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When does a function NOT have an antiderivative?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)How can you prove that a function has no closed form integral?Does L'Hôpital's work the other way?Finding volumes - when to use double integrals and triple integrals?Why does a function have to be bounded to be integrable?Proof that order of integration does not matter for non-continuous functionswhy we say this function have closed form while the other doesn't?Antiderivative of unbounded function?antiderivative of jump-discontinuous functionWhy is it legal to take the antiderivative of both sides of an equation?why $int sqrt(sin x)^2, mathrmdx = int |sin x| ,mathrmdx$Why does the antiderivative of $frac1x$ have to be $0$ at precisely $x=1$? (when $C = 0$)
$begingroup$
I know this question may sound naïve but why can't we write $int e^x^2 dx$ as $int e^2x dx$? The former does not have an antiderivative, while the latter has.
In light of this question, what are sufficient conditions for a function NOT to have an antiderivative. That is, do we need careful examination of a function to say it does not have an antiderivative or is there any way that once you see the function, you can right away say it does not have an antiderivative?
integration
asked 4 hours ago
RobRob
333110
$endgroup$
|
show 8 more comments
$begingroup$
I know this question may sound naïve but why can't we write $int e^x^2 dx$ as $int e^2x dx$? The former does not have an antiderivative, while the latter has.
In light of this question, what are sufficient conditions for a function NOT to have an antiderivative. That is, do we need careful examination of a function to say it does not have an antiderivative or is there any way that once you see the function, you can right away say it does not have an antiderivative?
integration
asked 4 hours ago
RobRob
333110
$endgroup$
6
$begingroup$
Maybe because $x^2$ isn't the same as $2x$?
$endgroup$
– Lord Shark the Unknown
4 hours ago
$begingroup$
Can't we write $a^b^c$ as $a^bc$?
$endgroup$
– Rob
4 hours ago
1
$begingroup$
$$e^x^2=e^xcdot xneq e^xcdot e^x = e^x+x=e^2x$$
$endgroup$
– Don Thousand
4 hours ago
$begingroup$
$$(a^b)^c=a^bcdot c$$$$a^(b^c)neq a^bcdot c$$
$endgroup$
– Don Thousand
4 hours ago
$begingroup$
$(e^x)^2$ would be where you use the rule that you're thinking of.
$endgroup$
– Tartaglia's Stutter
4 hours ago
|
show 8 more comments
$begingroup$
I know this question may sound naïve but why can't we write $int e^x^2 dx$ as $int e^2x dx$? The former does not have an antiderivative, while the latter has.
In light of this question, what are sufficient conditions for a function NOT to have an antiderivative. That is, do we need careful examination of a function to say it does not have an antiderivative or is there any way that once you see the function, you can right away say it does not have an antiderivative?
integration
asked 4 hours ago
RobRob
333110
$endgroup$
I know this question may sound naïve but why can't we write $int e^x^2 dx$ as $int e^2x dx$? The former does not have an antiderivative, while the latter has.
In light of this question, what are sufficient conditions for a function NOT to have an antiderivative. That is, do we need careful examination of a function to say it does not have an antiderivative or is there any way that once you see the function, you can right away say it does not have an antiderivative?
integration
integration
asked 4 hours ago
RobRob
333110
asked 4 hours ago
RobRob
333110
asked 4 hours ago
RobRob
333110
asked 4 hours ago
RobRob
333110
asked 4 hours ago
RobRob
333110
333110
6
$begingroup$
Maybe because $x^2$ isn't the same as $2x$?
$endgroup$
– Lord Shark the Unknown
4 hours ago
$begingroup$
Can't we write $a^b^c$ as $a^bc$?
$endgroup$
– Rob
4 hours ago
1
$begingroup$
$$e^x^2=e^xcdot xneq e^xcdot e^x = e^x+x=e^2x$$
$endgroup$
– Don Thousand
4 hours ago
$begingroup$
$$(a^b)^c=a^bcdot c$$$$a^(b^c)neq a^bcdot c$$
$endgroup$
– Don Thousand
4 hours ago
$begingroup$
$(e^x)^2$ would be where you use the rule that you're thinking of.
$endgroup$
– Tartaglia's Stutter
4 hours ago
|
show 8 more comments
6
$begingroup$
Maybe because $x^2$ isn't the same as $2x$?
$endgroup$
– Lord Shark the Unknown
4 hours ago
$begingroup$
Can't we write $a^b^c$ as $a^bc$?
$endgroup$
– Rob
4 hours ago
1
$begingroup$
$$e^x^2=e^xcdot xneq e^xcdot e^x = e^x+x=e^2x$$
$endgroup$
– Don Thousand
4 hours ago
$begingroup$
$$(a^b)^c=a^bcdot c$$$$a^(b^c)neq a^bcdot c$$
$endgroup$
– Don Thousand
4 hours ago
$begingroup$
$(e^x)^2$ would be where you use the rule that you're thinking of.
$endgroup$
– Tartaglia's Stutter
4 hours ago
6
6
$begingroup$
Maybe because $x^2$ isn't the same as $2x$?
$endgroup$
– Lord Shark the Unknown
4 hours ago
$begingroup$
Maybe because $x^2$ isn't the same as $2x$?
$endgroup$
– Lord Shark the Unknown
4 hours ago
$begingroup$
Can't we write $a^b^c$ as $a^bc$?
$endgroup$
– Rob
4 hours ago
$begingroup$
Can't we write $a^b^c$ as $a^bc$?
$endgroup$
– Rob
4 hours ago
1
1
$begingroup$
$$e^x^2=e^xcdot xneq e^xcdot e^x = e^x+x=e^2x$$
$endgroup$
– Don Thousand
4 hours ago
$begingroup$
$$e^x^2=e^xcdot xneq e^xcdot e^x = e^x+x=e^2x$$
$endgroup$
– Don Thousand
4 hours ago
$begingroup$
$$(a^b)^c=a^bcdot c$$$$a^(b^c)neq a^bcdot c$$
$endgroup$
– Don Thousand
4 hours ago
$begingroup$
$$(a^b)^c=a^bcdot c$$$$a^(b^c)neq a^bcdot c$$
$endgroup$
– Don Thousand
4 hours ago
$begingroup$
$(e^x)^2$ would be where you use the rule that you're thinking of.
$endgroup$
– Tartaglia's Stutter
4 hours ago
$begingroup$
$(e^x)^2$ would be where you use the rule that you're thinking of.
$endgroup$
– Tartaglia's Stutter
4 hours ago
|
show 8 more comments
4 Answers
4
active
oldest
votes
$begingroup$
As you might have realised, exponentiation is not associative:
$$left(a^bright)^c ne a^left(b^cright)$$
So what should $a^b^c$ mean? The convention is that exponentiation is right associative:
$$a^b^c = a^left(b^cright)$$
Because the otherwise left-associative exponentiation is just less useful and redundant, as it can be represented by multiplication inside the power (again as you might have realised):
$$a^bc = left(a^bright)^c$$
Wikipedia on associativity of exponentiation.
answered 3 hours ago
peterwhypeterwhy
12.4k21229
$endgroup$
add a comment |
$begingroup$
The exponential expression $a^b^c$ is equal to $a^(b^c)$. It is not equal to $(a^b)^c=a^bcdot c$ as you seem to think it is. In general an exponential is evaluated from right to left with the highest term evaluated first. That is to say
$$largex_0^x_1^x_2^dots^x_n=x_0^left(x_1^left(x_2^left(dots^(x_n)right)right)right)$$
For the second part of your question see this duplicate.
answered 4 hours ago
Peter ForemanPeter Foreman
8,5061321
$endgroup$
$begingroup$
I didn't think $a^(b^c) = (a^b)^c$. Once we put parentheses it is all clear. But I think one would not be wrong for interpreting $e^x^2$ as $e^2x$, since without parentheses it can mean both things . Am I correct?
$endgroup$
– Rob
4 hours ago
$begingroup$
NO! Without parentheses the expression $a^b^c$ is always equal to $a^left(b^cright)$. See here - en.wikipedia.org/wiki/Exponentiation#Identities_and_properties
$endgroup$
– Peter Foreman
3 hours ago
$begingroup$
I see! Well, I learnt something new that bothered me for a long time. So $e ^ x^2$ is understood as $e^(x^2)$. Also the link you sent, which I could not find at first, is very interesting but a little advanced for me.
$endgroup$
– Rob
3 hours ago
add a comment |
$begingroup$
To answer the titular question, there's a result in real analysis that shows that derivatives satisfy have the intermediate value property (just like continuous functions). It follows that a function that skips values cannot be the derivative of anything in the usual sense. This implies that functions with jump discontinuities (like the Heaviside step, for example) cannot be the derivative of anything.
answered 4 hours ago
AllawonderAllawonder
2,349617
$endgroup$
$begingroup$
Correct me if I am wrong, but that does not explain why $e^x^2$ has no antiderivative, right? I understand you are answering the titular question.
$endgroup$
– Rob
3 hours ago
$begingroup$
$e^x^2$ does have an antiderivative, but it can't be expressed in terms of elementary functions. Having an antiderivative and having an elementary antiderivative are not the same thing.
$endgroup$
– MathIsFun
3 hours ago
add a comment |
$begingroup$
Liouville's theorem:
In mathematics, Liouville's theorem, originally formulated by Joseph Liouville in 1833 to 1841, places an important restriction on antiderivatives that can be expressed as elementary functions.
The antiderivatives of certain elementary functions cannot themselves be expressed as elementary functions. A standard example of such a function is $e^-x^2$, whose antiderivative is (with a multiplier of a constant) the error function, familiar from statistics. Other examples include the functions $frac sin ( x ) x $ and $ x^x $.
From wikipedia. See the article for more details.
answered 33 mins ago
AccidentalFourierTransformAccidentalFourierTransform
1,462828
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As you might have realised, exponentiation is not associative:
$$left(a^bright)^c ne a^left(b^cright)$$
So what should $a^b^c$ mean? The convention is that exponentiation is right associative:
$$a^b^c = a^left(b^cright)$$
Because the otherwise left-associative exponentiation is just less useful and redundant, as it can be represented by multiplication inside the power (again as you might have realised):
$$a^bc = left(a^bright)^c$$
Wikipedia on associativity of exponentiation.
answered 3 hours ago
peterwhypeterwhy
12.4k21229
$endgroup$
add a comment |
$begingroup$
As you might have realised, exponentiation is not associative:
$$left(a^bright)^c ne a^left(b^cright)$$
So what should $a^b^c$ mean? The convention is that exponentiation is right associative:
$$a^b^c = a^left(b^cright)$$
Because the otherwise left-associative exponentiation is just less useful and redundant, as it can be represented by multiplication inside the power (again as you might have realised):
$$a^bc = left(a^bright)^c$$
Wikipedia on associativity of exponentiation.
answered 3 hours ago
peterwhypeterwhy
12.4k21229
$endgroup$
add a comment |
$begingroup$
As you might have realised, exponentiation is not associative:
$$left(a^bright)^c ne a^left(b^cright)$$
So what should $a^b^c$ mean? The convention is that exponentiation is right associative:
$$a^b^c = a^left(b^cright)$$
Because the otherwise left-associative exponentiation is just less useful and redundant, as it can be represented by multiplication inside the power (again as you might have realised):
$$a^bc = left(a^bright)^c$$
Wikipedia on associativity of exponentiation.
answered 3 hours ago
peterwhypeterwhy
12.4k21229
$endgroup$
As you might have realised, exponentiation is not associative:
$$left(a^bright)^c ne a^left(b^cright)$$
So what should $a^b^c$ mean? The convention is that exponentiation is right associative:
$$a^b^c = a^left(b^cright)$$
Because the otherwise left-associative exponentiation is just less useful and redundant, as it can be represented by multiplication inside the power (again as you might have realised):
$$a^bc = left(a^bright)^c$$
Wikipedia on associativity of exponentiation.
answered 3 hours ago
peterwhypeterwhy
12.4k21229
answered 3 hours ago
peterwhypeterwhy
12.4k21229
answered 3 hours ago
peterwhypeterwhy
12.4k21229
answered 3 hours ago
peterwhypeterwhy
12.4k21229
12.4k21229
add a comment |
add a comment |
$begingroup$
The exponential expression $a^b^c$ is equal to $a^(b^c)$. It is not equal to $(a^b)^c=a^bcdot c$ as you seem to think it is. In general an exponential is evaluated from right to left with the highest term evaluated first. That is to say
$$largex_0^x_1^x_2^dots^x_n=x_0^left(x_1^left(x_2^left(dots^(x_n)right)right)right)$$
For the second part of your question see this duplicate.
answered 4 hours ago
Peter ForemanPeter Foreman
8,5061321
$endgroup$
$begingroup$
I didn't think $a^(b^c) = (a^b)^c$. Once we put parentheses it is all clear. But I think one would not be wrong for interpreting $e^x^2$ as $e^2x$, since without parentheses it can mean both things . Am I correct?
$endgroup$
– Rob
4 hours ago
$begingroup$
NO! Without parentheses the expression $a^b^c$ is always equal to $a^left(b^cright)$. See here - en.wikipedia.org/wiki/Exponentiation#Identities_and_properties
$endgroup$
– Peter Foreman
3 hours ago
$begingroup$
I see! Well, I learnt something new that bothered me for a long time. So $e ^ x^2$ is understood as $e^(x^2)$. Also the link you sent, which I could not find at first, is very interesting but a little advanced for me.
$endgroup$
– Rob
3 hours ago
add a comment |
$begingroup$
The exponential expression $a^b^c$ is equal to $a^(b^c)$. It is not equal to $(a^b)^c=a^bcdot c$ as you seem to think it is. In general an exponential is evaluated from right to left with the highest term evaluated first. That is to say
$$largex_0^x_1^x_2^dots^x_n=x_0^left(x_1^left(x_2^left(dots^(x_n)right)right)right)$$
For the second part of your question see this duplicate.
answered 4 hours ago
Peter ForemanPeter Foreman
8,5061321
$endgroup$
$begingroup$
I didn't think $a^(b^c) = (a^b)^c$. Once we put parentheses it is all clear. But I think one would not be wrong for interpreting $e^x^2$ as $e^2x$, since without parentheses it can mean both things . Am I correct?
$endgroup$
– Rob
4 hours ago
$begingroup$
NO! Without parentheses the expression $a^b^c$ is always equal to $a^left(b^cright)$. See here - en.wikipedia.org/wiki/Exponentiation#Identities_and_properties
$endgroup$
– Peter Foreman
3 hours ago
$begingroup$
I see! Well, I learnt something new that bothered me for a long time. So $e ^ x^2$ is understood as $e^(x^2)$. Also the link you sent, which I could not find at first, is very interesting but a little advanced for me.
$endgroup$
– Rob
3 hours ago
add a comment |
$begingroup$
The exponential expression $a^b^c$ is equal to $a^(b^c)$. It is not equal to $(a^b)^c=a^bcdot c$ as you seem to think it is. In general an exponential is evaluated from right to left with the highest term evaluated first. That is to say
$$largex_0^x_1^x_2^dots^x_n=x_0^left(x_1^left(x_2^left(dots^(x_n)right)right)right)$$
For the second part of your question see this duplicate.
answered 4 hours ago
Peter ForemanPeter Foreman
8,5061321
$endgroup$
The exponential expression $a^b^c$ is equal to $a^(b^c)$. It is not equal to $(a^b)^c=a^bcdot c$ as you seem to think it is. In general an exponential is evaluated from right to left with the highest term evaluated first. That is to say
$$largex_0^x_1^x_2^dots^x_n=x_0^left(x_1^left(x_2^left(dots^(x_n)right)right)right)$$
For the second part of your question see this duplicate.
answered 4 hours ago
Peter ForemanPeter Foreman
8,5061321
answered 4 hours ago
Peter ForemanPeter Foreman
8,5061321
answered 4 hours ago
Peter ForemanPeter Foreman
8,5061321
answered 4 hours ago
Peter ForemanPeter Foreman
8,5061321
8,5061321
$begingroup$
I didn't think $a^(b^c) = (a^b)^c$. Once we put parentheses it is all clear. But I think one would not be wrong for interpreting $e^x^2$ as $e^2x$, since without parentheses it can mean both things . Am I correct?
$endgroup$
– Rob
4 hours ago
$begingroup$
NO! Without parentheses the expression $a^b^c$ is always equal to $a^left(b^cright)$. See here - en.wikipedia.org/wiki/Exponentiation#Identities_and_properties
$endgroup$
– Peter Foreman
3 hours ago
$begingroup$
I see! Well, I learnt something new that bothered me for a long time. So $e ^ x^2$ is understood as $e^(x^2)$. Also the link you sent, which I could not find at first, is very interesting but a little advanced for me.
$endgroup$
– Rob
3 hours ago
add a comment |
$begingroup$
I didn't think $a^(b^c) = (a^b)^c$. Once we put parentheses it is all clear. But I think one would not be wrong for interpreting $e^x^2$ as $e^2x$, since without parentheses it can mean both things . Am I correct?
$endgroup$
– Rob
4 hours ago
$begingroup$
NO! Without parentheses the expression $a^b^c$ is always equal to $a^left(b^cright)$. See here - en.wikipedia.org/wiki/Exponentiation#Identities_and_properties
$endgroup$
– Peter Foreman
3 hours ago
$begingroup$
I see! Well, I learnt something new that bothered me for a long time. So $e ^ x^2$ is understood as $e^(x^2)$. Also the link you sent, which I could not find at first, is very interesting but a little advanced for me.
$endgroup$
– Rob
3 hours ago
$begingroup$
I didn't think $a^(b^c) = (a^b)^c$. Once we put parentheses it is all clear. But I think one would not be wrong for interpreting $e^x^2$ as $e^2x$, since without parentheses it can mean both things . Am I correct?
$endgroup$
– Rob
4 hours ago
$begingroup$
I didn't think $a^(b^c) = (a^b)^c$. Once we put parentheses it is all clear. But I think one would not be wrong for interpreting $e^x^2$ as $e^2x$, since without parentheses it can mean both things . Am I correct?
$endgroup$
– Rob
4 hours ago
$begingroup$
NO! Without parentheses the expression $a^b^c$ is always equal to $a^left(b^cright)$. See here - en.wikipedia.org/wiki/Exponentiation#Identities_and_properties
$endgroup$
– Peter Foreman
3 hours ago
$begingroup$
NO! Without parentheses the expression $a^b^c$ is always equal to $a^left(b^cright)$. See here - en.wikipedia.org/wiki/Exponentiation#Identities_and_properties
$endgroup$
– Peter Foreman
3 hours ago
$begingroup$
I see! Well, I learnt something new that bothered me for a long time. So $e ^ x^2$ is understood as $e^(x^2)$. Also the link you sent, which I could not find at first, is very interesting but a little advanced for me.
$endgroup$
– Rob
3 hours ago
$begingroup$
I see! Well, I learnt something new that bothered me for a long time. So $e ^ x^2$ is understood as $e^(x^2)$. Also the link you sent, which I could not find at first, is very interesting but a little advanced for me.
$endgroup$
– Rob
3 hours ago
add a comment |
$begingroup$
To answer the titular question, there's a result in real analysis that shows that derivatives satisfy have the intermediate value property (just like continuous functions). It follows that a function that skips values cannot be the derivative of anything in the usual sense. This implies that functions with jump discontinuities (like the Heaviside step, for example) cannot be the derivative of anything.
answered 4 hours ago
AllawonderAllawonder
2,349617
$endgroup$
$begingroup$
Correct me if I am wrong, but that does not explain why $e^x^2$ has no antiderivative, right? I understand you are answering the titular question.
$endgroup$
– Rob
3 hours ago
$begingroup$
$e^x^2$ does have an antiderivative, but it can't be expressed in terms of elementary functions. Having an antiderivative and having an elementary antiderivative are not the same thing.
$endgroup$
– MathIsFun
3 hours ago
add a comment |
$begingroup$
To answer the titular question, there's a result in real analysis that shows that derivatives satisfy have the intermediate value property (just like continuous functions). It follows that a function that skips values cannot be the derivative of anything in the usual sense. This implies that functions with jump discontinuities (like the Heaviside step, for example) cannot be the derivative of anything.
answered 4 hours ago
AllawonderAllawonder
2,349617
$endgroup$
$begingroup$
Correct me if I am wrong, but that does not explain why $e^x^2$ has no antiderivative, right? I understand you are answering the titular question.
$endgroup$
– Rob
3 hours ago
$begingroup$
$e^x^2$ does have an antiderivative, but it can't be expressed in terms of elementary functions. Having an antiderivative and having an elementary antiderivative are not the same thing.
$endgroup$
– MathIsFun
3 hours ago
add a comment |
$begingroup$
To answer the titular question, there's a result in real analysis that shows that derivatives satisfy have the intermediate value property (just like continuous functions). It follows that a function that skips values cannot be the derivative of anything in the usual sense. This implies that functions with jump discontinuities (like the Heaviside step, for example) cannot be the derivative of anything.
answered 4 hours ago
AllawonderAllawonder
2,349617
$endgroup$
To answer the titular question, there's a result in real analysis that shows that derivatives satisfy have the intermediate value property (just like continuous functions). It follows that a function that skips values cannot be the derivative of anything in the usual sense. This implies that functions with jump discontinuities (like the Heaviside step, for example) cannot be the derivative of anything.
answered 4 hours ago
AllawonderAllawonder
2,349617
answered 4 hours ago
AllawonderAllawonder
2,349617
answered 4 hours ago
AllawonderAllawonder
2,349617
answered 4 hours ago
AllawonderAllawonder
2,349617
2,349617
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Correct me if I am wrong, but that does not explain why $e^x^2$ has no antiderivative, right? I understand you are answering the titular question.
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– Rob
3 hours ago
$begingroup$
$e^x^2$ does have an antiderivative, but it can't be expressed in terms of elementary functions. Having an antiderivative and having an elementary antiderivative are not the same thing.
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– MathIsFun
3 hours ago
add a comment |
$begingroup$
Correct me if I am wrong, but that does not explain why $e^x^2$ has no antiderivative, right? I understand you are answering the titular question.
$endgroup$
– Rob
3 hours ago
$begingroup$
$e^x^2$ does have an antiderivative, but it can't be expressed in terms of elementary functions. Having an antiderivative and having an elementary antiderivative are not the same thing.
$endgroup$
– MathIsFun
3 hours ago
$begingroup$
Correct me if I am wrong, but that does not explain why $e^x^2$ has no antiderivative, right? I understand you are answering the titular question.
$endgroup$
– Rob
3 hours ago
$begingroup$
Correct me if I am wrong, but that does not explain why $e^x^2$ has no antiderivative, right? I understand you are answering the titular question.
$endgroup$
– Rob
3 hours ago
$begingroup$
$e^x^2$ does have an antiderivative, but it can't be expressed in terms of elementary functions. Having an antiderivative and having an elementary antiderivative are not the same thing.
$endgroup$
– MathIsFun
3 hours ago
$begingroup$
$e^x^2$ does have an antiderivative, but it can't be expressed in terms of elementary functions. Having an antiderivative and having an elementary antiderivative are not the same thing.
$endgroup$
– MathIsFun
3 hours ago
add a comment |
$begingroup$
Liouville's theorem:
In mathematics, Liouville's theorem, originally formulated by Joseph Liouville in 1833 to 1841, places an important restriction on antiderivatives that can be expressed as elementary functions.
The antiderivatives of certain elementary functions cannot themselves be expressed as elementary functions. A standard example of such a function is $e^-x^2$, whose antiderivative is (with a multiplier of a constant) the error function, familiar from statistics. Other examples include the functions $frac sin ( x ) x $ and $ x^x $.
From wikipedia. See the article for more details.
answered 33 mins ago
AccidentalFourierTransformAccidentalFourierTransform
1,462828
$endgroup$
add a comment |
$begingroup$
Liouville's theorem:
In mathematics, Liouville's theorem, originally formulated by Joseph Liouville in 1833 to 1841, places an important restriction on antiderivatives that can be expressed as elementary functions.
The antiderivatives of certain elementary functions cannot themselves be expressed as elementary functions. A standard example of such a function is $e^-x^2$, whose antiderivative is (with a multiplier of a constant) the error function, familiar from statistics. Other examples include the functions $frac sin ( x ) x $ and $ x^x $.
From wikipedia. See the article for more details.
answered 33 mins ago
AccidentalFourierTransformAccidentalFourierTransform
1,462828
$endgroup$
add a comment |
$begingroup$
Liouville's theorem:
In mathematics, Liouville's theorem, originally formulated by Joseph Liouville in 1833 to 1841, places an important restriction on antiderivatives that can be expressed as elementary functions.
The antiderivatives of certain elementary functions cannot themselves be expressed as elementary functions. A standard example of such a function is $e^-x^2$, whose antiderivative is (with a multiplier of a constant) the error function, familiar from statistics. Other examples include the functions $frac sin ( x ) x $ and $ x^x $.
From wikipedia. See the article for more details.
answered 33 mins ago
AccidentalFourierTransformAccidentalFourierTransform
1,462828
$endgroup$
Liouville's theorem:
In mathematics, Liouville's theorem, originally formulated by Joseph Liouville in 1833 to 1841, places an important restriction on antiderivatives that can be expressed as elementary functions.
The antiderivatives of certain elementary functions cannot themselves be expressed as elementary functions. A standard example of such a function is $e^-x^2$, whose antiderivative is (with a multiplier of a constant) the error function, familiar from statistics. Other examples include the functions $frac sin ( x ) x $ and $ x^x $.
From wikipedia. See the article for more details.
answered 33 mins ago
AccidentalFourierTransformAccidentalFourierTransform
1,462828
answered 33 mins ago
AccidentalFourierTransformAccidentalFourierTransform
1,462828
answered 33 mins ago
AccidentalFourierTransformAccidentalFourierTransform
1,462828
answered 33 mins ago
AccidentalFourierTransformAccidentalFourierTransform
1,462828
1,462828
add a comment |
add a comment |
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6
$begingroup$
Maybe because $x^2$ isn't the same as $2x$?
$endgroup$
– Lord Shark the Unknown
4 hours ago
$begingroup$
Can't we write $a^b^c$ as $a^bc$?
$endgroup$
– Rob
4 hours ago
1
$begingroup$
$$e^x^2=e^xcdot xneq e^xcdot e^x = e^x+x=e^2x$$
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– Don Thousand
4 hours ago
$begingroup$
$$(a^b)^c=a^bcdot c$$$$a^(b^c)neq a^bcdot c$$
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– Don Thousand
4 hours ago
$begingroup$
$(e^x)^2$ would be where you use the rule that you're thinking of.
$endgroup$
– Tartaglia's Stutter
4 hours ago