# Taylor Polynomials?

• Nov 11th 2009, 07:43 PM
shah4u19
Taylor Polynomials?
Find the Taylor polynomial Tn(x) for the function f at the number a. Graph f and T3 on the same paper.

f(x)= e^-3x sin(3x), a= 0, n=3

f(x)= ln x/x, a= 1, x= 3

• Nov 11th 2009, 08:52 PM
Drexel28
Quote:

Originally Posted by shah4u19
Find the Taylor polynomial Tn(x) for the function f at the number a. Graph f and T3 on the same paper.

f(x)= e^-3x sin(3x), a= 0, n=3

f(x)= ln x/x, a= 1, x= 3

What have you tried? I'm guessing you don't know the Maclaurin series themselves and can just take the first three terms huh?

to calculate a third degree (I assume thats what the n means) Taylor polynomial at $x=a$ you must calcuate

$\sum_{n=0}^{3}\frac{f^{(n)}(a)(x-a)^n}{n!}$. Does that seem doable?
• Nov 11th 2009, 09:04 PM
shah4u19
Well, for the first term, I have tried whatever I could. I know maclaurin series for e^-3x is (-3x)^n/n! and maclaurin series for sin (3x) is (-1)^n (3x)^(2n+1)/(2n+1)!. I tried to divide sin(3x)/e^3x upto 3rd degree and submitted the answer I got, but it was wrong. So I don't know what to do now.

For ln x/x, I dont know where to start.

Thanks Anyways.
• Nov 11th 2009, 09:11 PM
Drexel28
Quote:

Originally Posted by shah4u19
Well, for the first term, I have tried whatever I could. I know maclaurin series for e^-3x is (-3x)^n/n! and maclaurin series for sin (3x) is (-1)^n (3x)^(2n+1)/(2n+1)!. I tried to divide sin(3x)/e^3x upto 3rd degree and submitted the answer I got, but it was wrong. So I don't know what to do now.

For ln x/x, I dont know where to start.

Thanks Anyways.

Ok. Since you know the MaClaurin series you don't have to do the derivatives (whew)

Merely take $\left(\sum_{n=0}^{3}\frac{x^n}{n!}\right)\cdot\lef t(\sum_{n=0}^{3}\frac{(-1)^nx^{2n+1}}{(2n+1)!}\right)$.

For the second one you could to the derivatives or notice that

$\frac{\ln(x)}{x}=\frac{d}{dx}\left[\frac{\ln^2(x)}{2}\right]$. The second one is well known...so then all you have to do is...