Sun K.

asked • 12/31/13

Find a series?

Find a series expansion of sin(t)/t.
 
Answer: 1-t^2/3!+t^4/5!-t^6/7!+...
 
Do I must memorize the formula for Maclaurin Series for problems like these?

Andre W.

tutor
You should at least memorize the MacLaurin series for ex, sin(x), cos(x), and the geometric series 1/(1-x). From the geometric series you can then quickly get the series for ln(1+x) and arctan(x) by integration. Ideally, you should also know the binomial series, (1+x)r.
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01/01/14

1 Expert Answer

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Inactive Tutor answered • 12/31/13

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Series for sin(t) is ∑t2k+1/(2k+1)!*(-1)k where sum is taken over k from 0 to ∞. Series sin(t)/t is given by ∑t2k/(2k+1)!*(-1)k
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Andre W.

tutor
Note that the series converges everywhere, even though the function is undefined at t=0.
 
Also, it's a simple factorial, (2k+1)!, not the double factorial, (2k+1)!!, which includes only odd integers.
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01/01/14

Inactive Tutor

Oop, thank you, Andre, it is simple factorial indeed. I corrected it. But sin(t)/t IS defined at t=0, its value is 1, limt→0 sin(t)/t=1. This is one of the special limits, along with limx→∞(1+k/x)mx=emk and some others.
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01/01/14

Inactive Tutor

The limit of sin(t)/t as t → 0 is 1, but isn't the function value indeterminate, 0/0? So the function has a hole at t = 0?
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01/03/14

Inactive Tutor

Agreed.
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01/03/14

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