My college textbook is not very clear on how to preform this procedure. It took a couple times of replaying the video but now I see what is going on. In places where the denominator, or perhaps the whole function, of 1/(1+x) is surrounded by a power we can take the derivative on the 'x' within the sum representation of the simpler, known, form of function to get the sum representation of function where the denominator of 1/(1+x) is surrounded by a power. (I hope that made sense) Its apparent that you love mathematics in that you started a web site so you can do what you love. Keep it up.
An easier way to differentiate 1/(1+x) is rewrite it as (1+x)^(-1), then d/dx[(1+x)^(-1)] becomes as simple chain rule problem. Which, would give -1(1+x)^(-2) or -1/(1+x)^2. I always prefer to use the chain rule whenever possible because it is so much simpler to use than the product rule or quotient rule
Oh my goodness gracious. Thank you so much for making these videos. I am struggling so much in calculus 2 class in college because I have my instructor is so bad and after watching your video, I finally understand all the steps! Thank you so much!
I am not sure if I am just dumb but since the infinite series starts at n=0, shouldn't it be (-1)^(n+1) instead of (-1)^(n)? What am I missing here? Respectively when you take the derivative , the signs inherently change in the series, wouldn't it then convert two in the derivative to (-1)^(n) because the series starts with a negative value?
Thanks for making these videos, they have helped me so much. One question though, once you get the value of the indefinite integral is it possible to find a value or assign a number to C? Or If you just leave C is okay? Thanks in advance
You can if you've been given an initial condition like y(0)=5. When you have an initial condition that you can use to find C, it's called an initial value problem. In this example, you'd just plug in 0 for x and 5 for y, and that would allow you to solve for a value of C. Hope that helps!
In the first example, it doesn't make any difference because the first term is zero. But in general, if you change n to n+1 you also need to change the index to n+1=0. So the summation should start at n=-1. Am I right?
You are right. The starting value of n should have been n=1 after the differentiation since the constant term became 0. Then after the re-indexing it should have gone back to 0. None of that was mentioned in the video.
integralCALC Keep the easy steps, I value the longer videos with more information and that's why I come here instead of PatrickJMT or khanacademy! If you want short "tip-clips", watch those guys noob.
I have a question though, why don't you change the bounds of the summation once you change all the n's to (n+1)
My college textbook is not very clear on how to preform this procedure. It took a couple times of replaying the video but now I see what is going on. In places where the denominator, or perhaps the whole function, of 1/(1+x) is surrounded by a power we can take the derivative on the 'x' within the sum representation of the simpler, known, form of function to get the sum representation of function where the denominator of 1/(1+x) is surrounded by a power. (I hope that made sense)
Its apparent that you love mathematics in that you started a web site so you can do what you love. Keep it up.
An easier way to differentiate 1/(1+x) is rewrite it as (1+x)^(-1), then d/dx[(1+x)^(-1)] becomes as simple chain rule problem. Which, would give -1(1+x)^(-2) or -1/(1+x)^2. I always prefer to use the chain rule whenever possible because it is so much simpler to use than the product rule or quotient rule
Great and thorough explanation. Much appreciated.
You took the derivative of the series but you didn't say one word about indexing the starting value of n.
Oh my goodness gracious. Thank you so much for making these videos. I am struggling so much in calculus 2 class in college because I have my instructor is so bad and after watching your video, I finally understand all the steps! Thank you so much!
You're welcome, I'm so glad that it's finally making sense!
integralCALC But I am confused though at one thing. I feel like you have to change the index when you want x^(n-1) to x^n.
Great Video! Love all of these thanks for all you do!
Aaron Chow Glad you like them!
I am not sure if I am just dumb but since the infinite series starts at n=0, shouldn't it be (-1)^(n+1) instead of (-1)^(n)? What am I missing here?
Respectively when you take the derivative , the signs inherently change in the series, wouldn't it then convert two in the derivative to (-1)^(n) because the series starts with a negative value?
Thanks for making these videos, they have helped me so much. One question though, once you get the value of the indefinite integral is it possible to find a value or assign a number to C? Or If you just leave C is okay? Thanks in advance
You can if you've been given an initial condition like y(0)=5. When you have an initial condition that you can use to find C, it's called an initial value problem. In this example, you'd just plug in 0 for x and 5 for y, and that would allow you to solve for a value of C. Hope that helps!
thanks so much always you do at your level best very helpful to me
Hamisi Wenge I'm happy to help!
that nyc f u thanks
In the first example, it doesn't make any difference because the first term is zero. But in general, if you change n to n+1 you also need to change the index to n+1=0. So the summation should start at n=-1. Am I right?
You are right. The starting value of n should have been n=1 after the differentiation since the constant term became 0. Then after the re-indexing it should have gone back to 0. None of that was mentioned in the video.
Yes, f(n) has an additive inverse sum to the index n=a. So, if f(n) becomes f(n+c) then the index n starts at n=a-c. The same works the other way.
not trying to be mean, but your demonstration/explanation is quite confusing.
you may have just saved my calculus grade
FUCKING THANK YOU
What a magic mam!!! Would you like to tell me how to find sum of any finite series??
omg you wonderful .. one of the best teachers ever ... very clear :)
There is no better than that. I hope to be as awesome as you are!
why don't you multiply -1 to nx^n-1 at 8:15?
is this the same as term by term differentiation?
Amazing explanation! Thank you sm!
You're very welcome, Bigger, I'm so glad you liked it! :D
This is just Awesome! Thank you so much. This topic is one of the hardest in Calculus.
Fatima Alamin You're welcome, I'm so glad it helped!
Finally, an explanation in plain English that actually makes sense \[T]/
:D
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Amazing very detailed thank you
Thanks Shawndel! 😀
Wow! You are awesome!
You too! :)
A piece of advice: you should not explain all the easy steps. It makes the video too long. Everything else is perfect.
Thanks for the feedback. It's always tough to balance... some people like the steps, some don't.
integralCALC Keep the easy steps, I value the longer videos with more information and that's why I come here instead of PatrickJMT or khanacademy! If you want short "tip-clips", watch those guys noob.
integralCALC The steps is what makes your teaching style stood out from others. Thanks for doing what you do right!
Daniel Rojas
you don't know basic algebra?
The videos are perfect with all the extra steps thanks! Some of us are brilliant at math and forget things! @intergal