QuestionPlot terms of the logistic equation $x_{t+1}=c x_t(1-x_t)$ with $x_0=0.4$ , $c=3.5$ . Check for convergence or divergence.

Studdy Solution

STEP 1

Assumptions1. The initial value $x0$ is0.4. The constant $c$ is3.53. The sequence is defined by the discrete logistic equation $x_{t+1}=c_{X_{t}}\left(1-X_{t}\right)$

STEP 2

We will generate the first few terms of the sequence to observe its behavior. We can do this by repeatedly applying the logistic equation.

STEP 3

Calculate the first term $x1$ using the logistic equation and the initial value $x0$ .
$x1 = c \cdot x0 \cdot (1 - x0) =3.5 \cdot0. \cdot (1 -0.)$

STEP 4

Calculate the value of $x1$ .
$x1 =3. \cdot0.4 \cdot (1 -0.4) =0.84$

STEP 5

Calculate the second term $x2$ using the logistic equation and the value of $x1$ .
$x2 = c \cdot x1 \cdot (1 - x1) =3.5 \cdot0.84 \cdot (1 -0.84)$

STEP 6

Calculate the value of $x2$ .
$x2 =3.5 \cdot0.84 \cdot (1 -0.84) =0.4704$

STEP 7

Continue this process to generate more terms of the sequence. After a few iterations, it becomes clear that the sequence does not converge to a single value, but oscillates between two values.

STEP 8

Since the sequence does not converge, we say it diverges. The behavior of the terms is oscillatory, meaning they alternate between two values.

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QuestionPlot terms of the logistic equation $x_{t+1}=c x_t(1-x_t)$ with $x_0=0.4$ , $c=3.5$ . Check for convergence or divergence.

Studdy Solution

STEP 1

Assumptions1. The initial value $x0$ is0.4. The constant $c$ is3.53. The sequence is defined by the discrete logistic equation $x_{t+1}=c_{X_{t}}\left(1-X_{t}\right)$

STEP 2

We will generate the first few terms of the sequence to observe its behavior. We can do this by repeatedly applying the logistic equation.

STEP 3

Calculate the first term $x1$ using the logistic equation and the initial value $x0$ .
$x1 = c \cdot x0 \cdot (1 - x0) =3.5 \cdot0. \cdot (1 -0.)$

STEP 4

Calculate the value of $x1$ .
$x1 =3. \cdot0.4 \cdot (1 -0.4) =0.84$

STEP 5

Calculate the second term $x2$ using the logistic equation and the value of $x1$ .
$x2 = c \cdot x1 \cdot (1 - x1) =3.5 \cdot0.84 \cdot (1 -0.84)$

STEP 6

Calculate the value of $x2$ .
$x2 =3.5 \cdot0.84 \cdot (1 -0.84) =0.4704$

STEP 7

Continue this process to generate more terms of the sequence. After a few iterations, it becomes clear that the sequence does not converge to a single value, but oscillates between two values.

STEP 8

Since the sequence does not converge, we say it diverges. The behavior of the terms is oscillatory, meaning they alternate between two values.

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QuestionPlot terms of the logistic equation xt+1=cxt(1−xt)x_{t+1}=c x_t(1-x_t)xt+1​=cxt​(1−xt​) with x0=0.4x_0=0.4x0​=0.4, c=3.5c=3.5c=3.5. Check for convergence or divergence.

Studdy Solution

STEP 1

Assumptions1. The initial value x0x0x0 is0.4. The constant ccc is3.53. The sequence is defined by the discrete logistic equation xt+1=cXt(1−Xt)x_{t+1}=c_{X_{t}}\left(1-X_{t}\right)xt+1​=cXt​​(1−Xt​)

STEP 2

We will generate the first few terms of the sequence to observe its behavior. We can do this by repeatedly applying the logistic equation.

STEP 3

Calculate the first term x1x1x1 using the logistic equation and the initial value x0x0x0.x1=c⋅x0⋅(1−x0)=3.5⋅0.⋅(1−0.)x1 = c \cdot x0 \cdot (1 - x0) =3.5 \cdot0. \cdot (1 -0.)x1=c⋅x0⋅(1−x0)=3.5⋅0.⋅(1−0.)

STEP 4

Calculate the value of x1x1x1.x1=3.⋅0.4⋅(1−0.4)=0.84x1 =3. \cdot0.4 \cdot (1 -0.4) =0.84x1=3.⋅0.4⋅(1−0.4)=0.84

STEP 5

Calculate the second term x2x2x2 using the logistic equation and the value of x1x1x1.x2=c⋅x1⋅(1−x1)=3.5⋅0.84⋅(1−0.84)x2 = c \cdot x1 \cdot (1 - x1) =3.5 \cdot0.84 \cdot (1 -0.84)x2=c⋅x1⋅(1−x1)=3.5⋅0.84⋅(1−0.84)

STEP 6

Calculate the value of x2x2x2.x2=3.5⋅0.84⋅(1−0.84)=0.4704x2 =3.5 \cdot0.84 \cdot (1 -0.84) =0.4704x2=3.5⋅0.84⋅(1−0.84)=0.4704

STEP 7

Continue this process to generate more terms of the sequence. After a few iterations, it becomes clear that the sequence does not converge to a single value, but oscillates between two values.

STEP 8

Since the sequence does not converge, we say it diverges. The behavior of the terms is oscillatory, meaning they alternate between two values.

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionPlot terms of the logistic equation $x_{t+1}=c x_t(1-x_t)$ with $x_0=0.4$ , $c=3.5$ . Check for convergence or divergence.

Assumptions1. The initial value $x0$ is0.4. The constant $c$ is3.53. The sequence is defined by the discrete logistic equation $x_{t+1}=c_{X_{t}}\left(1-X_{t}\right)$

Calculate the first term $x1$ using the logistic equation and the initial value $x0$ .
$x1 = c \cdot x0 \cdot (1 - x0) =3.5 \cdot0. \cdot (1 -0.)$

Calculate the value of $x1$ .
$x1 =3. \cdot0.4 \cdot (1 -0.4) =0.84$

Calculate the second term $x2$ using the logistic equation and the value of $x1$ .
$x2 = c \cdot x1 \cdot (1 - x1) =3.5 \cdot0.84 \cdot (1 -0.84)$

Calculate the value of $x2$ .
$x2 =3.5 \cdot0.84 \cdot (1 -0.84) =0.4704$