If you have any math discovery or fun problem that you want me to present in a video, please leave it here (with solutions hopefully). I will try my best to make it into a video for you and will credit you in the video.
Question
My formulas are: y=(x+a)^2 x=(-y+a)^2 I would like to solve for a where they touch (exact solution). Here is a graph demonstrating what I mean: As you can see, I have attempted to solve for a by making them closer and closer. I think 1.100917368760402698 ≤ a ≤ 1.100917368760402699, but floating point error may have reduced the accuracy of that range, but it should be near those numbers at least. There will also be a negative solution of a , which will be exactly negative I believe.
Sir, I have this new Idea what if you have trigonometric ratios to a base i.e. sin 90 to base π is 1 ? I'm working on it and I named it Trigology. Sir, I want you to help me out.
Ok! Thanks! I will be interested to take a look.
@blackpenredpen okay sir, will you make a video on it ?
Can you please explain why we add "2kpi" when finding complex roots? And also show the roots on the Argand diagram, giving us a detailed explanation? Thanks.
Hi,
Can you please solve the question in the image?
Can’t you just solve it graphically? Set theta equal to your equation, which is in terms of theta, in a graphing calculator (unless you are not allowed to use a calculator)
This is a question from a JEE Advanced paper.
Another thing, does this site have a glitch? Because the typed text is shown in wrong order.
I just wrote up a solution :) Here's a link to the pdf file: https://drive.google.com/file/d/1U3MJz4kklkZjAqbl6pbKg5w0bJ1wqqhr/view?usp=sharing
I suppose I should have posted my integral here... Anyway, I'd love to see a video about the alternate methods I posted in the discussions for solving the Putnam integral you just posted a video about. If you need more of a worked solution than the one I gave, then I can provide one :)
x*sin(x)=k
Find x in terms of k.
the roots of this polynomial
https://www.wolframalpha.com/input/?i=x*sin(x)+taylor+series
@michael einhorn Not looking for roots. I want to rearrange the equation so that I can isolate x.
@KING SOBIESKI That's impossible. You're asking to find the inverse, which in this case you can't do. Even if you could, it would not work on the whole real line, it would work in sections of length pi.
Simple Harmonic motion is governed by the differential equation F = -kx or d^2x/dt^2 = -kx/m.
This equation models the oscillation of an object on a spring.
The solution to this equation is a linear combination of sine cosine.
https://www.wolframalpha.com/input/?i=d%5E2x%2Fdt%5E2+%3D+-+kx
How is the integral evaluated, neither my AP physics nor multi variable calc teacher at my high school knew.
That is a second order linear differential equation with constant coefficients. This one is easier because there is no x'(t) term, so we can simply realize that both sin and cos are solutions, and they are linearly independent so their linear combination is a general solution (this is a classic theorem in differential equations). More generally to solve things like this, make the substitution x(t)=e^(at) where a is a constant that you will find. Here, you'll find that a is an imaginary number and you can use Euler's formula to turn that into the sin and cos (there are two values of a, and we generally look for real solutions so we throw out the solutions multiplied by i). You can read more about this at http://tutorial.math.lamar.edu/Classes/DE/SecondOrderConcepts.aspx
Could you make a video about why in some cases we have to simplify sqrt(x^2) to |x| instead of x and when we have to be more careful about that?
Do I have to be careful every time I simplify by 2 inside an exponent? For example if I’m simplifting x^(5/2)*x^(1/2) to x^3 do I have to put the absolute value?
Thank you for your time : )
http://mathserver.neu.edu/~lovett/teaching/math1140/challenge.html
The question is number 3. Using a general cubic function f(x) = ax^3 +bx^2 + cx + d,
find the tangent line to f(x) at x=x0, then find where that tangent line intersects the cubic function, other than at x0 of course.
Here is the link to my solution.
https://docs.google.com/document/d/1_qpDZEkmEnig9GGQQhLBziU_GjzN297AN3J29u4QD54/edit?usp=gmail
The integral of screwdriver, or how to turn English words into integrals
Hello ! You should try to determinate the domain of anything raised to anything (x^y). Here is my answer ! https://www.blackpenredpen.com/discussions/_math/domain-of-x-y
@blackpenredpen :D
please solve this equation
That would be x=3, since after simplifying, you get
(2-x)^2*(x-2)^2=1, and 1*1 is 1.
Integrate The following
(x^6057 + x^4038 + x^2019)* [(2x^4038 + 3x^2019 + 6)^1/6]
Daunting looking but has a very innovative solution
If a^b=b^a ,a and b>0
How to solve it by showing steps😂
There are already videos on this:
https://www.youtube.com/watch?v=PI1NeGtJo7s
https://www.youtube.com/watch?v=z65wrFB0W-Y
Remember that video you just did about all 3 cases of y' times y'' equalling y'''? Well, here's my solution to the general case 2 [Hopefully this comment pushes through unlike in the YouTube video]:
General Case 2: C2>0
1/(u^2+C2) du = 1/2 dx Integrate both sides: arctan(u/sqrt(C2))/sqrt(C2) = x/2 + C3 Multiply both sides by sqrt(C2): arctan(u/sqrt(C2)) = xsqrt(C2)/2 + C4 Take the tangent of both sides: u/sqrt(C2) = tan(xsqrt(C2)/2 + C4) Multiply by sqrt(C2): u = sqrt(C2) tan(xsqrt(C2)/2 + C4) Since u = dy/dx: dy/dx = sqrt(C2) tan(sqrt(C2)*x/2 + C4) Multiply by dx: dy = sqrt(C2) tan(sqrt(C2)*x/2 + C4) dx Integrate both sides: y = 2 sqrt(C2) Ln|sec[xsqrt(C2)/2 + C4]|/sqrt(C2) + C5 Cancel sqrt(C2) to get all answers from case 2: y = 2 Ln|sec[xsqrt(C2)/2+C4]| + C5
Lo and behold!
Oh, and here's case 3:
General Case 3 where C2<0:
Let -C2 = C3
Hence u^2+C2 can and will be written from here onwards as u^2-C3 where C3>0
special mini-case: u^2-C3=0, thus y= + or - sqrt(C3)+ C4 (this is where linear functions come from) [proof is trivial for this mini-case and is left as a mini-exercise for the reader]
Now, if u^2-C3 isn't 0:
1/sqrt(u^2-C3) du = 1/2 dx
Integrate both sides:
-1/sqrt(C3) arctanh(u/sqrt(C3)) = x/2 + C4 [where arctanh(w) is the inverse of tanh(w)]
Multiply both sides by -sqrt(C3):
arctanh(u/sqrt(C3)) = -xsqrt(C3)/2 + C5
Rearrange the right hand side:
arctanh(u/sqrt(C3)) = C5 - xsqrt(C3)/2
Taking the hyperbolic tangent of both sides:
u/sqrt(C3) = tanh(C5 - xsqrt(C3)/2)
Multiply both sides by sqrt(C3)
u = sqrt(C3) tanh(C5 - xsqrt(C3)/2)
Since u = dy/dx
dy/dx = sqrt(C3) tanh(C5 - xsqrt(C3)/2)
Multiply by dx
dy = sqrt(C3) tanh(C5 - xsqrt(C3)/2) dx
Integrate both sides:
y = sqrt(C3) * -2/sqrt(C3) * Ln(cosh(C5 - xsqrt(C3)/2)) + C6
Cancel sqrt(C3):
y = -2 Ln(cosh(C5 - xsqrt(C3)/2)) + C6
We can convert the negative sign outside the ln to taking the reciprocal of the input of ln. Since 1/cosh(w) = sech(w), the final result becomes:
y = 2 Ln(sech(C5 - xsqrt(C3)/2)) + C6
Lo and behold!
https://www.reddit.com/r/calculus/comments/axqy9p/need_help_with_this_integral/
I have tried this problem in a variety of ways and cannot seem to get it. I have messed around in Desmos and see that the integral is always pi/4, but cannot seem to get the trick, algebraically. I have tried in the complex world, as well, but the nth power seems to cause problems no matter how I try. And since nth powers of sine and cosine behave differently for odd and even powers, it appears to be a mess of an integral combined with infinite series or something. Attaching a pic of the problem, too. Would make a great video, I think.
Thanks for all you do. Best regards from Seattle.
Phil
Phil, here: https://youtu.be/9xaGYqiOkPM
@blackpenredpen OMG, thank you! So many crazy integration techniques that don't seem to be covered in many of the books I have. Any ideas where I might find more things like this? I have a ton of calculus books, and even a few that specialize in integration techniques, and nothing like this is covered. I know Brilliant.org has stuff, but since I joined, the advanced integration piece is still not done and public yet. I am doing a write-up of the solution to this on my math blog at mymathteacheristerrible.com. Thank you so much!
Hi BlackPenRedPen! Really enjoying your videos as I navigate Calc 2. I keep coming across what I think are telescoping series problems that I just can't figure out. They look like partial fractions but when I break 'em up I get three terms as opposed to two. Check this out:
Really enjoy your videos. Particularly the hard integrals. How about the double integral of
Hint, look at the derivatives of (sin x)^x and (x*cot x+ln sin x)
Hi BlackPenRedPen. I have here a very interesting integral, which shows that you cannot swap integration and limit (in general):
[Solution Spoilers:] The right-hand-side is infinity, the left-hand-side is the limes for h going to 0 of (log(4h+9)-log(4h+1))/2 , which is log(3).
Extra info: These are known as hyper singular integrals. More infors under Hilbert transform, principal value integral. Also: Hadamard finite part integral.
Could you limit this series above and belov for an error for 0.01%? Also it is interesting to illustrate how can geometry solve such things
i was curious if you can integrate for l'hopitals instead of taking the derivative. would that necessarily work?