# Thread: Partial Fractions with Repeated Roots

1. ## Partial Fractions with Repeated Roots

I had this in the calculus section in a different context. Although the calculus part of the question was answered (I new that anyway) I am still non the wiser regarding the partial fraction.

Can somebody show me the steps in converting this to a partial fraction?

(k-1)/2^(k+1)

2. Originally Posted by p75213
I had this in the calculus section in a different context. Although the calculus part of the question was answered (I new that anyway) I am still non the wiser regarding the partial fraction.

Can somebody show me the steps in converting this to a partial fraction?

(k-1)/2^(k+1)
Dear p75213,

$\displaystyle \frac{k-1}{2^{k+1}}=\frac{k+1-2}{2^{k+1}}=\frac{k+1}{2^{k+1}}-\frac{2}{2^{k+1}}=\frac{k+1}{2^{k+1}}-\frac{1}{2^{k}}$

3. Was that really what you intended? "Partial Fractions" are usually used to simplify fractions with polynomials in the denominator not exponentials.

4. Originally Posted by HallsofIvy
Was that really what you intended? "Partial Fractions" are usually used to simplify fractions with polynomials in the denominator not exponentials.
Dear HallsofIvy,

You are correct. But I remember trying to decompose these kind of partial fractions when I was doing my advanced level exams. In our case these were used in finding the sum of a series. For example a series with the general term $\displaystyle \frac{r-1}{2^{r+1}}$, we decomposed this into partial fractions as above and easily found the sum since there are similar terms which cancel off. Therefore I assumed that possibly p75213 want to decompose this for similar reasons. However if I am wrong many apologies.

5. Thanks Sudharaka. A neat way of breaking up the fraction.

6. Hi Sudharaka,
Actually (k-1)/2^(k+1) breaks down into
k/2^k - k+1/2^(k+1). Using your methodology on the k we get - k-1/2^k+1 = 2k-k-1/2^k+1 = 2k/2^k+1 - k+1/2^k+1 = k/2^k - k+1/2^k+1.

7. Originally Posted by p75213
Hi Sudharaka,
Actually (k-1)/2^(k+1) breaks down into
k/2^k - k+1/2^(k+1). Using your methodology on the k we get - k-1/2^k+1 = 2k-k-1/2^k+1 = 2k/2^k+1 - k+1/2^k+1 = k/2^k - k+1/2^k+1.
Dear p75213,

Either way it's correct, $\displaystyle \frac{k-1}{2^{k+1}}=\frac{k+1}{2^{k+1}}-\frac{1}{2^k}=\frac{k}{2^k}-\frac{k+1}{2^{k+1}}$