Answer:
[tex]f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}[/tex]
Step-by-step explanation:
The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by
[tex]f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...[/tex]
We first compute the n-th derivative of [tex]f(x)=\ln(1+2x)[/tex], note that
[tex]f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\[/tex]
Now, if we compute the n-th derivative at 0 we get
[tex]f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)![/tex]
and so the Maclaurin series for f(x)=ln(1+2x) is given by
[tex]f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}[/tex]
The Maclaurin series for f(x) = In (1+2x) is give as [tex]\sum^{\infty} _{n=1}[/tex] (-1)ⁿ⁻¹2ⁿ(xⁿ/n). This was derived by computing for the nth derivative of the Taylor Series function.
The Maclaurin Series refers to a form of series expansion in which all the terms are positive and nonnegative integers powers of the variable.
It can also be defined as the Taylor Series that is expanded proximal to the reference point zero. The function is given as:
f (x) = f (0) + (f'(0)x)/1! + (f''(0)x²)/2! + ... + (fⁿ(0)xⁿ)/n!
Step 1
Solve for the nth derivative of f(x) = In(1+2x)
f'(x) = 2 * (1+2x)⁻¹
f''(x) = 2² * (-1) * (1+2x)⁻²
f''(x) = 2³ * (-1)² * 2* (1+2x)⁻³ ...
fⁿ(x) = 2ⁿ * (-1)ⁿ⁻¹ * (n-1)! * (1+2x)⁻ⁿ
Step 2 -Following the above,
f(0) = ln (1+2 *0) = ln (1) = 0
f' (0) = 2*1 = 2
f'²(0) = 2²*(-1)
f³(0) = 2³ * (-1)² *2 ...
f⁽ⁿ⁾ (0) = 2ⁿ * (-1)⁽ⁿ⁻¹⁾ * (n-1)!
The Maclaurin Series for f(x) = In (1+2x) is given by:
f(x) = 0 +2x -2²(x²/2!) + 2³ *2!(x³/3!) + .... (-1)⁽ⁿ⁻¹⁾(n-1)! *2ⁿ (xⁿ/n!) + ...
= 0 +2x -2²(x²/2!) + 2³ *2!(x³/3!) + .... (-1)⁽ⁿ⁻¹⁾(n-1)! *2ⁿ (xⁿ/n!) + ...
Hence, we have
= [tex]\sum^{\infty} _{n=1}[/tex] (-1)ⁿ⁻¹2ⁿ(xⁿ/n).
Learn more about Maclaurin Series at:
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