Question

sin ax = ax - a^3 x^3/3! + a^5 x^5/5! - ... + a^n-1 x^n-1/n-1! sin(n-1)π/2 + a^n x^n/n! sin (nπ/2+a theta x),0<theta<1 by use maclaurin's theorem​

Answer 1

Answer:

This can be simplified to:

$\sin(ax) = \sum_{n=0}^\infty \frac{(-1)^{n+1} a^{2n+1}}{(2n+1)!} x^{2n+1} + \sum_{n=0}^\infty \frac{(-1)^{n} a^{2n+2}}{(2n+2)!} x^{2n+2} (-1)^{\frac{n}{2}}$

Grouping the terms with even and odd exponents together, we get:

Step-by-step explanation:

We can use Maclaurin's theorem to expand the function$\sin(ax)$ as a power series:

$\sin(ax) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n$

where $f(x) = \sin(ax)$ and $f^{(n)}(x)$ denotes the $n$ th derivative of $f(x)$ with respect to $x$.

To find the derivatives of $f(x)$, we can use the chain rule:

$f'(x) = a \cos(ax)$

$f''(x) = -a^2 \sin(ax)$

$f'''(x) = -a^3 \cos(ax)$

$\vdots$

$f^{(n)}(x) = (-1)^{n+1} a^n \sin(ax)$

Note that the derivatives of $\sin(ax)$ alternate between $\sin(ax)$ and $\cos(ax)$, with a factor of $a^n$ and a sign of $(-1)^{n+1}$.

Now, we can substitute these derivatives into the power series expansion of $\sin(ax)$to get:

$\sin(ax) = \sum_{n=0}^\infty \frac{(-1)^{n+1} a^n \sin(0)}{n!} x^n = \sum_{n=0}^\infty \frac{(-1)^{n+1} a^n}{n!} x^n$

Simplifying this expression, we get:

$\sin(ax) = \sum_{n=0}^\infty \frac{(-1)^{n+1} a^{2n+1}}{(2n+1)!} x^{2n+1}$

Next, we can use the identity $\sin(\frac{\pi}{2} + \theta) = \cos(\theta)$ to rewrite$\sin((n-1)\pi/2)$ as $\cos((n-1)\pi/2 - \frac{\pi}{2})$, which simplifies to $(-1)^{n/2}$.

Using this identity, we can rewrite the above expression as:

$\sin(ax) = \sum_{n=0}^\infty \frac{(-1)^{n+1} a^{2n+1}}{(2n+1)!} x^{2n+1} + \sum_{n=1}^\infty \frac{(-1)^{n/2} a^{2n}}{(2n)!} x^{2n} (-1)^{\frac{n-1}{2}}$

This can be simplified to:

$\sin(ax) = \sum_{n=0}^\infty \frac{(-1)^{n+1} a^{2n+1}}{(2n+1)!} x^{2n+1} + \sum_{n=0}^\infty \frac{(-1)^{n} a^{2n+2}}{(2n+2)!} x^{2n+2} (-1)^{\frac{n}{2}}$

Grouping the terms with even and odd exponents together, we get:

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