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
    1 hour 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
    1 hour 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
    1 hour 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
    1 hour 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
    1 hour 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
    1 hour 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
    1 hour 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
    1 hour 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
    1 hour 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
    1 hour 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 1 hour 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 1 hour 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
    1 hour 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
    1 hour 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
    1 hour 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
    1 hour 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
    1 hour ago


















  • For what it’s worth, you aren’t the first person to have asked about this. math.stackexchange.com/questions/2020837/…
    – Steve Kass
    1 hour 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
    1 hour 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
    1 hour 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
    1 hour 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
    1 hour ago
















For what it’s worth, you aren’t the first person to have asked about this. math.stackexchange.com/questions/2020837/…
– Steve Kass
1 hour ago




For what it’s worth, you aren’t the first person to have asked about this. math.stackexchange.com/questions/2020837/…
– Steve Kass
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour ago










1 Answer
1






active

oldest

votes


















4














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
    1 hour 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









4














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
    1 hour ago
















4














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
    1 hour ago














4












4








4






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 1 hour ago









Noble Mushtak

14.2k1734




14.2k1734












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


















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
















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




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










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










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Fotorealismo

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Guerrino Boatto, Venice's St. Barnaba Church and canal Il fotorealismo è un genere di pittura basato sull'uso di una o più fotografie dalle quali l'artista prende le informazioni necessarie da utilizzare poi nel processo della creazione dell'opera, in genere ad olio o ad acrilico, al fine di avere un aspetto finale il più simile possibile alla fotografia. Indice 1 Storia 2 Lista dei fotorealisti 3 Note 4 Voci correlate 5 Altri progetti Storia | Il fotorealismo è un movimento che nasce negli Stati Uniti d'America alla fine degli anni sessanta [1] sulla scia della Pop Art [2] [3] [4] , e insieme condividono il carattere reazionario verso la travolgente ascesa dei media, che alla metà del ventesimo secolo si trasforma in un fenomeno di massa talmente imponente da minacciare il valore dell'immagine nell'arte. [5] [6] Ne consegue così che, colto il gesto d'inizio che fu della Pop Art, i fotorealisti aprono ad un ...
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Sidney Franklin

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Questa voce sull'argomento registi statunitensi è solo un abbozzo . Contribuisci a migliorarla secondo le convenzioni di Wikipedia. Sidney Franklin, foto di Carl Van Vechten Oscar alla memoria Irving G. Thalberg 1943 Sidney Franklin , all'anagrafe Sidney Arnold Franklin , (San Francisco, 21 marzo 1893 – Santa Monica, 18 maggio 1972), è stato un regista, produttore cinematografico e attore statunitense. Il 20 giugno 1963, si sposò con l'attrice australiana Enid Bennett. Il matrimonio durò fino alla morte dell'attrice, il 14 maggio 1969 [1] Indice 1 Filmografia 1.1 Regista 1.2 Aiuto regia (parziale) 1.3 Produttore 1.4 Attore (parziale) 2 Note 3 Altri progetti 4 Collegamenti esterni Filmografia | Regista | The Baby , co-regia Chester M. Franklin (1915) The Rivals , co-regia Chester M. Franklin (1915) Little Dick's First Case , co-regia Chester M. Franklin (1915) Her Filmland Hero , co-...
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