How to solve a power/polynomial equation












0












$begingroup$


Is there a method to solve such equations : $AB^{x} + C x = D$ ? with $A$, $B$, $C$ and $D$ real parameters and $xin mathbb{R}$ and $B>0$?






real-analysis algebra-precalculus exponential-function







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    what is $x$? real?
    $endgroup$
    – José Alejandro Aburto Araneda
    Dec 29 '18 at 16:58






  • 1




    $begingroup$
    Yes indeed, I corrected the question thank you.
    $endgroup$
    – AlexC75
    Dec 29 '18 at 16:59










  • $begingroup$
    Is $B$ assumed to be positive? If not then $B^x$ becomes problematic...
    $endgroup$
    – Servaes
    Dec 29 '18 at 17:35










  • $begingroup$
    Yes B is assumed to B positive.
    $endgroup$
    – AlexC75
    Dec 29 '18 at 17:36
















0












$begingroup$


Is there a method to solve such equations : $AB^{x} + C x = D$ ? with $A$, $B$, $C$ and $D$ real parameters and $xin mathbb{R}$ and $B>0$?






real-analysis algebra-precalculus exponential-function







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    what is $x$? real?
    $endgroup$
    – José Alejandro Aburto Araneda
    Dec 29 '18 at 16:58






  • 1




    $begingroup$
    Yes indeed, I corrected the question thank you.
    $endgroup$
    – AlexC75
    Dec 29 '18 at 16:59










  • $begingroup$
    Is $B$ assumed to be positive? If not then $B^x$ becomes problematic...
    $endgroup$
    – Servaes
    Dec 29 '18 at 17:35










  • $begingroup$
    Yes B is assumed to B positive.
    $endgroup$
    – AlexC75
    Dec 29 '18 at 17:36














0












0








0





$begingroup$


Is there a method to solve such equations : $AB^{x} + C x = D$ ? with $A$, $B$, $C$ and $D$ real parameters and $xin mathbb{R}$ and $B>0$?






real-analysis algebra-precalculus exponential-function







share|cite|improve this question











$endgroup$




Is there a method to solve such equations : $AB^{x} + C x = D$ ? with $A$, $B$, $C$ and $D$ real parameters and $xin mathbb{R}$ and $B>0$?






real-analysis algebra-precalculus exponential-function




real-analysis algebra-precalculus exponential-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 23:49









Servaes

30.9k342101




30.9k342101










asked Dec 29 '18 at 16:57









AlexC75AlexC75

216




216








  • 2




    $begingroup$
    what is $x$? real?
    $endgroup$
    – José Alejandro Aburto Araneda
    Dec 29 '18 at 16:58






  • 1




    $begingroup$
    Yes indeed, I corrected the question thank you.
    $endgroup$
    – AlexC75
    Dec 29 '18 at 16:59










  • $begingroup$
    Is $B$ assumed to be positive? If not then $B^x$ becomes problematic...
    $endgroup$
    – Servaes
    Dec 29 '18 at 17:35










  • $begingroup$
    Yes B is assumed to B positive.
    $endgroup$
    – AlexC75
    Dec 29 '18 at 17:36














  • 2




    $begingroup$
    what is $x$? real?
    $endgroup$
    – José Alejandro Aburto Araneda
    Dec 29 '18 at 16:58






  • 1




    $begingroup$
    Yes indeed, I corrected the question thank you.
    $endgroup$
    – AlexC75
    Dec 29 '18 at 16:59










  • $begingroup$
    Is $B$ assumed to be positive? If not then $B^x$ becomes problematic...
    $endgroup$
    – Servaes
    Dec 29 '18 at 17:35










  • $begingroup$
    Yes B is assumed to B positive.
    $endgroup$
    – AlexC75
    Dec 29 '18 at 17:36








2




2




$begingroup$
what is $x$? real?
$endgroup$
– José Alejandro Aburto Araneda
Dec 29 '18 at 16:58




$begingroup$
what is $x$? real?
$endgroup$
– José Alejandro Aburto Araneda
Dec 29 '18 at 16:58




1




1




$begingroup$
Yes indeed, I corrected the question thank you.
$endgroup$
– AlexC75
Dec 29 '18 at 16:59




$begingroup$
Yes indeed, I corrected the question thank you.
$endgroup$
– AlexC75
Dec 29 '18 at 16:59












$begingroup$
Is $B$ assumed to be positive? If not then $B^x$ becomes problematic...
$endgroup$
– Servaes
Dec 29 '18 at 17:35




$begingroup$
Is $B$ assumed to be positive? If not then $B^x$ becomes problematic...
$endgroup$
– Servaes
Dec 29 '18 at 17:35












$begingroup$
Yes B is assumed to B positive.
$endgroup$
– AlexC75
Dec 29 '18 at 17:36




$begingroup$
Yes B is assumed to B positive.
$endgroup$
– AlexC75
Dec 29 '18 at 17:36










1 Answer
1






active

oldest

votes


















1












$begingroup$

First, three degenerate cases;




  1. If $C=0$ and $B=1$ then the equation reduces to $A=D$, and there is nothing to solve.

  2. If $C=0$ and $Bneq1$ then $AB^x=D$ and hence $x=frac{ln{frac{D}{A}}}{ln{B}}$.

  3. If $Cneq0$ and $B=1$ then $A+Cx=D$ and hence $x=frac{D-A}{C}$.


If $Cneq0$ and $Bneq1$ then setting $a:=frac{A}{C}$ and $d:=frac{D}{C}$ yields
$$aB^x=d-x.$$
Setting $y:=d-x$ and $v:=B^{-1}$ and $u:=aB^d$ then yields
$$y=aB^{d-y}=aB^dB^{-y}=uv^y=ue^{yln{v}},$$
where $e$ denotes Euler's constant. Finally setting $z:=-yln{v}$ and rearranging yields the equation
$$ze^z=-uln{v},$$
which has no 'elementary' closed form solution in general. The solution to this is precisely what the Lambert W function denotes. That is, by definition we have $z=W(-uln{v})$. Substituting everything back yields
$$-left(frac{D}{C}-xright)ln{B^{-1}}=Wleft(-frac{A}{C}B^{frac{D}{C}}ln{B^{-1}}right),$$
and isolating $x$ shows that
$$x=frac{D}{C}-frac{1}{ln{B}}Wleft(frac{A}{C}B^{frac{D}{C}}ln{B}right).$$
Finding exact values of the Lambert W function in terms of elementary functions is in general not possible, though of course there are practical algorithms for numerical approximations.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you very much for this, very interesting stuff. What would be the conditions on A,B,C and D for which $Wleft(frac{A}{C}B^{frac{D}{C}}ln{B^{-1}}right)$ could be expressed in terms of elementary function ?
    $endgroup$
    – AlexC75
    Dec 29 '18 at 17:59












  • $begingroup$
    See the wikipedia section on special values for a few known values of $W(z)$, which you can then translate to conditions on $A$, $B$, $C$ and $D$. One condition that particularly jumps out is $A=0$, and then obviously $x=frac{D}{C}$, but otherwise the expression seems too convoluted to say anything nice about. But I could be wrong.
    $endgroup$
    – Servaes
    Dec 29 '18 at 18:03












  • $begingroup$
    Thanks a lot. Do you know if this list is exhaustive ?
    $endgroup$
    – AlexC75
    Dec 29 '18 at 18:11












  • $begingroup$
    I have no doubt there are much more extensive lists, though I don't know where to find them. If there are particular values you are interested in, perhaps wolfram alpha can help you.
    $endgroup$
    – Servaes
    Dec 29 '18 at 18:21






  • 1




    $begingroup$
    Should it not be $ze^z = - u ln(v) $ ?
    $endgroup$
    – AlexC75
    Dec 30 '18 at 17:28














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1 Answer
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1 Answer
1






active

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active

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active

oldest

votes









1












$begingroup$

First, three degenerate cases;




  1. If $C=0$ and $B=1$ then the equation reduces to $A=D$, and there is nothing to solve.

  2. If $C=0$ and $Bneq1$ then $AB^x=D$ and hence $x=frac{ln{frac{D}{A}}}{ln{B}}$.

  3. If $Cneq0$ and $B=1$ then $A+Cx=D$ and hence $x=frac{D-A}{C}$.


If $Cneq0$ and $Bneq1$ then setting $a:=frac{A}{C}$ and $d:=frac{D}{C}$ yields
$$aB^x=d-x.$$
Setting $y:=d-x$ and $v:=B^{-1}$ and $u:=aB^d$ then yields
$$y=aB^{d-y}=aB^dB^{-y}=uv^y=ue^{yln{v}},$$
where $e$ denotes Euler's constant. Finally setting $z:=-yln{v}$ and rearranging yields the equation
$$ze^z=-uln{v},$$
which has no 'elementary' closed form solution in general. The solution to this is precisely what the Lambert W function denotes. That is, by definition we have $z=W(-uln{v})$. Substituting everything back yields
$$-left(frac{D}{C}-xright)ln{B^{-1}}=Wleft(-frac{A}{C}B^{frac{D}{C}}ln{B^{-1}}right),$$
and isolating $x$ shows that
$$x=frac{D}{C}-frac{1}{ln{B}}Wleft(frac{A}{C}B^{frac{D}{C}}ln{B}right).$$
Finding exact values of the Lambert W function in terms of elementary functions is in general not possible, though of course there are practical algorithms for numerical approximations.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you very much for this, very interesting stuff. What would be the conditions on A,B,C and D for which $Wleft(frac{A}{C}B^{frac{D}{C}}ln{B^{-1}}right)$ could be expressed in terms of elementary function ?
    $endgroup$
    – AlexC75
    Dec 29 '18 at 17:59












  • $begingroup$
    See the wikipedia section on special values for a few known values of $W(z)$, which you can then translate to conditions on $A$, $B$, $C$ and $D$. One condition that particularly jumps out is $A=0$, and then obviously $x=frac{D}{C}$, but otherwise the expression seems too convoluted to say anything nice about. But I could be wrong.
    $endgroup$
    – Servaes
    Dec 29 '18 at 18:03












  • $begingroup$
    Thanks a lot. Do you know if this list is exhaustive ?
    $endgroup$
    – AlexC75
    Dec 29 '18 at 18:11












  • $begingroup$
    I have no doubt there are much more extensive lists, though I don't know where to find them. If there are particular values you are interested in, perhaps wolfram alpha can help you.
    $endgroup$
    – Servaes
    Dec 29 '18 at 18:21






  • 1




    $begingroup$
    Should it not be $ze^z = - u ln(v) $ ?
    $endgroup$
    – AlexC75
    Dec 30 '18 at 17:28


















1












$begingroup$

First, three degenerate cases;




  1. If $C=0$ and $B=1$ then the equation reduces to $A=D$, and there is nothing to solve.

  2. If $C=0$ and $Bneq1$ then $AB^x=D$ and hence $x=frac{ln{frac{D}{A}}}{ln{B}}$.

  3. If $Cneq0$ and $B=1$ then $A+Cx=D$ and hence $x=frac{D-A}{C}$.


If $Cneq0$ and $Bneq1$ then setting $a:=frac{A}{C}$ and $d:=frac{D}{C}$ yields
$$aB^x=d-x.$$
Setting $y:=d-x$ and $v:=B^{-1}$ and $u:=aB^d$ then yields
$$y=aB^{d-y}=aB^dB^{-y}=uv^y=ue^{yln{v}},$$
where $e$ denotes Euler's constant. Finally setting $z:=-yln{v}$ and rearranging yields the equation
$$ze^z=-uln{v},$$
which has no 'elementary' closed form solution in general. The solution to this is precisely what the Lambert W function denotes. That is, by definition we have $z=W(-uln{v})$. Substituting everything back yields
$$-left(frac{D}{C}-xright)ln{B^{-1}}=Wleft(-frac{A}{C}B^{frac{D}{C}}ln{B^{-1}}right),$$
and isolating $x$ shows that
$$x=frac{D}{C}-frac{1}{ln{B}}Wleft(frac{A}{C}B^{frac{D}{C}}ln{B}right).$$
Finding exact values of the Lambert W function in terms of elementary functions is in general not possible, though of course there are practical algorithms for numerical approximations.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you very much for this, very interesting stuff. What would be the conditions on A,B,C and D for which $Wleft(frac{A}{C}B^{frac{D}{C}}ln{B^{-1}}right)$ could be expressed in terms of elementary function ?
    $endgroup$
    – AlexC75
    Dec 29 '18 at 17:59












  • $begingroup$
    See the wikipedia section on special values for a few known values of $W(z)$, which you can then translate to conditions on $A$, $B$, $C$ and $D$. One condition that particularly jumps out is $A=0$, and then obviously $x=frac{D}{C}$, but otherwise the expression seems too convoluted to say anything nice about. But I could be wrong.
    $endgroup$
    – Servaes
    Dec 29 '18 at 18:03












  • $begingroup$
    Thanks a lot. Do you know if this list is exhaustive ?
    $endgroup$
    – AlexC75
    Dec 29 '18 at 18:11












  • $begingroup$
    I have no doubt there are much more extensive lists, though I don't know where to find them. If there are particular values you are interested in, perhaps wolfram alpha can help you.
    $endgroup$
    – Servaes
    Dec 29 '18 at 18:21






  • 1




    $begingroup$
    Should it not be $ze^z = - u ln(v) $ ?
    $endgroup$
    – AlexC75
    Dec 30 '18 at 17:28
















1












1








1





$begingroup$

First, three degenerate cases;




  1. If $C=0$ and $B=1$ then the equation reduces to $A=D$, and there is nothing to solve.

  2. If $C=0$ and $Bneq1$ then $AB^x=D$ and hence $x=frac{ln{frac{D}{A}}}{ln{B}}$.

  3. If $Cneq0$ and $B=1$ then $A+Cx=D$ and hence $x=frac{D-A}{C}$.


If $Cneq0$ and $Bneq1$ then setting $a:=frac{A}{C}$ and $d:=frac{D}{C}$ yields
$$aB^x=d-x.$$
Setting $y:=d-x$ and $v:=B^{-1}$ and $u:=aB^d$ then yields
$$y=aB^{d-y}=aB^dB^{-y}=uv^y=ue^{yln{v}},$$
where $e$ denotes Euler's constant. Finally setting $z:=-yln{v}$ and rearranging yields the equation
$$ze^z=-uln{v},$$
which has no 'elementary' closed form solution in general. The solution to this is precisely what the Lambert W function denotes. That is, by definition we have $z=W(-uln{v})$. Substituting everything back yields
$$-left(frac{D}{C}-xright)ln{B^{-1}}=Wleft(-frac{A}{C}B^{frac{D}{C}}ln{B^{-1}}right),$$
and isolating $x$ shows that
$$x=frac{D}{C}-frac{1}{ln{B}}Wleft(frac{A}{C}B^{frac{D}{C}}ln{B}right).$$
Finding exact values of the Lambert W function in terms of elementary functions is in general not possible, though of course there are practical algorithms for numerical approximations.






share|cite|improve this answer











$endgroup$



First, three degenerate cases;




  1. If $C=0$ and $B=1$ then the equation reduces to $A=D$, and there is nothing to solve.

  2. If $C=0$ and $Bneq1$ then $AB^x=D$ and hence $x=frac{ln{frac{D}{A}}}{ln{B}}$.

  3. If $Cneq0$ and $B=1$ then $A+Cx=D$ and hence $x=frac{D-A}{C}$.


If $Cneq0$ and $Bneq1$ then setting $a:=frac{A}{C}$ and $d:=frac{D}{C}$ yields
$$aB^x=d-x.$$
Setting $y:=d-x$ and $v:=B^{-1}$ and $u:=aB^d$ then yields
$$y=aB^{d-y}=aB^dB^{-y}=uv^y=ue^{yln{v}},$$
where $e$ denotes Euler's constant. Finally setting $z:=-yln{v}$ and rearranging yields the equation
$$ze^z=-uln{v},$$
which has no 'elementary' closed form solution in general. The solution to this is precisely what the Lambert W function denotes. That is, by definition we have $z=W(-uln{v})$. Substituting everything back yields
$$-left(frac{D}{C}-xright)ln{B^{-1}}=Wleft(-frac{A}{C}B^{frac{D}{C}}ln{B^{-1}}right),$$
and isolating $x$ shows that
$$x=frac{D}{C}-frac{1}{ln{B}}Wleft(frac{A}{C}B^{frac{D}{C}}ln{B}right).$$
Finding exact values of the Lambert W function in terms of elementary functions is in general not possible, though of course there are practical algorithms for numerical approximations.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 1 at 21:09

























answered Dec 29 '18 at 17:48









ServaesServaes

30.9k342101




30.9k342101












  • $begingroup$
    Thank you very much for this, very interesting stuff. What would be the conditions on A,B,C and D for which $Wleft(frac{A}{C}B^{frac{D}{C}}ln{B^{-1}}right)$ could be expressed in terms of elementary function ?
    $endgroup$
    – AlexC75
    Dec 29 '18 at 17:59












  • $begingroup$
    See the wikipedia section on special values for a few known values of $W(z)$, which you can then translate to conditions on $A$, $B$, $C$ and $D$. One condition that particularly jumps out is $A=0$, and then obviously $x=frac{D}{C}$, but otherwise the expression seems too convoluted to say anything nice about. But I could be wrong.
    $endgroup$
    – Servaes
    Dec 29 '18 at 18:03












  • $begingroup$
    Thanks a lot. Do you know if this list is exhaustive ?
    $endgroup$
    – AlexC75
    Dec 29 '18 at 18:11












  • $begingroup$
    I have no doubt there are much more extensive lists, though I don't know where to find them. If there are particular values you are interested in, perhaps wolfram alpha can help you.
    $endgroup$
    – Servaes
    Dec 29 '18 at 18:21






  • 1




    $begingroup$
    Should it not be $ze^z = - u ln(v) $ ?
    $endgroup$
    – AlexC75
    Dec 30 '18 at 17:28




















  • $begingroup$
    Thank you very much for this, very interesting stuff. What would be the conditions on A,B,C and D for which $Wleft(frac{A}{C}B^{frac{D}{C}}ln{B^{-1}}right)$ could be expressed in terms of elementary function ?
    $endgroup$
    – AlexC75
    Dec 29 '18 at 17:59












  • $begingroup$
    See the wikipedia section on special values for a few known values of $W(z)$, which you can then translate to conditions on $A$, $B$, $C$ and $D$. One condition that particularly jumps out is $A=0$, and then obviously $x=frac{D}{C}$, but otherwise the expression seems too convoluted to say anything nice about. But I could be wrong.
    $endgroup$
    – Servaes
    Dec 29 '18 at 18:03












  • $begingroup$
    Thanks a lot. Do you know if this list is exhaustive ?
    $endgroup$
    – AlexC75
    Dec 29 '18 at 18:11












  • $begingroup$
    I have no doubt there are much more extensive lists, though I don't know where to find them. If there are particular values you are interested in, perhaps wolfram alpha can help you.
    $endgroup$
    – Servaes
    Dec 29 '18 at 18:21






  • 1




    $begingroup$
    Should it not be $ze^z = - u ln(v) $ ?
    $endgroup$
    – AlexC75
    Dec 30 '18 at 17:28


















$begingroup$
Thank you very much for this, very interesting stuff. What would be the conditions on A,B,C and D for which $Wleft(frac{A}{C}B^{frac{D}{C}}ln{B^{-1}}right)$ could be expressed in terms of elementary function ?
$endgroup$
– AlexC75
Dec 29 '18 at 17:59






$begingroup$
Thank you very much for this, very interesting stuff. What would be the conditions on A,B,C and D for which $Wleft(frac{A}{C}B^{frac{D}{C}}ln{B^{-1}}right)$ could be expressed in terms of elementary function ?
$endgroup$
– AlexC75
Dec 29 '18 at 17:59














$begingroup$
See the wikipedia section on special values for a few known values of $W(z)$, which you can then translate to conditions on $A$, $B$, $C$ and $D$. One condition that particularly jumps out is $A=0$, and then obviously $x=frac{D}{C}$, but otherwise the expression seems too convoluted to say anything nice about. But I could be wrong.
$endgroup$
– Servaes
Dec 29 '18 at 18:03






$begingroup$
See the wikipedia section on special values for a few known values of $W(z)$, which you can then translate to conditions on $A$, $B$, $C$ and $D$. One condition that particularly jumps out is $A=0$, and then obviously $x=frac{D}{C}$, but otherwise the expression seems too convoluted to say anything nice about. But I could be wrong.
$endgroup$
– Servaes
Dec 29 '18 at 18:03














$begingroup$
Thanks a lot. Do you know if this list is exhaustive ?
$endgroup$
– AlexC75
Dec 29 '18 at 18:11






$begingroup$
Thanks a lot. Do you know if this list is exhaustive ?
$endgroup$
– AlexC75
Dec 29 '18 at 18:11














$begingroup$
I have no doubt there are much more extensive lists, though I don't know where to find them. If there are particular values you are interested in, perhaps wolfram alpha can help you.
$endgroup$
– Servaes
Dec 29 '18 at 18:21




$begingroup$
I have no doubt there are much more extensive lists, though I don't know where to find them. If there are particular values you are interested in, perhaps wolfram alpha can help you.
$endgroup$
– Servaes
Dec 29 '18 at 18:21




1




1




$begingroup$
Should it not be $ze^z = - u ln(v) $ ?
$endgroup$
– AlexC75
Dec 30 '18 at 17:28






$begingroup$
Should it not be $ze^z = - u ln(v) $ ?
$endgroup$
– AlexC75
Dec 30 '18 at 17:28




















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Le Mesnil-Réaume

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Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
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Ida-Boy-Ed-Garten

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Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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web3.py web3.isConnected() returns false always

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1 I am trying an tutorial on web3.py so i created a profile in infura and used its project id.But it shows false always when i try web3.isConnected() PS C:UsersHP> python Python 3.6.3 (v3.6.3:2c5fed8, Oct 3 2017, 18:11:49) [MSC v.1900 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> from web3 import Web3 >>> infura_url = "mainnet.infura.io/v3/******************************" >>> web3 = Web3(Web3.HTTPProvider(infura_url)) >>> web3.isConnected() False web3.py share | improve this question
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