Trying to parse a Putnam solution from 1995
Clash Royale CLAN TAG#URR8PPP
$begingroup$
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:
beginalign*
sqrtx+a-sqrtx &= x^1/2(sqrt1+a/x - 1) \
&= x^1/2(1 + a/2x + O(x^-2)),
endalign*
It would appear that’s an error; it would appear the proof writer dropped the minus 1.
But it continues, hence
$sqrtsqrtx+a - sqrtx = x^1/4 (a/4x + O(x^-2))$
Which I’m at a loss how to arrive at.
binomial-theorem
asked Dec 30 '18 at 0:36
Thor KamphefnerThor Kamphefner
445
$endgroup$
|
show 2 more comments
$begingroup$
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:
beginalign*
sqrtx+a-sqrtx &= x^1/2(sqrt1+a/x - 1) \
&= x^1/2(1 + a/2x + O(x^-2)),
endalign*
It would appear that’s an error; it would appear the proof writer dropped the minus 1.
But it continues, hence
$sqrtsqrtx+a - sqrtx = x^1/4 (a/4x + O(x^-2))$
Which I’m at a loss how to arrive at.
binomial-theorem
asked Dec 30 '18 at 0:36
Thor KamphefnerThor Kamphefner
445
$endgroup$
$begingroup$
For what it’s worth, you aren’t the first person to have asked about this. math.stackexchange.com/questions/2020837/…
$endgroup$
– Steve Kass
Dec 30 '18 at 0:44
$begingroup$
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?
$endgroup$
– Kavi Rama Murthy
Dec 30 '18 at 0:45
$begingroup$
Note that on the right side of the first equation, x is in the denominator, whereas at the top, x is in the numerator.
$endgroup$
– Thor Kamphefner
Dec 30 '18 at 0:46
$begingroup$
@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 $O(x^-2)$
$endgroup$
– Noble Mushtak
Dec 30 '18 at 0:46
$begingroup$
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?
$endgroup$
– herb steinberg
Dec 30 '18 at 0:46
|
show 2 more comments
$begingroup$
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:
beginalign*
sqrtx+a-sqrtx &= x^1/2(sqrt1+a/x - 1) \
&= x^1/2(1 + a/2x + O(x^-2)),
endalign*
It would appear that’s an error; it would appear the proof writer dropped the minus 1.
But it continues, hence
$sqrtsqrtx+a - sqrtx = x^1/4 (a/4x + O(x^-2))$
Which I’m at a loss how to arrive at.
binomial-theorem
asked Dec 30 '18 at 0:36
Thor KamphefnerThor Kamphefner
445
$endgroup$
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:
beginalign*
sqrtx+a-sqrtx &= x^1/2(sqrt1+a/x - 1) \
&= x^1/2(1 + a/2x + O(x^-2)),
endalign*
It would appear that’s an error; it would appear the proof writer dropped the minus 1.
But it continues, hence
$sqrtsqrtx+a - sqrtx = x^1/4 (a/4x + O(x^-2))$
Which I’m at a loss how to arrive at.
binomial-theorem
binomial-theorem
asked Dec 30 '18 at 0:36
Thor KamphefnerThor Kamphefner
445
asked Dec 30 '18 at 0:36
Thor KamphefnerThor Kamphefner
445
edited Dec 30 '18 at 0:47
Thor Kamphefner
asked Dec 30 '18 at 0:36
Thor KamphefnerThor Kamphefner
445
asked Dec 30 '18 at 0:36
Thor KamphefnerThor Kamphefner
445
asked Dec 30 '18 at 0:36
Thor KamphefnerThor Kamphefner
445
445
$begingroup$
For what it’s worth, you aren’t the first person to have asked about this. math.stackexchange.com/questions/2020837/…
$endgroup$
– Steve Kass
Dec 30 '18 at 0:44
$begingroup$
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?
$endgroup$
– Kavi Rama Murthy
Dec 30 '18 at 0:45
$begingroup$
Note that on the right side of the first equation, x is in the denominator, whereas at the top, x is in the numerator.
$endgroup$
– Thor Kamphefner
Dec 30 '18 at 0:46
$begingroup$
@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 $O(x^-2)$
$endgroup$
– Noble Mushtak
Dec 30 '18 at 0:46
$begingroup$
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?
$endgroup$
– herb steinberg
Dec 30 '18 at 0:46
|
show 2 more comments
$begingroup$
For what it’s worth, you aren’t the first person to have asked about this. math.stackexchange.com/questions/2020837/…
$endgroup$
– Steve Kass
Dec 30 '18 at 0:44
$begingroup$
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?
$endgroup$
– Kavi Rama Murthy
Dec 30 '18 at 0:45
$begingroup$
Note that on the right side of the first equation, x is in the denominator, whereas at the top, x is in the numerator.
$endgroup$
– Thor Kamphefner
Dec 30 '18 at 0:46
$begingroup$
@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 $O(x^-2)$
$endgroup$
– Noble Mushtak
Dec 30 '18 at 0:46
$begingroup$
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?
$endgroup$
– herb steinberg
Dec 30 '18 at 0:46
$begingroup$
For what it’s worth, you aren’t the first person to have asked about this. math.stackexchange.com/questions/2020837/…
$endgroup$
– Steve Kass
Dec 30 '18 at 0:44
$begingroup$
For what it’s worth, you aren’t the first person to have asked about this. math.stackexchange.com/questions/2020837/…
$endgroup$
– Steve Kass
Dec 30 '18 at 0:44
$begingroup$
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?
$endgroup$
– Kavi Rama Murthy
Dec 30 '18 at 0:45
$begingroup$
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?
$endgroup$
– Kavi Rama Murthy
Dec 30 '18 at 0:45
$begingroup$
Note that on the right side of the first equation, x is in the denominator, whereas at the top, x is in the numerator.
$endgroup$
– Thor Kamphefner
Dec 30 '18 at 0:46
$begingroup$
Note that on the right side of the first equation, x is in the denominator, whereas at the top, x is in the numerator.
$endgroup$
– Thor Kamphefner
Dec 30 '18 at 0:46
$begingroup$
@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 $O(x^-2)$
$endgroup$
– Noble Mushtak
Dec 30 '18 at 0:46
$begingroup$
@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 $O(x^-2)$
$endgroup$
– Noble Mushtak
Dec 30 '18 at 0:46
$begingroup$
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?
$endgroup$
– herb steinberg
Dec 30 '18 at 0:46
$begingroup$
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?
$endgroup$
– herb steinberg
Dec 30 '18 at 0:46
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
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:
$$sqrtx+a-sqrtx=x^1/2(1+fraca2x+O(x^-2))$$
Now, take the square root:
$$sqrtsqrtx+a-sqrtx=[x^1/2(1+fraca2x+O(x^-2))]^1/2$$
Distribute the exponent:
$$sqrtsqrtx+a-sqrtx=x^1/4[(1+fraca2x+O(x^-2))]^1/2$$
Now, the second expression on the right is in the form of $(1+z)^1/2$, so I will use the following formula:
$$(1+z)^1/2=1+frac z 2+O(z^2)rightarrow \ (1+frac a2x+O(x^-2))^1/2=1+fraca4x+frac12O(x^-2)+O(fraca^24x^2)=1+fraca4x+O(x^-2)$$
Thus, we get:
$$sqrtsqrtx+a-sqrtx=x^1/4(1+fraca4x+O(x^-2))$$
Oh, and remember that $-1$ we dropped in the first step? Let's try adding that back in here, because...why not?
$$sqrtsqrtx+a-sqrtx=x^1/4(1+fraca4x+O(x^-2)-1)$$
$$sqrtsqrtx+a-sqrtx=x^1/4(fraca4x+O(x^-2))$$
Thus, we get the same equation they did, and we only had to make two mathematical errors to get there!
answered Dec 30 '18 at 0:52
Noble MushtakNoble Mushtak
15.2k1735
$endgroup$
$begingroup$
I like that you were able to decipher what the original proof writer most likely meant. Thank you for that.
$endgroup$
– Thor Kamphefner
Dec 30 '18 at 0:56
add a comment |
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1 Answer
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$begingroup$
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:
$$sqrtx+a-sqrtx=x^1/2(1+fraca2x+O(x^-2))$$
Now, take the square root:
$$sqrtsqrtx+a-sqrtx=[x^1/2(1+fraca2x+O(x^-2))]^1/2$$
Distribute the exponent:
$$sqrtsqrtx+a-sqrtx=x^1/4[(1+fraca2x+O(x^-2))]^1/2$$
Now, the second expression on the right is in the form of $(1+z)^1/2$, so I will use the following formula:
$$(1+z)^1/2=1+frac z 2+O(z^2)rightarrow \ (1+frac a2x+O(x^-2))^1/2=1+fraca4x+frac12O(x^-2)+O(fraca^24x^2)=1+fraca4x+O(x^-2)$$
Thus, we get:
$$sqrtsqrtx+a-sqrtx=x^1/4(1+fraca4x+O(x^-2))$$
Oh, and remember that $-1$ we dropped in the first step? Let's try adding that back in here, because...why not?
$$sqrtsqrtx+a-sqrtx=x^1/4(1+fraca4x+O(x^-2)-1)$$
$$sqrtsqrtx+a-sqrtx=x^1/4(fraca4x+O(x^-2))$$
Thus, we get the same equation they did, and we only had to make two mathematical errors to get there!
answered Dec 30 '18 at 0:52
Noble MushtakNoble Mushtak
15.2k1735
$endgroup$
$begingroup$
I like that you were able to decipher what the original proof writer most likely meant. Thank you for that.
$endgroup$
– Thor Kamphefner
Dec 30 '18 at 0:56
add a comment |
$begingroup$
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:
$$sqrtx+a-sqrtx=x^1/2(1+fraca2x+O(x^-2))$$
Now, take the square root:
$$sqrtsqrtx+a-sqrtx=[x^1/2(1+fraca2x+O(x^-2))]^1/2$$
Distribute the exponent:
$$sqrtsqrtx+a-sqrtx=x^1/4[(1+fraca2x+O(x^-2))]^1/2$$
Now, the second expression on the right is in the form of $(1+z)^1/2$, so I will use the following formula:
$$(1+z)^1/2=1+frac z 2+O(z^2)rightarrow \ (1+frac a2x+O(x^-2))^1/2=1+fraca4x+frac12O(x^-2)+O(fraca^24x^2)=1+fraca4x+O(x^-2)$$
Thus, we get:
$$sqrtsqrtx+a-sqrtx=x^1/4(1+fraca4x+O(x^-2))$$
Oh, and remember that $-1$ we dropped in the first step? Let's try adding that back in here, because...why not?
$$sqrtsqrtx+a-sqrtx=x^1/4(1+fraca4x+O(x^-2)-1)$$
$$sqrtsqrtx+a-sqrtx=x^1/4(fraca4x+O(x^-2))$$
Thus, we get the same equation they did, and we only had to make two mathematical errors to get there!
answered Dec 30 '18 at 0:52
Noble MushtakNoble Mushtak
15.2k1735
$endgroup$
$begingroup$
I like that you were able to decipher what the original proof writer most likely meant. Thank you for that.
$endgroup$
– Thor Kamphefner
Dec 30 '18 at 0:56
add a comment |
$begingroup$
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:
$$sqrtx+a-sqrtx=x^1/2(1+fraca2x+O(x^-2))$$
Now, take the square root:
$$sqrtsqrtx+a-sqrtx=[x^1/2(1+fraca2x+O(x^-2))]^1/2$$
Distribute the exponent:
$$sqrtsqrtx+a-sqrtx=x^1/4[(1+fraca2x+O(x^-2))]^1/2$$
Now, the second expression on the right is in the form of $(1+z)^1/2$, so I will use the following formula:
$$(1+z)^1/2=1+frac z 2+O(z^2)rightarrow \ (1+frac a2x+O(x^-2))^1/2=1+fraca4x+frac12O(x^-2)+O(fraca^24x^2)=1+fraca4x+O(x^-2)$$
Thus, we get:
$$sqrtsqrtx+a-sqrtx=x^1/4(1+fraca4x+O(x^-2))$$
Oh, and remember that $-1$ we dropped in the first step? Let's try adding that back in here, because...why not?
$$sqrtsqrtx+a-sqrtx=x^1/4(1+fraca4x+O(x^-2)-1)$$
$$sqrtsqrtx+a-sqrtx=x^1/4(fraca4x+O(x^-2))$$
Thus, we get the same equation they did, and we only had to make two mathematical errors to get there!
answered Dec 30 '18 at 0:52
Noble MushtakNoble Mushtak
15.2k1735
$endgroup$
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:
$$sqrtx+a-sqrtx=x^1/2(1+fraca2x+O(x^-2))$$
Now, take the square root:
$$sqrtsqrtx+a-sqrtx=[x^1/2(1+fraca2x+O(x^-2))]^1/2$$
Distribute the exponent:
$$sqrtsqrtx+a-sqrtx=x^1/4[(1+fraca2x+O(x^-2))]^1/2$$
Now, the second expression on the right is in the form of $(1+z)^1/2$, so I will use the following formula:
$$(1+z)^1/2=1+frac z 2+O(z^2)rightarrow \ (1+frac a2x+O(x^-2))^1/2=1+fraca4x+frac12O(x^-2)+O(fraca^24x^2)=1+fraca4x+O(x^-2)$$
Thus, we get:
$$sqrtsqrtx+a-sqrtx=x^1/4(1+fraca4x+O(x^-2))$$
Oh, and remember that $-1$ we dropped in the first step? Let's try adding that back in here, because...why not?
$$sqrtsqrtx+a-sqrtx=x^1/4(1+fraca4x+O(x^-2)-1)$$
$$sqrtsqrtx+a-sqrtx=x^1/4(fraca4x+O(x^-2))$$
Thus, we get the same equation they did, and we only had to make two mathematical errors to get there!
answered Dec 30 '18 at 0:52
Noble MushtakNoble Mushtak
15.2k1735
answered Dec 30 '18 at 0:52
Noble MushtakNoble Mushtak
15.2k1735
answered Dec 30 '18 at 0:52
Noble MushtakNoble Mushtak
15.2k1735
answered Dec 30 '18 at 0:52
Noble MushtakNoble Mushtak
15.2k1735
15.2k1735
$begingroup$
I like that you were able to decipher what the original proof writer most likely meant. Thank you for that.
$endgroup$
– Thor Kamphefner
Dec 30 '18 at 0:56
add a comment |
$begingroup$
I like that you were able to decipher what the original proof writer most likely meant. Thank you for that.
$endgroup$
– Thor Kamphefner
Dec 30 '18 at 0:56
$begingroup$
I like that you were able to decipher what the original proof writer most likely meant. Thank you for that.
$endgroup$
– Thor Kamphefner
Dec 30 '18 at 0:56
$begingroup$
I like that you were able to decipher what the original proof writer most likely meant. Thank you for that.
$endgroup$
– Thor Kamphefner
Dec 30 '18 at 0:56
add a comment |
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$begingroup$
For what it’s worth, you aren’t the first person to have asked about this. math.stackexchange.com/questions/2020837/…
$endgroup$
– Steve Kass
Dec 30 '18 at 0:44
$begingroup$
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?
$endgroup$
– Kavi Rama Murthy
Dec 30 '18 at 0:45
$begingroup$
Note that on the right side of the first equation, x is in the denominator, whereas at the top, x is in the numerator.
$endgroup$
– Thor Kamphefner
Dec 30 '18 at 0:46
$begingroup$
@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 $O(x^-2)$
$endgroup$
– Noble Mushtak
Dec 30 '18 at 0:46
$begingroup$
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?
$endgroup$
– herb steinberg
Dec 30 '18 at 0:46