Simplify 2/√5

1 answer:

4 0

The answer would be 2√5/5

As a rule, you cannot have any radical as a denominator. That's why you gotta multiply √5 with √5 which would be √25. The square root of this is 5, and now you have disposed of the radical! Hurray. However, keep in mind, that when you multiply √5, you also multiply it to the numerator. So, yeah, after this, you get the answer.

Hopefully, my explanation makes sense. :)

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Write the Taylor Series for f(x) = sin(x)center

tangare [24]

Answer:

Taylor series of sin(x) centered at x = 0.

Step-by-step explanation:

Taylor series expansion:

$\sum_{n=0}^N f^n(a)\displaystyle\frac{(x-a)^n}{n!}$

Here, f(x) = sin (x) and a = 0

$f(x) = \sin x, f(0) = 0\\f'(x) = \cos x, f'(0) = 1\\f''(x) = -\sin x, f''(0) = 0\\f'''(x) = -\cos x, f'''(0) = -1\\f^4(x) = \sin x, f^4(0) = 0\\f^5(x) = \cos x, f^5(0) = 0$

Putting all the values and expanding, we get,

$f(x) = f(a) + \displaystyle\frac{f'(a)(x-a)}{1!} + \displaystyle\frac{f''(a)(x-a)^2}{2!} + \displaystyle\frac{f'''(a)(x-a)^3}{3!} + \displaystyle\frac{f^4(a)(x-a)^4}{4!} + \displaystyle\frac{f^5(a)(x-a)^5}{5!} + ...\\\\= \sin 0 + \displaystyle\frac{x}{1!} + \displaystyle\frac{(0)x^2}{2!} + \displaystyle\frac{(-1)x^3}{3!} + \displaystyle\frac{(0)x^4}{4!} + \displaystyle\frac{(1)x^5}{5!} + ...$

Solving, we get

$\sin x = x - \displaystyle\frac{x^3}{3!} + \displaystyle\frac{x^5}{5!} - \displaystyle\frac{x^7}{7!} + ...\\\\\sin x = x - \displaystyle\frac{x^3}{6} + \displaystyle\frac{x^5}{120} - \displaystyle\frac{x^7}{5040} + ...$