Not only did I learn a more efficient method of expanding, it also made me wonder what Pascal's triangle is and search for it. excellent channel and video
As soon as we started calculus in Engineering maths i had no idea what was going on because i was never taught pascal's triangle (which was the method the teacher was using). This really helped. Cheers mate.
Liked and subscribed! Thanks - an adult relearning math as a hobby who found your videos thanks to a recommendation from the YT algorithm on my home page.
I accidentally found your website recently. Your method of teaching math is phenomenal. I wished I had you as a math teacher when in High School. Thank you so much.
oh my gosh, this s very easy when you started i didn't understand, but later i understood. thanks very much. God bless you for showing us your skills.we will subscribe and watch more.
Fantastic! I didn’t do this on my maths curriculum back in the day but was a 1st maths student and still enjoy it! This was perfect to take my panicking 16 year old through this topic. It was also a great opportunity to teach her the absolute joys of a scientific calculator. Thank you!
You're just superb. It's the nonchalant manner in which you present your explanation that puts the icing on the cake Being a mathematical genius is an obvious quality of yours. And the simplicity is just superb. Cheers ..
The coefficients found from Pascal's Triangle perfectly match the coefficients found from a binomial-theorem expansion (whenever the exponent is a positive integer)
is there a faster way to do this by any method like factoring or something else? This was good to know but idk if its good to do it this way in an exam where we only have like an hour and a half to finish 100+ questions Thank you for the answer in advance sir!
Request on Convergence vs. Divergence please ? I have to go over it but it will be a bit before I get to it as I'm doing all of Calc I again. Though I 4.0 the class still. Currently 1/2 way through online. I never like being rushed in College. Math is like a fine wine and should be enjoyed ! Though not many would agree with me on that. 🤣
I take it you are talking about convergence and divergence, of an infinite series, as there are other meanings of those words. A sequence (or progression) is a list of numbers, a series is a summation of a sequence. Usually k represents the index of the individual term, and n represents the total number of terms. There is usually a function of k that determines the sequence value, although a sequence can also be recursively defined like the Fibonacci sequence. Some series add up to a finite value, when you have an infinite number of terms, which is called a series that converges, or convergent. Other series either add up to infinity, or continuously flip/flop between the same two numbers and never get closer to zero or to infinity, which is called a divergent series. Some series include an alternator term, of (-1)^k. If an alternating series only converges when it is alternating, it's called conditionally convergent, for instance the harmonic series of 1/k. If a series converges, regardless of whether it alternates, it's called absolutely convergent. The substance of Calculus's work with series, is to analytically determine whether or not it converges, and if so, what value it converges to. There are various tests for convergence, based on properties of the series definition, that may use limits, powers, roots, and integrals, to test which kind of convergence a series has.
This guy has been helping me so much but i just got a question where theres only one letter thingy in the parentheses and the other is just a number (d+2)to the power of 8
You can put a letter in that number's place. For your example, let t = 2. Thus: (d + t)^8 = d^8 + 8*d^7*t + 28*d^6*t^2 + 56*d^5*t^3 + 70*d^4*t^4 + 56*d^3*t^5 + 28*d^2*t^6 + 8*d*t^7 + t^8 8th row of Pascal's triangle for reference: 1, 8, 28, 56, 70, 56, 28, 8, 1 Now evaluate each t-power at t=2, and have it join the coefficient. Put the d-power aside for later. 8*t = 16 28*t^2 = 28*4 = 112 56*t^3 = 56*8 = 448 70*t^4 = 70*16 = 1120 56*t^5 = 56*32 = 1792 28*t^6 = 28*64 = 1792 8*t^7 = 8*128 = 1024 t^8 = 256 Thus, the result is: d^8 + 16*d^7 + 112*d^6 + 448*d^5 + 1120*d^4 + 1792*d^3 + 1792*d^2 + 1024*d + 256
Yes. You just let beta = -b. Expand (a + beta)^5, exactly as you would with this method. Assuming a is positive: Every odd exponent on beta, will materialize as a negative term when translated back to b. Every even exponent on beta, will materialize as a positive term when translated back to b.
Here's an example with a negative: Suppose it starts as (u - v)^3, where u and v are both positive numbers. Let w = -v. Now we've reduced it to (u + w)^3. Expand it out accordingly, with u's power descending from 3 to 0, and w's power ascending from 0 to 3. Coefficients follow 3rd row of Pascal's triangle, which is 1, 3, 3, 1. u^3 + 3*u^2*w + 3*u*w^2 + w^3 Since squared negatives are positive, this means all terms with w to an even power, will equal the same term with v replacing w. The opposite is true for cubed negatives. All terms with w to an odd power, will have a negative coefficient, and otherwise equal the same term. This gives us the result: u^3 - 3*u^2*v + 3*u*v^2 - v^3
