Finding derivative of $sqrt[3]sin(2x)$ using only definition of derivative
Clash Royale CLAN TAG#URR8PPP
$begingroup$
First post here, so hello everyone.
Here's the problem:
Find the first derivative of: $$sqrt[3]sin(2x)$$
But, you can only use the difference quotient...
(i.e. the limit of $fracf(x+h)-f(x)h$ as $h rightarrow 0$)
This is a particular problem given by my prof. It's not for credit or even an assignment, she gave it as an interestingly difficult problem with some twist and turns. I've worked on it for several hours now and would like to see how others approach solving it as I'm wrapped around an axel after using double-angle ID and clearing the cubes in the numerator. Any advice is appreciated.
Thanks.
real-analysis calculus derivatives trigonometry
edited Feb 15 at 9:02
user21820
39.4k543155
asked Feb 14 at 14:24
Sawyer BensonSawyer Benson
719
$endgroup$
|
show 4 more comments
$begingroup$
First post here, so hello everyone.
Here's the problem:
Find the first derivative of: $$sqrt[3]sin(2x)$$
But, you can only use the difference quotient...
(i.e. the limit of $fracf(x+h)-f(x)h$ as $h rightarrow 0$)
This is a particular problem given by my prof. It's not for credit or even an assignment, she gave it as an interestingly difficult problem with some twist and turns. I've worked on it for several hours now and would like to see how others approach solving it as I'm wrapped around an axel after using double-angle ID and clearing the cubes in the numerator. Any advice is appreciated.
Thanks.
real-analysis calculus derivatives trigonometry
edited Feb 15 at 9:02
user21820
39.4k543155
asked Feb 14 at 14:24
Sawyer BensonSawyer Benson
719
$endgroup$
3
$begingroup$
Can you give some context on why you want to do that with only the difference quotient? Surely you can replicate the proof of the chain rule, but it's not going to be pretty (although pointless all the more).
$endgroup$
– Klaus
Feb 14 at 14:35
1
$begingroup$
Thanks, Cameron! Unfortunately I cannot use that either. I know how to solve it using the chain rule and even L'Hopital's rule, however, the challenge it to solving it algebraically using nothing more than trig IDs (half angle) and trig limit rules (lim of sinx/x = 1). I know it's a pain, but that's the game! Thanks for all the help!
$endgroup$
– Sawyer Benson
Feb 14 at 14:58
1
$begingroup$
It seems to work out straightforwardly enough if you clear the cube roots in the numerator (as you've already done), apply the identity for $sin A - sin B$ to the numerator of the resulting expression, and use the fact that $lim_hto0fracsin hh = 1$ (as you're allowed to).
$endgroup$
– Calum Gilhooley
Feb 14 at 15:09
1
$begingroup$
@JeanMarie Aww, I really liked that title! :)
$endgroup$
– Calum Gilhooley
Feb 15 at 11:51
1
$begingroup$
@JeanMarie, I believe you are conflating a title with a direct title. The original title, purposefully written to be indirect, was crafted to equally capture attention as it was to communicate the problem ahead. Despite its objective lack of creativity, I appreciate the specificity of your direct-form edited above.
$endgroup$
– Sawyer Benson
Feb 15 at 14:52
|
show 4 more comments
$begingroup$
First post here, so hello everyone.
Here's the problem:
Find the first derivative of: $$sqrt[3]sin(2x)$$
But, you can only use the difference quotient...
(i.e. the limit of $fracf(x+h)-f(x)h$ as $h rightarrow 0$)
This is a particular problem given by my prof. It's not for credit or even an assignment, she gave it as an interestingly difficult problem with some twist and turns. I've worked on it for several hours now and would like to see how others approach solving it as I'm wrapped around an axel after using double-angle ID and clearing the cubes in the numerator. Any advice is appreciated.
Thanks.
real-analysis calculus derivatives trigonometry
edited Feb 15 at 9:02
user21820
39.4k543155
asked Feb 14 at 14:24
Sawyer BensonSawyer Benson
719
$endgroup$
First post here, so hello everyone.
Here's the problem:
Find the first derivative of: $$sqrt[3]sin(2x)$$
But, you can only use the difference quotient...
(i.e. the limit of $fracf(x+h)-f(x)h$ as $h rightarrow 0$)
This is a particular problem given by my prof. It's not for credit or even an assignment, she gave it as an interestingly difficult problem with some twist and turns. I've worked on it for several hours now and would like to see how others approach solving it as I'm wrapped around an axel after using double-angle ID and clearing the cubes in the numerator. Any advice is appreciated.
Thanks.
real-analysis calculus derivatives trigonometry
real-analysis calculus derivatives trigonometry
edited Feb 15 at 9:02
user21820
39.4k543155
asked Feb 14 at 14:24
Sawyer BensonSawyer Benson
719
edited Feb 15 at 9:02
user21820
39.4k543155
asked Feb 14 at 14:24
Sawyer BensonSawyer Benson
719
edited Feb 15 at 9:02
user21820
39.4k543155
edited Feb 15 at 9:02
user21820
39.4k543155
edited Feb 15 at 9:02
user21820
39.4k543155
39.4k543155
asked Feb 14 at 14:24
Sawyer BensonSawyer Benson
719
asked Feb 14 at 14:24
Sawyer BensonSawyer Benson
719
asked Feb 14 at 14:24
Sawyer BensonSawyer Benson
719
719
3
$begingroup$
Can you give some context on why you want to do that with only the difference quotient? Surely you can replicate the proof of the chain rule, but it's not going to be pretty (although pointless all the more).
$endgroup$
– Klaus
Feb 14 at 14:35
1
$begingroup$
Thanks, Cameron! Unfortunately I cannot use that either. I know how to solve it using the chain rule and even L'Hopital's rule, however, the challenge it to solving it algebraically using nothing more than trig IDs (half angle) and trig limit rules (lim of sinx/x = 1). I know it's a pain, but that's the game! Thanks for all the help!
$endgroup$
– Sawyer Benson
Feb 14 at 14:58
1
$begingroup$
It seems to work out straightforwardly enough if you clear the cube roots in the numerator (as you've already done), apply the identity for $sin A - sin B$ to the numerator of the resulting expression, and use the fact that $lim_hto0fracsin hh = 1$ (as you're allowed to).
$endgroup$
– Calum Gilhooley
Feb 14 at 15:09
1
$begingroup$
@JeanMarie Aww, I really liked that title! :)
$endgroup$
– Calum Gilhooley
Feb 15 at 11:51
1
$begingroup$
@JeanMarie, I believe you are conflating a title with a direct title. The original title, purposefully written to be indirect, was crafted to equally capture attention as it was to communicate the problem ahead. Despite its objective lack of creativity, I appreciate the specificity of your direct-form edited above.
$endgroup$
– Sawyer Benson
Feb 15 at 14:52
|
show 4 more comments
3
$begingroup$
Can you give some context on why you want to do that with only the difference quotient? Surely you can replicate the proof of the chain rule, but it's not going to be pretty (although pointless all the more).
$endgroup$
– Klaus
Feb 14 at 14:35
1
$begingroup$
Thanks, Cameron! Unfortunately I cannot use that either. I know how to solve it using the chain rule and even L'Hopital's rule, however, the challenge it to solving it algebraically using nothing more than trig IDs (half angle) and trig limit rules (lim of sinx/x = 1). I know it's a pain, but that's the game! Thanks for all the help!
$endgroup$
– Sawyer Benson
Feb 14 at 14:58
1
$begingroup$
It seems to work out straightforwardly enough if you clear the cube roots in the numerator (as you've already done), apply the identity for $sin A - sin B$ to the numerator of the resulting expression, and use the fact that $lim_hto0fracsin hh = 1$ (as you're allowed to).
$endgroup$
– Calum Gilhooley
Feb 14 at 15:09
1
$begingroup$
@JeanMarie Aww, I really liked that title! :)
$endgroup$
– Calum Gilhooley
Feb 15 at 11:51
1
$begingroup$
@JeanMarie, I believe you are conflating a title with a direct title. The original title, purposefully written to be indirect, was crafted to equally capture attention as it was to communicate the problem ahead. Despite its objective lack of creativity, I appreciate the specificity of your direct-form edited above.
$endgroup$
– Sawyer Benson
Feb 15 at 14:52
3
3
$begingroup$
Can you give some context on why you want to do that with only the difference quotient? Surely you can replicate the proof of the chain rule, but it's not going to be pretty (although pointless all the more).
$endgroup$
– Klaus
Feb 14 at 14:35
$begingroup$
Can you give some context on why you want to do that with only the difference quotient? Surely you can replicate the proof of the chain rule, but it's not going to be pretty (although pointless all the more).
$endgroup$
– Klaus
Feb 14 at 14:35
1
1
$begingroup$
Thanks, Cameron! Unfortunately I cannot use that either. I know how to solve it using the chain rule and even L'Hopital's rule, however, the challenge it to solving it algebraically using nothing more than trig IDs (half angle) and trig limit rules (lim of sinx/x = 1). I know it's a pain, but that's the game! Thanks for all the help!
$endgroup$
– Sawyer Benson
Feb 14 at 14:58
$begingroup$
Thanks, Cameron! Unfortunately I cannot use that either. I know how to solve it using the chain rule and even L'Hopital's rule, however, the challenge it to solving it algebraically using nothing more than trig IDs (half angle) and trig limit rules (lim of sinx/x = 1). I know it's a pain, but that's the game! Thanks for all the help!
$endgroup$
– Sawyer Benson
Feb 14 at 14:58
1
1
$begingroup$
It seems to work out straightforwardly enough if you clear the cube roots in the numerator (as you've already done), apply the identity for $sin A - sin B$ to the numerator of the resulting expression, and use the fact that $lim_hto0fracsin hh = 1$ (as you're allowed to).
$endgroup$
– Calum Gilhooley
Feb 14 at 15:09
$begingroup$
It seems to work out straightforwardly enough if you clear the cube roots in the numerator (as you've already done), apply the identity for $sin A - sin B$ to the numerator of the resulting expression, and use the fact that $lim_hto0fracsin hh = 1$ (as you're allowed to).
$endgroup$
– Calum Gilhooley
Feb 14 at 15:09
1
1
$begingroup$
@JeanMarie Aww, I really liked that title! :)
$endgroup$
– Calum Gilhooley
Feb 15 at 11:51
$begingroup$
@JeanMarie Aww, I really liked that title! :)
$endgroup$
– Calum Gilhooley
Feb 15 at 11:51
1
1
$begingroup$
@JeanMarie, I believe you are conflating a title with a direct title. The original title, purposefully written to be indirect, was crafted to equally capture attention as it was to communicate the problem ahead. Despite its objective lack of creativity, I appreciate the specificity of your direct-form edited above.
$endgroup$
– Sawyer Benson
Feb 15 at 14:52
$begingroup$
@JeanMarie, I believe you are conflating a title with a direct title. The original title, purposefully written to be indirect, was crafted to equally capture attention as it was to communicate the problem ahead. Despite its objective lack of creativity, I appreciate the specificity of your direct-form edited above.
$endgroup$
– Sawyer Benson
Feb 15 at 14:52
|
show 4 more comments
4 Answers
4
active
oldest
votes
$begingroup$
Let's use $a^3-b^3=(a-b)(a^2+ab+b^2)$ for $a=sqrt[3]sin 2(x+h)$ and $b=sqrt[3]sin 2x$. Assume that $bneq 0$. This implies that $a^2+ab+b^2neq 0$ and we can divide:
$$fracsqrt[3]sin (2x+2h) - sqrt[3]sin 2xh = fracsin (2x+2h) - sin 2xh cdot frac1a^2+ab+b^2$$
Now using the identity $sin2(x+h)-sin 2x = 2sin h cos (2x+h)$, continuity of $sin$ and $cos$ and
$$lim_~hto 0fracsin hh=1$$
we get that the derivative is:
$$frac2cos 2x3sin^2/3 2x$$
answered Feb 14 at 15:20
Paweł CzyżPaweł Czyż
77411
$endgroup$
$begingroup$
nice approach !(+1)
$endgroup$
– G Cab
Feb 14 at 15:51
$begingroup$
@GCab "To praise it would amount to praising myself." (See comments on question - I was trying just to give a hint.) :)
$endgroup$
– Calum Gilhooley
Feb 14 at 15:56
add a comment |
$begingroup$
I have written down all steps in detail, and I have marked the main ingredients using bold face font. This is also a nice typing exercise in Latex.
Let $pgt0$ and consider more generally the difference quotient
$$d= frac1h left(sin(2x+2h)^p - sin(2x)^p right)$$
Now for the sine of the sum we can write
$$sin(2x+2h)= sin(2x)cos(2h)+cos(2x)sin(2h) \= cos(2h)sin(2x) (1+cot(2x) tan(2h))$$
Next for $hto 0$ we have approximately
$$cos(2h)= 1- 2h^2 + O(h^4)$$
$$tan(2h) = 2h + O(h^3)$$
so that, keeping only the first order in $h$, we get for the $p$-th power
$$sin(2x+2h)^p = sin(2x)^p(1+cot(2x)2h)^p\simeq sin(2x)^p+2 h p sin(2x)^pcot(2x)\=sin(2x)^p+2 h p sin(2x)^p-1cos(2x)$$
where we have use the binomial formula $(1+epsilon)^p simeq 1 + p; epsilon $.
Finally we get for $hto 0$
$$d = frac1h (sin(2x)^p+2 h p sin(2x)^p-1cos(2x) - sin(2x)^p) \= frac1h 2 h p sin(2x)^p-1cos(2x)\= 2 p sin(2x)^p-1cos(2x)$$
For $p=frac13$ this gives
$$d = frac23sin(2x)^-frac23cos(2x)$$
answered Feb 14 at 15:19
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,667619
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Can you also find the integral: $int_0^fracpi 2 sqrt[3]sin (2 x) , dx$ ?
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– Dr. Wolfgang Hintze
Feb 14 at 15:48
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...( by parts )
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– G Cab
Feb 14 at 15:53
2
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Your use of the Maclaurin expansion seems to sidestep the requirement to only rely on the difference quotient !
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– Jim
Feb 14 at 20:28
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@ Jim What would you propose instead? At a certain point you need to use this expansion in one form or other. Notice that all answer here use the same expansion, mostly even shorter, only as a limit.
$endgroup$
– Dr. Wolfgang Hintze
Feb 14 at 22:12
add a comment |
$begingroup$
Are you allowed to use the trigonometric limit identities of
$$lim limits_u to 0 fracsin uu = 1 hspace.5 in textand hspace.5 in lim limits_u to 0 frac1-cos uu = 0?$$
If not, the link above gives good hints. You'll need to prove these first. If you are allowed to use them, then let's proceed. The first step is to use some algebra to get rid of those pesky third roots:
beginalign*
&lim limits_h to 0 fracsqrt[3]sin(2x+2h) - sqrt[3]sin(2x)h\
=& lim limits_h to 0 fracsqrt[3]sin(2x+2h) - sqrt[3]sin(2x)h cdot fracsin(2x+2h)^2/3 + sin(2x+2h)^1/3sin(2x)^1/3 + sin(2x)^2/3sin(2x+2h)^2/3 + sin(2x+2h)^1/3sin(2x)^1/3 + sin(2x)^2/3\
=&lim limits_h to 0 fracsin(2x+2h) - sin(2x)h(sin(2x+2h)^2/3 + sin(2x+2h)^1/3sin(2x)^1/3 + sin(2x)^2/3)
endalign*
That denominator looks a little messy, so maybe we could call the part in parentheses
$$W(x,h) = sin(2x+2h)^2/3 + sin(2x+2h)^1/3sin(2x)^1/3 + sin(2x)^2/3.$$
Notice that $W(x,0) = 3sin(2x)^2/3$. That will come in handy later on. Now let's rewrite our limit using our $W$ shorthand and use the angle sum identity to move forward:
beginalign*
lim limits_h to 0 fracsin(2x+2h) - sin(2x)h cdot W(x,h) &= lim limits_h to 0 fracsin(2x)cos(2h) - cos(2h)sin(2x) - sin(2x)h cdot W(x,h)\
&= lim limits_h to 0 fraccos(2h)-1h cdot fracsin(2x)W(x,h) + limlimits_h to 0 frac-sin(2h)h cdot fraccos(2x)W(x,h)\
&= lim limits_h to 0 frac1-cos(2h)2h cdot frac-2sin(2x)W(x,h) + limlimits_h to 0 fracsin(2h)2h cdot frac-2cos(2x)W(x,h)
endalign*
Then, by the identities mentioned above, we have
beginalign*
lim limits_h to 0 fracsqrt[3]sin(2x+2h) - sqrt[3]sin(2x)h &= 0 cdot frac-2sin(2x)W(x,0) + 1 cdot frac-2cos(2x)W(x,0)\
&= frac-2cos(2x)3sin(2x)^2/3
endalign*
which is the answer you're looking for.
answered Feb 14 at 15:34
Ryan T JohnsonRyan T Johnson
1214
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Thanks Ryan. I’m not seeing the intuition of how the numerator is clearing to it’s simplified form. Are you distrubting?
$endgroup$
– Sawyer Benson
Feb 15 at 1:09
add a comment |
$begingroup$
Hint:
Use symmetry! :
$$sin(2x+2h)=sin((2x+h)+h) quad sin(2x)=sin((2x+h)-h)$$
answered Feb 14 at 15:39
G CabG Cab
19.9k31340
$endgroup$
$begingroup$
This is a good idea !
$endgroup$
– Dr. Wolfgang Hintze
Feb 14 at 15:44
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@Dr.WolfgangHintze: oh yes, to me, symmetry is the core of maths
$endgroup$
– G Cab
Feb 14 at 15:55
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's use $a^3-b^3=(a-b)(a^2+ab+b^2)$ for $a=sqrt[3]sin 2(x+h)$ and $b=sqrt[3]sin 2x$. Assume that $bneq 0$. This implies that $a^2+ab+b^2neq 0$ and we can divide:
$$fracsqrt[3]sin (2x+2h) - sqrt[3]sin 2xh = fracsin (2x+2h) - sin 2xh cdot frac1a^2+ab+b^2$$
Now using the identity $sin2(x+h)-sin 2x = 2sin h cos (2x+h)$, continuity of $sin$ and $cos$ and
$$lim_~hto 0fracsin hh=1$$
we get that the derivative is:
$$frac2cos 2x3sin^2/3 2x$$
answered Feb 14 at 15:20
Paweł CzyżPaweł Czyż
77411
$endgroup$
$begingroup$
nice approach !(+1)
$endgroup$
– G Cab
Feb 14 at 15:51
$begingroup$
@GCab "To praise it would amount to praising myself." (See comments on question - I was trying just to give a hint.) :)
$endgroup$
– Calum Gilhooley
Feb 14 at 15:56
add a comment |
$begingroup$
Let's use $a^3-b^3=(a-b)(a^2+ab+b^2)$ for $a=sqrt[3]sin 2(x+h)$ and $b=sqrt[3]sin 2x$. Assume that $bneq 0$. This implies that $a^2+ab+b^2neq 0$ and we can divide:
$$fracsqrt[3]sin (2x+2h) - sqrt[3]sin 2xh = fracsin (2x+2h) - sin 2xh cdot frac1a^2+ab+b^2$$
Now using the identity $sin2(x+h)-sin 2x = 2sin h cos (2x+h)$, continuity of $sin$ and $cos$ and
$$lim_~hto 0fracsin hh=1$$
we get that the derivative is:
$$frac2cos 2x3sin^2/3 2x$$
answered Feb 14 at 15:20
Paweł CzyżPaweł Czyż
77411
$endgroup$
$begingroup$
nice approach !(+1)
$endgroup$
– G Cab
Feb 14 at 15:51
$begingroup$
@GCab "To praise it would amount to praising myself." (See comments on question - I was trying just to give a hint.) :)
$endgroup$
– Calum Gilhooley
Feb 14 at 15:56
add a comment |
$begingroup$
Let's use $a^3-b^3=(a-b)(a^2+ab+b^2)$ for $a=sqrt[3]sin 2(x+h)$ and $b=sqrt[3]sin 2x$. Assume that $bneq 0$. This implies that $a^2+ab+b^2neq 0$ and we can divide:
$$fracsqrt[3]sin (2x+2h) - sqrt[3]sin 2xh = fracsin (2x+2h) - sin 2xh cdot frac1a^2+ab+b^2$$
Now using the identity $sin2(x+h)-sin 2x = 2sin h cos (2x+h)$, continuity of $sin$ and $cos$ and
$$lim_~hto 0fracsin hh=1$$
we get that the derivative is:
$$frac2cos 2x3sin^2/3 2x$$
answered Feb 14 at 15:20
Paweł CzyżPaweł Czyż
77411
$endgroup$
Let's use $a^3-b^3=(a-b)(a^2+ab+b^2)$ for $a=sqrt[3]sin 2(x+h)$ and $b=sqrt[3]sin 2x$. Assume that $bneq 0$. This implies that $a^2+ab+b^2neq 0$ and we can divide:
$$fracsqrt[3]sin (2x+2h) - sqrt[3]sin 2xh = fracsin (2x+2h) - sin 2xh cdot frac1a^2+ab+b^2$$
Now using the identity $sin2(x+h)-sin 2x = 2sin h cos (2x+h)$, continuity of $sin$ and $cos$ and
$$lim_~hto 0fracsin hh=1$$
we get that the derivative is:
$$frac2cos 2x3sin^2/3 2x$$
answered Feb 14 at 15:20
Paweł CzyżPaweł Czyż
77411
edited Feb 14 at 15:42
answered Feb 14 at 15:20
Paweł CzyżPaweł Czyż
77411
answered Feb 14 at 15:20
Paweł CzyżPaweł Czyż
77411
answered Feb 14 at 15:20
Paweł CzyżPaweł Czyż
77411
77411
$begingroup$
nice approach !(+1)
$endgroup$
– G Cab
Feb 14 at 15:51
$begingroup$
@GCab "To praise it would amount to praising myself." (See comments on question - I was trying just to give a hint.) :)
$endgroup$
– Calum Gilhooley
Feb 14 at 15:56
add a comment |
$begingroup$
nice approach !(+1)
$endgroup$
– G Cab
Feb 14 at 15:51
$begingroup$
@GCab "To praise it would amount to praising myself." (See comments on question - I was trying just to give a hint.) :)
$endgroup$
– Calum Gilhooley
Feb 14 at 15:56
$begingroup$
nice approach !(+1)
$endgroup$
– G Cab
Feb 14 at 15:51
$begingroup$
nice approach !(+1)
$endgroup$
– G Cab
Feb 14 at 15:51
$begingroup$
@GCab "To praise it would amount to praising myself." (See comments on question - I was trying just to give a hint.) :)
$endgroup$
– Calum Gilhooley
Feb 14 at 15:56
$begingroup$
@GCab "To praise it would amount to praising myself." (See comments on question - I was trying just to give a hint.) :)
$endgroup$
– Calum Gilhooley
Feb 14 at 15:56
add a comment |
$begingroup$
I have written down all steps in detail, and I have marked the main ingredients using bold face font. This is also a nice typing exercise in Latex.
Let $pgt0$ and consider more generally the difference quotient
$$d= frac1h left(sin(2x+2h)^p - sin(2x)^p right)$$
Now for the sine of the sum we can write
$$sin(2x+2h)= sin(2x)cos(2h)+cos(2x)sin(2h) \= cos(2h)sin(2x) (1+cot(2x) tan(2h))$$
Next for $hto 0$ we have approximately
$$cos(2h)= 1- 2h^2 + O(h^4)$$
$$tan(2h) = 2h + O(h^3)$$
so that, keeping only the first order in $h$, we get for the $p$-th power
$$sin(2x+2h)^p = sin(2x)^p(1+cot(2x)2h)^p\simeq sin(2x)^p+2 h p sin(2x)^pcot(2x)\=sin(2x)^p+2 h p sin(2x)^p-1cos(2x)$$
where we have use the binomial formula $(1+epsilon)^p simeq 1 + p; epsilon $.
Finally we get for $hto 0$
$$d = frac1h (sin(2x)^p+2 h p sin(2x)^p-1cos(2x) - sin(2x)^p) \= frac1h 2 h p sin(2x)^p-1cos(2x)\= 2 p sin(2x)^p-1cos(2x)$$
For $p=frac13$ this gives
$$d = frac23sin(2x)^-frac23cos(2x)$$
answered Feb 14 at 15:19
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,667619
$endgroup$
$begingroup$
Can you also find the integral: $int_0^fracpi 2 sqrt[3]sin (2 x) , dx$ ?
$endgroup$
– Dr. Wolfgang Hintze
Feb 14 at 15:48
$begingroup$
...( by parts )
$endgroup$
– G Cab
Feb 14 at 15:53
2
$begingroup$
Your use of the Maclaurin expansion seems to sidestep the requirement to only rely on the difference quotient !
$endgroup$
– Jim
Feb 14 at 20:28
$begingroup$
@ Jim What would you propose instead? At a certain point you need to use this expansion in one form or other. Notice that all answer here use the same expansion, mostly even shorter, only as a limit.
$endgroup$
– Dr. Wolfgang Hintze
Feb 14 at 22:12
add a comment |
$begingroup$
I have written down all steps in detail, and I have marked the main ingredients using bold face font. This is also a nice typing exercise in Latex.
Let $pgt0$ and consider more generally the difference quotient
$$d= frac1h left(sin(2x+2h)^p - sin(2x)^p right)$$
Now for the sine of the sum we can write
$$sin(2x+2h)= sin(2x)cos(2h)+cos(2x)sin(2h) \= cos(2h)sin(2x) (1+cot(2x) tan(2h))$$
Next for $hto 0$ we have approximately
$$cos(2h)= 1- 2h^2 + O(h^4)$$
$$tan(2h) = 2h + O(h^3)$$
so that, keeping only the first order in $h$, we get for the $p$-th power
$$sin(2x+2h)^p = sin(2x)^p(1+cot(2x)2h)^p\simeq sin(2x)^p+2 h p sin(2x)^pcot(2x)\=sin(2x)^p+2 h p sin(2x)^p-1cos(2x)$$
where we have use the binomial formula $(1+epsilon)^p simeq 1 + p; epsilon $.
Finally we get for $hto 0$
$$d = frac1h (sin(2x)^p+2 h p sin(2x)^p-1cos(2x) - sin(2x)^p) \= frac1h 2 h p sin(2x)^p-1cos(2x)\= 2 p sin(2x)^p-1cos(2x)$$
For $p=frac13$ this gives
$$d = frac23sin(2x)^-frac23cos(2x)$$
answered Feb 14 at 15:19
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,667619
$endgroup$
$begingroup$
Can you also find the integral: $int_0^fracpi 2 sqrt[3]sin (2 x) , dx$ ?
$endgroup$
– Dr. Wolfgang Hintze
Feb 14 at 15:48
$begingroup$
...( by parts )
$endgroup$
– G Cab
Feb 14 at 15:53
2
$begingroup$
Your use of the Maclaurin expansion seems to sidestep the requirement to only rely on the difference quotient !
$endgroup$
– Jim
Feb 14 at 20:28
$begingroup$
@ Jim What would you propose instead? At a certain point you need to use this expansion in one form or other. Notice that all answer here use the same expansion, mostly even shorter, only as a limit.
$endgroup$
– Dr. Wolfgang Hintze
Feb 14 at 22:12
add a comment |
$begingroup$
I have written down all steps in detail, and I have marked the main ingredients using bold face font. This is also a nice typing exercise in Latex.
Let $pgt0$ and consider more generally the difference quotient
$$d= frac1h left(sin(2x+2h)^p - sin(2x)^p right)$$
Now for the sine of the sum we can write
$$sin(2x+2h)= sin(2x)cos(2h)+cos(2x)sin(2h) \= cos(2h)sin(2x) (1+cot(2x) tan(2h))$$
Next for $hto 0$ we have approximately
$$cos(2h)= 1- 2h^2 + O(h^4)$$
$$tan(2h) = 2h + O(h^3)$$
so that, keeping only the first order in $h$, we get for the $p$-th power
$$sin(2x+2h)^p = sin(2x)^p(1+cot(2x)2h)^p\simeq sin(2x)^p+2 h p sin(2x)^pcot(2x)\=sin(2x)^p+2 h p sin(2x)^p-1cos(2x)$$
where we have use the binomial formula $(1+epsilon)^p simeq 1 + p; epsilon $.
Finally we get for $hto 0$
$$d = frac1h (sin(2x)^p+2 h p sin(2x)^p-1cos(2x) - sin(2x)^p) \= frac1h 2 h p sin(2x)^p-1cos(2x)\= 2 p sin(2x)^p-1cos(2x)$$
For $p=frac13$ this gives
$$d = frac23sin(2x)^-frac23cos(2x)$$
answered Feb 14 at 15:19
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,667619
$endgroup$
I have written down all steps in detail, and I have marked the main ingredients using bold face font. This is also a nice typing exercise in Latex.
Let $pgt0$ and consider more generally the difference quotient
$$d= frac1h left(sin(2x+2h)^p - sin(2x)^p right)$$
Now for the sine of the sum we can write
$$sin(2x+2h)= sin(2x)cos(2h)+cos(2x)sin(2h) \= cos(2h)sin(2x) (1+cot(2x) tan(2h))$$
Next for $hto 0$ we have approximately
$$cos(2h)= 1- 2h^2 + O(h^4)$$
$$tan(2h) = 2h + O(h^3)$$
so that, keeping only the first order in $h$, we get for the $p$-th power
$$sin(2x+2h)^p = sin(2x)^p(1+cot(2x)2h)^p\simeq sin(2x)^p+2 h p sin(2x)^pcot(2x)\=sin(2x)^p+2 h p sin(2x)^p-1cos(2x)$$
where we have use the binomial formula $(1+epsilon)^p simeq 1 + p; epsilon $.
Finally we get for $hto 0$
$$d = frac1h (sin(2x)^p+2 h p sin(2x)^p-1cos(2x) - sin(2x)^p) \= frac1h 2 h p sin(2x)^p-1cos(2x)\= 2 p sin(2x)^p-1cos(2x)$$
For $p=frac13$ this gives
$$d = frac23sin(2x)^-frac23cos(2x)$$
answered Feb 14 at 15:19
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,667619
edited Feb 14 at 15:39
answered Feb 14 at 15:19
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,667619
answered Feb 14 at 15:19
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,667619
answered Feb 14 at 15:19
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,667619
3,667619
$begingroup$
Can you also find the integral: $int_0^fracpi 2 sqrt[3]sin (2 x) , dx$ ?
$endgroup$
– Dr. Wolfgang Hintze
Feb 14 at 15:48
$begingroup$
...( by parts )
$endgroup$
– G Cab
Feb 14 at 15:53
2
$begingroup$
Your use of the Maclaurin expansion seems to sidestep the requirement to only rely on the difference quotient !
$endgroup$
– Jim
Feb 14 at 20:28
$begingroup$
@ Jim What would you propose instead? At a certain point you need to use this expansion in one form or other. Notice that all answer here use the same expansion, mostly even shorter, only as a limit.
$endgroup$
– Dr. Wolfgang Hintze
Feb 14 at 22:12
add a comment |
$begingroup$
Can you also find the integral: $int_0^fracpi 2 sqrt[3]sin (2 x) , dx$ ?
$endgroup$
– Dr. Wolfgang Hintze
Feb 14 at 15:48
$begingroup$
...( by parts )
$endgroup$
– G Cab
Feb 14 at 15:53
2
$begingroup$
Your use of the Maclaurin expansion seems to sidestep the requirement to only rely on the difference quotient !
$endgroup$
– Jim
Feb 14 at 20:28
$begingroup$
@ Jim What would you propose instead? At a certain point you need to use this expansion in one form or other. Notice that all answer here use the same expansion, mostly even shorter, only as a limit.
$endgroup$
– Dr. Wolfgang Hintze
Feb 14 at 22:12
$begingroup$
Can you also find the integral: $int_0^fracpi 2 sqrt[3]sin (2 x) , dx$ ?
$endgroup$
– Dr. Wolfgang Hintze
Feb 14 at 15:48
$begingroup$
Can you also find the integral: $int_0^fracpi 2 sqrt[3]sin (2 x) , dx$ ?
$endgroup$
– Dr. Wolfgang Hintze
Feb 14 at 15:48
$begingroup$
...( by parts )
$endgroup$
– G Cab
Feb 14 at 15:53
$begingroup$
...( by parts )
$endgroup$
– G Cab
Feb 14 at 15:53
2
2
$begingroup$
Your use of the Maclaurin expansion seems to sidestep the requirement to only rely on the difference quotient !
$endgroup$
– Jim
Feb 14 at 20:28
$begingroup$
Your use of the Maclaurin expansion seems to sidestep the requirement to only rely on the difference quotient !
$endgroup$
– Jim
Feb 14 at 20:28
$begingroup$
@ Jim What would you propose instead? At a certain point you need to use this expansion in one form or other. Notice that all answer here use the same expansion, mostly even shorter, only as a limit.
$endgroup$
– Dr. Wolfgang Hintze
Feb 14 at 22:12
$begingroup$
@ Jim What would you propose instead? At a certain point you need to use this expansion in one form or other. Notice that all answer here use the same expansion, mostly even shorter, only as a limit.
$endgroup$
– Dr. Wolfgang Hintze
Feb 14 at 22:12
add a comment |
$begingroup$
Are you allowed to use the trigonometric limit identities of
$$lim limits_u to 0 fracsin uu = 1 hspace.5 in textand hspace.5 in lim limits_u to 0 frac1-cos uu = 0?$$
If not, the link above gives good hints. You'll need to prove these first. If you are allowed to use them, then let's proceed. The first step is to use some algebra to get rid of those pesky third roots:
beginalign*
&lim limits_h to 0 fracsqrt[3]sin(2x+2h) - sqrt[3]sin(2x)h\
=& lim limits_h to 0 fracsqrt[3]sin(2x+2h) - sqrt[3]sin(2x)h cdot fracsin(2x+2h)^2/3 + sin(2x+2h)^1/3sin(2x)^1/3 + sin(2x)^2/3sin(2x+2h)^2/3 + sin(2x+2h)^1/3sin(2x)^1/3 + sin(2x)^2/3\
=&lim limits_h to 0 fracsin(2x+2h) - sin(2x)h(sin(2x+2h)^2/3 + sin(2x+2h)^1/3sin(2x)^1/3 + sin(2x)^2/3)
endalign*
That denominator looks a little messy, so maybe we could call the part in parentheses
$$W(x,h) = sin(2x+2h)^2/3 + sin(2x+2h)^1/3sin(2x)^1/3 + sin(2x)^2/3.$$
Notice that $W(x,0) = 3sin(2x)^2/3$. That will come in handy later on. Now let's rewrite our limit using our $W$ shorthand and use the angle sum identity to move forward:
beginalign*
lim limits_h to 0 fracsin(2x+2h) - sin(2x)h cdot W(x,h) &= lim limits_h to 0 fracsin(2x)cos(2h) - cos(2h)sin(2x) - sin(2x)h cdot W(x,h)\
&= lim limits_h to 0 fraccos(2h)-1h cdot fracsin(2x)W(x,h) + limlimits_h to 0 frac-sin(2h)h cdot fraccos(2x)W(x,h)\
&= lim limits_h to 0 frac1-cos(2h)2h cdot frac-2sin(2x)W(x,h) + limlimits_h to 0 fracsin(2h)2h cdot frac-2cos(2x)W(x,h)
endalign*
Then, by the identities mentioned above, we have
beginalign*
lim limits_h to 0 fracsqrt[3]sin(2x+2h) - sqrt[3]sin(2x)h &= 0 cdot frac-2sin(2x)W(x,0) + 1 cdot frac-2cos(2x)W(x,0)\
&= frac-2cos(2x)3sin(2x)^2/3
endalign*
which is the answer you're looking for.
answered Feb 14 at 15:34
Ryan T JohnsonRyan T Johnson
1214
$endgroup$
$begingroup$
Thanks Ryan. I’m not seeing the intuition of how the numerator is clearing to it’s simplified form. Are you distrubting?
$endgroup$
– Sawyer Benson
Feb 15 at 1:09
add a comment |
$begingroup$
Are you allowed to use the trigonometric limit identities of
$$lim limits_u to 0 fracsin uu = 1 hspace.5 in textand hspace.5 in lim limits_u to 0 frac1-cos uu = 0?$$
If not, the link above gives good hints. You'll need to prove these first. If you are allowed to use them, then let's proceed. The first step is to use some algebra to get rid of those pesky third roots:
beginalign*
&lim limits_h to 0 fracsqrt[3]sin(2x+2h) - sqrt[3]sin(2x)h\
=& lim limits_h to 0 fracsqrt[3]sin(2x+2h) - sqrt[3]sin(2x)h cdot fracsin(2x+2h)^2/3 + sin(2x+2h)^1/3sin(2x)^1/3 + sin(2x)^2/3sin(2x+2h)^2/3 + sin(2x+2h)^1/3sin(2x)^1/3 + sin(2x)^2/3\
=&lim limits_h to 0 fracsin(2x+2h) - sin(2x)h(sin(2x+2h)^2/3 + sin(2x+2h)^1/3sin(2x)^1/3 + sin(2x)^2/3)
endalign*
That denominator looks a little messy, so maybe we could call the part in parentheses
$$W(x,h) = sin(2x+2h)^2/3 + sin(2x+2h)^1/3sin(2x)^1/3 + sin(2x)^2/3.$$
Notice that $W(x,0) = 3sin(2x)^2/3$. That will come in handy later on. Now let's rewrite our limit using our $W$ shorthand and use the angle sum identity to move forward:
beginalign*
lim limits_h to 0 fracsin(2x+2h) - sin(2x)h cdot W(x,h) &= lim limits_h to 0 fracsin(2x)cos(2h) - cos(2h)sin(2x) - sin(2x)h cdot W(x,h)\
&= lim limits_h to 0 fraccos(2h)-1h cdot fracsin(2x)W(x,h) + limlimits_h to 0 frac-sin(2h)h cdot fraccos(2x)W(x,h)\
&= lim limits_h to 0 frac1-cos(2h)2h cdot frac-2sin(2x)W(x,h) + limlimits_h to 0 fracsin(2h)2h cdot frac-2cos(2x)W(x,h)
endalign*
Then, by the identities mentioned above, we have
beginalign*
lim limits_h to 0 fracsqrt[3]sin(2x+2h) - sqrt[3]sin(2x)h &= 0 cdot frac-2sin(2x)W(x,0) + 1 cdot frac-2cos(2x)W(x,0)\
&= frac-2cos(2x)3sin(2x)^2/3
endalign*
which is the answer you're looking for.
answered Feb 14 at 15:34
Ryan T JohnsonRyan T Johnson
1214
$endgroup$
$begingroup$
Thanks Ryan. I’m not seeing the intuition of how the numerator is clearing to it’s simplified form. Are you distrubting?
$endgroup$
– Sawyer Benson
Feb 15 at 1:09
add a comment |
$begingroup$
Are you allowed to use the trigonometric limit identities of
$$lim limits_u to 0 fracsin uu = 1 hspace.5 in textand hspace.5 in lim limits_u to 0 frac1-cos uu = 0?$$
If not, the link above gives good hints. You'll need to prove these first. If you are allowed to use them, then let's proceed. The first step is to use some algebra to get rid of those pesky third roots:
beginalign*
&lim limits_h to 0 fracsqrt[3]sin(2x+2h) - sqrt[3]sin(2x)h\
=& lim limits_h to 0 fracsqrt[3]sin(2x+2h) - sqrt[3]sin(2x)h cdot fracsin(2x+2h)^2/3 + sin(2x+2h)^1/3sin(2x)^1/3 + sin(2x)^2/3sin(2x+2h)^2/3 + sin(2x+2h)^1/3sin(2x)^1/3 + sin(2x)^2/3\
=&lim limits_h to 0 fracsin(2x+2h) - sin(2x)h(sin(2x+2h)^2/3 + sin(2x+2h)^1/3sin(2x)^1/3 + sin(2x)^2/3)
endalign*
That denominator looks a little messy, so maybe we could call the part in parentheses
$$W(x,h) = sin(2x+2h)^2/3 + sin(2x+2h)^1/3sin(2x)^1/3 + sin(2x)^2/3.$$
Notice that $W(x,0) = 3sin(2x)^2/3$. That will come in handy later on. Now let's rewrite our limit using our $W$ shorthand and use the angle sum identity to move forward:
beginalign*
lim limits_h to 0 fracsin(2x+2h) - sin(2x)h cdot W(x,h) &= lim limits_h to 0 fracsin(2x)cos(2h) - cos(2h)sin(2x) - sin(2x)h cdot W(x,h)\
&= lim limits_h to 0 fraccos(2h)-1h cdot fracsin(2x)W(x,h) + limlimits_h to 0 frac-sin(2h)h cdot fraccos(2x)W(x,h)\
&= lim limits_h to 0 frac1-cos(2h)2h cdot frac-2sin(2x)W(x,h) + limlimits_h to 0 fracsin(2h)2h cdot frac-2cos(2x)W(x,h)
endalign*
Then, by the identities mentioned above, we have
beginalign*
lim limits_h to 0 fracsqrt[3]sin(2x+2h) - sqrt[3]sin(2x)h &= 0 cdot frac-2sin(2x)W(x,0) + 1 cdot frac-2cos(2x)W(x,0)\
&= frac-2cos(2x)3sin(2x)^2/3
endalign*
which is the answer you're looking for.
answered Feb 14 at 15:34
Ryan T JohnsonRyan T Johnson
1214
$endgroup$
Are you allowed to use the trigonometric limit identities of
$$lim limits_u to 0 fracsin uu = 1 hspace.5 in textand hspace.5 in lim limits_u to 0 frac1-cos uu = 0?$$
If not, the link above gives good hints. You'll need to prove these first. If you are allowed to use them, then let's proceed. The first step is to use some algebra to get rid of those pesky third roots:
beginalign*
&lim limits_h to 0 fracsqrt[3]sin(2x+2h) - sqrt[3]sin(2x)h\
=& lim limits_h to 0 fracsqrt[3]sin(2x+2h) - sqrt[3]sin(2x)h cdot fracsin(2x+2h)^2/3 + sin(2x+2h)^1/3sin(2x)^1/3 + sin(2x)^2/3sin(2x+2h)^2/3 + sin(2x+2h)^1/3sin(2x)^1/3 + sin(2x)^2/3\
=&lim limits_h to 0 fracsin(2x+2h) - sin(2x)h(sin(2x+2h)^2/3 + sin(2x+2h)^1/3sin(2x)^1/3 + sin(2x)^2/3)
endalign*
That denominator looks a little messy, so maybe we could call the part in parentheses
$$W(x,h) = sin(2x+2h)^2/3 + sin(2x+2h)^1/3sin(2x)^1/3 + sin(2x)^2/3.$$
Notice that $W(x,0) = 3sin(2x)^2/3$. That will come in handy later on. Now let's rewrite our limit using our $W$ shorthand and use the angle sum identity to move forward:
beginalign*
lim limits_h to 0 fracsin(2x+2h) - sin(2x)h cdot W(x,h) &= lim limits_h to 0 fracsin(2x)cos(2h) - cos(2h)sin(2x) - sin(2x)h cdot W(x,h)\
&= lim limits_h to 0 fraccos(2h)-1h cdot fracsin(2x)W(x,h) + limlimits_h to 0 frac-sin(2h)h cdot fraccos(2x)W(x,h)\
&= lim limits_h to 0 frac1-cos(2h)2h cdot frac-2sin(2x)W(x,h) + limlimits_h to 0 fracsin(2h)2h cdot frac-2cos(2x)W(x,h)
endalign*
Then, by the identities mentioned above, we have
beginalign*
lim limits_h to 0 fracsqrt[3]sin(2x+2h) - sqrt[3]sin(2x)h &= 0 cdot frac-2sin(2x)W(x,0) + 1 cdot frac-2cos(2x)W(x,0)\
&= frac-2cos(2x)3sin(2x)^2/3
endalign*
which is the answer you're looking for.
answered Feb 14 at 15:34
Ryan T JohnsonRyan T Johnson
1214
answered Feb 14 at 15:34
Ryan T JohnsonRyan T Johnson
1214
answered Feb 14 at 15:34
Ryan T JohnsonRyan T Johnson
1214
answered Feb 14 at 15:34
Ryan T JohnsonRyan T Johnson
1214
1214
$begingroup$
Thanks Ryan. I’m not seeing the intuition of how the numerator is clearing to it’s simplified form. Are you distrubting?
$endgroup$
– Sawyer Benson
Feb 15 at 1:09
add a comment |
$begingroup$
Thanks Ryan. I’m not seeing the intuition of how the numerator is clearing to it’s simplified form. Are you distrubting?
$endgroup$
– Sawyer Benson
Feb 15 at 1:09
$begingroup$
Thanks Ryan. I’m not seeing the intuition of how the numerator is clearing to it’s simplified form. Are you distrubting?
$endgroup$
– Sawyer Benson
Feb 15 at 1:09
$begingroup$
Thanks Ryan. I’m not seeing the intuition of how the numerator is clearing to it’s simplified form. Are you distrubting?
$endgroup$
– Sawyer Benson
Feb 15 at 1:09
add a comment |
$begingroup$
Hint:
Use symmetry! :
$$sin(2x+2h)=sin((2x+h)+h) quad sin(2x)=sin((2x+h)-h)$$
answered Feb 14 at 15:39
G CabG Cab
19.9k31340
$endgroup$
$begingroup$
This is a good idea !
$endgroup$
– Dr. Wolfgang Hintze
Feb 14 at 15:44
$begingroup$
@Dr.WolfgangHintze: oh yes, to me, symmetry is the core of maths
$endgroup$
– G Cab
Feb 14 at 15:55
add a comment |
$begingroup$
Hint:
Use symmetry! :
$$sin(2x+2h)=sin((2x+h)+h) quad sin(2x)=sin((2x+h)-h)$$
answered Feb 14 at 15:39
G CabG Cab
19.9k31340
$endgroup$
$begingroup$
This is a good idea !
$endgroup$
– Dr. Wolfgang Hintze
Feb 14 at 15:44
$begingroup$
@Dr.WolfgangHintze: oh yes, to me, symmetry is the core of maths
$endgroup$
– G Cab
Feb 14 at 15:55
add a comment |
$begingroup$
Hint:
Use symmetry! :
$$sin(2x+2h)=sin((2x+h)+h) quad sin(2x)=sin((2x+h)-h)$$
answered Feb 14 at 15:39
G CabG Cab
19.9k31340
$endgroup$
Hint:
Use symmetry! :
$$sin(2x+2h)=sin((2x+h)+h) quad sin(2x)=sin((2x+h)-h)$$
answered Feb 14 at 15:39
G CabG Cab
19.9k31340
answered Feb 14 at 15:39
G CabG Cab
19.9k31340
answered Feb 14 at 15:39
G CabG Cab
19.9k31340
answered Feb 14 at 15:39
G CabG Cab
19.9k31340
19.9k31340
$begingroup$
This is a good idea !
$endgroup$
– Dr. Wolfgang Hintze
Feb 14 at 15:44
$begingroup$
@Dr.WolfgangHintze: oh yes, to me, symmetry is the core of maths
$endgroup$
– G Cab
Feb 14 at 15:55
add a comment |
$begingroup$
This is a good idea !
$endgroup$
– Dr. Wolfgang Hintze
Feb 14 at 15:44
$begingroup$
@Dr.WolfgangHintze: oh yes, to me, symmetry is the core of maths
$endgroup$
– G Cab
Feb 14 at 15:55
$begingroup$
This is a good idea !
$endgroup$
– Dr. Wolfgang Hintze
Feb 14 at 15:44
$begingroup$
This is a good idea !
$endgroup$
– Dr. Wolfgang Hintze
Feb 14 at 15:44
$begingroup$
@Dr.WolfgangHintze: oh yes, to me, symmetry is the core of maths
$endgroup$
– G Cab
Feb 14 at 15:55
$begingroup$
@Dr.WolfgangHintze: oh yes, to me, symmetry is the core of maths
$endgroup$
– G Cab
Feb 14 at 15:55
add a comment |
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3
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Can you give some context on why you want to do that with only the difference quotient? Surely you can replicate the proof of the chain rule, but it's not going to be pretty (although pointless all the more).
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– Klaus
Feb 14 at 14:35
1
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Thanks, Cameron! Unfortunately I cannot use that either. I know how to solve it using the chain rule and even L'Hopital's rule, however, the challenge it to solving it algebraically using nothing more than trig IDs (half angle) and trig limit rules (lim of sinx/x = 1). I know it's a pain, but that's the game! Thanks for all the help!
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– Sawyer Benson
Feb 14 at 14:58
1
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It seems to work out straightforwardly enough if you clear the cube roots in the numerator (as you've already done), apply the identity for $sin A - sin B$ to the numerator of the resulting expression, and use the fact that $lim_hto0fracsin hh = 1$ (as you're allowed to).
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– Calum Gilhooley
Feb 14 at 15:09
1
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@JeanMarie Aww, I really liked that title! :)
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– Calum Gilhooley
Feb 15 at 11:51
1
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@JeanMarie, I believe you are conflating a title with a direct title. The original title, purposefully written to be indirect, was crafted to equally capture attention as it was to communicate the problem ahead. Despite its objective lack of creativity, I appreciate the specificity of your direct-form edited above.
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– Sawyer Benson
Feb 15 at 14:52