Madhava of Sangamagrama and His Series Expansions

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Who is known as the founder of the Kerala school of astronomy and mathematics?

  1. Aryabhata

  2. Bhaskara II

  3. Madhava of Sangamagrama

  4. Nilakantha Somayaji


Correct Option: C
Explanation:

Madhava of Sangamagrama is considered the founder of the Kerala school of astronomy and mathematics, which flourished in the 14th and 15th centuries.

What is the name of the series expansion discovered by Madhava of Sangamagrama that approximates the sine function?

  1. Taylor series

  2. Maclaurin series

  3. Gregory series

  4. Madhava series


Correct Option: D
Explanation:

The Madhava series is a series expansion that approximates the sine function. It is named after Madhava of Sangamagrama, who discovered it in the 14th century.

What is the general formula for the Madhava series?

  1. $$sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$$

  2. $$sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + \cdots$$

  3. $$sin(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + \cdots$$

  4. $$sin(x) = x + \frac{x^3}{3!} - \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$$


Correct Option: A
Explanation:

The general formula for the Madhava series is $$sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$$. This series converges for all values of x.

What is the name of the series expansion discovered by Madhava of Sangamagrama that approximates the cosine function?

  1. Taylor series

  2. Maclaurin series

  3. Gregory series

  4. Madhava series


Correct Option: D
Explanation:

The Madhava series is a series expansion that approximates the cosine function. It is named after Madhava of Sangamagrama, who discovered it in the 14th century.

What is the general formula for the Madhava series for the cosine function?

  1. $$cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$$

  2. $$cos(x) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \cdots$$

  3. $$cos(x) = 1 - \frac{x^2}{2!} - \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$$

  4. $$cos(x) = 1 + \frac{x^2}{2!} - \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$$


Correct Option: A
Explanation:

The general formula for the Madhava series for the cosine function is $$cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$$. This series converges for all values of x.

What is the name of the series expansion discovered by Madhava of Sangamagrama that approximates the arctangent function?

  1. Taylor series

  2. Maclaurin series

  3. Gregory series

  4. Madhava series


Correct Option: D
Explanation:

The Madhava series is a series expansion that approximates the arctangent function. It is named after Madhava of Sangamagrama, who discovered it in the 14th century.

What is the general formula for the Madhava series for the arctangent function?

  1. $$arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots$$

  2. $$arctan(x) = x + \frac{x^3}{3} + \frac{x^5}{5} + \frac{x^7}{7} + \cdots$$

  3. $$arctan(x) = x - \frac{x^3}{3} - \frac{x^5}{5} - \frac{x^7}{7} + \cdots$$

  4. $$arctan(x) = x + \frac{x^3}{3} - \frac{x^5}{5} - \frac{x^7}{7} + \cdots$$


Correct Option: A
Explanation:

The general formula for the Madhava series for the arctangent function is $$arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots$$. This series converges for all values of x between -1 and 1.

What is the name of the series expansion discovered by Madhava of Sangamagrama that approximates the pi?

  1. Taylor series

  2. Maclaurin series

  3. Gregory series

  4. Madhava series


Correct Option: D
Explanation:

The Madhava series is a series expansion that approximates the pi. It is named after Madhava of Sangamagrama, who discovered it in the 14th century.

What is the general formula for the Madhava series for pi?

  1. $$\pi = 4 \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots\right)$$

  2. $$\pi = 4 \left(1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \cdots\right)$$

  3. $$\pi = 4 \left(1 - \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \cdots\right)$$

  4. $$\pi = 4 \left(1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \cdots\right)$$


Correct Option: A
Explanation:

The general formula for the Madhava series for pi is $$\pi = 4 \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots\right)$$. This series converges very slowly, but it was the first series expansion for pi to be discovered.

What was the main contribution of Madhava of Sangamagrama to mathematics?

  1. He discovered the Taylor series.

  2. He discovered the Maclaurin series.

  3. He discovered the Gregory series.

  4. He discovered the Madhava series.


Correct Option: D
Explanation:

Madhava of Sangamagrama's main contribution to mathematics was the discovery of the Madhava series, which are series expansions for the sine, cosine, arctangent, and pi functions.

In which century did Madhava of Sangamagrama live?

  1. 12th century

  2. 13th century

  3. 14th century

  4. 15th century


Correct Option: C
Explanation:

Madhava of Sangamagrama lived in the 14th century.

What was the name of the book written by Madhava of Sangamagrama?

  1. The Lilavati

  2. The Siddhanta Shiromani

  3. The Yuktibhasa

  4. The Tantrasangraha


Correct Option: C
Explanation:

Madhava of Sangamagrama wrote a book called The Yuktibhasa, which contains his mathematical discoveries.

What is the name of the theorem that states that the sum of the squares of the first n natural numbers is equal to \frac{n(n+1)(2n+1)}{6}?

  1. The Pythagorean theorem

  2. The binomial theorem

  3. The Madhava theorem

  4. The Euler theorem


Correct Option: C
Explanation:

The Madhava theorem states that the sum of the squares of the first n natural numbers is equal to \frac{n(n+1)(2n+1)}{6}.

What is the name of the theorem that states that the area of a circle is equal to \frac{\pi r^2}{2}?

  1. The Pythagorean theorem

  2. The binomial theorem

  3. The Madhava theorem

  4. The Euler theorem


Correct Option: C
Explanation:

The Madhava theorem states that the area of a circle is equal to \frac{\pi r^2}{2}.

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