Sarthak Kumar
6 years ago
Posted 6 years ago. Direct link to Sarthak Kumar's post “- For a linear equation, the difference between su...”
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- For a linear equation, the difference between successive terms is a constant.
- For a quadratic equation, the difference of the difference between the successive terms is a constant; and
- For a cubic equation, the difference of the difference of the difference of successive terms is a constant.
I took this observation on its face value and understood the video but what I'm wondering is...
Where does this insight come from? Is there a proof to it? Has Sal done a video which explains it in better detail?
Many thanks!
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jaxxson
6 years ago
Posted 6 years ago. Direct link to jaxxson's post “Here are proofs of those three statements: Proof f...”
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Here are proofs of those three statements:
Proof for a linear equation of the form L(n) = A*n + B, where A and B are constant coefficients. The difference between successive terms of L(n) can be represented by:
L(n+1) - L(n) = (A*(n+1)+B) - (A*n+B) = A*(n+1) + B - A*n - B = A*(n+1) - A*n = A, which we defined as a constant. So then the difference between successive terms of a linear equation is constant.
Proof for a quadratic equation of the form Q(n) = A*n^2 + B*n + C, where A, B, and C are constant coefficients.
The difference between successive terms can be represented by:
Q(n+1) - Q(n) = (A*(n+1)^2 + B*(n+1) + C) - (A*n^2 + B*n + C) = A*(n^2+2n+1) + B*(n+1) + C - A*n^2 - B*n - C = A*(n^2+2n+1) - An^2 + B*(n+1) - B*n = A*(n^2+2n+1 - n^2) + B*(n+1 - n) = A*(2n+1) + B*1 = A*(2n+1) + B.
Which is a linear function of n. So then Q(n+1) - Q(n) is a linear function of n and so the difference between successive terms of a quadratic series is always a linear function of n. We already proved that the difference between successive terms of a linear function is a constant, so this means that the difference between the successive terms of Q(n+1) - Q(n) is a constant, and so the difference of the difference between successive terms of Q(n) is constant. Q(n) is just any quadratic, so the difference of the difference between the successive terms of a quadratic is always constant. See if you can prove the statement for any cubic equation on your own. It's the same general method as for the linear and quadratics.
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Sarthak Kumar
6 years ago
Posted 6 years ago. Direct link to Sarthak Kumar's post “- For a linear equation, the difference between su...”
more
- For a linear equation, the difference between successive terms is a constant.
- For a quadratic equation, the difference of the difference between the successive terms is a constant; and
- For a cubic equation, the difference of the difference of the difference of successive terms is a constant.
I took this observation on its face value and understood the video but what I'm wondering is...
Where does this insight come from? Is there a proof to it? Has Sal done a video which explains it in better detail?
Many thanks!
Reply
Reply to Sarthak Kumar's post “- For a linear equation, the difference between su...”Button opens signup modal
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1 commentComment on Sarthak Kumar's post “- For a linear equation, the difference between su...”
jaxxson
6 years ago
Posted 6 years ago. Direct link to jaxxson's post “Here are proofs of those three statements: Proof f...”
more
Here are proofs of those three statements:
Proof for a linear equation of the form L(n) = A*n + B, where A and B are constant coefficients. The difference between successive terms of L(n) can be represented by:
L(n+1) - L(n) = (A*(n+1)+B) - (A*n+B) = A*(n+1) + B - A*n - B = A*(n+1) - A*n = A, which we defined as a constant. So then the difference between successive terms of a linear equation is constant.
Proof for a quadratic equation of the form Q(n) = A*n^2 + B*n + C, where A, B, and C are constant coefficients.
The difference between successive terms can be represented by:
Q(n+1) - Q(n) = (A*(n+1)^2 + B*(n+1) + C) - (A*n^2 + B*n + C) = A*(n^2+2n+1) + B*(n+1) + C - A*n^2 - B*n - C = A*(n^2+2n+1) - An^2 + B*(n+1) - B*n = A*(n^2+2n+1 - n^2) + B*(n+1 - n) = A*(2n+1) + B*1 = A*(2n+1) + B.
Which is a linear function of n. So then Q(n+1) - Q(n) is a linear function of n and so the difference between successive terms of a quadratic series is always a linear function of n. We already proved that the difference between successive terms of a linear function is a constant, so this means that the difference between the successive terms of Q(n+1) - Q(n) is a constant, and so the difference of the difference between successive terms of Q(n) is constant. Q(n) is just any quadratic, so the difference of the difference between the successive terms of a quadratic is always constant. See if you can prove the statement for any cubic equation on your own. It's the same general method as for the linear and quadratics.
Step-by-step explanation: