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Deffense [45]
3 years ago
5

Write down the range or the interval for set of your real numbers represented on the number line​

Mathematics
1 answer:
lapo4ka [179]3 years ago
3 0

Answer:

where is the picture? or number line

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Factor these expressions into an equivalent form.
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(c+8)(c-8)

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D not here

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<img src="https://tex.z-dn.net/?f=%20%20%5Crm%5Csum_%7Bn%20%3D%201%7D%5E%20%5Cinfty%20%28%20-%201%20%7B%29%7D%5E%7Bn%20-%201%7D%
victus00 [196]

Let

\displaystyle f(x) = \sum_{n=1}^\infty (-1)^{n-1} \frac{x^{2n-1}}{(2n-1)! (2n+1)}

The exponent is indeed 2n-1 - not a typo!

Take the antiderivative of f, denoted by F. This recovers a factor of 2n in the denominator, which lets us condense it to a single factorial.

\displaystyle F(x) = \int f(x) \, dx = C + \sum_{n=1}^\infty (-1)^{n-1} \frac{x^{2n}}{(2n+1)!}

Recall the series expansion of sine,

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

Then with a little algebraic manipulation, we get

\displaystyle F(x) = \int f(x) \, dx = C + 1 - \frac{\sin(x)}x

Differentiate to recover f.

f(x) = \dfrac{\sin(x) - x\cos(x)}{x^2}

Finally, f(\pi) = \frac1\pi, so our sum is

\displaystyle \pi^2 f(\pi) = \sum_{n=1}^\infty (-1)^{n-1} \frac{\pi^{2n+1}}{(2n-1)! (2n+1)} = \boxed{\pi}

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1 year ago
The length of a classroom is 34 feet in yard and frets
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The answer is 1156. I'm positive it is the answer.
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