Greetings from France. Thank you for all the videos you’ve created. I’m amazed of your skill in explaining clearly the solutions of the problems you give. I’m not a mathematician but I love the way you explain the resolution of ODE. That’s the kind of explanations that should be learnt by students in France. Many thanks !!!
The question doesn't say anything about 𝑓(𝑥) being decreasing over [0, 10]. In fact, we have basically no idea what 𝑓(𝑥) does between, say, 𝑥 = 0 and 𝑥 = 1. All we know is that it is continuous and positive. Over the interval (0, 1), 𝑓(𝑥) can be anywhere between 0 and ∞, which means that 0 < ∫[0, 1] 𝑓(𝑥)𝑑𝑥 < ∞. The same reasoning applies to the intervals (1, 2), (2, 3), ..., (9, 10) as well, and thus all we know is that 0 < ∫[0, 10] 𝑓(𝑥)𝑑𝑥 < ∞.
@@fuckingdumbo The discrete function 𝑓(𝑛) for 𝑛 ∈ {0, 1, 2, ..., 10} is decreasing, but that doesn't mean that the continuous function 𝑓(𝑥) for 𝑥 ∈ [0, 10] must be decreasing. Remember that 𝑓(𝑥) is decreasing over [0,10] if and only if 𝑓(𝑎) ≥ 𝑓(𝑏) for all 𝑎, 𝑏 ∈ [0, 10] such that 𝑎 < 𝑏. With the information given we can't say for sure that, for example, 𝑓(0.1) ≥ 𝑓(0.2), and therefore 𝑓(𝑥) is not necessarily decreasing over [0, 10].
I agree with others, we don't necessarily have a decreasing function. An example class would be all functions of the form f(x) = A * 1/5^x + B * sin^2(pi x) with A and B bigger than zero. The term involving B is always 0 at integer values of x and positive elsewhere since it multiplies a positive number by a square. So the given conditions are met. B can be arbitrarily large with respect to A, putting the integral outside any of those ranges if large enough.
The statement is true when f is decreasing. But without this assumption for fixed epsilon>0 we may take f(n)=(1/5)^n for any natural n, and f(x)=1 000 000 for x in [n-epsilono,n+epsilon] for any natural n.
Unfortunately, only valid when the function f is monotonic/decreasing. Otherwise I could have came up with something like f(x) = 5^-x + N(1 - cos(2pix)) for some arbitrarily MASSIVE N. Its value for f(0) is 1, but within the intervals between whole numbers the function explodes, as big as roughly 10N. Practically independent of f(0).
I solved it a different way, though it was due to the multiple choice options working out for me. f(x) = 5^(-x) satisfies this condition, with f(0) = 1 The antiderivative is -1/(ln(5)*5^(x)), so the integral becomes 1/(ln(5))-1/(ln(5)*5^(10)). Calculator allowed or not, I think it's clear that this value is between 0 and 1, so option A must be the answer.
I skipped a step by saying that 1-(1/5)^10 is less than 1. So, since we're trying to estimate the upper bound of the integral, just make the numerator 1. Then, 5/4f(0) becomes the upper bound of the integral.
You have no right to suppose that f is increasing or decreasing ( monotonic). Even if f is continuous. And even if it is differentiable on the interval the path could go any where. And the integral may admit any value. The distance between 0 and 1 is enough to draw the path any way and f is still contagious. The only thing I guarantee is that the integral exists and is limited as long as it is continuous on a compact area. What do you mean by geometric progression ? Sorry if I missed something. But u must have a condition on the derivative of f, or that f is continuous and has limited values on a set that is dense in the interval (0,10).
As already stated, you need to state that f(x) is decreasing. You also said that you don't care about the lower bound but that is not true. Let's suppose that ALL the choices had the lower bound as 0, than all the choices would be correct!
Most of the times, I like your videos a lot but this one is wrong and misleading. If you wanted to show the Riemann integral definition you should have taken a better proposal. Let takes these set of points. f(n) = 0 where n is an integer (0 * 1/5 = 0 so the geometric condition proportion is met) and f(n+1/2) = H with H>0.. Then join the points by strait lines and you have a "curve" which is a set of triangles of high H on the x axis. This is a continuous function defined from R into [0, +infini[ and integrable. Integral from 0 to 10 is 10*(H/2) which is >0 and all the supposed solution intervals are [0,0] ! Why are you so lazy with the initial conditions ?
Greetings from France. Thank you for all the videos you’ve created. I’m amazed of your skill in explaining clearly the solutions of the problems you give. I’m not a mathematician but I love the way you explain the resolution of ODE. That’s the kind of explanations that should be learnt by students in France. Many thanks !!!
The question doesn't say anything about 𝑓(𝑥) being decreasing over [0, 10].
In fact, we have basically no idea what 𝑓(𝑥) does between, say, 𝑥 = 0 and 𝑥 = 1. All we know is that it is continuous and positive.
Over the interval (0, 1), 𝑓(𝑥) can be anywhere between 0 and ∞, which means that 0 < ∫[0, 1] 𝑓(𝑥)𝑑𝑥 < ∞.
The same reasoning applies to the intervals (1, 2), (2, 3), ..., (9, 10) as well,
and thus all we know is that 0 < ∫[0, 10] 𝑓(𝑥)𝑑𝑥 < ∞.
the common ratio is 1/5 so the function has to decrease
@@fuckingdumbo The discrete function 𝑓(𝑛) for 𝑛 ∈ {0, 1, 2, ..., 10} is decreasing,
but that doesn't mean that the continuous function 𝑓(𝑥) for 𝑥 ∈ [0, 10] must be decreasing.
Remember that 𝑓(𝑥) is decreasing over [0,10] if and only if 𝑓(𝑎) ≥ 𝑓(𝑏) for all 𝑎, 𝑏 ∈ [0, 10] such that 𝑎 < 𝑏.
With the information given we can't say for sure that, for example, 𝑓(0.1) ≥ 𝑓(0.2),
and therefore 𝑓(𝑥) is not necessarily decreasing over [0, 10].
I agree with others, we don't necessarily have a decreasing function. An example class would be all functions of the form f(x) = A * 1/5^x + B * sin^2(pi x) with A and B bigger than zero. The term involving B is always 0 at integer values of x and positive elsewhere since it multiplies a positive number by a square. So the given conditions are met. B can be arbitrarily large with respect to A, putting the integral outside any of those ranges if large enough.
The statement is true when f is decreasing. But without this assumption for fixed epsilon>0 we may take f(n)=(1/5)^n for any natural n, and f(x)=1 000 000 for x in [n-epsilono,n+epsilon] for any natural n.
you assume f(0) is lower than 1 000 000 / 2
but yes I was thinking the same thing.
you only need f(x) = 3f(0) for x in [n+epsilon,n+1-epsilon]
I love how he say "Welcome to another video" and then smiles. 0:00
Unfortunately, only valid when the function f is monotonic/decreasing. Otherwise I could have came up with something like f(x) = 5^-x + N(1 - cos(2pix)) for some arbitrarily MASSIVE N. Its value for f(0) is 1, but within the intervals between whole numbers the function explodes, as big as roughly 10N. Practically independent of f(0).
Excellent explanation Sir. Thanks 🙏🙏🙏🙏
I solved it a different way, though it was due to the multiple choice options working out for me.
f(x) = 5^(-x) satisfies this condition, with f(0) = 1
The antiderivative is -1/(ln(5)*5^(x)), so the integral becomes 1/(ln(5))-1/(ln(5)*5^(10)).
Calculator allowed or not, I think it's clear that this value is between 0 and 1, so option A must be the answer.
I skipped a step by saying that 1-(1/5)^10 is less than 1. So, since we're trying to estimate the upper bound of the integral, just make the numerator 1. Then, 5/4f(0) becomes the upper bound of the integral.
I was just wondering where in the world you are from, Prime? I'm watching you from Melbourne, Australia!
Riemann
I must tell myself , I before E
You have no right to suppose that f is increasing or decreasing ( monotonic). Even if f is continuous. And even if it is differentiable on the interval the path could go any where. And the integral may admit any value. The distance between 0 and 1 is enough to draw the path any way and f is still contagious. The only thing I guarantee is that the integral exists and is limited as long as it is continuous on a compact area. What do you mean by geometric progression ? Sorry if I missed something. But u must have a condition on the derivative of f, or that f is continuous and has limited values on a set that is dense in the interval (0,10).
f(x) couldn't possibly be an increasing function as it is decreasing on the natural numbers.
It can if you let it be.
f(0) = 1
f(0.5) = 1024
f(1) = 1/5
f(1.5) = 600000
f(2) = 1/25
etc
@@nothingbutmathproofs7150
Lucky view count for me (1111)
As already stated, you need to state that f(x) is decreasing. You also said that you don't care about the lower bound but that is not true. Let's suppose that ALL the choices had the lower bound as 0, than all the choices would be correct!
Yep. I was cheating because I already know all the lower bounds were bigger than the first upper bound .
Hehe. The Germans are everywhere❤ And you have a typo in the name😅😊
Most of the times, I like your videos a lot but this one is wrong and misleading. If you wanted to show the Riemann integral definition you should have taken a better proposal. Let takes these set of points. f(n) = 0 where n is an integer (0 * 1/5 = 0 so the geometric condition proportion is met) and f(n+1/2) = H with H>0.. Then join the points by strait lines and you have a "curve" which is a set of triangles of high H on the x axis. This is a continuous function defined from R into [0, +infini[ and integrable. Integral from 0 to 10 is 10*(H/2) which is >0 and all the supposed solution intervals are [0,0] ! Why are you so lazy with the initial conditions ?