Hey, balckpenredpen. I have a long time watching your videos and I have a request for you, so I hope you make a video for it. I’ve been struggling with this succession for quite a week and still can’t find the answer. Yet the succession is not geometric neither arithmetic (I think), I think there’s a way to solve it. I’m a 12 Y/O (hope you take me serious) from Colombia. So the succession is: 4/5, 3/10, 1/2, 0... You have to find the nth value for this succession. Hope you read it!
@@blackpenredpen WAIT PLEASE CORRECT AND RESPOND YIU DID NOT JUSTOFY WHY YOU CAJ USE ONE OVER ONE MINUS X..YOU DONT KNOW THAT THE SERIES CONVERGES TO,THAT..WHERE,DID,YOU GET,THAT FROM?
1. Using Both Pen is really cool, 2. Very informative and Clear 3. Your Smile 4. The way of showing everything. 5. i still have a problem. ( sum of the series n*2^n , where nmax= 513 and nmin = 2)
This sum is related to Collatz conjecture. At every step of Collatz sequence: multiplication by 3 and adding 1 produces a number with random distribution of zeros and ones (in binary representation). Then you shift right to discard all least significant zeros. Probability of one rightmost zero is 1/2. Probability of two rightmost zeros is 1/4, three zeros 1/8, etc. On average, the shift step divides the result by 2^(1/2+2/4+3/8+4/16....)=4 To prove Collatz conjecture we only need to prove that for any starting number, the sequence always reaches a number smaller than the start. Average "reduction" being greater than 3 means the sequence is very likely to reach the smaller number.
I did it a third way, by saying {sum n=1 to infinity} ({sum k=1 to n} 1/2^n) = {sum k=1 to infinity} ({sum n=k to infinity} 1/2^n) = {sum k=1 to infinity} 1/2^(k-1) = {sum k=0 to infinity} (1/2)^k = 2
the numerator is 3(3^100), which is just 3^101. The middle term of the denominator can be written as 3^101/3 and the last term of the denominator can be written as 3^101/9. Factoring out 3^101 from the entire denominator we get 3^101(1-1/3-1/9) as the denominator. The numerator and the denominator can be simplified which leaves 1/(1-1/3-1/9) which is just 9/5.
I did it like this: 3^100+3^100+3^100=3*3^100=3^101; 3^101-3^100-3^99=3^101*(3^0-3^-1-3^-2)=3^101*(1-1/3-1/9); 3^101/(3^101*(1-1/3-1/9))=1/(1-1/3-1/9)=1/(5/9)=9/5
Picture the sum 1 + 1/2 +1/4 + 1/8 + ... by picturing the rectangle (square) with height 1 above the interval 0 to 1 on the x-axis, a rectangle with height 1/2 above the interval 1 to 2 on the x-axis, a rectangle with height 1/4 above the interval from 2 to 3, and so on. The total area of all the rectangles is 2. Another way to fill those same rectangles is to add the rectangles: 1/2
It is interesting to say that in the first method you can change terms' positions because the series converges absolutely; also, in the second method you can differentiate because it is a power series.
I think this can be generalized to arbitrary (arithmetic series)_n / (geometric series)_n. Say, S=a_1 + a_2*q +a_3*q^2 +...+a_n*q^(n-1) , then qS=a_1*q +a_2*q^2 + a_3*q^3 +...+a_n*q^n ; S-qS=a_1 + (a_2-a_1)q + (a_3-a_2)q^2 +...+(a_n - a_[n-1] )q^(n-1) - a_n*q^n . We know that a_n - a_[n-1] = d (the difference) and we can factor it out. The middle part is then d*(q +q^2 +...+q^(n-1) ) - just a geometric series, its sum is dq/(1-q), if |q|inf . Also we know that a_n=a_1 + d(n-1), so the last term splits into two: a_1*q^n and d*(n-1)*q^n. The first of these obviously goes to zero in the limit, the second goes to zero also, but L'hopital's rule is needed to show that. Finally, we are left with S-qS=a_1 +dq/(1-q) ; S=a_1/(1-q) +dq/(1-q)^2. If the geometric series doesn't start at 1, we can always factor that first term out, thus: if S=sum[n=1, n->inf] (a_1*b*q^(n-1) ) , then S=a_1*b/(1-q) +dqb/(1-q)^2 ; in our case a_1=1, b=1/2, q=1/2, d=1 ; S=1*0.5/0.5 +1*0.5*0.5/(0.5)^2 = 1 + 1 = 2.
I came here because of the bridge game in the Korean show "Squid Game". The expected value of winners is such an elegant real life example of this infinite series. I wonder if there was a mathematician involved with this show?
som from n=1 to if of (n/b^n) = b/(b-1)^2 b€R\(-1,1) you can easily prove it for every b if you replace 2 with b if you do the classical way, then you can test for which b it converges
Theory of undetermined coefficients may also be useful, viz. we can assume n/(2^n) = [ { a(n - 1) + b }/(2^(n - 1)) ] - (an + b)/2^n, where the coefficients a and b have to be determined
I really like this series. Actually the first one I looked at was the series of n^2/2^n. Which could be evaluated by applying the operator (xD)^2 (D means differentiation) to the geometric series 1/(1-x) and then evaluating at x=1/2. In general, applying the operator (xD)^k to the geometric series 1/(1-x), and then evaluating at x=1/k gives the value of the series of n^k/k^n, which I find to be an interesting looking series!
Felipe Rueda If k is a natural number, it is possible to breakdown the series of n^k/2^n into sums of series n^m/2^n, where m = 0, 1, ..., k - 1. It creates an important recurrence relation that defined a function with respect to k. I do not believe there is a closed-form explicit formula, however.
@@angelmendez-rivera351 I could find an explicit formula for the series of n^k/k^n using the method I described above. This lies on being able to give the formula for (xD)^k 1/(1-x)
I solved it by writing n = n-1 + 1 I get sum from n=1 to inf (n-1)/2^n + sum n=1 to inf 1/2^n The first is equal to 1/2 of the original sum, and the second is equal to 1 Letting S equals the original sum I get S = 1/2S + 1 => 1/2S = 1 => S = 2
Ok so I did the math and it turns out that the infinite summation of n²/2^n is equal to 6. The summation can be generalized like this: 1²(x¹)+2²(x²)+3²(x³)+.....+n²(x^n)+....=S Where x is the variable. Now for any value of x where |x| is smaller than one, the infinite sum is S= (x(1+x))/(1-x)³ [ x(1+x) divided by (1-x)³] In this case, the value of x is 1/2, so plugging into the equation, we get S(1/2)= {.5(1+.5)}/(1-.5)³ which is equal to 6. And for your second question, I'm not exactly sure what the answer is, although I will be looking into it, and if I find a way generalize to for n^a/b^n, I'll let you know
I'm a 17 and I want to be a mathematician or physicist. People say boys shouldn't be a physicist and to do a computer science yet I strive to do it. I hope I'm not the only boy.
@3:50 I thought that the geometric series starts with the index 0, thus with the term \frac{1}{2^{0}} = 1? In this case, the first of the splitted series starts with 1 / 2, so why don‘t we need a correction factor?
Let's imagine a cartesian plane with non-negative X, Y. Let every point on this plane with integer coordinates (a, b) have a weight of 1/2^(a + b + 1). Looking at diagonals (same a + b) we see that sum of weights is our needed sum. But we can count same sum of weights in columns - (1/2 + ...) + (1/4 + ...) + ... = + 1 + 1/2 + ... = 2.
S is given/ then multiplying s by .5 and subtracting the 2 series then it becomes 1/2+1/4+1/8............oo Again its a geometric series =.5s Solving s=2 Is it okey bprp pls reply (Didn't care about oo as its boring¥
When this channel first started, there were not blackboards or markers or whiteboards or chalk. It was all in pen and paper, and the only colors where black and red since those are the standard color for pens. Hence, the name. Of course, if we had to name the channel by today's standard, it would be named blackmarkerredmarkerbluemarkergreenmarkerpurplemarker.
This video reminds me of a more general problem, that of finding the summation of P(n)/a^n as n goes from 1 to infinity, where P(n) is a polynomial function and a is greater than 1. Since the polynomial can be broken up, it is sufficient to find solutions for the summation of n^k/a^n, where k is a positive integer. I seem to recall there were answers for this, but the only solutions I can find online are for k = 1 or 2. Aren't there solutions for larger values of k, or am I just imagining this?
Very nice, indeed! Before watching, the 1st of my 2 ways was your 2nd method - differentiating 1/(1-x) For my 2nd method, I squared the geometric series, S = ∑₀⁰⁰(x)ⁿ = 1 + x + x² + x³ + ... = 1/(1-x) to get S² = 1 + 2x + 3x² + 4x³ + ... [This can be seen by writing the original series, then below that, shifting 1 term to the right, x times the original series, then x² S below that, shifted 2ce, etc. Summing it all, gives the result.] So xS² = x + 2x² + 3x³ + 4x⁴ + ... = x/(1-x)² which is the sum we want for x = ½ , where xS² = x/(1-x)² = ½/(½)² = 2 Note: this multiply-shift-and-add trick is an infinite process, and as such, needs to be justified by a sort of convergence argument. Namely, that the terms of any absolutely convergent series, can be arbitrarily rearranged without affecting the sum. Fred
The takeaway of this is that the infinite series of n/2^n is equal to a multiple of the Cauchy product of the infinite series of 1/2^n with itself. Hence, this is why the breakdown of the series that you did worked for the first method. This was also reflected in your second method somewhat. EDIT: More explicitly, 1/2^1 + 2/2^2 + 3/2^3 + ••• = (1/2^1 + 1/2^2 + 1/2^3 + •••) + (1/2^2 + 1/2^3 + 1/2^4 + •••) + (1/2^3 + 1/2^4 + 1/2^5 + •••) + ••• = (1/2^1 + 1/2^2 + 1/2^3 + •••) + (1/2)(1/2^1 + 1/2^2 + 1/2^3 + •••) + (1/2^2)(1/2^1 + 1/2^2 + 1/2^3 + •••) + ••• = (1/2^1 + 1/2^2 + 1/2^3 + •••)(1/2^0 + 1/2^1 + 1/2^2 + 1/2^3) = 2(1/2^1 + 1/2^2 + 1/2^3 + •••)^2 = 2.
For that you can use the online graphing calculator at www.desmos.com/calculator. To help you more, it is the same as finding where sin(1/x^x) / cos(1/x^x) = tan(1/x^x) = 1, or 1/x^x = arctan(1) = pi/4, or x^x = 4 / pi and that is about 1.219, since 4/pi is about 1.273. Spoiler alert: if you look at the graphs, you will see that there is only 1 (real) solution!
That's not a big deal, this is just an airthmetico geometric progression, actually here in india, 11th standard students are presented with a general formula for the sum of first n terms and also sum of infinite terms
*A cow 3 years after its birth, gives birth to a new calf on the 3th year and every other year that follows. Every other newborn gives birth too on the 3th year after its birth and every other year that follows and so on. How many cows will there be after 20 years?*
I will announce the 3 finalists to the video editing contest this weekend!
Get ready to enjoy their videos & vote!
100/(1-x)
The sol is 9/5
x=3^100
S=(x+x+x)/(3x-x+x/3)=3x/(5x/3)=9x/5x=9/5
What is the song of end title?
Please advise
@blackpenredpen derivative of 1/(1-x) is -1/(1-x)^2 at 5:14
This classic sum can also be calculated by arithmetic geometric progression.... And it would be a single line solution though......
Hey, balckpenredpen. I have a long time watching your videos and I have a request for you, so I hope you make a video for it. I’ve been struggling with this succession for quite a week and still can’t find the answer. Yet the succession is not geometric neither arithmetic (I think), I think there’s a way to solve it. I’m a 12 Y/O (hope you take me serious) from Colombia. So the succession is: 4/5, 3/10, 1/2, 0... You have to find the nth value for this succession. Hope you read it!
8:14 Both sides are divisible by 3^99. So divide both sides by 3^99 and you get a more down to earth equation. (3+3+3)/(9-3-1) or 9/5.
Yep!
Teacher : Who is your best friend ?
Me : 1/(1-x)
Teacher : Why?
Me : It squares when we differentiate it.
Hahahhaha!!! Nice one!
@@blackpenredpen WAIT PLEASE CORRECT AND RESPOND YIU DID NOT JUSTOFY WHY YOU CAJ USE ONE OVER ONE MINUS X..YOU DONT KNOW THAT THE SERIES CONVERGES TO,THAT..WHERE,DID,YOU GET,THAT FROM?
e^x forever.
never changes and say something like "i don't like you anymore"
@@leif1075 holy shit calm down
8:21 the answear to your little calculation is 9/5
No, the right answer is 42.
The answer is actually 95 noobs 😎
Yes it's 9/5
Yes
I did some tremendous factorization and I agree.
1. Using Both Pen is really cool,
2. Very informative and Clear
3. Your Smile
4. The way of showing everything.
5. i still have a problem. ( sum of the series n*2^n , where nmax= 513 and nmin = 2)
This sum is related to Collatz conjecture.
At every step of Collatz sequence: multiplication by 3 and adding 1 produces a number with random distribution of zeros and ones (in binary representation). Then you shift right to discard all least significant zeros.
Probability of one rightmost zero is 1/2. Probability of two rightmost zeros is 1/4, three zeros 1/8, etc. On average, the shift step divides the result by 2^(1/2+2/4+3/8+4/16....)=4
To prove Collatz conjecture we only need to prove that for any starting number, the sequence always reaches a number smaller than the start. Average "reduction" being greater than 3 means the sequence is very likely to reach the smaller number.
I did it a third way, by saying {sum n=1 to infinity} ({sum k=1 to n} 1/2^n) = {sum k=1 to infinity} ({sum n=k to infinity} 1/2^n) = {sum k=1 to infinity} 1/2^(k-1) = {sum k=0 to infinity} (1/2)^k = 2
Oh wow double sum!! Very clever!
The second method was really nice and neatly done. It all came together. I couldn't stop smiling when I figured it out just before you finished :)
the numerator is 3(3^100), which is just 3^101. The middle term of the denominator can be written as 3^101/3 and the last term of the denominator can be written as 3^101/9. Factoring out 3^101 from the entire denominator we get 3^101(1-1/3-1/9) as the denominator. The numerator and the denominator can be simplified which leaves 1/(1-1/3-1/9) which is just 9/5.
Question at the end: Just factor out 3^99 on top and bottom. Then you get (3+3+3)/(9-3-1) = 9/5. Easy.
I did it like this:
3^100+3^100+3^100=3*3^100=3^101;
3^101-3^100-3^99=3^101*(3^0-3^-1-3^-2)=3^101*(1-1/3-1/9);
3^101/(3^101*(1-1/3-1/9))=1/(1-1/3-1/9)=1/(5/9)=9/5
Yup crct
Yes, that's why it says algebra 1 Herr Einstein
All hail Lelouch
Where does he get,one over one minus X from? He never justifies that..
Just found this channel, never thought I can look at math as being this fun!
thunder thank you! I am glad to hear it from you!
Calculus BC Teacher: WE ARE GONNA REVIEW SERIES!!
me: noooo
blackpenredpen: dont worry...
You are a lifesaver! I was losing my mind trying to understand this!
almost done. help calculate 4*(1/2)
= 4/2 = 2
@@gdash6925 Professor!
@@Archik4 i feel delighted
lol
Picture the sum 1 + 1/2 +1/4 + 1/8 + ...
by picturing the rectangle (square) with height 1 above the interval 0 to 1 on the x-axis, a rectangle with height 1/2 above the interval 1 to 2 on the x-axis, a rectangle with height 1/4 above the interval from 2 to 3, and so on. The total area of all the rectangles is 2.
Another way to fill those same rectangles is to add the rectangles:
1/2
Just messed up my Calculus exam, going to have to do a resit, never thought watching math videos would cheer me up :)
1:23 A.M. here and nothing feels more fun than math❤️
I love this channel !!
K1LL3R268 thank you!!
Amazing! Very elegant.
It is interesting to say that in the first method you can change terms' positions because the series converges absolutely; also, in the second method you can differentiate because it is a power series.
I think this can be generalized to arbitrary (arithmetic series)_n / (geometric series)_n.
Say, S=a_1 + a_2*q +a_3*q^2 +...+a_n*q^(n-1) ,
then qS=a_1*q +a_2*q^2 + a_3*q^3 +...+a_n*q^n ;
S-qS=a_1 + (a_2-a_1)q + (a_3-a_2)q^2 +...+(a_n - a_[n-1] )q^(n-1) - a_n*q^n .
We know that a_n - a_[n-1] = d (the difference) and we can factor it out. The middle part is then d*(q +q^2 +...+q^(n-1) ) - just a geometric series, its sum is dq/(1-q), if |q|inf . Also we know that a_n=a_1 + d(n-1), so the last term splits into two: a_1*q^n and d*(n-1)*q^n. The first of these obviously goes to zero in the limit, the second goes to zero also, but L'hopital's rule is needed to show that.
Finally, we are left with
S-qS=a_1 +dq/(1-q) ;
S=a_1/(1-q) +dq/(1-q)^2.
If the geometric series doesn't start at 1, we can always factor that first term out, thus:
if S=sum[n=1, n->inf] (a_1*b*q^(n-1) ) , then
S=a_1*b/(1-q) +dqb/(1-q)^2 ;
in our case a_1=1, b=1/2, q=1/2, d=1 ;
S=1*0.5/0.5 +1*0.5*0.5/(0.5)^2 = 1 + 1 = 2.
I like your utilisation of the best friend, very cool!
Thank you!
Similar to 1/2+1/4+1/8...series.
I made a formula generalising n, and also n.
I came here because of the bridge game in the Korean show "Squid Game".
The expected value of winners is such an elegant real life example of this infinite series. I wonder if there was a mathematician involved with this show?
This is so satisfying to watch!
Yay! Classic sum on Halloween!
som from n=1 to if of (n/b^n) = b/(b-1)^2 b€R\(-1,1) you can easily prove it for every b if you replace 2 with b if you do the classical way, then you can test for which b it converges
8:20 the answer is 9/5
Nice halloween headline!
Theory of undetermined coefficients may also be useful, viz. we can assume
n/(2^n) = [ { a(n - 1) + b }/(2^(n - 1)) ] - (an + b)/2^n,
where the coefficients a and b have to be determined
NGL I DIG THE RED AND BLACK PEN SWITCHING, IT'S HOW I DO MY MATH TOO, ONLY WITH 6 COLOURS AND ON PAPER
Black
Pen
Red
Pen
Yay
😆😅
I really like this series. Actually the first one I looked at was the series of n^2/2^n. Which could be evaluated by applying the operator (xD)^2 (D means differentiation) to the geometric series 1/(1-x) and then evaluating at x=1/2.
In general, applying the operator (xD)^k to the geometric series 1/(1-x), and then evaluating at x=1/k gives the value of the series of n^k/k^n, which I find to be an interesting looking series!
Felipe Rueda If k is a natural number, it is possible to breakdown the series of n^k/2^n into sums of series n^m/2^n, where m = 0, 1, ..., k - 1. It creates an important recurrence relation that defined a function with respect to k. I do not believe there is a closed-form explicit formula, however.
@@angelmendez-rivera351 I could find an explicit formula for the series of n^k/k^n using the method I described above. This lies on being able to give the formula for (xD)^k 1/(1-x)
Felipe Rueda I do not believe an explicit closed-form formula exists for [(xD)^k][1/(1 - x)], though, so my point would still stand.
I got a Brilliant annual subscription some months ago from here. I hope you will still be sponsored next year so I can get the discount again haha
Thank you so much for your support. I will continue to work hard!
This question was just like one I've seen in the OBMU(It's the Math Olympics for universities in Brazil)
Answer is 9/5
It's possible to give the general formula of the partial sum of this infinite sum: S_n = 2 - (n+2)/2^n. So no dirty infinite trick is needed.
I solved it by writing n = n-1 + 1
I get sum from n=1 to inf (n-1)/2^n + sum n=1 to inf 1/2^n
The first is equal to 1/2 of the original sum, and the second is equal to 1
Letting S equals the original sum I get S = 1/2S + 1 => 1/2S = 1 => S = 2
now do n^2/2^n
(is it possible to generalize a formula for n^a/b^n for integers a > 1 and b?)
I'm trying to figure out the answer to your question, if I find it I'll let you know immediately😃
Ok so I did the math and it turns out that the infinite summation of n²/2^n is equal to 6. The summation can be generalized like this:
1²(x¹)+2²(x²)+3²(x³)+.....+n²(x^n)+....=S
Where x is the variable.
Now for any value of x where |x| is smaller than one, the infinite sum is
S= (x(1+x))/(1-x)³ [ x(1+x) divided by (1-x)³]
In this case, the value of x is 1/2, so plugging into the equation, we get
S(1/2)= {.5(1+.5)}/(1-.5)³ which is equal to 6.
And for your second question, I'm not exactly sure what the answer is, although I will be looking into it, and if I find a way generalize to for n^a/b^n, I'll let you know
you are my best friend !
I like very much how you turn from a very nice smile to a terrifying classical math teacher stare in less than 0.1 second.
This is one of the most basic series to evaluate. Woopdy doo.
Calculate sum for 1 to infinity of 1/(n+i(n-1)) minus sum for 1 to infinity of 1/((n-1)+ni)
1:29 thats the new math meme
I'm a 17 and I want to be a mathematician or physicist. People say boys shouldn't be a physicist and to do a computer science yet I strive to do it. I hope I'm not the only boy.
Yes I am 17 now I also a boy like you
Enjoyed very much
I am also a math teacher and your videos are also great. Keep up the good work
Or use the identity that for zigma(n/a^n) is always equal to a/(a-1)^2
That's brilliant
Other way: use a python interpreter, get 1.99999999998, conclude that the result is 2.
sum( k/2**k for k in range(1,120))
Hahahahaha nice!
I am an engineer, i am allowed to do that. You are a mathematician, you should think this is a bad way because it does not prove that the result is 2.
@@happygimp0 lol
Very cool!
Thanks!
I took 1/2+1/4+1/8+1/16+... and squared it, getting 1/4+2/8+3/16+4/32+....
.....and then you multiplied it by 2?
Love the zooming!
Hahahaha thanks!
The answer is 9/5
3^100 + 3^100 + 3^100 = 3(3^100) = 9(3^99)
3^101 - 3^100 - 3^99 = 9(3^99) - 3(3^99) - 3^99 = 5(3^99)
9(3^99) / 5(3^99) = 9/5 (3^99 cancels)
Beautiful and clear, thak you so much sir!
@3:50 I thought that the geometric series starts with the index 0, thus with the term \frac{1}{2^{0}} = 1? In this case, the first of the splitted series starts with 1 / 2, so why don‘t we need a correction factor?
The correction factor is precisely the 1/2 at the top of the fraction:
(1/2)/(1-1/2)
This certainly qualifies as mathemagicks
Okay this question was so easy but I like the way you use our best friend in this
Why are you not done at 4:30? ( why isn't he done?)
8:20 The answer is 9/5. I solved it in my head in like 10 seconds.
It took me 32 seconds to do it in my head.
I'd like to join it, but ugh.. payment..
You can sign up for free and do their daily challenges
9/5 is the answer you just have too factorize by 3^99 then simplify
@Qwert Yuiop how did you find it?
@Qwert Yuiop not gonna lie i don't know what it is haha
In a way the first approach is better in that it is works from first principles.
The answer is 9/5. Pretty easy.
Let's imagine a cartesian plane with non-negative X, Y. Let every point on this plane with integer coordinates (a, b) have a weight of 1/2^(a + b + 1). Looking at diagonals (same a + b) we see that sum of weights is our needed sum. But we can count same sum of weights in columns - (1/2 + ...) + (1/4 + ...) + ... = + 1 + 1/2 + ... = 2.
Thanks bro!
Please make a video on Feynman's technique of integration!
Keep up the good work :)
S is given/ then multiplying s by .5 and subtracting the 2 series then it becomes 1/2+1/4+1/8............oo
Again its a geometric series =.5s
Solving s=2
Is it okey bprp pls reply
(Didn't care about oo as its boring¥
Has anyone ever wondered the reason why this channel is called blackpenredpen?
The answer is right in front of you
ok.
When this channel first started, there were not blackboards or markers or whiteboards or chalk. It was all in pen and paper, and the only colors where black and red since those are the standard color for pens. Hence, the name. Of course, if we had to name the channel by today's standard, it would be named blackmarkerredmarkerbluemarkergreenmarkerpurplemarker.
At 5:35 it was n=0 why he takes n=1..
So an infinite sum contains itself, plus extras. That's so weird.
This video reminds me of a more general problem, that of finding the summation of P(n)/a^n as n goes from 1 to infinity, where P(n) is a polynomial function and a is greater than 1. Since the polynomial can be broken up, it is sufficient to find solutions for the summation of n^k/a^n, where k is a positive integer. I seem to recall there were answers for this, but the only solutions I can find online are for k = 1 or 2. Aren't there solutions for larger values of k, or am I just imagining this?
and what about the sum of the (n^m)/(2^n) where m is a integer ?
Please, can you help me to evaluate the series of n^3/2^n ?
thanks
Thank you very much friend, I understood this even though I don't understand English
Which approach do you like more?
I like both honestly
thank you man, you can help me so much!!!!!
Very nice, indeed!
Before watching, the 1st of my 2 ways was your 2nd method - differentiating 1/(1-x)
For my 2nd method, I squared the geometric series, S = ∑₀⁰⁰(x)ⁿ = 1 + x + x² + x³ + ... = 1/(1-x) to get
S² = 1 + 2x + 3x² + 4x³ + ...
[This can be seen by writing the original series, then below that, shifting 1 term to the right, x times the original series, then x² S below that, shifted 2ce, etc. Summing it all, gives the result.]
So
xS² = x + 2x² + 3x³ + 4x⁴ + ... = x/(1-x)²
which is the sum we want for x = ½ , where xS² = x/(1-x)² = ½/(½)² = 2
Note: this multiply-shift-and-add trick is an infinite process, and as such, needs to be justified by a sort of convergence argument.
Namely, that the terms of any absolutely convergent series, can be arbitrarily rearranged without affecting the sum.
Fred
9/5 is answer
Dude Tomorrow I have a Linear Algebra and Circuit Theory exam Why am I here? but Worthwhile
Well explained 👌😇🙂
do the hard problems in brilliant
How to calculate cos(2π/17) without using WolframAlpha?
Beautiful
Can you do this summation to an nth term. Thank you,great video.
Sir please do integration of,
(1-log x)^2 dx /(1+(log x)^2)^2
Easy question
Ans:9/5
just take out common term 3^99 in num. and den.
tchu tchu the power!
The takeaway of this is that the infinite series of n/2^n is equal to a multiple of the Cauchy product of the infinite series of 1/2^n with itself. Hence, this is why the breakdown of the series that you did worked for the first method. This was also reflected in your second method somewhat.
EDIT: More explicitly, 1/2^1 + 2/2^2 + 3/2^3 + ••• = (1/2^1 + 1/2^2 + 1/2^3 + •••) + (1/2^2 + 1/2^3 + 1/2^4 + •••) + (1/2^3 + 1/2^4 + 1/2^5 + •••) + ••• = (1/2^1 + 1/2^2 + 1/2^3 + •••) + (1/2)(1/2^1 + 1/2^2 + 1/2^3 + •••) + (1/2^2)(1/2^1 + 1/2^2 + 1/2^3 + •••) + ••• = (1/2^1 + 1/2^2 + 1/2^3 + •••)(1/2^0 + 1/2^1 + 1/2^2 + 1/2^3) = 2(1/2^1 + 1/2^2 + 1/2^3 + •••)^2 = 2.
Wow
Ans is 9/5
blackpenredpen make a vid on, find the coordinates of the intersection between y=sin(1/x^x) and y=cos(1/x^x) pls.
For that you can use the online graphing calculator at www.desmos.com/calculator. To help you more, it is the same as finding where sin(1/x^x) / cos(1/x^x) = tan(1/x^x) = 1, or 1/x^x = arctan(1) = pi/4, or x^x = 4 / pi and that is about 1.219, since 4/pi is about 1.273. Spoiler alert: if you look at the graphs, you will see that there is only 1 (real) solution!
5:13 why we use derivative on series I can't understand why we use derivative on series please help me I am a beginner student
what will be answer to n?
That's not a big deal, this is just an airthmetico geometric progression, actually here in india, 11th standard students are presented with a general formula for the sum of first n terms and also sum of infinite terms
*A cow 3 years after its birth, gives birth to a new calf on the 3th year and every other year that follows. Every other newborn gives birth too on the 3th year after its birth and every other year that follows and so on. How many cows will there be after 20 years?*
holy shit i got surprised by the volume
8:06 9/5. Did it in my head though
woow you're so smart I did it faster than you though and you'll just have to take my word for it :)
@@GamerTheTurtle 🤗
Sir please solve this integral proble (sinx)^99 (cosx)^93 dx
I m unable to solve this and it is not available on any website
What is with Sn=1, infinity [n^2/2^n]. I can't find the answer... :(