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Math Help - Derivative Tutorial

  1. #16
    MHF Contributor Mathstud28's Avatar
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    Personally the best way to practice integrating in my opinion is all-variable expressions such as \int\ln|a+bx|dx....do this as you normally would it gives you a great concept of how to work these...the above statement is worked out below

    To evaluate this integral \int\ln|a+bx|dx the preferred method is Integration by parts(discussed earlier in the thread)

    so to start we define our individual parts

    u=ln|a+bx|

    dv=dx

    du=\frac{b}{a+bx}dx

    v=x

    therefore using the appropriate formula we get

    \int\ln(a+bx)dx=x\cdot\ln|a+bx|-\int\frac{bx}{a+bx}...

    now to evaluate the second integral we utilize the above technique to rewerite the second integral as \int\frac{bx+a-a}{a+bx}dx=\int\frac{bx+a}{bx+a}dx-\int\frac{a}{a+bx}dx...



    which with some constant manipulation x+\frac{a}{b}ln|a+bx|+c

    so combinging we get

    \int\ln|a+bx|dx=x\cdot\ln|a+bx|-(x-\frac{a}{b}\ln|a+bx|)+C=x\cdot\ln|a+bx|-x+\frac{a}{b}\ln|a+bx|+C
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  2. #17
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    Quote Originally Posted by Mathstud28 View Post
    To evaluate this integral \int\ln|a+bx|dx the preferred method is Integration by parts(discussed earlier in the thread)
    But don't apply integration by parts immediately, first substitute z=a+bx, then you'll end up with \frac1b\int\ln z\,dz and this is one of many integrals that one should have memorized.
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  3. #18
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Haha..three things...I like it like that actually(the way I did it).....secondly it is good to do that sort of non-memorized integral to work on your integration techniques...and thirdly weren't you the guy like two posts ago who said that some people dont have that summation formula memorized...you are right it is easier as such...but I like it that way...o by the way this isnt supposed to sound angry...I am grateful you are trying to help me!
    ok, so don't memorize it. it is easy to run integration by parts on ln(z), more so than ln(a + bx).

    hehe, this is not to sound angry or defend Krizalid, just a thought that came to my head. it's probably silly...
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  4. #19
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Jhevon View Post
    ok, so don't memorize it. it is easy to run integration by parts on ln(z), more so than ln(a + bx).

    hehe, this is not to sound angry or defend Krizalid, just a thought that came to my head. it's probably silly...
    Yes I would agree with both of you...I am just saying for educational purposes it is helpful to go through all the extra work...more integration practice
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  5. #20
    MHF Contributor Mathstud28's Avatar
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    This post is for those ubiquitous acceleration/velocity/etc. problems

    Ok we will start with the basics
    a(t)=v'(t)=s''(t)\Rightarrow{s(t)=\int{v(t)dt}=\in  t\int{a(t)dt}dt}

    Next up is this...

    Q.where does a particle change direction?
    A.Max or mins

    Q.when is a particle at rest?
    A. when v(t)=0

    Q.How do I find Max's and Min's for S(t)?
    A. Same method as you would for f(x)..utilizing the first derivative test or second derivative test(see earlier thread for more on those)

    Q.What is the total distance traveled by a particle from a to b?
    A. total distance traveled from a to b by a particle is \int_a^{b}|v(t)|dt

    Q.What does it mean when it says "Find the intervals when the particle is moving to the right on the x-axis"
    A. This is asking you to find increasing intervals which are the intervals for which every element of that interval f'(x)>0...to find these intervals use the first derivative test...but instead of finding max or min stop after evaluating points in the first derivative and just look for the intervals that f'(x)>0...when asking "...moving to the left on the x-axis" this applies to decreasing intervals which is the same concept as increasing except f'(x)<0

    Q.What exactly is speed?
    A.Speed is the absolute value of velocity? |v(t)|

    Q.How do I find whether or not speed is increasing at a point c?
    A. There are two methods to solving this:
    Physics Method...speed is increasing IFF a(c) and v(c) share the same sign then speed is increasing
    Math method:Easiest way to do this is graphically. Graph your point (c,V(c))..once this is done start plotting points based on your a(t)(slope of v(t)) if the line is approaching the x-axis it is decreasing...if it is going away from the x-axis it is increasing

    That is it for now
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  6. #21
    MHF Contributor Mathstud28's Avatar
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    The previous post brings to light an interesting topic...now I am sure you can do \int_{-3}^{2}x+2dx...but can you do \int_{-3}^{2}|x+2|dx?

    Here is how you evaluate \int_a^{b}|u(x)|dx

    first you set u(x)=0

    next set up test intervals based on the values that make u(x)=0 and test an element of each interval in u(x)

    if u(x)<0 on the interval (a,b) then on that interval |u(x)|=-u(x)

    and if u(x)>0 on the iterval (a,b) then on that interval |u(x)|=u(x)

    based on these splits break the integral apart and evaluate each change in sign of |u(x)| seperately

    Example:

    \int_{-3}^{2}|x+2|dx...you didnt excpect me to come up with another example did you?

    so first we solve x+2=0..obviously getting x=-2...if you can't get that you probably shouldnt be reading quick tip calculus notes

    ok so we set up those test intervals based on that number that makes the inside of the absolute value 0

    so we could have (-\infty,-2),(-2,\infty)...nah...why include points not between the limits of integration? Don't

    so we have (-3,-2),(-2,2)

    now we need to test an element of each interval in the function inside of the absolute value sign...so we select -2.5 for the one interval and 0 for the other...so (-2.5)+2<0 and 0+2>0

    thererfore on (-3,-2) |x+2|=-(x+2)

    and on (-2,2) |x+2|=x+2

    ok so now what do we do with this info?

    we break up the integral!

    We use the fact that \int_{a}^{c}f(x)dx=\int_{a}^{b}f(x)dx+\int_b^{c}f(  x)dx assuming a<b<c and f(x) is continous on (a,c)

    so we break up th integral to look like this \int_{-3}^{-2}|x+2|dx+\int_{-2}^{2}|x+2|dx

    now applying the knowledge that |x+2|=-(x+2),(-3,-2)

    and |x+2|=x+2,(-2,2) we can rewrite the integral to look like

    \int_{-3}^{2}|x+2|dx=\int_{-3}^{-2}-(x+2)dx+\int_{-2}^{2}(x+2)dx

    then by standard integration techniques we get \int_{-3}^{2}|x+2|dx=\frac{1}{2}+8=\frac{17}{2}
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  7. #22
    MHF Contributor Mathstud28's Avatar
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    Remember if you have ANY Calc I or II(Ab or BC also) questions feel free to ask!! I will help with the general questions on how to do something instead of the specific problem questions you usually ask! all are welcome
    Last edited by Mathstud28; April 21st 2008 at 06:52 PM.
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  8. #23
    MHF Contributor Mathstud28's Avatar
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    This time I will talk about partial fraction decomposition if conjuction with integrating...say you had the expression \int{\frac{x}{4x^2+4x+1}dx}...well I am sure that not many beginners would look at that and go "Hey thats obviously \frac{\ln|2x+1|}{4}+\frac{1}{8x+4}+C!"

    The way we go about this is the typical way we do partial fractions...if you don't know how to do partial fractions you might be able to use this but it will be kind of hard...I might come back later and do a segment on just how to do partial fractions...but to the task at hand!

    we want to decompose \frac{x}{4x^2+4x+1}=\frac{x}{(2x+1)^2}

    so we set up our partial fractions template for repeated quadratic factors...so we have

    \frac{x}{(2x+1)^2}=\frac{A}{2x+1}+\frac{B}{(2x+1)^  2}

    now multiplying through by (2x+1)^2 we get x=(2x+1)A+B

    expanding and grouping we get x=2Ax+A+B

    now by equating coefficents we get 2A=1 and A+B=0

    so 2A=1\Rightarrow{A=\frac{1}{2}}

    then A+B=0\Rightarrow{\frac{1}{2}+B=0}\Rightarrow{B=\fr  ac{-1}{2}}

    so now we know that \frac{x}{4x^2+4x+1}=\frac{1}{2(x+1)}-\frac{1}{2(2x+1)^2}

    therefore \int\frac{x}{4x^2+4x+1}dx=\frac{1}{2}\int\bigg[\frac{1}{2x+1}-\frac{1}{(2x+1)^2}dx\bigg]

    now by manipulating constants we get \frac{1}{4}\int\bigg[\frac{2}{2x+1}-\frac{2}{(2x+1)^2}dx\bigg]

    now it is obvious that because we have the derivative of the quantity the answer is \frac{1}{4}\bigg[\ln|2x+1|-\frac{-1}{2x+1}\bigg]+C=\frac{\ln|2x+1|}{4}+\frac{1}{8x+4}+C
    Last edited by Mathstud28; April 23rd 2008 at 03:49 AM.
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  9. #24
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    Quote Originally Posted by Krizalid View Post
    But don't apply integration by parts immediately, first substitute z=a+bx, then you'll end up with \frac1b\int\ln z\,dz and this is one of many integrals that one should have memorized.
    Hello,

    For this thing, we were learnt to make appear the derivative of what is in the logarithm...
    Kind of more rapid way sometimes
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  10. #25
    MHF Contributor Mathstud28's Avatar
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    Another general case integral so that you can see how they are done more in depth

    \int\frac{1}{x^2\sqrt{a^2-x^2}}dx,x>0

    First step make an "inverse substitution"

    Let x=\frac{1}{u}...

    Therefore dx=\frac{-1}{u^2}dx

    So our integral is rewritten as \int\frac{\frac{-1}{u^2}}{\bigg(\frac{1}{u}\bigg)^2\sqrt{a^2-\bigg(\frac{1}{u}\bigg)^2}}du=\int\frac{\frac{-1}{u^2}}{\frac{1}{u^2}\sqrt{a^2-\frac{1}{u^2}}}du

    now by obvious cancellation we have

    -\int\frac{1}{\sqrt{a^2-\frac{1}{u^2}}}du

    Now we multiply by \frac{\sqrt{u^2}}{\sqrt{u^2}}=\frac{u}{u}

    this is by the stipulation of the integral

    so we have -\int\frac{u}{\sqrt{au^2-1}}du

    and by constant manipulation we have \frac{-1}{2a}\int\frac{2au}{\sqrt{au^2-1}}du

    Then by using standard integration techniques we get

    \frac{-1}{2a}\cdot{2\sqrt{au^2-1}}+C=\frac{-\sqrt{au^2-1}}{a}+C

    Now by back substituting x=\frac{1}{u}\Rightarrow{u=\frac{1}{x}}

    So we have \frac{-\sqrt{\frac{a}{x^2}-1}}{a}+C

    Multipling by \frac{\sqrt{x^2}}{\sqrt{x^2}}=\frac{x}{x}

    we get

    \frac{-\sqrt{a-x^2}}{ax}+C
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  11. #26
    MHF Contributor Mathstud28's Avatar
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    Next we will discuss trigonomtric substitution

    First off the basics

    When you see the follwoing substitute with the following

    \sqrt{a^2-x^2}
    Let x=a\sin(\theta)

    When you have \sqrt{a^2+x^2}
    Let x=a\tan(\theta)

    And finally when we have \sqrt{x^2-a^2}
    Let x=a\sec(\theta)

    An optional approach is to memorize the following formulas

    \int\sqrt{a^2-x^2}dx=\frac{1}{2}\bigg(a^2\arcsin\bigg(\frac{x}{a  }\bigg)+x\sqrt{a^2-x^2}\bigg)+C

    \int\sqrt{x^2-a^2}dx=\frac{1}{2}\bigg(x\sqrt{x^2-a^2}-a^2\ln|x+\sqrt{x^2-a^2}|\bigg)+C

    \int\sqrt{a^2+x^2}dx=\frac{1}{2}\bigg(x\sqrt{a^2+x  ^2}-a^2\ln|x+\sqrt{a^2+x^2}|\bigg)+C

    But this is painful...instead try to just apply the substitutions and figure it out algebraically

    To cement how to do these...I will do...who would have guessed it another general case integral

    \int\frac{\sqrt{x^2-a^2}}{x}dx


    First identifying it is x=a\sec(\theta)

    now we must find dx dx=a\sec(\theta)tan(\theta)d{\theta}

    Now by substiuting we get

    \int\frac{\sqrt{a^2-a^2\sec^2(\theta)}}{a\sec(\theta)}\cdot{a\sec(\the  ta)\tan(\theta)}d{\theta}

    Now by simplifyig and cancelling we get

    \int\sqrt{a^2(\sec^2(\theta)-1)}\tan(\theta)d{\theta}=\int{a<br />
tan(\theta)\tan(\theta)d{\theta}}=a\int\tan^2(\the  ta)

    Now by making the useful substitution tan^2(x)=sec^2(x)-1

    we get

    a\int[\sec^2(\theta)-1]d{\theta}=a[\tan(\theta)-\theta]+C


    Now all we need to do is substitute x=a\sec(\theta)\Rightarrow\theta=arcsec\bigg(\frac  {x}{a}\bigg)

    therefore we have a\bigg[\tan\bigg(arcsec\bigg(\frac{x}{a}\bigg)\bigg)-arcsec\bigg(\frac{x}{a}\bigg)\bigg]+C

    and now by using reference triangles we get

    a\bigg[\frac{\sqrt{x^2-a^2}}{a}-arcsec\bigg(\frac{x}{a}\bigg)\bigg]+C

    and by distributing we get

    \int\frac{\sqrt{x^2-a^2}}{x}dx=\sqrt{x^2-a^2}-a\cdot{arcsec\bigg(\frac{x}{a}\bigg)}+C
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  12. #27
    MHF Contributor Mathstud28's Avatar
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    Next thing I will touch on is how to do trigonometric integrals...not trigonometric substitution but integrating trig functions

    Let's start with the basics
    \int\cos(x)dx=\sin(x)+c

    \int\sin(x)dx=-\cos(x)+C

    \int{-\cos(x)}dx=-\int\cos(x)dx=-\sin(x)+C

    \int{-\sin(x)}dx=-\int\sin(x)dx=-(-\cos(x))+C=\cos(x)+C

    \int\sec(x)dx=\ln|\sec(x)+\tan(x)|+C

    \int\sec^2(x)dx=\tan(x)+C
    \int\csc(x)dx=-\ln|\csc(x)+\cot(x)|+C

    \int\csc^2(x)dx=-\cot(x)+C

    \int\tan(x)dx=\int\frac{\sin(x)}{\cos(x)}dx=-\int\frac{-\sin(x)}{\cos(x)}dx=-\ln|\cos(x)|+C

    \int\cot(x)dx=\int\frac{\cos(x)}{\sin(x)}dx=\ln|\s  in(x)|+C

    Now that that is over with lets get on to some of the most useful trig indentities

    2\sin(x)\cos(x)=\sin(2x)

    \cos^2(x)=\frac{1+\cos(2x)}{2}

    \sin^2(x)=\frac{1-\cos(2x)}{2}

    1+\tan^2(x)=\sec^2(x)\Rightarrow\tan^2(x)=\sec^2(x  )-1

    1+\cot^2(x)=\csc^2(x)\Rightarrow\cot^2(x)=\csc^2(x  )-1

    \sin^2(x)+\cos^2(x)=1

    These are the most of the useful ones for INTEGRATION...all trig identites are useful but these are the best for integrating

    Next I will describe how to take the integral of things such as

    \int\sin^3(x)dx

    Since it is an integral with sine to an odd power make a substitution

    \int\sin^2(x)\cdot\sin(x)dx=\int(1-\cos^2(x))\sin(x)dx=\int\sin(x)-\cos^2(x)\sin(x)dx

    Then since the derivative of the quantity is wiht the second term we can just integrate to get

    \int\sin^3(x)dx=\int\sin(x)-\cos^2(x)\sin(x)dx=-\cos(x)+\frac{\cos^3(x)}{3}+C

    This works for all odd powers of sine or cosine(if it is cosine make the sub \cos^(x)=1-\sin^2(x))

    Next we will do

    \int\cos^4(x)dx

    Uh-oh the exponent is even...this means that we cant use the recently discussed method...so we must use our indentities

    \int\cos^4(x)dx=\int\bigg(\frac{1+\cos(2x)}{2}\big  g)^2dx=\frac{1}{4}\int(1+\cos(2x))^2dx =\frac{1}{4}\int[1+2\cos(2x)+\cos^2(2x)]dx

    Uh-oh...looks like we need to make another trig substitution

    \cos^2(2x)=\frac{1+\cos(4x)}{2}....so we have

    \frac{1}{4}\int\bigg[1+2\cos(2x)+\frac{1}{2}+\frac{\cos(4x)}{2}\bigg]dx=\frac{\sin(4x)}{32}+\frac{\sin(2x)}{4}+\frac{3x  }{8}+C

    To do odd powers of tan(x) or sec(x) make a similar move as with the sine and cosines

    and for even powers of tan(x) or sec(x) use the pythagorean indentitys with tan(x) and sec(x) as listed in the indentities...hope this has been helpful!

    It sure took a long time to write!!
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  13. #28
    MHF Contributor Mathstud28's Avatar
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    Next I will start series....for this I will assume you know sequences and all the terminology I am about to use...if not just post a question in this thread!

    In this thread I will provide you with a detailed explanation of each convergence test

    The first one and most overlooked is the n-th term test which states

    \sum_{n=0}^{\infty}a_n\text {diverges if}, \lim_{n \to {\infty}}a_n\ne0

    this test only proves divergence!!! If the series limit is 0 it could still be divergent

    Examples... \sum_{n=0}^{\infty}\frac{n}{2n+1}\Rightarrow{\lim_  {n \to {\infty}}\frac{n}{2n+1}}=\frac{1}{2}\ne{0}\therefo  re{divergent}

    Next example \sum_{n=0}^{\infty}\frac{1}{2^{n}}\Rightarrow\lim_  {n \to {\infty}}\frac{1}{2^n}=0..therefore the n-th term test tells us nothing

    The second test is Geometric series.....This is a special case where you have

    \sum_{n=0}^{\infty}x^{n}

    This series converges only if |x|<1

    Examples \sum_{n=0}^{\infty}\bigg(\frac{2}{3}\bigg)^{n} since \bigg|\frac{2}{3}\bigg|<1 series converges

    Another example \sum_{n=0}^{\infty}\frac{4^n}{3^{n+1}}=\frac{1}{3}  \sum_{n=0}^{\infty}\bigg(\frac{4}{3}\bigg)^{n}...which is divergent becaue \bigg|\frac{4}{3}\bigg|>{1}

    The next test is the integral test which states that if a_n is postive,continous, and decreasing then the series \sum_{n=c}^{\infty}a_n converges only if \int_c^{\infty}f(n)dn diverges...where f(n)=a_n

    Examples \sum_{n=2}^{\infty}\frac{1}{n(\ln(n))^4}...you can establish that it is decreasing and psotive....so the test does apply
    so
    \int_2^{\infty}\frac{1}{n(\ln(n))^4}dn=\frac{-1}{3(\ln(n))^3}\bigg|_2^{\infty}=\frac{1}{3(\ln(2)  )^3}..which converges since therefore the series converges

    To see an example of a divergent consider \sum_{n=0}^{\infty}\frac{1}{n\ln(n)}

    The next test is a P-series...similar in nature to the Geometric series in that it is a specific case...a P-series given by

    \sum_{n=0}^{\infty}\frac{1}{x^{p}} is divergent iff p>1...to validate this try using the integral test and consider values for p that are less than one

    \sum_{n=0}^{\infty}\frac{1}{x^2}...convergent

    \sum_{n=0}^{\infty}\frac{1}{x^{\frac{2}{3}}}...divergent

    The next test is the Direct comparison test...this test states that if 0\leq{a_n}\leq{b_n}

    If \sum_{n=1}^{\infty}b_n converges then \sum_{n=1}^{\infty}a_n converges

    and if \sum_{n=1}^{\infty}a_n diverges then \sum_{n=1}^{\infty}b_n diverges

    For example consider \sum_{n=0}^{\infty}\frac{1}{n^2+3n}....since \frac{1}{n^2+3n}<\frac{1}{n^2},\forall{n}\in\mathb  b{Z}^{+} we can condlude that \sum_{n=0}^{\infty}\frac{1}{n^2+3n} converges since \sum_{n=0}^{\infty}\frac{1}{n^2} converges by p-series

    Another Example would be \sum_{n=0}^{\infty}\frac{1}{2+sqrt{n}}...selecting \frac{1}{n} as our comparison series we see that \frac{1}{n}\leq\frac{1}{\sqrt{n}+2},\forall{n}\leq  {4} therefore since \frac{1}{n} is a divergent "harmonic series" \sum_{n=0}^{\infty}\frac{1}{2+\sqrt{n}} is divergent

    Next thread will include...Limit comparison test, Alternating series test, Ratio test, and Root test!!
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  14. #29
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    This time I will talk about partial fraction decomposition if conjuction with integrating...say you had the expression \int{\frac{x}{4x^{\color{red}4}+4x+1}dx}...well I am sure that not many beginners would look at that and go "Hey thats obviously \frac{\ln|2x+1|}{4}+\frac{1}{8x+4}+C!"

    The way we go about this is the typical way we do partial fractions...if you don't know how to do partial fractions you might be able to use this but it will be kind of hard...I might come back later and do a segment on just how to do partial fractions...but to the task at hand!

    we want to decompose \frac{x}{{\color{red}x^4}+4x+1}=\frac{x}{(2x+1)^2}
    In both cases you meant 4x^2.
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  15. #30
    MHF Contributor Mathstud28's Avatar
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    As promised I will continue with the rest of the convergence tests

    The next one that was in line was the limit comparison test that states that if

    0<\lim_{n \to {\infty}}\frac{a_n}{b_n}<\infty then both \sum_{n=1}^{\infty}a_n and \sum_{n=1}^{\infty}b_n converge or diverge

    Examples

    \sum_{n=2}^{\infty}\frac{n}{n^3-1}

    We will use \frac{1}{n^2} a convergent P-series as our comparison series...the best way to determine what series to use is as n\to{\infty} what does the series closely resemble?

    obviously as n goes to infinity the -1 is inconsequential leaving you with \frac{1}{n^2}

    so taking \lim_{n \to {\infty}}\frac{\frac{n}{n^3-1}}{\frac{1}{n^2}}=\lim_{n\to\infty}\frac{n^3}{n^3-1}=1

    therefore since it converges to value greater than one but less than ∞ they share convergence/divergence....and since our comparison series is converget this means our series in question is also convergent

    Another example would be \sum_{n=0}^{\infty}\frac{5^n}{7^n-n}

    Now since as we see as n\to\infty this resembles \bigg(\frac{5}{7}\bigg)^{n} a convergent geometric series

    so \lim_{n \to {\infty}}\frac{\frac{5^n}{7^n}}{\frac{5^n}{7^n-n}}=\lim_{n \to {\infty}}\frac{7^n-n}{7^n}=1

    Once again you use the fact that the limit is one to deduce that both series converget


    Next up is the Alternating Series Test which states that in \sum_{n=1}^{\infty}(-1)^{n}a_n...if a_{n+1}<a_n

    and \lim_{n\to\infty}a_n=0 then the series converges

    Examples:

    Determine the convergence of the series \frac{(-1)^n}{\ln(n)}

    We must test two things first off is \frac{1}{\ln(n+1)}<\frac{1}{\ln(n)}?

    The answer is yes...obviously because since the larger n is the larger \ln(n) is..and since the ln is on the bottom we use the fact that \text{if},a<b,\text{then}\frac{1}{a}>\frac{1}{b} to conclude that a_{n+1}<a_n.

    Next we must test to see if \lim_{n\to\infty}\frac{1}{\ln(n)}=0 which it obviously does

    Thererfore \sum_{n\to\infty}\frac{1}{\ln(n)} is conditionally convergent or that is at least what the test tells us! It is actually absolutely convergent but this test does not conclude that!


    Next test which in my opinion is the best is the Ratio Test which states that if \lim_{n\to\infty}\frac{a_{n+1}}{a_n}<1 the series converges...if \lim_{n\to\infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|<1 the series is absolutely convergent

    If the limit is greater than one the series is divergent

    if the limit is equal to one it is inconclusive

    Examples \sum_{n=0}^{\infty}\frac{4^n\cdot{n!}}{n\cdot(n+4)  !}

    Ok here is how you would do it since \frac{a_{n+1}}{a_n}=a_{n+1}\cdot\frac{1}{a_n}...so since thsi is fraction instead of dividing we multiply the reciporcal

    to get \lim_{n\to\infty}\frac{4\cdot{4^n}\cdot(n+1)\cdot{  n!}}{(n+1)(n+5)(n+4)!}\cdot\frac{n\cdot{(n+4)!}}{4  ^{n}n!}=\lim_{n\to\infty}\frac{4n}{n+5}=4>1

    Therefore the series diverges

    An easier method would have been simplifying (n+4)!=(n+4)(n+3)(n+2)(n+1)n! making the original series \sum_{n=0}^{\infty}\frac{4^{n}}{(n+4)(n+3)(n+2)(n+  1)n}...and using the n-th term test we see the series diverges

    The last thing to be discussed in this thread would be the Root test that states if \lim_{n\to\infty}\sqrt[n]{a_n}<1 the series converges if the limit is greater than one the series diverges..if the limit equals one the test is inconclusive

    Example \sum_{n=0}^{\infty}e^{-n}n^{n}

    So we apply our test to get \lim_{n\to\infty}\sqrt[n]{\frac{n^{n}}{e^{n}}}=\lim_{n\to\infty}\frac{n^{\f  rac{n}{n}}}{e^{\frac{n}{n}}}=\lim_{n\to\infty}\fra  c{n}{e}=\infty>1 therefore teh series diverges

    Thats it for this post....next I will do some derivative integral stuff...but I will get back and do taylor series and maclaurin series...especially their application to integration

    Remember any questions just ask!

    Thanks is appreciated!
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