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.
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














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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Puebla de Zaragoza

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Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc
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Musa

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Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
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