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Doing math in your head - Post Calculus

  1. Jan 17, 2007 #1
    I am finished with Calc III and now am in a Mathmatical Methods course. For the first day we were given a "Prerequisites Inventory", basically problems we should know. But I cant do most of them!

    If you can help me with any I would appreciate it.

    1. The problem statement, all variables and given/known data
    #4 If cos(24) (degrees) = 0.913, sin (24) = 0.407, cos(17) = 0.956 and sin (17) = 0.292 then what is cos (41)? Do not use a calculator.

    2. Relevant equations

    3. The attempt at a solution
    #4 Obviously 24+17 = 41, but I know that Cos(24) + cos(17) is not equal to cos(41).

    1. The problem statement, all variables and given/known data

    #6 Compute each of the following to 4 or 5 place accuracy as easily as you can without using a calculator or computer: (You should be able to do them in your head)
    (b) [tex]/sqrt{0.9997} [/tex]

    (d) cos(0.025) radians

    (e) ln(1.0001)

    (f) [tex] a^2 - b^2 [/tex], where a = 1.2543 and b = 1.2539

    2. Relevant equations

    3. The attempt at a solution

    (b) I know that the answer is going to be inbetween 1 and .09997, but I dont know how to get 4 to 5 digits of accuracy in my head.

    (d) I know that sin of a small angle can be approximated by that angle, but is there a similar rule for cosine? I thought maybe I can approximate the cos of a small angle by 1-(small angle), but that does not work.

    (e) Again, I know the ballpark (very close to zero, because ln(1) = 0), but I dont know how to get 4 to 5 digits of accuracy in my head.

    (f) Here I know that it could be represented by (a+b)(a-b). But I cant seem to do that in my head...

    1. The problem statement, all variables and given/known data
    #9 Compute the derivative of [tex] f(x) = x ln (\frac{x}{x^2 + a^2})[/tex]

    2. Relevant equations

    3. The attempt at a solution
    This derivative I can calculate on paper, but I cant do it quickly. In fact it took me quite a long time. The assignment says should be able to do in about 30mins, which makes me thinkg there must be some trick or mental algorithm or something to do it quickly.

    There are many other problems like this that I am struggling with. But they all have a similar theme, a problem I should be able to do in my head or at least without a calculator.

    I have never been good at math, but this assignment makes me feel real stupid. Any suggestions on how to approach these would be helpful.

    Thank you
  2. jcsd
  3. Jan 17, 2007 #2
    For #4, you could use the cosine addition formula: cos (x+y) = cos x cos y - sin x sin y.

    For #6b, consider using the binomial theorem and evaluating the first 2 terms. (1+x)^a = 1 + ax + ...
    #6d: For small angles theta, cos theta = 1-theta^2/2. It is just the first 2 terms from its Maclaurin series.
    #6e: I would use the series for ln(1+x) and evaluate the first 2 terms since it seems to converge very quickly.
    Last edited: Jan 17, 2007
  4. Jan 17, 2007 #3

    Gib Z

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    Ok Thank god but I know most of these :D so i can help i hope btw dont feel stupid if you can't compute these in your head, I know the easiest way how to, but I can't say I can, youll see what I mean later.

    q4. IMDerek got your solution there.
    q6.b) Okily, Heres the square rooting method I use, Its especially effective when you already have a good estimate, like in your case :D

    Basically, when you divide by a big number, you end up with a smaller one.
    When you divide by a smaller number, you end up with a bigger one.
    When you divide by the square root, you get the square root again. :D

    So, We get the number we want to divide by, in this case 0.9997. Then we divide by an estimate of its square root. Now, the closer your estimate, the more effective this is, but since dividing by say, 0.99985 is going to be a real ****, just use 1, its still very good :D.

    Now, We have one rough estimate of the square root. Becuase, we have the number, divided by something bigger than its root.

    When divided by bigger, we got a smaller number :D So the first bit got us something a bit smaller than the square root.

    So, we have 2 estimate, the first is 1, which is bigger than the real square root. and the 2nd is 0.9997/1, or 0.9997. One bigger, one smaller, to get a nice one in between we average them :D.

    Add em together, divide by 2. (You may have noticed in this case the entire thing was just averaging 1, your estimate, and the number, but thats only becuase in this case 1 is you estimate)

    So basically, from that, we get 0.99985, if we square that on our calculator, we get 0.999700022...., that looks accurate enough to me :D

    d) IMDerek got it right.

    e) He got it again, but in this case I would have just said 0 automatically. I know better to be sure by using The series for ln(1+x), but to me it looked small enough to be zero. by a calculator, 0.000009999...5 decimal places, good enough. Do you know the series, btw? It helps.

    f) You got it the right way. The addition and subtraction in your head in pretty easy, but i bet the multiplication isnt. Bad luck dude.

    9) This is isn't that hard actually. No offence, but im surprised they even gave 30 mins, after I post this ill do it and tell you how long I took.

    Have fun, good luck, (BTW, awesome way to square root stuff huh? I made it myself :D then I realised some people already knew that method...)
  5. Jan 17, 2007 #4

    Gib Z

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    Ok, In 6 minutes, I got
    [tex]f'(x)=ln(\frac{x}{x^2 + a^2}) + (\frac{a}{x})^4 -1[/tex]

    Fun Fun :D O btw, If you WERE able to do the best average, using the estimate 0.99985, using the method, you get exactly 0.9997 on your calculator :D.

    Hopefully you realised that the method can be used over and over again. By the end of it, you have a better estimate, which you can use as your estimate for the next time you use it. It works VERY fast when you have a good estimate like in this case.

    PS: 250 post!
  6. Jan 17, 2007 #5

    Gib Z

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    HMPH I checked my derivative...I have abit right lol. According to my answer, f(x)= Actual f(x) + some arctan mumbo jumbo.

    Lol Ill do it again later, nice work :D
  7. Jan 18, 2007 #6

    Gib Z

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    [tex]\frac{a^2-x^2}{a^2+x^2} + ln (\frac{x}{x^2+a^2})[/tex], 9 minutes :D
  8. Jan 18, 2007 #7
    You could simplify the log function:
    [tex]=x(lnx -ln(x^2+a^2))[/tex]
    [tex]f'(x)=lnx-ln(x^2+a^2) +x(\frac{1}{x} -\frac{2x}{x^2+a^2})[/tex]
  9. Jan 18, 2007 #8
    There is an identitiy for Cos(A+B).


    Last edited: Jan 18, 2007
  10. Jan 20, 2007 #9
    Thank you all for the help. I can now do #4 and #9.

    But I still cannot do any of #6 in my head. I tried getting help from my teacher on Fri, but he doesn't want to help up because this is a so called "weed out" class. So I have to figure it out on my own, lest I get weeded out.
  11. Jan 20, 2007 #10
    Given a couple of minutes or so, I could probably use the series for cosine to estimate 6.e to 4 decimal places.

    [tex]cos(x)=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}[/tex] will give you 4 decimals of accuracy if |x|<1. If I remember correctly anyway.

    Sine is easy to remember too, it starts the same way but has odd exponents and odd factorials in the denoms...


    Actually I think it is an error of 0.0002 (if |x|<1 and is in rad)... I was learning this stuff about a week ago, but am too lazy to go get textbook from upstairs.

    Maybe there is still an easier way... Definitely convert the 0.025 to 1/40 and then do the repeated multiplication, easier.
    Last edited: Jan 20, 2007
  12. Jan 20, 2007 #11

    Gib Z

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    Aww you dont like the way I square root stuff? its really simple for things like in this case. The method I described is far simpler when applied to a case where its close to 1, such as this time. That is probably why the teacher gave that value. Finding the average of your number and 1 isnt that hard :p
  13. Jan 20, 2007 #12
    sin x is x - x^3/3! + x^5/5!...

    For square root of 0.9997, you can do (1-0.0003)^(1/2), expand using binomial theorem, and take the first two terms.

    (1-0.0003)^(1/2)'s first 2 terms are 1-.0003/2 = .99985, which is pretty close to the actual value of .9998499887...
  14. Jan 20, 2007 #13
    good call, thanks.
  15. Jan 20, 2007 #14
    I like it, but I cant seem to do it in my head to 4 decimal places.

    Thx for the responses, Ill take a look at them in the morning!
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