Trying to parse a Putnam solution from 1995
I’m having trouble parsing the solution for the 1995 Putnam, question A2. The proof proceeds:
The easiest proof uses
``big-O'' notation and the fact that $(1+x)^{1/2} = 1 + x/2 +
O(x^{2})$ for $|x|<1$. (Here $O(x^{2})$ means bounded by a constant
times $x^{2}$.) Me: That’s all well and good, and I can arrive at that by the generalized binomial theorem.
The proof continues:
begin{align*}
sqrt{x+a}-sqrt{x} &= x^{1/2}(sqrt{1+a/x} - 1) \
&= x^{1/2}(1 + a/2x + O(x^{-2})),
end{align*}
It would appear that’s an error; it would appear the proof writer dropped the minus 1.
But it continues, hence
$sqrt{sqrt{x+a} - sqrt{x}} = x^{1/4} (a/4x + O(x^{-2}))$
Which I’m at a loss how to arrive at.
binomial-theorem
asked 2 hours ago
Thor Kamphefner
365
New contributor
Thor Kamphefner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
|
show 2 more comments
I’m having trouble parsing the solution for the 1995 Putnam, question A2. The proof proceeds:
The easiest proof uses
``big-O'' notation and the fact that $(1+x)^{1/2} = 1 + x/2 +
O(x^{2})$ for $|x|<1$. (Here $O(x^{2})$ means bounded by a constant
times $x^{2}$.) Me: That’s all well and good, and I can arrive at that by the generalized binomial theorem.
The proof continues:
begin{align*}
sqrt{x+a}-sqrt{x} &= x^{1/2}(sqrt{1+a/x} - 1) \
&= x^{1/2}(1 + a/2x + O(x^{-2})),
end{align*}
It would appear that’s an error; it would appear the proof writer dropped the minus 1.
But it continues, hence
$sqrt{sqrt{x+a} - sqrt{x}} = x^{1/4} (a/4x + O(x^{-2}))$
Which I’m at a loss how to arrive at.
binomial-theorem
asked 2 hours ago
Thor Kamphefner
365
New contributor
Thor Kamphefner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
For what it’s worth, you aren’t the first person to have asked about this. math.stackexchange.com/questions/2020837/…
– Steve Kass
2 hours ago
I am not sure if you have copied things properly. Why do we have $O(x^{2})$ in one place and $O(x^{-2})$ at others?
– Kavi Rama Murthy
2 hours ago
Note that on the right side of the first equation, x is in the denominator, whereas at the top, x is in the numerator.
– Thor Kamphefner
2 hours ago
@KaviRamaMurthy Not OP, but that is because the new "$x$" is now $a/x$, so when you substitute that in $O(x^2)$ becomes $O(a^2/x^2)$, which is the same as $O(x^{-2})$
– Noble Mushtak
2 hours ago
You should show the question. I looked up the reference and it doesn't appear to be what you want. Also, where is the proof that you refer to?
– herb steinberg
2 hours ago
|
show 2 more comments
I’m having trouble parsing the solution for the 1995 Putnam, question A2. The proof proceeds:
The easiest proof uses
``big-O'' notation and the fact that $(1+x)^{1/2} = 1 + x/2 +
O(x^{2})$ for $|x|<1$. (Here $O(x^{2})$ means bounded by a constant
times $x^{2}$.) Me: That’s all well and good, and I can arrive at that by the generalized binomial theorem.
The proof continues:
begin{align*}
sqrt{x+a}-sqrt{x} &= x^{1/2}(sqrt{1+a/x} - 1) \
&= x^{1/2}(1 + a/2x + O(x^{-2})),
end{align*}
It would appear that’s an error; it would appear the proof writer dropped the minus 1.
But it continues, hence
$sqrt{sqrt{x+a} - sqrt{x}} = x^{1/4} (a/4x + O(x^{-2}))$
Which I’m at a loss how to arrive at.
binomial-theorem
asked 2 hours ago
Thor Kamphefner
365
New contributor
Thor Kamphefner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I’m having trouble parsing the solution for the 1995 Putnam, question A2. The proof proceeds:
The easiest proof uses
``big-O'' notation and the fact that $(1+x)^{1/2} = 1 + x/2 +
O(x^{2})$ for $|x|<1$. (Here $O(x^{2})$ means bounded by a constant
times $x^{2}$.) Me: That’s all well and good, and I can arrive at that by the generalized binomial theorem.
The proof continues:
begin{align*}
sqrt{x+a}-sqrt{x} &= x^{1/2}(sqrt{1+a/x} - 1) \
&= x^{1/2}(1 + a/2x + O(x^{-2})),
end{align*}
It would appear that’s an error; it would appear the proof writer dropped the minus 1.
But it continues, hence
$sqrt{sqrt{x+a} - sqrt{x}} = x^{1/4} (a/4x + O(x^{-2}))$
Which I’m at a loss how to arrive at.
binomial-theorem
binomial-theorem
asked 2 hours ago
Thor Kamphefner
365
New contributor
Thor Kamphefner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Thor Kamphefner
365
New contributor
Thor Kamphefner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
asked 2 hours ago
Thor Kamphefner
365
New contributor
Thor Kamphefner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Thor Kamphefner
365
asked 2 hours ago
Thor Kamphefner
365
365
New contributor
Thor Kamphefner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Thor Kamphefner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Thor Kamphefner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
For what it’s worth, you aren’t the first person to have asked about this. math.stackexchange.com/questions/2020837/…
– Steve Kass
2 hours ago
I am not sure if you have copied things properly. Why do we have $O(x^{2})$ in one place and $O(x^{-2})$ at others?
– Kavi Rama Murthy
2 hours ago
Note that on the right side of the first equation, x is in the denominator, whereas at the top, x is in the numerator.
– Thor Kamphefner
2 hours ago
@KaviRamaMurthy Not OP, but that is because the new "$x$" is now $a/x$, so when you substitute that in $O(x^2)$ becomes $O(a^2/x^2)$, which is the same as $O(x^{-2})$
– Noble Mushtak
2 hours ago
You should show the question. I looked up the reference and it doesn't appear to be what you want. Also, where is the proof that you refer to?
– herb steinberg
2 hours ago
|
show 2 more comments
For what it’s worth, you aren’t the first person to have asked about this. math.stackexchange.com/questions/2020837/…
– Steve Kass
2 hours ago
I am not sure if you have copied things properly. Why do we have $O(x^{2})$ in one place and $O(x^{-2})$ at others?
– Kavi Rama Murthy
2 hours ago
Note that on the right side of the first equation, x is in the denominator, whereas at the top, x is in the numerator.
– Thor Kamphefner
2 hours ago
@KaviRamaMurthy Not OP, but that is because the new "$x$" is now $a/x$, so when you substitute that in $O(x^2)$ becomes $O(a^2/x^2)$, which is the same as $O(x^{-2})$
– Noble Mushtak
2 hours ago
You should show the question. I looked up the reference and it doesn't appear to be what you want. Also, where is the proof that you refer to?
– herb steinberg
2 hours ago
For what it’s worth, you aren’t the first person to have asked about this. math.stackexchange.com/questions/2020837/…
– Steve Kass
2 hours ago
For what it’s worth, you aren’t the first person to have asked about this. math.stackexchange.com/questions/2020837/…
– Steve Kass
2 hours ago
I am not sure if you have copied things properly. Why do we have $O(x^{2})$ in one place and $O(x^{-2})$ at others?
– Kavi Rama Murthy
2 hours ago
I am not sure if you have copied things properly. Why do we have $O(x^{2})$ in one place and $O(x^{-2})$ at others?
– Kavi Rama Murthy
2 hours ago
Note that on the right side of the first equation, x is in the denominator, whereas at the top, x is in the numerator.
– Thor Kamphefner
2 hours ago
Note that on the right side of the first equation, x is in the denominator, whereas at the top, x is in the numerator.
– Thor Kamphefner
2 hours ago
@KaviRamaMurthy Not OP, but that is because the new "$x$" is now $a/x$, so when you substitute that in $O(x^2)$ becomes $O(a^2/x^2)$, which is the same as $O(x^{-2})$
– Noble Mushtak
2 hours ago
@KaviRamaMurthy Not OP, but that is because the new "$x$" is now $a/x$, so when you substitute that in $O(x^2)$ becomes $O(a^2/x^2)$, which is the same as $O(x^{-2})$
– Noble Mushtak
2 hours ago
You should show the question. I looked up the reference and it doesn't appear to be what you want. Also, where is the proof that you refer to?
– herb steinberg
2 hours ago
You should show the question. I looked up the reference and it doesn't appear to be what you want. Also, where is the proof that you refer to?
– herb steinberg
2 hours ago
|
show 2 more comments
1 Answer
1
active
oldest
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OK, so the first step is clearly a mistake with them dropping the $-1$. However, I'm going to try to explain what they may have done, even though it doesn't make sense. First, let's start with their (albeit wrong) equation:
$$sqrt{x+a}-sqrt{x}=x^{1/2}(1+frac{a}{2x}+O(x^{-2}))$$
Now, take the square root:
$$sqrt{sqrt{x+a}-sqrt{x}}=[x^{1/2}(1+frac{a}{2x}+O(x^{-2}))]^{1/2}$$
Distribute the exponent:
$$sqrt{sqrt{x+a}-sqrt{x}}=x^{1/4}[(1+frac{a}{2x}+O(x^{-2}))]^{1/2}$$
Now, the second expression on the right is in the form of $(1+z)^{1/2}$, so I will use the following formula:
$$(1+z)^{1/2}=1+frac z 2+O(z^2)rightarrow \ (1+frac a{2x}+O(x^{-2}))^{1/2}=1+frac{a}{4x}+frac{1}{2}O(x^{-2})+O(frac{a^2}{4x^2})=1+frac{a}{4x}+O(x^{-2})$$
Thus, we get:
$$sqrt{sqrt{x+a}-sqrt{x}}=x^{1/4}(1+frac{a}{4x}+O(x^{-2}))$$
Oh, and remember that $-1$ we dropped in the first step? Let's try adding that back in here, because...why not?
$$sqrt{sqrt{x+a}-sqrt{x}}=x^{1/4}(1+frac{a}{4x}+O(x^{-2})-1)$$
$$sqrt{sqrt{x+a}-sqrt{x}}=x^{1/4}(frac{a}{4x}+O(x^{-2}))$$
Thus, we get the same equation they did, and we only had to make two mathematical errors to get there!
answered 2 hours ago
Noble Mushtak
14.3k1734
I like that you were able to decipher what the original proof writer most likely meant. Thank you for that.
– Thor Kamphefner
2 hours ago
add a comment |
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1 Answer
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1 Answer
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OK, so the first step is clearly a mistake with them dropping the $-1$. However, I'm going to try to explain what they may have done, even though it doesn't make sense. First, let's start with their (albeit wrong) equation:
$$sqrt{x+a}-sqrt{x}=x^{1/2}(1+frac{a}{2x}+O(x^{-2}))$$
Now, take the square root:
$$sqrt{sqrt{x+a}-sqrt{x}}=[x^{1/2}(1+frac{a}{2x}+O(x^{-2}))]^{1/2}$$
Distribute the exponent:
$$sqrt{sqrt{x+a}-sqrt{x}}=x^{1/4}[(1+frac{a}{2x}+O(x^{-2}))]^{1/2}$$
Now, the second expression on the right is in the form of $(1+z)^{1/2}$, so I will use the following formula:
$$(1+z)^{1/2}=1+frac z 2+O(z^2)rightarrow \ (1+frac a{2x}+O(x^{-2}))^{1/2}=1+frac{a}{4x}+frac{1}{2}O(x^{-2})+O(frac{a^2}{4x^2})=1+frac{a}{4x}+O(x^{-2})$$
Thus, we get:
$$sqrt{sqrt{x+a}-sqrt{x}}=x^{1/4}(1+frac{a}{4x}+O(x^{-2}))$$
Oh, and remember that $-1$ we dropped in the first step? Let's try adding that back in here, because...why not?
$$sqrt{sqrt{x+a}-sqrt{x}}=x^{1/4}(1+frac{a}{4x}+O(x^{-2})-1)$$
$$sqrt{sqrt{x+a}-sqrt{x}}=x^{1/4}(frac{a}{4x}+O(x^{-2}))$$
Thus, we get the same equation they did, and we only had to make two mathematical errors to get there!
answered 2 hours ago
Noble Mushtak
14.3k1734
I like that you were able to decipher what the original proof writer most likely meant. Thank you for that.
– Thor Kamphefner
2 hours ago
add a comment |
OK, so the first step is clearly a mistake with them dropping the $-1$. However, I'm going to try to explain what they may have done, even though it doesn't make sense. First, let's start with their (albeit wrong) equation:
$$sqrt{x+a}-sqrt{x}=x^{1/2}(1+frac{a}{2x}+O(x^{-2}))$$
Now, take the square root:
$$sqrt{sqrt{x+a}-sqrt{x}}=[x^{1/2}(1+frac{a}{2x}+O(x^{-2}))]^{1/2}$$
Distribute the exponent:
$$sqrt{sqrt{x+a}-sqrt{x}}=x^{1/4}[(1+frac{a}{2x}+O(x^{-2}))]^{1/2}$$
Now, the second expression on the right is in the form of $(1+z)^{1/2}$, so I will use the following formula:
$$(1+z)^{1/2}=1+frac z 2+O(z^2)rightarrow \ (1+frac a{2x}+O(x^{-2}))^{1/2}=1+frac{a}{4x}+frac{1}{2}O(x^{-2})+O(frac{a^2}{4x^2})=1+frac{a}{4x}+O(x^{-2})$$
Thus, we get:
$$sqrt{sqrt{x+a}-sqrt{x}}=x^{1/4}(1+frac{a}{4x}+O(x^{-2}))$$
Oh, and remember that $-1$ we dropped in the first step? Let's try adding that back in here, because...why not?
$$sqrt{sqrt{x+a}-sqrt{x}}=x^{1/4}(1+frac{a}{4x}+O(x^{-2})-1)$$
$$sqrt{sqrt{x+a}-sqrt{x}}=x^{1/4}(frac{a}{4x}+O(x^{-2}))$$
Thus, we get the same equation they did, and we only had to make two mathematical errors to get there!
answered 2 hours ago
Noble Mushtak
14.3k1734
I like that you were able to decipher what the original proof writer most likely meant. Thank you for that.
– Thor Kamphefner
2 hours ago
add a comment |
OK, so the first step is clearly a mistake with them dropping the $-1$. However, I'm going to try to explain what they may have done, even though it doesn't make sense. First, let's start with their (albeit wrong) equation:
$$sqrt{x+a}-sqrt{x}=x^{1/2}(1+frac{a}{2x}+O(x^{-2}))$$
Now, take the square root:
$$sqrt{sqrt{x+a}-sqrt{x}}=[x^{1/2}(1+frac{a}{2x}+O(x^{-2}))]^{1/2}$$
Distribute the exponent:
$$sqrt{sqrt{x+a}-sqrt{x}}=x^{1/4}[(1+frac{a}{2x}+O(x^{-2}))]^{1/2}$$
Now, the second expression on the right is in the form of $(1+z)^{1/2}$, so I will use the following formula:
$$(1+z)^{1/2}=1+frac z 2+O(z^2)rightarrow \ (1+frac a{2x}+O(x^{-2}))^{1/2}=1+frac{a}{4x}+frac{1}{2}O(x^{-2})+O(frac{a^2}{4x^2})=1+frac{a}{4x}+O(x^{-2})$$
Thus, we get:
$$sqrt{sqrt{x+a}-sqrt{x}}=x^{1/4}(1+frac{a}{4x}+O(x^{-2}))$$
Oh, and remember that $-1$ we dropped in the first step? Let's try adding that back in here, because...why not?
$$sqrt{sqrt{x+a}-sqrt{x}}=x^{1/4}(1+frac{a}{4x}+O(x^{-2})-1)$$
$$sqrt{sqrt{x+a}-sqrt{x}}=x^{1/4}(frac{a}{4x}+O(x^{-2}))$$
Thus, we get the same equation they did, and we only had to make two mathematical errors to get there!
answered 2 hours ago
Noble Mushtak
14.3k1734
OK, so the first step is clearly a mistake with them dropping the $-1$. However, I'm going to try to explain what they may have done, even though it doesn't make sense. First, let's start with their (albeit wrong) equation:
$$sqrt{x+a}-sqrt{x}=x^{1/2}(1+frac{a}{2x}+O(x^{-2}))$$
Now, take the square root:
$$sqrt{sqrt{x+a}-sqrt{x}}=[x^{1/2}(1+frac{a}{2x}+O(x^{-2}))]^{1/2}$$
Distribute the exponent:
$$sqrt{sqrt{x+a}-sqrt{x}}=x^{1/4}[(1+frac{a}{2x}+O(x^{-2}))]^{1/2}$$
Now, the second expression on the right is in the form of $(1+z)^{1/2}$, so I will use the following formula:
$$(1+z)^{1/2}=1+frac z 2+O(z^2)rightarrow \ (1+frac a{2x}+O(x^{-2}))^{1/2}=1+frac{a}{4x}+frac{1}{2}O(x^{-2})+O(frac{a^2}{4x^2})=1+frac{a}{4x}+O(x^{-2})$$
Thus, we get:
$$sqrt{sqrt{x+a}-sqrt{x}}=x^{1/4}(1+frac{a}{4x}+O(x^{-2}))$$
Oh, and remember that $-1$ we dropped in the first step? Let's try adding that back in here, because...why not?
$$sqrt{sqrt{x+a}-sqrt{x}}=x^{1/4}(1+frac{a}{4x}+O(x^{-2})-1)$$
$$sqrt{sqrt{x+a}-sqrt{x}}=x^{1/4}(frac{a}{4x}+O(x^{-2}))$$
Thus, we get the same equation they did, and we only had to make two mathematical errors to get there!
answered 2 hours ago
Noble Mushtak
14.3k1734
answered 2 hours ago
Noble Mushtak
14.3k1734
answered 2 hours ago
Noble Mushtak
14.3k1734
answered 2 hours ago
Noble Mushtak
14.3k1734
14.3k1734
I like that you were able to decipher what the original proof writer most likely meant. Thank you for that.
– Thor Kamphefner
2 hours ago
add a comment |
I like that you were able to decipher what the original proof writer most likely meant. Thank you for that.
– Thor Kamphefner
2 hours ago
I like that you were able to decipher what the original proof writer most likely meant. Thank you for that.
– Thor Kamphefner
2 hours ago
I like that you were able to decipher what the original proof writer most likely meant. Thank you for that.
– Thor Kamphefner
2 hours ago
add a comment |
Thor Kamphefner is a new contributor. Be nice, and check out our Code of Conduct.
Thor Kamphefner is a new contributor. Be nice, and check out our Code of Conduct.
Thor Kamphefner is a new contributor. Be nice, and check out our Code of Conduct.
Thor Kamphefner is a new contributor. Be nice, and check out our Code of Conduct.
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For what it’s worth, you aren’t the first person to have asked about this. math.stackexchange.com/questions/2020837/…
– Steve Kass
2 hours ago
I am not sure if you have copied things properly. Why do we have $O(x^{2})$ in one place and $O(x^{-2})$ at others?
– Kavi Rama Murthy
2 hours ago
Note that on the right side of the first equation, x is in the denominator, whereas at the top, x is in the numerator.
– Thor Kamphefner
2 hours ago
@KaviRamaMurthy Not OP, but that is because the new "$x$" is now $a/x$, so when you substitute that in $O(x^2)$ becomes $O(a^2/x^2)$, which is the same as $O(x^{-2})$
– Noble Mushtak
2 hours ago
You should show the question. I looked up the reference and it doesn't appear to be what you want. Also, where is the proof that you refer to?
– herb steinberg
2 hours ago