The easiest way to resolve the problem is as follows Let Sum = 1/5+2/5^2+3/5^3+… Then 5.Sum = 1+2/5+3/5^2+4/5^3+… 5.Sum - Sum = 4.Sum = 1+1/5+1/5^2+1/5^3+… = 1/(1-1/5) = 5/4 (a geometrical series) So Sum = (5//4)/4 = 5/16 No differentiation is required
i just thought one could split it up into (1/5 + 1/25 + ...) + (1/25 + 1/125 + ...) + ..., and theyd give a bunch of summations as 1/4, 1/20, 1/100 and the sum of them is 1/4 / 4/5 = 5/16
The correct answer is 5/16. It can be derived without differentiation or double infinite series. It's as simple as noticing that 5s-1-\sum_{n=1}^{\infty}{\frac{1}{5^n}} = s.
The differentiation is quite a nice approach - if you evaluate it correctly. I used the 2nd method although other comments show an even easier solution.
Here's how I did it. Let x = the given series. Then if you calculate 4x - 1, you'll see that equals the sum of 1 / 5^n. That geometric series sums to 1/4. Therefore, 4x - 1 = 1/4 x = 5/16
I guess, you meant 4 x - 1 = (5 x - 1) - x = sum (1/5)^n = 1/4 => x = 5/16. That's because 5 x - 1 = 2/5 + 3/25 + 4/125 + ..., and if you subtract the original series, you'll get the geometric series.
@@KrytenKoro First, I was confused as well, but he meant (5 x - 1) - x, which is 4 x - 1. That 1 isn't immediately subtracted, it's the superflous term when you sum 5 times x! (That is 5/5 + 10/25 + 15/125 + ... or 1 + 2/5 + 3/25 + ...)
First you can note that this sum converges absolutely by the ratio test. Next, let A be the sum, and let B be the related geometric sum, B = 1/5 + 1/25 + 1/125 + ..., which also converges absolutely. A - B = 0/5 + 1/25 + 2/125 + ... = 1/5 × (1/5 + 2/25 + 3/125 + ...) = A/5. 4/5 A = B = 1/(5-1) = 1/4. A = 5/4 × 1/4 = 5/16.
Not sure how you would go about proving convergence. My idea is to integrate n/5^n and get -n*5^-n/ln5-5^-n/(ln5)^2. Let f(X)=-x*5(-x)/ln5-5^(-x)/(ln5)^2 As an elementary function it is continuous and derivable on all intervals [k,k+1]. From T Laplace it follows that for each interval there exists kf(k+1)-f(k)>(k+1)/5^(k+1) So 1/5>f(2)-f(1)>2/25 2/25>f(3)-f(2)>3/125 .... Adding up to n 1/5+2/25+...+n/5^n>f(n)-f(1)>1/5+2/25+...+(n+1)/5^(n+1)-1/5 Taking the limit as n goes to infinity it results from the rightmost inequality -f(1)>Sum-1/5 1/(5ln5)+1/(5ln^2(5))>sum-1/5 so our sum has an upper bound. About calculating it I would go around a geometric series route as others have pointed out
Let S_1= 1/5 +1/5^2 +1/5^3 + ... = 1/5 [1/(1-1/5)] = 1/4. The given sum can be written as S= S_1 + 1/5 S and so, 4/5 S = S_1 , i.e., S = 5/4 S_1 = 5/16. So, S=5/16.
Yeah I also got the "correct" answer and didn't use any differentiation which is IMO too complicated for this problem. Like many others I used the much simpler approach of using S-Sr = r/(1-r) (where S is the original series). This is essentially like your 2nd method which frankliy is also making it look more complicated. However, thanks for presenting the differentiation tactic which I'm sure will come in handy in other cases where it's not so simple to break it down.
Others of calculated the correct answer, but here’s where you made your mistake: You computed f’(⅕) but not ⅕f’(⅕), but we were looking for r•f’(r) not f’(r). Thus, the correct answer for the first example is ⅕•25/16 and thus 5/16.
Cool way to use derivatives to get the desired coefficients! ❤👍🏼 problem Can you sum 1/5+2/25+3/125+... Call this S. We make use of the series P = 1+r+r² +r³ +... = 1/(1-r) , where r is the common ratio. It converges if |r| < 1. But S has the coefficients containing n. S = 1/5+2/25+3/125+... = ⅕ ( 1 + 2/5 + 3/25 + 4/125 + ... ) The series in the parentheses doesn't match the form of P. We factored out a ⅕ to get a leading 1 inside the parentheses. Consider differentiation of a function of x defined as ͚ f(x) = Σ x ⁿ ⁿ⁼⁰ This has the form of P, so for |x| < 1 we have ͚ f(x) = Σ x ⁿ = 1/(1-x) ⁿ⁼⁰ Differentiating term by term we get ͚ f'(x) = Σ n xⁿ⁻¹ ⁿ⁼⁰ At x = ⅕ ͚ f'(⅕) = Σ n ⅕ ⁿ⁻¹ ⁿ⁼⁰ , which matches the series in the parentheses. Note that the first term is 0 by virtue of the n coefficient being the derivative of the constant 1 in the first term's position, so the "real" sum starts at n=1. ͚ f'(⅕) = Σ n ⅕ ⁿ⁻¹ ⁿ⁼¹ S = ⅕ ( 1 + 2/5 + 3/25 + 4/125 + ...) S = ⅕ f'(⅕) Recall f(x) = 1/(1-x) = (1-x) ⁻¹ Take derivatives. f'(x) = -(1-x)⁻²(-1) = 1/[(1-x)²] At x = ⅕, we find for S S = ⅕ 1/[(1-⅕)²] = (1/5) 1/( 16/25 ) = (1/5) (25/16) = 5/16 answer 5/16
When I was in high school, I remember coming with what I thought was a fantastic way to calculate the summation for i = 1 to ∞ of the function f(i) = P(i)/aⁱ, where P is a polynomial function and a > 1 is a constant. Then I found out Euler had figured all this out about 220 years before me. Be that as it may, the solution was to break the summation into pieces of the form iᵏ/aⁱ. Each of these new summations could be resolved as follows: k = 0: 1/(a - 1) This is a standard geometric series. k = 1: a/(a - 1)² k = 2: (a + 1) × a/(a - 1)³ k = 3: (a² + 4k + 1) × a/(a - 1)⁴ k = 4: (a³ + 11a² + 11a + 1) × a/(a - 1)⁵ k = 5: (a⁴ + 26a³ + 66a² + 26a + 1) × a/(a - 1)⁶ etc. For all k > 0, the formula always has a polynomialial of degree k-1 multiplied by a/(a - 1)ᵏ⁺¹. The coefficients for the polynomial follow a pattern, which I found out correspond exactly to the values in Euler's Triangle (not to be confused with Euler's Triangle from geometry), and the polynomials themselves are known as the Eulerian Polynomials.
Note first that you can limit the range to 0 < x < 1 by enforcing positivity in the sqrts, eliminate the possibility of divide by 0, and check that x=0 is not a solution. You can then start by making the substitution u = 1-x. Subtract both sides by sqrt(u) to get sqrt(1/u - 1) = 5/2 - sqrt(u). Square both sides to get 1/u - 1 = 25/4 - 5 sqrt(u) + u Put everything on the lhs, divide both sides by u, and substitute v = u^ (-½) to get v^4 - 29/4 v^2 + 5 v - 1 = 0 This is in a depressed quartic form, which you can solve. Don't forget to return the solution to the x world and enforce boundary conditions.
@@FaerieDragonZook thats the problem bro cant solve this quartick eq and even if u solve it u will get 4 solutions and none of them satysfies the eq i managed to get its graph on desoms the intersection point represent the solution wich is x=0.81 but the problem is that i cant get the steps i should follow to get this sol
@@fahadalshammari3627 If this is from a math contest or an exam, you should check if you copied the equation correctly. As it is, this equation leads to a quartic polynomial which has no rational zeros since its cubic resolvent has no rational solutions. It is possible to express the (sole) real root in closed form, but as you can see in WolframAlpha its expression is prohibitively complicated. I suspect that the original equation may have been √(x/(1−x)) + √(1/(1−x)) = 5/2 which is easy to solve and has a nice (rational) solution.
@@NadiehFan nah the equation is correct as I checked it’s really challenging and complex unfortunately I think I will just write it in my black list as another equation that I couldn’t beat
The easiest way to resolve the problem is as follows
Let Sum = 1/5+2/5^2+3/5^3+…
Then 5.Sum = 1+2/5+3/5^2+4/5^3+…
5.Sum - Sum = 4.Sum = 1+1/5+1/5^2+1/5^3+… = 1/(1-1/5) = 5/4 (a geometrical series)
So Sum = (5//4)/4 = 5/16
No differentiation is required
The sum is 5/16, not 25/16
G(x) = 1 + x + x^2 + x^3 + ... = 1/(1-x)
G'(x) = 1 + 2x + 3x^2 + .... = 1/(1-x)^2
xG'(x) = x + 2x^2 + 3x^3 + ... = x/(1-x)^2
S = xG'(x) evaluated at x = 1/5 = (1/5) / (4/5)^2 = 5/16
I made a mistake!
He forgot to multiply f'(r) by r at the end which gives (25/16) x (1/5) =5/16.
@@SyberMaththen please amend the video
S=Σk/5^k(k=1,2,3...)...dalla teoria d/dx(Σx^k)=Σkx^(k-1)=(1/x)Σkx^k(k=1,2,3...)...risulta 1/(1-x)^2=(1/x)Σkx^k...per x=1/5...risulta 25/16=5Σk(1/5)^k=5S,risulta S=5/16
at 10:30 computed f'(r) but forgot to multiply by r to get the answer.
yes
@ that’s cool. Einstein made mistakes. Your videos are awesome. Thanks.
i just thought one could split it up into (1/5 + 1/25 + ...) + (1/25 + 1/125 + ...) + ..., and theyd give a bunch of summations as 1/4, 1/20, 1/100 and the sum of them is 1/4 / 4/5 = 5/16
Σ(1/5)^n*1/(1-1/5)=5/4*Σ(1/5)^n from n=1 to infinity.From geometric progress Σ(1/5)^n= 1/5*1/(1-1/5)=1/4 therefore the sum Σ=5/4*1/4=5/16.
The correct answer is 5/16. It can be derived without differentiation or double infinite series. It's as simple as noticing that 5s-1-\sum_{n=1}^{\infty}{\frac{1}{5^n}} = s.
Correct!
The differentiation is quite a nice approach - if you evaluate it correctly. I used the 2nd method although other comments show an even easier solution.
Here's how I did it.
Let x = the given series.
Then if you calculate 4x - 1, you'll see that equals the sum of 1 / 5^n. That geometric series sums to 1/4.
Therefore, 4x - 1 = 1/4
x = 5/16
4x-1=-1/5,3/25,27/125 vs 1/5,6/25,31/125?
I guess, you meant 4 x - 1 = (5 x - 1) - x = sum (1/5)^n = 1/4 => x = 5/16.
That's because 5 x - 1 = 2/5 + 3/25 + 4/125 + ..., and if you subtract the original series, you'll get the geometric series.
@@KrytenKoro First, I was confused as well, but he meant (5 x - 1) - x, which is 4 x - 1.
That 1 isn't immediately subtracted, it's the superflous term when you sum 5 times x!
(That is 5/5 + 10/25 + 15/125 + ... or 1 + 2/5 + 3/25 + ...)
First you can note that this sum converges absolutely by the ratio test. Next, let A be the sum, and let B be the related geometric sum, B = 1/5 + 1/25 + 1/125 + ..., which also converges absolutely.
A - B = 0/5 + 1/25 + 2/125 + ... = 1/5 × (1/5 + 2/25 + 3/125 + ...) = A/5.
4/5 A = B = 1/(5-1) = 1/4.
A = 5/4 × 1/4 = 5/16.
B=(1/5)/(1-1/5)=1/4
A-1/4=A/5
A-A/5=1/4
A=1/4/(1-1/5)=(1/4)/(4/5)=5/16
Let A=1/5+2/25+3/125.... (1/5)A=1/25+2/125+... (A)-(1/5A)=(4/5A)=1/5+1/25+1/125...=1/4, A=5/16
Excellent
I think, you forgot to multiply with r! r=1/5
sum = r / (1-r)² = r / (r - 1)² = 1/5 * (5 / 4)" = 5 / 4² = 5/16 (not 25/16).
You did not explain how you can apply differentiation here. r is a constant here.
Using the ratio test, the sum is convergent to 1/5
The recursive form for this type of sum, namely P(n,k)=Σ(p=1 to infty) [p^k)/(n^p)], is P(n,k) =(1/(n-1))Σ(m=0 to k-1) [nCr(k,m)P(n,m)].
Not sure how you would go about proving convergence. My idea is to integrate n/5^n and get
-n*5^-n/ln5-5^-n/(ln5)^2. Let f(X)=-x*5(-x)/ln5-5^(-x)/(ln5)^2
As an elementary function it is continuous and derivable on all intervals [k,k+1]. From T Laplace it follows that for each interval there exists kf(k+1)-f(k)>(k+1)/5^(k+1)
So
1/5>f(2)-f(1)>2/25
2/25>f(3)-f(2)>3/125
....
Adding up to n
1/5+2/25+...+n/5^n>f(n)-f(1)>1/5+2/25+...+(n+1)/5^(n+1)-1/5
Taking the limit as n goes to infinity it results from the rightmost inequality
-f(1)>Sum-1/5 1/(5ln5)+1/(5ln^2(5))>sum-1/5 so our sum has an upper bound.
About calculating it I would go around a geometric series route as others have pointed out
How come your two methods give different answers?
I made a mistake!
Arithmetico-Geometric progression
I used the second method.
This AGP series sums to 5/16.
If sum (where n = 1 to infinity) of n/(x^n) = x/((x - 1)^n), and x = -1, we get -1 + 2 - 3 + 4 - 5 + ... = -1/4. :D
5/16
Let S_1= 1/5 +1/5^2 +1/5^3 + ... = 1/5 [1/(1-1/5)] = 1/4. The given sum can be written as S= S_1 + 1/5 S and so, 4/5 S = S_1 , i.e., S = 5/4 S_1 = 5/16. So, S=5/16.
Yeah I also got the "correct" answer and didn't use any differentiation which is IMO too complicated for this problem. Like many others I used the much simpler approach of using S-Sr = r/(1-r) (where S is the original series). This is essentially like your 2nd method which frankliy is also making it look more complicated. However, thanks for presenting the differentiation tactic which I'm sure will come in handy in other cases where it's not so simple to break it down.
Others of calculated the correct answer, but here’s where you made your mistake:
You computed f’(⅕) but not ⅕f’(⅕), but we were looking for r•f’(r) not f’(r). Thus, the correct answer for the first example is ⅕•25/16 and thus 5/16.
That's right! Thank you
Cool way to use derivatives to get the desired coefficients! ❤👍🏼
problem
Can you sum
1/5+2/25+3/125+...
Call this S.
We make use of the series
P = 1+r+r² +r³ +... = 1/(1-r)
, where r is the common ratio. It converges if |r| < 1.
But S has the coefficients containing n.
S = 1/5+2/25+3/125+...
= ⅕ ( 1 + 2/5 + 3/25 + 4/125 + ... )
The series in the parentheses doesn't match the form of P. We factored out a ⅕ to get a leading 1 inside the parentheses.
Consider differentiation of a function of x defined as
͚
f(x) = Σ x ⁿ
ⁿ⁼⁰
This has the form of P, so for |x| < 1 we have
͚
f(x) = Σ x ⁿ = 1/(1-x)
ⁿ⁼⁰
Differentiating term by term we get
͚
f'(x) = Σ n xⁿ⁻¹
ⁿ⁼⁰
At x = ⅕
͚
f'(⅕) = Σ n ⅕ ⁿ⁻¹
ⁿ⁼⁰
, which matches the series in the parentheses.
Note that the first term is 0 by virtue of the n coefficient being the derivative of the constant 1 in the first term's position, so the "real" sum starts at n=1.
͚
f'(⅕) = Σ n ⅕ ⁿ⁻¹
ⁿ⁼¹
S = ⅕ ( 1 + 2/5 + 3/25 + 4/125 + ...)
S = ⅕ f'(⅕)
Recall
f(x) = 1/(1-x) = (1-x) ⁻¹
Take derivatives.
f'(x) = -(1-x)⁻²(-1)
= 1/[(1-x)²]
At x = ⅕, we find for S
S = ⅕ 1/[(1-⅕)²]
= (1/5) 1/( 16/25 )
= (1/5) (25/16)
= 5/16
answer
5/16
Nice!
This video is unlike any other before, it made me run in circles and eventually lost interest to continue the rest of it 🤕
sorry to hear that
Isnt this a simple agp?
To make the long story short ;), you made a mistake !
Anyway, thanks for the explaination.
For |r|
0.25>sum
When I was in high school, I remember coming with what I thought was a fantastic way to calculate the summation for i = 1 to ∞ of the function f(i) = P(i)/aⁱ, where P is a polynomial function and a > 1 is a constant. Then I found out Euler had figured all this out about 220 years before me. Be that as it may, the solution was to break the summation into pieces of the form iᵏ/aⁱ. Each of these new summations could be resolved as follows:
k = 0: 1/(a - 1) This is a standard geometric series.
k = 1: a/(a - 1)²
k = 2: (a + 1) × a/(a - 1)³
k = 3: (a² + 4k + 1) × a/(a - 1)⁴
k = 4: (a³ + 11a² + 11a + 1) × a/(a - 1)⁵
k = 5: (a⁴ + 26a³ + 66a² + 26a + 1) × a/(a - 1)⁶
etc. For all k > 0, the formula always has a polynomialial of degree k-1 multiplied by a/(a - 1)ᵏ⁺¹. The coefficients for the polynomial follow a pattern, which I found out correspond exactly to the values in Euler's Triangle (not to be confused with Euler's Triangle from geometry), and the polynomials themselves are known as the Eulerian Polynomials.
25/16 , resposta errada. A correta é 5/16.
yes
The answer is not correct.
f'(r).r = f'(1/5). 1/5
= 25/16 . 1/5 = 5/16 is correct answer
I made a mistake. Thanks
X,5
A: S(n) = 1/5 + 2/5^2 + 3/5^3 + ... + (n-1)/5^(n-1) + n/5^n
B: 5 S(n) = 1 + 2/5 + 3/5^2 + ... + n/5^(n-1). Subtract to get (B) - (A) as,
(5 - 1 = 4) S(n) = 1 + 1/5 + 1/25 + ... + 1/5^(n-1) - n/5^n = (1 - 1/5^n) / (1 - 1/5) - n / 5^n = 5/4 (1 - 1/5^n) - n/5^n
16/5 S(n) = 1 - 1/5^n - 4/5 x n/5^n = 1 - 1/5 (4n + 5)/5^n = 16/5 [ 5/16 - 1/16 (4n+5) / 5^n ]
So, S(n) = 5/16 - (4n+5) / (16 x 5^n) AND S(n = ∞) = 5/16. *Simple. Right* ?
S = 1/5(x + x^2 + x^3 + ...) = 1/5x/(1 - x)
S' = 1/5(1 + 2x + 3x^2 + ...) = d/dx(x/(5 (1 - x))) = 1/5/(1 - x)^2
x = 1/5 => S' = 1/5/(4/5)^2 = 1/5 * 25/16 = 5/16
Bro please can you try to solve this equation sqr(x/(1-x)) +sqr(1-x) =5/2
Note first that you can limit the range to 0 < x < 1 by enforcing positivity in the sqrts, eliminate the possibility of divide by 0, and check that x=0 is not a solution. You can then start by making the substitution u = 1-x. Subtract both sides by sqrt(u) to get
sqrt(1/u - 1) = 5/2 - sqrt(u).
Square both sides to get
1/u - 1 = 25/4 - 5 sqrt(u) + u
Put everything on the lhs, divide both sides by u, and substitute v = u^ (-½) to get
v^4 - 29/4 v^2 + 5 v - 1 = 0
This is in a depressed quartic form, which you can solve. Don't forget to return the solution to the x world and enforce boundary conditions.
@@FaerieDragonZook
thats the problem bro cant solve this quartick eq and even if u solve it u will get 4 solutions and none of them satysfies the eq i managed to get its graph on desoms the intersection point represent the solution wich is x=0.81 but the problem is that i cant get the steps i should follow to get this sol
@@FaerieDragonZook the range is 0
@@fahadalshammari3627 If this is from a math contest or an exam, you should check if you copied the equation correctly. As it is, this equation leads to a quartic polynomial which has no rational zeros since its cubic resolvent has no rational solutions. It is possible to express the (sole) real root in closed form, but as you can see in WolframAlpha its expression is prohibitively complicated.
I suspect that the original equation may have been
√(x/(1−x)) + √(1/(1−x)) = 5/2
which is easy to solve and has a nice (rational) solution.
@@NadiehFan nah the equation is correct as I checked it’s really challenging and complex unfortunately I think I will just write it in my black list as another equation that I couldn’t beat
S=∑[n=1,∞]n/5^n
f(x)=∑[n=1,∞]x^n=x/(1-x)
f‘(x)=1/(1-x)²
f‘(x)=∑[n=1,∞]nx^(n-1)
∴∑[n=1,∞]nx^n=xf‘(x)
x=1/5→
S=1/5·f‘(1/5)=1/5·1/(4/5)²=5/16
Do what you like ,but please, don't make a speech!!!
sorry
To make the long story short ;), you made a mistake !
Anyway, thanks for the explaination.
For |r|