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Step1 1996 q6



By using the formula, I did get to:
[br][br]2coskx=(sin(k+1/2)−sin(k−1/2))xsin(x/2)[br]\displaystyle[br]2coskx=\dfrac{(sin(k+1/2)-sin(k-1/2))x}{sin(x/2)}
Because other terms cancel out, I got
Unparseable latex formula:

[br]\displaystyle[br]\sum_{k=1}^{n}\(2coskx)=\dfrac{(sin(n+1/2)-sin(1/2))x}{sin(x/2)}[br]



But, if thats true, than:
Unparseable latex formula:

[br]1+\sum_{k=1}^{n}\(2coskx)=\dfrac{xsin(n+1/2)-xsin(1/2)+sin(x/2)}{sin(x/2)}=f(x)[br]


EDIT:
If the bit above is correct. Can anyone help me to prove sin(x/2)=xsin(1/2) for 0<x<pi?
Or show me mistakes in case it's wrong (what it's very likely to be)
(edited 11 years ago)

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Where it says sin(k+12)x sin(k + \frac{1}{2})x , it actually means sin((k+12)x) sin((k + \frac{1}{2})x) .

That is, the x x is actually inside the sin function.

Also, sin(x2)≠xsin(12) sin(\frac{x}{2}) \neq x sin(\frac{1}{2}) - just take x = 10, clearly sin(5)≠10sin(12)>1 sin(5) \neq 10 sin(\frac{1}{2}) > 1
Reply 2
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Dog4444
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Original post by dantheman1261
Where it says sin(k+12)x sin(k + \frac{1}{2})x , it actually means sin((k+12)x) sin((k + \frac{1}{2})x) .

That is, the x x is actually inside the sin function.

Also, sin(x2)≠xsin(12) sin(\frac{x}{2}) \neq x sin(\frac{1}{2}) - just take x = 10, clearly sin(5)≠10sin(12)>1 sin(5) \neq 10 sin(\frac{1}{2}) > 1


Fuuuuuuuuuuuuuuu, thanks. :biggrin:

Why didn't they put the brackets? Or are there any notational rules about it?
Original post by Dog4444
Fuuuuuuuuuuuuuuu, thanks. :biggrin:

Why didn't they put the brackets?


I looked at that, and aside from the fact that I couldn't work out what you were asking, I thought the original question was poorly formed - it just looks wrong.

However, from the "Using the formula" part, it's clear that the "x" must be within the function.
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davros
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Original post by Dog4444
Fuuuuuuuuuuuuuuu, thanks. :biggrin:

Why didn't they put the brackets? Or are there any notational rules about it?


It's standard notation - just as if you see sin 3x then you know it means sin (3x) and not (sin 3) x :smile:
Reply 5
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Dog4444
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Original post by davros
It's standard notation - just as if you see sin 3x then you know it means sin (3x) and not (sin 3) x :smile:


Right, I see. It's just the brackets what confused me.

And how would you write x times sin(n+1/2)? Just xsin(n+1/2), right?
Original post by Dog4444
Right, I see. It's just the brackets what confused me.

And how would you write x times sin(n+1/2)? Just xsin(n+1/2), right?


Yup!
Original post by Dog4444
Right, I see. It's just the brackets what confused me.

And how would you write x times sin(n+1/2)? Just xsin(n+1/2), right?


Yep. If the sine part is multiplied by anything, it is generally at the beginning, like you say. The notation isn't great but it's just generally accepted - I take it you managed the question? :smile:
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Dog4444
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Original post by dantheman1261
Yep. If the sine part is multiplied by anything, it is generally at the beginning, like you say. The notation isn't great but it's just generally accepted - I take it you managed the question? :smile:


I did the first part, but there's also the second part as well:


I managed to do it for f(x). But the way I did it, was to draw some sketches of cos x, cos 2x, cos 3x from 0 to pi, and show that the area of each is 0. (I'm not sure I would get away with it at an actual exam)
But I can't do this for f(x)cosx, and I feel there's another approach apart from sketches.
(edited 11 years ago)
Original post by Dog4444
I did the first part, but there's also the second part as well:


I managed to do it for f(x). But the way I did it, was to draw some sketches of cos x, cos 2x, cos 3x from 0 to pi, and show that the area of each is 0. (I'm not sure I would get away with it at an actual exam)
But I can't do this for f(x)cosx, and I feel there's another approach apart from sketches.


Do you know what the integral of cos(x) is? And similarly for cos(kx)? (Which function, I mean)
Original post by Dog4444
I did the first part, but there's also the second part as well:


I managed to do it for f(x). But the way I did it, was to draw some sketches of cos x, cos 2x, cos 3x from 0 to pi, and show that the area of each is 0. (I'm not sure I would get away with it at an actual exam)
But I can't do this for f(x)cosx, and I feel there's another approach apart from sketches.


For the first integral, I am not sure how you did the graphical approach but wouldn't just be simpler to actually integrate ∫0πcos⁡kx  dx?\displaystyle\int_0^{\pi} \cos kx\; dx? The result is quasi-immediate ∀k\forall k.
Reply 11
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Dog4444
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Original post by dantheman1261
Do you know what the integral of cos(x) is? And similarly for cos(kx)? (Which function, I mean)


Well, INT cos x = sin x, right? :biggrin:
But I dont know how to generalise it for cos kx, where k belongs to N.
Original post by Dog4444
Well, INT cos x = sin x, right? :biggrin:
But I dont know how to generalise it for cos kx, where k belongs to N.


Okay - what is ddx(sin(kx)) \frac{d}{dx}(sin(kx)) ? That should give you a clue how to generalise it :smile:
Reply 13
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Dog4444
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Original post by dantheman1261
Okay - what is ddx(sin(kx)) \frac{d}{dx}(sin(kx)) ? That should give you a clue how to generalise it :smile:


Didn't try inspection. Int cos kx = (1/k)sinkx. What makes f(x) part clear. Thanks.
But I still can't do f(x)cosx. The only idea I have is to integrate by parts 2cosx(sinx+1/2sin2x+1/3sin3x...1/n sin nx), what doesn't seem nice in my head(EDIT: actually, I will try this out).
(edited 11 years ago)
Original post by Dog4444
Didn't try inspection. Int cos kx = (1/k)sinkx.


Spoiler

Original post by Dog4444
Didn't try inspection. Int cos kx = (1/k)sinkx. What makes f(x) part clear. Thanks.
But I still can't do f(x)cosx. The only idea I have is to integrate by parts 2cosx(sinx+1/2sin2x+1/3sin3x...1/n sin nx), what doesn't seem nice in my head(EDIT: actually, I will try this out).


You would be looking at ∑k=1n∫0πcos⁡xcos⁡kx  dx\displaystyle\sum_{k=1}^n\int_0^{\pi} \cos x\cos kx\;dx not ∑k=1n∫0π1kcos⁡xsin⁡kx  dx\displaystyle\sum_{k=1}^n\int_0^{\pi} \frac{1}{k}\cos x\sin kx\;dx

Anyhow, remember that you don't have only one expression for f...f...

(a double IBP should work though - just seems a bit tedious!)
(edited 11 years ago)
Original post by Dog4444
Didn't try inspection. Int cos kx = (1/k)sinkx. What makes f(x) part clear. Thanks.
But I still can't do f(x)cosx. The only idea I have is to integrate by parts 2cosx(sinx+1/2sin2x+1/3sin3x...1/n sin nx), what doesn't seem nice in my head(EDIT: actually, I will try this out).


So we want ∫0π(f(x)cos(x))dx=∫0π(cos(x)+cos2(x)+⋯+cos(nx)cos(x))dx \int_{0}^{\pi} ( f(x)cos(x) ) dx = \int_{0}^{\pi} ( cos(x) + cos^{2}(x) + \dots + cos(nx)cos(x) ) dx

No doubt there is more than one way of doing this - do you have any ideas? If you've tried some stuff with no luck, then:

Spoiler

Reply 17
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Dog4444
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Original post by dantheman1261
So we want ∫0π(f(x)cos(x))dx=∫0π(cos(x)+cos2(x)+⋯+cos(nx)cos(x))dx \int_{0}^{\pi} ( f(x)cos(x) ) dx = \int_{0}^{\pi} ( cos(x) + cos^{2}(x) + \dots + cos(nx)cos(x) ) dx



I did integrate cosxcoskx by parts twice and got to:
[br][br]∫cosxcoskx=(1−k2)(coskxsinx−ksinkxcosx)[br]\displaystyle[br]\int cosxcoskx = (1-k^2)(coskxsinx-ksinkxcosx)
And from that I assume that
[br][br]∫0πcosxcoskx=0[br][br]\displaystyle[br]\int_{0}^{\pi} cosxcoskx = 0[br]
Is that right?
Original post by Dog4444
I did integrate cosxcoskx by parts twice and got to:
[br][br]∫cosxcoskx=(1−k2)(coskxsinx−ksinkxcosx)[br]\displaystyle[br]\int cosxcoskx = (1-k^2)(coskxsinx-ksinkxcosx)
And from that I assume that
[br][br]∫0πcosxcoskx=0[br][br]\displaystyle[br]\int_{0}^{\pi} cosxcoskx = 0[br]
Is that right?
What happens when k = 1?
Reply 19
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Dog4444
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Original post by DFranklin
What happens when k = 1?


Thanks, it seems to be true for K>1 though. :s-smilie:

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