When does a function NOT have an antiderivative? [duplicate]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 $C = 0$)
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When does a function NOT have an antiderivative? [duplicate]
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 $C = 0$)
$begingroup$
This question already has an answer here:
How can you prove that a function has no closed form integral?
7 answers
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 Apr 21 at 17:55
RobRob
358110
$endgroup$
marked as duplicate by Peter Foreman, Pedro, Xander Henderson, Brahadeesh, Billy Rubina Apr 22 at 3:32
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
|
show 7 more comments
$begingroup$
This question already has an answer here:
How can you prove that a function has no closed form integral?
7 answers
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 Apr 21 at 17:55
RobRob
358110
$endgroup$
marked as duplicate by Peter Foreman, Pedro, Xander Henderson, Brahadeesh, Billy Rubina Apr 22 at 3:32
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
14
$begingroup$
Maybe because $x^2$ isn't the same as $2x$?
$endgroup$
– Lord Shark the Unknown
Apr 21 at 17:56
3
$begingroup$
$$e^x^2=e^xcdot xneq e^xcdot e^x = e^x+x=e^2x$$
$endgroup$
– Don Thousand
Apr 21 at 17:57
1
$begingroup$
$(e^x)^2$ would be where you use the rule that you're thinking of.
$endgroup$
– Tartaglia's Stutter
Apr 21 at 17:58
1
$begingroup$
@DonThousand OP - "what are sufficient conditions for a function NOT to have an antiderivative"
$endgroup$
– Peter Foreman
Apr 21 at 18:01
2
$begingroup$
@DonThousand That is written within the question itself right after "In light of this question"
$endgroup$
– Peter Foreman
Apr 21 at 18:02
|
show 7 more comments
$begingroup$
This question already has an answer here:
How can you prove that a function has no closed form integral?
7 answers
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 Apr 21 at 17:55
RobRob
358110
$endgroup$
This question already has an answer here:
How can you prove that a function has no closed form integral?
7 answers
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?
This question already has an answer here:
How can you prove that a function has no closed form integral?
7 answers
integration
integration
asked Apr 21 at 17:55
RobRob
358110
asked Apr 21 at 17:55
RobRob
358110
asked Apr 21 at 17:55
RobRob
358110
asked Apr 21 at 17:55
RobRob
358110
asked Apr 21 at 17:55
RobRob
358110
358110
marked as duplicate by Peter Foreman, Pedro, Xander Henderson, Brahadeesh, Billy Rubina Apr 22 at 3:32
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Peter Foreman, Pedro, Xander Henderson, Brahadeesh, Billy Rubina Apr 22 at 3:32
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
14
$begingroup$
Maybe because $x^2$ isn't the same as $2x$?
$endgroup$
– Lord Shark the Unknown
Apr 21 at 17:56
3
$begingroup$
$$e^x^2=e^xcdot xneq e^xcdot e^x = e^x+x=e^2x$$
$endgroup$
– Don Thousand
Apr 21 at 17:57
1
$begingroup$
$(e^x)^2$ would be where you use the rule that you're thinking of.
$endgroup$
– Tartaglia's Stutter
Apr 21 at 17:58
1
$begingroup$
@DonThousand OP - "what are sufficient conditions for a function NOT to have an antiderivative"
$endgroup$
– Peter Foreman
Apr 21 at 18:01
2
$begingroup$
@DonThousand That is written within the question itself right after "In light of this question"
$endgroup$
– Peter Foreman
Apr 21 at 18:02
|
show 7 more comments
14
$begingroup$
Maybe because $x^2$ isn't the same as $2x$?
$endgroup$
– Lord Shark the Unknown
Apr 21 at 17:56
3
$begingroup$
$$e^x^2=e^xcdot xneq e^xcdot e^x = e^x+x=e^2x$$
$endgroup$
– Don Thousand
Apr 21 at 17:57
1
$begingroup$
$(e^x)^2$ would be where you use the rule that you're thinking of.
$endgroup$
– Tartaglia's Stutter
Apr 21 at 17:58
1
$begingroup$
@DonThousand OP - "what are sufficient conditions for a function NOT to have an antiderivative"
$endgroup$
– Peter Foreman
Apr 21 at 18:01
2
$begingroup$
@DonThousand That is written within the question itself right after "In light of this question"
$endgroup$
– Peter Foreman
Apr 21 at 18:02
14
14
$begingroup$
Maybe because $x^2$ isn't the same as $2x$?
$endgroup$
– Lord Shark the Unknown
Apr 21 at 17:56
$begingroup$
Maybe because $x^2$ isn't the same as $2x$?
$endgroup$
– Lord Shark the Unknown
Apr 21 at 17:56
3
3
$begingroup$
$$e^x^2=e^xcdot xneq e^xcdot e^x = e^x+x=e^2x$$
$endgroup$
– Don Thousand
Apr 21 at 17:57
$begingroup$
$$e^x^2=e^xcdot xneq e^xcdot e^x = e^x+x=e^2x$$
$endgroup$
– Don Thousand
Apr 21 at 17:57
1
1
$begingroup$
$(e^x)^2$ would be where you use the rule that you're thinking of.
$endgroup$
– Tartaglia's Stutter
Apr 21 at 17:58
$begingroup$
$(e^x)^2$ would be where you use the rule that you're thinking of.
$endgroup$
– Tartaglia's Stutter
Apr 21 at 17:58
1
1
$begingroup$
@DonThousand OP - "what are sufficient conditions for a function NOT to have an antiderivative"
$endgroup$
– Peter Foreman
Apr 21 at 18:01
$begingroup$
@DonThousand OP - "what are sufficient conditions for a function NOT to have an antiderivative"
$endgroup$
– Peter Foreman
Apr 21 at 18:01
2
2
$begingroup$
@DonThousand That is written within the question itself right after "In light of this question"
$endgroup$
– Peter Foreman
Apr 21 at 18:02
$begingroup$
@DonThousand That is written within the question itself right after "In light of this question"
$endgroup$
– Peter Foreman
Apr 21 at 18:02
|
show 7 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 Apr 21 at 18:36
peterwhypeterwhy
12.5k21229
$endgroup$
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 Apr 21 at 21:43
AccidentalFourierTransformAccidentalFourierTransform
1,492828
$endgroup$
add a comment |
$begingroup$
To answer the titular question, there's a result in real analysis that shows that derivatives 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 Apr 21 at 18:13
AllawonderAllawonder
2,441717
$endgroup$
$begingroup$
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
Apr 21 at 18:50
6
$begingroup$
$e^x^2$ 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
Apr 21 at 19:16
$begingroup$
@MathIsFun I see your point. I had the wrong understanding then! Thanks for pointing it out
$endgroup$
– Rob
Apr 21 at 22:28
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 Apr 21 at 18:11
Peter ForemanPeter Foreman
9,1891321
$endgroup$
$begingroup$
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
Apr 21 at 18:16
$begingroup$
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
Apr 21 at 18:17
$begingroup$
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
Apr 21 at 18:28
add a comment |
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 Apr 21 at 18:36
peterwhypeterwhy
12.5k21229
$endgroup$
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 Apr 21 at 18:36
peterwhypeterwhy
12.5k21229
$endgroup$
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 Apr 21 at 18:36
peterwhypeterwhy
12.5k21229
$endgroup$
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 Apr 21 at 18:36
peterwhypeterwhy
12.5k21229
answered Apr 21 at 18:36
peterwhypeterwhy
12.5k21229
answered Apr 21 at 18:36
peterwhypeterwhy
12.5k21229
answered Apr 21 at 18:36
peterwhypeterwhy
12.5k21229
12.5k21229
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 Apr 21 at 21:43
AccidentalFourierTransformAccidentalFourierTransform
1,492828
$endgroup$
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 Apr 21 at 21:43
AccidentalFourierTransformAccidentalFourierTransform
1,492828
$endgroup$
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 Apr 21 at 21:43
AccidentalFourierTransformAccidentalFourierTransform
1,492828
$endgroup$
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 Apr 21 at 21:43
AccidentalFourierTransformAccidentalFourierTransform
1,492828
answered Apr 21 at 21:43
AccidentalFourierTransformAccidentalFourierTransform
1,492828
answered Apr 21 at 21:43
AccidentalFourierTransformAccidentalFourierTransform
1,492828
answered Apr 21 at 21:43
AccidentalFourierTransformAccidentalFourierTransform
1,492828
1,492828
add a comment |
add a comment |
$begingroup$
To answer the titular question, there's a result in real analysis that shows that derivatives 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 Apr 21 at 18:13
AllawonderAllawonder
2,441717
$endgroup$
$begingroup$
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
Apr 21 at 18:50
6
$begingroup$
$e^x^2$ 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
Apr 21 at 19:16
$begingroup$
@MathIsFun I see your point. I had the wrong understanding then! Thanks for pointing it out
$endgroup$
– Rob
Apr 21 at 22:28
add a comment |
$begingroup$
To answer the titular question, there's a result in real analysis that shows that derivatives 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 Apr 21 at 18:13
AllawonderAllawonder
2,441717
$endgroup$
$begingroup$
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
Apr 21 at 18:50
6
$begingroup$
$e^x^2$ 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
Apr 21 at 19:16
$begingroup$
@MathIsFun I see your point. I had the wrong understanding then! Thanks for pointing it out
$endgroup$
– Rob
Apr 21 at 22:28
add a comment |
$begingroup$
To answer the titular question, there's a result in real analysis that shows that derivatives 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 Apr 21 at 18:13
AllawonderAllawonder
2,441717
$endgroup$
To answer the titular question, there's a result in real analysis that shows that derivatives 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 Apr 21 at 18:13
AllawonderAllawonder
2,441717
edited Apr 22 at 12:55
answered Apr 21 at 18:13
AllawonderAllawonder
2,441717
answered Apr 21 at 18:13
AllawonderAllawonder
2,441717
answered Apr 21 at 18:13
AllawonderAllawonder
2,441717
2,441717
$begingroup$
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
Apr 21 at 18:50
6
$begingroup$
$e^x^2$ 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
Apr 21 at 19:16
$begingroup$
@MathIsFun I see your point. I had the wrong understanding then! Thanks for pointing it out
$endgroup$
– Rob
Apr 21 at 22:28
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
Apr 21 at 18:50
6
$begingroup$
$e^x^2$ 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
Apr 21 at 19:16
$begingroup$
@MathIsFun I see your point. I had the wrong understanding then! Thanks for pointing it out
$endgroup$
– Rob
Apr 21 at 22:28
$begingroup$
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
Apr 21 at 18:50
$begingroup$
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
Apr 21 at 18:50
6
6
$begingroup$
$e^x^2$ 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
Apr 21 at 19:16
$begingroup$
$e^x^2$ 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
Apr 21 at 19:16
$begingroup$
@MathIsFun I see your point. I had the wrong understanding then! Thanks for pointing it out
$endgroup$
– Rob
Apr 21 at 22:28
$begingroup$
@MathIsFun I see your point. I had the wrong understanding then! Thanks for pointing it out
$endgroup$
– Rob
Apr 21 at 22:28
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 Apr 21 at 18:11
Peter ForemanPeter Foreman
9,1891321
$endgroup$
$begingroup$
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
Apr 21 at 18:16
$begingroup$
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
Apr 21 at 18:17
$begingroup$
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
Apr 21 at 18:28
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 Apr 21 at 18:11
Peter ForemanPeter Foreman
9,1891321
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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?
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– Rob
Apr 21 at 18:16
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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
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– Peter Foreman
Apr 21 at 18:17
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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.
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– Rob
Apr 21 at 18:28
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 Apr 21 at 18:11
Peter ForemanPeter Foreman
9,1891321
$endgroup$
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 Apr 21 at 18:11
Peter ForemanPeter Foreman
9,1891321
answered Apr 21 at 18:11
Peter ForemanPeter Foreman
9,1891321
answered Apr 21 at 18:11
Peter ForemanPeter Foreman
9,1891321
answered Apr 21 at 18:11
Peter ForemanPeter Foreman
9,1891321
9,1891321
$begingroup$
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?
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– Rob
Apr 21 at 18:16
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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
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– Peter Foreman
Apr 21 at 18:17
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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.
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– Rob
Apr 21 at 18:28
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
Apr 21 at 18:16
$begingroup$
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
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– Peter Foreman
Apr 21 at 18:17
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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.
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– Rob
Apr 21 at 18:28
$begingroup$
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
Apr 21 at 18:16
$begingroup$
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
Apr 21 at 18:16
$begingroup$
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
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– Peter Foreman
Apr 21 at 18:17
$begingroup$
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
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– Peter Foreman
Apr 21 at 18:17
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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.
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– Rob
Apr 21 at 18:28
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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.
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– Rob
Apr 21 at 18:28
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Maybe because $x^2$ isn't the same as $2x$?
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– Lord Shark the Unknown
Apr 21 at 17:56
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$$e^x^2=e^xcdot xneq e^xcdot e^x = e^x+x=e^2x$$
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– Don Thousand
Apr 21 at 17:57
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$(e^x)^2$ would be where you use the rule that you're thinking of.
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– Tartaglia's Stutter
Apr 21 at 17:58
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@DonThousand OP - "what are sufficient conditions for a function NOT to have an antiderivative"
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– Peter Foreman
Apr 21 at 18:01
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@DonThousand That is written within the question itself right after "In light of this question"
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– Peter Foreman
Apr 21 at 18:02