Find the coefficient of x^{21} without calculating the product

In summary, the conversation discusses finding the coefficient of $x^{21}$ in the product of two polynomials without actually calculating the product. The method used involves finding terms with indices that are multiples of both 2 and 3, and then summing them up to get the desired coefficient. This approach is deemed correct and satisfactory.
  • #1
mathmari
Gold Member
MHB
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Hey! :eek:

"Let the polynomials
$$f(x)=1+\sum_{k=1}^{8}{(2k)x^{2k}} \text{ and } g(x)=1+ \sum_{k=1}^{8}{(3k)x^{3k}}$$
of $\mathbb{Q}$. Without calculating $f(x)g(x)$, find the coefficient of $x^{21}$ at $f(x)g(x)$."
Let's consider $f(x)=\sum_{k=0}^{\infty}{a_kx^k}$ and $g(x)=\sum_{k=0}^{\infty}{b_kx^k}$

So the following coefficients exist:
\begin{matrix}
a_0 & b_0 \\
a_2 & b_3 \\
a_4 & b_6 \\
a_6 & b_9 \\
a_8 & b_{12} \\
a_{10} & b_{15} \\
a_{12} & b_{18} \\
a_{14} & b_{21} \\
a_{16} & b_{24} \\
\end{matrix}

Since $c_{21}=\sum_{i+j=21}{a_i b_j}$, we have to find each time two coefficients for which the sum of their indices is equal to $21$. So:
$a_0, b_{21}$
$a_6, b_{15}$
$a_{12}, b_9$

Therefore, $c_8=a_0 b_{21}+a_6 b_{15}+ a_{12} b_9=21+6 \cdot 15+ 12 \cdot 9=219$.

Is this correct?
Is the way I solved it ok, or is there a better way to write it?
 
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  • #2
That's fine.

What you are doing is solving:

$k = 0\text{ (mod }2)$
$21 - k = 0\text{ (mod }3)$

The second equation is clearly the same as:

$k = 0\text{ (mod }3)$ so together we get:

$k = 0\text{ (mod }6)$

Then we just find these values of $k$ with $0 \leq k \leq 16$ (since that's as high as the non-zero a's go).

*****

You should probably have used another index besides $k$, in your second formulation of $f$ and $g$. What you are interested in is the non-zero terms (technically ALL the terms "exist").
 
  • #3
Deveno said:
That's fine.

What you are doing is solving:

$k = 0\text{ (mod }2)$
$21 - k = 0\text{ (mod }3)$

The second equation is clearly the same as:

$k = 0\text{ (mod }3)$ so together we get:

$k = 0\text{ (mod }6)$

Then we just find these values of $k$ with $0 \leq k \leq 16$ (since that's as high as the non-zero a's go).

*****

You should probably have used another index besides $k$, in your second formulation of $f$ and $g$. What you are interested in is the non-zero terms (technically ALL the terms "exist").

Aha! Ok!
So is the way I formulated the solution nice? Or is there a better way to express it?
 
  • #4
What you have done is fine, and correct.
 
  • #5
Deveno said:
What you have done is fine, and correct.

Great! Thanks a lot! (Smirk)
 

FAQ: Find the coefficient of x^{21} without calculating the product

1. What is the coefficient of x^{21}?

The coefficient of x^{21} is the numerical value that multiplies the variable x^{21} in a polynomial expression.

2. How do I find the coefficient of x^{21} without calculating the product?

To find the coefficient of x^{21} without calculating the product, you can use the binomial theorem or the Pascal's triangle method to expand the polynomial expression.

3. Why is it important to find the coefficient of x^{21}?

Finding the coefficient of x^{21} is important because it allows us to determine the degree and leading term of a polynomial expression, which are essential in analyzing and solving equations.

4. Can I find the coefficient of x^{21} by simply counting the terms in the polynomial expression?

No, you cannot find the coefficient of x^{21} by simply counting the terms in the polynomial expression. The coefficient is the numerical value, not the number of terms in the expression.

5. Is there a specific formula for finding the coefficient of x^{21}?

There is no specific formula for finding the coefficient of x^{21}, but there are methods such as the binomial theorem and Pascal's triangle that can help you expand and determine the coefficient without calculating the product.

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