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    Hello,

    I am awful at maths and I have to take a compulsory calculus course at uni. Unfortunately while talking about taylor expansion my teacher focused more on quantum physics and particles in a box than maths (I've never studied physics so you can imagine the dumb-founded expression on my face during that class!).

    When would you need to use taylor expansion? And more so can someone explain to me how to do taylor series around a non-zero point? I really didn't understand this!

    Thank you in advance!
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    http://en.wikipedia.org/wiki/Taylor_series
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    (Original post by Scotty Bear)
    Hello,

    I am awful at maths and I have to take a compulsory calculus course at uni. Unfortunately while talking about taylor expansion my teacher focused more on quantum physics and particles in a box than maths (I've never studied physics so you can imagine the dumb-founded expression on my face during that class!).

    When would you need to use taylor expansion? And more so can someone explain to me how to do taylor series around a non-zero point? I really didn't understand this!

    Thank you in advance!
    The taylor series of an infinitely differentiable function about a non-zero point, a, is given by

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    (Original post by Scotty Bear)
    When would you need to use taylor expansion?
    Some applications of Taylor Series
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    (Original post by Indeterminate)
    The taylor series of an infinitely differentiable function about a non-zero point, a, is given by

    So if I had something like cosx (where x was near 3) would I begin like this:

    cosx=cos[3+(x-3)]= -3 + 3/2!(x-3)^2 - 3/4!(x-3)^4 ....
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    (Original post by Scotty Bear)
    So if I had something like cosx (where x was near 3) would I begin like this:

    cosx=cos[3+(x-3)]= -3 + 3/2!(x-3)^2 - 3/4!(x-3)^4 ....
    No

    f(x) = \cos(x)

    Now use what I posted.

    Some notation:

    f^{(n)} (x)

    means "f differentiated n times", and this is can also be denoted by n dashes
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    (Original post by Indeterminate)
    No

    f(x) = \cos(x)

    Now use what I posted.

    Some notation:

    f^{(n)} (x)

    means "f differentiated n times", and this is can also be denoted by n dashes
    So I have to differentiate cosx first? so then

    cosx=cos[3-(x-3)]= cos3cos(x-3)-sin3sin(x-3)= -cos(x-3)

    And f '(a) = -cos(x-3) ?

    Then the formula filled in would be:

    3 - cos(x-3)/2! (x-3)^2 + cos (x-3)/4! (x-3)^4
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    (Original post by Scotty Bear)
    So I have to differentiate cosx first? so then

    cosx=cos[3-(x-3)]= cos3cos(x-3)-sin3sin(x-3)= -cos(x-3)

    And f '(a) = -cos(x-3) ?

    Then the formula filled in would be:

    3 - cos(x-3)/2! (x-3)^2 + cos (x-3)/4! (x-3)^4
    You don't write

    x=3+(x-3)

    Just keep differentiating

    \cos(x)

    and evaluate derivatives at x=3 (keep the exact value).

    Now can you do it?
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    (Original post by Scotty Bear)
    So I have to differentiate cosx first? so then

    cosx=cos[3-(x-3)]= cos3cos(x-3)-sin3sin(x-3)= -cos(x-3)

    And f '(a) = -cos(x-3) ?

    Then the formula filled in would be:

    3 - cos(x-3)/2! (x-3)^2 + cos (x-3)/4! (x-3)^4
    for  \cos x you'd simply input 3 into the equation at the start. From then on you'd differentiate it and put 3 into the differentiate that. You'd then multiply that by  (x-3) and divide by  1!

    You differentiate it again put 3 into the second differential and multiply the value by (x-3)^2 and divide it by  2!
    So the  r^t^h is  f^{(r)}(3)\frac{(x-3)^r}{r!} .

    Cos x would be a fairly simple one because the constant in front of the term cycles between 4 values.
    So the expansion for cos x about 3 would be:
     \cos3 + \sin3(x-3) - \cos3\frac{(x-3)^2}{2!} - \sin3\frac{(x-3)^2}{3!} + \cos3\frac{(x-3)^4}{4!}+\ldots
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    Lets hope this is third time lucky lol...

    f(3)= cos3
    f ' (3)= -sin3
    f ''(3)= -cos3

    so then the series is

    cos3 - sin3/1! (x-3) -sin3/2! (x-3)^2 -cos3/3! (x-3) ^3
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    (Original post by Scotty Bear)
    Lets hope this is third time lucky lol...

    f(3)= cos3
    f ' (3)= -sin3
    f ''(3)= -cos3

    so then the series is

    cos3 - sin3/1! (x-3) -sin3/2! (x-3)^2 -cos3/3! (x-3) ^3
    yes, though that's just the first three terms of the expansion. Because the expansion carries on infinitely, unless it only asks for 3 terms, always add  \ldots at the end.
    See if you can give the first 5 terms in the expansion of  \sin x about 5.
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    (Original post by Rainingshame)
    yes, though that's just the first three terms of the expansion. Because the expansion carries on infinitely, unless it only asks for 3 terms, always add  \ldots at the end.
    See if you can give the first 5 terms in the expansion of  \sin x about 5.
    Great thanks

    so sinx at 5 would be:

    Sin5 + cos5/1! (x-5) - sin5/2! (x-5) ^2 - cos5/3! (x-5)^3 + sin5/4! (x-5)^4 + cos5 (x-5)^5
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    (Original post by Scotty Bear)
    Great thanks

    so sinx at 5 would be:

    Sin5 + cos5/1! (x-5) - sin5/2! (x-5) ^2 - cos5/3! (x-5)^3 + sin5/4! (x-5)^4 + cos5 (x-5)^5
    You missed the divided by  5! on the SIXTH term lol. Sorry if that seemed a bit harsh pointing it out but the last thing you want is a mistake like that costing you mark on an exam that matters. But other than that it's correct.
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    Thanks for all the help from all of you! Much appreciated! and yepp definately don't want to lose a mark on an exam for such a silly mistake!
 
 
 
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