Algebraic Manipulation (1 Viewer)

[hurikai]

This seems ridiculously simple and I feel like an idiot for asking it. I may have done this at some point but I can't remember...

How do you solve simultaneously:
y = x^2 + 1
y = 2^x

I can find (0,1) and (1,2) on inspection, but envisioning the graph I'm fairly sure there's a third point... somewhere.

 

[kurt.physics]

I have attached a picture of the graph. The first one is the graph, the second is the close up of the co-ordinate and the third is a zoomed out one.

 

[Slidey]

This seems ridiculously simple and I feel like an idiot for asking it. I may have done this at some point but I can't remember...

How do you solve simultaneously:
y = x^2 + 1
y = 2^x

I can find (0,1) and (1,2) on inspection, but envisioning the graph I'm fairly sure there's a third point... somewhere.

2^x-x^2-1=0
Use:
2^x=x^2+1, and
y=2^x-x^2-1

On inspection, 2^0=0^2+1 and 2^1=1^2+1 (roots at x=0 and x=1)
y'=2^x-2x=0
By inspection, two stationary points at x=1 and x=2 and that's it.
y(-1)=-3/2 < 0
y(0)=0
y(1/2)=sqrt(2)-5/4 > 0
y(1)=0
y(3/2)=2sqrt(2)-13/4 < 0
But as x -> infinity, y-> infinity, thus one more root exists after x=1, yes.
Let's test x=3:

~x=3-y(3)/y'(3)=3-(-2)/(2)=4
~x=4-y(4)/y'(4)=4-(-1)/(8)=4.125
~x=4.125-(-0.57)/(9.2)=4.187
Since this could go on forever, let's test near it: y(4.25)=-0.035
Hmm. The root is actually closer to x=4.25. I wonder why it's taking so long to approach it.

You could of skipped straight to Newton's method, but I just wanted to more rigourously show why another root exists.

 

[hurikai]

I know the third root exists - couldn't you just differentiate both equations and show 2^x is steeper?

But yeah, nice solution - we tried Newton's method in class and got 4.25 as well, it subbed in decently.



Turns out it's not actually solvable - our maths teacher said they were "transcendental equations" and the only way you could get close was through the iterative methods (approximations).

So that solves that one. Which leads me to ask - is there some way to apply limits to Newton's equation or to the method of Halving the Integral to find the value the approximation approaches as you apply the equation infinite times?

 

[Slidey]

I know the third root exists - couldn't you just differentiate both equations and show 2^x is steeper?

Huh. I suppose so. Cool. I probably used to know that.

Turns out it's not actually solvable - our maths teacher said they were "transcendental equations" and the only way you could get close was through the iterative methods (approximations).

lol. Sorry, I assumed you knew that. Any function which is defined by e is transcendental, thus normal methods can't be used to solve it. I think transcendental equations can be solved if you express the answer as things like infinite series of integrals or derivatives and such but I'm not sure. Trigonometric functions are defined by e (through the relationship e^i*(@+2kpi)=cis(@) thus they are also transcendental. Logarithmic functions are also defined by e (by virtue of being the inverse of the exponential).

So that solves that one. Which leads me to ask - is there some way to apply limits to Newton's equation or to the method of Halving the Integral to find the value the approximation approaches as you apply the equation infinite times?

A very good question. Because the algorithms are recursive in nature (self-referential), probably not (much like you can't find an explicit inverse for many functions). These days we just use computers to solve them in a split second.

There is an entire field of mathematics devoted to this called 'numerical analysis'.

Example: solve for x:
2^x+3^x=5
There is a very simple solution to this (it shouldn't be hard to see), but the only way to find it is to use numerical analysis (bisection or Newton's method will do).