Ln and exp

  1. hi every one ..
    i have some question
    1 - why the value of exp = 2.71 ?? were from they get this value ? any derive or rule ?

    2 - who i can fiend the value of ln and exp and sin or cos with out using calculator ?

    like cos 56
    sin 28 ??


    ln 24 ?
    exp 33 ?
    without using calculator ?

    thank you
  2. jcsd
  3. jedishrfu

    Staff: Mentor

    e is a most curious number as a base.

    e^x is the only curve whose derivative with respect to x is itself. This fact is used a lot in solving differential equations by assuming solutions based on e^x for certain types of differential equations.

    You can read more about it at:


    Its true that its not an easy number to work with numerically but the insight gained in using it is far more profound.
    Last edited by a moderator: May 10, 2013
  4. Borek

    Staff: Mentor

    You can calculate e by yourself, using this sum:

    [tex]e = \sum\limits_{n=0}^\infty \frac 1 {n!}[/tex]

    Edit: and a spreadsheet.

    Attached Files:

  5. Mentallic

    Mentallic 3,783
    Homework Helper

    You can find some relatively reasonable approximations to these values using your basic rules, but if you need any sort of precision or are short on time, you'd definitely want to use a calculator.

    For the logs, you can split them up into a sum of smaller logs:


    Now, you just need to have ln(2) and ln(3) memorized to make the calculation.

    For exponentials:

    [tex]e^{33}=10^{\log_{10}(3)\cdot 33}[/tex]



    So, if you memorize the natural log of 2,3,5 and 7, you'll have most values covered, so then you can find an approximation to [itex]\log_{10}(3)[/itex] and thus will have the order of magnitude of e33.

    The trigs are trickier and I wouldn't recommend making a calculated approximation for them. As long as you understand the graphs, you should be able to make reasonable approximations in your head.
    If you still want further precision though, in degrees,



    Now, [itex]\cos(45)=\sin(45)=1/\sqrt{2}[/itex]
    and for small angles x in radians, [itex]\cos(x)\approx 1[/itex] and [itex]\sin(x)\approx x[/itex].

    Thus we will take [itex]\cos(11)\approx 1[/itex]
    and (remembering to convert 11o to radians), [itex]\sin(11^o)=\sin(\frac{11\pi}{180})\approx \frac{11\pi}{180}\approx \frac{35}{180}\approx 1/5[/itex]

    So finally,

    [tex]\cos(56)\approx \frac{1}{\sqrt{2}}(1-1/5)=\frac{4\sqrt{2}}{10}\approx \frac{2\cdot \frac{7}{5}}{5}=14/25= 56/100 = 0.56[/tex]

    Where the true value is [itex]\cos(56)=0.559...[/itex]

    As you can see, it's a lot of work for this approximation.

    Also, I'd like to note that you shouldn't expect your approximations to come this close in future (0.56 compared to 0.559... is less than 0.1% off), it's just that my over-approximation of cos(11o) and taking away my over-approximation of [itex]\frac{11\pi}{180}\approx 1/5[/itex] lead to a very close result by pure luck.
    Last edited: May 10, 2013
  6. HallsofIvy

    HallsofIvy 41,262
    Staff Emeritus
    Science Advisor

    Another way of looking at it:

    For any positive number, a, if [itex]y= a^x[/itex] then dy/dx is given by
    [tex]\lim_{h\to 0}\frac{a^{x+h}- a^x}{h}= \lim_{h\to 0}\frac{a^xa^h- a^x}{h}= a^x\lim_{h\to 0}\frac{a^h- 1}{h}[/tex]

    That is, the derivative of [itex]a^x[/itex] is just [itex]a^x[/itex] itself times a number- that number being [itex]\lim_{h\to 0}(a^h- 1)/h[/itex].

    It is easy to see that if a= 2, [itex]\lim_{h\to 0}(2^h- 1)/h[/itex] is less than 1. For example, with h= 0.0001, that fraction is [itex](2^{0.0001}- 1)/.0001= (1.000069317- 1)/.0001[/itex]= 0.000069317/.0001= .69317

    And if a= 3, [itex]\lim_{h\to 0}(3^h- 1)/h[/itex] is larger than 1. With h= 0.0001, again, we have [itex](3^{0.0001}- 1)/.0001= (1.00010987- 1)/.0001[/itex]= .00010987/.0001= 1.0987.

    If you do the same thing with, say, a= 2.5 and a= 2.8, you would again get that constant less than 1 for 2.5, larger than 1 for 2.8 but both closer to 1. There exist a value of a in between 2 and 3, and between 2.5 and 2.8, such that the constant is exactly equal to 1: that is, for that a, the derivative of [itex]a^x[/itex] is just [itex]a^x[/itex] itself. We define "e" to be that number.
    Last edited: May 10, 2013
  7. Integral

    Integral 7,288
    Staff Emeritus
    Science Advisor
    Gold Member

    Before calculators we used tables for trig functions and logs. Fact is there is no easy way to compute these function values without some form of an aid.
  8. what is "n" ?

    pleas can you fiend for me the value of e^4 by this equation
    [tex]e = \sum\limits_{n=0}^\infty \frac 1 {n!}[/tex]
  9. The sum process (with n being the element index) does not compute powers of 3, it computes e. You'll have to multiply it by itself 4 times to get e^4

    What grade are you in? I get the impression that you are asking about things you understand so little that you can't even understand the answers. That just means you have to study the basics a bit more to get rid of your confusion.
  10. thank you very much
  11. i am in college :biggrin: .. but i am not good in maths :frown:

    and tell now i dont know how i fiend e^4 :frown:
  12. e^4 is e*e*e*e
  13. micromass

    micromass 20,052
    Staff Emeritus
    Science Advisor
    Education Advisor

    [tex]e^4 = \sum_{n=0}^{+\infty} \frac{4^n}{n!}[/tex]
  14. Borek

    Staff: Mentor

    n takes all possible integer values, starting with 0.

    Not by this equation, but you can use this one:

    [tex]e^x = \sum\limits_{n=0}^\infty \frac {x^n} {n!}[/tex]

    The first equation was just a special case, for x=1 (and obviously e1 = e).
  15. I'm confused here. Why have both of you used a "3", when surely you mean "e"? If it was just one of you then I would simply put it down to a typo, but as two different people have done this, then maybe I've completely misunderstood something. Apologies if that is the case!
  16. HallsofIvy

    HallsofIvy 41,262
    Staff Emeritus
    Science Advisor

    To err is human- to really screw up requires a computer! I suspect Mentalic made a typo then phinds "copied and pasted".
  17. Haha! I've known that proverb all my life and it's still one of my favourites. :cool:

    I agree that would be a reasonable explanation normally, but it looked to me that Mentallic's wasn't just a typo as he/she continued to to say log(3) etc later on. And phinds's message certainly doesn't appear to have been of any copy'n'paste effect to me.

    Like I say, it really made me question myself. But glad we've agreed that it should've been an "e" in both cases - it certainly has satisfied me that I'm not going mad.

    I wasn't trying to be a pedant, but on sites such as this, little things like these can confuse a person who is trying to learn - not just the original poster. Sorry if I caused any offence by pointing it out.
  18. Uh ... stupidity ? Wait, no that's not it. It's a typo I tell you, a typo. Or a senior moment. I forget.
  19. Haha! My vote is for "senior". I consider myself senior too, but by no means consider myself "senior" to any of your maths knowledge. I'm well over half my way to my late 80s.
  20. Mentallic

    Mentallic 3,783
    Homework Helper

    Ahaha yeah that was a terrible [STRIKE]typo[/STRIKE] senior moment :redface:

    Of course, my post should have said

    [tex]e^{33}=10^{\log_{10}(e)\cdot 33}[/tex]

    And everything else should be changed accordingly. I'd edit it, but that option is gone now.

  21. Good job I'm here then! :smile:
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