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Correct use of smash with math and root signs

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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

5

How can I get the same height on the second root-sign as the first?;

$$sqrttfrac1+2cdot4a^2+(4a^2)^216a^2
=sqrttfrac1+2cdot4a^2+(4a^2)^2(4a)^2$$

I've tried different methods using smash, vphantom and rule[] but could not work out which was the best and most 'proper' way of solving this 'problem'. TIA.

equations

share|improve this question

edited Jul 23 at 21:49

Mico

300k3232 gold badges411411 silver badges816816 bronze badges

asked Jul 23 at 21:42

mf67mf67

6533 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks for all help, but I cannot understand why the ^2 cannot go under the smash command in the solution suggested by Mico. The following two examples does not result in the same display: smash(4a)^2 and smash(4a)^2 (Sorry, did not know how to type in the comment box to get better display of codes.)
  
  – mf67
  Jul 24 at 10:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  If you compare the outputs of smash(4a)^2 and smash(4a)^2, you'll notice that the exponent is placed higher relative to the baseline if the scope of smash includes the exponent -- not by a huge amount, for sure, but by about 1 or 2 points. This difference results in a slight increase in the overall height of the denominator which, in turn, explains why LaTeX sees fit to employ a taller (and deeper) square root symbol when it processes smash(4a)^2.
  
  – Mico
  Jul 24 at 12:31
  
  

  
  
  
  

add a comment | 

5

How can I get the same height on the second root-sign as the first?;

$$sqrttfrac1+2cdot4a^2+(4a^2)^216a^2
=sqrttfrac1+2cdot4a^2+(4a^2)^2(4a)^2$$

I've tried different methods using smash, vphantom and rule[] but could not work out which was the best and most 'proper' way of solving this 'problem'. TIA.

equations

share|improve this question

edited Jul 23 at 21:49

Mico

300k3232 gold badges411411 silver badges816816 bronze badges

asked Jul 23 at 21:42

mf67mf67

6533 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks for all help, but I cannot understand why the ^2 cannot go under the smash command in the solution suggested by Mico. The following two examples does not result in the same display: smash(4a)^2 and smash(4a)^2 (Sorry, did not know how to type in the comment box to get better display of codes.)
  
  – mf67
  Jul 24 at 10:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  If you compare the outputs of smash(4a)^2 and smash(4a)^2, you'll notice that the exponent is placed higher relative to the baseline if the scope of smash includes the exponent -- not by a huge amount, for sure, but by about 1 or 2 points. This difference results in a slight increase in the overall height of the denominator which, in turn, explains why LaTeX sees fit to employ a taller (and deeper) square root symbol when it processes smash(4a)^2.
  
  – Mico
  Jul 24 at 12:31
  
  

  
  
  
  

add a comment | 

How can I get the same height on the second root-sign as the first?;

$$sqrttfrac1+2cdot4a^2+(4a^2)^216a^2
=sqrttfrac1+2cdot4a^2+(4a^2)^2(4a)^2$$

I've tried different methods using smash, vphantom and rule[] but could not work out which was the best and most 'proper' way of solving this 'problem'. TIA.

equations

share|improve this question

edited Jul 23 at 21:49

Mico

300k3232 gold badges411411 silver badges816816 bronze badges

asked Jul 23 at 21:42

mf67mf67

6533 bronze badges

$$sqrttfrac1+2cdot4a^2+(4a^2)^216a^2
=sqrttfrac1+2cdot4a^2+(4a^2)^2(4a)^2$$

share|improve this question

edited Jul 23 at 21:49

Mico

300k3232 gold badges411411 silver badges816816 bronze badges

asked Jul 23 at 21:42

mf67mf67

6533 bronze badges

share|improve this question

edited Jul 23 at 21:49

Mico

300k3232 gold badges411411 silver badges816816 bronze badges

asked Jul 23 at 21:42

mf67mf67

6533 bronze badges

edited Jul 23 at 21:49

Mico

300k3232 gold badges411411 silver badges816816 bronze badges

edited Jul 23 at 21:49

Mico

300k3232 gold badges411411 silver badges816816 bronze badges

asked Jul 23 at 21:42

mf67mf67

6533 bronze badges

asked Jul 23 at 21:42

mf67mf67

6533 bronze badges

3 Answers
3

active

oldest

votes

6

What you need to do is replace (4a)^2 in the second denominator with either smash(4a)^2 or smash[b](4a)^2. This yields compact-looking square root terms, and it works with both tfrac and dfrac.

Observe that if you, alternatively, replaced 16a^2 in the first denominator with 16a^2mathstrut, the two square root symbols would also have equal sizes. However, they would be much taller -- excessively and unnecessarily so, IMNSHO -- than with the adjustment suggested above.

documentclassarticle
usepackageamsmath 
begindocument
[
 sqrtdfrac1+2cdot4a^2+(4a^2)^216a^2
=sqrtdfrac1+2cdot4a^2+(4a^2)^2smash(4a)^2
]

[
 sqrttfrac1+2cdot4a^2+(4a^2)^216a^2
=sqrttfrac1+2cdot4a^2+(4a^2)^2smash(4a)^2
]
enddocument

share|improve this answer

edited Jul 24 at 12:28

answered Jul 23 at 21:57

MicoMico

300k3232 gold badges411411 silver badges816816 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Please see comment in the question, I might have put it in the wrong place… Perhaps it should have been here.
  
  – mf67
  Jul 24 at 10:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @mf67 - Please see the comment I left below your follow-up comment. Did I answer your question? Please advise.
  
  – Mico
  Jul 24 at 12:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I don't fully understand the effect of smash it seems. I thought that smash(4a)^2 would remove any height and 'equal' it to, e.g. an "a" or "x" and thus not extend the root sign any deeper. Instead it seems like it, as you write, places the exponent higher, but normal, than in smash(4a)^2 and thus forces down the base and therefore the depth of the root sign. A 0-height box, as I thought smash(4a)^2 was, would not do that? The smash(4a)^2 'compress' the exponent and, although not extremely disturbingly 'ugly', it looks a bit different from the a^2 in the numerator.
  
  – mf67
  Jul 24 at 17:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @mf67 - I don't thinks it's entirely correct to call smash(4a)^2 a "zero-height box". Please compile [ sqrtfrac1+2cdot4a^2+(4a^2)^2smash(4a)^2 = sqrtfrac1+2cdot4a^2+(4a^2)^2smash(4a)^2 = sqrtfrac1+2cdot4a^2+(4a^2)^2smashsmash(4a)^2 ]. You'll see that the square root term in the middle, which contains smash(4a)^2, is taller than the other two. It looks like it's necessary to smash the term (4a)^2 twice in order to obtain the compact square root expression shown on the left. That's why I recommended writing smash(4a)^2...
  
  – Mico
  Jul 24 at 17:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @mf67 - If this topic continues to puzzle you, may I suggest that you post a new query, in which you would ask for an explanation of how exactly smash works? There are some amazing TeX wizards on this site; they can explain how smash works in all gory detail -- and much much better than I ever could.
  
  – Mico
  Jul 24 at 17:50
  
  

  
  
  
  

add a comment | 

11

Equalizing radicals is something of a black art.

The difference is due to the right hand side having parentheses. We can cope with this by adding mathstrut in the left hand side denominator. But this makes TeX choose the next size for the radical. Using smash[b]... for the denominator doesn't help.

The problem is that tfrac imposes textstyle, which has raised denominators. One could use cramped from mathtools, but there's a slicker solution:

documentclassarticle
usepackageamsmath

begindocument

[
textstyle
sqrtfrac1+2cdot4a^2+(4a^2)^2mathstrut 16a^2
=sqrtfrac1+2cdot4a^2+(4a^2)^2(4a)^2
]

enddocument

share|improve this answer

answered Jul 23 at 22:32

egregegreg

759k9090 gold badges19861986 silver badges33323332 bronze badges

add a comment | 

5

A simple vphantom will do the trick. And, please, don't use the plain TeX construct $$ ... $$ for unnumbered displayed equations. Use [ ... ] instead.

documentclass[11pt, a4paper]article
usepackageamsmath

begindocument

[ sqrttfrac1+2cdot4a^2+(4a^2)^2vphantom)16a^2 =sqrttfrac1+2cdot4a^2+(4a^2)^2(4a)^2 ]

enddocument

share|improve this answer

edited Jul 24 at 17:59

answered Jul 23 at 21:55

BernardBernard

186k77 gold badges8484 silver badges220220 bronze badges

add a comment | 

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6

What you need to do is replace (4a)^2 in the second denominator with either smash(4a)^2 or smash[b](4a)^2. This yields compact-looking square root terms, and it works with both tfrac and dfrac.

Observe that if you, alternatively, replaced 16a^2 in the first denominator with 16a^2mathstrut, the two square root symbols would also have equal sizes. However, they would be much taller -- excessively and unnecessarily so, IMNSHO -- than with the adjustment suggested above.

documentclassarticle
usepackageamsmath 
begindocument
[
 sqrtdfrac1+2cdot4a^2+(4a^2)^216a^2
=sqrtdfrac1+2cdot4a^2+(4a^2)^2smash(4a)^2
]

[
 sqrttfrac1+2cdot4a^2+(4a^2)^216a^2
=sqrttfrac1+2cdot4a^2+(4a^2)^2smash(4a)^2
]
enddocument

share|improve this answer

edited Jul 24 at 12:28

answered Jul 23 at 21:57

MicoMico

300k3232 gold badges411411 silver badges816816 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Please see comment in the question, I might have put it in the wrong place… Perhaps it should have been here.
  
  – mf67
  Jul 24 at 10:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @mf67 - Please see the comment I left below your follow-up comment. Did I answer your question? Please advise.
  
  – Mico
  Jul 24 at 12:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I don't fully understand the effect of smash it seems. I thought that smash(4a)^2 would remove any height and 'equal' it to, e.g. an "a" or "x" and thus not extend the root sign any deeper. Instead it seems like it, as you write, places the exponent higher, but normal, than in smash(4a)^2 and thus forces down the base and therefore the depth of the root sign. A 0-height box, as I thought smash(4a)^2 was, would not do that? The smash(4a)^2 'compress' the exponent and, although not extremely disturbingly 'ugly', it looks a bit different from the a^2 in the numerator.
  
  – mf67
  Jul 24 at 17:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @mf67 - I don't thinks it's entirely correct to call smash(4a)^2 a "zero-height box". Please compile [ sqrtfrac1+2cdot4a^2+(4a^2)^2smash(4a)^2 = sqrtfrac1+2cdot4a^2+(4a^2)^2smash(4a)^2 = sqrtfrac1+2cdot4a^2+(4a^2)^2smashsmash(4a)^2 ]. You'll see that the square root term in the middle, which contains smash(4a)^2, is taller than the other two. It looks like it's necessary to smash the term (4a)^2 twice in order to obtain the compact square root expression shown on the left. That's why I recommended writing smash(4a)^2...
  
  – Mico
  Jul 24 at 17:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @mf67 - If this topic continues to puzzle you, may I suggest that you post a new query, in which you would ask for an explanation of how exactly smash works? There are some amazing TeX wizards on this site; they can explain how smash works in all gory detail -- and much much better than I ever could.
  
  – Mico
  Jul 24 at 17:50
  
  

  
  
  
  

add a comment | 

6

What you need to do is replace (4a)^2 in the second denominator with either smash(4a)^2 or smash[b](4a)^2. This yields compact-looking square root terms, and it works with both tfrac and dfrac.

Observe that if you, alternatively, replaced 16a^2 in the first denominator with 16a^2mathstrut, the two square root symbols would also have equal sizes. However, they would be much taller -- excessively and unnecessarily so, IMNSHO -- than with the adjustment suggested above.

documentclassarticle
usepackageamsmath 
begindocument
[
 sqrtdfrac1+2cdot4a^2+(4a^2)^216a^2
=sqrtdfrac1+2cdot4a^2+(4a^2)^2smash(4a)^2
]

[
 sqrttfrac1+2cdot4a^2+(4a^2)^216a^2
=sqrttfrac1+2cdot4a^2+(4a^2)^2smash(4a)^2
]
enddocument

share|improve this answer

edited Jul 24 at 12:28

answered Jul 23 at 21:57

MicoMico

300k3232 gold badges411411 silver badges816816 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Please see comment in the question, I might have put it in the wrong place… Perhaps it should have been here.
  
  – mf67
  Jul 24 at 10:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @mf67 - Please see the comment I left below your follow-up comment. Did I answer your question? Please advise.
  
  – Mico
  Jul 24 at 12:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I don't fully understand the effect of smash it seems. I thought that smash(4a)^2 would remove any height and 'equal' it to, e.g. an "a" or "x" and thus not extend the root sign any deeper. Instead it seems like it, as you write, places the exponent higher, but normal, than in smash(4a)^2 and thus forces down the base and therefore the depth of the root sign. A 0-height box, as I thought smash(4a)^2 was, would not do that? The smash(4a)^2 'compress' the exponent and, although not extremely disturbingly 'ugly', it looks a bit different from the a^2 in the numerator.
  
  – mf67
  Jul 24 at 17:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @mf67 - I don't thinks it's entirely correct to call smash(4a)^2 a "zero-height box". Please compile [ sqrtfrac1+2cdot4a^2+(4a^2)^2smash(4a)^2 = sqrtfrac1+2cdot4a^2+(4a^2)^2smash(4a)^2 = sqrtfrac1+2cdot4a^2+(4a^2)^2smashsmash(4a)^2 ]. You'll see that the square root term in the middle, which contains smash(4a)^2, is taller than the other two. It looks like it's necessary to smash the term (4a)^2 twice in order to obtain the compact square root expression shown on the left. That's why I recommended writing smash(4a)^2...
  
  – Mico
  Jul 24 at 17:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @mf67 - If this topic continues to puzzle you, may I suggest that you post a new query, in which you would ask for an explanation of how exactly smash works? There are some amazing TeX wizards on this site; they can explain how smash works in all gory detail -- and much much better than I ever could.
  
  – Mico
  Jul 24 at 17:50
  
  

  
  
  
  

add a comment | 

What you need to do is replace (4a)^2 in the second denominator with either smash(4a)^2 or smash[b](4a)^2. This yields compact-looking square root terms, and it works with both tfrac and dfrac.

Observe that if you, alternatively, replaced 16a^2 in the first denominator with 16a^2mathstrut, the two square root symbols would also have equal sizes. However, they would be much taller -- excessively and unnecessarily so, IMNSHO -- than with the adjustment suggested above.

documentclassarticle
usepackageamsmath 
begindocument
[
 sqrtdfrac1+2cdot4a^2+(4a^2)^216a^2
=sqrtdfrac1+2cdot4a^2+(4a^2)^2smash(4a)^2
]

[
 sqrttfrac1+2cdot4a^2+(4a^2)^216a^2
=sqrttfrac1+2cdot4a^2+(4a^2)^2smash(4a)^2
]
enddocument

share|improve this answer

edited Jul 24 at 12:28

answered Jul 23 at 21:57

MicoMico

300k3232 gold badges411411 silver badges816816 bronze badges

documentclassarticle
usepackageamsmath 
begindocument
[
 sqrtdfrac1+2cdot4a^2+(4a^2)^216a^2
=sqrtdfrac1+2cdot4a^2+(4a^2)^2smash(4a)^2
]

[
 sqrttfrac1+2cdot4a^2+(4a^2)^216a^2
=sqrttfrac1+2cdot4a^2+(4a^2)^2smash(4a)^2
]
enddocument

share|improve this answer

edited Jul 24 at 12:28

answered Jul 23 at 21:57

MicoMico

300k3232 gold badges411411 silver badges816816 bronze badges

answered Jul 23 at 21:57

MicoMico

300k3232 gold badges411411 silver badges816816 bronze badges

answered Jul 23 at 21:57

MicoMico

300k3232 gold badges411411 silver badges816816 bronze badges

11

Equalizing radicals is something of a black art.

The difference is due to the right hand side having parentheses. We can cope with this by adding mathstrut in the left hand side denominator. But this makes TeX choose the next size for the radical. Using smash[b]... for the denominator doesn't help.

The problem is that tfrac imposes textstyle, which has raised denominators. One could use cramped from mathtools, but there's a slicker solution:

documentclassarticle
usepackageamsmath

begindocument

[
textstyle
sqrtfrac1+2cdot4a^2+(4a^2)^2mathstrut 16a^2
=sqrtfrac1+2cdot4a^2+(4a^2)^2(4a)^2
]

enddocument

share|improve this answer

answered Jul 23 at 22:32

egregegreg

759k9090 gold badges19861986 silver badges33323332 bronze badges

add a comment | 

11

Equalizing radicals is something of a black art.

The difference is due to the right hand side having parentheses. We can cope with this by adding mathstrut in the left hand side denominator. But this makes TeX choose the next size for the radical. Using smash[b]... for the denominator doesn't help.

The problem is that tfrac imposes textstyle, which has raised denominators. One could use cramped from mathtools, but there's a slicker solution:

documentclassarticle
usepackageamsmath

begindocument

[
textstyle
sqrtfrac1+2cdot4a^2+(4a^2)^2mathstrut 16a^2
=sqrtfrac1+2cdot4a^2+(4a^2)^2(4a)^2
]

enddocument

share|improve this answer

answered Jul 23 at 22:32

egregegreg

759k9090 gold badges19861986 silver badges33323332 bronze badges

add a comment | 

Equalizing radicals is something of a black art.

The difference is due to the right hand side having parentheses. We can cope with this by adding mathstrut in the left hand side denominator. But this makes TeX choose the next size for the radical. Using smash[b]... for the denominator doesn't help.

The problem is that tfrac imposes textstyle, which has raised denominators. One could use cramped from mathtools, but there's a slicker solution:

documentclassarticle
usepackageamsmath

begindocument

[
textstyle
sqrtfrac1+2cdot4a^2+(4a^2)^2mathstrut 16a^2
=sqrtfrac1+2cdot4a^2+(4a^2)^2(4a)^2
]

enddocument

share|improve this answer

answered Jul 23 at 22:32

egregegreg

759k9090 gold badges19861986 silver badges33323332 bronze badges

documentclassarticle
usepackageamsmath

begindocument

[
textstyle
sqrtfrac1+2cdot4a^2+(4a^2)^2mathstrut 16a^2
=sqrtfrac1+2cdot4a^2+(4a^2)^2(4a)^2
]

enddocument

share|improve this answer

answered Jul 23 at 22:32

egregegreg

759k9090 gold badges19861986 silver badges33323332 bronze badges

answered Jul 23 at 22:32

egregegreg

759k9090 gold badges19861986 silver badges33323332 bronze badges

answered Jul 23 at 22:32

egregegreg

759k9090 gold badges19861986 silver badges33323332 bronze badges

5

A simple vphantom will do the trick. And, please, don't use the plain TeX construct $$ ... $$ for unnumbered displayed equations. Use [ ... ] instead.

documentclass[11pt, a4paper]article
usepackageamsmath

begindocument

[ sqrttfrac1+2cdot4a^2+(4a^2)^2vphantom)16a^2 =sqrttfrac1+2cdot4a^2+(4a^2)^2(4a)^2 ]

enddocument

share|improve this answer

edited Jul 24 at 17:59

answered Jul 23 at 21:55

BernardBernard

186k77 gold badges8484 silver badges220220 bronze badges

add a comment | 

5

A simple vphantom will do the trick. And, please, don't use the plain TeX construct $$ ... $$ for unnumbered displayed equations. Use [ ... ] instead.

documentclass[11pt, a4paper]article
usepackageamsmath

begindocument

[ sqrttfrac1+2cdot4a^2+(4a^2)^2vphantom)16a^2 =sqrttfrac1+2cdot4a^2+(4a^2)^2(4a)^2 ]

enddocument

share|improve this answer

edited Jul 24 at 17:59

answered Jul 23 at 21:55

BernardBernard

186k77 gold badges8484 silver badges220220 bronze badges

add a comment | 

A simple vphantom will do the trick. And, please, don't use the plain TeX construct $$ ... $$ for unnumbered displayed equations. Use [ ... ] instead.

documentclass[11pt, a4paper]article
usepackageamsmath

begindocument

[ sqrttfrac1+2cdot4a^2+(4a^2)^2vphantom)16a^2 =sqrttfrac1+2cdot4a^2+(4a^2)^2(4a)^2 ]

enddocument

share|improve this answer

edited Jul 24 at 17:59

answered Jul 23 at 21:55

BernardBernard

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documentclass[11pt, a4paper]article
usepackageamsmath

begindocument

[ sqrttfrac1+2cdot4a^2+(4a^2)^2vphantom)16a^2 =sqrttfrac1+2cdot4a^2+(4a^2)^2(4a)^2 ]

enddocument

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edited Jul 24 at 17:59

answered Jul 23 at 21:55

BernardBernard

186k77 gold badges8484 silver badges220220 bronze badges

answered Jul 23 at 21:55

BernardBernard

186k77 gold badges8484 silver badges220220 bronze badges

answered Jul 23 at 21:55

BernardBernard

186k77 gold badges8484 silver badges220220 bronze badges