Mixing
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Solve given equation to a approximate value
$begingroup$
Can some explain to me how the relation shown below gives $log k$ on approximation?
$$x = sum_{i=0}^{k-1} frac{1 }{k-i}$$
sequences-and-series approximation
edited Jan 20 at 18:46
Daniele Tampieri
2,3272922
asked Jan 19 at 16:51
KivtasKivtas
275
$endgroup$
add a comment |
$begingroup$
Can some explain to me how the relation shown below gives $log k$ on approximation?
$$x = sum_{i=0}^{k-1} frac{1 }{k-i}$$
sequences-and-series approximation
edited Jan 20 at 18:46
Daniele Tampieri
2,3272922
asked Jan 19 at 16:51
KivtasKivtas
275
$endgroup$
$begingroup$
letting $j=k-i$ gives $sum limits_{j=1}^k frac1j$ as an expression for the harmonic numbers
$endgroup$
– Henry
Jan 19 at 17:00
add a comment |
$begingroup$
Can some explain to me how the relation shown below gives $log k$ on approximation?
$$x = sum_{i=0}^{k-1} frac{1 }{k-i}$$
sequences-and-series approximation
edited Jan 20 at 18:46
Daniele Tampieri
2,3272922
asked Jan 19 at 16:51
KivtasKivtas
275
$endgroup$
Can some explain to me how the relation shown below gives $log k$ on approximation?
$$x = sum_{i=0}^{k-1} frac{1 }{k-i}$$
sequences-and-series approximation
sequences-and-series approximation
edited Jan 20 at 18:46
Daniele Tampieri
2,3272922
asked Jan 19 at 16:51
KivtasKivtas
275
edited Jan 20 at 18:46
Daniele Tampieri
2,3272922
asked Jan 19 at 16:51
KivtasKivtas
275
edited Jan 20 at 18:46
Daniele Tampieri
2,3272922
edited Jan 20 at 18:46
Daniele Tampieri
2,3272922
edited Jan 20 at 18:46
Daniele Tampieri
2,3272922
2,3272922
asked Jan 19 at 16:51
KivtasKivtas
275
asked Jan 19 at 16:51
KivtasKivtas
275
asked Jan 19 at 16:51
KivtasKivtas
275
275
$begingroup$
letting $j=k-i$ gives $sum limits_{j=1}^k frac1j$ as an expression for the harmonic numbers
$endgroup$
– Henry
Jan 19 at 17:00
add a comment |
$begingroup$
letting $j=k-i$ gives $sum limits_{j=1}^k frac1j$ as an expression for the harmonic numbers
$endgroup$
– Henry
Jan 19 at 17:00
$begingroup$
letting $j=k-i$ gives $sum limits_{j=1}^k frac1j$ as an expression for the harmonic numbers
$endgroup$
– Henry
Jan 19 at 17:00
$begingroup$
letting $j=k-i$ gives $sum limits_{j=1}^k frac1j$ as an expression for the harmonic numbers
$endgroup$
– Henry
Jan 19 at 17:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that $$x = sum_{i=0}^{k-1} frac{1 }{k-i}= 1+1/2+1/3+...+1/k$$
Also note that the above sum is an approximation to $$ int _1 ^k (1/x)dx = ln k $$
That is reason behind the approximation.
answered Jan 19 at 17:03
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that $$x = sum_{i=0}^{k-1} frac{1 }{k-i}= 1+1/2+1/3+...+1/k$$
Also note that the above sum is an approximation to $$ int _1 ^k (1/x)dx = ln k $$
That is reason behind the approximation.
answered Jan 19 at 17:03
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
Note that $$x = sum_{i=0}^{k-1} frac{1 }{k-i}= 1+1/2+1/3+...+1/k$$
Also note that the above sum is an approximation to $$ int _1 ^k (1/x)dx = ln k $$
That is reason behind the approximation.
answered Jan 19 at 17:03
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
Note that $$x = sum_{i=0}^{k-1} frac{1 }{k-i}= 1+1/2+1/3+...+1/k$$
Also note that the above sum is an approximation to $$ int _1 ^k (1/x)dx = ln k $$
That is reason behind the approximation.
answered Jan 19 at 17:03
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
Note that $$x = sum_{i=0}^{k-1} frac{1 }{k-i}= 1+1/2+1/3+...+1/k$$
Also note that the above sum is an approximation to $$ int _1 ^k (1/x)dx = ln k $$
That is reason behind the approximation.
answered Jan 19 at 17:03
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 19 at 17:03
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 19 at 17:03
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 19 at 17:03
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
41.6k42061
add a comment |
add a comment |
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$begingroup$
letting $j=k-i$ gives $sum limits_{j=1}^k frac1j$ as an expression for the harmonic numbers
$endgroup$
– Henry
Jan 19 at 17:00