Mnemonic for cubic discriminant?
$begingroup$
I sort of doubt there is a good one out there, but I thought I'd ask. I'm looking for a mnemonic for the general discriminant of the polynomial $ax^3+bx^2+cx+d$, which is
$$b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$$
Obviously if necessary, one can use different letters and / or assume a=1. However, the mnemonic is homogeneous with respect to the coefficients, so it might actually be easier to leave $a$ in.
A much easier case is the discriminant of $x^3+px+q$, which has discriminant $-4p^3-27q^2$.
mnemonic
asked Dec 7 '18 at 0:24
Elliot GElliot G
10.2k21645
$endgroup$
add a comment |
$begingroup$
I sort of doubt there is a good one out there, but I thought I'd ask. I'm looking for a mnemonic for the general discriminant of the polynomial $ax^3+bx^2+cx+d$, which is
$$b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$$
Obviously if necessary, one can use different letters and / or assume a=1. However, the mnemonic is homogeneous with respect to the coefficients, so it might actually be easier to leave $a$ in.
A much easier case is the discriminant of $x^3+px+q$, which has discriminant $-4p^3-27q^2$.
mnemonic
asked Dec 7 '18 at 0:24
Elliot GElliot G
10.2k21645
$endgroup$
$begingroup$
Not a mnemonic, but a different way to approach the cubic: nickalls.org/dick/papers/maths/cubic1993.pdf
$endgroup$
– Andy Walls
Dec 7 '18 at 2:13
add a comment |
$begingroup$
I sort of doubt there is a good one out there, but I thought I'd ask. I'm looking for a mnemonic for the general discriminant of the polynomial $ax^3+bx^2+cx+d$, which is
$$b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$$
Obviously if necessary, one can use different letters and / or assume a=1. However, the mnemonic is homogeneous with respect to the coefficients, so it might actually be easier to leave $a$ in.
A much easier case is the discriminant of $x^3+px+q$, which has discriminant $-4p^3-27q^2$.
mnemonic
asked Dec 7 '18 at 0:24
Elliot GElliot G
10.2k21645
$endgroup$
I sort of doubt there is a good one out there, but I thought I'd ask. I'm looking for a mnemonic for the general discriminant of the polynomial $ax^3+bx^2+cx+d$, which is
$$b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$$
Obviously if necessary, one can use different letters and / or assume a=1. However, the mnemonic is homogeneous with respect to the coefficients, so it might actually be easier to leave $a$ in.
A much easier case is the discriminant of $x^3+px+q$, which has discriminant $-4p^3-27q^2$.
mnemonic
mnemonic
asked Dec 7 '18 at 0:24
Elliot GElliot G
10.2k21645
asked Dec 7 '18 at 0:24
Elliot GElliot G
10.2k21645
asked Dec 7 '18 at 0:24
Elliot GElliot G
10.2k21645
asked Dec 7 '18 at 0:24
Elliot GElliot G
10.2k21645
asked Dec 7 '18 at 0:24
Elliot GElliot G
10.2k21645
10.2k21645
$begingroup$
Not a mnemonic, but a different way to approach the cubic: nickalls.org/dick/papers/maths/cubic1993.pdf
$endgroup$
– Andy Walls
Dec 7 '18 at 2:13
add a comment |
$begingroup$
Not a mnemonic, but a different way to approach the cubic: nickalls.org/dick/papers/maths/cubic1993.pdf
$endgroup$
– Andy Walls
Dec 7 '18 at 2:13
$begingroup$
Not a mnemonic, but a different way to approach the cubic: nickalls.org/dick/papers/maths/cubic1993.pdf
$endgroup$
– Andy Walls
Dec 7 '18 at 2:13
$begingroup$
Not a mnemonic, but a different way to approach the cubic: nickalls.org/dick/papers/maths/cubic1993.pdf
$endgroup$
– Andy Walls
Dec 7 '18 at 2:13
add a comment |
2 Answers
2
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I don't have a nnemonic to suggest. However, after my sixth grade gym teacher made us write down our locker combination and the serial number on the back 200 times on the first day of school, I have never forgotten either. I still use the lock and I'm in my forties.
answered Dec 7 '18 at 2:11
BillBill
194
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add a comment |
$begingroup$
I know this is barely a mnemonic, but I am doubtful there will ever be a nice one. I managed to memorize this in one day with the following method:
Break the expression into two parts: one with letters and one with numbers.
Letters: $bcac, bdad, abcd$.
Numbers: $122-413-431-17 22-18$.
Explanation: if you already have an idea of what the discriminant should look like, and remember that each term has degree four, it is easy to reconstruct from this. The last term is a multiple of $abcd$, which is easy to remember, so we start with
$$bc ac bd ad abcd$$
and add in the coefficients and exponents. These come from the number sequence. So $122$ and $bc$ become$ b^2c^2$. Then $-413$ and $ac$ become $-4ac^3$, etc. The last term is $-18abcd$, although I didn't write $181111$.
I showed this to someone and they thought it was about as useful as just memorizing the expression from scratch, but somehow it is easier to remember these two sequences. The letters sound like words "buhCACK, buDAD," and "abcd" is easy to remember. Then the sequence of numbers is about as easy as memorizing a phone number.
Like I said, its pretty dumb and hardly a mnemonic, but it worked for me.
answered Dec 7 '18 at 23:12
Elliot GElliot G
10.2k21645
$endgroup$
add a comment |
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2 Answers
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$begingroup$
I don't have a nnemonic to suggest. However, after my sixth grade gym teacher made us write down our locker combination and the serial number on the back 200 times on the first day of school, I have never forgotten either. I still use the lock and I'm in my forties.
answered Dec 7 '18 at 2:11
BillBill
194
$endgroup$
add a comment |
$begingroup$
I don't have a nnemonic to suggest. However, after my sixth grade gym teacher made us write down our locker combination and the serial number on the back 200 times on the first day of school, I have never forgotten either. I still use the lock and I'm in my forties.
answered Dec 7 '18 at 2:11
BillBill
194
$endgroup$
add a comment |
$begingroup$
I don't have a nnemonic to suggest. However, after my sixth grade gym teacher made us write down our locker combination and the serial number on the back 200 times on the first day of school, I have never forgotten either. I still use the lock and I'm in my forties.
answered Dec 7 '18 at 2:11
BillBill
194
$endgroup$
I don't have a nnemonic to suggest. However, after my sixth grade gym teacher made us write down our locker combination and the serial number on the back 200 times on the first day of school, I have never forgotten either. I still use the lock and I'm in my forties.
answered Dec 7 '18 at 2:11
BillBill
194
answered Dec 7 '18 at 2:11
BillBill
194
answered Dec 7 '18 at 2:11
BillBill
194
answered Dec 7 '18 at 2:11
BillBill
194
194
add a comment |
add a comment |
$begingroup$
I know this is barely a mnemonic, but I am doubtful there will ever be a nice one. I managed to memorize this in one day with the following method:
Break the expression into two parts: one with letters and one with numbers.
Letters: $bcac, bdad, abcd$.
Numbers: $122-413-431-17 22-18$.
Explanation: if you already have an idea of what the discriminant should look like, and remember that each term has degree four, it is easy to reconstruct from this. The last term is a multiple of $abcd$, which is easy to remember, so we start with
$$bc ac bd ad abcd$$
and add in the coefficients and exponents. These come from the number sequence. So $122$ and $bc$ become$ b^2c^2$. Then $-413$ and $ac$ become $-4ac^3$, etc. The last term is $-18abcd$, although I didn't write $181111$.
I showed this to someone and they thought it was about as useful as just memorizing the expression from scratch, but somehow it is easier to remember these two sequences. The letters sound like words "buhCACK, buDAD," and "abcd" is easy to remember. Then the sequence of numbers is about as easy as memorizing a phone number.
Like I said, its pretty dumb and hardly a mnemonic, but it worked for me.
answered Dec 7 '18 at 23:12
Elliot GElliot G
10.2k21645
$endgroup$
add a comment |
$begingroup$
I know this is barely a mnemonic, but I am doubtful there will ever be a nice one. I managed to memorize this in one day with the following method:
Break the expression into two parts: one with letters and one with numbers.
Letters: $bcac, bdad, abcd$.
Numbers: $122-413-431-17 22-18$.
Explanation: if you already have an idea of what the discriminant should look like, and remember that each term has degree four, it is easy to reconstruct from this. The last term is a multiple of $abcd$, which is easy to remember, so we start with
$$bc ac bd ad abcd$$
and add in the coefficients and exponents. These come from the number sequence. So $122$ and $bc$ become$ b^2c^2$. Then $-413$ and $ac$ become $-4ac^3$, etc. The last term is $-18abcd$, although I didn't write $181111$.
I showed this to someone and they thought it was about as useful as just memorizing the expression from scratch, but somehow it is easier to remember these two sequences. The letters sound like words "buhCACK, buDAD," and "abcd" is easy to remember. Then the sequence of numbers is about as easy as memorizing a phone number.
Like I said, its pretty dumb and hardly a mnemonic, but it worked for me.
answered Dec 7 '18 at 23:12
Elliot GElliot G
10.2k21645
$endgroup$
add a comment |
$begingroup$
I know this is barely a mnemonic, but I am doubtful there will ever be a nice one. I managed to memorize this in one day with the following method:
Break the expression into two parts: one with letters and one with numbers.
Letters: $bcac, bdad, abcd$.
Numbers: $122-413-431-17 22-18$.
Explanation: if you already have an idea of what the discriminant should look like, and remember that each term has degree four, it is easy to reconstruct from this. The last term is a multiple of $abcd$, which is easy to remember, so we start with
$$bc ac bd ad abcd$$
and add in the coefficients and exponents. These come from the number sequence. So $122$ and $bc$ become$ b^2c^2$. Then $-413$ and $ac$ become $-4ac^3$, etc. The last term is $-18abcd$, although I didn't write $181111$.
I showed this to someone and they thought it was about as useful as just memorizing the expression from scratch, but somehow it is easier to remember these two sequences. The letters sound like words "buhCACK, buDAD," and "abcd" is easy to remember. Then the sequence of numbers is about as easy as memorizing a phone number.
Like I said, its pretty dumb and hardly a mnemonic, but it worked for me.
answered Dec 7 '18 at 23:12
Elliot GElliot G
10.2k21645
$endgroup$
I know this is barely a mnemonic, but I am doubtful there will ever be a nice one. I managed to memorize this in one day with the following method:
Break the expression into two parts: one with letters and one with numbers.
Letters: $bcac, bdad, abcd$.
Numbers: $122-413-431-17 22-18$.
Explanation: if you already have an idea of what the discriminant should look like, and remember that each term has degree four, it is easy to reconstruct from this. The last term is a multiple of $abcd$, which is easy to remember, so we start with
$$bc ac bd ad abcd$$
and add in the coefficients and exponents. These come from the number sequence. So $122$ and $bc$ become$ b^2c^2$. Then $-413$ and $ac$ become $-4ac^3$, etc. The last term is $-18abcd$, although I didn't write $181111$.
I showed this to someone and they thought it was about as useful as just memorizing the expression from scratch, but somehow it is easier to remember these two sequences. The letters sound like words "buhCACK, buDAD," and "abcd" is easy to remember. Then the sequence of numbers is about as easy as memorizing a phone number.
Like I said, its pretty dumb and hardly a mnemonic, but it worked for me.
answered Dec 7 '18 at 23:12
Elliot GElliot G
10.2k21645
answered Dec 7 '18 at 23:12
Elliot GElliot G
10.2k21645
answered Dec 7 '18 at 23:12
Elliot GElliot G
10.2k21645
answered Dec 7 '18 at 23:12
Elliot GElliot G
10.2k21645
10.2k21645
add a comment |
add a comment |
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$begingroup$
Not a mnemonic, but a different way to approach the cubic: nickalls.org/dick/papers/maths/cubic1993.pdf
$endgroup$
– Andy Walls
Dec 7 '18 at 2:13