Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to calculate a Taylor expansion which is : $\cos(x). exp(x)$ in the neighborhood of 0 in order 3

this is the result I got :

$$\cos(x). exp(x) = \left(1-\frac{x²}{2}+\epsilon(x)x^3\right) . \left(1+x+\frac{x²}{2}+\frac{x^3}{6}+\epsilon(x)x^3\right)$$

And now I need to multiply the two expressions.

I think that there is a method where we use a table to multiply to Taylor expansion, but I don't know hw to do it.

share|cite|improve this question
Can you multiply polynomials? think series expansions as polynomials with possibly infinite degree – Federica Maggioni May 1 '13 at 16:36
@FedericaMaggioni sorry I didn't understand what you mean – Aimad Majdou May 1 '13 at 16:41
Throw away the tiny terms at the end. Just multiply the binomial $(1-x^2/2)$ and the tetranomial $(1+x+x^2/2+x^3/6)$. Since you're not considered with any terms of degree $>3$, they won't matter. Following that multiplication, throw out the terms you get from that product of degree $>3$. – Ian Coley May 1 '13 at 16:45
up vote 0 down vote accepted

Why don't you use regular definition $$\hat f(x)=f(x=0)+f'(x=0)x+\frac 12 f''(x=0)x^2+\frac 16 f'''(x=0)x^3+O(x^4)$$ where $$f(x=0)=e^x \cos(x)=1$$ $$f'(x=0)=e^x \cos(x)-e^x \sin(x)=1$$ $$f''(x=0)=-2e^x \sin(x)=0$$ $$f'''(x=0)=-2e^x \cos(x)-2e^x \sin(x)=-2$$ and $$\hat f(x)=1+x-\frac 26 x^3+O(x^4)$$

share|cite|improve this answer

You use the distributive property, so the expansion starts off $1\cdot 1 +1\cdot x +1 \cdot \frac {x^2}2 -\frac {x^2}2\cdot 1$, then collect the terms of the same degree

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.