Question

Find the first four nonzero terms of the Taylor series about 0 for the function tsin(2t).

Answer 1

First write out the Taylor Series for sin(x)...and thats a classical series.

Then replace each "x" in that Taylor Series by "2x".

Then multiply that series, (for sin (2x)), by x.

Straightforward.

Answer 2

Use t instead of x. No big deal what variable you use.

Answer 3

I assume you know the Taylor Series for sin(x).

Answer 4

sin(x)=x-x^3/3!+x^5/5!-...

Answer 5

so then by replacing x with 2x it would become....2x-2x^3/3!+2x^5/5!-...

Answer 6

Exactly, now just repace all x's with (2t)...and then multiply the entire series (term by term) by t, and you have the Taylor Series for t sin(2t).

Answer 7

t(2t-2t^3/3!+2t^5/5!-...)

Answer 8

Two things...the second term is (2t)^3..parenthesis is a MUST...and you see why thats is needed...and same for all others.

Second, you should multiply t by each of the terms, and not leave it as t(.......)

Answer 9

okay thank you:)

Answer 10

2t^2-(((2t^3)/(3!))*t)+(((2t^5)/(5!))*t)-(((2t^7)/(7!))*t)

this is what i entered and its wrong
@Easyaspi314

Answer 11

2t^2 - (4/3)t^4 + (4/15)t^6 - (8/315)t^8

thats the answer

Answer 12

sin (x) = x - x^3/3! + x^5/5! - .....

sin (2t) = (2t) - [(2t)^3/3!] + [(2t)^5/5!] - ....

which simplifies to, sin (2t) = (2t) - (8t^3)/3! + (32t^5)/5! - ...

t sin (2t) = t[ (2t) - (8t^3)/3! + (32t^5/5! - ...

which simplifies to, 2t^2 - 8t^4/5! + 32t^6/5! - ....

@megannicole51

Answer 13

thank you!

Answer 14

@megannicole51

Is it crystal clear now?