What is the interval of convergence of Sinx?

Andrea S. sinxx=∞∑n=0(−1)nx2n(2n+1)! with radius of convergence R=∞ .

Which is correct in the expansion of sin x?

The Maclaurin expansion of sinx is given by Sinx=x1!

How many non zero terms are needed in the Maclaurin series expansion of sin x in order to maintain an error below 0.1 within the interval of − π 2 π 2 )?

g ( a ) = f ( x ) − f ( a ) − f ′ ( a ) ( x − a ) − f ″ ( a ) 2 ! ( x − a ) 2 + ⋯ + f ( n ) ( a ) n ! ( x − a ) n − R n ( x ) = f ( x ) − p n ( x ) − R n ( x ) = 0 , g ( x ) = f ( x ) − f ( x ) − 0 − ⋯ − 0 = 0. g ( a ) = f ( x ) − f ( a ) − f ′ ( a ) ( x − a ) − f ″ ( a ) 2 !

What is the binomial expansion of sin x?

Sinx=x- x^3/3!

What does Sinx converge to?

You cannot talk about a limit of a function without specifying where the limit is to be taken. It is trivial that sin(x) and cox(x) converge as, say, x goes to 0 or, for that matter to any real number. Yes, both sin(x) and cos(x) diverge as x goes to infinity or -infinity.

What is the series of Sinx?

The Maclaurin series of sin(x) is only the Taylor series of sin(x) at x = 0. If we wish to calculate the Taylor series at any other value of x, we can consider a variety of approaches. Suppose we wish to find the Taylor series of sin(x) at x = c, where c is any real number that is not zero.

Which one of the following is the Taylor series for sinx at x 0 *?

Maclaurin series of sin
The Maclaurin series of sin(x) is only the Taylor series of sin(x) at x = 0.

How does the number of terms in its Maclaurin series expansion affect the error of approximation?

For an exponential function, like f(x)=ex or g(x)=10x, how does the number of terms in its Maclaurin series expansion affect the error of approximation? The number of terms depends on the base of the exponential function. The number of terms does not affect the error.

How do you use Maclaurin series to approximate?

A Maclaurin series can be used to approximate a function, find the antiderivative of a complicated function, or compute an otherwise uncomputable sum. Partial sums of a Maclaurin series provide polynomial approximations for the function. ∑ n = 0 ∞ f ( n ) ( 0 ) x n n ! = f ( 0 ) + f ′ ( 0 ) x + f ′ ′ ( 0 ) 2 !

Why is Sinx divergent?

Because the sum of the infinite series of sin(x) alternates between -1 and 1 it does not go a specific value, meaning it will diverge.

Does the series sin n converge or diverge?

Since 2 > 1, the Ratio Test says that the series diverges. converge or diverge? sinn does not exist, so the Divergence Test says that the series diverges.