CALCU-Solve these problems for meIn this question

Solve these problems for me

Solution by Steps

step 1

We start with the given series expansion for

\ln(1-x)

:

\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \cdots

step 2

Choose

x = -\frac{1}{2}

, which satisfies |x| < 1. Substituting

x = -\frac{1}{2}

into the series:

\ln\left(1 - \left(-\frac{1}{2}\right)\right) = \ln\left(\frac{3}{2}\right) = \frac{1}{2} + \frac{1}{2 \times 2^2} + \frac{1}{3 \times 2^3} + \frac{1}{4 \times 2^4} + \cdots

step 3

Since

\ln(2) = \ln\left(\frac{3}{2}\right) + \ln\left(\frac{4}{3}\right)

, we can write:

\ln(2) = \frac{1}{2} + \frac{1}{2 \times 2^2} + \frac{1}{3 \times 2^3} + \frac{1}{4 \times 2^4} + \cdots

Part (ii)

step 1

Given \frac{1}{n 2^n} < \frac{1}{3 \times 2^n} for

n \geq 4

, we need to find

k

such that \frac{k}{24} < \ln(2) < \frac{k+1}{24}

step 2

Using the approximation

\ln(2) \approx 0.69315

, we solve: \frac{k}{24} < 0.69315 < \frac{k+1}{24}

step 3

Multiplying through by 24: k < 24 \times 0.69315 < k + 1 \implies k < 16.6356 < k + 1

step 4

Therefore,

k = 16

Part (iii)

step 1

Using the series expansion for

\ln(1-x)

with

x = \frac{1}{2}

:

\ln\left(\frac{3}{2}\right) = \frac{1}{2} - \frac{1}{2 \times 2^2} + \frac{1}{3 \times 2^3} - \frac{1}{4 \times 2^4} + \cdots

step 2

To find

\ln(3)

, we use:

\ln(3) = \ln\left(\frac{3}{2}\right) + \ln(2)

step 3

Substituting the series expansions:

\ln(3) = \left(\frac{1}{2} - \frac{1}{2 \times 2^2} + \frac{1}{3 \times 2^3} - \cdots\right) + \left(\frac{1}{2} + \frac{1}{2 \times 2^2} + \frac{1}{3 \times 2^3} + \cdots\right)

step 4

Simplifying, we get:

\ln(3) = 1 + \frac{1}{3 \times 2^2} + \frac{1}{5 \times 2^4} + \frac{1}{7 \times 2^6} + \cdots

Part (iv)

step 1

Using the approximation

\ln(3) \approx 1.0986

, we need to show: \frac{13}{12} < \ln(3) < \frac{11}{10}

step 2

Converting to decimal:

\frac{13}{12} \approx 1.0833 \quad \text{and} \quad \frac{11}{10} = 1.1

step 3

Since 1.0833 < 1.0986 < 1.1, the inequality holds

Part (v)

step 1

To compare

3^{17}

and

4^{13}

, we use logarithms:

\ln(3^{17}) = 17 \ln(3) \quad \text{and} \quad \ln(4^{13}) = 13 \ln(4)

step 2

Using approximations

\ln(3) \approx 1.0986

and

\ln(4) = 2 \ln(2) \approx 1.3863

:

17 \times 1.0986 \approx 18.6762 \quad \text{and} \quad 13 \times 1.3863 \approx 18.0219

step 3

Since 18.6762 > 18.0219,

3^{17}

is larger than

4^{13}

Answer

Part (i):

\ln(2) = \frac{1}{2} + \frac{1}{2 \times 2^2} + \frac{1}{3 \times 2^3} + \frac{1}{4 \times 2^4} + \cdots

Part (ii):

k = 16

Part (iii):

\ln(3) = 1 + \frac{1}{3 \times 2^2} + \frac{1}{5 \times 2^4} + \frac{1}{7 \times 2^6} + \cdots

Part (iv): \frac{13}{12} < \ln(3) < \frac{11}{10}

Part (v):

3^{17}

is larger than

4^{13}

Key Concept

Series Expansion and Logarithms

Explanation

Using series expansions and properties of logarithms, we can approximate and compare values effectively.

please explain the series expansions for me

To explain series expansions, let's consider the function

f(x) = \ln(1-x)

. We can expand this function as a Taylor series around

x=0

. The general formula for the Taylor series expansion of a function

f(x)

around

x=a

is:

f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots

In our case,

f(x) = \ln(1-x)

and we want to expand around

x=0

. First, let's find the derivatives of

f(x)

: -

f'(x) = \frac{-1}{1-x}

-

f''(x) = \frac{1}{(1-x)^2}

-

f'''(x) = \frac{-2}{(1-x)^3}

-

f^{(4)}(x) = \frac{6}{(1-x)^4}

- and so on... Now, we evaluate these derivatives at

x=0

to find the coefficients of the series expansion. Substitute these values into the general formula to get the series expansion of

\ln(1-x)

around

x=0

. I hope this helps in understanding series expansions involving logarithmic functions! Let me know if you need further clarification.

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