Finding An Interesting Sum | 2 Ways
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- Опубліковано 15 вер 2024
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You can solve this without differentiating too. If you multiply the whole series by X and then subtract from the original series the difference is a summable series. The sum of this series can then be divided by 1-X to get the result.
Yes, that's exactly what I did, which gives you S(1 - x) = 1 + GP starting 1 + x + .....
The differentiation is clever, too, to get those coefficients all the same. Nice problem, neat solutions, happiness.
First method modified: Let our sum be T = 2 + 3x + 4x^2 + ... Then xT = 2x + 3x^2 + 4x^3 which is the differential of (x^2 + x^3 + x^4 + ... ). The sum of x^2 + x^3 + x^4 is x^2/(1-x) by G.P formula.
So xT = differential of x^2/(1-x) = ( (1-x).2x - x^2.(-1) ) / (1-x)^2 because D(U/V) = (V.DU - U.DV)/V^2 differential of a quotient formula. Notice we didn't divide by x yet, so no worries about x=0.
Then xT = (2x - 2x^2 + x^2) / (1-x)^2 = (2x - x^2)/(1-x)^2 = x(2-x)/(1-x)^2. Finally we divide both sides by x giving T = (2-x)/(1-x)^2. Note that it remains true for x=0, i.e. T(0) = 2.
2+3x+4x²+...=2x⁰+3x+4x²+...
The nth element is
(n+1)x^(n-1)=nx^(n-1)+x^(x-1)
=(dx^n/dx)+x^(n-1).
Summing all n for the differential term is integral of (dx^n/dx) from 0 to infinity, while sum of the 2nd term is sum a geometric series with the 1st element a=1 and ratio r=1/x
only 0
S = 2 + 3x + 4x^2 + ...
S*x = 2x + 3x^2 + ...
S - S*x = 2 + x + x^2 + x^3 ... = 1 + Sum(x^n) = 1 + (1/(1-x)) = ( (1 - x + 1) / (1 - x) ) = (2-x)/(1-x)
Thus:
S*(1-x) = (2-x)/(1-x)
S = (2-x)/(1-x)^2
Same as I did. ;)
Multiply by (1-x), and you get 2+x+x^2+x^3+...
Do it again and you get 2-x. So the sum is (2-x)/(1-2x+x^2), with a radius of convergence of 1.
Of course, we have to assume that |x|
Can I refer you to 0:06 and again to 1:02 where the bounds of x are stated?
Can we not do it like an AGP?
If the given expression to be evaluated is S
We subtract xS from X
and get a simple 1 + (1 + x + x^2 + x^3 +....)
yes
Wow I actually tried doing this last week and I managed to find a multi variable equation that has input x (real numbers) and the number of terms in the sequence
I got this by starting with (1-x^(n+1))/1-x and differentiating with respect to x.
@@jasimmathsandphysics very good!
@@SyberMath 👍
I've been waiting for months where we get to do the second method first. And here I got the second method second which is called the first method. So, now I really don't know if I got what I was waiting for. 🤣🤣🤣
You got it!!! 😍😄
Nice!
✌️👍😃😃👍✌️
😍😍😄😄
(Sinx+sin^2x)^2+(sinx-sinx^2)^2=? Please give me answer it's my homework
Have you tried wolfram alpha?
Also try setting u=sin(x) and then expand the two terms. Then you use the quadratic formula to find u^2 and from there it’s not too bad
@@jasimmathsandphysics I have tried but couldn't find my absolute answer .in my homework answer sheet the answer is given 4 but I can't solve it
@@jasimmathsandphysics can you please solve the full question
@@monshamorsalin oh sorry I thought it was solving for x. Do I have to simplify it?
Hey bro I would be greatly obliged if you solve the following
If x^2+x=1 then x^7+34/x^5+1=??
nothing nice is coming up. did you mean (x^7+34)/(x^5+13)?
3:56 from where you get that formula
1 + x + x² +x³ + x⁴ + ... = 1 + x(1 + x + x² +x³ + x⁴ + ...)
1(1 + x + x² +x³ + x⁴ + ...) - x(1 + x + x² +x³ + x⁴ + ...) = 1
(1 - x)(1 + x + x² +x³ + x⁴ + ...) = 1
1 + x + x² +x³ + x⁴ + ... = 1/(1 - x)
@@SyberMath thanks..
You make awesome math videos..
There are already 3 years that i found a similar question of the type: A = 1 + 2x² + 3x³ + .... + 2021 x ²⁰²¹
your first term does not match the pattern
@@SyberMath actually it fully matches, because the initial problem was not with x, but of the form A = 1 + 2 ×4² + ...+ 2021×4²⁰²¹, and thr question was to find the resulting sum A, which is a very simple problem :)
(d/d x( x ^2 + x ^ 3 + x ^4 ..))/ x
= d /dx ( ( x^2 /( 1 - x)) / x
= 2 / ( 1 - x) + x /( 1 - x)^2
= ( 2 - 2 x + x)/( 1 - x)^2
= (2 - x)/( 1- x)^2
f(x) = ∑[n=1, ∞](n+1)x^(n-1)
= g(x)+h(x) (|x|
=Σ(k+2)x^k=Σ2x^k+Σkx^k=2/(1-x)+x(d/dx)(1/(1-x))=2/(1-x)+x/(1-x)^2=(2-x)/(1-x)^2..ovviamente x
pretty good!
Basically it makes ♾️
Unless x is between -1 and 1. Which he also stated in the video
@@wanyekivincent2883 ok , the question was a bit missed at first then. All is well 👍🏻👍🏻