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Sum of n squares (part 1)
What is the sum of the first n squares, 1 + 4 + 9 + 16 + ... + n²? In this video we begin the journey towards finding a formula for this sum. Created by Sal Khan.
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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!(40 votes)
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.(57 votes)
"You'll appreciate this even more when you get into calculus."-Sal,2:53
Was he referring to how the second derivative of x^2 is 2?(29 votes)- I believe what he was getting at is that the second derivative of a quadratic equation is always a constant, and because his second step in looking at the change between adjacent numbers did not produce a constant change, the formula can't be quadratic, which is why we go on to determine that it must be a cubic function.(28 votes)
- hi Sal, I don't understand why this general function? especially why n^3, I mean I didn't get the connection between the difference of the terms and where comes the function. please explain to me(13 votes)
Sal figures out that the function that results in the behavior of this series is a third degree function by looking at the differences in the series when n is 0, 1, 2, 3, and 4. If those differences were constant, that would mean that the function to model the behavior of this series would be a linear function. But the differences aren't constant. So he looks at the differences of the differences. If those were constant, which they aren't, that would mean that the function to model the behavior of this series would be a quadratic function. When he looks at the differences of the differences of the differences he does find a constant value. That means this is a cubic function. The general form of a cubic function is An^3 + Bn^2 + Cn^1 + Dn^0 or, written more simply: An^3 + Bn^2 + Cn + D, where A, B, C, and D are real numbers (including zero).(21 votes)
- At3:19sal says the sum of 1^2 + 2^2 + ... is equal to An^3 + Bn^2 + Cn + D.
I understand why he used An^3 + Bn^2 + Cn + D but any chance there's a video which explain that in further details?(4 votes)- A^3 Bn^2 +Cn^ + D is just the general form of a cubic equation. if he had determined that the series was quadratic he would have used An^2 + Bn + C .. does that help?(2 votes)
At3:10Sal determines that the function that gives the sum of the first n terms of the series must be cubic because the differences of the difference of the differences is constant. Is there any reason why it couldn't be some flukey quartic or quintic, or is it simply an educated guess?(3 votes)
Am I correct in saying that Sal finds the differences from the sequence of the partial sums?
If so, is the notation {Sn} = 0, 1, 5, 14, 30... correct to use instead of the gaps and sigma notation Sal used from1:10onwards?(3 votes)
Ok so i was wondering how would I solve something like ((2i)^2)+i from 1 to 10?
I got 10((5+410)/2). Which evaluates to 2075. I know it is wrong, so how would I solve it?
Ps. This is not a homework question(2 votes)- Hello people, so I realize that there are quite a few number of concepts introduced in this video, all of which seem to be large in size.
First, is the concept of finding the difference of the difference of something, or the d of the d of the d of the d etc. Now, knowing mathematicians, I presume there is a shorthand for this?
Second is the fact that the difference of a quadratic can be modelled by a cubic one. Is there a proof for this somewhere, which may provide some intuition?
Third, I never knew that a systems of equations could be used to solve something such as this. Where does the terms A, B, and C come from, and why is there coefficients n^3, n^2, and n? What if the formula was only in the form of An^3+Cn, or A^n3+Bn^2, and so on.
Any help would be much appreciated ^.^(2 votes)
The set up of the problem has to do with whether you are working backwards, graph to equation, or forwards, from equation to graph. Each variable represents the different value that is given by a graph, or in a n equation to transfer to a graph. The reason why the equation is with the A,B, and C terms along with the coefficients n^3, n^2, and n are because in the computing of the problem. With the example of . . .
Area PQRS = the summation of x^2 and x^1 times dx/V = 20 Sumation x^2 and x^1 dx/x = 18.33 hr.
In this equation f(x) is a function of x [b, f(x) =1/V] . Plot x and measure the area under the curve between the vertical lines at x = x^1 and x= x^2.
Pqrs is approx. 36.4 squares and the square LMRO is 100 squares and has an area of o.10 X 500 = 50 (hr.).
Thusly, the integral is 36.4/100 X 50 = 18.2 hr.
Compared with the correct value of 18.33 hr.
The error is very little. Does this response help?(1 vote)
How about if you have to find the sum of the same number but it's exponent continuously increases? Like 2/3 + (2/3)^2 + (2/3)^3 + (2/3)^4 + (2/3)^5? Is there a video for something similar to that?(1 vote)
That is called a geometric series and the study of them is coming up after the section you are currently on. Here is the video that talks about your particular question.
https://www.khanacademy.org/math/integral-calculus/sequences_series_approx_calc/seq_series_review/v/geometric-series(2 votes)
- So what would the sum from i=0 to n of i^pi be? How does this work with irrational powers?(1 vote)
- That is an exceedingly advanced problem. I think you'd have to use a Generalized Harmonic for that. I'm not sure, that is probably graduate school level calculus.(2 votes)
2:53Video transcript
What I want to do in this video
is come up with an expression for finding the sum from i
equals 0 to n of i squared. So if I were to
expand this out, this is equal to 0 squared
plus 1 squared plus 2 squared plus 3 squared. And we're going to keep on
going all the way to n squared. So my goal is to find
some type of a function that you give me
the n and I will find the sum from 0
squared, 1 squared, 2 squared, all the
way to n squared. And so you can imagine
that'd be useful, because this might be OK
if n is reasonably small. But if n is a big
number, this is going to take you forever to do. So let's first study this. Let's study what the input and
the output of this function needs to be. So the input is
going to be our n. So here we're starting with--
so n can go from 0 all the way-- and we'll just try
up a bunch of values. So n could be 0. We could go from 0
all the way to 0. n could be 1. n could be 2. n could be 3. And we could just keep on
going on and on and on. But I'll just stop
there for now. Actually, let's just
go to 4, just for fun. And now, for each
of these, let's see what the output of our
function should be. The output of the function
should be this thing. It should be the sum from i
equals 0 to n of i squared. So when n is 0-- well, that's
just going to be 0 squared. We'd just stop right over there. So that's just 0. When n is 1, it's 0
squared plus 1 squared. So that is 1. When n is 2, it's 0 squared
plus 1 squared plus 2 squared. So that's 1 plus 4, which is 5. When n is 3, now we
go all the way to 3. So it's going to be 1 plus
4, which is 5, plus 9. So 5 plus 9 is 14. And then when n is 4, we're
going to add the 16, 4 squared, to this. So this gets us to 30. And of course, we could
keep going on and on and on. So let's study this
a little bit to think about what type of a function
that, for each of these inputs, might give us this
type of an output. So let's first look at the
difference between these terms. So the difference here is 1. The difference here is 4. And this is obvious. We added 1 here. We added 2 squared here. We added 3 squared, or 9, here. We added 4 squared, or 16, here. And the reason
why I'm doing this is if this was a
linear function, then the difference
between successive terms would be the same. Now, if this is a
quadratic function, then the differences between the
differences would be the same. Let's see if that's the case. So the difference here is 1. The difference here is 4. So the difference
between those is 3. The difference here is 5. The difference here is 7. So even the difference of the
differences is increasing. But if this is a cubic
function, then the differences of the difference of the
difference should be constant. So let's see if that's the case. And you'll appreciate
this even more when you start
learning calculus. So let's see. The difference
between 3 and 5 is 2. The difference
between 5 and 7 is 2. And so we keep having
a constant shift of 2. So the fact that the
difference of the difference of the difference
is fixed tells us that we should be
able to express this as some type of
a cubic function. So this we could
write as this should be equal to some
function in terms of n. And we could write it as An
to the third plus Bn squared plus C times n
plus D. And now we can just use what the inputs
are and the outputs are of these to solve for A, B, C, and D.
And I encourage you to do that. Well, let's first think
about when n is equal to 0. When n is equal to 0, this
function evaluates to D. So this function evaluates
to D. But that function needs to evaluate to
0, so D needs to be 0. So I'm just trying to
fix these letters here to get the right outputs. So when n is 0, this
expression evaluates to D. And it needs to evaluate to 0,
so D needs to be equal to 0. So D is equal to 0, or
we could just ignore it. So that helps us a little bit. We know from this
data point we're able to whittle it
down to it having this form right over here. And so now we can take
each of these inputs and figure out what their
corresponding output is. So let's do that--
I'll do that over here. So when n is 1,
this thing evaluates to-- let me do this in
a new color-- this thing evaluates to A times 1 to the
third power, which is just 1, plus B times 1 squared, which
is just 1, plus C times 1, which is just C. And this
needs to be equal to 1. Now, when n is 2, we have
A times n to the third. So that's 8A plus
2 squared is 4, plus 4B plus 2C needs
to be equal to 5. And I need to set
up three equations if I want to solve
for three unknowns. And so let's go to 3. So A times 3 to the third power,
so that's going to be 27A, plus 9B plus 3C is
going to be equal to 14. So I've set up three
equations in three unknowns. Now I just have to solve
these for A, B, and C. And I will have a generalized
formula for finding this sum right over here, the sum of
the first n numbers squared, I guess you could call it. So what I want to do now is
I'm going to stop this video. I encourage you to try to
solve the simultaneous equation on your own. In the next video, I'll
actually go and solve it.