I believe there was some confusion among the viewers on whether "Riemann Integral" is an appropriate term to use. The answer is yes--a Riemann integral is defined to be the limit of a Riemann sum, and it is a correct term to use to describe the limit of sums in our problem. For more information, please consult mathworld.wolfram.com/RiemannIntegral.html.
That is more than fine! It is never too late to learn. I believe you do not need to understand sigma notation for Calculus I; on the other hand, for Calculus II, it is very important to understand sigma notation inside out because of its close relationship with the definite integral. The notation is not as complex as it may seem: for example, the sum from i = 1 to 3 of i^2 is 1^2+2^2+3^2 (basically, you are plugging in i = 1, 2, 3 into i^2 and adding all of them up). After learning the basic definition, it is best to focus on some tactics (such as geometric series and telescoping series) and some properties of summation that may help you evaluate the given summation--for the problem in this video, you may connect the summation to the limit definition of definite integral to evaluate the expression. I encourage you to come back to this video and try out the problem again once you learn some Calculus II. =)
LetsSolveMathProblems I’ll definitely come back to this once I’m ready! I watch your videos all the time (especially the integral ones from MIT bees), because the solutions and processes are always fascinating to me! Once I properly learn these areas, I’ll come back to this video and take a stab at it. Thanks for the inspiring reply, and as always, please keep making your awesome videos!
The answer's 2018. Here's how I derived the solution. The first summation and limit can be evaluated by applying the GP formula (to simplify the exponential term in the numerator). After doing so, you end up with a limit, which can be made easier by substituting 1/n as t (all the terms obtained after evaluating the sum contain 1/n, so the substitution simplifies the limit to a great extent). Now t tebds to 0, as n approaches infinity. It's clear that after the substitution is made, you made an infinity/infinity form limit, and hence L' Hospital's rule is applicable. The final result for the first term comes out to be (2^11-1). The second limit can be evaluated by applying the limit as a sum approach, which after simplification yields 29. Hence, the final answer is 2047-29=2018.
stop aplying l’hospital like a mad man,there are more elegant solutions than just aplying l’h non stop,plus you can’t aply l’h when doing limits of sequences,if you want the equivalent of that is the stoltz-cesaro theorem for infinity/infinity or 0/0,you won’t become a good mathematician by using methods which are not rigouros
Riemann integral is defined to be the limit of a Riemann sum, and it is thus appropriate in the context. For more information, please consult mathworld.wolfram.com/RiemannIntegral.html.
I believe there was some confusion among the viewers on whether "Riemann Integral" is an appropriate term to use. The answer is yes--a Riemann integral is defined to be the limit of a Riemann sum, and it is a correct term to use to describe the limit of sums in our problem. For more information, please consult mathworld.wolfram.com/RiemannIntegral.html.
2018...nice one!
I feel like some people just say the majority answer to sound smart
LASER BEAR ASSAULT UNIT They definitely do!!
I’m still in Calculus I and I haven’t actually studied sigma notation, so this is pretty scary.
That is more than fine! It is never too late to learn. I believe you do not need to understand sigma notation for Calculus I; on the other hand, for Calculus II, it is very important to understand sigma notation inside out because of its close relationship with the definite integral. The notation is not as complex as it may seem: for example, the sum from i = 1 to 3 of i^2 is 1^2+2^2+3^2 (basically, you are plugging in i = 1, 2, 3 into i^2 and adding all of them up). After learning the basic definition, it is best to focus on some tactics (such as geometric series and telescoping series) and some properties of summation that may help you evaluate the given summation--for the problem in this video, you may connect the summation to the limit definition of definite integral to evaluate the expression. I encourage you to come back to this video and try out the problem again once you learn some Calculus II. =)
LetsSolveMathProblems I’ll definitely come back to this once I’m ready! I watch your videos all the time (especially the integral ones from MIT bees), because the solutions and processes are always fascinating to me! Once I properly learn these areas, I’ll come back to this video and take a stab at it. Thanks for the inspiring reply, and as always, please keep making your awesome videos!
You could use Riemann sums, but he wrote the year in blue to emphasize, and that's why I think that the answer is 2018.
Haha! =) Yes! Writing the year in blue was completely intentional. I'm glad you noticed it.
I found a truly remarkable solution to this problem, but the word limit in this comment thread is too small to contain it.
Happy new year everyone!
I do not know the maths behind it, however I used a computer program I wrote for the problem and got 2017.9997 so my answer is 2018.
Samantha here davson
The answer's 2018. Here's how I derived the solution. The first summation and limit can be evaluated by applying the GP formula (to simplify the exponential term in the numerator). After doing so, you end up with a limit, which can be made easier by substituting 1/n as t (all the terms obtained after evaluating the sum contain 1/n, so the substitution simplifies the limit to a great extent). Now t tebds to 0, as n approaches infinity. It's clear that after the substitution is made, you made an infinity/infinity form limit, and hence L' Hospital's rule is applicable. The final result for the first term comes out to be (2^11-1). The second limit can be evaluated by applying the limit as a sum approach, which after simplification yields 29. Hence, the final answer is 2047-29=2018.
Obtain*. I apologise for making those grammatical errors :(
It's a 0/0 form for the first limit. Sorry!!
stop aplying l’hospital like a mad man,there are more elegant solutions than just aplying l’h non stop,plus you can’t aply l’h when doing limits of sequences,if you want the equivalent of that is the stoltz-cesaro theorem for infinity/infinity or 0/0,you won’t become a good mathematician by using methods which are not rigouros
The answer is the sum of two riemann sums which is 2^11-30=2018. My entire working out can be found here:
imgur.com/a/k4J7K
Two Riemann integrals multiplied by constants: 2018.
William Wen it’s riemann sums jeez,most of don’t even solve them rigurously,you aply a formula like an inginier
U mean I should say "Definite integral as the limit of a Riemann sum"?
William Wen or just riemann sum
2018 :-D (converted these to definite integrals)
2018.
2018
ans is 2018 riemann integrals converted to definite integrals
Max Jackson riemann sums not integrals
Riemann integral is defined to be the limit of a Riemann sum, and it is thus appropriate in the context. For more information, please consult mathworld.wolfram.com/RiemannIntegral.html.
They are just definate integrals so ans : 2018
The answer is 2018
2018