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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.

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closed as off-topic by M. Vinay, Shaun, Shailesh, Math1000, Daniel W. Farlow Jun 29 at 1:05

This question appears to be off-topic. The users who voted to close gave this specific reason:

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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
    
Possible duplicate of Multiplying Taylor series and composition – PHPirate Jun 28 at 9:31
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 $$\hat f(x)=1+x-\frac 26 x^3+O(x^4)$$

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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

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