How to using the method of undetermined coefficients in this equation?
3
$begingroup$
I want to find the rational numbers $a$, $b$, $c$ satisfying the condition
$$displaystyle int _ { 5 } ^ { 21 } frac { mathrm { d } x } { x cdot sqrt { x + 4 } } = a ln 3 + b ln 5 + c ln 7.$$
I solve by hand.
Integrate[1/(x Sqrt[x + 4]), {x, 5, 21}]
1/2 Log[15/7]
From here, I got, $ a = dfrac{1}{2} $, $ b = dfrac{1}{2} $, $ c = -dfrac{1}{2} .$
How can I tell Mathematica to do that?
With Maple, I got the answer directly
equation-solving coefficients
asked Feb 12 at 7:23
minhthien_2016minhthien_2016
573311
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add a comment |
3
$begingroup$
I want to find the rational numbers $a$, $b$, $c$ satisfying the condition
$$displaystyle int _ { 5 } ^ { 21 } frac { mathrm { d } x } { x cdot sqrt { x + 4 } } = a ln 3 + b ln 5 + c ln 7.$$
I solve by hand.
Integrate[1/(x Sqrt[x + 4]), {x, 5, 21}]
1/2 Log[15/7]
From here, I got, $ a = dfrac{1}{2} $, $ b = dfrac{1}{2} $, $ c = -dfrac{1}{2} .$
How can I tell Mathematica to do that?
With Maple, I got the answer directly
equation-solving coefficients
asked Feb 12 at 7:23
minhthien_2016minhthien_2016
573311
$endgroup$
3
$begingroup$
You're just calculating the integral rather than solving the equation in Maple. If such results are desired for you, simplyIntegrate[1/(x Sqrt[x + 4]), {x, 5, 21}] // PowerExpand // Expand.
$endgroup$
– xzczd
Feb 12 at 8:46
$begingroup$
Thank you very much.
$endgroup$
– minhthien_2016
Feb 12 at 8:52
add a comment |
3
3
3
$begingroup$
I want to find the rational numbers $a$, $b$, $c$ satisfying the condition
$$displaystyle int _ { 5 } ^ { 21 } frac { mathrm { d } x } { x cdot sqrt { x + 4 } } = a ln 3 + b ln 5 + c ln 7.$$
I solve by hand.
Integrate[1/(x Sqrt[x + 4]), {x, 5, 21}]
1/2 Log[15/7]
From here, I got, $ a = dfrac{1}{2} $, $ b = dfrac{1}{2} $, $ c = -dfrac{1}{2} .$
How can I tell Mathematica to do that?
With Maple, I got the answer directly
equation-solving coefficients
asked Feb 12 at 7:23
minhthien_2016minhthien_2016
573311
$endgroup$
I want to find the rational numbers $a$, $b$, $c$ satisfying the condition
$$displaystyle int _ { 5 } ^ { 21 } frac { mathrm { d } x } { x cdot sqrt { x + 4 } } = a ln 3 + b ln 5 + c ln 7.$$
I solve by hand.
Integrate[1/(x Sqrt[x + 4]), {x, 5, 21}]
1/2 Log[15/7]
From here, I got, $ a = dfrac{1}{2} $, $ b = dfrac{1}{2} $, $ c = -dfrac{1}{2} .$
How can I tell Mathematica to do that?
With Maple, I got the answer directly
equation-solving coefficients
equation-solving coefficients
asked Feb 12 at 7:23
minhthien_2016minhthien_2016
573311
asked Feb 12 at 7:23
minhthien_2016minhthien_2016
573311
edited Feb 12 at 8:40
minhthien_2016
asked Feb 12 at 7:23
minhthien_2016minhthien_2016
573311
asked Feb 12 at 7:23
minhthien_2016minhthien_2016
573311
asked Feb 12 at 7:23
minhthien_2016minhthien_2016
573311
573311
3
$begingroup$
You're just calculating the integral rather than solving the equation in Maple. If such results are desired for you, simplyIntegrate[1/(x Sqrt[x + 4]), {x, 5, 21}] // PowerExpand // Expand.
$endgroup$
– xzczd
Feb 12 at 8:46
$begingroup$
Thank you very much.
$endgroup$
– minhthien_2016
Feb 12 at 8:52
add a comment |
3
$begingroup$
You're just calculating the integral rather than solving the equation in Maple. If such results are desired for you, simplyIntegrate[1/(x Sqrt[x + 4]), {x, 5, 21}] // PowerExpand // Expand.
$endgroup$
– xzczd
Feb 12 at 8:46
$begingroup$
Thank you very much.
$endgroup$
– minhthien_2016
Feb 12 at 8:52
3
3
$begingroup$
You're just calculating the integral rather than solving the equation in Maple. If such results are desired for you, simply
Integrate[1/(x Sqrt[x + 4]), {x, 5, 21}] // PowerExpand // Expand.$endgroup$
– xzczd
Feb 12 at 8:46
$begingroup$
You're just calculating the integral rather than solving the equation in Maple. If such results are desired for you, simply
Integrate[1/(x Sqrt[x + 4]), {x, 5, 21}] // PowerExpand // Expand.$endgroup$
– xzczd
Feb 12 at 8:46
$begingroup$
Thank you very much.
$endgroup$
– minhthien_2016
Feb 12 at 8:52
$begingroup$
Thank you very much.
$endgroup$
– minhthien_2016
Feb 12 at 8:52
add a comment |
3 Answers
3
active
oldest
votes
5
$begingroup$
{a/d, b/e, c/f} /.
FindInstance[Integrate[1/(x Sqrt[x + 4]), {x, 5, 21}] ==
a Log[3]/d + b Log[5]/e + c Log[7]/f, {a, b, c, d, e, f}, Integers]
{{1/2, 1/2, -(1/2)}}
answered Feb 12 at 7:32
kglrkglr
190k10206424
$endgroup$
$begingroup$
How about equation? I don't get the result?Clear[a, b, c, d, e, f] {a/d, b/e} /. FindInstance[ Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log[3]/d + b Log[11]/e, {a, d, b, e}, Integers]
$endgroup$
– minhthien_2016
Feb 12 at 8:01
3
$begingroup$
@minhthien_2016, as the message says, The methods available to FindInstance are insufficient to find the requested instances or prove they do not exist. You can tryReduce: for example,Reduce[{Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log[3]/d + b Log[11]/e, And @@ Thread[-5 <= {a, d, b, e} <= 5]}, {a, d, b, e}, Integers]gives a solution.
$endgroup$
– kglr
Feb 12 at 8:23
$begingroup$
1/4 Log[27/11]can be written3/4 Log[3] - 1/4 Log[11]
$endgroup$
– minhthien_2016
Feb 12 at 8:32
add a comment |
2
$begingroup$
Not as automated as kglr's solution, but the following works:
eq = Integrate[1/(x Sqrt[x + 4]), {x, 5, 21}] == a Log@3 + b Log@5 + c Log@7
(* 1/2 Log[15/7] == a Log[3] + b Log[5] + c Log[7] *)
PowerExpand@eq
(* 1/2 (Log[3] + Log[5] - Log[7]) == a Log[3] + b Log[5] + c Log[7] *)
Collect[Subtract @@ %, Log[_]] == 0
(* (1/2 - a) Log[3] + (1/2 - b) Log[5] + (-(1/2) - c) Log[7] == 0 *)
Cases[%, coe_ Log[_] :> Solve[coe == 0], Infinity] // Flatten
(* {a -> 1/2, b -> 1/2, c -> -(1/2)} *)
This method also handles the new added example Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log[3] + b Log[11].
answered Feb 12 at 8:29
xzczdxzczd
27.7k574257
$endgroup$
$begingroup$
When I triedeq1 = Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log@3 + b Log@5; PowerExpand@eq1; Collect[Subtract @@ %, Log[_]] == 0; Cases[%, coe_ Log[_] :> Solve[coe == 0], Infinity] // FlattenI got{a -> 3/4, b -> 0}. I think, It is wrong.
$endgroup$
– minhthien_2016
Feb 12 at 9:02
$begingroup$
@minhthien_2016 The equation itself is wrong. TheLog@5should beLog@11.
$endgroup$
– xzczd
Feb 12 at 9:08
$begingroup$
Yes. I think, the answer, It can not be found.
$endgroup$
– minhthien_2016
Feb 12 at 9:09
$begingroup$
As already mentioned, this solution is not a fully automated solution, error handling should be done manually.
$endgroup$
– xzczd
Feb 12 at 9:13
$begingroup$
I am looking forward to a genenral method.
$endgroup$
– minhthien_2016
Feb 12 at 9:14
|
show 1 more comment
0
$begingroup$
I am interpreting your question to mean that you want logarithms of rational numbers to be expressed purely in terms of logarithms of primes. If so, one can use a replacement rule:
{Integrate[1/(x Sqrt[x + 4]), {x, 5, 21}],
Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}]} // Simplify
{1/2 Log[15/7], 1/4 Log[27/11]}
% /. Log[r_Rational] :> (Total[#2 Log[#1] & @@@ FactorInteger[Numerator[r]]] -
Total[#2 Log[#1] & @@@ FactorInteger[Denominator[r]]]) // Expand
{Log[3]/2 + Log[5]/2 - Log[7]/2, 3 Log[3]/4 - Log[11]/4}
Alternatively, one can use FindIntegerNullVector, similar to what was done in this answer:
-Rest[#]/First[#] &[FindIntegerNullVector[{1/2 Log[15/7], Log[3], Log[5], Log[7]}]]
{1/2, 1/2, -1/2}
-Rest[#]/First[#] &[FindIntegerNullVector[{1/4 Log[27/11], Log[3], Log[11]}]]
{3/4, -1/4}
answered Feb 18 at 1:26
J. M. is away♦J. M. is away
98.9k10311467
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
5
$begingroup$
{a/d, b/e, c/f} /.
FindInstance[Integrate[1/(x Sqrt[x + 4]), {x, 5, 21}] ==
a Log[3]/d + b Log[5]/e + c Log[7]/f, {a, b, c, d, e, f}, Integers]
{{1/2, 1/2, -(1/2)}}
answered Feb 12 at 7:32
kglrkglr
190k10206424
$endgroup$
$begingroup$
How about equation? I don't get the result?Clear[a, b, c, d, e, f] {a/d, b/e} /. FindInstance[ Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log[3]/d + b Log[11]/e, {a, d, b, e}, Integers]
$endgroup$
– minhthien_2016
Feb 12 at 8:01
3
$begingroup$
@minhthien_2016, as the message says, The methods available to FindInstance are insufficient to find the requested instances or prove they do not exist. You can tryReduce: for example,Reduce[{Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log[3]/d + b Log[11]/e, And @@ Thread[-5 <= {a, d, b, e} <= 5]}, {a, d, b, e}, Integers]gives a solution.
$endgroup$
– kglr
Feb 12 at 8:23
$begingroup$
1/4 Log[27/11]can be written3/4 Log[3] - 1/4 Log[11]
$endgroup$
– minhthien_2016
Feb 12 at 8:32
add a comment |
5
$begingroup$
{a/d, b/e, c/f} /.
FindInstance[Integrate[1/(x Sqrt[x + 4]), {x, 5, 21}] ==
a Log[3]/d + b Log[5]/e + c Log[7]/f, {a, b, c, d, e, f}, Integers]
{{1/2, 1/2, -(1/2)}}
answered Feb 12 at 7:32
kglrkglr
190k10206424
$endgroup$
$begingroup$
How about equation? I don't get the result?Clear[a, b, c, d, e, f] {a/d, b/e} /. FindInstance[ Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log[3]/d + b Log[11]/e, {a, d, b, e}, Integers]
$endgroup$
– minhthien_2016
Feb 12 at 8:01
3
$begingroup$
@minhthien_2016, as the message says, The methods available to FindInstance are insufficient to find the requested instances or prove they do not exist. You can tryReduce: for example,Reduce[{Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log[3]/d + b Log[11]/e, And @@ Thread[-5 <= {a, d, b, e} <= 5]}, {a, d, b, e}, Integers]gives a solution.
$endgroup$
– kglr
Feb 12 at 8:23
$begingroup$
1/4 Log[27/11]can be written3/4 Log[3] - 1/4 Log[11]
$endgroup$
– minhthien_2016
Feb 12 at 8:32
add a comment |
5
5
5
$begingroup$
{a/d, b/e, c/f} /.
FindInstance[Integrate[1/(x Sqrt[x + 4]), {x, 5, 21}] ==
a Log[3]/d + b Log[5]/e + c Log[7]/f, {a, b, c, d, e, f}, Integers]
{{1/2, 1/2, -(1/2)}}
answered Feb 12 at 7:32
kglrkglr
190k10206424
$endgroup$
{a/d, b/e, c/f} /.
FindInstance[Integrate[1/(x Sqrt[x + 4]), {x, 5, 21}] ==
a Log[3]/d + b Log[5]/e + c Log[7]/f, {a, b, c, d, e, f}, Integers]
{{1/2, 1/2, -(1/2)}}
answered Feb 12 at 7:32
kglrkglr
190k10206424
answered Feb 12 at 7:32
kglrkglr
190k10206424
answered Feb 12 at 7:32
kglrkglr
190k10206424
answered Feb 12 at 7:32
kglrkglr
190k10206424
190k10206424
$begingroup$
How about equation? I don't get the result?Clear[a, b, c, d, e, f] {a/d, b/e} /. FindInstance[ Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log[3]/d + b Log[11]/e, {a, d, b, e}, Integers]
$endgroup$
– minhthien_2016
Feb 12 at 8:01
3
$begingroup$
@minhthien_2016, as the message says, The methods available to FindInstance are insufficient to find the requested instances or prove they do not exist. You can tryReduce: for example,Reduce[{Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log[3]/d + b Log[11]/e, And @@ Thread[-5 <= {a, d, b, e} <= 5]}, {a, d, b, e}, Integers]gives a solution.
$endgroup$
– kglr
Feb 12 at 8:23
$begingroup$
1/4 Log[27/11]can be written3/4 Log[3] - 1/4 Log[11]
$endgroup$
– minhthien_2016
Feb 12 at 8:32
add a comment |
$begingroup$
How about equation? I don't get the result?Clear[a, b, c, d, e, f] {a/d, b/e} /. FindInstance[ Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log[3]/d + b Log[11]/e, {a, d, b, e}, Integers]
$endgroup$
– minhthien_2016
Feb 12 at 8:01
3
$begingroup$
@minhthien_2016, as the message says, The methods available to FindInstance are insufficient to find the requested instances or prove they do not exist. You can tryReduce: for example,Reduce[{Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log[3]/d + b Log[11]/e, And @@ Thread[-5 <= {a, d, b, e} <= 5]}, {a, d, b, e}, Integers]gives a solution.
$endgroup$
– kglr
Feb 12 at 8:23
$begingroup$
1/4 Log[27/11]can be written3/4 Log[3] - 1/4 Log[11]
$endgroup$
– minhthien_2016
Feb 12 at 8:32
$begingroup$
How about equation? I don't get the result?
Clear[a, b, c, d, e, f] {a/d, b/e} /. FindInstance[ Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log[3]/d + b Log[11]/e, {a, d, b, e}, Integers]$endgroup$
– minhthien_2016
Feb 12 at 8:01
$begingroup$
How about equation? I don't get the result?
Clear[a, b, c, d, e, f] {a/d, b/e} /. FindInstance[ Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log[3]/d + b Log[11]/e, {a, d, b, e}, Integers]$endgroup$
– minhthien_2016
Feb 12 at 8:01
3
3
$begingroup$
@minhthien_2016, as the message says, The methods available to FindInstance are insufficient to find the requested instances or prove they do not exist. You can try
Reduce: for example, Reduce[{Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log[3]/d + b Log[11]/e, And @@ Thread[-5 <= {a, d, b, e} <= 5]}, {a, d, b, e}, Integers] gives a solution.$endgroup$
– kglr
Feb 12 at 8:23
$begingroup$
@minhthien_2016, as the message says, The methods available to FindInstance are insufficient to find the requested instances or prove they do not exist. You can try
Reduce: for example, Reduce[{Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log[3]/d + b Log[11]/e, And @@ Thread[-5 <= {a, d, b, e} <= 5]}, {a, d, b, e}, Integers] gives a solution.$endgroup$
– kglr
Feb 12 at 8:23
$begingroup$
1/4 Log[27/11] can be written 3/4 Log[3] - 1/4 Log[11]$endgroup$
– minhthien_2016
Feb 12 at 8:32
$begingroup$
1/4 Log[27/11] can be written 3/4 Log[3] - 1/4 Log[11]$endgroup$
– minhthien_2016
Feb 12 at 8:32
add a comment |
2
$begingroup$
Not as automated as kglr's solution, but the following works:
eq = Integrate[1/(x Sqrt[x + 4]), {x, 5, 21}] == a Log@3 + b Log@5 + c Log@7
(* 1/2 Log[15/7] == a Log[3] + b Log[5] + c Log[7] *)
PowerExpand@eq
(* 1/2 (Log[3] + Log[5] - Log[7]) == a Log[3] + b Log[5] + c Log[7] *)
Collect[Subtract @@ %, Log[_]] == 0
(* (1/2 - a) Log[3] + (1/2 - b) Log[5] + (-(1/2) - c) Log[7] == 0 *)
Cases[%, coe_ Log[_] :> Solve[coe == 0], Infinity] // Flatten
(* {a -> 1/2, b -> 1/2, c -> -(1/2)} *)
This method also handles the new added example Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log[3] + b Log[11].
answered Feb 12 at 8:29
xzczdxzczd
27.7k574257
$endgroup$
$begingroup$
When I triedeq1 = Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log@3 + b Log@5; PowerExpand@eq1; Collect[Subtract @@ %, Log[_]] == 0; Cases[%, coe_ Log[_] :> Solve[coe == 0], Infinity] // FlattenI got{a -> 3/4, b -> 0}. I think, It is wrong.
$endgroup$
– minhthien_2016
Feb 12 at 9:02
$begingroup$
@minhthien_2016 The equation itself is wrong. TheLog@5should beLog@11.
$endgroup$
– xzczd
Feb 12 at 9:08
$begingroup$
Yes. I think, the answer, It can not be found.
$endgroup$
– minhthien_2016
Feb 12 at 9:09
$begingroup$
As already mentioned, this solution is not a fully automated solution, error handling should be done manually.
$endgroup$
– xzczd
Feb 12 at 9:13
$begingroup$
I am looking forward to a genenral method.
$endgroup$
– minhthien_2016
Feb 12 at 9:14
|
show 1 more comment
2
$begingroup$
Not as automated as kglr's solution, but the following works:
eq = Integrate[1/(x Sqrt[x + 4]), {x, 5, 21}] == a Log@3 + b Log@5 + c Log@7
(* 1/2 Log[15/7] == a Log[3] + b Log[5] + c Log[7] *)
PowerExpand@eq
(* 1/2 (Log[3] + Log[5] - Log[7]) == a Log[3] + b Log[5] + c Log[7] *)
Collect[Subtract @@ %, Log[_]] == 0
(* (1/2 - a) Log[3] + (1/2 - b) Log[5] + (-(1/2) - c) Log[7] == 0 *)
Cases[%, coe_ Log[_] :> Solve[coe == 0], Infinity] // Flatten
(* {a -> 1/2, b -> 1/2, c -> -(1/2)} *)
This method also handles the new added example Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log[3] + b Log[11].
answered Feb 12 at 8:29
xzczdxzczd
27.7k574257
$endgroup$
$begingroup$
When I triedeq1 = Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log@3 + b Log@5; PowerExpand@eq1; Collect[Subtract @@ %, Log[_]] == 0; Cases[%, coe_ Log[_] :> Solve[coe == 0], Infinity] // FlattenI got{a -> 3/4, b -> 0}. I think, It is wrong.
$endgroup$
– minhthien_2016
Feb 12 at 9:02
$begingroup$
@minhthien_2016 The equation itself is wrong. TheLog@5should beLog@11.
$endgroup$
– xzczd
Feb 12 at 9:08
$begingroup$
Yes. I think, the answer, It can not be found.
$endgroup$
– minhthien_2016
Feb 12 at 9:09
$begingroup$
As already mentioned, this solution is not a fully automated solution, error handling should be done manually.
$endgroup$
– xzczd
Feb 12 at 9:13
$begingroup$
I am looking forward to a genenral method.
$endgroup$
– minhthien_2016
Feb 12 at 9:14
|
show 1 more comment
2
2
2
$begingroup$
Not as automated as kglr's solution, but the following works:
eq = Integrate[1/(x Sqrt[x + 4]), {x, 5, 21}] == a Log@3 + b Log@5 + c Log@7
(* 1/2 Log[15/7] == a Log[3] + b Log[5] + c Log[7] *)
PowerExpand@eq
(* 1/2 (Log[3] + Log[5] - Log[7]) == a Log[3] + b Log[5] + c Log[7] *)
Collect[Subtract @@ %, Log[_]] == 0
(* (1/2 - a) Log[3] + (1/2 - b) Log[5] + (-(1/2) - c) Log[7] == 0 *)
Cases[%, coe_ Log[_] :> Solve[coe == 0], Infinity] // Flatten
(* {a -> 1/2, b -> 1/2, c -> -(1/2)} *)
This method also handles the new added example Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log[3] + b Log[11].
answered Feb 12 at 8:29
xzczdxzczd
27.7k574257
$endgroup$
Not as automated as kglr's solution, but the following works:
eq = Integrate[1/(x Sqrt[x + 4]), {x, 5, 21}] == a Log@3 + b Log@5 + c Log@7
(* 1/2 Log[15/7] == a Log[3] + b Log[5] + c Log[7] *)
PowerExpand@eq
(* 1/2 (Log[3] + Log[5] - Log[7]) == a Log[3] + b Log[5] + c Log[7] *)
Collect[Subtract @@ %, Log[_]] == 0
(* (1/2 - a) Log[3] + (1/2 - b) Log[5] + (-(1/2) - c) Log[7] == 0 *)
Cases[%, coe_ Log[_] :> Solve[coe == 0], Infinity] // Flatten
(* {a -> 1/2, b -> 1/2, c -> -(1/2)} *)
This method also handles the new added example Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log[3] + b Log[11].
answered Feb 12 at 8:29
xzczdxzczd
27.7k574257
answered Feb 12 at 8:29
xzczdxzczd
27.7k574257
answered Feb 12 at 8:29
xzczdxzczd
27.7k574257
answered Feb 12 at 8:29
xzczdxzczd
27.7k574257
27.7k574257
$begingroup$
When I triedeq1 = Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log@3 + b Log@5; PowerExpand@eq1; Collect[Subtract @@ %, Log[_]] == 0; Cases[%, coe_ Log[_] :> Solve[coe == 0], Infinity] // FlattenI got{a -> 3/4, b -> 0}. I think, It is wrong.
$endgroup$
– minhthien_2016
Feb 12 at 9:02
$begingroup$
@minhthien_2016 The equation itself is wrong. TheLog@5should beLog@11.
$endgroup$
– xzczd
Feb 12 at 9:08
$begingroup$
Yes. I think, the answer, It can not be found.
$endgroup$
– minhthien_2016
Feb 12 at 9:09
$begingroup$
As already mentioned, this solution is not a fully automated solution, error handling should be done manually.
$endgroup$
– xzczd
Feb 12 at 9:13
$begingroup$
I am looking forward to a genenral method.
$endgroup$
– minhthien_2016
Feb 12 at 9:14
|
show 1 more comment
$begingroup$
When I triedeq1 = Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log@3 + b Log@5; PowerExpand@eq1; Collect[Subtract @@ %, Log[_]] == 0; Cases[%, coe_ Log[_] :> Solve[coe == 0], Infinity] // FlattenI got{a -> 3/4, b -> 0}. I think, It is wrong.
$endgroup$
– minhthien_2016
Feb 12 at 9:02
$begingroup$
@minhthien_2016 The equation itself is wrong. TheLog@5should beLog@11.
$endgroup$
– xzczd
Feb 12 at 9:08
$begingroup$
Yes. I think, the answer, It can not be found.
$endgroup$
– minhthien_2016
Feb 12 at 9:09
$begingroup$
As already mentioned, this solution is not a fully automated solution, error handling should be done manually.
$endgroup$
– xzczd
Feb 12 at 9:13
$begingroup$
I am looking forward to a genenral method.
$endgroup$
– minhthien_2016
Feb 12 at 9:14
$begingroup$
When I tried
eq1 = Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log@3 + b Log@5; PowerExpand@eq1; Collect[Subtract @@ %, Log[_]] == 0; Cases[%, coe_ Log[_] :> Solve[coe == 0], Infinity] // Flatten I got {a -> 3/4, b -> 0}. I think, It is wrong.$endgroup$
– minhthien_2016
Feb 12 at 9:02
$begingroup$
When I tried
eq1 = Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}] == a Log@3 + b Log@5; PowerExpand@eq1; Collect[Subtract @@ %, Log[_]] == 0; Cases[%, coe_ Log[_] :> Solve[coe == 0], Infinity] // Flatten I got {a -> 3/4, b -> 0}. I think, It is wrong.$endgroup$
– minhthien_2016
Feb 12 at 9:02
$begingroup$
@minhthien_2016 The equation itself is wrong. The
Log@5 should be Log@11.$endgroup$
– xzczd
Feb 12 at 9:08
$begingroup$
@minhthien_2016 The equation itself is wrong. The
Log@5 should be Log@11.$endgroup$
– xzczd
Feb 12 at 9:08
$begingroup$
Yes. I think, the answer, It can not be found.
$endgroup$
– minhthien_2016
Feb 12 at 9:09
$begingroup$
Yes. I think, the answer, It can not be found.
$endgroup$
– minhthien_2016
Feb 12 at 9:09
$begingroup$
As already mentioned, this solution is not a fully automated solution, error handling should be done manually.
$endgroup$
– xzczd
Feb 12 at 9:13
$begingroup$
As already mentioned, this solution is not a fully automated solution, error handling should be done manually.
$endgroup$
– xzczd
Feb 12 at 9:13
$begingroup$
I am looking forward to a genenral method.
$endgroup$
– minhthien_2016
Feb 12 at 9:14
$begingroup$
I am looking forward to a genenral method.
$endgroup$
– minhthien_2016
Feb 12 at 9:14
|
show 1 more comment
0
$begingroup$
I am interpreting your question to mean that you want logarithms of rational numbers to be expressed purely in terms of logarithms of primes. If so, one can use a replacement rule:
{Integrate[1/(x Sqrt[x + 4]), {x, 5, 21}],
Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}]} // Simplify
{1/2 Log[15/7], 1/4 Log[27/11]}
% /. Log[r_Rational] :> (Total[#2 Log[#1] & @@@ FactorInteger[Numerator[r]]] -
Total[#2 Log[#1] & @@@ FactorInteger[Denominator[r]]]) // Expand
{Log[3]/2 + Log[5]/2 - Log[7]/2, 3 Log[3]/4 - Log[11]/4}
Alternatively, one can use FindIntegerNullVector, similar to what was done in this answer:
-Rest[#]/First[#] &[FindIntegerNullVector[{1/2 Log[15/7], Log[3], Log[5], Log[7]}]]
{1/2, 1/2, -1/2}
-Rest[#]/First[#] &[FindIntegerNullVector[{1/4 Log[27/11], Log[3], Log[11]}]]
{3/4, -1/4}
answered Feb 18 at 1:26
J. M. is away♦J. M. is away
98.9k10311467
$endgroup$
add a comment |
0
$begingroup$
I am interpreting your question to mean that you want logarithms of rational numbers to be expressed purely in terms of logarithms of primes. If so, one can use a replacement rule:
{Integrate[1/(x Sqrt[x + 4]), {x, 5, 21}],
Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}]} // Simplify
{1/2 Log[15/7], 1/4 Log[27/11]}
% /. Log[r_Rational] :> (Total[#2 Log[#1] & @@@ FactorInteger[Numerator[r]]] -
Total[#2 Log[#1] & @@@ FactorInteger[Denominator[r]]]) // Expand
{Log[3]/2 + Log[5]/2 - Log[7]/2, 3 Log[3]/4 - Log[11]/4}
Alternatively, one can use FindIntegerNullVector, similar to what was done in this answer:
-Rest[#]/First[#] &[FindIntegerNullVector[{1/2 Log[15/7], Log[3], Log[5], Log[7]}]]
{1/2, 1/2, -1/2}
-Rest[#]/First[#] &[FindIntegerNullVector[{1/4 Log[27/11], Log[3], Log[11]}]]
{3/4, -1/4}
answered Feb 18 at 1:26
J. M. is away♦J. M. is away
98.9k10311467
$endgroup$
add a comment |
0
0
0
$begingroup$
I am interpreting your question to mean that you want logarithms of rational numbers to be expressed purely in terms of logarithms of primes. If so, one can use a replacement rule:
{Integrate[1/(x Sqrt[x + 4]), {x, 5, 21}],
Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}]} // Simplify
{1/2 Log[15/7], 1/4 Log[27/11]}
% /. Log[r_Rational] :> (Total[#2 Log[#1] & @@@ FactorInteger[Numerator[r]]] -
Total[#2 Log[#1] & @@@ FactorInteger[Denominator[r]]]) // Expand
{Log[3]/2 + Log[5]/2 - Log[7]/2, 3 Log[3]/4 - Log[11]/4}
Alternatively, one can use FindIntegerNullVector, similar to what was done in this answer:
-Rest[#]/First[#] &[FindIntegerNullVector[{1/2 Log[15/7], Log[3], Log[5], Log[7]}]]
{1/2, 1/2, -1/2}
-Rest[#]/First[#] &[FindIntegerNullVector[{1/4 Log[27/11], Log[3], Log[11]}]]
{3/4, -1/4}
answered Feb 18 at 1:26
J. M. is away♦J. M. is away
98.9k10311467
$endgroup$
I am interpreting your question to mean that you want logarithms of rational numbers to be expressed purely in terms of logarithms of primes. If so, one can use a replacement rule:
{Integrate[1/(x Sqrt[x + 4]), {x, 5, 21}],
Integrate[1/(x Sqrt[x + 16]), {x, 9, 33}]} // Simplify
{1/2 Log[15/7], 1/4 Log[27/11]}
% /. Log[r_Rational] :> (Total[#2 Log[#1] & @@@ FactorInteger[Numerator[r]]] -
Total[#2 Log[#1] & @@@ FactorInteger[Denominator[r]]]) // Expand
{Log[3]/2 + Log[5]/2 - Log[7]/2, 3 Log[3]/4 - Log[11]/4}
Alternatively, one can use FindIntegerNullVector, similar to what was done in this answer:
-Rest[#]/First[#] &[FindIntegerNullVector[{1/2 Log[15/7], Log[3], Log[5], Log[7]}]]
{1/2, 1/2, -1/2}
-Rest[#]/First[#] &[FindIntegerNullVector[{1/4 Log[27/11], Log[3], Log[11]}]]
{3/4, -1/4}
answered Feb 18 at 1:26
J. M. is away♦J. M. is away
98.9k10311467
edited Feb 18 at 1:44
answered Feb 18 at 1:26
J. M. is away♦J. M. is away
98.9k10311467
answered Feb 18 at 1:26
J. M. is away♦J. M. is away
98.9k10311467
answered Feb 18 at 1:26
J. M. is away♦J. M. is away
98.9k10311467
98.9k10311467
add a comment |
add a comment |
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$begingroup$
You're just calculating the integral rather than solving the equation in Maple. If such results are desired for you, simply
Integrate[1/(x Sqrt[x + 4]), {x, 5, 21}] // PowerExpand // Expand.$endgroup$
– xzczd
Feb 12 at 8:46
$begingroup$
Thank you very much.
$endgroup$
– minhthien_2016
Feb 12 at 8:52