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Math Help - Three problems

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Three problems

    Here are three problems I have done on other websites recently, if anyone would care to give them a go, that would be great.

    Problem one

    Compute

    L=\lim_{n\to\infty}\int_{a}^{\infty}\frac{n}{1+n^2  x^2}dx


    Problem two

    Let a_1>a_2>a_3\cdots>a_n>0

    Furthermore let p_1>p_2>p_3\cdots>p_n

    and p_1+p_2+p_3+\cdots+{p_n}=1



    So then if F(x)=\left(p_1a_1^x+p_2a_2^x+p_3a_3^x+\cdots+p_na_  n^x\right)^{\frac{1}{x}}

    Compute

    L=\lim_{x\to{0^+}}F(x)

    L_=\lim_{x\to\infty}F(x)

    L_2=\lim_{x\to-\infty}F(x)


    Problem three


    Prove that if a>1 that

    0\leq\int_a^{\infty}\frac{\sin(x)}{x}\leq\int_0^{\  infty}\frac{\sin(x)}{x}~dx


    Note

    If you have seen my solution, please don't duplicate it, for that wouldn't be fun. Plus, I always want to see other points of view!
    Last edited by Mathstud28; July 15th 2008 at 09:16 AM.
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  2. #2
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    I'll attempt problem 1.

    L=\lim_{n\to\infty}\int_{a}^{\infty}\frac{n}{1+n^2  x^2}dx

    (1) I divide the numerator and denominator by n which yields
     \int_{a}^{\infty} \frac{1}{1/n+nx^2}dx
    (2) There's a formula if a>0, then
     \int \frac{1}{x^2+a^2}dx = \frac{1}{a} tan^{-1} \frac{x}{a} + C
    (3) Since ( \sqrt{n}x)^2=nx^{2} from (1), I use u-substitution. I let u = \sqrt{n}x
    (4) By u substitution the integral becomes
    \int \frac{1}{1/u^2+u^2}du . Edit:No, this is not correct. Scratch this.
    (5) Applying the formula (2) with (4) with get tan^{-1}(nx)+C
    (6) As for the limit, \lim_{n\to\infty} tan^{-1}(nx) = \frac{\pi}{2}

    For problem 3, I believe there's a typo.
    0\leq\int_a^{\infty}\frac{\sin(x)}{x}\leq\int_0^{\  infty}~dx\frac{\sin(x)}{x}~dxshould it be
    0 \leq \int_a^{\infty} \frac{\sin(x)}{x}~dx \leq \int_0^{\infty}\frac{\sin(x)}{x}~dx?
    I'm sure I have to use the Squeezing Principle for this. I can't seem to integrate sin(x)/x in terms of elementary functions. I'm guessing I can integrate this using the Taylor series for the sine function, but I have no idea after that.
    Last edited by Paperwings; July 15th 2008 at 08:26 AM.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Paperwings View Post
    I'll attempt problem 1.

    L=\lim_{n\to\infty}\int_{a}^{\infty}\frac{n}{1+n^2  x^2}dx

    (1) I divide the numerator and denominator by n which yields
     \int_{a}^{\infty} \frac{1}{1/n+nx^2}dx
    (2) There's a formula if a>0, then
     \int \frac{1}{x^2+a^2}dx = \frac{1}{a} tan^{-1} \frac{x}{a} + C
    (3) Since ( \sqrt{n}x)^2=nx^{2} from (1), I use u-substitution. I let u = \sqrt{n}x
    (4) By u substitution the integral becomes
    \int \frac{1}{1/u^2+u^2}du . Edit:No, this is not correct. Scratch this.
    (5) Applying the formula (2) with (4) with get tan^{-1}(nx)+C
    (6) As for the limit, \lim_{n\to\infty} tan^{-1}(nx) = \frac{\pi}{2}

    For problem 3, I believe there's a typo.
    0\leq\int_a^{\infty}\frac{\sin(x)}{x}\leq\int_0^{\  infty}~dx\frac{\sin(x)}{x}~dxshould it be
    0 \leq \int_a^{\infty} \frac{\sin(x)}{x}~dx \leq \int_0^{\infty}\frac{\sin(x)}{x}~dx?
    I'm sure I have to use the Squeezing Principle for this. I can't seem to integrate sin(x)/x in terms of elementary functions. I'm guessing I can integrate this using the Taylor series for the sine function, but I have no idea after that.
    Your work for the first one is hard to follow (consider using white space better), but assuming you did do it partly correct (which no offense, I do no think you did), you only got a third of the answer.


    As for the second one, yes thank you for the typo alert.


    And \frac{\sin(x)}{x}=\frac{\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}}{x}=\sum_{n=0}^{\infty}\fra  c{(-1)^nx^{2n}}{(2n+1)!}

    So

    \int\frac{\sin(x)}{x}~dx=\int\sum_{n=0}^{\infty}\f  rac{(-1)^nx^{2n}}{(2n+1)!}~dx=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}+C

    But that is actually not how you have to do this problem, consider that it is a definite integral, and consider the ramifications of the methods able to be applied now.
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  4. #4
    Lord of certain Rings
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    Quote Originally Posted by Mathstud28 View Post
    Here are three problems I have done on other websites recently, if anyone would care to give them a go, that would be great.

    Problem one

    Compute

    L=\lim_{n\to\infty}\int_{a}^{\infty}\frac{n}{1+n^2  x^2}dx
    L=\lim_{n\to\infty}\int_{a}^{\infty}\frac{n}{1+n^2  x^2}dx = \int_{a}^{\infty}\lim_{n\to\infty}\dfrac{\frac1{n}  }{\frac1{n^2}+x^2}dx = \int_{a}^{\infty}\dfrac{0}{0+x^2}dx = 0

    Another Way:

    L=\lim_{n\to\infty}\int_{a}^{\infty}\frac{n}{1+n^2  x^2}dx =\lim_{n\to\infty}(\tan^{-1} (nx)) \bigg{|}_{a}^{\infty} = \lim_{n\to\infty}\left(\frac{\pi}2 -\tan^{-1} na\right) = \frac{\pi}2 - \lim_{n\to\infty}\tan^{-1} na = \frac{\pi}2 - \frac{\pi}2 = 0
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Isomorphism View Post
    L=\lim_{n\to\infty}\int_{a}^{\infty}\frac{n}{1+n^2  x^2}dx = \int_{a}^{\infty}\lim_{n\to\infty}\dfrac{\frac1{n}  }{\frac1{n^2}+x^2}dx = \int_{a}^{\infty}\dfrac{0}{0+x^2}dx = 0

    Another Way:

    L=\lim_{n\to\infty}\int_{a}^{\infty}\frac{n}{1+n^2  x^2}dx =\lim_{n\to\infty}(\tan^{-1} (nx)) \bigg{|}_{a}^{\infty} = \lim_{n\to\infty}\left(\frac{\pi}2 -\tan^{-1} na\right) = \frac{\pi}2 - \lim_{n\to\infty}\tan^{-1} na = \frac{\pi}2 - \frac{\pi}2{\color{red}\leftarrow\text{here}} = 0
    That is only one-thrid of the solution. Sorry, would you like me to post my solution? Or Pm it to you? I'll give you a hint, look above
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  6. #6
    Lord of certain Rings
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    Quote Originally Posted by Mathstud28 View Post
    Problem two

    Let a_1>a_2>a_3\cdots>a_n>0

    Furthermore let p_1>p_2>p_3\cdots>p_n

    and p_1+p_2+p_3+\cdots+{p_n}=1



    So then if F(x)=\left(p_1a_1^x+p_2a_2^x+p_3a_3^x+\cdots+p_na_  n^x\right)^{\frac{1}{x}}

    Compute

    L=\lim_{x\to{0^+}}F(x)

    L_=\lim_{x\to\infty}F(x)

    L_2=\lim_{x\to-\infty}F(x)
    I will post my solution if the answers are right..

    L=\lim_{x\to{0^+}}F(x) = a_1^{p_1}a_2 ^{p_2} \cdots a_n^{p_n}

    L_=\lim_{x\to\infty}F(x) = a_1

    L_2=\lim_{x\to-\infty}F(x) = a_n


    Quote Originally Posted by Mathstud28 View Post
    That is only one-thrid of the solution. Sorry, would you like me to post my solution? Or Pm it to you? I'll give you a hint, look above
    Hmm... you mean the limit is not 0?
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Isomorphism View Post
    I will post my solution if the answers are right..

    L=\lim_{x\to{0^+}}F(x) = a_1^{p_1}a_2 ^{p_2} \cdots a_n^{p_n}

    L_=\lim_{x\to\infty}F(x) = a_1

    L_2=\lim_{x\to-\infty}F(x) = a_n




    Hmm... you mean the limit is not 0?
    Here is a huge hint, your answer should look like this

    L=\left\{\begin{array}{c}?~~\text{if}~~a?\\?~~\tex  t{if}~~a?\\?~~\text{if}~~a?\end{array}\right\}


    Even bigger hint in white
    arctan is odd
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  8. #8
    Lord of certain Rings
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    Quote Originally Posted by Mathstud28 View Post






    Here is a huge hint, your answer should look like this

    L=\left\{\begin{array}{c}?~~\text{if}~~a?\\?~~\tex  t{if}~~a?\\?~~\text{if}~~a?\end{array}\right\}


    Even bigger hint in white
    arctan is odd
    I understand what you are trying to imply but I was wondering whats wrong with the first method

    Cant I interchange the limit and the integral there?
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Isomorphism View Post
    I understand what you are trying to imply but I was wondering whats wrong with the first method

    Cant I interchange the limit and the integral there?
    It leaves out two answers first off, and secondly no, I do not think we can since we do not know a.

    Because don't you have to have that

    \lim_{n\to\infty}\frac{n}{1+n^2x^2}\quad\forall{x}  \in(a,\infty)?

    What if 0\in(a,\infty)

    Then \lim_{n\to\infty}\frac{n}{1+n^2x^2}\ne{0}\quad\for  all{x}\in(a,\infty)

    That is just one complication I see.


    So I do not think an interchanging of limit and integral is permissable here.
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Here are three problems I have done on other websites recently, if anyone would care to give them a go, that would be great.

    Problem one

    Compute

    L=\lim_{n\to\infty}\int_{a}^{\infty}\frac{n}{1+n^2  x^2}dx


    Problem two

    Let a_1>a_2>a_3\cdots>a_n>0

    Furthermore let p_1>p_2>p_3\cdots>p_n

    and p_1+p_2+p_3+\cdots+{p_n}=1



    So then if F(x)=\left(p_1a_1^x+p_2a_2^x+p_3a_3^x+\cdots+p_na_  n^x\right)^{\frac{1}{x}}

    Compute

    L=\lim_{x\to{0^+}}F(x)

    L_=\lim_{x\to\infty}F(x)

    L_2=\lim_{x\to-\infty}F(x)


    Problem three


    Prove that if a>1 that

    0\leq\int_a^{\infty}\frac{\sin(x)}{x}\leq\int_0^{\  infty}\frac{\sin(x)}{x}~dx


    Note

    If you have seen my solution, please don't duplicate it, for that wouldn't be fun. Plus, I always want to see other points of view!

    Since no one is biting I will solve the first and give hints on the others

    One

    L=\lim_{n\to\infty}\int_0^{\infty}\frac{n}{1+n^2x^  2}~dx

    So if we let

    nx=\varphi\Rightarrow\frac{d\varphi}{n}=dx

    and when x=a\Rightarrow{\varphi=na}

    and as x\to\infty\Rightarrow\varphi\to\infty

    So we have

    L=\lim_{n\to\infty}n\cdot\int_{na}^{\infty}\frac{\  frac{d\varphi}{n}}{1+\varphi^2}

    =\lim_{n\to\infty}n\cdot\frac{1}{n}\int_0^{\infty}  \frac{d\varphi}{1+\varphi^2}

    =\lim_{n\to\infty}\arctan\left(\varphi\right)\bigg  |_{na}^{\infty}

    =\lim_{n\to\infty}\bigg[\frac{\pi}{2}-\arctan(na)\bigg]

    \text{If }a>0\Rightarrow\lim_{n\to\infty}\bigg[\frac{\pi}{2}-\arctan(na)\bigg]

    =\frac{\pi}{2}-\frac{\pi}{2}=0

    \text{If }a=0\Rightarrow\lim_{n\to\infty}\bigg[\frac{\pi}{2}-\arctan(na)\bigg]

    =\frac{\pi}{2}-0=\frac{\pi}{2}

    \text{If }a<0\Rightarrow\lim_{n\to\infty}\bigg[\frac{\pi}{2}-\arctan(na)\bigg]

    =\frac{\pi}{2}\underbrace{+}_{\text{oddness of arctan}}\frac{\pi}{2}=\pi


    \therefore\quad\boxed{L= \left\{ \begin{array}{rcl}<br />
0 & \mbox{if} & a>0 \\ \frac{\pi}{2} & \mbox{if} & a=0 \\ \pi & \mbox{if} & a<0<br />
\end{array}\right.}\quad\blacksquare

    Two

    \to{0} Use L'hops

    \to\infty and \to-\infty Use a sub

    Three

    This ones a given
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  11. #11
    Moo
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    Hello,

    Similarly to the first one, compute :

    \lim_{n \to \infty} \int_a^\infty \frac{n+\sin(x)}{1+n^2x^2} ~ dx
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    Similarly to the first one, compute :

    \lim_{n \to \infty} \int_a^\infty \frac{n+\sin(x)}{1+n^2x^2} ~ dx
    Is this going the right way? Let a>1. Since \frac{n+\sin(x)}{1+(nx)^2} converges pointwise to 0 and \left|\frac{n+\sin(x)}{1+(nx)^2}\right|\leqslant\f  rac{1}{1+(nx)^2} is integrable we can conclude by LDCT that \lim \int_a^{\infty}\frac{n+\sin(x)}{1+(nx)^2}=0.

    Sorry if that is incorrect
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  13. #13
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    Quote Originally Posted by Moo View Post
    Hello,

    Similarly to the first one, compute : \lim_{n \to \infty} \int_a^\infty \frac{n+\sin(x)}{1+n^2x^2} ~ dx
    or more generally if f(x) is bounded, say |f(x)| \leq M, and a>0, then: \lim_{n\to\infty}\int_a^{\infty} \frac{n+f(x)}{1+n^2x^2} \ dx =0, because: 0 \leq \left|\int_a^{\infty} \frac{n+f(x)}{1+n^2x^2} \ dx \right| \leq \int_a^{\infty} \frac{n + M}{1+n^2x^2} \ dx=\frac{n+M}{n} \left[\frac{\pi}{2}-\tan^{-1}(na) \right].

    now apply the squeeze theorem to finish the proof. i think the case a \leq 0 is more intersting. you guys might be intersted in discussing it?
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  14. #14
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    or more generally if f(x) is bounded, say |f(x)| \leq M, and a>0, then: \lim_{n\to\infty}\int_a^{\infty} \frac{n+f(x)}{1+n^2x^2} \ dx =0, because: 0 \leq \left|\int_a^{\infty} \frac{n+f(x)}{1+n^2x^2} \ dx \right| \leq \int_a^{\infty} \frac{n + M}{1+n^2x^2} \ dx{\color{red}(*)}=\frac{n+M}{n} \left[\frac{\pi}{2}-\tan^{-1}(na) \right].

    now apply the squeeze theorem to finish the proof. i think the case a \leq 0 is more intersting. you guys might be intersted in discussing it?
    I believe that we can conclude the same result. Noting that at \color{red}(*) we have eveness...I think we can conjecture that based on the eveness we must just show that \lim\int_a^{b}\frac{n+f(x)}{1+(nx)^2}dx converges for all finite a,b. Am I missing something or is this trivial?
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    ok, actually it's not as interesting as i thought! the answer is \frac{\pi}{2} if a=0 and \pi if a <0. here's why: we have \lim_{n\to\infty} \int_a^{\infty} \frac{n}{1+n^2x^2} \ dx=\frac{\pi}{2} - \lim_{n\to\infty} \tan^{-1}(na)=\begin{cases} 0 & \text{if} \ \ a >0 \\ \frac{\pi}{2} & \text{if} \ \ a = 0 \\ \pi & \text{if} \ \ a < 0 \end{cases}.

    on the other hand: 0 \leq \left|\int_a^{\infty} \frac{f(x)}{1+n^2x^2} \ dx \right| \leq M \int_a^{\infty} \frac{dx}{1+n^2x^2}=\frac{M}{n}\left[\frac{\pi}{2}- \tan^{-1}(na) \right]. thus by the squeeze theorem: \lim_{n\to\infty} \int_a^{\infty} \frac{f(x)}{1+n^2x^2} \ dx = 0, \ \forall \ a \in \mathbb{R}, and we're done!
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