@@stefanobilleter2556 Statement of the problem appears to be incomplete, or ambiguous. When you say, "... and b being also a prime ..." - did you forget to say that p is prime? Or a? Or is b the only one required to be prime? In which case, why use "p" rather than just, "c"? Fred
I assumed the translation values as a and b. So (x+a)(y+b)=k --> xy + bx + ay + (ab - k) = 0 (eqn 1) Also, y = (x - 3)/(1 - 3x) --> -3xy - x + y + 3 = 0 Dividing by -3 to make xy coefficient 1: --> xy + (1/3)x + (-1/3)y + (-1) = 0 (eqn 2) Comparing like terms in eqns 1 & 2: a = -1/3, b = 1/3, ab - k = -1 --> -1/9 - k = -1 --> k = -1/9 + 1 --> k = 8/9
yup ... that is how I was taught to do it. But it was in a college 200 level dynamics class. How to translate an origin to a new reference by a constant factor. The lesson started off ... 'remember in high school when they showed you this' ... and every one was like ... NO!.
First idea of attack (meaning it's probably not the best way): You can get xy = k, by first getting y = k/x. So let's "translate" that denominator like so: x' = x-⅓; x = x'+⅓ Then we will have y = (x-3)/(1-3x) = (x'-8/3)/(-3x') = -⅓(1 - 8/[3x']) = -⅓ + 8/[9x'] y + ⅓ = 8/[9x'] Now just let y' = y + ⅓ and we get x'y' = 8/9 so k = 8/9 . . . not something I would have anticipated, without doing the work Now to see how bprp does it... Fred
So if you make the denominator 1-3x equal to x' by translation, then x = (x'-1)/-3 = (1-x')/3 When x' → 0 ←, x = 1/3. So x - 1/3 is a natural choice because we know neither x or y can be zero because xy=k>0.
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The way I did it was by reasoning, that the graph of y can be written in the form (x-a)(y-b)=k (just by the question) So, multiplying this thnig out yields xy - xb - ya + ab = k If you multiply the actual function of the graph by the denominator, you get y(1-3x) = x - 3, rearranging a little bit and you have -3xy -x + y = -3 We want no konstant term in front of the xy term, so divide by -3: xy + x/3 - y/3 = 1 We can now easily choose a and be as the factors for the single x and y term, so that they match the general formula at the top: a = -1/3, b = 1/3 => xy - (-1/3)x - (1/3)y = 1 Now adding on both sides the ab = (-1/3)(1/3) = -1/9 term will give us the solution: xy - (-1/3)x - (1/3)y + (-1/3)(1/3) = 1 + (-1/3)(1/3) = 8/9 Now, we not only have our k (k=8/9) but also the x and y-shift (a is x-shift and b is y-shift)
Yes, the question is actually really simple. The hint especially almost gives the answer, just fill in x' and y' as x+a and y+b respectively and do a little algebra. Because of the strict form of both the answer and the transform (only translation) you can really only do it one way. When I started doing it I thought it might possibly be a range or multiple values of k, but you really don't need any insight in the problem to solve it. Literally did it on the back (ok, the front) of an envelope.
I use the definition of a translation: (x-3)/(1-3x) = f(x) f(x) translated is k/x because xy = k becomes y = k/x. f(x-a)+b = k/x (x-a-3)/(1-3(x-a)) + b = k/x (x-a-3+b(1-3x+3a))/(1-3x+3a) = k/x Multiplying it out, rearranging, and then factoring, I got: ((1-3b)x-a+b-3ab-3)/(-3x+3a+1) = k/x To get it into a form with x in the denominator, I took out a factor of -1/3 and put it in the numerator. Then, the coefficient of x in the numerator had to be 0, so I had: -(1-3b)/3 = 0 3b = 1 b = 1/3 The denominator is now (x-a-1/3), so (-a-1/3) must equal 0. -a-1/3 = 0 a = -1/3 Now that the coefficient of x in the numerator is 0, all terms with a & b in the numerator come out to the constant k. -(-a+b+3ab-3)/3 = k This simplifies to 8/9.
I think a simpler way to state the problem would be to write "The graph of y = (x-3)/(1-3x) can be transformed into the graph of xy = k by a translation. For what value of k is this possible?"
Without knowing the answer... y = (x - 3) / (1 - 3x) We want to make the numerator on the right hand side a constant. Add 1/3 to both sides y + 1/3 = [(x - 3) + (1 - 3x)/3)] / (1 - 3 x) y +1/3 = (-3 + 1/3) / (1 - 3x) y + 1/3 = (-8/3) / (1 - 3x) y +1/3 = (8/3) / (3x - 1) y +1/3 = (8/9) / (x - 1/3) (x - 1/3)(y + 1/3) = 8/9 The left hand side shows the translation. The right hand side is k
You're gonna lose most people without showing the cancellations and substitutions. You rewrote 3(x-3) + (1-3x) as straight to (x - 3) + (1 - 3x)/3), that's a lot of jumps to follow. (x-3) 1 3(x-3) + (1-3x) 3(-3+1/3) 1/3 - 3 (-8/3) / -3 8/9 → k ____ + ---- = ------------------------ = ---------------- = -------------- = ---------------- = ------ ----- = y + 1/3 → y' (1- 3x) 3 3(1-3x) 3(1-3x) 1-3x (1-3x) / -3 x-1/3 → x' There ... so they can follow. It cancels x in the numerator? I guess that is saying you want to make the numerator constant again like you originally said. What you meant was the variable x in the numerator is 1x and the variable in the denominator is -3x so you add 1/3 to get the numerator as a constant? I know you know what you are talking about ... just trying to make it easier for everyone else to understand the method ... it's called partial fraction decomposition if the numerator has an x value with a lower power than the numerator or in this case polynomial long division ... just like he did in the video.
@@abds8842 just as he did in the video ... he divided (x-3) by (-3x-1) ... -3x goes into 1x, -1/3 times and the x cancels, so you have to add 1/3 to both sides ... which cancels the -1/3 on the right hand side making the numerator a constant and y+1/3 on the left hand side. Hope that helps.
Pay attention to my method, it is very simple. I see this function as a ratio and therefore use the properties of ratios. y=(x-3)/(1-3x) I multiply both sides of equality by 3. The goal is to eliminate x from the right-hand side of the ratio. 3y=(3x-9)/(1-3x) We add the denominators of the fractions to the ratios. (3y+1)=-8/(1-3x) So (3y+1)(1-3x)=-8 (y+1/3)(1/3-x)=-8/9
The best way, find dy/dx, then generate its scaling first order symmetries i.e. ^x = x+ce, ^y = y + ce, which define ^y=y exp(e). Sorry, it’s using Lie group theory is too complex a procedure for your viewers, but it does solve ALL the scaling transformation.
I think my way was better. Firstly, I wan't sure what was meant by a translation. Did it mean just horizontal shift , but when I saw the question on the paper BPRP showed to camera, I realised it meant horizontal and vertical, as in linear algebra. So I substituted (x-a) for x , and (y-b) for y in the original equation and expanded it out to see what values of a and be were needed to make the unwanted terms in x and y disappear.
Just find the vertical and horizontal asymptotes first so we know how much the graph is translated. Set the denominator to zero to find the vertical asymptote at x=1/3. Set the limit x goes to infinity to find the horizontal asymptote at y=-1/3. So we need to translate xy=k right by 1/3 and down by -1/3, which means we solve for k from (x-1/3)(y+1/3)=k and y=(x-3)/(1-3x) (two equations are identical). Expanding both equations, we get xy+x/3-y/3-1/9=k and -3xy+y-x=-3, then divide the second equation by -3 to become xy+x/3-y/3=1. Now it is obvious k=8/9.
How I tackled this wasn't at all this smooth, but it does show deductive reasoning so I would like to share it. The first thing I did was move all the variables to one side and all the constants on the other side and got y - 3xy - x = -3. I noticed the lone y on the left hand side is because I first multiplied both sides by 1-3x. So I think I need a horizontal shift such that the 1 disappears in the denominator, so I shift the graph to the left by 1/3. With that I got y=(x-8/9)/(-3x) = -1/3 + 8/9x. Okay, so now the -1/3 is an issue, no problem I'll shift the graph up by 1/3 and it's gone. So now the transformed graph is y = 8/9x, multiply both sides by x and you get xy = 8/9.
Wow that was alot simpler. I first solved for x/y intercept, asymptote of the post translated equation and plugged in the values to solve for k in a rewritten form of the function before translation. Asm: (1/3,∞) X-int: (3,0) Y-int: (0,-3) y=k/(x-A) + B From Asm => A = 1/3 From X-int => B = -3k/8 From Y-int => k = 24/27 = 8/9
I had an easy time understnading the question but never though to do polynomial division, because usually polynomial division just complicates the expression and gets you nowhere. Some great rules I follor for math: * don't divide polynomials * don't substitute * only trust the formulas you memeorized for the specific scenarrio So, even though I always have smart ideas, I never actually succeed because my rules prevent me from using anything cool.
My method - Manipulate the original equations so that they are in the same form: y = (x - 3) / (1 - 3x) and (x + a)(y + b) = k (baking the translations as additional variables in the equation involving k). Starting from (x + a)(y + b) = k: 1. (y + b) = k / (x + a) 2. y = k / (x + a) - b 3. y = (k - b(x + a)) / (x + a) 4. y = (k - bx - ba) / (x + a) 5. y = ((k - ba) - bx) / (x + a) 6. y = (-3(k - ba) - (-3bx)) / (-3x - 3a) 7. y = (3bx - 3(k - ba)) / (-3a - 3x) Note, (7) this is in the same form as the original equation (y = (px - q) / (r - sx)), so we can equate the coefficients between them: p: 3b = 1; q: 3(k - ba) = 3; r: -3a = 1; s: 3 = 3. I already handled this one by multiplying by -3/-3 in (5)->(6) :) A system of equations in 3 variables, where it is very easy to solve for a=-1/3 and b=1/3, and substitute in the third equation to find k=8/9.
Revisiting this video after a while, I did it this way, with comparatively minimal algebra: xy=k means y=k/x, so the translation has to have a vertical asymptote at x=0 and a horizontal asymptote at y=0. f(x)=(x-3)/(1-3x) has a VA at x=1/3. To get it to 0, f(x+ 1/3)= (x- 8/3)/(1-3x-1) f(x+ 1/3)=g(x)=(x- 8/3)/(-3x) g(x) has an HA at y=-1/3. To get it to 0, g(x)+ 1/3= (x- 8/3)/(-3x)+ 1/3 = (8/3 -x)/(3x) + x/(3x) = (8/3)/(3x) = (8/9)/x y = (8/9)/x k = 8/9.
Wait, how did you end up with y + 1/3? Shouldn't it be y - 1/3 if we are trying to translate the horizontal asymptote toward y=0? You subtract the amount you want it to move up. Same with x, you move the vertical asymptote toward x = 0, meaning you subtract the amount you want to move right, meaning you get x + 1/3. In fact, replacing y with y + 1/3 and x with x -1/3 in the original function simplifies to y = (2x - 4) / (2 - 3x) which has asymptotes at x = 2/3 and y = -2/3, meaning it cannot match the graph yx=k which would have asymptotes at y=0 and x=0.
A logic sequence might go like this ... y is undefined when the denominator 1-3x = 0. So we know there is an asymptote when x = 1/3. As x→ ∞, ( x- 3 ) / (1 - 3x ) = y → -1/3. So we know that is an asymptote. The translation must have it's asymptotes shifted. x = 1/3, so x - 1/3 = 0, & y = -1/3, so y + 1/3 = 0. Therefore, ( x - 1/3 ) ( y + 1/3 ) = k. Substitute for y. k = ( x - 1/3 ) ((( x - 3 ) / ( 1 - 3x )) + 1/3 ). When x'=1, y'=k, x' = x - 1/3, so when x = 4/3. k = (4/3 - 3) / (1 - 12/3) + 1/3 = -5/3 / -3 + 1/3 = 5/9 + 3/9 = 8/9 k = 8/9. Logic.
I rewrote the denominator as 3(1/3 - x) and did partial fraction decomposition 😂 That gave -1/3 - (8/9) / (1/3 -x). Then x -> x + a and adding b yields: y = -1/3 - (8/9) / (1/3 - x - a) + b. Now it is easy to see that a and b must be 1/3, respectively. Therefore y = 8/(9x) and k = 8/9.
I went with the approach (x+a)(y+b)=k, multiplied it out to get xy+bx+ay+ab-k=0 Similarly for the other expression I got y-3xy-x+3=0 Then a term by term comparison (starting with multiplying the first equation by -3 to get the roght xy-term-coefficient) to yield the result k=8/9
Hi blackpenredpen; Will you solve a problem of differential equation with the Frobenius Method. Problem: Find two linear independent series solutions in powers of x of DE x^2y" - (x^2 + x )y′ + y = 0 I am waiting.. 🙂
Did it entirely in head, took some 2 mins to massage the equation and it yielded (3y+1)(3x-1)=8. Divided by 9 and yes, got the correct answer, in head.
Before you solve ans factorise thé équation you should explain what a translation means for a graph and how you see it in a function. If you do not do that, once you have factorise correctly Yon Can still not understand that you have answered the question
perhaps a more straightforward method is the following. Suppose we have the translation (x,y) --> (x+A, y+B) for A, B reals, on the original equation y = (x-3)/(1-3x). Then, the equation becomes y = (x + A - 3) / (1 - 3x - 3A) - B. Next, we desire to find the values of A and B (ultimately to find k) such that y = (x + A - 3) / (1 - 3x - 3A) - B = k/x. Getting rid of fractions and regrouping based on coefficients of powers of x. Comparing coefficients we can get a system of three equations in A, B and k. Finding k is then trivial (finding A and B is not even necessary). The value of k is then found to be 8/9.
How do you compare coefficients to get the system of equations? I was trying to work this out myself but got stuck after I grouped the coefficients by powers of x
@@cameronkhanpour3002 basically bring everything to the LHS and because the RHS is 0 all coefficients of the powers of x must be 0 and also the constant term must be 0.
I love this question because it involves absolutely nothing very technical to solve it but it's really hard to understand it properly I didn't understand what were we supposed to get with K, but seeing the end result makes perfect sense and makes the question look almost easy. The trick here was really understanding what we were looking for and being really handy with number manipulations to make the translation formula appear
If you want to solve this by graphical intuition, I'd say the way as presented is pretty good. If you want the most efficient way, a better question might be _How can I manipulate the first equation to get to the form of the second equation?_ The answer is to multiply both sides by *(x - 1/3)* with *x ≠ 1/3* to get rid of the denominator: *(x - 1/3) * y = 1 - x/3 => (x - 1/3) * (y + 1/3) = 1 - 1/3^2 = 8/9 =: k*
I rewrote the goal as y = k/x, so I figured that I needed some sort of translation to get rid of the 1 in the denominator of (x - 3)/(1 - 3x). This can be done with a translation x › x + 1/3, which turns it into (x + 1/3 - 9/3)/(1 - 3x - 3/3) = (x - 8/3)/(-3x) = -1/3 + (8/9)/x. This is now a vertically shifted version of xy = 8/9 = k, as required.
I didn’t really understand the question till I actually just went at it, I started simply by rearranging the RHS to -(1/3) + (-8/3)/(-3x+1). Then after playing around with it a bit I got y+(1/3) = 8/9(x-(1/3)). From here you I just kinda saw the answer to equal 8/9 after I multiplied out part of the RHS denominator.
I was confused about what "translation" meant. So I pondered over the comments and someone said it basically is a shift of axes, like x becomes x-a and y becomes y-b. I simply put that into the xy=k equation, and compared it with the given equation. k is easily found out to be 8/9
As seen at the end of the video; if we draw the graph of f function which is given . Vertical is 1/3, horizontal is -1/3 it means that to get y=k/x x--->(x+1/3) and y--->(y+1/3) can be written at f and k is obtained.
My approach was to write a generic function g(x) as a vertical and horizontal translation of f(x) and create 3 equations (well, one was finding that the horizontal shift is trivially by -1/3) with 3 unknowns. Basic substitution/elimination from there.
to elaborate further, as this wasn't the easiest method but I think a fairly easy method to understand... g(x) = f(x-d)+c (1) f(0) = -3 which means that g(d) = -3 + c = k/d (2) f(3) = 0 which means that g(3+d) = 0+c = k/(3+d) and finally (3) the horizontal asymptote was moved by 1/3, giving d = -1/3 or using the same style f(1/3) is undefined therefore g(1/3 + d) = k / (1/3 + d) is undefined --> d = -1/3 So from there it's simply 2 equations with 2 unknowns
It's pretty straightforward to just do x -> x-a and y -> y-b, then figure out which a and b give you y = k/x. After expanding everything, the coefficient of x on top is 1-3b, so let b = 1/3; the constant in the denominator is 1+3a, so let a = -1/3. Do the rest of the expansion with these numbers and you get y = 8/9x, so xy = 8/9 = k. Long division and factoring are definitely the smart way to do it, but in the heat of the moment, I think the plug-n-chug method is what I would reach for.
You can alternatively do that by mapping out the transformations and then letting y = k/x and solving for k. If you get a constant for k, there's a unique solution. If you get an expression involving x for k, it would probably best to do a domain and range map.
that's basically what my solution was. A few more lines of writing compared to the one in the video but it was easier to follow and think up in advance, I think
@@robertveith6383 I found my solution. Write me if the link doesn't work. drive.google.com/file/d/1ITcdchJ90XU_BF9es718LV4TXRdGTons/view?usp=drivesdk In my class such problems were something that we used to solve every day. Sorry for my bad English)
Your problem is off subject, *and* you wrote it incorrectly in at least one way. Whatever it might be (if it were correct), you supposedly intend to be dividing by 2x. You need to have 2x inside grouping symbols.
The first way that I thought of doing this was, following the thumbnail, to write (x+a)(y+b) = k. Then use the original equation to substitute for y, getting an equation with only x (along with the unknown constants a, b, and k). Now plug in any 3 values for x (or 4 if you accidentally use x = -1/3 for one of them and get a division by zero), getting 3 equations for the 3 unknowns. But then I decided that this would be too much work and did it your way. (Although I made a sign error and got -8/9. 🫤)
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Try this!
IMO, 5th problem:
Finds the triplets (a, b, p) of integer positive numbers, and b being also a prime number, s.t. : a^p=b!+p
@@stefanobilleter2556 Statement of the problem appears to be incomplete, or ambiguous.
When you say, "... and b being also a prime ..." - did you forget to say that p is prime? Or a?
Or is b the only one required to be prime? In which case, why use "p" rather than just, "c"?
Fred
That’s an example of “Understanding the question is half of the solution”
exam marking rubrics: about that
@@ghsdftfm 😂😂
8/9 of the difficulty
@@trueriver1950 Ikr
I assumed the translation values as a and b.
So (x+a)(y+b)=k
--> xy + bx + ay + (ab - k) = 0 (eqn 1)
Also, y = (x - 3)/(1 - 3x)
--> -3xy - x + y + 3 = 0
Dividing by -3 to make xy coefficient 1:
--> xy + (1/3)x + (-1/3)y + (-1) = 0 (eqn 2)
Comparing like terms in eqns 1 & 2:
a = -1/3, b = 1/3,
ab - k = -1 --> -1/9 - k = -1
--> k = -1/9 + 1 --> k = 8/9
yup ... that is how I was taught to do it. But it was in a college 200 level dynamics class. How to translate an origin to a new reference by a constant factor. The lesson started off ... 'remember in high school when they showed you this' ... and every one was like ... NO!.
First idea of attack (meaning it's probably not the best way):
You can get xy = k, by first getting y = k/x. So let's "translate" that denominator like so:
x' = x-⅓; x = x'+⅓
Then we will have
y = (x-3)/(1-3x) = (x'-8/3)/(-3x') = -⅓(1 - 8/[3x']) = -⅓ + 8/[9x']
y + ⅓ = 8/[9x']
Now just let y' = y + ⅓ and we get
x'y' = 8/9
so k = 8/9 . . . not something I would have anticipated, without doing the work
Now to see how bprp does it...
Fred
So if you make the denominator 1-3x equal to x' by translation, then x = (x'-1)/-3 = (1-x')/3
When x' → 0 ←, x = 1/3. So x - 1/3 is a natural choice because we know neither x or y can be zero because xy=k>0.
Slowly and unnoticed, our little channel is about to reach a million subscribers. I remember i first found this channel in 2020 and it really helped me cope with lockdown and learn new things. It always has been a beautiful experience being with this community
The way I did it was by reasoning, that the graph of y can be written in the form (x-a)(y-b)=k (just by the question)
So, multiplying this thnig out yields xy - xb - ya + ab = k
If you multiply the actual function of the graph by the denominator, you get y(1-3x) = x - 3, rearranging a little bit and you have -3xy -x + y = -3
We want no konstant term in front of the xy term, so divide by -3:
xy + x/3 - y/3 = 1
We can now easily choose a and be as the factors for the single x and y term, so that they match the general formula at the top:
a = -1/3, b = 1/3 => xy - (-1/3)x - (1/3)y = 1
Now adding on both sides the ab = (-1/3)(1/3) = -1/9 term will give us the solution:
xy - (-1/3)x - (1/3)y + (-1/3)(1/3) = 1 + (-1/3)(1/3) = 8/9
Now, we not only have our k (k=8/9) but also the x and y-shift (a is x-shift and b is y-shift)
Did the same thing. I stopped watching the video when he said "let's do polynomial long division" :D
Yes, the question is actually really simple. The hint especially almost gives the answer, just fill in x' and y' as x+a and y+b respectively and do a little algebra. Because of the strict form of both the answer and the transform (only translation) you can really only do it one way.
When I started doing it I thought it might possibly be a range or multiple values of k, but you really don't need any insight in the problem to solve it. Literally did it on the back (ok, the front) of an envelope.
Please please do more of these. My school took these when I was in high school and they have some really interesting and difficult questions.
When you don't know what the teacher is teaching and just want to pretend that you are paying attention: 0:08
I use the definition of a translation:
(x-3)/(1-3x) = f(x)
f(x) translated is k/x because xy = k becomes y = k/x.
f(x-a)+b = k/x
(x-a-3)/(1-3(x-a)) + b = k/x
(x-a-3+b(1-3x+3a))/(1-3x+3a) = k/x
Multiplying it out, rearranging, and then factoring, I got:
((1-3b)x-a+b-3ab-3)/(-3x+3a+1) = k/x
To get it into a form with x in the denominator, I took out a factor of -1/3 and put it in the numerator. Then, the coefficient of x in the numerator had to be 0, so I had:
-(1-3b)/3 = 0
3b = 1
b = 1/3
The denominator is now (x-a-1/3), so (-a-1/3) must equal 0.
-a-1/3 = 0
a = -1/3
Now that the coefficient of x in the numerator is 0, all terms with a & b in the numerator come out to the constant k.
-(-a+b+3ab-3)/3 = k
This simplifies to 8/9.
I think a simpler way to state the problem would be to write "The graph of y = (x-3)/(1-3x) can be transformed into the graph of xy = k by a translation. For what value of k is this possible?"
That's the same question in a slightly different order. I don't understand why this would be simpler
Yeah the transformed by translation is redundant. They could have said "be translated in to the graph xy=k."
Without knowing the answer...
y = (x - 3) / (1 - 3x)
We want to make the numerator on the right hand side a constant. Add 1/3 to both sides
y + 1/3 = [(x - 3) + (1 - 3x)/3)] / (1 - 3 x)
y +1/3 = (-3 + 1/3) / (1 - 3x)
y + 1/3 = (-8/3) / (1 - 3x)
y +1/3 = (8/3) / (3x - 1)
y +1/3 = (8/9) / (x - 1/3)
(x - 1/3)(y + 1/3) = 8/9
The left hand side shows the translation. The right hand side is k
how do you know to add 1/3 to both sides?
@@abds8842 Because it cancels the x in the numerator
You're gonna lose most people without showing the cancellations and substitutions. You rewrote 3(x-3) + (1-3x) as straight to (x - 3) + (1 - 3x)/3), that's a lot of jumps to follow.
(x-3) 1 3(x-3) + (1-3x) 3(-3+1/3) 1/3 - 3 (-8/3) / -3 8/9 → k
____ + ---- = ------------------------ = ---------------- = -------------- = ---------------- = ------ ----- = y + 1/3 → y'
(1- 3x) 3 3(1-3x) 3(1-3x) 1-3x (1-3x) / -3 x-1/3 → x'
There ... so they can follow.
It cancels x in the numerator? I guess that is saying you want to make the numerator constant again like you originally said. What you meant was the variable x in the numerator is 1x and the variable in the denominator is -3x so you add 1/3 to get the numerator as a constant?
I know you know what you are talking about ... just trying to make it easier for everyone else to understand the method ... it's called partial fraction decomposition if the numerator has an x value with a lower power than the numerator or in this case polynomial long division ... just like he did in the video.
@@abds8842 just as he did in the video ... he divided (x-3) by (-3x-1) ... -3x goes into 1x, -1/3 times and the x cancels, so you have to add 1/3 to both sides ... which cancels the -1/3 on the right hand side making the numerator a constant and y+1/3 on the left hand side. Hope that helps.
Btw is there a black pen red pen's discord ?
Pay attention to my method, it is very simple. I see this function as a ratio and therefore use the properties of ratios.
y=(x-3)/(1-3x)
I multiply both sides of equality by 3. The goal is to eliminate x from the right-hand side of the ratio.
3y=(3x-9)/(1-3x)
We add the denominators of the fractions to the ratios.
(3y+1)=-8/(1-3x)
So (3y+1)(1-3x)=-8 (y+1/3)(1/3-x)=-8/9
Well done!
Multiply through by -1 to get the final answer, that supports out the coefficient of x at the same time :)
The best way, find dy/dx, then generate its scaling first order symmetries i.e. ^x = x+ce, ^y = y + ce, which define ^y=y exp(e). Sorry, it’s using Lie group theory is too complex a procedure for your viewers, but it does solve ALL the scaling transformation.
I think my way was better. Firstly, I wan't sure what was meant by a translation. Did it mean just horizontal shift , but when I saw the question on the paper BPRP showed to camera, I realised it meant horizontal and vertical, as in linear algebra. So I substituted (x-a) for x , and (y-b) for y in the original equation and expanded it out to see what values of a and be were needed to make the unwanted terms in x and y disappear.
Idk my mind instantly just wanted to divide the numerator
Just find the vertical and horizontal asymptotes first so we know how much the graph is translated.
Set the denominator to zero to find the vertical asymptote at x=1/3. Set the limit x goes to infinity to find the horizontal asymptote at y=-1/3. So we need to translate xy=k right by 1/3 and down by -1/3, which means we solve for k from (x-1/3)(y+1/3)=k and y=(x-3)/(1-3x) (two equations are identical). Expanding both equations, we get xy+x/3-y/3-1/9=k and -3xy+y-x=-3, then divide the second equation by -3 to become xy+x/3-y/3=1. Now it is obvious k=8/9.
How I tackled this wasn't at all this smooth, but it does show deductive reasoning so I would like to share it. The first thing I did was move all the variables to one side and all the constants on the other side and got y - 3xy - x = -3. I noticed the lone y on the left hand side is because I first multiplied both sides by 1-3x. So I think I need a horizontal shift such that the 1 disappears in the denominator, so I shift the graph to the left by 1/3. With that I got y=(x-8/9)/(-3x) = -1/3 + 8/9x. Okay, so now the -1/3 is an issue, no problem I'll shift the graph up by 1/3 and it's gone. So now the transformed graph is y = 8/9x, multiply both sides by x and you get xy = 8/9.
Wow that was alot simpler. I first solved for x/y intercept, asymptote of the post translated equation and plugged in the values to solve for k in a rewritten form of the function before translation.
Asm: (1/3,∞)
X-int: (3,0)
Y-int: (0,-3)
y=k/(x-A) + B
From Asm => A = 1/3
From X-int => B = -3k/8
From Y-int => k = 24/27 = 8/9
I had an easy time understnading the question but never though to do polynomial division, because usually polynomial division just complicates the expression and gets you nowhere.
Some great rules I follor for math:
* don't divide polynomials
* don't substitute
* only trust the formulas you memeorized for the specific scenarrio
So, even though I always have smart ideas, I never actually succeed because my rules prevent me from using anything cool.
My method - Manipulate the original equations so that they are in the same form: y = (x - 3) / (1 - 3x) and (x + a)(y + b) = k (baking the translations as additional variables in the equation involving k).
Starting from (x + a)(y + b) = k:
1. (y + b) = k / (x + a)
2. y = k / (x + a) - b
3. y = (k - b(x + a)) / (x + a)
4. y = (k - bx - ba) / (x + a)
5. y = ((k - ba) - bx) / (x + a)
6. y = (-3(k - ba) - (-3bx)) / (-3x - 3a)
7. y = (3bx - 3(k - ba)) / (-3a - 3x)
Note, (7) this is in the same form as the original equation (y = (px - q) / (r - sx)), so we can equate the coefficients between them:
p: 3b = 1;
q: 3(k - ba) = 3;
r: -3a = 1;
s: 3 = 3. I already handled this one by multiplying by -3/-3 in (5)->(6) :)
A system of equations in 3 variables, where it is very easy to solve for a=-1/3 and b=1/3, and substitute in the third equation to find k=8/9.
Revisiting this video after a while, I did it this way, with comparatively minimal algebra:
xy=k means y=k/x, so the translation has to have a vertical asymptote at x=0 and a horizontal asymptote at y=0.
f(x)=(x-3)/(1-3x) has a VA at x=1/3. To get it to 0,
f(x+ 1/3)= (x- 8/3)/(1-3x-1)
f(x+ 1/3)=g(x)=(x- 8/3)/(-3x)
g(x) has an HA at y=-1/3. To get it to 0,
g(x)+ 1/3= (x- 8/3)/(-3x)+ 1/3
= (8/3 -x)/(3x) + x/(3x)
= (8/3)/(3x)
= (8/9)/x
y = (8/9)/x
k = 8/9.
Awesome!!!
Wait, how did you end up with y + 1/3? Shouldn't it be y - 1/3 if we are trying to translate the horizontal asymptote toward y=0? You subtract the amount you want it to move up. Same with x, you move the vertical asymptote toward x = 0, meaning you subtract the amount you want to move right, meaning you get x + 1/3.
In fact, replacing y with y + 1/3 and x with x -1/3 in the original function simplifies to y = (2x - 4) / (2 - 3x) which has asymptotes at x = 2/3 and y = -2/3, meaning it cannot match the graph yx=k which would have asymptotes at y=0 and x=0.
A logic sequence might go like this ...
y is undefined when the denominator 1-3x = 0. So we know there is an asymptote when x = 1/3.
As x→ ∞, ( x- 3 ) / (1 - 3x ) = y → -1/3. So we know that is an asymptote.
The translation must have it's asymptotes shifted. x = 1/3, so x - 1/3 = 0, & y = -1/3, so y + 1/3 = 0.
Therefore, ( x - 1/3 ) ( y + 1/3 ) = k. Substitute for y.
k = ( x - 1/3 ) ((( x - 3 ) / ( 1 - 3x )) + 1/3 ).
When x'=1, y'=k, x' = x - 1/3, so when x = 4/3.
k = (4/3 - 3) / (1 - 12/3) + 1/3 = -5/3 / -3 + 1/3 = 5/9 + 3/9 = 8/9
k = 8/9.
Logic.
I rewrote the denominator as 3(1/3 - x) and did partial fraction decomposition 😂 That gave -1/3 - (8/9) / (1/3 -x). Then x -> x + a and adding b yields: y = -1/3 - (8/9) / (1/3 - x - a) + b. Now it is easy to see that a and b must be 1/3, respectively. Therefore y = 8/(9x) and k = 8/9.
I went with the approach (x+a)(y+b)=k, multiplied it out to get
xy+bx+ay+ab-k=0
Similarly for the other expression I got
y-3xy-x+3=0
Then a term by term comparison (starting with multiplying the first equation by -3 to get the roght xy-term-coefficient) to yield the result k=8/9
how you find Circumference of an ellipse? this is hasnwt formula!!!!! how!///
Hi blackpenredpen; Will you solve a problem of differential equation with the Frobenius Method.
Problem: Find two linear independent series solutions in powers of x of DE x^2y" - (x^2 + x )y′ + y = 0
I am waiting.. 🙂
Did it entirely in head, took some 2 mins to massage the equation and it yielded (3y+1)(3x-1)=8. Divided by 9 and yes, got the correct answer, in head.
After how many years of schooling do you have this task offered in your country?
More correctly, for what age of pupils? 13+?
Thank you for clearly differentiating your content from the sponsor. I really appreciate that.
This is a typical problem of changing coordinates.
Is this what the "by translation" means?
Before you solve ans factorise thé équation you should explain what a translation means for a graph and how you see it in a function. If you do not do that, once you have factorise correctly Yon Can still not understand that you have answered the question
hi can you plz teach us how to solve national Taiwan university transfer exam about calculus a problem
Awesome explanation thank you
Waiting for bprp launch channel for Analysis
perhaps a more straightforward method is the following. Suppose we have the translation (x,y) --> (x+A, y+B) for A, B reals, on the original equation y = (x-3)/(1-3x). Then, the equation becomes y = (x + A - 3) / (1 - 3x - 3A) - B. Next, we desire to find the values of A and B (ultimately to find k) such that y = (x + A - 3) / (1 - 3x - 3A) - B = k/x. Getting rid of fractions and regrouping based on coefficients of powers of x. Comparing coefficients we can get a system of three equations in A, B and k. Finding k is then trivial (finding A and B is not even necessary). The value of k is then found to be 8/9.
How do you compare coefficients to get the system of equations? I was trying to work this out myself but got stuck after I grouped the coefficients by powers of x
@@cameronkhanpour3002 basically bring everything to the LHS and because the RHS is 0 all coefficients of the powers of x must be 0 and also the constant term must be 0.
@@ethanthiebaut2596 oh so you just set each coefficient of the polynomial to 0 and solved each each equation from there?
Your explanation is not straight forward.
@@ejrupp9555 I agree that the last part is rather sloppy, but that was simply because I was in a rush
I got k=sqrt(8/9). It's so close it looks right!
I love this question because it involves absolutely nothing very technical to solve it but it's really hard to understand it properly
I didn't understand what were we supposed to get with K, but seeing the end result makes perfect sense and makes the question look almost easy. The trick here was really understanding what we were looking for and being really handy with number manipulations to make the translation formula appear
Pa cuando el vídeo de 100 límites?
What I did:
Notice that y goes to infinity when x = 1/3, so x' = x-1/3 must be the x translation.
Plug that in and the problem is basically solved.
You know, I would’ve got this right within like 5 minutes, except my brain thought that 3*3=27…
Try this!
IMO, 5th problem:
Finds the triplets (a, b, p) of integer positive numbers, and b being also a prime number, s.t. : a^p=b!+p
If you want to solve this by graphical intuition, I'd say the way as presented is pretty good. If you want the most efficient way, a better question might be
_How can I manipulate the first equation to get to the form of the second equation?_
The answer is to multiply both sides by *(x - 1/3)* with *x ≠ 1/3* to get rid of the denominator:
*(x - 1/3) * y = 1 - x/3 => (x - 1/3) * (y + 1/3) = 1 - 1/3^2 = 8/9 =: k*
Just cross multiply and solve like a quadratic equation with xy in place of x^2. This will make it little bit faster
I rewrote the goal as y = k/x, so I figured that I needed some sort of translation to get rid of the 1 in the denominator of (x - 3)/(1 - 3x). This can be done with a translation x › x + 1/3, which turns it into (x + 1/3 - 9/3)/(1 - 3x - 3/3) = (x - 8/3)/(-3x) = -1/3 + (8/9)/x. This is now a vertically shifted version of xy = 8/9 = k, as required.
Bruh, at first I got 8/9. But I thought that was to simple and tried some other way and got -1.
Hey bprp, love your content, you should try to do a video where you prove the pythagorean theorem blindfolded.
I didn’t really understand the question till I actually just went at it, I started simply by rearranging the RHS to -(1/3) + (-8/3)/(-3x+1). Then after playing around with it a bit I got y+(1/3) = 8/9(x-(1/3)). From here you I just kinda saw the answer to equal 8/9 after I multiplied out part of the RHS denominator.
I was confused about what "translation" meant. So I pondered over the comments and someone said it basically is a shift of axes, like x becomes x-a and y becomes y-b. I simply put that into the xy=k equation, and compared it with the given equation. k is easily found out to be 8/9
As seen at the end of the video; if we draw the graph of f function which is given . Vertical is 1/3, horizontal is -1/3 it means that to get
y=k/x x--->(x+1/3) and y--->(y+1/3) can be written at f and k is obtained.
Difficulty level :
Eaaaasssyyyyyyyyy
What does translation mean?
This question was surprisingly easy. I decided to solve it myself for once
My approach was to write a generic function g(x) as a vertical and horizontal translation of f(x) and create 3 equations (well, one was finding that the horizontal shift is trivially by -1/3) with 3 unknowns. Basic substitution/elimination from there.
to elaborate further, as this wasn't the easiest method but I think a fairly easy method to understand...
g(x) = f(x-d)+c
(1)
f(0) = -3
which means that g(d) = -3 + c = k/d
(2)
f(3) = 0
which means that g(3+d) = 0+c = k/(3+d)
and finally
(3)
the horizontal asymptote was moved by 1/3, giving d = -1/3 or using the same style
f(1/3) is undefined
therefore g(1/3 + d) = k / (1/3 + d) is undefined --> d = -1/3
So from there it's simply 2 equations with 2 unknowns
a ver con qué sorpresa llegas hoy …
Nope. No idea what I just watched.
It's pretty straightforward to just do x -> x-a and y -> y-b, then figure out which a and b give you y = k/x. After expanding everything, the coefficient of x on top is 1-3b, so let b = 1/3; the constant in the denominator is 1+3a, so let a = -1/3. Do the rest of the expansion with these numbers and you get y = 8/9x, so xy = 8/9 = k. Long division and factoring are definitely the smart way to do it, but in the heat of the moment, I think the plug-n-chug method is what I would reach for.
For better readability, leave a space between addition and subtraction signs: x - a, y - b, 1 - 3b,
1 + 3a, etc.
Can you do this with a matrix and Eugene vectors?
No, both functions aren't linear. Besides, translation is an affine transformation and not a linear one.
Cool problem
You can alternatively do that by mapping out the transformations and then letting y = k/x and solving for k.
If you get a constant for k, there's a unique solution. If you get an expression involving x for k, it would probably best to do a domain and range map.
that's basically what my solution was. A few more lines of writing compared to the one in the video but it was easier to follow and think up in advance, I think
Damn
last
I'm last now :)
first
I think @nurani chandra beat you
That one was easy
*Thumbs-down.* Please describe your "easy" solution here instead of boasting about it with an otherwise empty post.
@@robertveith6383 I found my solution. Write me if the link doesn't work.
drive.google.com/file/d/1ITcdchJ90XU_BF9es718LV4TXRdGTons/view?usp=drivesdk
In my class such problems were something that we used to solve every day. Sorry for my bad English)
@@robertveith6383 And this called "comment". Don't be so toxic)
Prove →Sinh ( ln(x) ) = [x² - 1]/2x
Your problem is off subject, *and* you wrote it incorrectly in at least one way. Whatever it might be (if it were correct), you supposedly intend to be dividing by 2x. You need to have 2x inside grouping symbols.
The first way that I thought of doing this was, following the thumbnail, to write (x+a)(y+b) = k. Then use the original equation to substitute for y, getting an equation with only x (along with the unknown constants a, b, and k). Now plug in any 3 values for x (or 4 if you accidentally use x = -1/3 for one of them and get a division by zero), getting 3 equations for the 3 unknowns. But then I decided that this would be too much work and did it your way. (Although I made a sign error and got -8/9. 🫤)