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disa [49]
3 years ago
6

find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

ot show that Rn(x) → 0.] Find the associated radius of convergence R. f(x) = 6(1 − x)−2 Step 1 The Maclaurin series formula is f(0) + f '(0)x + f ''(0) 2! x2 + f '''(0) 3! x3 + f (4)(0) 4! x4 + .
Mathematics
1 answer:
guajiro [1.7K]3 years ago
6 0

Answer:

= ∑ 6*n*x^n-1

Radius of convergence = 1

Step-by-step explanation:

f(x) = 6(1-x)^-2

Maclaurin series can be expressed using the formula

f(x) =  f(0) + f '(0)x + f ''(0)/ 2!  (x)^2 + f '''(0)/3! (x)^3 + f (4)(0) 4! x4 + .

attached below is the detailed solution

Radius of convergence = 1

The Maclaurin series for f(x) = 6 / (1 - x )^2  = ∑ 6*n*x^n-1  ( boundary ; ∞ and n = 1 )

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Verify that y1(t) = 1 and y2(t) = t ^1/2 are solutions of the differential equation:
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Answer: it is verified that:

* y1 and y2 are solutions to the differential equation,

* c1 + c2t^(1/2) is not a solution.

Step-by-step explanation:

Given the differential equation

yy'' + (y')² = 0

To verify that y1 solutions to the DE, differentiate y1 twice and substitute the values of y1'' for y'', y1' for y', and y1 for y into the DE. If it is equal to 0, then it is a solution. Do this for y2 as well.

Now,

y1 = 1

y1' = 0

y'' = 0

So,

y1y1'' + (y1')² = (1)(0) + (0)² = 0

Hence, y1 is a solution.

y2 = t^(1/2)

y2' = (1/2)t^(-1/2)

y2'' = (-1/4)t^(-3/2)

So,

y2y2'' + (y2')² = t^(1/2)×(-1/4)t^(-3/2) + [(1/2)t^(-1/2)]² = (-1/4)t^(-1) + (1/4)t^(-1) = 0

Hence, y2 is a solution.

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Let us differentiate this twice, and verify if it satisfies the differential equation.

y = c1 + c2t^(1/2)

y' = (1/2)c2t^(-1/2)

y'' = (-1/4)c2t(-3/2)

yy'' + (y')² = [c1 + c2t^(1/2)][(-1/4)c2t(-3/2)] + [(1/2)c2t^(-1/2)]²

= (-1/4)c1c2t(-3/2) + (-1/4)(c2)²t(-3/2) + (1/4)(c2)²t^(-1)

= (-1/4)c1c2t(-3/2)

≠ 0

This clearly doesn't satisfy the differential equation, hence, it is not a solution.

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3 years ago
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