# Thread: Derivative of scalar/vector field

1. I have noticed that many different areas of physics and engineering make use of 'scalar and vector fields'. I just want someone to tell me whether my understanding is right:

A scalar field will have some scalar value 'A' at every point in space, and can be represented by (in 3 dimensions):

A=f(x,y,z)

The rate of change of this field with respect to time at any point is given by:

∂A/∂t = ∂x/∂t + ∂y/∂t + ∂z/∂t

For a vector field, there will be some vector value 'v' at every point in space, then each component of the vector will be a function of x, y and z independently. so:

v = [V1, V2, V3] = [ f(x,y,z), g(x,y,z), h(x,y,z) ]

Then for the rate of change of the vector field at any point, we will have a partial differential equation for each component (3 in this case), for example:

∂V1/∂t = f '(x) + f '(y) + f '(z)

2.

3. For the scalar field, you should apply the chain rule for multiple variables. This gives:

dA/dt = ∂f/∂x dx/dt + ∂f/∂y dy/dt + ∂f/∂z dz/dt

For the vector field, treat each vector component like a scalar field, so if:

[V<sub>1</sub>,V<sub>2</sub>,V<sub>3</sub>] = [f<sub>1</sub>(x,y,z),f<sub>2</sub>(x,y,z),f<sub>3</sub>(x,y,z)]

you should have for i =1, 2, 3:

dV<sub>i</sub>/dt = ∂f<sub>i</sub>/∂x dx/dt + ∂f<sub>i</sub>/∂y dy/dt + ∂f<sub>i</sub>/∂z dz/dt

4. Hi Serpicojr, thanks for the reply.

Originally Posted by serpicojr
For the scalar field, you should apply the chain rule for multiple variables. This gives:

dA/dt = ∂f/∂x dx/dt + ∂f/∂y dy/dt + ∂f/∂z dz/dt
Why do you have to use a chain rule if the three variables are different? I thought the idea of partial differentiation, was to differentiate each variable in turn while 'holding' the others constant. So effectively you have...

A = f(x, y, z) = f(x) + f(y) + f(z)

dA/dt = ∂x/∂t + ∂y/∂t + ∂z/∂t

...because each function is actually independent of each other isn't it? A is just the sum of their values. Doesn't the chain rule only need to come into play if the variable is the same?...for example within f(x) itself?

Although, now you mention it I have heard of 'the multi variable chain rule', so I must be missing something.

For the vector field, treat each vector component like a scalar field, so if:

[V<sub>1</sub>,V<sub>2</sub>,V<sub>3</sub>] = [f<sub>1</sub>(x,y,z),f<sub>2</sub>(x,y,z),f<sub>3</sub>(x,y,z)]

you should have for i =1, 2, 3:

dV<sub>i</sub>/dt = ∂f<sub>i</sub>/∂x dx/dt + ∂f<sub>i</sub>/∂y dy/dt + ∂f<sub>i</sub>/∂z dz/dt
Ok so I was right in thinking that each component of the vector has its own, independent function in x, y, and z?...as opposed to V1, V2, and V3 all being the same function, f( x, y, z).

Then you have differentiated each function individually, as I attempted, but again you seem to have used this multivariable chain ruale again, and got a different expression to me.

Do you mind explaining why this has been used?

Thanks as always.

5. We have to use the chain rule whenever we're composing two maps. In your case, we have two functions:

p: R -> R<sup>3</sup>

where p(t) = [x,y,z] is the position of whatever at time t. Then we have the scalar field:

f: R<sup>3</sup> -> R

where f(x,y,z) is some scalar associated the the position [x,y,z]. When we wish to find the change in the scalar field with respect to time, we are really thinking of f as a function of t: f(x,y,z) = f(p(t)). This is where the chain rule comes in. Now, if p spit out and f took in single variable arguments, then we'd jst use the regular chain rule. But they don't, so we need to use the multivariable chain rule.

In general, we can't think of f(x,y,z) as the sum of three independent functions, each of which depends on exactly one of x, y, and z. And, even if they were, your expression for the change in the scalar field with respect to time doesn't make sense: if f(x,y,z) = F(x) + G(y) + H(z), then we'd have:

df/dt = dF/dx dx/dt + dG/dy dy/dt + dH/dz dz/dt

(Again applying the chain rule!) dx/dt, dy/dt, and dz/dt represent the change in each positional direction with respect to time, so it doesn't make sense to add them up to get the change in the scalar field, as they're quantities or functions which are independent of the chosen scalar field.

6. Originally Posted by serpicojr
We have to use the chain rule whenever we're composing two maps. In your case, we have two functions:

p: R -> R<sup>3</sup>

where p(t) = [x,y,z] is the position of whatever at time t. Then we have the scalar field:

f: R<sup>3</sup> -> R
Ok, You've lost me with that notation there. I have seen R<sup>3</sup> used to express normal 3 dimension space, so I assume R by itself is one dimensional space? I'm not sure what the operation is or how it fits in. Is this what you call linear transformation? From what I have seen of that before, I didn't really grasp it....Was too much abstract mathematics for me... seemed a bit pointless like defining vectors/matrices over sets/fields as discussed in a previous post.

Aside from that, when we have...

f(x, y, z)

...Then all this represents is that for every [x, y, z] co-ordinate within the field, we have a value for f.... i.e. a scalar field. But from what I've gathered you are saying that x, y, and z, are all functions of t...i.e:

x = f<sub>1</sub>(t)
y = f<sub>2</sub>(t)
z = f<sub>3</sub>(t)

This is what I don't understand. If for instance we were describing the motion of a particle through the field, then the particles position at any point in time would be a function of time (as denoted above), but if we consider the field by itself, then we can pick any co-ordinate [x,y,z], and have a value for f. Are you saying that because this chosen co-ordinate would change with time, then each x, y, and z must be a function of time?...so effectively we have:

A = f(x, y, z) = f(g<sub>1</sub>(t), g<sub>2</sub>(t), g<sub>3</sub>(t))

So in order to differentiate the field, we must not only know the function f(x, y, z), but also the functions, g<sub>1</sub>(t), g<sub>2</sub>(t), and g<sub>3</sub>(t)?

What about writing the function as:

A = f(x, y, z, t)

or simply,

A = f(t) ?

Thanks

7. I agree that I clouded things up with mathematical notation and that I didn't explain what I was trying to do. Before I go down this path again, let me ask exactly what you're trying to do. I think we may have been talking past each other, as neither of us explicitly laid out the scenario we're trying to describe.

So what I assumed you were asking about in the first place is this: we have some scalar field on 3-dimensional space. This field is constant with respect to time. An object moves through space, and so we can then ask how does the scalar field (as experienced by the object) changes with respect to time. Is this the original scenario? Some of your posts suggest to me that you're actually considering the scalar field as something that changes with time and that you're not considering an object moving through the field.

8. Yeah I see what you mean, I think we're speaking about different scenarios.

So what I actually meant was how the field itself changes with respect to time. For example, the scalar field of density (ρ), where at every point [x, y, z], we have value for ρ:

ρ = f (x, y, z)

Now suppose we have some turbulent fluid flow, with compressible fluids - The density throughout the field will not be constant (at any [x, y, z]), and the density at any individual point will also change over time.

So in order to find the expression for rate of change of density with respect to time, we need to take the derivative of f(x, y, z) with respect to t.

This scenario would be different to the case of a particle moving through the fluid, where the actual position of interest is a function of time.

So how do we represent that?

Thanks

9. Alright, now we're getting somewhere! If we're looking at a scalar field which changes with time, then it's really a function of four variables, as you had suggested. We might denote this f(x,y,z,t) or even f(x,y,z;t) to distinguish the variable t from the others. In this case, the change with respect to time is just the partial derivative with respect to time, ∂f/∂t. We can't get a more detailed expression without more information. For the vector field, we have F(x,y,z;t) = [f(x,y,z;t), g(x,y,z;t), h(x,y,z;t)], and the chain with respect to time is just the vector of partial derivatives ∂F/∂t = [∂f/∂t, ∂g/∂t, ∂h/∂t].

If we're looking at a particle moving in the field (and the field is constant), then it looks like what I was describing at first, and you have to use the chain rule. We could also combine the two scenarios, say look at a particle moving through a field which changes with respect to time. Then we'd have something like (for a scalar field):

df/dt = ∂f/∂t + ∂f/∂x dx/dt + ∂f/∂y dy/dt + ∂f/∂z dz/dt

This all assumes that the path of the particle doesn't depend on the field, which is kind of unrealistic. If you want to analyze this situation, we'll have to find someone else, as I don't know a whole lot about it...

10. ahhh, of course!....so if the scalar field changes over time, then the value of density is not just a function of space (co-ordinate), but also a function of time, so that to find the value of density at any point in space and time we have to plug in each of the four values:

ρ = f(x, y, z, t)

Ok, thats simple enough. So for the rate of change of the field with respect to time, then we just ignore the functions in x, y, and z, and take the derivative of 'f(t)' (not sure if that notation is accurate) so we could say:

dρ/dt = d f(t)/dt ?

This means that for any value of x, y, and z, that we pick, the value for the rate of change of density at that point will be the same, because if we are simply differentiating the time component of the function f(x, y, z, t), then x, y, and,z changes are irrelevant... But what comes immediately to mind, is that for a real fluid flow, the rate of change of density will be different at different (spatial) points too.

So somehow we need to express the rate of change of density with respect to x, y, z, and t. I'm only familiar with taking the derivative with respect to a single variable....so I'm not too sure how we would do this. I'm guessing we have to differentiate each function f(x), f(y), f(z) and f(t) individually, with respect to each variable. Getting my head round this is a real brain bender though.

As for a particle moving through a field, It makes sense that this would be an uncommon scenario, as most of the time I suppose the change of the field itself would dictate the particles motion. This would seem to be the case of a magnetic or electric field anyway, but I suppose it might arise if the magnetic/electric effects are only one component of the total forces acting on the particle....it might have some other force acting on it too. For now though, I'm gonna stick with the basics.

11. ...Oh hang on a minute... can we just differentiate each function with respect to its own variable?... for example:

ρ = f(x, y, z, t)

ρ = f(x) + f(y) + f(z) +f(t) (still not sure if that expression is accurate?)

dρ/d(x, y, z, t) = ∂/∂x + ∂/∂y + ∂/∂z + ∂/∂t

Where each function is simply differentiated with respect to is own variable.... f(t) with respect to t for example?

12. If f(x,y,z,t) is a scalar field that depends on both position and time, then the change with respect to time still depends on the position: the change with respect to time at the point [x,y,z] is (∂f/∂t)(x,y,z,t). For example, if f(x,y,z,t) = xyzt, then (∂f/∂t)(x,y,z,t) = xyz--this is certainly something that depends upon the point in space.

Again, if you have a function of several variables, you cannot assume that it is the sum of a bunch of functions of one variable. If we could, life would be a lot easier--we'd be able to separate all variables from each other and study them independently. As it stands, that's not the case, and life is harder but more interesting.

I think you're a bit confused with the notation of derivatives. What calculus courses have you taken?

13. As far as calculus goes, I have had formal education for; basic differentiation, chain rule, product rule, quotient rule, implicit differentiation, trig diff' exponential/logarithm diff', basic integration, int' by substitution, int' by parts, trig int', exp/log int'. int of revolution, and applied calculus questions regarding physics/basic electronics (well...electricity)/mechanics. I must admit alot of that is rusty as it is.

For multivariable calc/partial diff and scalar/vector fields etc.. I have no formal education. I have been reading 'Engineering mathematics' by K.A Shroud, and this page. Also, I already have formal education of vectors including inner product/ cross product etc, and lots of applied problems with vectors.

You can pretty much see my standard of calculus, which probably explains my confused notation with this problem.

I think I see what you mean with the above problem though. If we are differentiating with respect to time only, then the resulting function (the derivative) will still be a function of x, y, and, y, so by plugging different values of [x,y,z] into it we do get a different rate of change at different positions within the fluid...so would this be correct for instance?...

f(x, y, z, t) = xyt<sup>4</sup> - e<sup>2cos(yt)</sup> + e<sup>(2zt-3t)</sup>

df/dt = 4xyt<sup>3</sup> + 2ysin(yt)e<sup>2cos(yt)</sup> + (2z - 3)e<sup>(2zt - 3t)</sup>

...in other words, all we do is treat x,z,y as constants, just as we would 1, 2, or 3?

Then if we want to know the rate of change of density at a given point, we just plug [x,y,z] from that point into the above function.

I think my confusion is with the notation. In particular with the ∂ sign for partial derivative.

14. The partial derivative is precisely what we're taking when we differentiate f(x,y,z,t) with respect to t while treated x, y, z as constants. It assumes that there is no interdependence among the variables, i.e. t can vary without changing x, y, or z. We express this partial derivative of f with respect to t as ∂f/∂t. The notation is just like that for regular derivatives, except we use ∂ instead of d.

15. Ok so in the case above, I should have wrote ∂f/∂t instead of df/dt? Thanks, that seems simple enough to understand.

I'm also trying to understand about the Nabla/Del operator, and for a scalar field the multivariate derivative ( or just 'Gradient' as they call it) is given by:

So i, j, and k are unit vectors forming orthogonal basis. This expression is for 3 variables x, y, z, as you see but I think it has been extended for more dimensions (variables) too:

For the case of the 4-D scalar field above, we would have

∇= i∂/∂x + j∂/∂y + k∂/∂z + l∂/∂t

So basically we are differentiating with respect to all variables in turn, and multiplying each one by the corresponding unit vector. So ∇f would actually be a vector? And this is different to what we did above, because we are taking the partial derivative with respect to all variables? (individually)

The reason I'm looking at this is because I want to try and understand the gradient/curl/divergence/laplacian (whatever they are) of scalar vector fields.

16. Indeed, the gradient is a vector. It basically collates all of the information about the partial derivatives of a function, and it allows you to do a few neat things. First, there's the notion of a directional derivative. Intuitively, this is saying "what is the change in f as we move away from a fixed point p in the direction of some vector v?" This generalizes the partial derivatives, in which v is taken to be one of our standard basis vectors. Formally, the directional derivative is defined by the limit:

D<sub>v</sub>f(p) = lim<sub>h->0</sub> (1/h)(f(p+hv)-f(p))

Now let's let (v<sub>1</sub>,v<sub>2</sub>) denote the usual dot product of two vectors in Euclidean space. It turns out we can calculate the directional derivative very easily from the gradient--we have:

D<sub>v</sub>f(p) = (∇f(p),v)

In other words, the directional derivative is just the sum of the partial derivatives "weighted" by the coordinates of v with respect to the standard basis vectors.

The second cool piece of information that the gradient gives us is the "direction of steepest ascent". By this I mean it finds the unit vector u (i.e., |u| = 1) that maximizes D<sub>u</sub>f(p). We can find this using the above information. Recall that:

(v<sub>1</sub>,v<sub>2</sub>) = |v<sub>1</sub>||v<sub>2</sub>| cos θ

where θ is the angle between v<sub>1</sub> and v<sub>2</sub>. So if we want to maximize:

(∇f(p),u) = |∇f(p)||u| cos θ = |∇f(p)| cos θ

Cosine is maximized when θ = 0, i.e. when ∇f(p) and u lie on the same line. This means that u = ∇f(p)/|∇f(p)| maximizes this quantity, and so ∇f(p) is itself the direction of steepest ascent. (Note that this doesn't make sense when ∇f(p) = 0.)

Third, the gradient helps us find local minima and maxima. For one variable, we know the derivative vanishes at a local maximum or minimum. The same thing happens in multiple variables, except we replaces the derivative by the gradient. We can use this to maximize a function on, say, a hypercube in Euclidean space--we find the points where the gradient vanishes and then find the maxima and minima along the boundary (surface) of the hypercube. For example, to maximize a function f(x,y) on the unit square [0,1]x[0,1], we take its gradient, find when it's 0, and write down those points. Then we have to look at the boundary--all points (x,0), (x,1), (0,y), (1,y). We can look at f(x,y) as a function of one variable along these four segments and use single variable calculus to find the critical points. And we still have to consider (0,0), (0,1), (1,0), and (1,1), as these are the boundary points of the segments above. Then we just plug in these points and compare values of f(x,y).

I'm sure I could go on, but that's all that's coming off the top of my head at the moment.

17. Thanks, alot to digest there so I'm gonna tackle it bit by bit. So the idea of a directional derivative is to give us the rate of change of 'f' as we move along some specified direction? Basically extending the idea of derivatives with respect to x, y, z, to any given direction? So we represent this direction by a vector. If we have a vector, v= <a, b>, pointing in the direction of travel, then why must this be a unit vector? (I read it must have a length of 1)

I read that the rate of change of f(x,y) in the direction <a,b> will be given by:

∇<sub>v</sub>f(x, y) = a(∂f/∂x)+b(∂f/∂y)

..since x, and y components are independent of each other...but still why must v be of length 1?...meaning a<sup>2</sup> + b<sup>2</sup> = 1 ?

Also, where you have expressed the directional derivative in terms of the limit, as h tends to 0, what exactly is h?

BTW, I just invested in a new book (reading off a monitor is a pain)..."Advanced Calculus Demystified", and I've slowly started reading through it today. It starts of pretty basic, but I've flicked to have a look at the section about directional derivatives anyway.

18. Your description of directional derivatives is spot on! However, the directional derivative doesn't have to be with respect to a unit vector (which indeed has length 1)--note that I didn't restrict myself to unit vectors in defining it. I only used unit vectors in the "steepest ascent" interpretation of the gradient so as to have a one-to-one correspondence between the vectors I was considering and directions. And, in the definition of directional derivatives, h is a real parameter (i.e., a real number) tending to 0. This should look a lot like the usual definition of derivative.

19. Ok, I see what you mean, so 'h' is basically the very small (approximately linear) change in the vector? So for a normal derivative, we can define a secant line on the curve, passing through points [x, f(x)], and [x+h, f(x+h)], where 'h' is a very small magnitude, so that:

f '(x) = lim<sub>h→0</sub> ( f(x+h) - f(x) ) / h

So when expressing the directive derivative form the limit, 'p' will be a point vector (the starting point of the vector...or simply, a co-ordinate), 'v' will be the direction vector, (e.g. <a, b> for two dimensions, as above), and h will be a very small magnitude (scalar) that when multiplied by v, will produce a very small change in it. <ha, hb>

This will also stand for 3 dimensions, so in the case of the scalar field of density if we pick a point [a, b, c] (or P), in the field then:

ρ = f(a, b, c)
..or
ρ = f(P).

So then if we have a vector, V = <δx, δy, δz> starting at P, then, if we go a small distance along it, we will be at another point (M), so that;

M = P + hV
ρ = f(M)
ρ = f(P + hV)

So basically as we move such a small direction along V, the change in density will be given by:

f(P + hV) - f(P)

So then the change in density with respect to V is given by:

dρ / dV = lim<sub>h→0</sub> [f(P + hV) - f(P)] / h

Strictly speaking, the distance moved along V won't be h, but rather,
= √ [(hδx)<sup>2</sup> + (hδy)<sup>2</sup> + (hδz)<sup>2</sup>]. I'm guessing that this will be close enough to h to still stand though, so we can just write the change along V as 'h'.

So, I think thats more-a-less right?... but I still can't see the exact relationship between this and the Gradient of a function.

Thanks

20. it looks like you've got everything down pretty well: h indeed is a scalar which allows us to represent an infinitesimal change in the direction of v. Let me just clear a few things up:

1. h does not represent the distance moved along v. If v = [x,y,z], then hv = [hx,hy,hz], and we have:

|hv| = ((hx)<sup>2</sup>+(hy)<sup>2</sup>+(hz)<sup>2</sup>)<sup>1/2</sup> = h(x<sup>2</sup>+y<sup>2</sup>+z<sup>2</sup>)<sup>1/2</sup> = h|v|

So, in fact, this distance is always proportional to |v|, and h can be interpreted as a "parameter of proportion", I suppose.

This fact is important! It means if we take the directional derivative with respect to two vectors which are scalar multiples of each other, then the corresponding directional derivatives will differ by the same scalar! So, if we think of a particle moving through a scalar field f at velocity v through point p, the change in the scalar field is D<sub>v</sub>f(p). If another particle moves twice as fact (2v) through the same point, the change is D<sub>2v</sub>f(p), which we would expect to be 2D<sub>v</sub>f(p)--its change in position is twice as fast, so it should experience the change in the field twice as fast.

2. Why does the gradient give us the directional derivative? Because the directional derivative is linear--it satisfies for any vectors v, w, any scalar k, any scalar field f, and any point p:

-D<sub>v+w</sub>f(p) = D<sub>v</sub>f(p) + D<sub>w</sub>f(p)
-D<sub>kv</sub>f(p) = kD<sub>v</sub>f(p)

The second fact is what I was saying about. This can be shown via the chain rule. The first fact can be shown from the definition but the only way I can think to show it right now requires that the derivative be continuous (and I'm not sure if I can loosen this). In any case, illustrating what happens in two dimensions, we have (with standard basis vectors e<sub>i</sub>, i = 1, 2):

D<sub>v<sub>1</sub>e<sub>1</sub>+v<sub>2</sub>e<sub>2</sub></sub>f(p) = v<sub>1</sub>D<sub>e<sub>1</sub></sub>f(p) + v<sub>2</sub>D<sub>e<sub>2</sub></sub>f(p) = v<sub>1</sub>(∂f/∂x)(p)+v<sub>2</sub>(∂f/∂y)(p) = (v,∇f(p))

21. 1. h does not represent the distance moved along v. If v = [x,y,z], then hv = [hx,hy,hz], and we have:

|hv| = ((hx)2+(hy)2+(hz)2)1/2 = h(x2+y2+z2)1/2 = h|v|

So, in fact, this distance is always proportional to |v|, and h can be interpreted as a "parameter of proportion", I suppose.
Thanks that makes sense. The bigger vector |v|, then the bigger the vector |hv|, but the respective magnitudes of |v| and |hv| will be proportionally the same for any value of |v|, and so with respect to |v|, the change |hv| can still be considered small enough?

This fact is important! It means if we take the directional derivative with respect to two vectors which are scalar multiples of each other, then the corresponding directional derivatives will differ by the same scalar! So, if we think of a particle moving through a scalar field f at velocity v through point p, the change in the scalar field is Dvf(p). If another particle moves twice as fact (2v) through the same point, the change is D2vf(p), which we would expect to be 2Dvf(p)--its change in position is twice as fast, so it should experience the change in the field twice as fast.
But If two vectors are scalar multiples of each other then why would the directional derivatives differ at all? Won't the two vectors have differing magnitudes, but still be in the same direction?...so the rate of change of the field along that particular direction will still be the same?..unless the scalar multiple is negative, in which case the derivative would be negative?

I think it might be best to go through with an example. So say we have a field:

f(x,y,z) = x<sup>2</sup> + y<sup>3</sup> + z<sup>4</sup>

Then we have a vector in that field, v = <2,4,6>, the directional derivative along that vector would be:

D<sub>v</sub>f(x,y,z) = v • ∇f
= <2, 4, 6> • <(∂f/∂x), (∂f/∂y), (∂f/∂z)>
= 2(2x + y<sup>3</sup> + z<sup>4</sup>) + 4(x<sup>2</sup> + 3y<sup>2</sup> + z<sup>4</sup>) + 6(x<sup>2</sup> + y<sup>3</sup> + 4z<sup>3</sup>)
=4x + 2y<sup>3</sup> + 2z<sup>4</sup> + 4x<sup>2</sup> + 12y<sup>2</sup> + 4z<sup>4</sup> + 6x<sup>2</sup> + 6y<sup>3</sup> + 24z<sup>3</sup>
= 4x + 10x<sup>2</sup> + 12y<sup>2</sup> +8y<sup>3</sup> + 24z<sup>3</sup> + 6z<sup>4</sup>

So then at the point [1,2,3] the rate of change would be 1260, by plugging in the values. I can already see that by doubling the vector v, we would get a larger rate of change, but would the difference between these derivatives, D<sub>v</sub>f(x,y,z) and D<sub>2v</sub>f(x,y,z) necessarily be of order two?

2. Why does the gradient give us the directional derivative? Because the directional derivative is linear--it satisfies for any vectors v, w, any scalar k, any scalar field f, and any point p:

-Dv+wf(p) = Dvf(p) + Dwf(p)
-Dkvf(p) = kDvf(p)

The second fact is what I was saying about. This can be shown via the chain rule. The first fact can be shown from the definition but the only way I can think to show it right now requires that the derivative be continuous (and I'm not sure if I can loosen this). In any case, illustrating what happens in two dimensions, we have (with standard basis vectors ei, i = 1, 2):

Dv1e1+v2e2f(p) = v1De1f(p) + v2De2f(p) = v1(∂f/∂x)(p)+v2(∂f/∂y)(p) = (v,∇f(p))
I'm afraid I'm a little lost again with this bit. What is this notation for 'e'?

Thanks, and sorry the posting is a little sparse, I'm really busy this time of year.

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