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Part of my first year university degree, I need to work with the Maple software.

I have this question.

Find an approximation to the smallest positive solution to

$$(-38/25)*sin(9/10*x)=(13/25)*exp(-16*x/25)$$

Correct to 10 sig.figs

No matter what I do, I don't seem to get the correct answer.

Initially, I got a negative answer but then I sent the bound of x from 0 to infinity. Still getting the wrong answer.

This is my best attempt so far.

fsolve((-38/25)*sin(9/10*x)=(13/25)*exp(-16*x/25),x,x=0..infinity);

10.47244230

All help is greatly appreciated

Aaron
  • 21

1 Answers1

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restart;
expr1 := -38/25*sin(9/10*x):
expr2 := 13/25*exp(-16/25*x):

P1 := plot(expr1,x=0..10,color=blue):
P2 := plot(expr2,x=0..10,color=red):
plots:-display(P1,P2,view=-2..2);

enter image description here

The purpose of creating the plot is to allow you visually discover a range for x between which the two curves intersect only once.

For example you can see that they intersect only once between x=0 and x=5.

Thus you can figure out, by visual inspection of the plot, a suitable range to use when calling fsolve.

fsolve(expr1-expr2, x=0..5);

                      3.530355063

Here are some other ways,

RootFinding:-NextZero(unapply(expr1-expr2,x), 0.0);

                      3.530355062

Student:-Calculus1:-Roots(expr1-expr2, x=0..10, numeric);

               [3.530355063, 6.976944937]

min(%);

                      3.530355063
acer
  • 5,293
  • I see what you mean. When I set the bound to infinity, I didn't recognise there were 2 points of intersection, hence I was not getting the smallest value. Thank you for the graph. Really explained it well. In fact now trying it again, there was nothing wrong with my input, just bounds were incorrect. – Aaron Jul 11 '19 at 15:37
  • For the range x=0..infinity there are an infinite number of intersections, not just 2. Also, the given plot (or a plot of the difference) shows that the roots of expr1-expr2 are close to the roots of expr1. And it turns out that using a root of expr1 as an initial-point for fsolve will allow it to find the corresponding close root of expr1-expr2. But we know the roots of expr1, which are k*10/9*Pi, k=1,2,.... And indeed calling fsolve(expr1-expr2,x=10/9*Pi); returns the desired first point of intersection. – acer Jul 11 '19 at 19:55