Mixing
Is there any way to simplify the following fractions? Thank you!
$begingroup$
Is there any way to simplify the following fractions?
$[n^{3}-1]/[n^{4}-1]$
$n!/n^{n}$
fractions
asked Jan 24 at 3:37
JennyJenny
625
$endgroup$
add a comment |
$begingroup$
Is there any way to simplify the following fractions?
$[n^{3}-1]/[n^{4}-1]$
$n!/n^{n}$
fractions
asked Jan 24 at 3:37
JennyJenny
625
$endgroup$
$begingroup$
You can factor an $n-1$ from both numerator and denominator in your first expression.
$endgroup$
– coreyman317
Jan 24 at 3:40
$begingroup$
As for the second, pick a random integer and write out the numerator and denominator as $frac{1cdot 2 cdot...cdot n}{n cdot n cdot...cdot n}$ and see which number can cancel.
$endgroup$
– coreyman317
Jan 24 at 3:43
add a comment |
$begingroup$
Is there any way to simplify the following fractions?
$[n^{3}-1]/[n^{4}-1]$
$n!/n^{n}$
fractions
asked Jan 24 at 3:37
JennyJenny
625
$endgroup$
Is there any way to simplify the following fractions?
$[n^{3}-1]/[n^{4}-1]$
$n!/n^{n}$
fractions
fractions
asked Jan 24 at 3:37
JennyJenny
625
asked Jan 24 at 3:37
JennyJenny
625
asked Jan 24 at 3:37
JennyJenny
625
asked Jan 24 at 3:37
JennyJenny
625
asked Jan 24 at 3:37
JennyJenny
625
625
$begingroup$
You can factor an $n-1$ from both numerator and denominator in your first expression.
$endgroup$
– coreyman317
Jan 24 at 3:40
$begingroup$
As for the second, pick a random integer and write out the numerator and denominator as $frac{1cdot 2 cdot...cdot n}{n cdot n cdot...cdot n}$ and see which number can cancel.
$endgroup$
– coreyman317
Jan 24 at 3:43
add a comment |
$begingroup$
You can factor an $n-1$ from both numerator and denominator in your first expression.
$endgroup$
– coreyman317
Jan 24 at 3:40
$begingroup$
As for the second, pick a random integer and write out the numerator and denominator as $frac{1cdot 2 cdot...cdot n}{n cdot n cdot...cdot n}$ and see which number can cancel.
$endgroup$
– coreyman317
Jan 24 at 3:43
$begingroup$
You can factor an $n-1$ from both numerator and denominator in your first expression.
$endgroup$
– coreyman317
Jan 24 at 3:40
$begingroup$
You can factor an $n-1$ from both numerator and denominator in your first expression.
$endgroup$
– coreyman317
Jan 24 at 3:40
$begingroup$
As for the second, pick a random integer and write out the numerator and denominator as $frac{1cdot 2 cdot...cdot n}{n cdot n cdot...cdot n}$ and see which number can cancel.
$endgroup$
– coreyman317
Jan 24 at 3:43
$begingroup$
As for the second, pick a random integer and write out the numerator and denominator as $frac{1cdot 2 cdot...cdot n}{n cdot n cdot...cdot n}$ and see which number can cancel.
$endgroup$
– coreyman317
Jan 24 at 3:43
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since $n^k-1 = 0$ when $n=1$, $(n-1)$ is clearly a factor. Then you can factor out $(n-1)$ by long division for example. This answer also gives a factorization of $n^k-1$.
Not much information is given except the common $n$.
answered Jan 24 at 3:48
MonkeyKingMonkeyKing
2,3211029
$endgroup$
add a comment |
$begingroup$
For the first:
$$frac{n^3-1}{n^4-1}=frac{(n-1)(1+n+n^2)}{(n-1)(1+n+n^2+n^3)}=1-frac{n^3}{1+n+n^2+n^3}$$
For the second:
$$frac{n!}{n^n}=frac{(n-1)!}{n^{n-1}}=frac{(1-frac 1n)^{n-1}(n-2)!}{(n-1)^{n-2}}=frac{(n-k-1)!prod_{r=1}^{k}(1-frac{1}{n-r+1})^{n-r}}{(n-k)^{n-k-1}}$$ for $k in Bbb N$
answered Jan 24 at 4:12
Rhys HughesRhys Hughes
7,0201630
$endgroup$
$begingroup$
If you're wondering how I got this, note that $$n(1-frac 1n)=n-1$$
$endgroup$
– Rhys Hughes
Jan 24 at 4:17
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since $n^k-1 = 0$ when $n=1$, $(n-1)$ is clearly a factor. Then you can factor out $(n-1)$ by long division for example. This answer also gives a factorization of $n^k-1$.
Not much information is given except the common $n$.
answered Jan 24 at 3:48
MonkeyKingMonkeyKing
2,3211029
$endgroup$
add a comment |
$begingroup$
Since $n^k-1 = 0$ when $n=1$, $(n-1)$ is clearly a factor. Then you can factor out $(n-1)$ by long division for example. This answer also gives a factorization of $n^k-1$.
Not much information is given except the common $n$.
answered Jan 24 at 3:48
MonkeyKingMonkeyKing
2,3211029
$endgroup$
add a comment |
$begingroup$
Since $n^k-1 = 0$ when $n=1$, $(n-1)$ is clearly a factor. Then you can factor out $(n-1)$ by long division for example. This answer also gives a factorization of $n^k-1$.
Not much information is given except the common $n$.
answered Jan 24 at 3:48
MonkeyKingMonkeyKing
2,3211029
$endgroup$
Since $n^k-1 = 0$ when $n=1$, $(n-1)$ is clearly a factor. Then you can factor out $(n-1)$ by long division for example. This answer also gives a factorization of $n^k-1$.
Not much information is given except the common $n$.
answered Jan 24 at 3:48
MonkeyKingMonkeyKing
2,3211029
answered Jan 24 at 3:48
MonkeyKingMonkeyKing
2,3211029
answered Jan 24 at 3:48
MonkeyKingMonkeyKing
2,3211029
answered Jan 24 at 3:48
MonkeyKingMonkeyKing
2,3211029
2,3211029
add a comment |
add a comment |
$begingroup$
For the first:
$$frac{n^3-1}{n^4-1}=frac{(n-1)(1+n+n^2)}{(n-1)(1+n+n^2+n^3)}=1-frac{n^3}{1+n+n^2+n^3}$$
For the second:
$$frac{n!}{n^n}=frac{(n-1)!}{n^{n-1}}=frac{(1-frac 1n)^{n-1}(n-2)!}{(n-1)^{n-2}}=frac{(n-k-1)!prod_{r=1}^{k}(1-frac{1}{n-r+1})^{n-r}}{(n-k)^{n-k-1}}$$ for $k in Bbb N$
answered Jan 24 at 4:12
Rhys HughesRhys Hughes
7,0201630
$endgroup$
$begingroup$
If you're wondering how I got this, note that $$n(1-frac 1n)=n-1$$
$endgroup$
– Rhys Hughes
Jan 24 at 4:17
add a comment |
$begingroup$
For the first:
$$frac{n^3-1}{n^4-1}=frac{(n-1)(1+n+n^2)}{(n-1)(1+n+n^2+n^3)}=1-frac{n^3}{1+n+n^2+n^3}$$
For the second:
$$frac{n!}{n^n}=frac{(n-1)!}{n^{n-1}}=frac{(1-frac 1n)^{n-1}(n-2)!}{(n-1)^{n-2}}=frac{(n-k-1)!prod_{r=1}^{k}(1-frac{1}{n-r+1})^{n-r}}{(n-k)^{n-k-1}}$$ for $k in Bbb N$
answered Jan 24 at 4:12
Rhys HughesRhys Hughes
7,0201630
$endgroup$
$begingroup$
If you're wondering how I got this, note that $$n(1-frac 1n)=n-1$$
$endgroup$
– Rhys Hughes
Jan 24 at 4:17
add a comment |
$begingroup$
For the first:
$$frac{n^3-1}{n^4-1}=frac{(n-1)(1+n+n^2)}{(n-1)(1+n+n^2+n^3)}=1-frac{n^3}{1+n+n^2+n^3}$$
For the second:
$$frac{n!}{n^n}=frac{(n-1)!}{n^{n-1}}=frac{(1-frac 1n)^{n-1}(n-2)!}{(n-1)^{n-2}}=frac{(n-k-1)!prod_{r=1}^{k}(1-frac{1}{n-r+1})^{n-r}}{(n-k)^{n-k-1}}$$ for $k in Bbb N$
answered Jan 24 at 4:12
Rhys HughesRhys Hughes
7,0201630
$endgroup$
For the first:
$$frac{n^3-1}{n^4-1}=frac{(n-1)(1+n+n^2)}{(n-1)(1+n+n^2+n^3)}=1-frac{n^3}{1+n+n^2+n^3}$$
For the second:
$$frac{n!}{n^n}=frac{(n-1)!}{n^{n-1}}=frac{(1-frac 1n)^{n-1}(n-2)!}{(n-1)^{n-2}}=frac{(n-k-1)!prod_{r=1}^{k}(1-frac{1}{n-r+1})^{n-r}}{(n-k)^{n-k-1}}$$ for $k in Bbb N$
answered Jan 24 at 4:12
Rhys HughesRhys Hughes
7,0201630
answered Jan 24 at 4:12
Rhys HughesRhys Hughes
7,0201630
answered Jan 24 at 4:12
Rhys HughesRhys Hughes
7,0201630
answered Jan 24 at 4:12
Rhys HughesRhys Hughes
7,0201630
7,0201630
$begingroup$
If you're wondering how I got this, note that $$n(1-frac 1n)=n-1$$
$endgroup$
– Rhys Hughes
Jan 24 at 4:17
add a comment |
$begingroup$
If you're wondering how I got this, note that $$n(1-frac 1n)=n-1$$
$endgroup$
– Rhys Hughes
Jan 24 at 4:17
$begingroup$
If you're wondering how I got this, note that $$n(1-frac 1n)=n-1$$
$endgroup$
– Rhys Hughes
Jan 24 at 4:17
$begingroup$
If you're wondering how I got this, note that $$n(1-frac 1n)=n-1$$
$endgroup$
– Rhys Hughes
Jan 24 at 4:17
add a comment |
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$begingroup$
You can factor an $n-1$ from both numerator and denominator in your first expression.
$endgroup$
– coreyman317
Jan 24 at 3:40
$begingroup$
As for the second, pick a random integer and write out the numerator and denominator as $frac{1cdot 2 cdot...cdot n}{n cdot n cdot...cdot n}$ and see which number can cancel.
$endgroup$
– coreyman317
Jan 24 at 3:43