How to, using Taylor series approximation , estimate the value of π, when arctan(x) ≈ x-x3/3+x5/5-x7/7 ?
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Using the specified TS approximation we get
With the provided truncation (approximation) of the Taylor Series, we have:
Applying the well-known outcome:
Thus:
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To estimate the value of ( \pi ) using the given Taylor series approximation for ( \arctan(x) ), which is ( \arctan(x) \approx x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} ), we can substitute ( x = 1 ) into the series:
[ \arctan(1) \approx 1 - \frac{1^3}{3} + \frac{1^5}{5} - \frac{1^7}{7} ]
Simplify the expression:
[ \arctan(1) \approx 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} ]
[ \arctan(1) \approx \frac{49}{105} ]
Since ( \arctan(1) = \frac{\pi}{4} ), we can solve for ( \pi ):
[ \frac{\pi}{4} = \frac{49}{105} ]
[ \pi \approx \frac{49 \times 4}{105} ]
[ \pi \approx \frac{196}{105} ]
So, using the given Taylor series approximation, the estimated value of ( \pi ) is approximately ( \frac{196}{105} ).
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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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