*_CHALLENGE_* Calculate the integral of *log₂₇ 9 + log₈ 4* from *−(log₂ √6 + log₂ √⅔)* to *log₅ 50 − log₅ 2* . Please upload this sum’s solution video. I have specially made it for you to solve.
I suspect you’d get the exact same result, given the geometric demonstration at the end. This should apply to all n-polynomials taking the limit at the x^n coefficient goes to infinity
@@heinrich.hitzinger you don’t need such formulae. The leading coefficient always determines the steepness of the curve and thus as long as the y intercept is below 0, the x intercepts to the polynomials will always approach (0,0)
@@frimi8593 ..actually no... consider ax³+bx²+cx+d ... as a and b goes towards 0 or infinity, x intercept x goes towards d, as well as some extremes for 'ax³+bx²+cx+d' its at d/2 as ab goes to infinity, u actually get a very odd triangle(infinite high) connected to a line, actually a very long smoth parabola with so low curvature we can consider it a triangle, sloope goes toward some constant depending on abc -> it approaches -(∞/2)x + d edit: as löong as c in the ' cx ' term does not apprach infinity, in that case x intercept → =0
@@Patrik6920 so I mean I haven’t formally proven it, but I’m quite confident that as a goes towards infinity, all x intercepts approach 0, provided the y intercept is below 0. Consider, as all the other coefficients remain constant, they quickly become irrelevant and we end up looking for the solutions of lim a->inf ax^n-d=0 where n is the degree of the polynomial. With all the other terms removed it becomes much more obvious that x is approaching 0, no?
This makes geometric sense. As a -> oo, the parabola "tightens up" onto the y-axis. An infinitely "tight" parabola, that also crosses the x-axis (hence c < 0), is like a ray, and when at the y-axis, that means it intersects the x-axis at x = 0.
Check out Dr. Barker's video on the alternate quadratic formula: x = [ 2c ] / [ -b plus/minus sqrt(b^2 - 4ac) ] As a ----> inf with c0, one from the right and one from the right. ruclips.net/video/kOrTSW0_394/видео.html
We can do this by another approach, so (b^2-4ac)^1/2 we can apply binomial theorem and we will see that we will get the highest power of a as 1/2 now in the denominator we have a^1 so when we will divide by a^(1/2) in numerator and denominator as well, we will get constant in numerator and in denominator we will have (a) ^1/2 so there fore we have constant/infinity thus it tends to zero
@blackpenredpen Hello sir, can you explain how to solve y'' + ky = 0 w/o trial and error? Some solutions i found online use laplace transform but i dont really get it! An explanation would really help of the solution and if required, what laplace transform is! Thank you!
@5:50. That graph becomes so hard to draw without losing meaning without context, could you just draw the y=x^2 graph, but have the first mark on the x axis at 1/[infinity] and placed wide.
"Stay positive, don't be so negative all the time!" While certainly not bad advice, c has to flout it, at least this time. Because today, the motto is, "Keep things real!"
Can you try letting a approaches to zero? I found it difficult to do so and result was quite strange when I enter in on wolfram alpha, it was not just bx+c=0 Later I found that the first step of getting the quadratic formula is diving a for all terms, that's why the result is strange
You can rationalize the numerator, to derive this variant on the quadratic formula: x = 2*c/[-b +/- sqrt(b^2 - 4*a*c)] Now you can let a = 0 directly: x = 2*c/[-b +/- sqrt(b^2)] x = 2*c/(-b +/- b) Of the two solutions, only -b-b is legitimate. The other solution would have us dividing by zero. This gives us: x = -c/b And this is consistent with the solution to the linear equation, b*x + c = 0.
See the Michael Penn video on that - see link in the description. He has a very different style than BPRP but I find the maths just as interesting from both channels. I'm really glad that they've started to reference each others work 😊
I think Gabriel is just trying to make an easier version of calculus. He doesn't hate derivatives and integrals, he only hates the idea of limits. And to his defense, a lot of people struggle defining a limit during their college years.
Well well, you are taking a limit of a given function so you are supposed to deal with limit and then put the value. You'll learn in class 11 limits and derivatives, hope this helps
B = Inf: (sqrt(a) times x times sqrt(c)) times 2 = inf times x; but x irrelevantly smaler than a and or c ? What Happens if you put in inf for b? Is this a stupid question?
The f(x) = ax^2 + bx + c can be thought of as a family of functions parametrised by a. There is no uniform limit of this family of functions. There is a point-wise limit existing at x = 0, which is just a singular point (0,c), but that's all I'm afraid, a point wise limit doesn't exist at any other point. Really you need to explore the surface g(x, a) = ax^2 + bx + c where b and c are fixed. If you consider g(1/sqrt(a), a) = 1 + b/sqrt(a) + c which tends to a finite value as a tends to infinity...
Kind of a casual diss of a→0, considering that there are _two_ expressions in the quadratic formula and only _one_ of them approaches the solution to bx+c=0. (But I suppose that the video by Michael Penn covers that.)
Same result if lim a -> -inf by inspection if c also negative. Considering the cases where there's a negative sign under the root, the roots are complex for finite a: but as |a| --> inf the roots tend to 0.i which is the same limit but approached from another direction in the complex plane.
The graph you sketched should have a vertex that lies on x = 0!! I was going crazy for a bit after looking at your graph and imagining the change in A lol
Should be 0, I think, and perhaps makes sense, since if the first coefficient, the square's coefficient is made larger and larger and the others are kept constant, perhaps the only thing that could make the equation 0 would be 0, lol...I mean, should be zero because the power of the numerator is 1/2, and the power of the denominator is 1...
Great video! Now please take the limit of the Thin Lens Formula. 1/f = 1/p + 1/q where f = Focal Length of the lens, p = Object distance from the lens, and q = Image distance from the lens, take the limit of this formula as f approaches 0,
In all of this man’s videos I get from his energy that he is flexing math skills rather than actually teaching to the audience. Although the content is still great. I just wish he wouldn’t be an arrogant ass
We know that for a general parabola, the vertex is at x= -b/2a. Plug that back into the general function, and after simplifying we find the y value of the vertex is always at -(b^2)/4a+c Take the limit as a goes to infinity, and we get y=c
as in sqrt(a^2+b^2)? if you take it as a (or b) approaches 0, itd just be 0, and if you take it as a (or b) approaches infinity itd just be infinity And if you take a multivariable limit where both of them approach zero, itd still just be 0, and if you take a multivariable limit where both of them approach infinity, itd be infinity again so he probably wouldnt cover it because no matter how you do it its very boring and straightfoward
@@ChessBros-ic5tz In that case youd be taking the same limit of 2 equations with different values which simply isnt possible the only way you can do it is like the one i showed
Yes, you are right. But it has been already assumed that x is tending to POSITIVE infinity. So, the modulus opens with a positive sign, thus, |2a| = 2a
I mean, a lot of his videos aren't necessarily "useful." That doesn't mean it isn't interesting or worth watching, at least, it doesn't given the right target audience.
Integral of ln(x) from 1 to "what" =2? ruclips.net/video/GPfl8khHGYg/видео.html
Stay amazing and thank you.
Sorry you have to deal with horrible people saying nasty things.
@@T-Rex_Prime I would but my arms are to short to reach the pencil
@@T-Rex_Prime 😁 They probably have longer arms outside of tyrannosaurid family.
I totally understand why you need to ask for help with the math.
*_CHALLENGE_*
Calculate the integral of *log₂₇ 9 + log₈ 4* from *−(log₂ √6 + log₂ √⅔)* to *log₅ 50 − log₅ 2* .
Please upload this sum’s solution video. I have specially made it for you to solve.
@@Maths_Wonderland answer is 8 eazy bro
If he could marry the quadratic formula, he'd definitely do so lmfao
Nah, he'd probably marry taking the limit of every possible smoothstep.
He'd marry the Lambert w function
Grow up, childish and hateful.
no he would marry the Alpha fish
I saw how he was looking at that equation, it was love eyes lmfao
Now, take limit of a cubic formula.
I suspect you’d get the exact same result, given the geometric demonstration at the end. This should apply to all n-polynomials taking the limit at the x^n coefficient goes to infinity
@@frimi8593There are no formulae for quintic and higher polynomials. 😔
@@heinrich.hitzinger you don’t need such formulae. The leading coefficient always determines the steepness of the curve and thus as long as the y intercept is below 0, the x intercepts to the polynomials will always approach (0,0)
@@frimi8593 ..actually no...
consider ax³+bx²+cx+d ... as a and b goes towards 0 or infinity, x intercept x goes towards d, as well as some extremes for 'ax³+bx²+cx+d' its at d/2 as ab goes to infinity, u actually get a very odd triangle(infinite high) connected to a line, actually a very long smoth parabola with so low curvature we can consider it a triangle, sloope goes toward some constant depending on abc -> it approaches -(∞/2)x + d
edit: as löong as c in the ' cx ' term does not apprach infinity,
in that case x intercept → =0
@@Patrik6920 so I mean I haven’t formally proven it, but I’m quite confident that as a goes towards infinity, all x intercepts approach 0, provided the y intercept is below 0. Consider, as all the other coefficients remain constant, they quickly become irrelevant and we end up looking for the solutions of lim a->inf ax^n-d=0 where n is the degree of the polynomial. With all the other terms removed it becomes much more obvious that x is approaching 0, no?
Is there an application for using geometric formulas in calculus?
look into analytic geometry
@@soumakirimoto2195 something new to look into! thanks
@@soumakirimoto2195 Yes. Analytic geometry is the main bridge to calculus, by combining the cartesian plane, functions and geometry.
Aod also provides some numericals based on co ordinate geometry and derivatives
Also in Related Rates and Optimization problems
5:33 He's either talking about the dirac delta function or sin 1/x which oscillates infinitely
I also was expecting the dirac delta (or one of its cousins, unit doublet/triplet/etc)
This makes geometric sense. As a -> oo, the parabola "tightens up" onto the y-axis. An infinitely "tight" parabola, that also crosses the x-axis (hence c < 0), is like a ray, and when at the y-axis, that means it intersects the x-axis at x = 0.
bro what is this
😂😂😂
Upar se nikal gaya kya 😂😂
Stand proud, you are strong
@@utkarshjain861 😭😭😭
@@tilkkone4257 lol💀💀
@@savitatawade2403 😂😂😂
Check out Dr. Barker's video on the alternate quadratic formula: x = [ 2c ] / [ -b plus/minus sqrt(b^2 - 4ac) ]
As a ----> inf with c0, one from the right and one from the right.
ruclips.net/video/kOrTSW0_394/видео.html
u should now do lim a->0 and show that the quadratic root (sometimes!) tends to the linear root in different cases of (a,b,c)
Did you not watch the video? He said Michael Penn published a video doing exactly that, so he himself won't.
@@Sir_Isaac_Newton_i skipped the first minute until when he actually started
And it can't work because there is then a division by zero which makes the formula useless for a=0.
That graph looks a bit like an inverted Dirac-delta, which is a sufficiently cursed concept that you should do a video!
Only Steve can integrate, differentiate, and take limit(s) of the quadratic and not be diagnosed with insanity.
Is his name really Steve?
Day 10 of asking if you can do the intergal of 1/(x^i) with π and e as its limits.
"The bottom is going to dominate the top." Well there's a twist :P
You mean a switch, surely?
Now you only need to take the infinite series of the quadratic formula to complete the saga
Bro a request to make videos on functional equations you haven't made videos on it, also i only better understand maths from you only ❤
We can do this by another approach, so (b^2-4ac)^1/2 we can apply binomial theorem and we will see that we will get the highest power of a as 1/2 now in the denominator we have a^1 so when we will divide by a^(1/2) in numerator and denominator as well, we will get constant in numerator and in denominator we will have (a) ^1/2 so there fore we have constant/infinity thus it tends to zero
@blackpenredpen Hello sir, can you explain how to solve y'' + ky = 0 w/o trial and error? Some solutions i found online use laplace transform but i dont really get it! An explanation would really help of the solution and if required, what laplace transform is! Thank you!
Now try an asymptotic expantion aproach
@5:50. That graph becomes so hard to draw without losing meaning without context, could you just draw the y=x^2 graph, but have the first mark on the x axis at 1/[infinity] and placed wide.
If we take lim a➛0, quadratic equation will become linear and it holds, i.e. x=-c/b, very funny.
"Stay positive, don't be so negative all the time!" While certainly not bad advice, c has to flout it, at least this time. Because today, the motto is, "Keep things real!"
Can this result be obtained from root locus
Sir please provide lectures on sieve theory
1:#3 "When keeping it real goes wrong". I wonder if anyone will get this reference?
Can you try letting a approaches to zero?
I found it difficult to do so and result was quite strange when I enter in on wolfram alpha, it was not just bx+c=0
Later I found that the first step of getting the quadratic formula is diving a for all terms, that's why the result is strange
You can rationalize the numerator, to derive this variant on the quadratic formula:
x = 2*c/[-b +/- sqrt(b^2 - 4*a*c)]
Now you can let a = 0 directly:
x = 2*c/[-b +/- sqrt(b^2)]
x = 2*c/(-b +/- b)
Of the two solutions, only -b-b is legitimate. The other solution would have us dividing by zero. This gives us:
x = -c/b
And this is consistent with the solution to the linear equation, b*x + c = 0.
See the Michael Penn video on that - see link in the description. He has a very different style than BPRP but I find the maths just as interesting from both channels. I'm really glad that they've started to reference each others work 😊
Cant you use l'hopital's rule?
Hi blackpenredpen, since you do a lot of calculus, what do you think of John Gabriel's New Calculus? Is it rubbish or is there some truth in it?
I think Gabriel is just trying to make an easier version of calculus. He doesn't hate derivatives and integrals, he only hates the idea of limits. And to his defense, a lot of people struggle defining a limit during their college years.
if we put infinity in the formula thet directly gets zero. is that wrong? (im in class 10 but have knowledge of some things of class 11-12)
Well well, you are taking a limit of a given function so you are supposed to deal with limit and then put the value. You'll learn in class 11 limits and derivatives, hope this helps
Is it even possible to find a general solution for x for the equation a^(x ln(x))+bx+c=0, Wolfram Alpha doesn't really give an answer.
Can you do the Spanish math college entrance exam of 2024, it’s rising a lot of controversy because of the difficulty?
B = Inf: (sqrt(a) times x times sqrt(c)) times 2 = inf times x; but x irrelevantly smaler than a and or c ? What Happens if you put in inf for b? Is this a stupid question?
Does anything weird happen when you allow c to be greater than 0?
Like, I get that it becomes a complex number, but what is wrong with that?
Distance between line and point:
ruclips.net/video/nvLpz3Ek0Gg/видео.htmlsi=50J8Vb_TRSCXUXRG
Why if a approaches zero does the quadratic formula not give the solution x= -c/b?
b: i'm only constant after all, i'm only constant after alo don't put the blame on me
Bro once come to INDIA we study all these in class 11 for the world's 2nd most toughest exam 😂
1:15 why not use complex numbers? Surely inf*i isn’t too bad to work with. 😁
Interesting that the limit is 0 for a formula that is designed to give the zeroes of a quadratic function.
hey i have a challenge for u : find the range of the function , f(x)=(3x-x^3)/(1-3x^2)
All real numbers.
ruclips.net/video/eNrM2rFMmQo/видео.html : "And keep things real" - always with the gangsta talk
-c/b and -b/a
Excellent Sir
plz also take some good questios from ventor and 3d geometry plz
Wake up babe new delta function just dropped 🗣️🗣️
So the limit as a approaches infinity of a parabola is a delta function?
The f(x) = ax^2 + bx + c can be thought of as a family of functions parametrised by a. There is no uniform limit of this family of functions. There is a point-wise limit existing at x = 0, which is just a singular point (0,c), but that's all I'm afraid, a point wise limit doesn't exist at any other point. Really you need to explore the surface g(x, a) = ax^2 + bx + c where b and c are fixed. If you consider g(1/sqrt(a), a) = 1 + b/sqrt(a) + c which tends to a finite value as a tends to infinity...
Outstanding 👍
Kind of a casual diss of a→0, considering that there are _two_ expressions in the quadratic formula and only _one_ of them approaches the solution to bx+c=0. (But I suppose that the video by Michael Penn covers that.)
Can you try finding the exact solution of this?
“(e^x)+(3x^3)-(1/x)=0”,
Thanks so much for your time at looking this.
nx^2+bx+c=f_n(x), there Is not uniform convergention on by Moore Osgood theorem. There Is the problem in zero with changes limits.
Same result if lim a -> -inf by inspection if c also negative.
Considering the cases where there's a negative sign under the root, the roots are complex for finite a: but as |a| --> inf the roots tend to 0.i which is the same limit but approached from another direction in the complex plane.
Can you prove that why
(-)(+)=-
(-)(-)=+
This should be the same as the limit as p -> 0 and q -> 0-. Your manipulation just turned it into the p, q formula.
And now the limes of the pq formula.
Big fan but a boring conclusion considering what a high school algebra student should know about what a does to a quadratic
I've come up with a cool geometry problem, I don't know if you would be interested
Of course he's interested.
Now intergrate the cubic formula
The graph you sketched should have a vertex that lies on x = 0!! I was going crazy for a bit after looking at your graph and imagining the change in A lol
Bro invented dirac delta function part 2
Singular perturbation. Use method of strained variables.
Hi blackpenredpen!
Could you do a video on non-linear regression? Talking about using the jacobian and hessian will be cool!
you've integrated, differentiated, and now taken the limit of the quadratic formula!!
Should be 0, I think, and perhaps makes sense, since if the first coefficient, the square's coefficient is made larger and larger and the others are kept constant, perhaps the only thing that could make the equation 0 would be 0, lol...I mean, should be zero because the power of the numerator is 1/2, and the power of the denominator is 1...
now raise the quadratic formula to the quadratic formula power
becomes ax^2 = 0 or x^2=0 i.e. x=0
Great video! Now please take the limit of the Thin Lens Formula.
1/f = 1/p + 1/q
where f = Focal Length of the lens, p = Object distance from the lens, and q = Image distance from the lens, take the limit of this formula as f approaches 0,
As f approaches zero, either p or q approaches zero or both of them approach zero.
I have only recently been watching Michel Penn and now you reference him lol. Nice coincidence.
What if you make “a” approaching a complex or a irrational number 🤔
Tou dont even need a limit, just ser a to the number and it will work!
what about videos on elliptic curves?
Please make a video on Fourier transform.
Now take the limit of the quartic formula
Bprp always keeping things real 😤✊
what are you after
Dirac delta function?!
Manx where do we use 0^+
????
Goat
In all of this man’s videos I get from his energy that he is flexing math skills rather than actually teaching to the audience. Although the content is still great. I just wish he wouldn’t be an arrogant ass
The thumbnail is fire 🔥🔥🔥😂
where can i buy this T-shirt
nvm
Can we be sure that for x=0, y = c, when a goes infinite?
We know that for a general parabola, the vertex is at x= -b/2a. Plug that back into the general function, and after simplifying we find the y value of the vertex is always at
-(b^2)/4a+c
Take the limit as a goes to infinity, and we get y=c
جمييل جداا❤
This guy is so good
We love content like this.
Love from India
nerd
hey
Cool
1
What about if you take the limit of the phytagoras
as in sqrt(a^2+b^2)? if you take it as a (or b) approaches 0, itd just be 0, and if you take it as a (or b) approaches infinity itd just be infinity
And if you take a multivariable limit where both of them approach zero, itd still just be 0, and if you take a multivariable limit where both of them approach infinity, itd be infinity again
so he probably wouldnt cover it because no matter how you do it its very boring and straightfoward
I mean a^2+b^2-c^2 like when you take the c^2 to the other side
Also what happens if you equate phytagoras in the form(a^2+b^2-c^2) to = the quadratic formula
@@ChessBros-ic5tz In that case youd be taking the same limit of 2 equations with different values which simply isnt possible the only way you can do it is like the one i showed
First 😂
Ew... 🤢
Isn’t sqrt(4a^2)=|2a| ?
a---> ∞ so no need of the modulus
Yes, you are right. But it has been already assumed that x is tending to POSITIVE infinity. So, the modulus opens with a positive sign, thus, |2a| = 2a
Divide numerator and denominator by a. Numerator goes to zero and denominator is two. The limit is zero
Limit DNE. The domain of the quadratic formula in a is bounded above.
what a waste of a video
I mean, a lot of his videos aren't necessarily "useful." That doesn't mean it isn't interesting or worth watching, at least, it doesn't given the right target audience.
1) Write a sentence. 2) It is a terrible thing to waste your terrible mind for not appreciating the video.