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Math Help - Some Revision Help

  1. #1
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    Some Revision Help

    Hi

    I got a calculus exam soon and during revision I noticed that some ofthe solutions were unclear to me. Could you please explain what is happening on areas marked red such as what rules are used.

    1. Find the derivative of the following functions


    d/dy Artanh (arctan (y^2))
    = (Artanh)' (arctan (y^2)) * d/dy * arctan(y^2)
    = [1/(1 - (arctan(y^2))^2] * [(2y)/(1 + y^4)]

    I do not understand how did the 3rd line was made, I know that:
    arctan dfferentiated is 1/(1+x^2). I presume the numerator of the 2nd part is 1 + y^2 and denom' is ((1 + y^2)^2)

    2. Prove that the function x -> 1/x is an asymptotics to x -> x/[(1 + x^4)^(1/2)] for

    x -> infin {x/[(1 + x^4)^(1/2)] } / (1/x) = x^2 / [(1 + x^4)^(1/2)]

    and

    = Lim x -> infin x^2 / [(1 + x^4)^(1/2)]
    = Lim x -> infin x^2 / [(1 + [1/x^4])^(1/2)] = 1

    At this stage I presumed you divide all the x's by the highest power of x which seems to be x^2 but the denominator does not show that it seems.

    3. Give a short argument based on symmetry considerations that

    Integ (lims of 2(pi) by -(pi)) : (x + (sin x)(cos^4x))dx =
    Integ (lims of (pi) by (pi)) : (x + (sin x)(cos^4x))dx

    Integ (lims of 2(pi) by -(pi)) : (x + (sin x)(cos^4x))dx +Integ (lims of (pi) by (pi)) : (x + (sin x)(cos^4x))dx

    First part is equal to zero since this is odd. How do you confirm if the function is odd or even?


    4. When is the Sandwich (Squish) Rule is used? Is it just for limits only?

    5. Use comparison test to determine the convergence or divergence of

    {Sum of}_infin(n=1) [1/( n + (n)^(1/2))]

    Quite a few subjects im doing OK on just these seems unclear to me any help would be great.

    Thanks

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  2. #2
    MHF Contributor Reckoner's Avatar
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    Hello, NeloAngelo! A fan of Devil May Cry, I take it?

    Quote Originally Posted by NeloAngelo View Post
    1. Find the derivative of the following functions

    d/dy Artanh (arctan (y^2))
    = (Artanh)' (arctan (y^2)) * d/dy * arctan(y^2)
    = [1/(1 - (arctan(y^2))^2] * [(2y)/(1 + y^4)]

    I do not understand how did the 3rd line was made, I know that:
    arctan dfferentiated is 1/(1+x^2). I presume the numerator of the 2nd part is 1 + y^2 and denom' is ((1 + y^2)^2)
    Make sure you notice that we are actually differentiating the inverse hyperbolic tangent, and \frac d{dx}\left[\text{artanh}\,x\right] = \frac1{1-x^2}. We have

    \frac d{dy}\left[\text{artanh}\left({\color{blue}\arctan y^2}\right)\right]

    = \frac1{1 - \left({\color{blue}\arctan y^2}\right)^2}\,\left[\frac d{dy}\left[{\color{blue}\arctan y^2}\right]\right]

    = \frac1{1 - \left(\arctan y^2\right)^2}\,\left[\frac d{dy}\left[\arctan {\color{red}y^2}\right]\right]

    = \frac1{1 - \left(\arctan y^2\right)^2}\,\left[\frac1{1 + \left({\color{red}y^2}\right)^2}\right]\left[\frac d{dy}\left[{\color{red}y^2}\right]\right]

    = \frac1{1 - \left(\arctan y^2\right)^2}\,\left[\frac1{1 + \left(y^2\right)^2}\right]\left(2y\right)

    = \frac1{1 - \left(\arctan y^2\right)^2}\,\left[\frac{2y}{1 + y^4}\right]

    Quote Originally Posted by NeloAngelo View Post
    2. Prove that the function x -> 1/x is an asymptotics to x -> x/[(1 + x^4)^(1/2)] for

    x -> infin {x/[(1 + x^4)^(1/2)] } / (1/x) = x^2 / [(1 + x^4)^(1/2)]

    and

    = Lim x -> infin x^2 / [(1 + x^4)^(1/2)]
    = Lim x -> infin x^2 / [(1 + [1/x^4])^(1/2)] = 1

    At this stage I presumed you divide all the x's by the highest power of x which seems to be x^2 but the denominator does not show that it seems.
    \lim_{x\to\infty}\frac{x^2}{\sqrt{1 + x^4}}

    =\lim_{x\to\infty}\frac{x^2}{\sqrt{x^4\left(\frac1  {x^4} + 1\right)}}

    =\lim_{x\to\infty}\frac{x^2}{\sqrt{x^4}\sqrt{\frac  1{x^4} + 1}}

    =\lim_{x\to\infty}\frac{x^2}{|x^2|\sqrt{\frac1{x^4  } + 1}}

    =\lim_{x\to\infty}\frac{x^2}{x^2\sqrt{\frac1{x^4} + 1}}\text{ (since }x > 0\text{)}

    =\lim_{x\to\infty}\frac{1}{\sqrt{\frac1{x^4} + 1}}

    =\frac{1}{\sqrt{{\color{red}0} + 1}}

    =\frac{1}{\sqrt1} = 1

    Quote Originally Posted by NeloAngelo View Post
    3. Give a short argument based on symmetry considerations that

    Integ (lims of 2(pi) by -(pi)) : (x + (sin x)(cos^4x))dx =
    Integ (lims of (pi) by (pi)) : (x + (sin x)(cos^4x))dx

    Integ (lims of 2(pi) by -(pi)) : (x + (sin x)(cos^4x))dx +Integ (lims of (pi) by (pi)) : (x + (sin x)(cos^4x))dx

    First part is equal to zero since this is odd. How do you confirm if the function is odd or even?
    I'm not sure I understand what you have written here, but you can tell if a function f is odd by checking that f(-x) = -f(x) and if it's even by checking that f(-x) = f(x). For example, f(x)=x^3 is odd because f(-x) = (-x)^3 = -\left(x^3\right) = -f(x). Note that sine is odd and cosine is even. We have

    \int_{-\pi}^{2\pi}\left(x + \sin x\cos^4x\right)dx

    =\int_{-\pi}^{\pi}\left(x + \sin x\cos^4x\right)dx + \int_{\pi}^{2\pi}\left(x + \sin x\cos^4x\right)dx \text{\quad (splitting the integral at }\pi\text{)}

    =0 + \int_{\pi}^{2\pi}\left(x + \sin x\cos^4x\right)dx = \int_{\pi}^{2\pi}\left(x + \sin x\cos^4x\right)dx

    We know that the integrand is odd because

    (-x) + \sin(-x)\cos^4(-x) = (-x) - \sin x\cos^4x = -\left(x + \sin x\cos^4x\right)

    Quote Originally Posted by NeloAngelo View Post
    4. When is the Sandwich (Squish) Rule is used? Is it just for limits only?
    The squeeze/sandwich theorem regards limits. It is used when you have two functions with known equal limits and want to find the limit of a function that lies between them (on an interval).

    Quote Originally Posted by NeloAngelo View Post
    5. Use comparison test to determine the convergence or divergence of

    {Sum of}_infin(n=1) [1/( n + (n)^(1/2))]
    We have \sum_{n=1}^\infty\frac1{n+\sqrt n}.

    p-series often work well in limit comparisons. In this case, the highest degree of n is 1, so let us compare with the known divergent harmonic series, \sum_{n=0}^\infty\frac1n:

    \lim_{n\to\infty}\frac{1/\left(n+\sqrt n\right)}{1/n}

    =\lim_{n\to\infty}\frac n{n+\sqrt n}

    =\lim_{n\to\infty}\frac n{n+n^{1/2}}

    =\lim_{n\to\infty}\frac n{n\left(1+n^{-1/2}\right)}

    =\lim_{n\to\infty}\frac1{1+n^{-1/2}}

    =\lim_{n\to\infty}\frac1{1+\frac1{\sqrt n}} = \frac1{1+{\color{red}0}} = 1

    Now, what can you conclude?
    Last edited by Reckoner; May 26th 2008 at 01:38 PM. Reason: Added the rest of the problems
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  3. #3
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    Nice one Reckoner the step by step stages have helped me a lot.
    And yes I like Devil May Cry lol
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