# Thread: Another Taylor Polynomials problem

1. ## Another Taylor Polynomials problem

I still not getting how even start to answer a question related to Taylor Polynomials so please I need some strong help and advice in how to tackle such questions as this ones above:

i) Use Taylor polynomials p1,p3,p5,... about 0, successively, to evaluate sin(pi/10) to four decimal place showing all your working.

ii) Use multiplication of Taylor series to find the quartic Taylor polynomial about 0 for the function f(x) = e^x - 1 / sqr of 1+x
evaluating the coefficients

I don't need just an answer I also need I follow trough with explanations of what is happening.

xxx

2. ## Re: Another Taylor Polynomials problem

Well what is the MacLaurin Series for sin(x)? What do you get when you truncate it to 1, 3, 5 places? What do you get when you substitute pi/10?

3. ## Re: Another Taylor Polynomials problem

Hi Prove It,

First I got the Taylor series for sin x ( why this series does not use the even terms?)

Then I started to calculate :

p1 = x - 1/3! x^3

p3 = p1 + 1/5! x^5

p5 = p3 - 1/7! x^7

p7 = p5 + 1/9! x^9

when x = pi/10

It is correct?

4. ## Re: Another Taylor Polynomials problem

ii) Use multiplication of Taylor series to find the quartic Taylor polynomial about 0 for the function f(x) = e^x - 1 / sqr of 1+x
evaluating the coefficients

I'm not sure if I have done it right

First I obtained the Taylor series for e^x and the subtracted -1 from each term and ended up with the following series:

e^x - 1 = x - 1 + (1/2)x^2 - (5/6)x^3 - (23/24)x^4 - ....

Second I used the binomial series to find the taylor serie about 0 for the function 1/sqr 1+x = 1/(1+x)^1/2 = (1+x)^-1/2

1/sqr 1+x = 1 - (1/2)x + (3/8)x^2 - (5/16)x^3 + (35/128)x^4 - ...

Third I started to multiply both series term by term:

e^x - 1/sqr 1+x = (x - 1 + (1/2)x^2 - (5/6)x^3 - (23/24)x^4 - ....) x (1 - (1/2)x + (3/8)x^2 - (5/16)x^3 + (35/128)x^4 - ...)

=x(1 - (1/2)x + (3/8)x^2 - (5/16)x^3 + (35/128)x^4 - ...) - 1(1 - (1/2)x + (3/8)x^2 - (5/16)x^3 + (35/128)x^4 - ...) +((1/2)x^2)(1 - (1/2)x + (3/8)x^2 - (5/16)x^3 + (35/128)x^4 - ...) - ((5/6)x^3)(1 - (1/2)x + (3/8)x^2 - (5/16)x^3 + (35/128)x^4 - ...) - ((23/24)x^4)(1 - (1/2)x + (3/8)x^2 - (5/16)x^3 + (35/128)x^4 - ...) - ...

Does it makes any sense or I totally wrong?

5. ## Re: Another Taylor Polynomials problem

Originally Posted by michele
Hi Prove It,

First I got the Taylor series for sin x ( why this series does not use the even terms?)

Then I started to calculate :

p1 = x - 1/3! x^3

p3 = p1 + 1/5! x^5

p5 = p3 - 1/7! x^7

p7 = p5 + 1/9! x^9

when x = pi/10

It is correct?
So what do you get when you sub in x = pi/10?

6. ## Re: Another Taylor Polynomials problem

Originally Posted by Prove It
So what do you get when you sub in x = pi/10?
I got:
p1=0.30899155
p3=0.30901705
p5=0.30901699
p7=0.30901699

How p5 and p7 agree to 6 decimal places sin(pi/10)=0.3090 to 4 decimal places.

Right?