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The Proof is Trivial!

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Reply 60
Original post by bogstandardname
Problem 11* (if you have never seen substitution before, **)

Find the general solution, by a suitable substitution or otherwise of this differential equation.

dydx=sinx(cosx+y)cosxy\displaystyle \dfrac{dy}{dx}=\dfrac{sinx(cosx+ y)}{cosx-y}



Solution 12 (a bit rough-and-ready, I'm afraid, and I feel like there should be a nice "otherwise"):

Spoiler

(edited 10 years ago)
Reply 61
Original post by Lord of the Flies
Problem 15*/**

Evaluate limn1πn1+xndx\displaystyle \lim_{n\to\infty} \int_1^{\pi} \frac{n}{1+x^n}\,dx


Hmm not totally sure how to do this (although I have an idea)

If you let fn(x)=n1+xnf_n(x) = \dfrac{n}{1+x^n} then limnfn(x)=0\lim_{n \rightarrow \infty} f_n(x) = 0 for all x(1,π]x \in (1, \pi]. I believe the fact it doesn't converge to 0 at one point isn't a problem with this (Lebesgue)

So by Lebesgue

limn1πfn(x)dx=1π0dx=0\displaystyle \lim_{n\to\infty} \int_1^{\pi} f_n(x)\,dx = \int_1^{\pi} 0\,dx = 0

Disclaimer: There's a very good chance this is wrong (I haven't covered Lebesgue :frown: )
Problem 16 **/***

Let ff be a function such that f(x) dx<\displaystyle \int_{-\infty}^{\infty} f(x)\ dx < \infty.

Define f±(x)=f(x+x2+1)±f(xx2+1)f_{\pm}(x) = f(x + \sqrt{x^2 + 1}) \pm f(x - \sqrt{x^2 + 1}) and I(f)(x)=f+(x)+xx2+1f(x)\displaystyle \mathfrak{I}(f)(x) = f_{+}(x) + \frac{x}{\sqrt{x^2 + 1}} f_{-}(x).

Show that f(x) dx=I(f)(x) dx\displaystyle \int_{-\infty}^{\infty} f(x)\ dx = \int_{-\infty}^{\infty} \mathfrak{I}(f)(x)\ dx.


Find cos(x)sin(x2+1)x2+1 dx\displaystyle \int_{-\infty}^{\infty} \frac{\cos(x)\sin(\sqrt{x^2 + 1})}{\sqrt{x^2 + 1}}\ dx.
Original post by Noble.
Hmm not totally sure how to do this (although I have an idea)

If you let fn(x)=n1+xnf_n(x) = \dfrac{n}{1+x^n} then limnfn(x)=0\lim_{n \rightarrow \infty} f_n(x) = 0 for all x(1,π]x \in (1, \pi]. I believe the fact it doesn't converge to 0 at one point isn't a problem with this (Lebesgue)

So by Lebesgue

limn1πfn(x)dx=1π0dx=0\displaystyle \lim_{n\to\infty} \int_1^{\pi} f_n(x)\,dx = \int_1^{\pi} 0\,dx = 0

Disclaimer: There's a very good chance this is wrong (I haven't covered Lebesgue :frown: )


Spoiler

Reply 64
I'm getting these from somewhere but I think some of these are beautiful problems which hopefully someone with C1-4 can do (except 18, which ironically might be the easiest). Hope you don't mind the influx!

Problem 17*

Let an{a_n} be an arithmetic sequence and gn{g_n} be a geometric sequence. The first four terms of sn=gn+an{s_n}={g_n + a_n} is 0, 0, 1 and 0.

Find s10s_{10}.

Problem 18**

Evaluate limx0(xxxxx)\lim_{x \rightarrow 0} (x^{x^x}-x^x)

Problem 19*

If x+y+z=0x+y+z=0, evaluate xy+yz+zxx2+y2+z2\frac{xy+yz+zx}{x^2+y^2+z^2}

Problems 20 and 21 are below. I told you I'd spam this thread :tongue:

Problem 22*

Evaluate 1(logxx)ndx\int^{\infty}_1 (\frac{\log x}{x})^n dx

Problem 23*

Let SS denote the set of triples (i,j,k)(i,j,k) such that i+j+k=ni+j+k=n.

Evaluate (i,j,k)Sijk\sum_{(i,j,k)\in S} ijk

(I wanted to edit this to restrict the choice of n, but actually it makes little difference. Do people need a hint?)
(edited 10 years ago)
Problem 20 **

By using a suitable substitution, or otherwise, show that

r2x2 dx=12(r2arctan(xr2x2)+xr2x2)+C\displaystyle \int \sqrt{r^2 - x^2} \ dx = \frac{1}{2}\left(r^2 \arctan \left(\frac{x}{\sqrt{r^2 - x^2}}\right) + x\sqrt{r^2 - x^2}\right) + C

Show further that the area of a circle, A, satisfies

A=πr2A=\pi r^2
(edited 10 years ago)
Problem 21 - */**

The uniqueness theorem of anti-derivatives states that, if f(x)=g(x) f’(x) = g’(x) , then f(x)=g(x)+c f(x) = g(x)+c .

By considering the derivatives of cos2x cos^2x and sin2x sin^2x , verify that cos2x+sin2x=1 cos^2x+sin^2x = 1 .

By considering suitable derivatives, prove the following identities:

i) lnxn=nlnx lnx^n = n lnx

ii) ln(f(x)g(x))=lnf(x)+lng(x) ln(f(x)g(x)) = lnf(x) + lng(x)

iii) lnx=logaxlogae lnx = \dfrac{log_ax}{log_ae} .

Deduce that the results in parts i) and ii) hold independently of the base of the logarithm.

Spoiler

(edited 10 years ago)
Reply 67
Solution 17

Let an=a+(n1)da_n=a+(n-1)d and gn=grn1g_n=gr^{n-1}. Then:

a+g=0[br]a+d+gr=0[br]a+2d+gr2=1[br]a+3d+gr3=0a+g=0[br]a+d+gr=0[br]a+2d+gr^2=1[br]a+3d+gr^3=0

Solving simultaneously, we obtain a=19a=-\frac{1}{9}, g=19g=\frac{1}{9}, d=13d=\frac{1}{3} and r=2r=-2.

Hence S10=19+(101)13+19(2)101=54S_{10}=-\frac{1}{9}+(10-1) \cdot\frac{1}{3}+\frac{1}{9} \cdot (-2)^{10-1}=-54.
(edited 10 years ago)
Reply 68
Solution 19

x+y+z=0(x+y+z)2=x2+y2+z2+2(xy+xz+yz)=0[br]xy+xz+yzx2+y2+z2=12x+y+z=0\Rightarrow (x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)=0[br]\Rightarrow \frac{xy+xz+yz}{x^2+y^2+z^2}=-\frac{1}{2}.
Reply 69
Original post by und
Solution 17

Let an=a+(n1)da_n=a+(n-1)d and gn=grn1g_n=gr^{n-1}. Then:

a+g=0[br]a+d+gr=0[br]a+2d+gr2=1[br]a+3d+gr3=0a+g=0[br]a+d+gr=0[br]a+2d+gr^2=1[br]a+3d+gr^3=0

Solving simultaneously, we obtain a=19a=-\frac{1}{9}, g=19g=\frac{1}{9}, d=13d=\frac{1}{3} and r=2r=-2.

Hence S100=19+(1001)13+19(2)1001=70425033346012744527594622488S_{100}=-\frac{1}{9}+(100-1) \cdot\frac{1}{3}+\frac{1}{9} \cdot (-2)^{100-1}=-70425033346012744527594622488.

Right, where have I gone wrong??


I didn't say the answer was pretty! But I meant s_10 :tongue:
(edited 10 years ago)
Reply 70
Original post by shamika


Problem 18**

Evaluate limx(xxxxx)\lim_{x \rightarrow \infty} (x^{x^x}-x^x)

Solution 18

Spoiler

Reply 71
Original post by Smaug123
Solution 18

Spoiler



I really need to stop posting when I'm exhausted - I've corrected the problem, sorry.
Reply 72
Original post by Lord of the Flies
Problem 15*/**


Evaluate limn1nn1+xndx\displaystyle \lim_{n\to\infty} \int_1^{n} \frac{n}{1+x^n}\,dx


What if the upper limit in the integral were fixed, say π?\pi?



Hi :smile:


Take the binomial expansion of n/(1+x^2)^-1 (far too tedious to write out all the working!)
Then simplify and integrate, eliminate the n terms, and you're left with 1 - 1/2 + 1/3 - 1/4... which is of course the taylor expansion of ln(2) :biggrin:


sorry I don't know how to use LaTex so it would be a lot of effort to type out the working, but it should be straightforward to follow the steps
Reply 73
Solution 22

Let Im=1(logx)mxndxI_m=\int^{\infty}_1 \frac{(logx)^m}{x^n} dx. Then by parts, Im=mn1Im1I_m=\frac{m}{n-1}I_{m-1}. Hence In=n!(n1)n[11nx1n]1=n!(n1)n+1I_n=\frac{n!}{(n-1)^n}\left[ \frac{1}{1-n}x^{1-n} \right]_1^{\infty}=\frac{n!}{(n-1)^{n+1}}
(edited 10 years ago)
Reply 74
Original post by und
Solution 19

x+y+z=0(x+y+z)2=x2+y2+z2+2(xy+xz+yz)=0[br]xy+xz+yzx2+y2+z2=12x+y+z=0\Rightarrow (x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)=0[br]\Rightarrow \frac{xy+xz+yz}{x^2+y^2+z^2}=-\frac{1}{2}.


:smile: yep!
Reply 75
Original post by shamika
I really need to stop posting when I'm exhausted - I've corrected the problem, sorry.

Heh, I did wonder; you did say it might be the easiest :P

Problem 18**

Evaluate limx0(xxxxx)\lim_{x \rightarrow 0} (x^{x^x}-x^x)

Solution 18

Spoiler

Reply 76
Original post by und
Solution 22

Let Im=1(logx)mxndxI_m=\int^{\infty}_1 \frac{(logx)^m}{x^n} dx. Then by parts, Im=mn1Im1I_m=\frac{m}{n-1}I_{m-1}. Hence In=n!(n1)n[11nx1n]1=n!(n1)n+1I_n=\frac{n!}{(n-1)^n}\left[ \frac{1}{1-n}x^{1-n} \right]_1^{\infty}=\frac{n!}{(n-1)^{n+1}}


Yep!

Original post by Smaug123
Heh, I did wonder; you did say it might be the easiest :P

Problem 18**

Evaluate limx0(xxxxx)\lim_{x \rightarrow 0} (x^{x^x}-x^x)

Solution 18

Spoiler



Yep!
Reply 77
Original post by und
Solution 17

Let an=a+(n1)da_n=a+(n-1)d and gn=grn1g_n=gr^{n-1}. Then:

a+g=0[br]a+d+gr=0[br]a+2d+gr2=1[br]a+3d+gr3=0a+g=0[br]a+d+gr=0[br]a+2d+gr^2=1[br]a+3d+gr^3=0

Solving simultaneously, we obtain a=19a=-\frac{1}{9}, g=19g=\frac{1}{9}, d=13d=\frac{1}{3} and r=2r=-2.

Hence S10=19+(101)13+19(2)101=54S_{10}=-\frac{1}{9}+(10-1) \cdot\frac{1}{3}+\frac{1}{9} \cdot (-2)^{10-1}=-54.


Yep. This isn't as nice as I'd tried to make it, but oh well, it was meant to entice some lurkers to solve something which is essentially C1/2 standard
Original post by _Izzy
...


Well done :wink: I'll type it out for you (credit to Izzy in the OP):

1nn1+xndx=1nnxn(11+(1/x)n)dx=1nnxn(11xn+1x2n)dx=n1n1xn1x2n+1x3ndx=n[1n112n1+]n[n1nn1n12n2n1+]0\displaystyle\begin{aligned}\int_1^n \frac{n}{1+x^n}\,dx &=\int_{1}^n \frac{n}{x^n}\left(\frac{1 }{1+(1/x)^n}\right)dx\\&=\int_1^n\frac{n}{x^n}\left(1-\frac{1}{x^n}+\frac{1}{x^{2n}}-\cdots\right)dx\\&=n\int_1^n \frac{1}{x^n}-\frac{1}{x^{2n}}+\frac{1}{x^{3n}}-\cdots\,dx\\&=n\left[\frac{1}{n-1}-\frac{1}{2n-1}+\cdots\right]-\underbrace{n\left[\frac{n^{1-n}}{n-1}-\frac{n^{1-2n}}{2n-1}+\cdots\right]}_{\to\,0}\end{aligned}

Hence the limit is ln2\ln 2

Replacing nn with π\pi obviously makes no difference.
(edited 10 years ago)
Reply 79
Solution 21

Let f(x)=sin2xf(x)=sin^{2}x and g(x)=cos2xg(x)=cos^{2}x. Then f(x)+g(x)=sin2xsin2x=0sin2x+cos2x=cf'(x)+g'(x)=sin2x-sin2x=0\Rightarrow sin^{2}x+cos^{2}x=c. Letting x=0x=0 gives c=1c=1 as required.

i) Let f(x)=ln(xn)f(x)=ln(x^n) and g(x)=nln(x)g(x)=nln(x). Then f(x)g(x)=nxn1xnnx=0ln(xn)nln(x)=cf'(x)-g'(x)=\frac{nx^{n-1}}{x^n}-\frac{n}{x}=0\Rightarrow ln(x^n)-nln(x)=c. Letting x=1x=1 gives c=0c=0 as required.

ii) Let p(x)=ln(f(x)g(x))p(x)=ln(f(x)g(x)) and q(x)=ln(f(x))+ln(g(x))q(x)=ln(f(x))+ln(g(x)). Then p(x)q(x)=f(x)g(x)+f(x)g(x)f(x)g(x)(f(x)f(x)+g(x)g(x))=0[br]ln(f(x)g(x))ln(f(x))+ln(g(x))=cp'(x)-q'(x)=\frac{f'(x)g(x)+f(x)g'(x)}{f(x)g(x)}-(\frac{f'(x)}{f(x)}+\frac{g'(x)}{g(x)})=0 [br]\Rightarrow ln(f(x)g(x))-ln(f(x))+ln(g(x))=c. Letting x=1x=1 gives c=0c=0 as required.

iii) Let f(x)=logxf(x)=logx and g(x)=logaxlogaeg(x)=\frac{log_a{x}}{log_{a}e}. Then f(x)g(x)=1x1xlogealogae=0lnxlogaxlogae=cf'(x)-g'(x)=\frac{1}{x}-\frac{1}{xlog_{e}alog_{a}e}=0 \Rightarrow lnx-\frac{log_a{x}}{log_{a}e}=c. Letting x=1x=1 gives c=0c=0 as required.

Any logarithm can be mapped onto the natural logarithm using a linear transformation, hence the results still hold.
(edited 10 years ago)

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