How do you use a power series to find the exact value of the sum of the series #pi/4-(pi/4)^3/(3!)+(pi/4)^5/(5!)-(pi/4)^7/(7!) + …# ?

Answer 1

Since

#x-x^3/{3!}+x^5/{5!}-x^7/{7!}+cdots=sinx#,
#pi/4-(pi/4)^3/{3!}+(pi/4)^5/{5!}-(pi/4)^7/{7!}+cdots=sin(pi/4)=1/sqrt{2}#
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Answer 2

To find the exact value of the sum of the series ( \frac{\pi}{4} - \frac{(\pi/4)^3}{3!} + \frac{(\pi/4)^5}{5!} - \frac{(\pi/4)^7}{7!} + \ldots ) using a power series, we can recognize that it resembles the Taylor series expansion of the trigonometric function ( \arctan(x) ).

The Taylor series expansion of ( \arctan(x) ) is:

[ \arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \ldots ]

Comparing this with the given series, we see that it matches with ( x = \frac{\pi}{4} ).

So, the sum of the given series is ( \arctan\left(\frac{\pi}{4}\right) ).

Using the known value of ( \arctan(1) = \frac{\pi}{4} ), we conclude that the exact value of the sum of the series is ( \boxed{\frac{\pi}{4}} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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