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inn [45]
3 years ago
7

1 point) Are the functions f,g, and h given below linearly independent? f(x)=e3x+cos(5x), g(x)=e3x−cos(5x), h(x)=cos(5x). If the

y are independent, enter all zeroes. If they are not linearly independent, find a nontrivial solution to the equation below. Be sure you can justify your answer. 1 (e3x+cos(5x))+ -1 (e3x−cos(5x))+ 0 (cos(5x))=0.
Mathematics
1 answer:
Fynjy0 [20]3 years ago
8 0

Answer:

Functions are linearly dependent (are not linearly independent.)

Step-by-step explanation:

Remember that two functions f(x), g(x) and h(x) are said linearly independent on an interval I if the <em>only solution</em> to the equation

\alpha f(x)+\beta g(x)+\omega h(x)=0, \ \text{for all } x\in I

is the trivial one: α = 0, β = 0, ω = 0. If they are not linearly independent, they are called linearly dependent.

Now, let f(x), g(x) and h(x) be the functions:

f(x)=e^{3x}+\cos(5x),

g(x)=e^{3x}-\cos(5x),

h(x)=\cos(5x).

Then, letting α = 1, β= -1 and ω = -2, we see that:

\alpha f(x)+\beta g(x)+ \omega h(x)=e^{3x}+\cos(5x)-e^{3x}+\cos(5x)+2\cos(5x)=0.

Hence, the functions f(x), g(x) and h(x) are not linearly independent, or equivalently, are linearly dependent.

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Now we change 100,000,000 into a power of 10. A 1 followed by a number of zeros is the same as 10 to the power equal to the number of zeros. In 100,000,000 the 1 is followed by 8 zeros, so 100,000,000 = 10^8.

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