Question

Challenge problem/discussion for you guys:
Let say number \(e\) was never calculated...

How can we find the value of \(a\) where \(f(x) = a^x = f'(x)\)?

Answer 1

we are given that f(x) = f ' (x)

now by definition
f ' (x)
= lim [f(x + h) - f(x)] /h
= lim [a ^ ( x + h) - a^x ] / h
= lim [a^x * a^h - a^x] / h
= lim [a^x ( a^h -1 ) / h ]
= a^x * [lim a^h -1 ] /h


so we have so far

a^x = a^x lim [ a^h -1 ] / h

Answer 2

usually e is computed with an infinite series

Answer 3

a=0

Answer 4

Ok, is that only possible value of a?

Answer 5

well of course e, and actually ae^x

do we know that e^x is the inverse of ln?

Answer 6

e? ln? What are you talking about? ;)

Answer 7

\(\frac{dy}{dx}=y\implies \int \frac{1}{y}dy=\int dx\implies y\ne 0\)

Answer 8

ahh ok will then that don't work:)

Answer 9

micah , one moment

Answer 10

haha

Answer 11

ok so you agree with the derivative of f(x) = a^x

Answer 12

I think you will need series:(

Answer 13

Yeah, I agree?

Answer 14

do you agree that for f(x) = a^x

f ' (x)
= lim [f(x + h) - f(x)] /h
= lim [(a ^ ( x + h) - a^x ) / h]
= lim [(a^x * a^h - a^x) / h ]
= lim [a^x ( a^h -1 ) / h ]
= a^x * lim [( a^h -1 ) / h ]

Answer 15

I do

Answer 16

Yeah @perl

Answer 17

lets define a function
m(a) = lim ((a^h - 1) /h) , as h->0

m(2) <1 , by evaluation

m(3)>1

so by intermediate value theorem, and since m is a continuous function
there exists a value , lets call it e, such that m(e) = 1

this simplifies the expression

Answer 18

e is between 2 and 3

Answer 19

now
f ' (x) = m(a) * a^x ,

and if a = e , we have

f ' (x) = m(e) * a^x = 1*a^x

Answer 20

' by evaluation ' i mean numerical evaluation

Answer 21

so yes, i actually was write before.

we need to find a value 'a' such that

lim (a^h -1 / h) = 1 , as h->0

Answer 22

Yep, we know that a is in between 2 and 3, but I am sure we can do better than that.

But I am stuck on \(\lim_{h\to0}\dfrac{a^h-1}{h}=1\) lol...

Answer 23

you can use intermediate value theorem

Answer 24

since that limit is continuous , lets call that limit m(a) , since it changes with different 'a'.

there is some decimal 'a' such that

we know that m(a) is continuous and

m(2) < 1 < m(3)

so there must be come decimal between 2 and 3 such that

m(2) < 1= m(e) < m (3)

Answer 25

Well, yeah, but it looks like it takes numerous of numerical evaluation for us to find good approximate of a.

plus how did you evaluate m(x)? like you did with m(2) and m(3)

Answer 26

you can make a table

Answer 27

also i should add, a = 0 is undefined for m(a).

but m(a) is continuous on [2,3]

Answer 28

I am sorta new to calculus and I was just wondered how e came in.

*groan* guess we would have to do that stuffs to calculate e, like how most of mathematicians in past calculated well known number such pi or gold ratio lol...

Thanks, I am satisfied.

Answer 29

Gonna left this question open, in case someone have different approaches.

That's ok @perl ?

Answer 30

ill make a table with a calculator, ok ?

Answer 31

sure

Answer 32

ok cool

Answer 33

my draw button seems to be failing, so ill be back later when i get a chance

Answer 34

basically you do

h = .01
(2^.01 -1 ) / .01=

h = .001
(2^.001 -1 ) / .001=

Answer 35

that will converge to m(2), otherwise known as ln 2

Answer 36

ah, and I takes that we should do evaluate on m(2.5) and see whether it is smaller than 1 or not, then evaluation m(2.75)? and so on, then we will approaches to e, right?

Answer 37

correct, 2.75 is too big though

Answer 38

so you will have to go back to 2.25

Answer 39

m(2.5) should be smaller than 1, so we should then go next to m(2.625)

Answer 40

basically just go half and half all the way.

Answer 41

m(2) < 1
m(3) >1

then by bisection method, we choose the halfway point

m(2.5) <1

then
m(2.75 ) > 1 oops, too big

then halfway between 2.5 and 2.75

Answer 42

right

Answer 43

yeah

Answer 44

ok i see what youre doing, you want to actually determine e with some degree of accuracy.

Answer 45

Another approach if you are still interested @micahwood50:

Since \(f(x) = f'(x)\), it is true that \(f'(x) = f''(x)\) and \(f''(x) = f'''(x)\) and so on...

So we can see that \(f(x) = f^{(n)}(x)\quad\forall n\in\mathbb{N}\), right?

Ok, we can expand \(a^x\) into infinite polynomial using Maclaurin Series.

\[f(x) = \sum_{n=0}^\infty f^{(n)}(0)\dfrac{x^n}{n!}\]
So...
\[a^x = \sum_{n=0}^\infty a^0\dfrac{x^n}{n!} = \sum_{n=0}^\infty \dfrac{x^n}{n!}\]

Note that \(\displaystyle \dfrac{\mathrm d}{\mathrm d x}\sum_{n=0}^\infty \dfrac{x^n}{n!} = \sum_{n=0}^\infty \dfrac{x^n}{n!}\)

So finally we can calculate \(a\) by letting x=1: \[f(1) = a = \sum_{n=0}^\infty \dfrac{1}{n!}\]
Add up to, like, first 7 terms, and you should have good approximate value of a.

Answer 46

This limit actually is another definition for \(e\).

Answer 47

And it just happens to be equivalent to other definitions of \(e\).

Answer 48

But hypothetically, if it was a different value, then we would have defined another constant just for the sake of this limit, since this limit allows us to differentiate exponential functions.