- #1

Kumar8434

- 127

- 5

Hi, I've got this:

$$\sin{(A*B)}\approx \frac{Si(B^2)-Si(A^2)}{2(\ln{B}-ln{A})}$$, whenever the RHS is defined and B is close to A ( I don't know how close).

Here ##Si(x)## is the integral of ##\frac{\sin{x}}{x}##

But, to check it, I need to evaluate the ##Si(x)## function. I'm new with Taylor series, so, how am I supposed to do it? The only confusion is that should I work with the Taylor series of ##Si(x)## around ##x=A^2## or ##x=B^2## to check it? How do I evaluate ##Si(B^2)-Si(A^2)## approximately?

$$\sin{(A*B)}\approx \frac{Si(B^2)-Si(A^2)}{2(\ln{B}-ln{A})}$$, whenever the RHS is defined and B is close to A ( I don't know how close).

Here ##Si(x)## is the integral of ##\frac{\sin{x}}{x}##

But, to check it, I need to evaluate the ##Si(x)## function. I'm new with Taylor series, so, how am I supposed to do it? The only confusion is that should I work with the Taylor series of ##Si(x)## around ##x=A^2## or ##x=B^2## to check it? How do I evaluate ##Si(B^2)-Si(A^2)## approximately?

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