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 $C = 0$)
Noise in Eigenvalues plot
Table formatting with tabularx?
Can gravitational waves pass through a black hole?
What criticisms of Wittgenstein's philosophy of language have been offered?
Is there a verb for listening stealthily?
Why is there so little support for joining EFTA in the British parliament?
How can I prevent/balance waiting and turtling as a response to cooldown mechanics
newbie Q : How to read an output file in one command line
Why do C and C++ allow the expression (int) + 4*5;
"Destructive power" carried by a B-52?
What is "Lambda" in Heston's original paper on stochastic volatility models?
Does a random sequence of vectors span a Hilbert space?
Where and when has Thucydides been studied?
Why did Bronn offer to be Tyrion Lannister's champion in trial by combat?
One-one communication
How do Java 8 default methods hеlp with lambdas?
Understanding piped commands in GNU/Linux
Weaponising the Grasp-at-a-Distance spell
How can I list files in reverse time order by a command and pass them as arguments to another command?
Why complex landing gears are used instead of simple, reliable and light weight muscle wire or shape memory alloys?
How to make an animal which can only breed for a certain number of generations?
New Order #6: Easter Egg
Adapting the Chinese Remainder Theorem (CRT) for integers to polynomials
My mentor says to set image to Fine instead of RAW — how is this different from JPG?
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 6 hours ago
RobRob
331110
$endgroup$
|
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 6 hours ago
RobRob
331110
$endgroup$
10
$begingroup$
Maybe because $x^2$ isn't the same as $2x$?
$endgroup$
– Lord Shark the Unknown
6 hours ago
2
$begingroup$
$$e^x^2=e^xcdot xneq e^xcdot e^x = e^x+x=e^2x$$
$endgroup$
– Don Thousand
6 hours ago
1
$begingroup$
$(e^x)^2$ would be where you use the rule that you're thinking of.
$endgroup$
– Tartaglia's Stutter
6 hours ago
1
$begingroup$
Possible duplicate of How can you prove that a function has no closed form integral?
$endgroup$
– Peter Foreman
6 hours ago
1
$begingroup$
@DonThousand OP - "what are sufficient conditions for a function NOT to have an antiderivative"
$endgroup$
– Peter Foreman
6 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 6 hours ago
RobRob
331110
$endgroup$
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 6 hours ago
RobRob
331110
asked 6 hours ago
RobRob
331110
asked 6 hours ago
RobRob
331110
asked 6 hours ago
RobRob
331110
asked 6 hours ago
RobRob
331110
331110
10
$begingroup$
Maybe because $x^2$ isn't the same as $2x$?
$endgroup$
– Lord Shark the Unknown
6 hours ago
2
$begingroup$
$$e^x^2=e^xcdot xneq e^xcdot e^x = e^x+x=e^2x$$
$endgroup$
– Don Thousand
6 hours ago
1
$begingroup$
$(e^x)^2$ would be where you use the rule that you're thinking of.
$endgroup$
– Tartaglia's Stutter
6 hours ago
1
$begingroup$
Possible duplicate of How can you prove that a function has no closed form integral?
$endgroup$
– Peter Foreman
6 hours ago
1
$begingroup$
@DonThousand OP - "what are sufficient conditions for a function NOT to have an antiderivative"
$endgroup$
– Peter Foreman
6 hours ago
|
show 8 more comments
10
$begingroup$
Maybe because $x^2$ isn't the same as $2x$?
$endgroup$
– Lord Shark the Unknown
6 hours ago
2
$begingroup$
$$e^x^2=e^xcdot xneq e^xcdot e^x = e^x+x=e^2x$$
$endgroup$
– Don Thousand
6 hours ago
1
$begingroup$
$(e^x)^2$ would be where you use the rule that you're thinking of.
$endgroup$
– Tartaglia's Stutter
6 hours ago
1
$begingroup$
Possible duplicate of How can you prove that a function has no closed form integral?
$endgroup$
– Peter Foreman
6 hours ago
1
$begingroup$
@DonThousand OP - "what are sufficient conditions for a function NOT to have an antiderivative"
$endgroup$
– Peter Foreman
6 hours ago
10
10
$begingroup$
Maybe because $x^2$ isn't the same as $2x$?
$endgroup$
– Lord Shark the Unknown
6 hours ago
$begingroup$
Maybe because $x^2$ isn't the same as $2x$?
$endgroup$
– Lord Shark the Unknown
6 hours ago
2
2
$begingroup$
$$e^x^2=e^xcdot xneq e^xcdot e^x = e^x+x=e^2x$$
$endgroup$
– Don Thousand
6 hours ago
$begingroup$
$$e^x^2=e^xcdot xneq e^xcdot e^x = e^x+x=e^2x$$
$endgroup$
– Don Thousand
6 hours ago
1
1
$begingroup$
$(e^x)^2$ would be where you use the rule that you're thinking of.
$endgroup$
– Tartaglia's Stutter
6 hours ago
$begingroup$
$(e^x)^2$ would be where you use the rule that you're thinking of.
$endgroup$
– Tartaglia's Stutter
6 hours ago
1
1
$begingroup$
Possible duplicate of How can you prove that a function has no closed form integral?
$endgroup$
– Peter Foreman
6 hours ago
$begingroup$
Possible duplicate of How can you prove that a function has no closed form integral?
$endgroup$
– Peter Foreman
6 hours ago
1
1
$begingroup$
@DonThousand OP - "what are sufficient conditions for a function NOT to have an antiderivative"
$endgroup$
– Peter Foreman
6 hours ago
$begingroup$
@DonThousand OP - "what are sufficient conditions for a function NOT to have an antiderivative"
$endgroup$
– Peter Foreman
6 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 6 hours ago
peterwhypeterwhy
12.4k21229
$endgroup$
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 2 hours ago
AccidentalFourierTransformAccidentalFourierTransform
1,482828
$endgroup$
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 6 hours ago
Peter ForemanPeter Foreman
8,5171321
$endgroup$
$begingroup$
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
6 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
6 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
6 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 6 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
5 hours ago
2
$begingroup$
$e^x^2$ 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
5 hours ago
$begingroup$
@MathIsFun I see your point. I had the wrong understanding then! Thanks for pointing it out
$endgroup$
– Rob
2 hours ago
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3196128%2fwhen-does-a-function-not-have-an-antiderivative%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
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 6 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 6 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 6 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 6 hours ago
peterwhypeterwhy
12.4k21229
answered 6 hours ago
peterwhypeterwhy
12.4k21229
answered 6 hours ago
peterwhypeterwhy
12.4k21229
answered 6 hours ago
peterwhypeterwhy
12.4k21229
12.4k21229
add a comment |
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 2 hours ago
AccidentalFourierTransformAccidentalFourierTransform
1,482828
$endgroup$
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 2 hours ago
AccidentalFourierTransformAccidentalFourierTransform
1,482828
$endgroup$
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 2 hours ago
AccidentalFourierTransformAccidentalFourierTransform
1,482828
$endgroup$
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 2 hours ago
AccidentalFourierTransformAccidentalFourierTransform
1,482828
answered 2 hours ago
AccidentalFourierTransformAccidentalFourierTransform
1,482828
answered 2 hours ago
AccidentalFourierTransformAccidentalFourierTransform
1,482828
answered 2 hours ago
AccidentalFourierTransformAccidentalFourierTransform
1,482828
1,482828
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 6 hours ago
Peter ForemanPeter Foreman
8,5171321
$endgroup$
$begingroup$
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
6 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
6 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
6 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 6 hours ago
Peter ForemanPeter Foreman
8,5171321
$endgroup$
$begingroup$
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
6 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
6 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
6 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 6 hours ago
Peter ForemanPeter Foreman
8,5171321
$endgroup$
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 6 hours ago
Peter ForemanPeter Foreman
8,5171321
answered 6 hours ago
Peter ForemanPeter Foreman
8,5171321
answered 6 hours ago
Peter ForemanPeter Foreman
8,5171321
answered 6 hours ago
Peter ForemanPeter Foreman
8,5171321
8,5171321
$begingroup$
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
6 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
6 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
6 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
6 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
6 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
6 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
6 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
6 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
6 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
6 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
6 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
6 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 6 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
5 hours ago
2
$begingroup$
$e^x^2$ 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
5 hours ago
$begingroup$
@MathIsFun I see your point. I had the wrong understanding then! Thanks for pointing it out
$endgroup$
– Rob
2 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 6 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
5 hours ago
2
$begingroup$
$e^x^2$ 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
5 hours ago
$begingroup$
@MathIsFun I see your point. I had the wrong understanding then! Thanks for pointing it out
$endgroup$
– Rob
2 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 6 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 6 hours ago
AllawonderAllawonder
2,349617
answered 6 hours ago
AllawonderAllawonder
2,349617
answered 6 hours ago
AllawonderAllawonder
2,349617
answered 6 hours ago
AllawonderAllawonder
2,349617
2,349617
$begingroup$
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
5 hours ago
2
$begingroup$
$e^x^2$ 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
5 hours ago
$begingroup$
@MathIsFun I see your point. I had the wrong understanding then! Thanks for pointing it out
$endgroup$
– Rob
2 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
5 hours ago
2
$begingroup$
$e^x^2$ 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
5 hours ago
$begingroup$
@MathIsFun I see your point. I had the wrong understanding then! Thanks for pointing it out
$endgroup$
– Rob
2 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
5 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
5 hours ago
2
2
$begingroup$
$e^x^2$ 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
5 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
5 hours ago
$begingroup$
@MathIsFun I see your point. I had the wrong understanding then! Thanks for pointing it out
$endgroup$
– Rob
2 hours ago
$begingroup$
@MathIsFun I see your point. I had the wrong understanding then! Thanks for pointing it out
$endgroup$
– Rob
2 hours ago
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3196128%2fwhen-does-a-function-not-have-an-antiderivative%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
10
$begingroup$
Maybe because $x^2$ isn't the same as $2x$?
$endgroup$
– Lord Shark the Unknown
6 hours ago
2
$begingroup$
$$e^x^2=e^xcdot xneq e^xcdot e^x = e^x+x=e^2x$$
$endgroup$
– Don Thousand
6 hours ago
1
$begingroup$
$(e^x)^2$ would be where you use the rule that you're thinking of.
$endgroup$
– Tartaglia's Stutter
6 hours ago
1
$begingroup$
Possible duplicate of How can you prove that a function has no closed form integral?
$endgroup$
– Peter Foreman
6 hours ago
1
$begingroup$
@DonThousand OP - "what are sufficient conditions for a function NOT to have an antiderivative"
$endgroup$
– Peter Foreman
6 hours ago