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blackstreet23
 one year ago
Approximate tan inverse (0.1) to three decimal place accuracy using the Maclaurin series tan inverse x
blackstreet23
 one year ago
Approximate tan inverse (0.1) to three decimal place accuracy using the Maclaurin series tan inverse x

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blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0I did this so far The book says that the error to 3 decimal places is \[(0.1) \frac{ 0.1^3 }{ 3 }\]

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0however when i put (0.1^3)/3 in my calculator the answer is 3.3333 * 10^4 that is less that 0.5 * 10^3

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0so why could that be the answer? wouldn't it just be 0.1 ?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1the question asks you to do it using the maclaurin for cos x is that what you did?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442179486992:dw terrible graphics i know but it seems you plugged your numbers into the expansion for\(\tan^{1}x\).

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I wonder why it says to approximate arctan(0.1) using the maclaurin series for cos(x) and here it says use maclaurin series for arctan(x) I think we would want to use the series for arctan(x) since we are apporximating arctan( something)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2but it really does say cos(x)?

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0tan inverse (0.1)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2You know I'm trying to tell you that there is conflicting information here right?

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0my bad the one i copied was wrong

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0but the question here is right

freckles
 one year ago
Best ResponseYou've already chosen the best response.2ok this question makes more sense to me :) Approximate tan inverse (0.1) to three decimal place accuracy using the Maclaurin series tan inverse x \[0.1\frac{0.1^3}{3} \text{ is a pretty good approximation for } \arctan(0.1) \\ 0.1\frac{0.1^3}{3} \approx 0.099\overline{6} \\ \text{ while } \arctan(0.1) \text{ to 6 decimal spots is } 0.09966\]

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0Why is the the answer they have on the back of the book is (0.1)−(0.1^3/3)

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0but the way I learned is by taking each monomial of the series and substituting the value of x

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0and (0.1^3)/3 is less that 5*10^4

freckles
 one year ago
Best ResponseYou've already chosen the best response.2yep but then you have to add the terms in the series

freckles
 one year ago
Best ResponseYou've already chosen the best response.2where does 5*10^(4) come from ?

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0from a rule that says that the decimal places they are asking you for is equal to 0.5*10^n where n is the number of decimal places

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0so i just simplified to 5*10^4

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0and i need to find a term that is smaller than 5*10^4 in the series

freckles
 one year ago
Best ResponseYou've already chosen the best response.2oh I don't remember that I would have honestly found arctan(0.1) first which is approximately 0.0996 from the calculator then added the terms in the series until I got with in digits of that number \[\arctan(x)=\sum_{k=0}^\infty (1)^k \frac{x^{2k+1}}{2k+1} \\ \arctan(x)=(1)^0 \frac{x^{2(0)+1}}{2(0)+1}+(1)^{1}\frac{x^{2(1)+1}}{2(1)+1}+(1)^{2}\frac{x^{2(2)+1}}{2(2)+1} \\ + \cdots +(1)^k \frac{x^{2k+1}}{2k+1}+ \cdots \\ \text{ so let's see if one term is enough } \\ \arctan(x) \approx x \\ \arctan(0.1) \approx 0.1 \text{ this is a really bad approximation } \\ \text{ how about two terms } \\ \arctan(x) \approx x \frac{x^3}{3}\] \[\arctan(0.1) \approx 0.1\frac{0.1^3}{3} =.099\overline{6} \text{ this is an awesome approximation }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2the approximation only gets better if we add more terms onto that

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442182723650:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@blackstreet23 Hello... Freckles and irishboy have it right here... That formula you have listed (0.5*10^n)I dont recall (though I think I can guess where it comes from), but if your issue is you want more accuracy simply add one more term to the approximating series no?

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0well i guess i will ask this one to my professor. Thanks guys for your help!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry I couldn't help Im not sure what you are hung up on as far as the problem is concerned

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0ohh no problem! i know you did your best :)
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