stealth_writer
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Can anyone illustrate the method for how to do this question.

Reputation points to any person who can show me a fully worked solution, on the method to solve this question. I can the first few parts. But beyond that i get stuck.

Cheers


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TenOfThem
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(Original post by stealth_writer)
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Can anyone illustrate the method for how to do this question.

Reputation points to any person who can show me a fully worked solution.

Cheers


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You start with Sin(y) = x

You differentiate

You spot that Cos(y) = the root of (1-sin^2x)



I could give you a fully worked answer so I expect the Rep points in spite of the fact that I will not since I think you will learn more if you think it through yourself
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PerArduaAdAstra
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(Original post by TenOfThem)
You start with Sin(y) = x

You differentiate

You spot that Cos(y) = the root of (1-sin^2x)



I could give you a fully worked answer so I expect the Rep points in spite of the fact that I will not since I think you will learn more if you think it through yourself
This. Maths is all about the struggle man.
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stealth_writer
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(Original post by TenOfThem)
You start with Sin(y) = x

You differentiate

You spot that Cos(y) = the root of (1-sin^2x)



I could give you a fully worked answer so I expect the Rep points in spite of the fact that I will not since I think you will learn more if you think it through yourself
Hi,

i can do all of that bit my friend. I can get up to the bit where we show that y=ksin-1x satisfies the differential equation.
It is the bit beyond (in other word beyond the "in this particular case" part) that i cannot do despite struggling for hours.
I too also agree with your sentiment in your reply; however I have exhausted all options on this question, and tsr has been my last resort. Please could you show me how to do this question ...... so that I can work backwards. Sometimes one can learn more working backwards than they did working forwards if you know what i mean.


P. s this isn't coursework or homework- rather independent study/revision.
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SamKeene
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(Original post by stealth_writer)
Hi,

i can do all of that bit my friend. I can get up to the bit where we show that y=ksin-1x satisfies the differential equation.
It is the bit beyond (in other word beyond the "in this particular case" part) that i cannot do despite struggling for hours.
I too also agree with your sentiment in your reply; however I have exhausted all options on this question, and tsr has been my last resort. Please could you show me how to do this question ...... so that I can work backwards. Sometimes one can learn more working backwards than they did working forwards if you know what i mean.


P. s this isn't coursework or homework- rather independent study/revision.
Judging by the question style and content... this is STEP III question?

If so, you really should be spending at least 2 days on the question before seeking any guidance.
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stealth_writer
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(Original post by SamKeene)
Judging by the question style and content... this is STEP III question?

If so, you really should be spending at least 2 days on the question before seeking any guidance.
It's just a question i found in one my independent study text books.

Can you do this question?
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felamaslen
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Here's a clue:

y = sin(3arcsin(x)) = -4x^{3} + 3x

Try substituting u = arcsin(x).

(Justification for the above statement: because the solution is assumed to be unique [a proof of that would be a dependency of any further results regarding sin(3x)], the statement is true for all x).
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stealth_writer
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Can you elaborate a bit more/actually show me.
The best way to show me would be for you to do the worked solution on a piece of paper and then take a photo of your working. It's pretty easy to do in the mobile phone if you the tsr app
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felamaslen
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(Original post by stealth_writer)
Can you elaborate a bit more/actually show me.
The best way to show me would be for you to do the worked solution on a piece of paper and then take a photo of your working. It's pretty easy to do in the mobile phone if you the tsr app
Since you asked:

Spoiler:
Show
Name:  B7djYyh.jpg
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Size:  360.3 KB

Name:  AXzaf5p.jpg
Views: 151
Size:  347.7 KB


Apologies for the lack of clarity (hastily writing it down as well as a rubbish phone camera ). The first image contains the first part (with the second bit at the top right corner), and the second image contains the bit which you were asking about (the last part).

EDIT:

I put the solution in latex because it's better like that:

Spoiler:
Show
\mbox{Let }sin(y) = x \implies y = sin^{-1}(x)
\implies cos(y)\frac{dy}{dx} = 1
\frac{dy}{dx} = \frac{1}{cos(y)} = \frac{1}{\sqrt{1-sin^{2}(y)}}
\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} \mbox{ as required.}

\mbox{Let }y = sin(ksin^{-1}x)
\implies \frac{dy}{dx} = \frac{k}{\sqrt{1-x^2}}cos(ksin^{-1}(x))
\frac{d^{2}y}{dx^{2}} = \frac{\frac{-k^{2}sin(ksin^{-1}(x))}{\sqrt{1-x^{2}}}(\sqrt{1-x^{2}})-(-2x(\frac{1}{2}){(1-x^{2})}^{-\frac{1}{2}} \times kcos(ksin^{-1}(x)))}{1-x^{2}}
\frac{d^{2}y}{dx^{2}} = \frac{-k^{2}sin(ksin^{-1}(x))+\frac{x}{\sqrt{1-x^{2}}}kcos(ksin^{-1}(x)}{1-x^{2}}
(1-x^{2})\frac{d^{2}y}{dx^{2}} = -k^{2}sin(ksin^{-1}(x))+\frac{kxcos(ksin^{-1}(x))}{\sqrt{1-x^{2}}}
(1-x^{2})\frac{d^{2}y}{dx^{2}} = x\frac{dy}{dx} - k^{2}y
\therefore (1-x^{2})\frac{d^{2}y}{dx^{2}} - x\frac{dy}{dx} + k^{2}y = 0 \mbox{ as required.}

\mbox{When k = 3: }(1-x^{2})\frac{d^{2}y}{dx^{2}} - x\frac{dy}{dx} + 9y = 0
\mbox{If }y = -4x^{3} + 3x\mbox{:}
\frac{dy}{dx} = -12x^{2} + 3
\frac{d^{2}y}{dx^{2}} = -24x
\implies (1-x^{2})\frac{d^{2}y}{dx^{2}} - x\frac{dy}{dx} + 9y = (1-x^{2})(-24x) - x(-12x^{2}+3) + 9(-4x^{3}+3x)
= -24x + 24x^{3} + 12x^{3} - 3x - 36x^{3} + 27x
= 0\mbox{ as required.}
\mbox{When x = 0: }y = 0\mbox{ and }\frac{dy}{dx} = 3\mbox{ as required.}

y = sin(3sin^{-1}(x)) = -4x^{3} + 3x
\mbox{Let }u = sin^{-1}(x) \implies x = sin(u)
\implies sin(3u) = -4sin^{3}(u) + 3sin(u)
\therefore sin(3x) \equiv 3sin(x) - 4sin^{3}(x)
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davros
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(Original post by SamKeene)
Judging by the question style and content... this is STEP III question?

If so, you really should be spending at least 2 days on the question before seeking any guidance.
It's almost, but not quite, identical to STEP I 1999 Q7
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stealth_writer
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(Original post by felamaslen)
Since you asked:

Spoiler:
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Name:  AXzaf5p.jpg
Views: 151
Size:  347.7 KB


Apologies for the lack of clarity (hastily writing it down as well as a rubbish phone camera ). The first image contains the first part (with the second bit at the top right corner), and the second image contains the bit which you were asking about (the last part).

EDIT:

I put the solution in latex because it's better like that:

Spoiler:
Show
\mbox{Let }sin(y) = x \implies y = sin^{-1}(x)
\implies cos(y)\frac{dy}{dx} = 1
\frac{dy}{dx} = \frac{1}{cos(y)} = \frac{1}{\sqrt{1-sin^{2}(y)}}
\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} \mbox{ as required.}

\mbox{Let }y = sin(ksin^{-1}x)
\implies \frac{dy}{dx} = \frac{k}{\sqrt{1-x^2}}cos(ksin^{-1}(x))
\frac{d^{2}y}{dx^{2}} = \frac{\frac{-k^{2}sin(ksin^{-1}(x))}{\sqrt{1-x^{2}}}(\sqrt{1-x^{2}})-(-2x(\frac{1}{2}){(1-x^{2})}^{-\frac{1}{2}} \times kcos(ksin^{-1}(x)))}{1-x^{2}}
\frac{d^{2}y}{dx^{2}} = \frac{-k^{2}sin(ksin^{-1}(x))+\frac{x}{\sqrt{1-x^{2}}}kcos(ksin^{-1}(x)}{1-x^{2}}
(1-x^{2})\frac{d^{2}y}{dx^{2}} = -k^{2}sin(ksin^{-1}(x))+\frac{kxcos(ksin^{-1}(x))}{\sqrt{1-x^{2}}}
(1-x^{2})\frac{d^{2}y}{dx^{2}} = x\frac{dy}{dx} - k^{2}y
\therefore (1-x^{2})\frac{d^{2}y}{dx^{2}} - x\frac{dy}{dx} + k^{2}y = 0 \mbox{ as required.}

\mbox{When k = 3: }(1-x^{2})\frac{d^{2}y}{dx^{2}} - x\frac{dy}{dx} + 9y = 0
\mbox{If }y = -4x^{3} + 3x\mbox{:}
\frac{dy}{dx} = -12x^{2} + 3
\frac{d^{2}y}{dx^{2}} = -24x
\implies (1-x^{2})\frac{d^{2}y}{dx^{2}} - x\frac{dy}{dx} + 9y = (1-x^{2})(-24x) - x(-12x^{2}+3) + 9(-4x^{3}+3x)
= -24x + 24x^{3} + 12x^{3} - 3x - 36x^{3} + 27x
= 0\mbox{ as required.}
\mbox{When x = 0: }y = 0\mbox{ and }\frac{dy}{dx} = 3\mbox{ as required.}

y = sin(3sin^{-1}(x)) = -4x^{3} + 3x
\mbox{Let }u = sin^{-1}(x) \implies x = sin(u)
\implies sin(3u) = -4sin^{3}(u) + 3sin(u)
\therefore sin(3x) \equiv 3sin(x) - 4sin^{3}(x)
You deserve a medal!. Thank you so much my dear friend! God bless you! Now i will learn much more efficiently and effectively.
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stealth_writer
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My friends on TSR.

Due to the success of the last solution of my last problems could you guys help me out with another tricky question please. In the same manner as before (a worked solution).

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Once again. Working backwards is much more fruitful IMO sometimes than working forwards
rep points to anyone who helps

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stealth_writer
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(Original post by davros)
It's almost, but not quite, identical to STEP I 1999 Q7
Hmmm. thanks for pointing this out. Is my above post a step paper question as well?
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davros
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(Original post by stealth_writer)
Hmmm. thanks for pointing this out. Is my above post a step paper question as well?
I don't recognize it, but then I only recognized the last one because I remembered doing it a couple of months ago!

(Original post by stealth_writer)
My friends on TSR.

Due to the success of the last solution of my last problems could you guys help me out with another tricky question please. In the same manner as before (a worked solution).

Name:  ImageUploadedByStudent Room1390051692.209124.jpg
Views: 132
Size:  134.0 KB

Once again. Working backwards is much more fruitful IMO sometimes than working forwards
rep points to anyone who helps

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That's not really how TSR works! You should have a go at this question and post your working when you get stuck so that someone can help you move forwards. How far have you got with this particular question?
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stealth_writer
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(Original post by davros)
I don't recognize it, but then I only recognized the last one because I remembered doing it a couple of months ago!



That's not really how TSR works! You should have a go at this question and post your working when you get stuck so that someone can help you move forwards. How far have you got with this particular question?
Can do the first 1/3. But no further. And tsr needs to understand that not all people learn the same way.

I understand your sentiment but i must stress i have tried this question and reached a brick wall. Hence, why i am on tsr.
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davros
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(Original post by stealth_writer)
Can do the first 1/3. But no further. And tsr needs to understand that not all people learn the same way.

I understand your sentiment but i must stress i have tried this question and reached a brick wall. Hence, why i am on tsr.
TSR doesn't make any assumptions about how people learn - that's why we ask people to show what they've attempted and where they're stuck so that we can give constructive feedback and appropriate levels of help.

Post your working so far and we'll take it from there
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Mr M
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(Original post by stealth_writer)
..
Desperation is not an attractive quality.

Show your working. TSR is not a homework service.
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stealth_writer
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(Original post by Mr M)
Desperation is not an attractive quality.

Show your working. TSR is not a homework service.
this isn't homework .
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Mr M
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(Original post by stealth_writer)
this isn't homework .
Of course it isn't.

So I imagine the bit you can't do is to derive the reduction formula.

\displaystyle I_n = \int_1^e x (\ln x)^n \, dx

Integrate by parts.

\displaystyle I_n = \left[ \frac{x^2 (\ln x)^n}{2} \right]_1^e - \frac{n}{2} \int_1^e x (\ln x)^{(n-1)} \, dx
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stealth_writer
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(Original post by Mr M)
Of course it isn't.

So I imagine the bit you can't do is to derive the reduction formula.

\displaystyle I_n = \int_1^e x (\ln x)^n \, dx

Integrate by parts.

\displaystyle I_n = \left[ \frac{x^2 (\ln x)^n}{2} \right]_1^e - \frac{n}{2} \int_1^e x (\ln x)^{(n-1)} \, dx
I did that bit on my own. However, how times do i need to integrate the bit on the right.... surely this goes on n times ?
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