Trying to parse a Putnam solution from 1995












3














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







share|cite|improve this question









New contributor




Thor Kamphefner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
    3 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
    3 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
    3 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
    3 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
    3 hours ago
















3














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







share|cite|improve this question









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
    3 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
    3 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
    3 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
    3 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
    3 hours ago














3












3








3







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







share|cite|improve this question









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






share|cite|improve this question









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.











share|cite|improve this question









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.









share|cite|improve this question




share|cite|improve this question








edited 3 hours ago





















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 3 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
    3 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
    3 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
    3 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
    3 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
    3 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
    3 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
    3 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
    3 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
    3 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
    3 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
3 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
3 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
3 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
3 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
3 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
3 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
3 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
3 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
3 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
3 hours ago










1 Answer
1






active

oldest

votes


















7














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!






share|cite|improve this answer





















  • I like that you were able to decipher what the original proof writer most likely meant. Thank you for that.
    – Thor Kamphefner
    3 hours ago











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









7














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!






share|cite|improve this answer





















  • I like that you were able to decipher what the original proof writer most likely meant. Thank you for that.
    – Thor Kamphefner
    3 hours ago
















7














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!






share|cite|improve this answer





















  • I like that you were able to decipher what the original proof writer most likely meant. Thank you for that.
    – Thor Kamphefner
    3 hours ago














7












7








7






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!






share|cite|improve this answer












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!







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 3 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
    3 hours ago


















  • I like that you were able to decipher what the original proof writer most likely meant. Thank you for that.
    – Thor Kamphefner
    3 hours ago
















I like that you were able to decipher what the original proof writer most likely meant. Thank you for that.
– Thor Kamphefner
3 hours ago




I like that you were able to decipher what the original proof writer most likely meant. Thank you for that.
– Thor Kamphefner
3 hours ago










Thor Kamphefner is a new contributor. Be nice, and check out our Code of Conduct.










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數位音樂下載

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數位音樂下載 為非以「實體」方式來販賣音樂的相關產品，而使用數位格式如mp3、AAC等文件格式來進行銷售販賣的一种音乐产业形式。著名的數位音樂下載服務商有iTunes Store、RecoChoku、Google Play音乐和微软Zune等。数字音乐下载在二十一世纪初一度风靡全球， [1] [2] 使得传统唱片产业走向衰落，但其自身也在2010年代流媒体音乐服务如Apple Music、Spotify等的挤压下迅速失去市场份额。 [3] [4] 数字音乐下载有便于携带、传播迅速、制作简单成本低等优点，与流媒体音乐相比也能更好保障音乐人权益； [5] 但它同时很可能带来侵犯版权和滥用的情形， [6] 也给音乐产业带来了诸多不利影响。数字音乐下载为音乐人和乐迷提供了更加便捷直接的交流渠道，也一定程度上推动了网络音乐人的出现和独立音乐的迅速发展。 [7] 目录 1 历史 1.1 技术积淀（1987-2000） 1.2 出现与兴起（2000-2004） 1.3 唱片市场的颠覆者（2004-2007） 1.4 巅峰和衰落（2005-2013） 1.5 摇摇欲坠的现状与不可预知的明天（2014至今） 2 主要文件格式 2.1 PCM编码格式 2.1.1 WAV 2.1.2 APE 2.1.3 FLAC 2.2 苹果推出的无损文件格式 2.2.1 AIFF 2.2.2 ALAC 2.3 有损压缩格式 2.3.1 MP3 2.3.2 AAC 3 主要市场 3.1 iTunes Store 3.2 Google Play音乐商店 3.3 亚马逊音乐商店 3.4 部分其他音乐商店 4 影响与评价 4.1 对音乐产业的影响 4.2 对科技行业的影响 4.3 正面评价 4.4 负面评价 4.4.1 侵权问题 4.4.2 经营模式不明晰 4.4.3 不适应移动时代 4.5 文化现象 5 延伸阅读 6 参考资料 历史 苹果iTunes+iPod的软硬件搭配让数字音...
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When can things happen in Etherscan, such as the picture below?

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2 I don't know why balance of the address is 0. when can this situation? etherscan balances share | improve this question asked yesterday Ru Hi Ru Hi 11 1 New contributor Ru Hi is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  ...
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格利澤436b

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格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP...
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