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seraphim [82]
2 years ago
5

The graph shows the location of P in point hour. Point are is on the Y axis and has the same Y coordinate of point PPoint Q is

graft at (N, -2). The distance from P to Q is equal to the distance from P point R. What is the distance from point P to point Q? What is the value of N? Explain how you determine the distance from point B to point Q, and the value in. The graph shows the location of P in point R. Point are is on the Y axis and has the same Y coordinate of point P

Mathematics
1 answer:
sergejj [24]2 years ago
6 0

Answer:

I cant understand read ur lesson and ask help from ur teacher

#CarryOnLearning

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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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