x,uo(x))$ , G is the foot of the perpendicular of F on CB, then the triangle AHB of FGB are similar triangles. So the ratio of corresponding sides of the two triangles are equal.

$FGAH=GBHB$ (V)

From the Figure 2, use $u0(x)$ for $FG$ , $1$ for $AH$ , $5−x$ for $CE$ , and $1$ for $CH$ .

$u0(x)1=5−x1u0(x)=5−x for 4<x<5$ (VI)

Use equation (IV) and (VI) in the equation (II).

$Bn=14[∫34(x−3)sinnπx8dx+∫45(5−x)sinnπx8dx]$ (VII)

Integrate the first part of the equation (VII).

$∫34(x−3)sinnπx8dx=∫34xsinnπx8dx−3∫34sinnπx8dx={[x(−cosnπx8)nπ8]34−∫341⋅(−cosnπx8)nπ8}−3[(−cosnπx8)nπ8]34=−8nπ(4cos4nπ8−3cos3nπ8)+8nπ∫34cosnπx8dx+38nπ(cos4nπ8−cos3nπ8)$ (VIII)

Solve the equation (VIII).

$∫34(x−3)sinnπx8dx=−32nπcosnπ2+24nπcos3nπ8+8nπ(sin(nπx8)nπ8)34+24nπcosnπ2−24nπcos3nπ8=−8nπcosnπ2+(8nπ)2(sin4nπ8−sin3nπ8)=−8nπcosnπ2+(8nπ)2(sinnπ2−sin3nπ8)$ (IX)

Integrate the second part of the equation (VII).

$∫34(5−x)sinnπx8dx=∫455sinnπx8dx−∫45xsinnπx8dx=[(−cosnπx8)nπ8]45−{[x⋅(−cosnπx8)nπ8]45−∫451⋅(−cosnπx8)nπ8dx}=−40nπ(cos5nπ8−cos4nπ8)+8nπ(5cos5nπ8−4cos4nπ8)−8nπ∫45cosnπx8dx$ (X)

Solve the equation (X).

$∫34(5−x)sinnπx8dx=−40nπcos5nπ2+40nπcosnπ2+40nπcos5nπ8−32nπcosnπ2−8nπ(sin(nπx8)nπ8)45=8nπcosnπ2−(8nπ)2(sin5nπ8−sin4nπ8)=8nπcosnπ2−(8nπ)2(sin5nπ2−sinnπ2)$ (XI)

Use equation (X) and (XI) in (VII).

$Bn=14[−8nπcosnπ2+(8nπ)2(sinnπ2−sin3nπ8)+8nπcosnπ2−(8nπ)2(sin5nπ2−sinnπ2)]=14[(8nπ)2(sinnπ2−sin3nπ8−sin5nπ2+sinnπ2)]=16n2π2(2sinnπ2−sin3nπ8−sin5nπ2)=16n2π2(2sinnπ2−[sin3nπ8+sin5nπ2])$ (XII)

Solve equation (XII).

$Bn=16n2π2(2sinnπ2−[sin3nπ8+sin5nπ2])=16n2π2(2sinnπ2−2sin(5nπ2+3nπ82)cos(5nπ2−3nπ82))=16n2π2(2sinnπ2−2sinnπ2cosnπ8)=16n2π22sinnπ2(1−cosnπ8)=32n2π2sinnπ2(1−cosnπ8)$ (XIII)

If $n$ is even, $sinnπx2=0$ , and if $n$ is odd, $sinnπx2=(−1)m where m=0,1,2..$ . So equation (XIII) becomes zero for even $n$ . Then the equation (XIII) can be written as for odd $n$

$Bn=(−1)m32(2m+1)2π2[1−cos(2m+1)π8]$ .

Conclusion:

Therefore, the Fourier coefficients of the triangular wave of Figure 16.7 are zero for $n$ even and $(−1)m32(2m+1)2π2(1−cos(2m+1)π8)$ for $n$ odd.

Answer 7

Chapter 16, Problem 16.10P

To determine

Show that the Fourier coefficients of the triangular wave of Figure 16.7 are zero for $n$ even and $(−1)m32(2m+1)2π2(1−cos(2m+1)π8)$ for $n$ odd.

Answer 8

Expert Solution & Answer

Answer to Problem 16.10P

Showed that the Fourier coefficients of the triangular wave of Figure 16.7 are zero for $n$ even and $(−1)m32(2m+1)2π2(1−cos(2m+1)π8)$ for $n$ odd.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Consider the Figure shown below.

Classical Mechanics, Chapter 16, Problem 16.10P , additional homework tip 1

From the Figure, it is clear that, $u0(x)=0$ for $0<x<3&5<x<8$ , $u0(x)$ is increasing in the interval $x=3 to 4$ from $0 to1$ , and decreasing from $1 to 0$ in the interval $x=4 to 5$ .

Write the equation 16.33.

$Bn=2L∫0Lu0(x)sinnπxLdx$ (I)

Divide the integral into four parts, use $8$ for $L$ , and solve.

$Bn=28[∫03u0(x)sinnπx8dx+∫34u0(x)sinnπx8dx+∫45u0(x)sinnπx8dx+∫58u0(x)sinnπx8dx]=14[0+∫34u0(x)sinnπx8dx+∫45u0(x)sinnπx8dx+0]=14[∫34u0(x)sinnπx8dx+∫45u0(x)sinnπx8dx]$ (II)

Consider the enlarged view of triangle in the Figure 1 is shown below.

Classical Mechanics, Chapter 16, Problem 16.10P , additional homework tip 2

Let D is a point on straight line CA, the coordinate of this point is $(x,uo(x))$ , E is the foot of the perpendicular of D on CB, then the triangle CAH of CDE are similar triangles. So the ratio of corresponding sides of the two triangles are equal.

$DEAH=CECH$ (III)

From the Figure 2, use $u0(x)$ for $DE$ , $1$ for $AH$ , $x−3$ for $CE$ , and $1$ for $CH$ .

$u0(x)1=x−31u0(x)=x−3 for 3<x<4$ (IV)

Let F is a point on straight line AB, the coordinate of this point is $(x,uo(x))$ , G is the foot of the perpendicular of F on CB, then the triangle AHB of FGB are similar triangles. So the ratio of corresponding sides of the two triangles are equal.

$FGAH=GBHB$ (V)

From the Figure 2, use $u0(x)$ for $FG$ , $1$ for $AH$ , $5−x$ for $CE$ , and $1$ for $CH$ .

$u0(x)1=5−x1u0(x)=5−x for 4<x<5$ (VI)

Use equation (IV) and (VI) in the equation (II).

$Bn=14[∫34(x−3)sinnπx8dx+∫45(5−x)sinnπx8dx]$ (VII)

Integrate the first part of the equation (VII).

$∫34(x−3)sinnπx8dx=∫34xsinnπx8dx−3∫34sinnπx8dx={[x(−cosnπx8)nπ8]34−∫341⋅(−cosnπx8)nπ8}−3[(−cosnπx8)nπ8]34=−8nπ(4cos4nπ8−3cos3nπ8)+8nπ∫34cosnπx8dx+38nπ(cos4nπ8−cos3nπ8)$ (VIII)

Solve the equation (VIII).

$∫34(x−3)sinnπx8dx=−32nπcosnπ2+24nπcos3nπ8+8nπ(sin(nπx8)nπ8)34+24nπcosnπ2−24nπcos3nπ8=−8nπcosnπ2+(8nπ)2(sin4nπ8−sin3nπ8)=−8nπcosnπ2+(8nπ)2(sinnπ2−sin3nπ8)$ (IX)

Integrate the second part of the equation (VII).

$∫34(5−x)sinnπx8dx=∫455sinnπx8dx−∫45xsinnπx8dx=[(−cosnπx8)nπ8]45−{[x⋅(−cosnπx8)nπ8]45−∫451⋅(−cosnπx8)nπ8dx}=−40nπ(cos5nπ8−cos4nπ8)+8nπ(5cos5nπ8−4cos4nπ8)−8nπ∫45cosnπx8dx$ (X)

Solve the equation (X).

$∫34(5−x)sinnπx8dx=−40nπcos5nπ2+40nπcosnπ2+40nπcos5nπ8−32nπcosnπ2−8nπ(sin(nπx8)nπ8)45=8nπcosnπ2−(8nπ)2(sin5nπ8−sin4nπ8)=8nπcosnπ2−(8nπ)2(sin5nπ2−sinnπ2)$ (XI)

Use equation (X) and (XI) in (VII).

$Bn=14[−8nπcosnπ2+(8nπ)2(sinnπ2−sin3nπ8)+8nπcosnπ2−(8nπ)2(sin5nπ2−sinnπ2)]=14[(8nπ)2(sinnπ2−sin3nπ8−sin5nπ2+sinnπ2)]=16n2π2(2sinnπ2−sin3nπ8−sin5nπ2)=16n2π2(2sinnπ2−[sin3nπ8+sin5nπ2])$ (XII)

Solve equation (XII).

$Bn=16n2π2(2sinnπ2−[sin3nπ8+sin5nπ2])=16n2π2(2sinnπ2−2sin(5nπ2+3nπ82)cos(5nπ2−3nπ82))=16n2π2(2sinnπ2−2sinnπ2cosnπ8)=16n2π22sinnπ2(1−cosnπ8)=32n2π2sinnπ2(1−cosnπ8)$ (XIII)

If $n$ is even, $sinnπx2=0$ , and if $n$ is odd, $sinnπx2=(−1)m where m=0,1,2..$ . So equation (XIII) becomes zero for even $n$ . Then the equation (XIII) can be written as for odd $n$