MATH1251 Questions HELP (3 Viewers)

leehuan

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Thanks guys.

How'd you do this one?



I'm guessing it's really simple comparison like the one I posted above, just can think of the proper one...


I don't think it says it's safe to assume that for this question.... Here's the proper screenshot:

Oh whoops. Ok then just stick to absolute convergence.
 

InteGrand

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Thanks guys.

How'd you do this one?



I'm guessing it's really simple comparison like the one I posted above, just can think of the proper one...
Sorry made an error before.

 

Paradoxica

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Thanks guys.

How'd you do this one?



I'm guessing it's really simple comparison like the one I posted above, just can think of the proper one...
To expand on Leehuan's comment, here's the proof:

We know that ∑(a_n)² converges, and it's well known that ∑ 1/n² converges.

By the AM-GM inequality, it follows that ∑(a_n)² + ∑ 1/n² ≥ 2∑(a_n)/n

Thus by the comparison test, it converges.
 

leehuan

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I'm just trying to understand my lecture slides here.

Is this like total differential approximation?
 
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1008

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I'm just trying to understand my lecture slides here.

Is this like total differential approximation?

Yes, I'm pretty sure this incorporates the total differential approximation in computing the tangent plane equation to the graph of f. Notice the upside down triangle in the equation, that's the total differential approximation we're computing. It's the same as the tangent line you're used to using in 1 dimension to approximate values of a curve near the region where the line touches tangentially with the curve (i.e. the gradient of the curve = gradient of the line at that point). This idea has now been generalised to higher dimensions with planes instead of lines and surfaces instead of curves.
 

1008

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How'd you do part (d) and (e) of this question?

 

InteGrand

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(It is easily checked that (1,1,1) lies on the surface and on the curve.) At (1,1,1), this is t = 1 on the curve. So the curve here has tangent vector r’(1) = (2, 1, 5). A normal to the surface at (1,1,1) can be found using grad(2x^2 + y^2 + 5z^2) = (4x, 2y, 10z) to be (4,2,10), which is proportional to r’(1) as we can see. Hence r(t) curve is normal to the surface at (1,1,1).
 
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InteGrand

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(To get the general tangent vector r’(t), differentiate r(t) component-wise wrt t. Then sub. in t = 1 for the tangent vector at t = 1.)
 

leehuan

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Ah I see. Didn't know how to deal with that "vector" function. Makes sense.
_____________________________________



Answer to a):




Also a quick question: Since a field technically satisfies the relevant axioms, it's also a vector space right?
 

InteGrand

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Ah I see. Didn't know how to deal with that "vector" function. Makes sense.
_____________________________________



Answer to a):




Also a quick question: Since a field technically satisfies the relevant axioms, it's also a vector space right?
Yes, any field F forms a vector space over itself (meaning that if F is a field, then taking the elements of F as the 'vectors' with the field of scalars also being F yields a vector space, where the vector addition rule is the addition rule of F and the scalar multiplication rule is the multiplication rule for elements from F).
 

InteGrand

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Wait, is the Taylor polynomial of degree 1 essentially the equation of the tangent plane?
Basically you already found the Taylor polynomial about the point P in the earlier part (up to degree 2, but just take it up to degree 1 as the Q. asks for the approximation). Just sub. x = 1.01 and y = 0.99 into that polynomial to get the answer.

 
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leehuan

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So I know it is not. But I was going to provide a counterexample to prove the map's bound to fail the addition condition somehow. But the answers say that Z is not a vector space - which axiom is it breaking?
 

InteGrand

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So I know it is not. But I was going to provide a counterexample to prove the map's bound to fail the addition condition somehow. But the answers say that Z is not a vector space - which axiom is it breaking?
For one thing, the codomain isn't a vector space. So it's not a linear map, since linear maps are generally defined from one vector space to another.

Anyway, one way to show it fails to preserve addition (it is the floor function):

floor(1.5 + 1.5) = floor(3) = 3 =/= floor(1.5) + floor(1.5) = 1 + 1 = 2.
 
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leehuan

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For one thing, the codomain isn't a vector space. So it's not a linear map, since linear maps are defined from one vector space to another.
Except I thought Z was. But I just realised something

Anyway, one way to show it fails to preserve addition (it is the floor function):

floor(1.5 + 1.5) = floor(3) = 3 =/= floor(1.5) + floor(1.5) = 1 + 1 = 2.
Yeah sweet that's what I had in mind
 
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InteGrand

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Except I thought Z was. But I just realised something

Z isn't a field (can't in general find multiplicative inverses in Z, e.g. 2 doesn't have a multiplicative inverse in Z), so we can't call it a vector space over itself like we can with say R (or general fields F).

Z equipped with a field of scalars R still wouldn't be a vector space, with ordinary multiplication, since αn is not always an integer for α real and n integer, e.g. take α = 1/2 and n = 1 (i.e. we wouldn't have closure under scalar multiplication).
 

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