Math

QuestionFind fixed points of the sequence defined by an+1=12(an+3an)a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{3}{a_{n}}\right).

Studdy Solution

STEP 1

Assumptions1. We are looking for fixed points of the sequence {an}\left\{a_{n}\right\}. . A fixed point of a sequence is a value aa such that an+1=an=aa_{n+1} = a_n = a for some nn.
3. The sequence is defined by the recursive formula an+1=1(an+3an)a_{n+1}=\frac{1}{}\left(a_{n}+\frac{3}{a_{n}}\right).

STEP 2

To find the fixed points, we need to solve the equation an+1=an=aa_{n+1} = a_n = a for aa. This gives us the equationa=12(a+a)a = \frac{1}{2}\left(a + \frac{}{a}\right)

STEP 3

Multiply both sides of the equation by 2a2a to clear the fraction.
2a2=a2+32a^2 = a^2 +3

STEP 4

Subtract a2a^2 from both sides of the equation to isolate the terms involving aa on one side.
a2=3a^2 =3

STEP 5

Take the square root of both sides of the equation to solve for aa. Remember that the square root has two solutions, one positive and one negative.
a=3,a=3a = \sqrt{3},\, a = -\sqrt{3} The fixed points of the sequence are 3\sqrt{3} and 3-\sqrt{3}.

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