Anna Tomskova

Anna Tomskova

I am a Ph.D. student in Mathematics at UNSW Australia. I love Maths and I believe that Maths helps us to see the real beauty of this world.

Location Sydney, Australia

Activity

  • When we say inverse proportionality we mean that y=a/x. If we say inverse proportionality to the square, of course it means y=a/x^2.
    But we do not consider the second case to be a particular case of the first one, these two cases are different, that's why for Q3 you have to answer no, y=a/x^2 is not an inverse proportionality.
    Hope it makes more sense now :-)

  • Hi everyone, it's an interesting discussion! I think it is all started with the matrix. There are many ways to find these numbers, one of them, using this matrix. This method came from linear algebra.
    Now, as we have the matrix, we can identify it with mobius transformation, as disscussed earlier.
    And yes, you can use composition of Mobius transformations...

  • Hi Liz, you are right, this condition ad=bc should be mentioned in lots of questions on the Mobius transformation.
    But I think we do not consider any of them here, since we do not go so deep in the subject.

  • I think you have lost the denominator in the second expression.
    Open the brackets here
    a(ex+f/gx+h)+b=(aex+ad)/(gx+h)+b
    Now find the common denominator
    =((aex+ad)+b(gx+h))/(gx +h)
    Opening the brackets then rearranging the numerator and taking x out, we have
    =(aex+ad+bgx+hb)/(gx +h)=
    ((ae+bg)x+ad+bh))/(gx +h)
    Hope this helps :-)

  • Assuming that x and y are positive. Taking the ln of two parts we have
    lnx/x=lny/y
    Now we may look at the graph
    y=lnt/t and find two points for which the values of y are equal.
    That is we need a horizontal line which intersects this graph in two points then this intersections would have t-coordinates say t1 and t2
    And the same y-coordinate
    So...

  • Hi Suzi, if we have a data set (number of points on the coordinate plane) you can connect them somehow to make a continuous curve. We can try to find the equation of the curve which fits best using matlab. I mean that the error between the real points and the points on the curve can be minimized.
    In this paragraph we have the data set about English words of...

  • Hi William,
    you uses f2 notation for different purpose, as I have understood from your comment Fn or fn is nth
    Fibonacci number. But Reino asked about the power of a function f^n which we do not use in the lecture (or it seems so since I could not find any)
    Hope this makes sense.

  • Hi Aaron,
    There's no deadline for the certificate - you will qualify for a certificate as soon as you have met the requirements (even if that's after the course finishes) and can opt to buy one at any time after qualifying.

  • Hi Reino, in this lecture we use the upper index for the power of the multiplication of matrices.
    So for the matrix F, F^2=F*F, F^3=F*F*F and so on.
    We do not use f^2 for functions at all and f(f(x)) is always denoted as f o f, not f^2.
    Probably I missed something....
    Let me know if I did.

  • Hi Keith, zooming out the screen may help.
    Please, let me know if it does not work.

  • Hi Reino,
    We do not consider complex valued functions in this course.
    But concerning your question, there is nothing wrong if some values of a complex valued function are real, since real numbers are also complex numbers with imaginary part equals zero.
    Hope this helps.

  • As soon as you understand the idea of f o g you will feel that they (f o g and g o f) are absolutely different things.
    It is like "to fill a pool with water (function f) and then jump (function g)" or "first jump to the pool and then to fill it with water"
    Sometimes f o g is well defined but g o f can not be defined at all.
    For example take f(x)=-x^2-1 and...

  • Yes, you are right.
    Fixed now
    Thank you!

  • Hi Evanthia, this is a very good question.
    It depends what you call "dimension"
    The analogy in my comment above is the following:
    The Mobius band is made from two-dimensional object (strip) but Klein bottle made using the same idea but from three dimensional object (cylinder)

    In a usual sense of a dimension, both objects are 3-dimensional.

    Going to...

  • Using properties of log (by log I mean log10):
    P(2i)+P(2i+1)=log(2i+1)-log(2i)+log(2i+2)-log(2i+1)=
    -log(2i)+log(2i+2)=log(i+1)-log(i)=P(i)

  • Hi Dai, In this course e is the number such that the area under y=1/x from 1 to e equals 1. However, I can not find now where and how we defined e here ((((
    If you find it, please, let me know.

  • Ah really, that's great :-)
    We have a street named The Great Silk Road))
    It is indeed a beautiful country, and uzbek people are well-known by their kindness and hospitality, so everyone is welcome to Uzbekistan :-)

  • Hi Tony,
    give it another shot, it is quite easy to get :-)
    When we compose two functions we apply a function to the value of another function.
    Let's take something easy to have a look at:
    for example f(x)=5x+3
    we may look at this as at a linear function, but on the other hand it is a composition of two functions:
    First we take x, multiply it by 5 (first...

  • I think all these questions are for an interesting and quite big research!
    The starting point can be to find some software to calculate frequencies easily
    and then use different texts to answer these questions.
    I have found this one just for fun
    http://www.writewords.org.uk/word_count.asp

  • Hi Richie, I see your point, but when we defined the inverse proportionality (it was in 1.2) as y=a/x for a constant 'a'.
    Therefore, y=x^(-2) is not an inverse proportionality by definition we have given.

  • Haha, yes, Norman is great!

  • Hi Foteini, yes, the 1st one is also correct since it is normal to assume constant resistance in simple circuit, unless otherwise is said.

  • Thanks, Antonio, try to do my best :-)

  • You are right Jonathan, it should be 150 :-) Thank you for pointing this out!

  • I should be now, since it's around 1 am)))
    5 hours ago... no, I am currently in Uzbekistan :-)

  • You are right Chris, it would be very useful to cover the general equation of the hyperbola and to explain all numbers appearing in the equation. Thank you for pointing this out! :-)

  • The reason is we do not have to make it equal to 1 (do not need the standard form) if we want to find the center.
    As soon as you get the equation like
    d(x-a)^-e(y-b)^2=c
    where a,b,c,d are numbers, you can say that [a,b] is the center without additional rearranging.

  • Hi Tony, I think a closely related (3D) object is the Klein bottle. A Klein bottle can be produced by gluing two Mobius strips together along their edges
    or we glue not the ends of a strip (as we did for Mobius band) but the ends of a cylinder.
    see what we have here
    https://en.wikipedia.org/wiki/Klein_bottle
    and for more...

  • This trick is quite common, when we want to cancel out two expressions but one of them has different sign we do this trick.
    If we open the brackets in -(u-t) we get -u+t and then we just change the order of the summands to get +t-u=t-u.
    Hope this makes sense now)))

  • Hi everyone,
    About the pair of lines: we can get this if the plane passes the vertex of the cone, however in this case we may also have a point and a single line depending on the angle at the vertex.
    See e.g. http://math2.org/math/algebra/conics.htm
    To Jim's questions:
    "Does that mean that the resulting two sides of a triangle is a hyperbola, even though...

  • Hi Benedict,
    Unfortunately if you did not take the previous course then you missed quite a lot of information and techniques, so you have to work a bit harder.
    Try this link for completing the square
    https://en.wikipedia.org/wiki/Completing_the_square
    There is a nice animation)))
    Also try to understand
    1.1 Background
    1.2 Basic example
    1.3 General...

  • Hi Antonio,
    when two springs (k1 and k2) are in series then the overall constant is k1+k2
    and if they are in parallel then the constant is (k1xk2)/(k1+k2).

  • Hi Benedict, I am afraid that this is the easiest explanation I can give.
    Can you indicate the step you do not understand?

  • Haha, that's a nice reminder!
    Yes, it is definitely like this
    C is a capacitance (the ability of a body to store an electrical charge)
    Q is an electrical charge
    V is a voltage

  • Hi Tony,
    Great observations!
    Just a minor thing:
    In this course we use square brackets to identify the point,
    say [1,2] is the point with x-coordinate 1 and y-coordinate 2.
    and one more thing:
    the curve X=1/Y has also the second branch when X and Y are negative.
    the same for X=3/Y

  • Hi Tony,
    The curve can be anything,
    like
    1/x
    1/ (any polynomial)
    1/ ln x
    1/ ln (ln x) and so on

    I am sure someone may find the precise curve (using maple or Matlab)
    1/(polynomial) which will fit this particular problem)))

  • Let's consider the hyperbola y=1/x.
    and let X=[t,1/t] be the point on this hyperbola.
    We know (will know from 1.6) that the equation of the tangent line at X is x+t^2y=2t.
    Let's prove Q(O,P) x Q(O,R)=CONST where O=[0,0] and P=[0,p] and R=[r,0] two points of intersection of the tangent with y and x -axes respectively (since the axes are asymptotes for...

  • Hi Rhoda, sorry for taking so long time to response.
    I think the easiest way to find the directrix is to find first the point of symmetry, since the first directrix is known and the second one is located symmetrically.

    I hope I understand your question correctly.

  • Hi Jitka, thank you for pointing this out,
    will be fixed soon!

  • Hi Tahir,
    The phrase
    "where A is THE point on the hyperbola"
    in Q5 means that A is the same point we considered in Q4.
    So in the solution of Q5 we take exactly the point A with coordinates (t,1/t).
    Sorry if this confused you!

  • Hi Benedict,
    I am sure you can find good hints on Q2 in the comments below)))

  • Hi Lesley, I am very sorry that you have to struggle a lot.
    But I think you should not worry so much.
    Just go through the course and try to catch a general idea as much as you can
    do not struggle a lot on a particular question.
    We are here to help you, and we do our best :-)
    Unfortunately, we have a lot of information already discussed in the first...

  • Hi Ian, please read the comments below.
    We have already discussed this question deeply at the beginning of this week)))

  • It is okay, Mary, it is always good when you ask questions)))
    I see your point now))
    We construct the right triangle to use pythagoras theorem, but as you said, we may just use the distance formula :-)
    Thank you for pointing this out!

  • Hi Lesley, it is a pity that you have to spend so much time on this :-(
    It is better to remember this notions picturing them.
    Please have a look at the picture:
    https://en.wikipedia.org/wiki/Hyperbola#/media/File:Hyperbola_properties.svg
    You can see there two foci F1 and F2
    The center is the point C (center of symmetry of the hyperbola)
    and two...

  • No worries! :-)

  • Hi Mary,
    you have to find the distances anyway, so it is not alternative.
    The quadrance is a notation to write/use more conveniently the square of a distance.
    Let's have a right triangle ABC with the hypotenuse AB
    then the pythagoras theorem says that |AB|^2=|AC|^2+|BC|^2.
    The quadrance Q(A,B) is a squared distance between points A and B, that is...

  • Hi Tyseer,
    Let's have a look at the picture above Q1 (in Q1 we obtain the same hyperbola, only x and y coordinates are interchanged, therefore, we will work with this picture as with an example):
    We know that this hyperbola is symmetric with respect to some vertical line.
    If we can find this line, then we can find the other focus the other directrix using...

  • Hi Richie, yes, this is quite rough, but still nice :-)
    Remember that these numbers are avarage values, you will not have exactly 6.8 percent for "the" in a particular text anyway.
    The important information here, not in particular numbers or value, but in the overall distribution of these words. Hope this helps :-)

  • Hi Devin, the quadrance is a square of a distance, they are involved to avoid square roots.
    Let's have a right triangle ABC with the hypotenuse AB
    then the pythagoras theorem says that |AB|^2=|AC|^2+|BC|^2.
    The quadrance Q(A,B) is a squared distance between points A and B, that is
    Q(A,B)=|AB|^2 and so on.
    in Q1 (see A1) we use the same theorem but in...

  • Hi Tony,
    as I understand
    your question can be reduced to the following:
    why (-5)^1=-5 and (-5)^2/2=5=|-5| are not equal?
    this is because the function
    x^1=x and the function sq rt of (x^2)=|x|
    are different.
    we define an absolute value
    |x|=sq rt of (x^2)
    like this to avoid such questions and confusions.
    and the function x^(6/4) is a combination of...

  • Hi Monty,
    I have never heard about this before. If you have a link, please let us know! it seems interesting :-)

  • Please have a look at comments below. We had a lot of discussion on this question, hope they help :-)

  • Hi Benedict, the constant "a" can be anything and this is a scaling constant which roughly shows how far the hyperbola is from its center.
    Try to draw several hyperbolas to feel this meaning,
    e.g.
    y=1/x, y=2/x, y=1/2x
    and with negative a
    y=-1/x, y=-2/x, y=-1/2x
    (all this hyperbolas have centers at [0,0])

  • Let's try to work on 1/u - 1/t first
    what would we do if we get numbers say
    1/5-1/7=(7-5)/7*5
    the same is here
    1/u - 1/t=(t-u)/tu
    next step
    dividing by (u-t) the same as multiplying by 1/(u-t), let's use it
    (1/u - 1/t)/(u - t)= (1/u - 1/t)*1/(u-t)=(t-u)/tu*1/(u-t)
    to cancel out the numerator of the first fraction and the denominator of the second one...

  • Hi Gerry, I do not think that we may state the precise values of coordinates of X, since they are not given in the problem. We may use only numbers which are given and the picture is always given to help us to understand an approximate situation.

  • Hi Lukombo.
    I will try to explain this one more time from the begining, hope that this will be useful.
    Look at the picture and first find the quadrance from X to the line l
    The distance from the line l to the point X in x-1 , where x is x-coordinate of X, so the quadrance is
    Q(X,l)=(x-1)^2
    Now to find Q(X,F) take a perpendicular from the point X to the...

  • Hi Dave, please try to read/understand the comments below.
    There is a lot of discussion on this question :-)

  • But if you want to finish
    you also have to have h^2 = 4/9, so we have to add and substract h^2, so proceeding from the previous comment we get
    3y^2-4y=3 (y^2 - 4y/3)=3 (y^2 - 4y/3+4/9-4/9)= 3 (y^2 - 4y/3+4/9)-4/3=
    3(y-2/3)^2-4/3

    Overall we obtain
    (x^2)-(3y^2-4y)=(x-0)^2-3(y-2/3)^2+4/3=0
    or
    (x-0)^2-3(y-2/3)^2=-4/3
    We do not care about a and b here,...

  • Hi Lorena! You are right, we have studied compliting the square in the previous course.
    However, it is not difficult, I am sure you can figure this out easily.

    We take the equation x^2-3y^2+4y=0
    to find the center we need to rewrite it in the form
    (x - k)^2/a^2 - (y - h)^2/b^2 =1
    for some numbers h,k,a,b
    this is the general form of a hyperbola...

  • Good job! :-)

  • About the factor: you are absolutely right.
    If we rotate h1, we obtain h2, which has the equation x^2-y^2=2 (it is not easy to see in fact)
    (check this: the point of intersections of h2 with x-axis is [sqrt(2),0])
    so, to obtain h3 that is x^2-y^2=1 we have to scale in both x and y directions by the factor
    1/sqrt(2).
    Then the point [sqrt(2),0] becomes...

  • Hi Dai.
    "Or (x-y) = 1/(x+y), hard to believe there could be a relationship there."
    There is a relationship indeed)))
    Passing from h1 to h3 is in some sense moving from (x,y)-plane to a new coordinate system, say (x',y')-plane.
    Consider the hyperbola h3 in (x,y)-plane
    If you now make a substitution
    x'=x-y
    y'=x+y
    then in (x',y')-plane you obtain the...

  • Yes Jana, you are right.
    If we take y=1/x (let's call it h1) and just rotate it, then the point [1,1] will be moved to the point [square root of 2,0], which is the intersection of this new hyperbola (h2) with x-axis. But the hyperbola (h3) we want to obtain intersects x-axis at [1,0]. Thus, to obtain h3 we also have to scale h2.
    What do you think is the...

  • My answers are the same )))

  • Hi Dai, yes, you are right, this is another way to find roots of the quadratic equation
    x^2 +px+q=0.
    from my previous comment you may see that
    r+s=-p
    rs=q
    From this system for given p and q you may guess the roots.
    However, it is not always applicable, since the roots can be irrational or fractional, in this case it is hard to guess the...

  • Hi Chris, the idea in question 3 from the quiz is that you do not have to solve the quadratic equation, just expand
    x^2+4x-77=(x-r)(x-s)=x^2-(r+s)x+rs
    and then compare the coefficients in front of x on the both sides of the equality
    so you get
    4=-(r+s) which implies r+s=-4

    and even if you solve the equation
    x^2+4x-77=0, the roots are
    -11 and 7, so...

  • Hi Tomas, the idea to find vertex and focus is in translating and dilating if necessary the parabola y=x^2 and see where the vertex and focus are moved.
    You have already learned how to complete the square. This is the first step to find vertex and focus.
    Let's consider an example
    y=x^2-6x+5=(x-3)^2-4
    It means that the vertex is [3,-4], see solution for...

  • Hi Tomas, the solution is written at the end, if you cannot follow it, please, let me know what is your question, which step is not clear.....otherwise I have to rewrite the same)))

  • Hi Tony, the explanation from the lecture is that every parabola has this point F and a line l such that if you take a point on the parabola, the distance from this point to F and to l will be the same. To determine the position of F and l is not so easy task. But you know for sure that F lies on the axis of the parabola and if you look at the vertex, you see...

  • Hi Antonio, thank you for nice words, it is a pleasure :-)
    If you want to get a certificate, at the top of the course page go to "progress" your achievements will be calculated there and if it's 70% or more at the end of this page you will be suggested to buy the certificate.
    Let me know how it goes.

  • Hi Sling!
    Thank you for your question, it is indeed not clear from the lecture.
    Answering your first question, no, it is not the case. We always have to say the range for the parameter t that is used in parametrization.
    And you are right, if you restrict t as you did, you will obtain a segment of a line but not the whole line.
    To obtain the line, we need...

  • Hi everyone, Ursula has given above a very detailed explanation what the function is. Thank you, Ursula!
    However, the phrase in Q2 that this is not a function assumes that x is an independent variable (the domain) and y (the range) depends on x.
    Personally, as soon as I see x and y in an expression, it is usually automatic (for me)
    to look at this as y...

  • Goog job :-)
    What about the other way around? If AB passes through the focus then the tangents are perpendicular.

  • Yes, that's right, Jana :-)

  • Hi Gerry!Good job :-)
    Yes, d=2x, c=1 is right.

    We multiply by 4a to make the same denominator.
    This is a usual trick in adding fractions, when we find a common denominator.
    E.g. if we want to add
    2/3 + 1/12
    The trick is to multiply the numerator and denominator of the first fraction by 4, this does not change the fraction but it makes easier to add...

  • Hi Ursula! If you consider x as a function of y then, you are right))
    Usually x is an independent variable and so we consider y as a function of x.
    In this case y(x) is not a function, this is what it meant in Q2, but I agree it should be clarified))

  • Hi everyone :-) Nice to see this interesting discussion!
    I like this clever idea to see a tangent as a degenerate secant,
    since it is indeed a limiting case of a secant.
    Like we discussed the pair-of-line and considered this situation as a degenerate case of a hyperbola))))

  • Hi Clare. Adding to the explanation given by Suzi, you may also think about t as about a moment of time and x and y are coordinates of some object (say a car) which moves along the curve (see the animation given in the lecture, the car is the point moving).
    Then the parametrization [t, t^2/4] says that at the moment t=2 the car was at the position [2,1]...

  • Hi Gerry!
    Multiplying by 4a we have
    c/a=(4ac)/4a^2
    and so
    c/a - b^2/4a^2=(4ac)/4a^2-b^2/4a^2=(4ac - b^2)/4a^2

  • Hi Ursula! Yes, it means here that there are no real solutions :-) Sorry for the confusion!

  • Hi Gerry! You indeed need some experience/practice to see how this works.
    As soon as you see the expression like d^2+2cd+c^2 you can write it as (d+c)^2
    (this is the main formula which we use here, and it's very easy to prove just opening the brackets (d+c)(d+c)).
    As soon as you remember the formula the problem is to determine c and d
    in some particular...

  • Hi Suzi! It appeared that I have answered your question too in my previous comment :-) hope it helps!

  • Hi Nadia! In Q4 f) it is claimed that x=y and x=-y is a conic section and this is indeed true.
    You can obtain two lines when a plane intersects the double cone and passes through the apex.
    Please follow the link below to visualize this situation
    http://math2.org/math/algebra/conics.htm

    Thus, two-lines is a conic section, but it is not a...

  • I am glad that it helped :-) yes, you are absolutely right, this is a limiting case. However, we try to avoid such words here to not confuse people, who are not familiar with the concept of limit. I also tried to find a simpler example. Let me know if you find something ;-)

  • Yes, I see your point!!! This is a degenerate case of the hyperbola, but it is still not a hyperbola.
    The same as e.g. take a right polygon say with 100 sides then increase the number of sides, you may consider the circle as a degenerate case of such polygons, but you would never say that circle is a polygon. May be it is not the best example, but hope this...

  • Hi Nadia! The equation x^2=y^2 is the equation of two lines y=x and y=-x. This is degenerate case of conic sections. This appears in 3.4 "Pairs of lines". Hope this helps!

  • Hi Louis! This angle can be useful if you want to change the coordinate system to put ellipse at the "right" place (in such a way that the origin becomes the center of the ellipse and the axis of the ellipse becomes the x- and y-axis). To do this you have to rotate and move the original coordinate system . This angle will be the angle of rotation. There is...

  • Hi Reino! It is so good that you have noticed this problem! It is not because you have taken perpendicular lines, the problem here in this special point [0,0]. In this correspondence between points and lines if you take a point which is closer and closer to the center [0,0] you get a line which is further and further from the ellipse. and so for [0,0] the...

  • Anna Tomskova replied to [Learner left FutureLearn]

    Q1: Sorry that the information about the axis is not here. The axis of the parabola is parallel to the y-axis and goes trough the vertex. Try to finish the solution to get the idea.
    You can find an alternative solution for Q4 in comments above e.g. see the solution suggested by Nick James)

  • Yes, you are right, (1,4) in round brackets is used to denote a vector. You may think that this vector starts at [0,0] and ends at [1,4]. However, remember that vector has fixed length and a direction but its position is not fixed.

  • Great solution, Kevin. Just a minor typo: (2t)(2u)=-1 not -1/4.

  • Hi everyone! Yes, in Q4 it is asked about real solutions. We do not assume that you know complex numbers)))

  • Yes, William, it is correct. The vertex is above x-axis and the parabola is in the upper half plane, so it does not have intercepts with x-axis.

  • Yes, Andrew, that's correct! Sorry to hear that it was so difficult for you. Hope it will be getting easier.

  • I do not think that it is indirect. The factor of proportionality is negative in this case. But it is still a direct proportionality.

  • Yes, if the price of one drink is fixed)

  • Hi Laura, good examples! yes, in the third example the factor of proportionality is pi.

  • Hi Georgette! Yes, it is a very nice example to demonstrate such a problem!
    To everyone: have fun solving this one too))))