For the functional equation g(x) = αg 2 ( x α ) , arise in renormalization analysis of period doubling. If g ( x ) ≈ 1 + c 2 x 2 for small x, then solve for c 2 and α For g ( x ) ≈ 1 + c 2 x 2 + c 4 x 4 , solve equation for c 2 , c 4 and α , and compare the results with exact values c 2 ≈ -1 .527 ... , c 4 ≈ 0 .1048 ... and α ≈ -2 .5029 ... Concept Introduction: The power series expansion of the g ( x α ) is g ( x α ) = 1+c 2 ( x α ) 2 + c 4 ( x α ) 4 + ...

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Answer 1

Question

Chapter 10.7, Problem 1E

Interpretation Introduction

Interpretation:

For the functional equation $g(x) = αg2(xα)$ , arise in renormalization analysis of period doubling.

If $g(x)≈1+c2x2$ for small x, then solve for $c2 and α$

For $g(x)≈1+c2x2+c4x4$ , solve equation for $c2, c4 and α$ , and compare the results with exact values $c2≈-1.527..., c4≈0.1048... and α≈-2.5029...$

Concept Introduction:

The power series expansion of the $g(xα)$ is

$g(xα) = 1+c2(xα)2+c4(xα)4+...$

Expert Solution & Answer

Answer to Problem 1E

Solution:

$α=−2.73205 and c2=−1.36603$
$α=−2.534... c2=−1.52224... c4=0.12761...$ and these are compared with exact values of $α,c2 and c4$

Explanation of Solution

The function equation arose in renormalization equation is,

$g(x) = αg2(xa)$

The power series expansion of the above equation is

$g(xα) = 1+c2(xα)2+c4(xα)4+...$

$g2(xα)=g(g(xα))=g(1+c2(xα)2)=1+c2+2c22(xα)2+O(x4)$

$∴αg2(xa)=α(1+c2+2c22(xα)2+O(x4))=α(1+c2)+2c22x2α+αO(x4)$

The given approximation is

$g(x)=1+c2x2$

Comparing this with $αg2(xa)$

$1 + c2x2 = α(1+c2)+2c22x2α + O(x3)$

Comparing the coefficients of $x0 and x2$

$1= α(1+c2)$

$c2=2c22α$

Solving above two equations

$⇒(α,c2)={(1,0)(−1−3,−1−32)≈(−2.73205,−1.36603)(−1+3,−1+32)≈(0.732051,0.366025)$

Here, take the second solution because in that solution $α$ is closer to Feigenbaum $α=−2.5...$
$g(x) = αg2(xa)=α(1+c2+c4+2(c22+2c2c4)(xα)2+(c23+6c22c4+2c2c4+c42)(xα)4+O(x5))$

The given approximation is

$g(x)=1+c2x2+c4x4$

Equating this with $αg2(xa)$ $1+c2x2+c4x4=α(1+c2+c4)+2(c22+2c2c4)x2α+(c23+6c22c4+2c2c4+c42)x4α3+O(x5)$

Comparing the coefficients of $x0,x2and x4$ , we get

$1= α(1+c2+c4)$

$c2=2(c22+2c2c4)α$

$c4=(c23+6c22c4+2c2c4+c42)α3$

Using first and second equations for solving $c2 and c4$ in terms of $α$

$c2=−2+2α−α2$

$c4=1−1α+α2$

Substituting these values of $c2 and c4$ in third equation, then it becomes

$4α6+3α5−60α4−104α3+168α2+240α−384+128α=0$

The root of above equation closest to Feigenbaum is $α=−2.534...$

Substituting this value of $α$ in $c2=−2+2α−α2$ and $c4=1−1α+α2$

$α=−2.534... c2=−1.52224... c4=0.12761...$

Comparing the results with exact values $c2≈-1.527..., c4≈0.1048... and α≈-2.5029...$ they are nearly equal.

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Answer 2

Question

Chapter 10.7, Problem 1E

Interpretation Introduction

Interpretation:

For the functional equation $g(x) = αg2(xα)$ , arise in renormalization analysis of period doubling.

If $g(x)≈1+c2x2$ for small x, then solve for $c2 and α$

For $g(x)≈1+c2x2+c4x4$ , solve equation for $c2, c4 and α$ , and compare the results with exact values $c2≈-1.527..., c4≈0.1048... and α≈-2.5029...$

Concept Introduction:

The power series expansion of the $g(xα)$ is

$g(xα) = 1+c2(xα)2+c4(xα)4+...$

Expert Solution & Answer

Answer to Problem 1E

Solution:

$α=−2.73205 and c2=−1.36603$
$α=−2.534... c2=−1.52224... c4=0.12761...$ and these are compared with exact values of $α,c2 and c4$

Explanation of Solution

The function equation arose in renormalization equation is,

$g(x) = αg2(xa)$

The power series expansion of the above equation is

$g(xα) = 1+c2(xα)2+c4(xα)4+...$

$g2(xα)=g(g(xα))=g(1+c2(xα)2)=1+c2+2c22(xα)2+O(x4)$

$∴αg2(xa)=α(1+c2+2c22(xα)2+O(x4))=α(1+c2)+2c22x2α+αO(x4)$

The given approximation is

$g(x)=1+c2x2$

Comparing this with $αg2(xa)$

$1 + c2x2 = α(1+c2)+2c22x2α + O(x3)$

Comparing the coefficients of $x0 and x2$

$1= α(1+c2)$

$c2=2c22α$

Solving above two equations

$⇒(α,c2)={(1,0)(−1−3,−1−32)≈(−2.73205,−1.36603)(−1+3,−1+32)≈(0.732051,0.366025)$

Here, take the second solution because in that solution $α$ is closer to Feigenbaum $α=−2.5...$
$g(x) = αg2(xa)=α(1+c2+c4+2(c22+2c2c4)(xα)2+(c23+6c22c4+2c2c4+c42)(xα)4+O(x5))$

The given approximation is

$g(x)=1+c2x2+c4x4$

Equating this with $αg2(xa)$ $1+c2x2+c4x4=α(1+c2+c4)+2(c22+2c2c4)x2α+(c23+6c22c4+2c2c4+c42)x4α3+O(x5)$

Comparing the coefficients of $x0,x2and x4$ , we get

$1= α(1+c2+c4)$

$c2=2(c22+2c2c4)α$

$c4=(c23+6c22c4+2c2c4+c42)α3$

Using first and second equations for solving $c2 and c4$ in terms of $α$

$c2=−2+2α−α2$

$c4=1−1α+α2$

Substituting these values of $c2 and c4$ in third equation, then it becomes

$4α6+3α5−60α4−104α3+168α2+240α−384+128α=0$

The root of above equation closest to Feigenbaum is $α=−2.534...$

Substituting this value of $α$ in $c2=−2+2α−α2$ and $c4=1−1α+α2$

$α=−2.534... c2=−1.52224... c4=0.12761...$

Comparing the results with exact values $c2≈-1.527..., c4≈0.1048... and α≈-2.5029...$ they are nearly equal.

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Answer 3

Question

Chapter 10.7, Problem 1E

Interpretation Introduction

Interpretation:

For the functional equation $g(x) = αg2(xα)$ , arise in renormalization analysis of period doubling.

If $g(x)≈1+c2x2$ for small x, then solve for $c2 and α$

For $g(x)≈1+c2x2+c4x4$ , solve equation for $c2, c4 and α$ , and compare the results with exact values $c2≈-1.527..., c4≈0.1048... and α≈-2.5029...$

Concept Introduction:

The power series expansion of the $g(xα)$ is

$g(xα) = 1+c2(xα)2+c4(xα)4+...$

Expert Solution & Answer

Answer to Problem 1E

Solution:

$α=−2.73205 and c2=−1.36603$
$α=−2.534... c2=−1.52224... c4=0.12761...$ and these are compared with exact values of $α,c2 and c4$

Explanation of Solution

The function equation arose in renormalization equation is,

$g(x) = αg2(xa)$

The power series expansion of the above equation is

$g(xα) = 1+c2(xα)2+c4(xα)4+...$

$g2(xα)=g(g(xα))=g(1+c2(xα)2)=1+c2+2c22(xα)2+O(x4)$

$∴αg2(xa)=α(1+c2+2c22(xα)2+O(x4))=α(1+c2)+2c22x2α+αO(x4)$

The given approximation is

$g(x)=1+c2x2$

Comparing this with $αg2(xa)$

$1 + c2x2 = α(1+c2)+2c22x2α + O(x3)$

Comparing the coefficients of $x0 and x2$

$1= α(1+c2)$

$c2=2c22α$

Solving above two equations

$⇒(α,c2)={(1,0)(−1−3,−1−32)≈(−2.73205,−1.36603)(−1+3,−1+32)≈(0.732051,0.366025)$

Here, take the second solution because in that solution $α$ is closer to Feigenbaum $α=−2.5...$
$g(x) = αg2(xa)=α(1+c2+c4+2(c22+2c2c4)(xα)2+(c23+6c22c4+2c2c4+c42)(xα)4+O(x5))$

The given approximation is

$g(x)=1+c2x2+c4x4$

Equating this with $αg2(xa)$ $1+c2x2+c4x4=α(1+c2+c4)+2(c22+2c2c4)x2α+(c23+6c22c4+2c2c4+c42)x4α3+O(x5)$

Comparing the coefficients of $x0,x2and x4$ , we get

$1= α(1+c2+c4)$

$c2=2(c22+2c2c4)α$

$c4=(c23+6c22c4+2c2c4+c42)α3$

Using first and second equations for solving $c2 and c4$ in terms of $α$

$c2=−2+2α−α2$

$c4=1−1α+α2$

Substituting these values of $c2 and c4$ in third equation, then it becomes

$4α6+3α5−60α4−104α3+168α2+240α−384+128α=0$

The root of above equation closest to Feigenbaum is $α=−2.534...$

Substituting this value of $α$ in $c2=−2+2α−α2$ and $c4=1−1α+α2$

$α=−2.534... c2=−1.52224... c4=0.12761...$

Comparing the results with exact values $c2≈-1.527..., c4≈0.1048... and α≈-2.5029...$ they are nearly equal.

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Answer 4

Question

Chapter 10.7, Problem 1E

Interpretation Introduction

Interpretation:

For the functional equation $g(x) = αg2(xα)$ , arise in renormalization analysis of period doubling.

If $g(x)≈1+c2x2$ for small x, then solve for $c2 and α$

For $g(x)≈1+c2x2+c4x4$ , solve equation for $c2, c4 and α$ , and compare the results with exact values $c2≈-1.527..., c4≈0.1048... and α≈-2.5029...$

Concept Introduction:

The power series expansion of the $g(xα)$ is

$g(xα) = 1+c2(xα)2+c4(xα)4+...$

Expert Solution & Answer

Answer to Problem 1E

Solution:

$α=−2.73205 and c2=−1.36603$
$α=−2.534... c2=−1.52224... c4=0.12761...$ and these are compared with exact values of $α,c2 and c4$

Explanation of Solution

The function equation arose in renormalization equation is,

$g(x) = αg2(xa)$

The power series expansion of the above equation is

$g(xα) = 1+c2(xα)2+c4(xα)4+...$

$g2(xα)=g(g(xα))=g(1+c2(xα)2)=1+c2+2c22(xα)2+O(x4)$

$∴αg2(xa)=α(1+c2+2c22(xα)2+O(x4))=α(1+c2)+2c22x2α+αO(x4)$

The given approximation is

$g(x)=1+c2x2$

Comparing this with $αg2(xa)$

$1 + c2x2 = α(1+c2)+2c22x2α + O(x3)$

Comparing the coefficients of $x0 and x2$

$1= α(1+c2)$

$c2=2c22α$

Solving above two equations

$⇒(α,c2)={(1,0)(−1−3,−1−32)≈(−2.73205,−1.36603)(−1+3,−1+32)≈(0.732051,0.366025)$

Here, take the second solution because in that solution $α$ is closer to Feigenbaum $α=−2.5...$
$g(x) = αg2(xa)=α(1+c2+c4+2(c22+2c2c4)(xα)2+(c23+6c22c4+2c2c4+c42)(xα)4+O(x5))$

The given approximation is

$g(x)=1+c2x2+c4x4$

Equating this with $αg2(xa)$ $1+c2x2+c4x4=α(1+c2+c4)+2(c22+2c2c4)x2α+(c23+6c22c4+2c2c4+c42)x4α3+O(x5)$

Comparing the coefficients of $x0,x2and x4$ , we get

$1= α(1+c2+c4)$

$c2=2(c22+2c2c4)α$

$c4=(c23+6c22c4+2c2c4+c42)α3$

Using first and second equations for solving $c2 and c4$ in terms of $α$

$c2=−2+2α−α2$

$c4=1−1α+α2$

Substituting these values of $c2 and c4$ in third equation, then it becomes

$4α6+3α5−60α4−104α3+168α2+240α−384+128α=0$

The root of above equation closest to Feigenbaum is $α=−2.534...$

Substituting this value of $α$ in $c2=−2+2α−α2$ and $c4=1−1α+α2$

$α=−2.534... c2=−1.52224... c4=0.12761...$

Comparing the results with exact values $c2≈-1.527..., c4≈0.1048... and α≈-2.5029...$ they are nearly equal.

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Answer 5

Chapter 10.7, Problem 1E

Interpretation Introduction

Interpretation:

For the functional equation $g(x) = αg2(xα)$ , arise in renormalization analysis of period doubling.

If $g(x)≈1+c2x2$ for small x, then solve for $c2 and α$

For $g(x)≈1+c2x2+c4x4$ , solve equation for $c2, c4 and α$ , and compare the results with exact values $c2≈-1.527..., c4≈0.1048... and α≈-2.5029...$

Concept Introduction:

The power series expansion of the $g(xα)$ is

$g(xα) = 1+c2(xα)2+c4(xα)4+...$

Expert Solution & Answer

Answer to Problem 1E

Solution:

$α=−2.73205 and c2=−1.36603$
$α=−2.534... c2=−1.52224... c4=0.12761...$ and these are compared with exact values of $α,c2 and c4$

Explanation of Solution

The function equation arose in renormalization equation is,

$g(x) = αg2(xa)$

The power series expansion of the above equation is

$g(xα) = 1+c2(xα)2+c4(xα)4+...$

$g2(xα)=g(g(xα))=g(1+c2(xα)2)=1+c2+2c22(xα)2+O(x4)$

$∴αg2(xa)=α(1+c2+2c22(xα)2+O(x4))=α(1+c2)+2c22x2α+αO(x4)$

The given approximation is

$g(x)=1+c2x2$

Comparing this with $αg2(xa)$

$1 + c2x2 = α(1+c2)+2c22x2α + O(x3)$

Comparing the coefficients of $x0 and x2$

$1= α(1+c2)$

$c2=2c22α$

Solving above two equations

$⇒(α,c2)={(1,0)(−1−3,−1−32)≈(−2.73205,−1.36603)(−1+3,−1+32)≈(0.732051,0.366025)$

Here, take the second solution because in that solution $α$ is closer to Feigenbaum $α=−2.5...$
$g(x) = αg2(xa)=α(1+c2+c4+2(c22+2c2c4)(xα)2+(c23+6c22c4+2c2c4+c42)(xα)4+O(x5))$

The given approximation is

$g(x)=1+c2x2+c4x4$

Equating this with $αg2(xa)$ $1+c2x2+c4x4=α(1+c2+c4)+2(c22+2c2c4)x2α+(c23+6c22c4+2c2c4+c42)x4α3+O(x5)$

Comparing the coefficients of $x0,x2and x4$ , we get

$1= α(1+c2+c4)$

$c2=2(c22+2c2c4)α$

$c4=(c23+6c22c4+2c2c4+c42)α3$

Using first and second equations for solving $c2 and c4$ in terms of $α$

$c2=−2+2α−α2$

$c4=1−1α+α2$

Substituting these values of $c2 and c4$ in third equation, then it becomes

$4α6+3α5−60α4−104α3+168α2+240α−384+128α=0$

The root of above equation closest to Feigenbaum is $α=−2.534...$

Substituting this value of $α$ in $c2=−2+2α−α2$ and $c4=1−1α+α2$

$α=−2.534... c2=−1.52224... c4=0.12761...$

Comparing the results with exact values $c2≈-1.527..., c4≈0.1048... and α≈-2.5029...$ they are nearly equal.

Answer 6

Chapter 10.7, Problem 1E

Interpretation Introduction

Interpretation:

For the functional equation $g(x) = αg2(xα)$ , arise in renormalization analysis of period doubling.

If $g(x)≈1+c2x2$ for small x, then solve for $c2 and α$

For $g(x)≈1+c2x2+c4x4$ , solve equation for $c2, c4 and α$ , and compare the results with exact values $c2≈-1.527..., c4≈0.1048... and α≈-2.5029...$

Concept Introduction:

The power series expansion of the $g(xα)$ is

$g(xα) = 1+c2(xα)2+c4(xα)4+...$

Expert Solution & Answer

Answer to Problem 1E

Solution:

$α=−2.73205 and c2=−1.36603$
$α=−2.534... c2=−1.52224... c4=0.12761...$ and these are compared with exact values of $α,c2 and c4$

Explanation of Solution

The function equation arose in renormalization equation is,

$g(x) = αg2(xa)$

The power series expansion of the above equation is

$g(xα) = 1+c2(xα)2+c4(xα)4+...$

$g2(xα)=g(g(xα))=g(1+c2(xα)2)=1+c2+2c22(xα)2+O(x4)$

$∴αg2(xa)=α(1+c2+2c22(xα)2+O(x4))=α(1+c2)+2c22x2α+αO(x4)$

The given approximation is

$g(x)=1+c2x2$

Comparing this with $αg2(xa)$

$1 + c2x2 = α(1+c2)+2c22x2α + O(x3)$

Comparing the coefficients of $x0 and x2$

$1= α(1+c2)$

$c2=2c22α$

Solving above two equations

$⇒(α,c2)={(1,0)(−1−3,−1−32)≈(−2.73205,−1.36603)(−1+3,−1+32)≈(0.732051,0.366025)$

Here, take the second solution because in that solution $α$ is closer to Feigenbaum $α=−2.5...$
$g(x) = αg2(xa)=α(1+c2+c4+2(c22+2c2c4)(xα)2+(c23+6c22c4+2c2c4+c42)(xα)4+O(x5))$

The given approximation is

$g(x)=1+c2x2+c4x4$

Equating this with $αg2(xa)$ $1+c2x2+c4x4=α(1+c2+c4)+2(c22+2c2c4)x2α+(c23+6c22c4+2c2c4+c42)x4α3+O(x5)$

Comparing the coefficients of $x0,x2and x4$ , we get

$1= α(1+c2+c4)$

$c2=2(c22+2c2c4)α$

$c4=(c23+6c22c4+2c2c4+c42)α3$

Using first and second equations for solving $c2 and c4$ in terms of $α$

$c2=−2+2α−α2$

$c4=1−1α+α2$

Substituting these values of $c2 and c4$ in third equation, then it becomes

$4α6+3α5−60α4−104α3+168α2+240α−384+128α=0$

The root of above equation closest to Feigenbaum is $α=−2.534...$

Substituting this value of $α$ in $c2=−2+2α−α2$ and $c4=1−1α+α2$

$α=−2.534... c2=−1.52224... c4=0.12761...$

Comparing the results with exact values $c2≈-1.527..., c4≈0.1048... and α≈-2.5029...$ they are nearly equal.

Answer 7

Chapter 10.7, Problem 1E

Interpretation Introduction

Interpretation:

For the functional equation $g(x) = αg2(xα)$ , arise in renormalization analysis of period doubling.

If $g(x)≈1+c2x2$ for small x, then solve for $c2 and α$

For $g(x)≈1+c2x2+c4x4$ , solve equation for $c2, c4 and α$ , and compare the results with exact values $c2≈-1.527..., c4≈0.1048... and α≈-2.5029...$

Concept Introduction:

The power series expansion of the $g(xα)$ is

$g(xα) = 1+c2(xα)2+c4(xα)4+...$

Answer 8

Expert Solution & Answer

Answer to Problem 1E

Solution:

$α=−2.73205 and c2=−1.36603$
$α=−2.534... c2=−1.52224... c4=0.12761...$ and these are compared with exact values of $α,c2 and c4$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The function equation arose in renormalization equation is,

$g(x) = αg2(xa)$

The power series expansion of the above equation is

$g(xα) = 1+c2(xα)2+c4(xα)4+...$

$g2(xα)=g(g(xα))=g(1+c2(xα)2)=1+c2+2c22(xα)2+O(x4)$

$∴αg2(xa)=α(1+c2+2c22(xα)2+O(x4))=α(1+c2)+2c22x2α+αO(x4)$

The given approximation is

$g(x)=1+c2x2$

Comparing this with $αg2(xa)$

$1 + c2x2 = α(1+c2)+2c22x2α + O(x3)$

Comparing the coefficients of $x0 and x2$

$1= α(1+c2)$

$c2=2c22α$

Solving above two equations

$⇒(α,c2)={(1,0)(−1−3,−1−32)≈(−2.73205,−1.36603)(−1+3,−1+32)≈(0.732051,0.366025)$

Here, take the second solution because in that solution $α$ is closer to Feigenbaum $α=−2.5...$
$g(x) = αg2(xa)=α(1+c2+c4+2(c22+2c2c4)(xα)2+(c23+6c22c4+2c2c4+c42)(xα)4+O(x5))$

The given approximation is

$g(x)=1+c2x2+c4x4$

Equating this with $αg2(xa)$ $1+c2x2+c4x4=α(1+c2+c4)+2(c22+2c2c4)x2α+(c23+6c22c4+2c2c4+c42)x4α3+O(x5)$

Comparing the coefficients of $x0,x2and x4$ , we get

$1= α(1+c2+c4)$

$c2=2(c22+2c2c4)α$

$c4=(c23+6c22c4+2c2c4+c42)α3$

Using first and second equations for solving $c2 and c4$ in terms of $α$

$c2=−2+2α−α2$

$c4=1−1α+α2$

Substituting these values of $c2 and c4$ in third equation, then it becomes

$4α6+3α5−60α4−104α3+168α2+240α−384+128α=0$

The root of above equation closest to Feigenbaum is $α=−2.534...$

Substituting this value of $α$ in $c2=−2+2α−α2$ and $c4=1−1α+α2$

$α=−2.534... c2=−1.52224... c4=0.12761...$

Comparing the results with exact values $c2≈-1.527..., c4≈0.1048... and α≈-2.5029...$ they are nearly equal.

Answer 12

Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E Ch. 10.1 - Prob. 7E Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E

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