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The Proof is Trivial!

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Original post by joostan
Problem 72: *

Bit of an effort: - But hey ho.
Prove by considering:  I=01x4(1x4)1+x2 dx\ I =\displaystyle\int^1_0 \dfrac{x^4(1 - x^4)}{1+x^2}\ dx that π<227\pi < \frac{22}{7}

Find  I=1+ex dx\ I = \displaystyle\int \sqrt{1+e^x}\ dx


Solution 72


After finding that

x4(1x)41+x2 dx=x5+4x+x772x634x334arctanx+C\displaystyle\int \dfrac{x^4(1 - x)^4}{1+x^2}\ dx = x^5 + 4x +\frac{x^7}{7} - \frac{2x^6}{3} - \frac{4x^3}{3} - 4\arctan x + C

we get that

01x4(1x)41+x2 dx=227π\displaystyle\int^1_0 \dfrac{x^4(1 - x)^4}{1+x^2}\ dx = \frac{22}{7} - \pi

x(0,1),x4(1x)41+x2>0\forall x \in (0,1) , \dfrac{x^4(1-x)^4}{1+x^2} > 0

so

227π>0π<227\frac{22}{7} - \pi > 0 \Rightarrow \pi < \frac{22}{7}

(the area must be above the x axis, and certainly positive).

For the next part, we have

I=1+ex dxI=\displaystyle \int \sqrt{1+e^x} \ dx

Use

u=ex+1u = \sqrt{e^x + 1}

to get

I=2u2(u1)(u+1) duI=\displaystyle \int \dfrac{2u^2}{(u-1)(u+1)} \ du

Using partial fractions:

I=2+1u11u+1 du=2uln(u1)ln(u+1)+CI = \displaystyle \int 2 + \dfrac{1}{u-1} - \dfrac{1}{u+1} \ du = 2u - \ln(u-1) - \ln(u+1) + C

Substituting back gives

I=21+exln(1+ex1)ln(ex+1+1)+KI = 2\sqrt{1+e^x} - \ln(\sqrt{1+e^x} - 1) - \ln(\sqrt{e^x + 1} + 1) + K

Simplifying further:

I=2(1+extanh1(1+ex))+KI = 2\left(\sqrt{1+e^x} - \tanh^{-1} (\sqrt{1+e^x})\right) + K
(edited 10 years ago)
Reply 541
LoL OMG how did I end up here? Still at least I know it is possible to do equations on computers.
Can computers give you answers to those equations????
*probably a stupid question*
Reply 542
Solution 26

Notice that H1×H2={(g1,g2)G1×G2(g1,e)K,(e,g2)K}H_{1} \times H_{2} = \{(g_{1},g_{2}) \in G_{1} \times G_{2}| (g_{1},e) \in K, (e,g_{2}) \in K\}
As KK is a group, it follows that H1G1H_{1} \le G_{1}, H2G2H_{2} \le G_{2}
Moreover, if (g1,g2)H1×H2(g_{1},g_{2}) \in H_{1} \times H_{2}, then, obviously, (g1,g2)K(g_{1},g_{2}) \in K. It is easy to see that (g11,e)(g_{1}^{-1},e) and (e,g21)(e,g_{2}^{-1}) are in KK, and therefore (g11,g21)H1×H2(g_{1}^{-1},g_{2}^{-1})\in H_{1} \times H_{2}. Furthermore, (g1,g2),(g3,g4)H1×H2(g_{1},g_{2}), (g_{3},g_{4}) \in H_{1} \times H_{2} implies (g1g3,g2g4)H1×H2(g_{1}g_{3},g_{2}g_{4}) \in H_{1} \times H_{2}. Trivially, (e,e)H1×H2(e,e) \in H_{1} \times H_{2}. We can conclude that H1×H2H_{1} \times H_{2} is a subgroup of KK.
Suppose that there is at least one element (g1,g2)K(g_{1},g_{2}) \in K, which is not in H1×H2H_{1} \times H_{2}.
The numbers card(G1)card (G_{1}) and card(G2)card(G_{2}) are relatively prime. Consequently, we can choose n1,n2n_{1}, n_{2} such that n1card(G1)1(modcard(G2))n_{1}card(G_{1}) \equiv 1 \pmod {card(G_{2})} and n2card(G2)1(modcard(G1))n_{2}card(G_{2}) \equiv 1 \pmod {card(G_{1})}.
Then, (g1n1card(G1),g2n1card(G1))=(e,g2)K(g_{1}^{n_{1}card(G_{1})},g_{2}^{n_{1}card(G_{1})})= (e,g_{2}) \in K and (g1n2card(G2),g2n2card(G2))=(g1,e)K(g_{1}^{n_{2}card(G_{2})},g_{2}^{n_{2}card(G_{2})})= (g_{1},e) \in K. Hence, (g1,g2)H1×H2(g_{1},g_{2}) \in H_{1} \times H_{2}, which is clearly a contradiction.
Reply 543
Original post by Mladenov
You mean x4(1x)41+x2\displaystyle \frac{x^{4}(1-x)^{4}}{1+x^{2}}, don't you?


good spot :tongue:
Reply 544
Original post by Indeterminate
Solution 72


After finding that

x4(1x)41+x2 dx=x5+4x+x772x634x334arctanx+C\displaystyle\int \dfrac{x^4(1 - x)^4}{1+x^2}\ dx = x^5 + 4x +\frac{x^7}{7} - \frac{2x^6}{3} - \frac{4x^3}{3} - 4\arctan x + C

we get that

01x4(1x)41+x2 dx=227π\displaystyle\int^1_0 \dfrac{x^4(1 - x)^4}{1+x^2}\ dx = \frac{22}{7} - \pi

x(0,1),x4(1x)41+x2>0\forall x \in (0,1) , \dfrac{x^4(1-x)^4}{1+x^2} > 0

so

227π>0π<227\frac{22}{7} - \pi > 0 \Rightarrow \pi < \frac{22}{7}

(the area must be above the x axis, and certainly positive).

For the next part, we have

I=1+ex dxI=\displaystyle \int \sqrt{1+e^x} \ dx

Use

u=ex+1u = \sqrt{e^x + 1}

to get

I=2u2(u1)(u+1) duI=\displaystyle \int \dfrac{2u^2}{(u-1)(u+1)} \ du

Using partial fractions:

I=2+1u11u+1 du=2uln(u1)ln(u+1)+CI = \displaystyle \int 2 + \dfrac{1}{u-1} - \dfrac{1}{u+1} \ du = 2u - \ln(u-1) - \ln(u+1) + C

Substituting back gives

I=21+exln(1+ex1)ln(ex+1+1)+KI = 2\sqrt{1+e^x} - \ln(\sqrt{1+e^x} - 1) - \ln(\sqrt{e^x + 1} + 1) + K


This simplfies a bit more I believe :smile:
Reply 545
(You'll probably all find this easy, but perhaps its a good one to make you think about initial approach. Plus i made it myself!)

Problem 73*


i) Find the number of ordered sets of positive integers x1,x2,x3,...,xn{x_{1}, x_{2}, x_{3}, ... , x_{n}} such that i=1nxii=1nxi=n1\displaystyle\sum_{i=1}^n x_{i} - \displaystyle\prod_{i=1}^n x_{i} = n-1 and xim x_{i} \leq m where m is a positive integer such that mn m \geq n . Justify your answer.
ii) Extend your result to include all ordered sets of non-negative integers.
(edited 10 years ago)
Original post by joostan
This simplfies a bit more I believe :smile:


If you insist, I'll edit it in :smile:
Original post by metaltron
Is that the simplest it can be? It looks like you can get to:

Spoiler



Surely

2tanh1(1+ex)=xln(1+1+ex11+ex)=x2\tanh^{-1} (\sqrt{1+e^x}) = x \Rightarrow \ln\left(\frac{1+\sqrt{1+e^x}}{1-\sqrt{1+e^x}}\right) = x

which isn't true in general.
Original post by Indeterminate
Surely

2tanh1(1+ex)=xln(1+1+ex11+ex)=x2\tanh^{-1} (\sqrt{1+e^x}) = x \Rightarrow \ln\left(\frac{1+\sqrt{1+e^x}}{1-\sqrt{1+e^x}}\right) = x

which isn't true in general.


No i deleted it, as I considered how inadequate my knowledge of the subject was having not looked at hyperboilc functions. It just seems to come straight from the line above that, but I'm probably missing something. How did you manage to reply to that, i thought I deleted it in the nick of time.
Reply 549
Original post by metaltron
No i deleted it, as I considered how inadequate my knowledge of the subject was having not looked at hyperboilc functions. It just seems to come straight from the line above that, but I'm probably missing something. How did you manage to reply to that, i thought I deleted it in the nick of time.


If you have a message quoted, I think it posts it regardless of whether the other person deletes it. When you delete a message, it only really removes it from view for the majority of the members of the forum, but it still has a unique post code and can still be quoted - so if someone quotes you before you delete it, it will have all the text you posted as well.
Original post by Noble.
If you have a message quoted, I think it posts it regardless of whether the other person deletes it. When you delete a message, it only really removes it from view for the majority of the members of the forum, but it still has a unique post code and can still be quoted - so if someone quotes you before you delete it, it will have all the text you posted as well.


Yeah, I know I've been humiliated many a time.
Reply 551
Problem 74**

Let a,b,ca,b,c be positive real numbers. Then we have the following inequality: a(b+c)2+b(c+a)2+c(a+b)2+2(ab+bc+ca)(a+b)(b+c)(c+a)3(a+b+c)2(ab+bc+ca)\displaystyle \frac{a}{(b+c)^{2}}+ \frac{b}{(c+a)^{2}}+ \frac{c}{(a+b)^{2}} + \frac{2(ab+bc+ca)}{(a+b)(b+c)(c+a)} \ge \frac{3(a+b+c)}{2(ab+bc+ca)}.

Spoiler

Reply 552
Original post by Mladenov
Problem 74**

Let a,b,ca,b,c be positive real numbers. Then we have the following inequality: a(b+c)2+b(c+a)2+c(a+b)2+2(ab+bc+ca)(a+b)(b+c)(c+a)3(a+b+c)2(ab+bc+ca)\displaystyle \frac{a}{(b+c)^{2}}+ \frac{b}{(c+a)^{2}}+ \frac{c}{(a+b)^{2}} + \frac{2(ab+bc+ca)}{(a+b)(b+c)(c+a)} \ge \frac{3(a+b+c)}{2(ab+bc+ca)}.

Spoiler

Spoiler



Noticing that either side on the equation is symmetrical in terms of the interchangeability of a, b and c, we can assume, W.L.O.G, that 0abc 0 \leq a \leq b \leq c (*).

Simplifying we get: LHS= a(b+c)2+b(c+a)2+c(a+b)2+a(a+b)(c+a)+b(a+b)(b+c)+c(b+c)(c+a)\displaystyle \frac{a}{(b+c)^{2}}+ \frac{b}{(c+a)^{2}}+ \frac{c}{(a+b)^{2}} + \frac{a}{(a+b)(c+a)}+\frac{b}{(a+b)(b+c)}+\frac{c}{(b+c)(c+a)}

As each term is positive and clearly the denominator of each term in less than or equal to (c+c)2=4c2(c+c)^2=4c^2 by (*).
Simplifying we get... LHS(a+b+c)2c2 \geq \frac{(a+b+c)}{2c^2} for ALL a, b and c that satisfy (*).
The maximum value of this expression occurs when a=b=c (i.e: a+b+c=3c) so we get LHS32c=32a \geq \frac{3}{2c} = \frac{3}{2a} (by the fact a=b=c had to be assumed in order to get the maximum).

Using the same technique, RHS3(3c)2(3a2)=3c2a2 \leq \frac{3(3c)}{2(3a^2)} =\frac{3c}{2a^2} . This clearly occurs when a=b=c also so RHS32a \leq \frac{3}{2a}

Therefore LHS32a \geq \frac{3}{2a} \geq RHS.
The result follows.
(edited 10 years ago)
Reply 553
Original post by Jkn
...


We have 32aLHSa+b+c2a2\displaystyle \frac{3}{2a} \le LHS \le \frac{a+b+c}{2a^{2}} for all a,b,ca,b,c. Hence 32aLHS32a\frac{3}{2a} \le LHS \le \frac{3}{2a} and equality occurs only when a=b=ca=b=c?!

Edit: I mean, what makes you believe that LHS32a\displaystyle LHS \ge \frac{3}{2a} is always true? You have LHSa+b+c2c2\displaystyle LHS \ge \frac{a+b+c}{2c^{2}} for all a,b,ca,b,c, but does this imply LHSmax(a+b+c2c2)\displaystyle LHS \ge \max(\frac{a+b+c}{2c^{2}}) for all a,b,ca,b,c? Bear in mind that LHSLHS changes as you change a,b,ca,b,c.
(edited 10 years ago)
Reply 554
Original post by Mladenov
We have 32aLHSa+b+c2a2\displaystyle \frac{3}{2a} \le LHS \le \frac{a+b+c}{2a^{2}} for all a,b,ca,b,c. Hence 32aLHS32a\frac{3}{2a} \le LHS \le \frac{3}{2a} and equality occurs only when a=b=ca=b=c?!

I haven't got an upper bound for the LHS (I said it was greater that a+b+c2a2 \frac{a+b+c}{2a^2} I think you misread!)

Edit: I mean, what makes you believe that LHS32a\displaystyle LHS \ge \frac{3}{2a} is always true? You have LHSa+b+c2c2\displaystyle LHS \ge \frac{a+b+c}{2c^{2}} for all a,b,ca,b,c, but does this imply LHSmax(a+b+c2c2)\displaystyle LHS \ge \max(\frac{a+b+c}{2c^{2}}) for all a,b,ca,b,c? Bear in mind that LHSLHS changes as you change a,b,ca,b,c.

Hmm I see your point. Well, short of using partial differentiation to justify this is a minimum, would it be sufficient to note that...
3c2a2\frac{3c}{2a^2} \geq LHS 3a2c2 \geq \frac{3a}{2c^2} by (*) and note that as a approaches c, each inequality approaches equality and hence demonstrate that the minimum value occurs at a=b=c? (if this works then the same could be easily done for the RHS). Cheers for the feedback btw, I wasn't sure it worked but I thought I'd give it go!
(edited 10 years ago)
Reply 555
Original post by Jkn
I haven't got an upper bound for the LHS (I said it was greater that a+b+c2a2 \frac{a+b+c}{2a^2} I think you misread!)

I gave you the upper bound just to make you think that you need to justify your inequalities.


Original post by Jkn
Hmm I see your point. Well, short of using partial differentiation to justify this is a minimum, would it be sufficient to note that...
3c2a2\frac{3c}{2a^2} \leq LHS 3a2c2 \geq \frac{3a}{2c^2} by (*) and note that as a approaches c, each inequality approaches equality and hence demonstrate that the minimum value occurs at a=b=c? (if this works then the same could be easily done for the RHS). Cheers for the feedback btw, I wasn't sure it worked but I thought I'd give it go!


Your argument proves only that LHS=32aLHS=\frac{3}{2a}, when a=b=ca=b=c. Also, should it not be 3c2a2LHS\frac{3c}{2a^{2}} \ge LHS?

If we denote f(a,b,c)=LHSf(a,b,c) = LHS, then we have to prove that there are no a,b,ca,b,c such that 32a>f(a,b,c)a+b+c2c2\displaystyle \frac{3}{2a} > f(a,b,c) \ge \frac{a+b+c}{2c^{2}}; in other words, we have to prove that ff reaches its minimum when a=b=ca=b=c.

Partial differentiation would work, but it is quite tough task.
(edited 10 years ago)
Reply 556
Original post by Mladenov
I gave you the upper bound just to make you think that you need to justify your inequalities.

Your argument proves only that LHS=32aLHS=\frac{3}{2a}, when a=b=ca=b=c. Also, should it not be 3c2a2LHS\frac{3c}{2a^{2}} \ge LHS?

Sorry, I'll change that now

If we denote f(a,b,c)=LHSf(a,b,c) = LHS, then we have to prove that there are no a,b,ca,b,c such that 32a>f(a,b,c)a+b+c2c2\displaystyle \frac{3}{2a} > f(a,b,c) \ge \frac{a+b+c}{2c^{2}}; in other words, we have to prove that ff reaches its minimum when a=b=ca=b=c.

Partial differentiation would work, but it is quite tough task.

Well using partial derivatives on the 6-term expression with respect to a, b and c in turn we get that...
LHSa,LHSb,LHSc<0 \frac{\partial LHS}{\partial a} , \frac{\partial LHS}{\partial b} , \frac{\partial LHS}{\partial c} < 0 a,b,c>0\forall a,b,c>0 which implies f(a,b,c) approaches a stationary point as all three variables coalesce to zero. Choosing a few values above and below shows you that this is a minimum point and hence, the minima occurs when and only when a=b=c.
(edited 10 years ago)
Reply 557
Original post by Jkn

Well using partial derivatives on the 6-term expression with respect to a, b and c in turn we get that...
LHSa,LHSb,LHSc<0 \frac{\partial LHS}{\partial a} , \frac{\partial LHS}{\partial b} , \frac{\partial LHS}{\partial c} < 0 a,b,c>0\forall a,b,c>0 which implies f(a,b,c) approaches a stationary point as all three variables coalesce to zero. Choosing a few values above and below shows you that this is a minimum point and hence, the minima occurs when and only when a=b=c.


You are right, the points, at which the partial derivatives fail to exist, are good candidates for extremums. Yet, this does not help us at all.
I tried several methods to prove your inequalities, including some brute-force techniques such as Lagrange Multipliers, but nothing works. Also, when a=1,b=7,c=8a=1,b=7,c=8, we have f(1,7,8)<32=f(1,1,1)f(1,7,8)< \frac{3}{2} = f(1,1,1). Hence f(a,b,c)32af(a,b,c) \ge \frac{3}{2a} for all a,b,c>0a,b,c > 0 is clearly not true. I have not tried f(a,b,c)32cf(a,b,c) \ge \frac{3}{2c}, but even if it is true, then you have to prove that RHS32cRHS \le \frac{3}{2c} is order your solution to work.

The problem is that you are doing the following:
Say you have correctly proven that for each xSx \in S, f(x)g(x)f(x) \ge g(x). Then, you are claiming that f(x)max(g(x)xS)f(x) \ge \max(g(x)_{x\in S}), which is generally not true.

Spoiler

Reply 558
Original post by Mladenov

Spoiler



Which ones? I know cauchy-schwarz and AM-GM inequalities, is that enough?
Reply 559
Original post by james22
Which ones? I know cauchy-schwarz and AM-GM inequalities, is that enough?


Precisely. There is a solution using exactly these two inequalities.

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