Mixing
Definition of real exponent
4
$begingroup$
Is there some article or chapter in some book which covers proving all usual properties of exponential function with definition of real exponentiation via showing that sequence ${a^{x_n}}$ converges when sequence ${x_n}$ of rationals converges and the value of the limit of ${a^{x_n}}$ is independent of the sequence ${x_n}$, where $a>0$ ?
Thanks in advance.
reference-request exponential-function definition
asked Dec 30 '18 at 13:08
Юрій ЯрошЮрій Ярош
1,071615
$endgroup$
add a comment |
4
$begingroup$
Is there some article or chapter in some book which covers proving all usual properties of exponential function with definition of real exponentiation via showing that sequence ${a^{x_n}}$ converges when sequence ${x_n}$ of rationals converges and the value of the limit of ${a^{x_n}}$ is independent of the sequence ${x_n}$, where $a>0$ ?
Thanks in advance.
reference-request exponential-function definition
asked Dec 30 '18 at 13:08
Юрій ЯрошЮрій Ярош
1,071615
$endgroup$
add a comment |
4
4
4
1
$begingroup$
Is there some article or chapter in some book which covers proving all usual properties of exponential function with definition of real exponentiation via showing that sequence ${a^{x_n}}$ converges when sequence ${x_n}$ of rationals converges and the value of the limit of ${a^{x_n}}$ is independent of the sequence ${x_n}$, where $a>0$ ?
Thanks in advance.
reference-request exponential-function definition
asked Dec 30 '18 at 13:08
Юрій ЯрошЮрій Ярош
1,071615
$endgroup$
Is there some article or chapter in some book which covers proving all usual properties of exponential function with definition of real exponentiation via showing that sequence ${a^{x_n}}$ converges when sequence ${x_n}$ of rationals converges and the value of the limit of ${a^{x_n}}$ is independent of the sequence ${x_n}$, where $a>0$ ?
Thanks in advance.
reference-request exponential-function definition
reference-request exponential-function definition
asked Dec 30 '18 at 13:08
Юрій ЯрошЮрій Ярош
1,071615
asked Dec 30 '18 at 13:08
Юрій ЯрошЮрій Ярош
1,071615
asked Dec 30 '18 at 13:08
Юрій ЯрошЮрій Ярош
1,071615
asked Dec 30 '18 at 13:08
Юрій ЯрошЮрій Ярош
1,071615
asked Dec 30 '18 at 13:08
Юрій ЯрошЮрій Ярош
1,071615
1,071615
add a comment |
add a comment |
1 Answer
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4
+50
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This is in Terence Tao's book "Analysis I". See section 6.7, p152, third edition. Lemma 6.7.1 captures the essence of your question.
answered Jan 1 at 14:04
Math_QEDMath_QED
7,32931450
$endgroup$
$begingroup$
Out of interest: How does he define $a^x$ in the first place for rational $x$, but probably irrational $a$?
$endgroup$
– Behnam Esmayli
Jan 2 at 17:49
1
$begingroup$
Assume $x$ is positive. First, $x^n$ for naturals in the obvious way: repeated multiplication. Then $x^n$ for integers by $x^{-n} = 1/x^n$ where $n$ is again a positive natural number. Then $x^{1/n}$ as the unique root of the equation $a^n = x$. Then $x^{p/q}$ by $(x^{1/q})^p$ and eventually for reals by considering limits as in the OP's question.
$endgroup$
– Math_QED
Jan 2 at 19:08
$begingroup$
At each step, some work has to be done. For example, showing that such equations have unique roots takes some work and showing that $x^{p/q}$ makes sense takes work as well.
$endgroup$
– Math_QED
Jan 2 at 19:09
$begingroup$
In my honest opinion, it is better to develop some real analysis first and leave exponentiation undefined until one has treaten (power) series and continuïty. Then the existence of roots will follow by intermediate value theorem and one can define the exponential using power series in a much cleaner way. But it may obscure a little bit what's going on.
$endgroup$
– Math_QED
Jan 2 at 19:11
$begingroup$
thanks for your expansive reply. I agree that probably the most elegant way is to develop some analysis first. Actually, the most convincing definition I find is to define $ln x := int_1 ^x frac{1}{t}dt $ for positive $x$, then some properties are immediate, such as monotone increase, or product to sum rule, etc (via integration by sub/parts…) Then exponential $e^x$ is defined as its inverse function. Now, any $a^b$ is defined to be $e^{b ln a}$, where $a >0 $.
$endgroup$
– Behnam Esmayli
Jan 2 at 20:26
|
show 2 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
4
+50
$begingroup$
This is in Terence Tao's book "Analysis I". See section 6.7, p152, third edition. Lemma 6.7.1 captures the essence of your question.
answered Jan 1 at 14:04
Math_QEDMath_QED
7,32931450
$endgroup$
$begingroup$
Out of interest: How does he define $a^x$ in the first place for rational $x$, but probably irrational $a$?
$endgroup$
– Behnam Esmayli
Jan 2 at 17:49
1
$begingroup$
Assume $x$ is positive. First, $x^n$ for naturals in the obvious way: repeated multiplication. Then $x^n$ for integers by $x^{-n} = 1/x^n$ where $n$ is again a positive natural number. Then $x^{1/n}$ as the unique root of the equation $a^n = x$. Then $x^{p/q}$ by $(x^{1/q})^p$ and eventually for reals by considering limits as in the OP's question.
$endgroup$
– Math_QED
Jan 2 at 19:08
$begingroup$
At each step, some work has to be done. For example, showing that such equations have unique roots takes some work and showing that $x^{p/q}$ makes sense takes work as well.
$endgroup$
– Math_QED
Jan 2 at 19:09
$begingroup$
In my honest opinion, it is better to develop some real analysis first and leave exponentiation undefined until one has treaten (power) series and continuïty. Then the existence of roots will follow by intermediate value theorem and one can define the exponential using power series in a much cleaner way. But it may obscure a little bit what's going on.
$endgroup$
– Math_QED
Jan 2 at 19:11
$begingroup$
thanks for your expansive reply. I agree that probably the most elegant way is to develop some analysis first. Actually, the most convincing definition I find is to define $ln x := int_1 ^x frac{1}{t}dt $ for positive $x$, then some properties are immediate, such as monotone increase, or product to sum rule, etc (via integration by sub/parts…) Then exponential $e^x$ is defined as its inverse function. Now, any $a^b$ is defined to be $e^{b ln a}$, where $a >0 $.
$endgroup$
– Behnam Esmayli
Jan 2 at 20:26
|
show 2 more comments
4
+50
$begingroup$
This is in Terence Tao's book "Analysis I". See section 6.7, p152, third edition. Lemma 6.7.1 captures the essence of your question.
answered Jan 1 at 14:04
Math_QEDMath_QED
7,32931450
$endgroup$
$begingroup$
Out of interest: How does he define $a^x$ in the first place for rational $x$, but probably irrational $a$?
$endgroup$
– Behnam Esmayli
Jan 2 at 17:49
1
$begingroup$
Assume $x$ is positive. First, $x^n$ for naturals in the obvious way: repeated multiplication. Then $x^n$ for integers by $x^{-n} = 1/x^n$ where $n$ is again a positive natural number. Then $x^{1/n}$ as the unique root of the equation $a^n = x$. Then $x^{p/q}$ by $(x^{1/q})^p$ and eventually for reals by considering limits as in the OP's question.
$endgroup$
– Math_QED
Jan 2 at 19:08
$begingroup$
At each step, some work has to be done. For example, showing that such equations have unique roots takes some work and showing that $x^{p/q}$ makes sense takes work as well.
$endgroup$
– Math_QED
Jan 2 at 19:09
$begingroup$
In my honest opinion, it is better to develop some real analysis first and leave exponentiation undefined until one has treaten (power) series and continuïty. Then the existence of roots will follow by intermediate value theorem and one can define the exponential using power series in a much cleaner way. But it may obscure a little bit what's going on.
$endgroup$
– Math_QED
Jan 2 at 19:11
$begingroup$
thanks for your expansive reply. I agree that probably the most elegant way is to develop some analysis first. Actually, the most convincing definition I find is to define $ln x := int_1 ^x frac{1}{t}dt $ for positive $x$, then some properties are immediate, such as monotone increase, or product to sum rule, etc (via integration by sub/parts…) Then exponential $e^x$ is defined as its inverse function. Now, any $a^b$ is defined to be $e^{b ln a}$, where $a >0 $.
$endgroup$
– Behnam Esmayli
Jan 2 at 20:26
|
show 2 more comments
4
+50
4
+50
4
+50
$begingroup$
This is in Terence Tao's book "Analysis I". See section 6.7, p152, third edition. Lemma 6.7.1 captures the essence of your question.
answered Jan 1 at 14:04
Math_QEDMath_QED
7,32931450
$endgroup$
This is in Terence Tao's book "Analysis I". See section 6.7, p152, third edition. Lemma 6.7.1 captures the essence of your question.
answered Jan 1 at 14:04
Math_QEDMath_QED
7,32931450
answered Jan 1 at 14:04
Math_QEDMath_QED
7,32931450
answered Jan 1 at 14:04
Math_QEDMath_QED
7,32931450
answered Jan 1 at 14:04
Math_QEDMath_QED
7,32931450
7,32931450
$begingroup$
Out of interest: How does he define $a^x$ in the first place for rational $x$, but probably irrational $a$?
$endgroup$
– Behnam Esmayli
Jan 2 at 17:49
1
$begingroup$
Assume $x$ is positive. First, $x^n$ for naturals in the obvious way: repeated multiplication. Then $x^n$ for integers by $x^{-n} = 1/x^n$ where $n$ is again a positive natural number. Then $x^{1/n}$ as the unique root of the equation $a^n = x$. Then $x^{p/q}$ by $(x^{1/q})^p$ and eventually for reals by considering limits as in the OP's question.
$endgroup$
– Math_QED
Jan 2 at 19:08
$begingroup$
At each step, some work has to be done. For example, showing that such equations have unique roots takes some work and showing that $x^{p/q}$ makes sense takes work as well.
$endgroup$
– Math_QED
Jan 2 at 19:09
$begingroup$
In my honest opinion, it is better to develop some real analysis first and leave exponentiation undefined until one has treaten (power) series and continuïty. Then the existence of roots will follow by intermediate value theorem and one can define the exponential using power series in a much cleaner way. But it may obscure a little bit what's going on.
$endgroup$
– Math_QED
Jan 2 at 19:11
$begingroup$
thanks for your expansive reply. I agree that probably the most elegant way is to develop some analysis first. Actually, the most convincing definition I find is to define $ln x := int_1 ^x frac{1}{t}dt $ for positive $x$, then some properties are immediate, such as monotone increase, or product to sum rule, etc (via integration by sub/parts…) Then exponential $e^x$ is defined as its inverse function. Now, any $a^b$ is defined to be $e^{b ln a}$, where $a >0 $.
$endgroup$
– Behnam Esmayli
Jan 2 at 20:26
|
show 2 more comments
$begingroup$
Out of interest: How does he define $a^x$ in the first place for rational $x$, but probably irrational $a$?
$endgroup$
– Behnam Esmayli
Jan 2 at 17:49
1
$begingroup$
Assume $x$ is positive. First, $x^n$ for naturals in the obvious way: repeated multiplication. Then $x^n$ for integers by $x^{-n} = 1/x^n$ where $n$ is again a positive natural number. Then $x^{1/n}$ as the unique root of the equation $a^n = x$. Then $x^{p/q}$ by $(x^{1/q})^p$ and eventually for reals by considering limits as in the OP's question.
$endgroup$
– Math_QED
Jan 2 at 19:08
$begingroup$
At each step, some work has to be done. For example, showing that such equations have unique roots takes some work and showing that $x^{p/q}$ makes sense takes work as well.
$endgroup$
– Math_QED
Jan 2 at 19:09
$begingroup$
In my honest opinion, it is better to develop some real analysis first and leave exponentiation undefined until one has treaten (power) series and continuïty. Then the existence of roots will follow by intermediate value theorem and one can define the exponential using power series in a much cleaner way. But it may obscure a little bit what's going on.
$endgroup$
– Math_QED
Jan 2 at 19:11
$begingroup$
thanks for your expansive reply. I agree that probably the most elegant way is to develop some analysis first. Actually, the most convincing definition I find is to define $ln x := int_1 ^x frac{1}{t}dt $ for positive $x$, then some properties are immediate, such as monotone increase, or product to sum rule, etc (via integration by sub/parts…) Then exponential $e^x$ is defined as its inverse function. Now, any $a^b$ is defined to be $e^{b ln a}$, where $a >0 $.
$endgroup$
– Behnam Esmayli
Jan 2 at 20:26
$begingroup$
Out of interest: How does he define $a^x$ in the first place for rational $x$, but probably irrational $a$?
$endgroup$
– Behnam Esmayli
Jan 2 at 17:49
$begingroup$
Out of interest: How does he define $a^x$ in the first place for rational $x$, but probably irrational $a$?
$endgroup$
– Behnam Esmayli
Jan 2 at 17:49
1
1
$begingroup$
Assume $x$ is positive. First, $x^n$ for naturals in the obvious way: repeated multiplication. Then $x^n$ for integers by $x^{-n} = 1/x^n$ where $n$ is again a positive natural number. Then $x^{1/n}$ as the unique root of the equation $a^n = x$. Then $x^{p/q}$ by $(x^{1/q})^p$ and eventually for reals by considering limits as in the OP's question.
$endgroup$
– Math_QED
Jan 2 at 19:08
$begingroup$
Assume $x$ is positive. First, $x^n$ for naturals in the obvious way: repeated multiplication. Then $x^n$ for integers by $x^{-n} = 1/x^n$ where $n$ is again a positive natural number. Then $x^{1/n}$ as the unique root of the equation $a^n = x$. Then $x^{p/q}$ by $(x^{1/q})^p$ and eventually for reals by considering limits as in the OP's question.
$endgroup$
– Math_QED
Jan 2 at 19:08
$begingroup$
At each step, some work has to be done. For example, showing that such equations have unique roots takes some work and showing that $x^{p/q}$ makes sense takes work as well.
$endgroup$
– Math_QED
Jan 2 at 19:09
$begingroup$
At each step, some work has to be done. For example, showing that such equations have unique roots takes some work and showing that $x^{p/q}$ makes sense takes work as well.
$endgroup$
– Math_QED
Jan 2 at 19:09
$begingroup$
In my honest opinion, it is better to develop some real analysis first and leave exponentiation undefined until one has treaten (power) series and continuïty. Then the existence of roots will follow by intermediate value theorem and one can define the exponential using power series in a much cleaner way. But it may obscure a little bit what's going on.
$endgroup$
– Math_QED
Jan 2 at 19:11
$begingroup$
In my honest opinion, it is better to develop some real analysis first and leave exponentiation undefined until one has treaten (power) series and continuïty. Then the existence of roots will follow by intermediate value theorem and one can define the exponential using power series in a much cleaner way. But it may obscure a little bit what's going on.
$endgroup$
– Math_QED
Jan 2 at 19:11
$begingroup$
thanks for your expansive reply. I agree that probably the most elegant way is to develop some analysis first. Actually, the most convincing definition I find is to define $ln x := int_1 ^x frac{1}{t}dt $ for positive $x$, then some properties are immediate, such as monotone increase, or product to sum rule, etc (via integration by sub/parts…) Then exponential $e^x$ is defined as its inverse function. Now, any $a^b$ is defined to be $e^{b ln a}$, where $a >0 $.
$endgroup$
– Behnam Esmayli
Jan 2 at 20:26
$begingroup$
thanks for your expansive reply. I agree that probably the most elegant way is to develop some analysis first. Actually, the most convincing definition I find is to define $ln x := int_1 ^x frac{1}{t}dt $ for positive $x$, then some properties are immediate, such as monotone increase, or product to sum rule, etc (via integration by sub/parts…) Then exponential $e^x$ is defined as its inverse function. Now, any $a^b$ is defined to be $e^{b ln a}$, where $a >0 $.
$endgroup$
– Behnam Esmayli
Jan 2 at 20:26
|
show 2 more comments
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