2.8 Q-1

Question Statement

1. Prove the following series expansions using the Maclaurin series:

   • i. $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots$
   • ii. $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots$
   • iii. $\sqrt{1-x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} + \ldots$
   • iv. $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots$
   • v. $e^{2x} = 1 + 2x + \frac{4x^2}{2!} + \frac{8x^3}{3!} + \ldots$

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Background and Explanation

The Maclaurin series is a special case of the Taylor series expanded about $x=0$. For a function $f(x)$, the series is given by:

$$f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f^{(3)}(0)x^3}{3!} + \ldots$$

Here, $f'(x)$, $f''(x)$, etc., denote successive derivatives of $f(x)$. To compute the series:

1. Evaluate the function and its derivatives at $x=0$.
2. Substitute these values into the Maclaurin series formula.

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Solution

i. Prove $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots$

1. Let $f(x) = \ln(1+x)$. Compute derivatives:

   • $f'(x) = \frac{1}{1+x}$
   • $f''(x) = -\frac{1}{(1+x)^2}$
   • $f^{(3)}(x) = \frac{2}{(1+x)^3}$
   • $f^{(4)}(x) = -\frac{6}{(1+x)^4}$

2. Evaluate at $x=0$:

   • $f(0) = \ln(1) = 0$
   • $f'(0) = 1$
   • $f''(0) = -1$
   • $f^{(3)}(0) = 2$
   • $f^{(4)}(0) = -6$

3. Substitute into the Maclaurin series:

$$f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f^{(3)}(0)x^3}{3!} + \frac{f^{(4)}(0)x^4}{4!} + \ldots$$ $$\ln(1+x) = 0 + x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots$$

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ii. Prove $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots$

1. Let $f(x) = \cos x$. Compute derivatives:

   • $f'(x) = -\sin x$
   • $f''(x) = -\cos x$
   • $f^{(3)}(x) = \sin x$
   • $f^{(4)}(x) = \cos x$

2. Evaluate at $x=0$:

   • $f(0) = 1$
   • $f'(0) = 0$
   • $f''(0) = -1$
   • $f^{(3)}(0) = 0$
   • $f^{(4)}(0) = 1$

3. Substitute into the Maclaurin series:

$$\cos x = f(0) + \frac{f''(0)x^2}{2!} + \frac{f^{(4)}(0)x^4}{4!} + \ldots$$ $$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots$$

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iii. Prove $\sqrt{1-x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} + \ldots$

1. Let $f(x) = (1-x)^{\frac{1}{2}}$. Compute derivatives:

   • $f'(x) = -\frac{1}{2}(1-x)^{-\frac{1}{2}}$
   • $f''(x) = \frac{1}{4}(1-x)^{-\frac{3}{2}}$
   • $f^{(3)}(x) = -\frac{3}{8}(1-x)^{-\frac{5}{2}}$
   • $f^{(4)}(x) = \frac{15}{16}(1-x)^{-\frac{7}{2}}$

2. Evaluate at $x=0$:

   • $f(0) = 1$
   • $f'(0) = -\frac{1}{2}$
   • $f''(0) = -\frac{1}{8}$
   • $f^{(3)}(0) = \frac{3}{16}$
   • $f^{(4)}(0) = -\frac{15}{128}$

3. Substitute into the Maclaurin series:

$$\sqrt{1-x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} - \frac{x^4}{128} + \ldots$$

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iv. Prove $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots$

1. Let $f(x) = e^x$. Compute derivatives:

   • All derivatives are $f^{(n)}(x) = e^x$.

2. Evaluate at $x=0$:

   • $f(0) = f'(0) = f''(0) = \ldots = 1$

3. Substitute into the Maclaurin series:

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots$$

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v. Prove $e^{2x} = 1 + 2x + \frac{4x^2}{2!} + \frac{8x^3}{3!} + \ldots$

1. Let $f(x) = e^{2x}$. Compute derivatives:

   • $f'(x) = 2e^{2x}$, $f''(x) = 4e^{2x}$, $f^{(3)}(x) = 8e^{2x}$, etc.

2. Evaluate at $x=0$:

   • $f(0) = 1$, $f'(0) = 2$, $f''(0) = 4$, $f^{(3)}(0) = 8$, etc.

3. Substitute into the Maclaurin series:

$$e^{2x} = 1 + 2x + \frac{4x^2}{2!} + \frac{8x^3}{3!} + \ldots$$

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Key Formulas or Methods Used

1. Maclaurin series expansion:

$$f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \ldots$$

2. Successive differentiation and substitution at $x=0$.

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Summary of Steps

1. Define $f(x)$ for the given series.
2. Compute the necessary derivatives of $f(x)$.
3. Evaluate derivatives at $x=0$.
4. Substitute into the Maclaurin series formula.
5. Simplify and confirm the series expansion.