# Ln and exp

by eng.mustafaS
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 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
 P: 2,784 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.
 Admin P: 23,369 You can calculate e by yourself, using this sum: $$e = \sum\limits_{n=0}^\infty \frac 1 {n!}$$ Edit: and a spreadsheet. Attached Thumbnails
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Ln and exp

 Quote by eng.mustafaS 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
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:

$$\ln(24)=\ln(2^33)=\ln(2^3)+\ln(3)=3\ln(2)+\ln(3)$$

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

For exponentials:

$$e^{33}=10^{\log_{10}(3)\cdot 33}$$

Where

$$\log_{10}(3)=\frac{\ln(3)}{\ln(10)}=\frac{\ln(3)}{\ln(5)+\ln(2)}$$

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 $\log_{10}(3)$ 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,

$$\cos(56)=\cos(45+11)$$

$$=\cos(45)\cos(11)-\sin(45)\sin(11)$$

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

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

So finally,

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

Where the true value is $\cos(56)=0.559...$

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 $\frac{11\pi}{180}\approx 1/5$ lead to a very close result by pure luck.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,339 Another way of looking at it: For any positive number, a, if $y= a^x$ then dy/dx is given by $$\lim_{h\to 0}\frac{a^{x+h}- a^x}{h}= \lim_{h\to 0}\frac{a^xa^h- a^x}{h}= a^x\lim_{h\to 0}\frac{a^h- 1}{h}$$ That is, the derivative of $a^x$ is just $a^x$ itself times a number- that number being $\lim_{h\to 0}(a^h- 1)/h$. It is easy to see that if a= 2, $\lim_{h\to 0}(2^h- 1)/h$ is less than 1. For example, with h= 0.0001, that fraction is $(2^{0.0001}- 1)/.0001= (1.000069317- 1)/.0001$= 0.000069317/.0001= .69317 And if a= 3, $\lim_{h\to 0}(3^h- 1)/h$ is larger than 1. With h= 0.0001, again, we have $(3^{0.0001}- 1)/.0001= (1.00010987- 1)/.0001$= .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 $a^x$ is just $a^x$ itself. We define "e" to be that number.
 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.
P: 9
 Quote by Borek You can calculate e by yourself, using this sum: $$e = \sum\limits_{n=0}^\infty \frac 1 {n!}$$ Edit: and a spreadsheet.
what is "n" ?

pleas can you fiend for me the value of e^4 by this equation
$$e = \sum\limits_{n=0}^\infty \frac 1 {n!}$$
PF Gold
P: 6,099
 Quote by eng.mustafaS what is "n" ? pleas can you fiend for me the value of e^4 by this equation $$e = \sum\limits_{n=0}^\infty \frac 1 {n!}$$
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.
P: 9
 Quote by HallsofIvy Another way of looking at it: For any positive number, a, if $y= a^x$ then dy/dx is given by $$\lim_{h\to 0}\frac{a^{x+h}- a^x}{h}= \lim_{h\to 0}\frac{a^xa^h- a^x}{h}= a^x\lim_{h\to 0}\frac{a^h- 1}{h}$$ That is, the derivative of $a^x$ is just $a^x$ itself times a number- that number being $\lim_{h\to 0}(a^h- 1)/h$. It is easy to see that if a= 2, $\lim_{h\to 0}(2^h- 1)/h$ is less than 1. For example, with h= 0.0001, that fraction is $(2^{0.0001}- 1)/.0001= (1.000069317- 1)/.0001$= 0.000069317/.0001= .69317 And if a= 3, $\lim_{h\to 0}(3^h- 1)/h$ is larger than 1. With h= 0.0001, again, we have $(3^{0.0001}- 1)/.0001= (1.00010987- 1)/.0001$= .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 2.6, such that the constant is exactly equal to 1: that is, for that a, the derivative of $a^x$ is just $a^x$ itself. We define "e" to be that number.
thank you very much
P: 9
 Quote by phinds 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.
i am in college .. but i am not good in maths

and tell now i dont know how i fiend e^4
PF Gold
P: 6,099
 Quote by eng.mustafaS i am in college .. but i am not good in maths and tell now i dont know how i fiend e^4
e^4 is e*e*e*e
Mentor
P: 18,036
 Quote by eng.mustafaS i am in college .. but i am not good in maths and tell now i dont know how i fiend e^4
$$e^4 = \sum_{n=0}^{+\infty} \frac{4^n}{n!}$$
P: 23,369
 Quote by eng.mustafaS what is "n" ?
n takes all possible integer values, starting with 0.

 pleas can you fiend for me the value of e^4 by this equation $$e = \sum\limits_{n=0}^\infty \frac 1 {n!}$$
Not by this equation, but you can use this one:

$$e^x = \sum\limits_{n=0}^\infty \frac {x^n} {n!}$$

The first equation was just a special case, for x=1 (and obviously e1 = e).
P: 229
 Quote by Mentallic For exponentials: $$e^{33}=10^{\log_{10}(3)\cdot 33}$$ Where $$\log_{10}(3)=\frac{\ln(3)}{\ln(10)}=\frac{\ln(3)}{\ln(5)+\ln(2)}$$ 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 $\log_{10}(3)$ and thus will have the order of magnitude of e33.
 Quote by phinds The sum process (with n being the element index) does not compute powers of 3, it computes e.
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!
Math
Emeritus
Thanks
PF Gold
P: 39,339
 Quote by oay 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!
To err is human- to really screw up requires a computer! I suspect Mentalic made a typo then phinds "copied and pasted".
P: 229
 Quote by HallsofIvy To err is human- to really screw up requires a computer!
Haha! I've known that proverb all my life and it's still one of my favourites.

 I suspect Mentalic made a typo then phinds "copied and pasted".
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.
PF Gold
P: 6,099
 Quote by oay 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!
Uh ... stupidity ? Wait, no that's not it. It's a typo I tell you, a typo. Or a senior moment. I forget.
P: 229
 Quote by phinds Uh ... stupidity ? Wait, no that's not it. It's a typo I tell you, a typo. Or a senior moment. I forget.
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.