How to, using Taylor series approximation , estimate the value of π, when arctan(x) ≈ x-x3/3+x5/5-x7/7 ?

Answer 1
let #x=1# #pi/4 = arctan(1) = (1)-(1)^3/3+(1)^5/5-(1)^7/7...# #pi/4 = 1-1/3+1/5-1/7...# #pi = 4(1-1/3+1/5-1/7...)#
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Answer 2

Using the specified TS approximation we get # ~~= 2.90 \ \ \ # (3sf)

With the provided truncation (approximation) of the Taylor Series, we have:

# arctanx ~~ x - x^3/3 + x^5/5 - x^7/7 #

Applying the well-known outcome:

# tan (pi/4) =1 => arctan1=pi/4 #
So, substituting #x=1# into the given TS we have:
# pi/4 ~~ 1 - 1/3 + 1/5 - 1/7 #
# \ \ \ \ = (3*5*7 - 5*7 + 3*7 - 3.5)/(3*5*7) #
# \ \ \ \ = (105 - 35 + 21-15)/(105) #
# \ \ \ \ = (76)/(105) #

Thus:

# pi ~~ 4 * (76)/(105) #
# \ \ \ \ = 304/105 #
# \ \ \ \ ~~= 2.90 \ \ \ # (3sf)
A (not so) interesting fact is that using this particular Taylor Series and method to approximate #pi# tp 3 significant figures (or 2 decimal places) requires #147# terms of the sequence, as is therefore a particularly inefficient method to estimate #pi#.
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Answer 3

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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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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