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Prime Newtons
United States
Registrace 14. 11. 2019
This channel is to help College and High school students master essential math skills in order to be ready for higher mathematics . I present and solve select questions in each video.
Regional Math Olympiad Problem
This is from the Regional Math Olympiad. The trick here is that not all polynomials are solved in terms of x. Sometimes the other unknown variable becomes the key.
zhlédnutí: 4 062
Video
Invoking the Gamma Function
zhlédnutí 4,4KPřed 4 hodinami
This definite integral could not be evaluated using the integration techniques learned in calculus2. I showed that the problem could be modified by appropriate substitution to facilitate the use of the gamma function for its evaluation.
An exponential trig equation
zhlédnutí 2,6KPřed 7 hodinami
This was a problem from the Canadian Euclid Math Contest from 2024 . A 12 grader is expected to easily answer this question without requiring any special knowledge or reasoning.
x - 1/x² = (rad2)i , Find x^2187 - 1/x^2187.
zhlédnutí 3,1KPřed 7 hodinami
This was a multiple choice problem from Jee Admanced sent in by a subscriber. The required skills were basic algebra and a drop of keen observation.
2024 Canada Euclid Math Contest
zhlédnutí 5KPřed 9 hodinami
The main idea was to use the laws of logarithms and then some multiplication and division. It was a good problem to wreslte with for any 12th-grade student.
The Last 3 digits of sqrt( 1^3 + 2^3 + ...+ 2024^3)
zhlédnutí 4,4KPřed 12 hodinami
The main idea was to use the fact that the sum of n natural cubes is the square of the sum of n natural numbers. This is the video on the proof referred to in the video czcams.com/video/VgwLVxLoLz0/video.htmlsi=DLy8qqhh0NMOJDvE
Team Selection Test (Ecuador 2008)
zhlédnutí 5KPřed 14 hodinami
This system of equations could be easily solved by inspection or algebraic substitution. It is only necessary to find all solutions.
A problem from Denmark 2006 (Georg Mohr)
zhlédnutí 4,9KPřed 14 hodinami
This problem is quite easy considering the algebra and level of reasoning involved in the solution I gave. After finding one set of solution, It was required to show that there were no other solutions.
Position, Jerk, Pop and Other Derivatives of The Position Function
zhlédnutí 3,4KPřed 19 hodinami
Whenever we discuss motion, we often talk about position, velocity and acceleration. in this video, i introduced the audience to higher derivatives of the position function such as the jerk, which is the change in acceleration with time. Other derivatives were the snap/jounce, crackle and the pop.
How to depress a cubic
zhlédnutí 7KPřed dnem
The cubic formula is rarely used and rarely talked about. This video explains how to depress a cubic polynomial into a form that works with the cubic formula. Watch Cube-root of Unity here: czcams.com/video/lHe6iieqzBw/video.htmlsi=n7MZFf6ALc4giVnU
Cubic Formula for Depressed Cubic
zhlédnutí 7KPřed dnem
The cubic formula is rarely used and rarely talked about. This is a very effective formula fpr computing the roots of a depressed cubic equation - A cubic missing the quadratic term. In this, video I showed a simple derivation of the formula by reverse engineering.
JEE Advanced 2022 #2
zhlédnutí 6KPřed dnem
This this limit problem is the limit of a composition of trig and logarithmic function. The main idea is to know how to compose the functions and take the limit of the composed function as the function of the limit.
1961 IMO #1
zhlédnutí 4,5KPřed 14 dny
This was question 1 in the 1961 IMO in Hungary. I have made an attempt at solving this algebra problem using basic high school reasoning. Hope it makes sense.
Belphegor's Prime
zhlédnutí 5KPřed 14 dny
This is a palindrome beastly prime number. it contains the number 666 and strange forms of the number 13. It was discovered by Harvey Dubner and named by Clifford Pickover after one of the Seven Princes of Hell in his book.
A set Theory problem from JEE Advanced 2022
zhlédnutí 3,1KPřed 14 dny
A set Theory problem from JEE Advanced 2022
Third International Mathematics Olympiad #2
zhlédnutí 8KPřed 14 dny
Third International Mathematics Olympiad #2
Find the last digit of (1! + 2! +...+ 1982!)^1982
zhlédnutí 10KPřed 14 dny
Find the last digit of (1! 2! ... 1982!)^1982
Evaluate z^2024 + 1/z^2024 Given that z+1/z=1
zhlédnutí 33KPřed 21 dnem
Evaluate z^2024 1/z^2024 Given that z 1/z=1
Identifying the graphs of a function and its derivatives
zhlédnutí 3,1KPřed 21 dnem
Identifying the graphs of a function and its derivatives
Solving a radical polynomial with trig substitution
zhlédnutí 9KPřed 21 dnem
Solving a radical polynomial with trig substitution
Sum of the roots of a 2001st power polynomial
zhlédnutí 10KPřed měsícem
Sum of the roots of a 2001st power polynomial
listened again getting clearer we assume a delta less than 1 what if the limit of the function dne then this would be false
watching this video makes me feel smart in mathematic but when i tried to solve an easy looking algebra, i don't even know what to do first 😂😂
Excellent
Use sec(x) squared = 1 + tan(x) squared to get sec(x) squared is always greater than 1
Great❤
Sir can you find all real and complex solutions for this polynomial equation which has fractional powers that is X^{4/3} - 4X^{2} + 4 = 0
2^2^2=16 lmao I'm now mathematician😅😅
It is easier to optimize the square of the area
I think jp maths made a video on this recently...
The microphone was not good in this video. I noticed because my volume was turned up all the way.
I found the value of XY from equation 1 and then substitute it in equation 2, that gives equation similar to hyperboloid: x^2 + y^2 +2Z^2 = 6 - cut by a plane x+ y = 2. No Other point than (1 1 0).
Like seriously thx vry much
As an Indian, we do this in 11th STD
I found this right after watcching your second video from four years ago about finding the last digit (as opposed to multiple last digits as in this video). I learned two new concepts here that were not in the other video. Presentation was really polished too.
I was looking over your videos and saw this one. Your current presentations have really become polished in contrast.
What if it's a cube root or even a fourth root
Thanks teacher... U relaly have a good heart
Prime Newtoon's owner, how to find d^i/dx^i, when i is imaginary number?
16
2 to the power of 2 to the power of 2 2 to the power of 2 = 4 then 2 to the power of 4 = 2x2x2x2 = 16 ohhhh.... I like this sort of maths!
Please, can you explain what exactly the Feynman technique, with simple exercise
I solved this problem using derivation. We know that the graph of f(x) needs to intersect the x-axis at 4 points for there to be 4 real solutions. Between each of these points, there is a local minimum or maximum, which means that the derivative of the function has to intersect the x-axis at at least 2 points. To find out under which conditions this is true, we find an expression for the x-value of the local minimum of f'(x) by setting the second derivative to 0: f''(x)=0. This yields an x-value of sqrt(a/3). Because the local minimum of the first derivative has to lie on or below the x-axis, we can solve f'(sqrt(a/3))<0 for a and find that a>3/4
I don't think your reasoning is correct. If the graph of the quartic intersects the x-axis at 4 distinct points (meaning we have 4 distinct real zeros) then the first derivative has _three_ distinct zeros, one in between each two consecutive zeros of the quartic. In fact, this gives yet another way to approach this problem. The condition for the first derivative to be zero is 4x³ − 4ax − 1 = 0 or x³ − ax − ¼ = 0 This is a depressed cubic x³ + px + q = 0 which has three distinct real zeros if and only if (½q)² + (⅓p)³ < 0 so a must then satisfy (−⅛)² + (−⅓a)³ < 0 which indeed gives a > ¾. This is a _necessary_ condition for the quartic x⁴ − 2ax² − x + (a² − a) to have four distinct real zeros but we would still need to prove that this is also a _sufficient_ condition. In fact we must prove that the two local minima of x⁴ − 2ax² − x + (a² − a) are both negative for a > ¾ to have four distinct real zeros and I don't see you doing that. The two zeros of the second derivative 12x² − 4a give the positions of the inflection points at x = √(a/3) and x = −√(a/3) but your claim that the point where the first derivative reaches a local minimum, that is, at x = √(a/3), has to lie on or below the x-axis makes no sense. In fact, both inflection points can lie above the x-axis and then the graph of the quartic can still cross the x-axis at four distinct points. And in fact if we substitute x = √(a/3) in f'(x) = 4x³ − 4ax − 1 and solve for f'(√(a/3)) < 0 we do _not_ get a > ¾.
@@NadiehFan Thank you for your observations. First, I agree with your point about the first derivative needing to have at least 3 zeroes for the function to have 4 distinct real roots. The reason I said there need to be at least two roots, is because I wanted to include identical roots. Your other point is more interesting, because I wrongly assumed the two local minima of the function to always be negative, which seems to be true for this particular function, but not for a general quartic. In a general case, one could remove all the real roots by adding a constant to the function, which would not be visible from the derivative. My question then becomes, what is it about the particular constant of a^2-a that ensures that the local minima are negative?
first of all you just know is not an argument. I just know because you state a reason is an argument. Like I now because i have seen this form of limit and thew general equation has a vertical asymptote and the limit from the left and right do not converge is an argument. "Just trust me bro" is not used in math and science. Funny? maybe.
but on a more serious note. this is why we should have never dropped the infinitesimal part from calculus. it would be accurate to say that as we approach the limit as x approaches 3 from the left or right that we are adding or taking away an infinitesimal. such that the equation does not need concrete examples this would show that on either side of the asymptote that the signs of the graph changes and thus do not converge. We could also look at converge tests to prove that the limit does not converge and thus is DNE.
16
Please solve this "the limit as x tends to 1/3 3x-1/5x+1=0"
It is entirely possible to do without the use of lambert W. Just need to perform the same operations you did for converting -ln(√2) for converting into the W function, and then compare both sides. You end up with: ln(1/2)e^(ln(1/2)) = -ln(x)e^(-ln(x)) By comparing both sides you find: ln(1/2)=-ln(x) x=2
You are best teacher in differentiation
Please I want u to help me with a matrix question
You are a gifted instructor. But tell us about yourself. Where did you get your math background. You are a mystery man.
Very good video, and I like your t-shirt.
Mind blowingly good video! Small mistake at 14:38. Denominator should be x-0. I love your videos though! Any chance you could explain why “every power series is a Taylor series” without all of the heavy analysis stuff in Borel’s proof?! I never took analysis and this “every power series is a Taylor series” is really bothering me !
So well explained. If you had been my teacher in school I would’ve had a way different life.
You actually don't have to check all the conditions of inequality to mach. Just pick the two minor sides and add 'em then compare to the greater side.
hey man your voice is so cool.....i was kinnda like dancing ....
That's not correct . X^2187 = (X^2188)/X = 1/X . The answer is -(sqr2)i
To the point, with perfect explanation, telling exactly what we need to know without stretching it way too long nor just doing the solutions! Subscribed
OMG amazing . Love from ♥️🇮🇳
I too am unclear why you think that a>-1/4 AND a>3/4. You might be correct, but can you please explain why it's not a.-1/4 OR a>3/4?
When a variable satisfies two inequalities, it satisfies the intersection.
16
Sir I really enjoy you videos Whenever I watch your videos I Learn with enjoys❤️ Really loves your videos, you are amazing 😻
Thank you
Are you familiar with the discriminant (b^2 - 4ac)?
Don't think so
@@PrimeNewtonsIt's the part of the quadratic formula inside the square root.
awesome
Tremendous! What a splendid strategy to attack this problem.
This was so very helpful. Thank you!
One thing I get confused is Newton's f prime notation drops the dx. But Liebnez always keeps the dx. Then when we try to do integral we need to write dx back to f prime. Does anybody know why that is? And
legend
You teach really nice but ill suggest you give copius example to strengthen what we already know 😊 ....my cadid advise😅
*@ Prime Newtons* -- As with another comment in another video of yours, please *stop* using the implication symbol. The equals symbols needs to be used instead.
Noted