What's My Favorite Transcendental Number?
For a positive constant b , I draw a graph of y = b x , on the first quadrant.
I noticed that the gradient of this curve at any point is equal to the value of the function itself. In other words,
d x d y = y .
How many possible values of b satisfy this scenario?
Image Credit: Wikimedia Dmcq .
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3 solutions
Thank you for your solution
If b=0 , then also slope is 0 and y=0 ..So we get 2 values
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Well, no.
0 is not a positive constant. It is a whole number.
y
=
b
x
d
x
d
y
=
b
x
ln
b
=
y
ln
b
=
y
ln
b
=
1
b
=
e
e
is Sir
@Pi Han Goh
's favorite transcendental number! =D
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Of course!
d
x
d
b
x
=
d
x
d
e
x
ln
b
=
e
x
ln
b
ln
b
=
b
x
ln
b
At the red equality sign I used chain rule.
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Then this begs the question:
How do you prove that d/dx (e^x) = e^x?
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@Pi Han Goh
–
Well......
It depends on which definition of
e
x
you pick.
If you pick:
1)
e
x
is the solution to the ODE
d
x
d
y
=
y
with initial condition
y
∣
x
=
0
=
1
2) , or
e
x
=
1
+
1
!
x
+
2
!
x
2
+
3
!
x
3
+
4
!
x
4
+
⋯
then the proof should be easy.
But I think there is still some other equivalent definitions......
Your first proof is just the rephrasing of d/dx (e^x) = e^x.
Note that your second proof is the Maclaurin Series of e^x, so the proof is circular and not valid.
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You have to pick a definition before talking about its properties. Which definition did you pick?
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lim(n>infinity) (1+1/n)^n
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@Pi Han Goh – I think you mean (1+x/n)^n. Use binomial expansion and this becomes the Maclaurin Series of e^x.
Its e^x. The derivative of e^x is e^x. Sorry I do not have an actual proof.
Okay, I will show you the differentiation of e x .
Let y = e x
Then the definition of its derivative:
d x d y = lim Δ x → 0 Δ x Δ y = lim Δ x → 0 Δ x e x + Δ x − e x = lim Δ x → 0 Δ x e x . e Δ x − e x = e x lim Δ x → 0 Δ x e Δ x − 1 = e x . 1 = e x
Given that, y = b x
Differentiating both sides with respect to x, we have:
d x d y = b x l n b
Condition: d x d y = y
Subsequently, b x l n b = b x ⟹ l n b = 1 ⟹ b = e
Hence, only one possible solution.