Never seen anything like it

[plush minnow]

Any help would be great, thanks a lot.

[daring needle]

1. Wait patiently for a helper to come along.
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[knotty ether]

let's see

[plush minnow]

Thank you so much

hmmm i'll see ;)

[still hill]

@plush minnow has given 1 rep to @knotty ether

[knotty ether]

i'm thinking about the problem right now. it seems like this simplifies down nicely

[plush minnow]

I think it should, as it should be done in less than 4 minutes in an exam, seems impossible

[knotty ether]

integrating f(2^(r-1)*x) from 0 to 1 is 0.5 for any positive integer r, if i'm not mistaken

but because you're summing 10 such integrals, the final result would be 5

[plush minnow]

yes the final answer is 5. however i really don't know how to work that out or anything

[cerulean sapphire]

may I ask if this is a university or high school question? @plush minnow because I just finish with my A-level mathematic but never see anything like this

[plush minnow]

It's high school, it's meant to be at the level of A level.

why do you think thats's my title... lol

[cerulean sapphire]

Is it from a Cambridge Coursebook?? lol?

[cerulean sapphire]

hahahahaha I know right

[knotty ether]

did you get why it is five? if not, i can go over, with more detail, why the answer is 5. if you did, i can tell you how i came up with the solution.

[plush minnow]

I don't know where it's from, but it's for a test at a level level.

No i didn't get why it's five, please could you let me know how you got the solution @knotty ether

[knotty ether]

first part: taking the definitive integral of g here is equivalent to taking the definitive integrals of the functions g is a sum of, and adding them together. do you understand this step?

[plush minnow]

No sorry

[knotty ether]

basically, you know how, when you have a function that is a sum of two functions, you can take its integral by taking the integrals of the two functions, and then summing them up?

the function g is the sum of different functions of the form f(2^(r-1)*x), where r is a positive integer.

so you can just take integrals of these functions individually, and then sum them up to get the final answer

[plush minnow]

Yes that I understand. So how do you do that final step?

[plush minnow]

How would i go about calculating that as f(x) is very weird??

[knotty ether]

now, here is where the trick lies

when r is any positive integer, the definitive integral of f(2^(r-1)*x) from 0 to 1, is in fact 0.5

i'll prove it soon

[knotty ether]

because of this, and because you're adding 10 such functions, you can just multiply 0.5 by 10, and get 5

[knotty ether]

i think the most intuitive way to convice you of this is to just show you the graphs of the functions. while solving, i graphed them in my head. however, it can be done a lot more rigorously too

[plush minnow]

Yes please i'd like to see that, it would help a lot. Then yes I think i may finally understand it

[knotty ether]

here is the graph of y=f(x)

next:

graph of f(2x), which is the same thing as f(2^(r-1)*x) when r is 2

now, here is f(16x), which is the same thing as f(2^(r-1)*x) when r is 5

basically, they're all just stretched versions of f(x)

the spikes, which are perfectly triangular, always occupy half of the area of the 1x1 square with the bottom left corner at 0,0

do you get why the area is 0.5

[knotty ether]

this works, because the spikes always with perfectly inside the square

the definite integral of f(a*x) from 0 to 1 is 0.5 for any positive integer a, in fact

when a is not a positive integer, it doesn't work, but 2^(r-1) is always a positive integer when r is

did you get what i'm trying to communicate?

@plush minnow

[plush minnow]

Wow that's dedication building those graphs somehow.

Yes i understand, thanks a lot. But how would I see this in an exam when I only have 5 minutes for the question?

[knotty ether]

i used a graphing tool called desmos, i can recommend it

i used a little trick to graph f(x)

(x doesn't need to be in brackets)

[knotty ether]

honestly, i'm not sure. it's hard for me to explain my thought process. however, what you can take away from this is that when you see a question like this (you have a integrate a function that is a sum of other functions), looking individually at the functions it is a sum of can often help. another tip you can take away is that when integrating a periodic function, there are sometimes tricks possible. also, if a question that looks difficult on the surface appears in an exam, you might be able to reduce it down somehow, so you could think about ways of potentially reducing it.

[plush minnow]

well thanks a lot anyway

[still hill]

@plush minnow has given 1 rep to @knotty ether

[plush minnow]

+close