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Ln and expby eng.mustafaS
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#1
May1013, 04:53 AM

P: 9

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 ?? or ln 24 ? exp 33 ? without using calculator ? thank you 


#2
May1013, 05:18 AM

P: 2,777

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: http://en.wikipedia.org/wiki/E_(mathematical_constant) Its true that its not an easy number to work with numerically but the insight gained in using it is far more profound. 


#3
May1013, 05:43 AM

Admin
P: 23,363

You can calculate e by yourself, using this sum:
[tex]e = \sum\limits_{n=0}^\infty \frac 1 {n!}[/tex] Edit: and a spreadsheet. 


#4
May1013, 05:49 AM

HW Helper
P: 3,508

Ln and exp
For the logs, you can split them up into a sum of smaller logs: [tex]\ln(24)=\ln(2^33)=\ln(2^3)+\ln(3)=3\ln(2)+\ln(3)[/tex] 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] Where [tex]\log_{10}(3)=\frac{\ln(3)}{\ln(10)}=\frac{\ln(3)}{\ln(5)+\ln(2)}[/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 e^{33}. 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, [tex]\cos(56)=\cos(45+11)[/tex] [tex]=\cos(45)\cos(11)\sin(45)\sin(11)[/tex] 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 11^{o} 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}}(11/5)=\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 overapproximation of cos(11^{o}) and taking away my overapproximation of [itex]\frac{11\pi}{180}\approx 1/5[/itex] lead to a very close result by pure luck. 


#5
May1013, 07:55 AM

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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. 


#6
May1013, 10:11 AM

Mentor
P: 7,315

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.



#7
May1013, 01:01 PM

P: 9

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] 


#8
May1013, 01:21 PM

PF Gold
P: 6,083

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. 


#9
May1013, 02:27 PM

P: 9




#10
May1013, 02:33 PM

P: 9

and tell now i dont know how i fiend e^4 


#13
May1013, 03:26 PM

Admin
P: 23,363

[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 e^{1} = e). 


#14
May1013, 05:23 PM

P: 229




#15
May1013, 05:53 PM

Math
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PF Gold
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#16
May1013, 06:23 PM

P: 229

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. 


#17
May1013, 08:08 PM

PF Gold
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#18
May1013, 08:20 PM

P: 229




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